Triangulated Categories. (AM-148), Volume 148 9781400837212

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Annals of Mathematics Studies Number 148

Triangulated Categories by AMNON NEEMAN

PRINCETON UNIVERSITY PRESS PRINCETON and OXFORD 2001

Copyright © 2001 by Princeton University Press Published by Princeton University Press, 41 William Street, Princeton, New Jersey 08540 In the United Kingdom: Princeton University Press, 3 Market Place, Woodstock, Oxfordshire OX20 1SY All Rights Reserved The Annals of Mathematics Studies are edited by Luis A. Caffarelli, John N. Mather, and Elias M. Stein Library of Congress Catalog Card Number 99-068088 ISBN 0-691-08685-0(cloth) ISBN 0-691-08686-9 (pbk.) The publisher would like to acknowledge the authors of this volume for providing the camera-ready copy from which this book was printed The paper used in this publication meets the minimum requirements of ANSIINISO Z39.48-1992 (R 1997) (Permanence of Paper)

www.pup.princeton.edu Printed in the United States of America

3 5 7 9 10 8 6 4 2

3 5 7 9

w

(Pbk.)

8 6 4 2

Contents 0. 1.

Acknowledgements Introduction

3 3

Chapter 1. 1.1. 1.2. 1.3. 1.4. 1.5. 1.6. 1.7. 1.8.

Definition and elementary properties of triangulated categories Pre-triangulated categories Corollaries of Proposition 1.1.20 Mapping cones, and the definition of triangulated categories Elementary properties of triangulated categories Triangulated subcategories Direct sums and products, and homotopy limits and colimits Some weak "functoriality" for homotopy limits and colimits History of the results in Chapter 1

29 29 37 45 52 60 63 68 70

Chapter 2. 2.1. 2.2. 2.3.

Triangulated functors and localizations of triangulated categories 73 Verdier localization and thick subcategories 73 Sets and classes 99 100 History of the results in Chapter 2

Chapter 3. Perfection of classes 3.1. Cardinals 3.2. Generated subcategories 3.3. Perfect classes 3.4. History of the results in Chapter 3

103 103 103 110 122

Chapter 4. Small objects, and Thomason's localisation theorem 4.1. Small objects 4.2. Compact objects 4.3. Maps factor through (8)(3 4.4. Maps in the quotient 4.5. A refinement in the countable case 4.6. History of the results in Chapter 4

123 123 128 130 135 144 150

Chapter 5.

153

The category A(S) v

5.1. 5.2. 5.3. 5.4.

The abelian category A(g) Subobjects and quotient objects in A(g) The functoriality of A(g) History of the results in Chapter 5

153 172 177 182

Chapter 6. The category C.x (gop, Ab) 183 6.1. C.x(goP,Ab) is an abelian category satisfying [AB3] and [AB3*]183 6.2. The case of g = 'J 201 6.3. t.x(goP,Ab) satisfies [AB4] and [AB4*], but not [AB5] or [AB5*] 206 6.4. Projectives and injectives in the category C.x (gop, Ab) 211 6.5. 6.6.

The relation between A('J) and C.x ( {'J} op, Ab) History of the results of Chapter 6

Chapter 7. Homological properties of C.x(goP,Ab) 7.1. C.x(goP,Ab) as a locally presentable category 7.2. Homological objects in C.x(goP,Ab) 7.3. A technical lemma and some consequences 7.4. The derived functors of colimits in C.x(goP,Ab) 7.5. The adjoint to the inclusion of C.x(goP,Ab) 7.6. History of the results in Chapter 7

214 220 221 221 224 230 253 266 271

Chapter 8. Brown representability 273 8.1. Preliminaries 273 8.2. Brown representability 275 8.3. The first representability theorem 280 8.4. Corollaries of Brown representability 285 8.5. Applications in the presence of injectives 288 8.6. The second representability theorem: Brown representability for the dual 303 8. 7. History of the results in Chapter 8 306 Chapter 9. Bousfield localisation 9.1. Basic properties 9.2. The six gluing functors 9.3. History of the results in Chapter 9

309 309 318 319

Appendix A. Abelian categories A.l. Locally presentable categories A.2. Formal properties of quotients A.3. Derived functors of limits A.4. Derived functors of limits via injectives A.5. A Mittag-Leffier sequence with non-vanishing limn A.6. History of the results of Appendix A

321 321 327 345 354 361 366

vi

Appendix B. Homological functors into [AB5'-"] categories B.l. A filtration B.2. Abelian categories satisfying [AB5'-"] B.3. History of the results in Appendix B

369 369 378 385

Appendix C. Counterexamples concerning the abelian category A('J) 387 C.l. The submodules piM 387 C.2. A large R-module 392 C.3. The category A(S) is not well-powered 393 C.4. A category cx(S 0 P,Ab) without a cogenerator 395 C.5. History of the results of Appendix C 405 Appendix D. Where 'J is the homotopy category of spectra D .1. Localisation with respect to homology D.2. The lack of injectives D.3. History ofthe results in Appendix D

407 407 420 426

Appendix E. Examples of non-perfectly-generated categories 427 E.l. If 'J is ~ 0 -compactly generated, 'J0 P is not even well-generated427 E.2. An example of a non ~ 1 -perfectly generated 'J 432 E.3. For 'J = K(Z), neither 'J nor 'J0 P is well-generated. 437 EA. History of the results in Appendix E 442 Bibliography

443

Index

445

vii

Triangulated Categories

0. Acknowledgements The author would like to thank Andrew Brooke-Taylor, Daniel Christensen, Pierre Deligne, Jens Franke, Bernhard Keller, Shun-ichi Kimura, Henning Krause, Jack Morava, Saharon Shelah and Vladimir Voevodsky for helpful discussions, and for their valuable contributions to the book. 1. Introduction

Before describing the contents of this book, let me explain its origins. The book began as a joint project between the author and Voevodsky. The idea was to assemble coherently the facts about triangulated categories, that might be relevant in the applications to motives. Since the presumed reader would be interested in applications, Voevodsky suggested that we keep the theory part of the book free of examples. The interested reader should have an example in mind, and read the book to find out what the general theory might have to say about the example. The theory should be presented cleanly, and the examples kept separate. The division of labor was that I should write the theory, Voevodsky the applications to motives. What then happened was that my part of this joint project mushroomed out of proportion. This book consists just of the formal theory of triangulated categories. In a sequel, we hope to discuss the motivic applications. The project was initially intended to be purely expository. We meant to cover many topics, but had no new results. This was to be an exposition of the known facts about Brown representability, Bousfield localisation, tstructures and triangulated categories with tensor products. The results should be presented in a unified, clear way, with the exposition accessible to a graduate student wishing to learn the theory. The catch was that the theory should be developed in the generality one would need motivically. The motivic examples, unlike the classical ones, are not compactly generated triangulated categories (whatever this means). The classical literature basically does not treat the situation in the generality required. My job amounted to modifying the classical arguments, to work in the greater generality. As I started doing this, I quickly came to the conclusion that both the statements and the proofs given classically are very unsatisfactory. The proofs in the literature frequently rely on lifting problems about triangulated categories to problems about more rigid models. Right at the outset I decided that in this book, I will do everything to avoid models. Part of the challenge was to see how much of the theory can be developed without the usual crutch. But there was a far more serious problem. Many of the statements were known only in somewhat special cases, decidedly not including the sort that come up in motives. Thomason once told me that "compact objects are as necessary to this theory as air to 3

4

1. INTRODUCTION

breathe". In his words, I was trying to develop the theory in the absence of oxygen. The book is the result of my work on the subject. It treats a narrower scope of topics than initially planned; we deal basically only with Brown representability and Bousfield localisation. But in some sense we make great progress on the problems. In the process of setting up the theory in the right generality and without lifting to models, we end up with some new and surprising theorems. The book was meant to be an exposition of known results. The way it turned out, it develops a completely new theory. And this theory gives interesting, new applications to very old problems. Now it is time to summarise the mathematical content of the book. The first two chapters of the book are nothing more than a selfcontained exposition of known results. Chapter 1 is the definitions and elementary properties of triangulated categories, while Chapter 2 gives Verdier's construction of the quotient of a triangulated category by a triangulated subcategory. This book was after all intended as a graduate textbook, and therefore assumes little prior knowledge. We assume that the reader is familiar with the language of categories and functors. The reader should know Yoneda's Lemma, the general facts about adjoint functors between categories, units and counits of adjunction, products and coproducts. It is also assumed that the reader has had the equivalent of an elementary course on homological algebra. We assume familiarity with abelian categories, exact sequences, the snake lemma and the 5-lemma. But this is all we assume. In particular, the reader is not assumed to have ever seen the definition of a triangulated category. In practice, since we give no examples, the reader might wish to find one elsewhere, to be able to keep it in mind as an application of the general theory. One place to find a relatively simple, concrete exposition of one example, is the first chapter of Hartshorne's book [19]. This first chapter develops the derived category. Note that, since we wish to study mostly triangulated categories closed under all small coproducts, the derived category of most interest is the unbounded derived category. In [19], this is the derived category that receives probably the least attention. There is another short account of the derived category in Chapter 10, pages 369-415 of Weibel's [37]. There are, of course, many other excellent accounts. But they tend to be longer. Anyway, if the reader is willing to forget the examples, begin with the axioms, and see what can be proved using them, then this book is relatively self-contained. Chapters 1 and 2 are an account of the very classical theory. There are some expository innovations in these two chapters, but otherwise little new. If the reader wants to be able to compare the treatment given here with the older treatments, at the end of each chapter there is a historical summary. In the body of the chapters, I rarely give references to older works. The

1. INTRODUCTION

5

historical surveys at the end of each chapter contain references to other expositions. They also try to point out what, if anything, distinguishes the exposition given here from older ones. Starting with Chapter 3, little of what is in the book may be found in the literature. For the reader who has some familiarity with triangulated categories, it seems only fair that the introduction summarise what, if anything, he or she can expect to find in this book which they did not already know. It is inevitable, however, that such an explanation will demand from the reader some prior knowledge of triangulated categories. The graduate student, who has never before met triangulated categories, is advised to skip the remainder ofthe introduction and proceed to Chapter 1. After reading Chapters 1 and 2, the rest of the introduction will make a lot more sense. Let me begin with the concrete. Few people have a stomach strong enough for great generalities. Sweeping, general theorems about arbitrary 2-categories tend to leave us cold. We become impressed only when we learn that these theorems teach us something new. Preferably something new about an old, concrete example that we know and love. Before I state the results in the book in great generality, let me tell the reader what we may conclude from them about a special case. Let us look at the special case, where 'J is the homotopy category of spectra. Let 'J be the homotopy category of spectra. Let E be a spectrum (ie. an object of 'J). Following Bousfield, the full subcategory 'JE C 'J, whose objects are called the E-acyclic spectra, is defined by

{x E Ob('J) I x 1\ E

=

0}.

The full subcategory j_'JE C 'J, whose objects are called the E-local spectra, is defined by

{y

E Ob('J)

I Vx

E Ob('JE),

'J(x,y)

= 0}.

An old theorem of Bousfield (see [6]) asserts that one can localise spectra with respect to any homology theory E. In the notation above, Bousfield's theorem asserts THEOREM (BousFIELD, 1979). Let E be a spectrum, that is E is an object of 'J. Let 'JE C 'J and j_'JE C 'J be defined as above. Suppose x is an object of 'J. Then there is a triangle in 'J

XE ~ X ~ j_XE ~ ~XE,

with XE

E 'JE,

and j_XE

E

j_'JE.

Bousfield's theorem has been known for a long time. What this book has to add, are surprising structure theorems about the categories 'JE and j_'JE. We prove the following representability theorems

1. INTRODUCTION

6

THEOREM (NEW, THIS BOOK). Let E be a spectrum. Let 'JE C 'J and _i'JE C 'J be defined as above. The representable functors

'JE(-,h)

_l_'JE(-,h)

and

can be characterised as the homological functors H : 'rJ ----+ Ab (respectively H : j_'rl ----+ Ab} taking coproducts in 'JE {respectively _j_'JE) to products in Ab. The representable functors 'JE(h,-) can be characterised as the homological functors H : 'JE products to products.

----+

Ab, taking

Proof: The characterisation of the functors 'JE (-, h) and j_'JE (-, h) may be found in Theorem D.l.l2. More precisely, for 'JE( -,h) see D.l.l2.1, while the statement for _i'JE( -,h) is contained in D.l.l2.5. The characterisation of the functors 'JE(h,-) may be found in Lemma D.l.l4. 0 Representability theorems are central to this subject. What we have achieved here, is to extend Brown's old representability theorem of [7]. Brown proved that the functors 'J( -,h) can be characterised as the homological functors 'J0 P ----+ Ab taking coproducts to products. We have generalised this to 'JE, rr; and j_'JE, but unfortunately not to j_rr;. We do not know, whether the the functors _i'JE(h,-) can be characterised as the homological functors taking products to products. Another amusing fact we learn in this book, is that the categories 'JE and _i'JE are not equivalent to 'J0 P. There are, in fact, many more amusing facts we prove. Let us give one more. We begin with a definition. DEFINITION. Let a be a regular cardinal. A morphism f : x ---+ y in 'J is called an a-phantom map if, for any spectrum s with fewer than a cells, any composite

s----+x~y vanishes. With this definition, we are ready to state another fun fact that we learn in this book. THEOREM (NEW, THIS BOOK). Let a > No be a regular cardinal. There is an object z E 'J, which admits no maximal a-phantom map y ---+ z. That is, given any a-phantom map y ---+ z, there is at least one aphantom map x ----+ z not factoring as X

----+

y

-----+ Z.

1. INTRODUCTION

7

Proof: The proof of this fact follows from Proposition D.2.5, coupled with Lemma 8.5.20. D REMARK 1.1. It should be noted that the above is surprising. If a = No, the a-phantom maps are the maps vanishing on all finite spectra. These are very classical, and have been extensively studied in the literature. Usually, they go by the name phantom maps; the reference to a = No is new to this book, where we study the naturallargl:M:ardinal generalisation. From the work of Christensen and Strickland (9], we know that every object z E 'J admits a maximal N0 -phantom map y ---+ z. There is an Nophantom map y---+ z, so that all other N0 -phantom maps x---+ z factor as x~y~z.

What is quite surprising is that this is very special to a

= N0 .

So far, we have given the reader a sampling of facts about the homotopy category of spectra, which follow from the more general results of this book. I could give more; but it is perhaps more instructive to indicate the broad approach. The idea of this book is to study a certain class of triangulated categories, the well-generated triangulated categories. And the thrust is to prove great facts about them. We will show, among many other things THEOREM

1.2. The following facts are true:

1.2.1. Let 'J be the homotopy category of spectra. Let E be an object of 'J. Then both the category 'JE and the category j_'JE are well-generated triangulated categories. 1.2.2. Suppose 'J is a well-generated triangulated category. The representable functors 'J(-, h) can be characterised as the homological functors H : 'J0 P ---+ Ab, taking coproducts in 'J to products in Ab. In other words, we will prove a vast generalisation of Brown's representability theorem. Not only does it generalise to 'JE and j_'JE• but to very many other categories as well. The categories that typically come up in the study of motives are examples. And now it is probably time to tell the reader what a well-generated triangulated category is. It turns out to be quite a deep fact that this structure even makes sense. Let 'J be a triangulated category. We remind the reader: a homological functor 'J ---+ A is a functor from 'J to an abelian category A, taking triangles to long exact sequences. We can consider the collection of all homological functors 'J---+ A. An old theorem of Freyd's (see [13]) asserts that

1. INTRODUCTION

8

THEOREM (FREYD, 1966). Among all the homological functors 'J ---+ A there is a universal one. There is an abelian category A('J) and a homological functor 'J ---+ A('J), so that any other homological functor 'J ---+ A factors as

'J

--------+

A('J) ~ A

where the exact functor A('J) ---+ A is unique up to canonical equivalence. Any natural tranformation of homological functors 'J ---+ A factors uniquely through a natural transformation of the (unique) exact functors A('J)---+ A. This theorem tells us that, associated naturally to every triangulated category 'J, there is an abelian category A('J). The association is easily seen to be functorial. It takes the 2-category of triangulated categories and triangulated functors to the 2-category of abelian categories and exact functors, and is a lax functor. One can wonder about the homological algebra of the abelian category A('J). Freyd proves also PROPOSITION (FREYD, 1966). Let 'J be a triangulated category. The abelian category A('J) of the previous theorem has enough projectives and enough injectives. In fact, the projectives and injectives in A('J) are the same. An object a E A('J) is projective (equivalently, injective) if and only if there exists an object b E A('J), so that

a EB b E 'J c A('J). That is, a is a direct summand of an object aEBb, and aEBb is in the image of the universal homological functor 'J ---+ A('J). This universal homological functor happens to be a fully faithful embedding; hence I allow myself to write 'J C A('J). It turns out to be easy to deduce the following corollary: COROLLARY 1.3. Let F : S ---+ 'J be a triangulated functor. IfF has a right adjoint G: 'J---+ S, then G is also triangulated, and A(G) : A('J) ---+ A(S) is right adjoint to A(F) : A(S) ---+ A('J). But more interesting is the following. If every idempotent in S splits, then F : S ---+ 'J has a right adjoint if and only if A(F) : A(S) ---+ A('J) does. That is, if A(F) : A(S) ---+ A('J) has a right adjoint G: A('J) ---+ A(S), then F: S---+ 'J has a right adjoint G: 'J---+ S, and of course A(G) is naturally isomorphic to

c.

Proof: Lemma 5.3.6 shows that the adjoint of a triangulated functor is triangulated, Lemma 5.3.8 proves that if G is right adjoint to F then A(G) is right adjoint to A(F), while Proposition 5.3.9 establishes that if A(F) has a right adjoint G, then F has a right adjoint G. D

1. INTRODUCTION

9

REMARK 1.4. It turns out that many of the deepest and most interesting questions about triangulated categories involve the existence of adjoints. This suggests that Corollary 1.3 should be great. It tells us that finding adjoints to triangulated functors between triangulated categories, a difficult problem, is equivalent to finding adjoints to exact functors between abelian categories. We feel much more comfortable with abelian categories, so the Corollary should make us very happy. The problem is that the abelian categories that arise are terrible. For example, let 'J be the category D(Z), the derived category of the category of all abelian groups. Then the abelian group Z can be viewed as an object of D(Z); it is the complex which is the group Z in dimension 0, and zero elsewhere. The universal homological functor D(Z)

----+

A(D(Z))

takes Z to an object of A(D(Z)). I assert that this object, in the abelian category A (D(Z)), has a proper class of subobjects. The collection of subobjects of Z E A(D(Z)) is not a set; it is genuinely only a class. The proof may be found in Appendix C. In the light of Remark 1.4, it is natural to look for approximations to the abelian category A('J). It seems reasonable to try to find other abelian categories A, together with exact functors A('J) ---+ A, which are "reasonable" approximations. It is natural to want the objects of A to only have sets (not classes) of subobjects. But otherwise it would be nice if A is as close as possible to the universal abelian category A('J). The universal property of A('J) asserts that exact functors A('J) ---+A are in 1-1 correspondence with homological functors 'J ---+ A. We therefore want to find reasonable homological functors 'J ---+ A, for suitable A. Let 'J be a triangulated category. It is said to satisfy [TR5] if the coproduct of any small set of objects in 'J exists in 'J. If the dual category 'J0 P satisfies [TR5], then 'J is said to satisfy [TR5*]. Let a be an infinite cardinal. Let 'J be a triangulated category satisfying [TR5] and g a triangulated subcategory. We say that S is a-localising if any coproduct of fewer than a objects of S lies in S. We call g C 'J localising if it is a-localising for every infinite cardinal a. Given a triangulated category 'J satisfying [TR5], and an a-localising subcategory g C 'J, there is a God-given abelian category one can construct out of S, and a homological functor

'J

----+

cx(goP,Ab).

Now it is time to define these. The category ex (gop, Ab) is the abelian category of all functors gop ---+ Ab which preserve products of fewer than a objects. Recall that g is alocalising. Given fewer than a objects in g, their coproduct exists in 'J

10

1.

INTRODUCTION

because 'J satisfies [TR5], and is contained in g because g is a-localising. That is, the product exists in the dual gap, and we look at functors to abelian groups gap ----+ Ab

preserving all such products. The reader can easily check (see Lemma 6.1:4 for details) that ex(gaP,Ab) is an abelian category. Furthermore, there is a homological functor 'J

----+

ex(gaP,Ab).

It is the functor that takes an object t E 'J to the representable functor 'J( -, t), restricted to g C 'J. We denote this restriction

'J(-,t)ls· This construction depends on the choice of an infinite cardinal a, and an a-localising subcategory g C 'J. In what follows, it is convenient to assume that the cardinal a is regular. That is, a is not the sum of fewer than a cardinals, all smaller than a. Starting with any regular cardinal a and any a-localising subcategory g C 'J, we have constructed a homological functor 'J

----+

ex(gaP,Ab).

It factors uniquely through the universal homological functor, to give an exact functor

A('J) ~ ex(gaP,Ab). It is very easy to show that the functor 1r has a left adjoint; we denote the left adjoint F: ex(gaP,Ab) ---+ A('J). We deduce a unit of adjunction 'fJ: 1 ---+ 1rF.

We prove PROPOSITION 1.5. Suppose 'fJ : 1 ---+ 1r F is the unit of adjunction 7r : A('J) ---+ ex(gaP,Ab) preserves coproducts, then

above. If the functor 'fJ is an isomorphism.

Proof: See Poposition 6.5.3.

D

By Gabriel's theory of localisations of abelian categories (see Appendix A), the unit of adjunction 'fJ: 1 ---+ 1rF is an isomorphism if and only if ex(gaP,Ab) is a quotient of A('J), with 7r being the quotient map. It is therefore natural to want to study the g c 'J for which this happens. By Proposition 1.5, a particularly interesting case is when the map

A('J) ~ ex(gaP,Ab)

1. INTRODUCTION

11

preserves coproducts. It turns out that, for a given regular cardinal o:, this depends on a choice of the o:-localising subcategory S c 'J'. We proceed now to describe the complete answer. DEFINITION 1.6. Let o: be a regular cardinal. Let 'J be a triangulated category satisfying [TR5]. An object t E 'J is called o:-small if any morphism from t to a coproduct

t---+

II X;..

>--EA

factors through a coproduct of fewer than o: objects. C A, A' of cardinality < o:, and a factorisation

A'

t

---+

II X;..

c

There is a subset

II X;...

The full subcategory of all o:-small objects in 'J is denoted 'J(a:). Next we need DEFINiTION 1.7'. Let o: be a regular cardinal. Let 'J be a triangulated category satisfying [TR5]. A class T, containing 0, of objects in 'J is called o:-perfect if, for any collection {X;..,.\ E A} of fewer than o: objects of 'J, any object t E T, and any map

t---+

II X;..

>--EA

there is a factorisation

>-.EA

>-.EA

with t;.. in T. Furthermore, if the composite

t---+

II t;..

>--EA

vanishes, then each of the maps factors as

tA with u;..

E

--:-----t

uA

---+

T, so that the composite

t

--:-----t

II t

>-.EA

already vanishes.

A ---+

XA

II u

AEA

A

12

1. INTRODUCTION

With these two definitions, we have a theorem THEOREM 1.8. Let a be a regular cardinal, 'J a triangulated category satisfying [TR5j. Let S C 'J be an a-localising subcategory. The natuml functor

A('J) ~ £x(S 0 P,Ab) preserves coproducts if and only if 1.8.1. The objects of S are all a-small; that isS c 'J(). 1.8.2. The class of all objects in S is a-perfect, as in Definition 1. 7.

Proof: Lemma 6.2.5. In the statement of Lemma 6.2.5, we only assert the sufficiency; if S satisfies 1.8.1 and 1.8.2, then 1r preserves coproducts. But the proof immediately also gives the converse. D It is therefore of interest to study a-localising subcategories S C 'J(a), whose collection of objects form an a-perfect class. The remarkable fact is that there is a biggest one. For any regular cardinal a, we can define a canonical, God-given a-localising subcategory 'J"'. It is given by DEFINITION 1.9. The full subcategory 'J"' C 'J, of all a-compact objects in 'J, is defined as follows. The class of objects in Ob('J"') is the unique maximal a-perfect class in 'J(). It of course needs to be shown that such a maximal a-perfect class exists. It is also relevant to know that 'J"' C 'J is an a-localising triangulated subcategory. All this is proved in Chapters 3 and 4.

The categories 'J"' are, in a certain sense, the optimal choices for S. For each a, we have an exact functor

A('J) ~ t:x({'J"'yv,Ab) preserving coproducts and products, having a left adjoint F, and so that t:x({'J"'tv,Ab) is the Gabriel quotient of A('J) by the Serre subcategory of all objects on which the functor 1r vanishes. It is natural to study the subcategories 'J"' C 'J. EXAMPLE 1.10. The previous paragraphs may be a little confusing. But the upshot is the following. Suppose we are given a triangulated category 'J closed under coproducts, and a regular cardinal a. There is some mysterious, canonical way to define an a-localising triangulated subcategory, denoted 'J"' c 'J. The reader might naturally be curious to know what 'J"' is, in some simple examples. If 'J is the homotopy category of spectra, then 'J"' is the full subcategory of spectra with fewer than a cells. If 'J is the derived category of an

1. INTRODUCTION

13

associative ring R, then the objects of 'Ja turn out to be chain complexes of projective R-modules, whose total rank( =rank of the sum of all the modules) is a:. They are just the closure of 'J"' with respect to coproducts of fewer than f3 objects, and triangles. Call this statement zero of Thomason's localisation theorem. The rest of the theorem concerns the situation of a Verdier quotient. THEOREM 1.14. LetS be a triangulated category satisfying [TR5], ~ c S a localising subcategory. Write 'J for the Verdier quotient Sf~. Suppose there is a regular cardinal a:, a class of objects S C S"' and another class of objects R c ~ n S"', so that

~= (R)

S = (S).

and

Then for any regular f3 2: a:,

(R)f3 = ~f3 = ~ n s!3, (s)f3

=

s!3.

The natural map factors as and the functor sf3 1~(3 is fully faithful. If f3

-+

:r!3

> No, the functor sf3 1~(3

-+

:r!3

is an equivalence of categories. If f3 = No, then every object of 'Jf3 is a direct summand of an object in Sf3 j~f3.

Proof: Theorem 4.4.9.

0

Since I have been telling the reader that this theory is quite new, the reader may well wonder why Thomason's name is attached to it. Thomason proved the special case where a: = f3 = N0 , and S is the derived category of the category of quasi-coherent sheaves on a quasi-compact, separated scheme. (Actually, he studies the slightly more general situation of a semiseparated scheme. A scheme is semi-separated if it has an open cover by affine open subsets with affine intersections.) For details, the reader is referred to Thomason's [34]. In all fairness to Thomason, his wonderful observation was that this fact had great applications in K-theory. In

1. INTRODUCTION

15

any case, what is really new here is the generalisation to arbitrary regular cardinals a. This raises, of course, the question of why one cares. Thomason proved the theorem where a = f3 = N0 , and S is the derived category of the category of quasi-coherent sheaves on a quasi-compact, separated scheme. The author gave a simpler proof, which also generalised the result to all S, as long as a= f3 = N0 . This may be found in [23]. The obvious question is: who cares about the case of large regular cardinals? The short answer is that everybody should. First of all, it has already been mentioned that in the applications to motives, the case a = f3 = No does not apply. Only rarely is there a set of objects T c 'J'No, with (T) = 'J'. But even the people with both their feet firmly on the ground, the ones who could not care less about motives, should be interested in the case of large cardinals. The reason is the following. If 'J' is the homotopy category of spectra, it has been known for a long time that ']'No generates 'J'. But now let E be a homology theory. Following Bousfield, let 'J'E C 'J' be the subcategory of E-acyclic spectra, and let .l'J'E be the subcategory of E-local spectra. In general, {'J'E}No and {.l'J'E} No are small and very uninteresting. It is only for sufficiently large a that the categories {'J'E }'' and { .l'J'E }"' start generating. See Remark D.l.15, for an estimate on how large a must be. The moral is very simple. Suppose the main object of interest is a triangulated category to which Thomason's theorem, or my old generalisation of it, apply. That is, the main object of study is a category for which the case a = f3 = N0 is non-trivial. As soon as we Bousfield localise it, we get a category for which we are naturally forced into the large cardinal a generalisation. So far we have seen that, for each regular cardinal a, it is possible to attach to 'J' a canonically defined a-localising subcategory 'J'. We have also seen Thomason's localisation theorem, which says that the subcategories 'J' behave well with respect to Verdier quotients. But to convince the reader that the exercise is worthwhile, I must use the subcategories 'J' C 'J' to prove a statement not directly involving them. First we need a key definition. DEFINITION 1.15. Let a be a regular cardinal. Let 'J' be a triangulated category with small Hom-sets, satisfying [TR5j. If the subcategory 'J' is essentially small, and if ('J'"') = 'J', we say that 'J' is a-compactly generated. It turns out that if'J' is a-compactly generated, then it is also /3-compactly generated for any /3 > a. A triangulated category 'J' is said to be well generated if

1.15.1. 'J' has small Hom-sets. 1.15.2. 'J' satisfies [TR5j.

16

1.

INTRODUCTION

1.15.3. For some regular a, 'J' is a-compactly generated.

Of course, a well generated triangulated category is in fact (3-compactly generated for all sufficiently large f3. REMARK 1.16. It might be worth restating Thomason's localisation theorem (Theorem 1.14) in the above terms. Let S be a triangulated category satisfying [TR5], let :R c S be a localising subcategory, and let 'J' = S/:R be the Verdier quotient. Thomason's localisation theorem asserts that if S is well generated then, under mild hypotheses, so are :R and 'J'. The hypotheses are that :R be generated by a set of its objects. The precise statement keeps track of the cardinals f3 for which these categories are (3compactly generated. The reader is refered to Theorem 1.14 for the more refined statement. Now we begin the main theorems of the book. THEOREM 1.17. (BROWN REPRESENTABILITY). Let 'J' be any wellgenerated triangulated category. Let H be a contravariant functor H : 'J'0 P ----+ Ab, where Ab is the category of abelian groups. The functor H is representable if and only if it is homological, and takes coproducts in 'J' to products of abelian groups. In other words, the representable functors 'J'(-, t) can be characterised as the co homological functors taking coproducts in 'J' to products in Ab.

Proof: Theorem 8.3.3.

D

This theorem has several immediate corollaries. One of them is COROLLARY 1.18. Let 'J' be any well-generated triangulated category. Then 'J' satisfies [TRS* ]; all small products exist in 'J'.

Proof: Given a set {X>., A E A} of objects in 'J', the following functor 'J'OP ----+ Ab

H(-)

II 'J'( -,X>.)

.\EA

is homological, and takes coproducts in 'J' to products of abelian groups. By Theorem 1.17 it is representable. The representing object is then the product, in 'J', of {X>., A E A}. See also Proposition 8.4.6. D In other words, we learn that a well-generated triangulated category is closed under products. This leads naturally to the question of whether the dual of Brown's representability theorem holds. We prove THEOREM 1.19. (BROWN REPRESENTABILITY FOR THE DUAL). Let'J' be any well-generated triangulated category. Choose some regular cardinal

1.

INTRODUCTION

17

a, for which 'J' is a-compactly generated. Suppose for that a, the abelian category ex ({'J'"'Yp' Ab) has enough injectives. Let H be a covariant functor H : 'J' - - t Ab. Then H will be representable if and only if it is homological, and takes products in 'J' to products of abelian groups. In other words, the representable functors 'J'(t,-) can be characterised as the homological functors respecting products. Proof: Theorem 8.6.1.

0

We can formalise this DEFINITION 1.20. A triangulated category 'J' satisfying [TR5j is said to satisfy the representability theorem if the (contravariant) representable functors 'J'( -, t) are precisely the homological functors H : 'J'0 P - - t Ab taking coproducts to products. The content of Theorems 1.17 and 1.19 is that well-generated categories 'J' satisfy the representability theorem, as do their duals, provided {'J'"' Ab) has enough injectives. An easy proposition states

ex (

YP,

PROPOSITION 1.21. Let F : S - - t 'J' be a triangulated functor of triangulated categories. Suppose S satisfies the representability theorem, as in Definition 1.20. The functor F has a righ;t adjoint G: 'J' - - t S if and only ifF respects coproducts.

Proof: Theorem 8.4.4.

0

REMARK 1.22. In Remark 1.4 we noted that the deep questions about triangulated categories involve the existence of adjoints. Let F : S - - t 'J' be a triangulated functor. By Corollary 1.3, F has a right adjoint if and only if A( F) : A(S) ---+ A('J') does. But this amounts to reducing a difficult problem to an impossible one. The categories A(S) are terrible, and the author does not know a single example where one can show directly that A(F) : A(S) - - t A('J') has an adjoint. By contrast, Proposition 1.21 is practical to apply. If S or gop is wellgenerated, F will have a right adjoint if and only if it preserves coproducts. REMARK 1.23. It should be noted that Franke has independently obtained a representability theorem strongly reminiscent of Theorem 1.17, Franke's theorem also assumes that 'J' can be written as a union of T"' satisfying suitable hypotheses. But this is where the similarity becomes confusing. It is not clear whether the 'J'"''s studied here in general satisfy the hypotheses placed on Franke's T""s. It also is not clear whether there could be some other choice for Franke's T"''s. In his application, to the derived category of a Grothendieck abelian category, Franke's T"' is just the 'J'"' we have been studying here. See Franke's [11].

18

1. INTRODUCTION

Franke's method does not generalise to the dual of a well-generated triangulated category. So far in the Introduction, we have presented the main results of the book. We ordered them in a way that motivated the definitions. We began with Freyd's construction of the universal homological functor 'J----> A('J). We discussed its properties, and the fact that, in general, A('J) is terrible. Then we spoke about approximations to A('J), in particular approximations of the form ex(S 0 P,Ab), for an a-localising subcategory S c 'J. We reasoned that, for every regular cardinal a, there is a canonical best choice for S; the largest possible S is 'J"'. Thomason's localisation theorem is the statement that 'J"' behaves reasonably well with respect to Verdier localisations. Our main theorems give an application of the 'J"''s. The first major theorem asserts that the representability theorem holds for 'J whenever 'J"' is essentially small and generates 'J. The second asserts that the representability theorem holds for 'J0 P if 'J is a-compactly generated, and furthermore ex ( {'J"'Vp' Ab) has enough injectives. Now it is time to explain the way the exposition of these facts is organised in the book, and to discuss some of the less major theorems that we prove on the way, or as consequences. The order in which the results are presented in the book is the Bourbaki order. It is the logical order, not the order that would motivate the constructions. Chapters 1 and 2 give the elementary properties of triangulated categories. Chapters 3 and 4 give the definitions of the categories 'J"', and their formal properties. This culminates in Thomason's localisation theorem, which asserts 'J"' passes to Verdier quotients. This is quite unmotivated. We define the categories 'J"', and study their formal properties, before we have any indication that they might be of some use. Only in Chapter 5 do we treat Freyd's classical theorem, concerning the universal homological functor. In Chapter 6 we finally come around to the categories ex ( {'J"'} op, Ab). We develop the elementary properties of the categories ex(S 0 P,Ab), and of the functor A('J) ----t £x(S 0 P,Ab). Chapter 6 should help clarify, somewhat belatedly, the point of studying the categories 'J"'. In Theorem 1.19, we saw that Brown representability sometimes holds for the dual of 'J; in particular, it holds if ex (gav, Ab) has enough injectives. It becomes interesting to study whether there are enough injectives. The general answer is No. Counterexamples may be found in Sections C.4 and D.2. Nevertheless, it is possible that Brown representability for the dual could be proved with less than the existence of injectives. The homological algebra of the categories ex(S 0 P,Ab) is interesting, and its careful study might yield great results. In Chapter 7, I assemble assorted facts I know. These do not really lead anywhere yet, but I thought they

1. INTRODUCTION

19

might be useful to future researchers. The Chapter may safely be skipped by all but the truly committed. The category ex(S 0 P,Ab) does not satisfy [AB5]. We remind the reader: this means direct limits of exact sequences need not be exact. It follows that ex (gop, Ab) is not a Grothendieck abelian category, and hence the classical proofs of the existence of injectives break down. Of course, the proofs must break down, since we know that there are ncit, in general, enough injectives. But it is interesting to analyse just where the breakdown occurs. We will analyse this for the argument that appears in Grothendieck's Tohoku paper; see Theoreme 1.10.1, on page 135 of [18]. I do not know the origin of the argument; Grothendieck said that it had been well-known, and he was merely sketching it. The argument is based on adding cells. We should perhaps remind the reader. Let A be an abelian category, x an object of A. We wish to embed x in an injective I. This means that given any extension in Ext 1 (z,x), that is any exact sequence 0 the map x factor as

--+

-----+

X

-----+ y -----+ Z -----+

0,

I should kill it. In other words, the map x X

-----+

y

-----+

--+

I should

I.

This suggests a natural way to try to construct I. If x is not injective, it has an extension

·o

-----+

x -- y -- z --

o.

If y is not injective, we can repeat the process. We can construct a sequence of monomorphisms

x = x0

-----+ x 1 -----+ x 2 -----+ · · ·

and hope that the colimit of the xi will be injective. This is the process of adding cells, and the proof in [18] was based on a slightly refined version of this construction.· The first and most serious problem with this construction is that, for an abelian category not satisfying [AB5], it is not clear that x injects into the colimxi. It might well be that colimxi = 0. We say that an abelian cat--+

-+

egory satisfies [AB4.5] if, for any (transfinite) sequence of monomorphisms as above, the map x 0 --+ colim xi is injective. ~

Actually, for the purpose of the proofs given in the article,. it is convenient to give an equivalent statement in terms of the derived functors of the colimit. An abelian category satisfies [AB4.5]· if, for any (transfinite) sequence of monomorphisms as above, and for any n 2:: 1, the nth derived

20

1. INTRODUCTION

functor of the colimit vanishes. That is,

We do not prove the equivalence of the two statements; we use the second as a definition, and we use the fact that it implies the first. The converse is true, but of no importance to us. One can easily show (Lemma A.3.15) that if an abelian category satisfies [AB3] (has coproducts) and has enough injectives, then it satisfies [AB4.5]. It is instructive to know that, if g is sufficiently ridiculous, the category ex(goP,Ab) need not satisfy [AB4.5]. For this reason, in Chapter 6 we begin by defining ex(goP,Ab) for fairly arbitrary g. For the g•s for which we define it, ex(goP,Ab) always satisfies (AB4]; coproducts are exact. See Lemma 6.3.2. However, if g is the category of normed, nonarchimedean, complete topological abelian groups, then ex(goP,Ab) does not satisfy [AB4.5]. See Proposition A.5.12. But in this book, we are mostly interested in the case where the category g is triangulated. I have no example of a triangulated category g, for which I can show that ex(goP,Ab) does not satisfy [AB4.5]. For a while, I thought I could prove [AB4.5] for such ex(goP,Ab). But there is a gap. Included in Chapter 7, is the part of the argument that is correct. The study of derived functors of co-Mittag-Leffier sequences in abelian categories is of some independent interest, and the existence of an abelian category ex (gop, Ab), satisfying [AB4] but not [AB4.5], is new and surprising. The reader is referred to Proposition 1 in [29], or Lemma 1.15 on page 213 of (20], to see just how striking it is. Since the results are about abelian, as opposed to triangulated, categories, they have been put in an appendix; see Appendix A. The property [AB4.5] is extensively studied, for the abelian categories ex(goP,Ab), in Chapter 7 and Appendix A. The study is inconclusive, but might be helpful to others. This occupies most of Chapter 7. But the final section, Section 7.5, is quite unrelated. Let g be a triangulated category dosed under coproducts of < a of its objects. The category ex (gop, Ab) is the category of functors gop ----+ Ab, which respect products of fewer than a objects. It is contained in the category eat(goP,Ab), of all additive functors gop ----+ Ab. Let i be the inclusion

In Section 7.5, we prove that i has a left adjoint j. So far, this is a special case of a theorem of Gabriel and Ulmer (16]. But more interestingly, the functor j has left derived functors Ln j. And most remarkably, if F is an object of ex(goP,Ab), then iF is an object of eat(goP,Ab), and we prove

1. INTRODUCTION

21

that, for n 2: 1,

Lnj{iF}

0.

An easy consequence is that, given objects F and Gin the abelian category ex(S 0 P,Ab), the groups Extn(F,G) agree, whether we compute them in ex(S 0 P,Ab) or in the larger eat(S 0 P,Ab). Chapter 8 has the proof of the two Brown representability theorems. It shows how the earlier theory can be used to prove practical theorems. It is now time to give fairly precise statements of what we prove, and explain the consequences. In Theorems 1.17 and 1.19, we saw that if 'J' is well-generated, then the representability theorem holds for 'J', and also for 'J'0 P, as long as some abelian category has enough injectives. See Definition 1.20 for what it means for triangulated category 'J to satisfy the representability theorem. These are true statements, but we prove more. Now it is time to be precise about what we prove. To say that 'J is well-generated asserts that, for some regular cardinal a, S = 'J is essentially small, and the natural homological functor

a

'J

---t

ex(S 0 P,Ab)

does not annihilate any object. A more precise statement of the theorem we prove would be THEOREM 1.24. Suppose 'J' is a triangulated category satisfying [TRS}. Suppose a is a regular cardinal, and suppose S C 'J is an a-localising subcategory. Suppose S is essentially small, and suppose the natural homological functor

'J

---t

ex(S 0 P,Ab)

does not annihilate any object, and respects countable coproducts. Then 'J' satisfies the representability theorem. Note that we do not assume, in the statement of the theorem, that S = 'J. If S = 'J, the conditions placed on S amount to saying that 'J is acompactly generated. For S = 'J, Theorem 1.24 reduces to Theorem 1.17. The fact that we allow other S's is a generalisation. Now let us analyse this. If we assume that the map 'J

---t

ex(S 0 P,Ab)

preserves all coproducts, then the generalisation is in fact very minor. Let us explain why. The point is that if 'J

------->

ex(S 0 P,Ab)

respects all coproducts, then S C 'J. Recall that 'J is the largest of all the S's for which coproducts are preserved. It contains all others, in particular

22

1. INTRODUCTION

it contains S. It turns out that if a 2:: N1 , S C 'J is a-localising, and the map 'J

----t

cx(S 0 P,Ab)

does not annihilate any object, then S = 'J. We know this because in Theorem 8.3.3 we prove 'J = (S), and Thomason's localisation theorem (Theorem 4.4.9) then tells us that has a . . t" as a ng a JOin mJec IVes cogenerator And in the counterexamples of Sections C.4 and D.2, we see that in general the category £x(S 0 P,Ab) need not have a cogenerator. Thus the right adjoint G need not exist, and C:x (gop, Ab) may fail to have enough injectives. Injective objects and right adjoints are abstract and perhaps uninviting. It is therefore illuminating to rephrase everything in terms of phantom maps. A morphism f : x - + y in 'J is called a-phantom if its image vanishes in C:x(goP,Ab). That is,

'J(-,f)ls: 'J(-,x)ls

---->

'J(-,y)ls

is the zero map. In Lemma 8.5.20, we prove that the right adjoint to A('J)-+ C:x(goP,Ab) exists if and only if, for every object z E 'J, there is a maximal a-phantom map y - + z. That is, every a-phantom x - + z must factor, non-uniquely, as X

-----+

y

----> Z.

In Lemma 8.5.17, we see that the category C:x(goP,Ab) will have enough injectives if and only if, for every object z E 'J, the maximal a-phantom map y-+ z may be so chosen that, in the triangle y

---t

z

---t

t

- - - t ~y

the object t is orthogonal to the a-phantom maps. Every a-phantom map x-+ t vanishes. If the category £x (gop, Ab) has enough injectives, there is a right adjoint G to the functor 1r : A('J) - + C:x(goP,Ab). Let I be an injective cogenerator of C:x(goP,Ab). We may form the object GI. Since G has an exact left adjoint 1r, GI must be injective in A('J). That is, GI is really an object in 'J C A('J); the injective objects are direct summands of objects in 'J, and as 'J satisfies [TR5], idempotents split in 'J. See Proposition 1.6.8. We call the object GI a Brown-Comen(itz object, and denote it lffiC. The Brown-Comenetz objects are somehow crucial to our proof that the dual of 'J satisfies the representability theorem. Since C:x (gop, Ab) need not have enough injectives, it would be nice to have another proof, which does not so critically hinge on the existence of injectives in ex (gop, Ab). The last chapter before the appendices is Chapter 9. It discusses Bousfield localisation. It is relatively short, and exposes no new results. Let me briefly tell the reader the contents of the Chapter. Let 'J be a triangulated category, g c 'J a triangulated subcategory. We say that a Bousfield localisation functor exists for the pair g c 'J if the

1. INTRODUCTION

25

map 'J ---+ 'J/S has a right adjoint. Chapter 9 explores in some detail what happens. It turns out that the adjoint 'J/S ~ 'J is always fully faithful, which allows us to think of 'J/S as a subcategory of 'J. The Verdier quotient 'J 'J/S is naturally isomorphic to S, and the embedding

TJs

= S

is left adjoint to the quotient map 'J

--+

--+

TJs

'J

= S.

There are some parallels with Gabriel's constructions of quotients of abelian categories. Anyway, we hope the reader finds the exposition of Chapter 9 amusing, even if the results are basically all known. In the second volume, which Voevodsky and the author still promise to write, there will hopefully be more about Bousfield localisation. There are also some appendices. The appendices contain two types of results. The first is background which the reader will need, and which is assembled here for convenience. There are several facts we want to use about abelian categories, which go beyond the elementary homological algebra that is a prerequisite for the book. These results maybe found elsewhere, scattered around the literature. But Appendix A offers ~he reader a condensed summary. See Section A.l for locally presentable categories, Section A.2 for Gabriel's treatment of localisation in abelian categories, and Sections A.3 and A.4 for the derived functors of limits. The remaining material in the appendices is new, often of independent interest. Generally, if a result has no strong, direct bearing on the development of the theory, it is left to appendix. For example, Appendix A contains more than just a summary of known facts about abelian categories. The treatment of Mittag-Leffler sequences is new, as is [AB4.5]. And the example of the category S for which ex (gop, Ab) satisfies [AB4] but not [AB4.5] is not only new, it is quite surprising. It goes against the expectations in the literature. Since it is only tangentially related to the main subject of the book, the reader will find it consigned to Section A.5. Appendix A is about abelian categories, with some of the material being old, some new. The remaining appendices offer results about triangulated categories. These also divide into two types. In Appendix B, we show that the functor 'J---+ ex({'JayP,Ab) can be characterised by a universal property. An abelian category A is said to satisfy [AB5] if a-filtered colimits are exact in A. We prove

an

26

1. INTRODUCTION

THEOREM 1.25. Let 'Y be an a-compactly generated triangulated category. The coproduct-preserving homological functors H : 'Y ----; A, where A is an abelian category satisfying [AB5cv. ], factor uniquely, up to canonical equivalence, as

'Y

----+

ex( {'Ycv.Vv,Ab)

~ A,

with the functor ex ( {'Ycv.VP, Ab) ----; A coproduct-preserving and exact. Natural transformations between coproduct-preserving homological functors 'Y ----; A are in 1-to-1 correspondence with natural transformations of coproduct-preserving exact functors ex ( {'Ycv.VP, Ab) ----;A. Proof: Theorem B.2.5.

D

This result is the type that perhaps merits inclusion in the body of the book. But we do not really use it elsewhere. It gives more evidence that the construction of the categories 'Jcv. is natural. Beyond this, it does not have a strong bearing on the development of the theory. At least, not yet. For this reason, it was put in an appendix. I would like to thank Christensen, who kept asking me about such results. Christensen and Strickland, in [9], proved the assertion when 'Y is the category of spectra, and a = N0 . Their methods do not seem to generalise, even to other 'Y but with a still N0 . If not for Christensen's prodding, I would probably never have obtained the result. Now for the remaining three appendices. These are basically examples. In Appendix D, we work out in some detail what the general theory says in the special case, where 'Y is the homotopy category of spectra. We began the Introduction with this, hence we will not repeat it. There are two remaining Appedices, C and E. These mostly are about pathological behavior. The reader is expected to know a little bit about the derived category to read these examples. The body of the book does not discuss examples, and does not depend on knowing any. But in the appendices, we assume some acquaintance with the derived category. Appendix C has two results. First it proves that, in general, the objects of Freyd's universal category A('J) have classes, not sets, of subobjects. Very concretely, we show it for the object Z E D(Z) c A [D(Z)]. This result is certainly known to the experts, and in fact seems to have been independently rediscovered several times. The earliest reference I could find is Freyd's [14]. The second result is that the category ex(S 0 P,Ab) need not have a cogenerator. As we have seen above, this also means that there need not be a right adjoint to 7r: A('J)----; ex(S 0 P,Ab), and that ex(Sop,Ab) need not have enough injectives.

1. INTRODUCTION

27

Appendix E offers examples of categories which are not well-generated. In Corollary E.1.3, we see that if 'J is No-compactly generated, then cyop cannot be well-generated. For this result, the reader does not need to know any examples. In Section E.3, more specifically Summary E.3.3, we see that if K(Z) is the homotopy category of chain complexes of abelian groups, then neither K(Z) nor K(ztP is well-generated. Section E.2 treats a condition possibly weaker than well-generation. It shows that the dual of D(Q), the derived category of vector spaces over the field Q, is not even N1-perfectly generated. One thing should be noted. In Corollary E.l.3, we prove that the dual of an N0 -compactly generated 'J cannot be well-generated. If 'J is the homotopy category of spectra, we deduce that 'J0 P cannot be wellgenerated. Hence 'J and cyop cannot be equivalent. The assertion that the homotopy category is not self-dual is an old theorem of Boardman, [2]. What we have here is a generalisation. It is not the best generalisation one can prove, but in the interest of simplicity, it is the one we give.

CHAPTER 1

Definition and elementary properties of triangulated categories 1.1. Pre-triangulated categories DEFINITION 1.1.1. Let e be an additive category and E : e ---+ e be an additive endofunctor of e. Assume throughout that the endofunctor E is invertible. A candidate triangle in e {with respect to E) is a diagram of the form:

X~Y~Z~EX

such that the composites v o u, w o v and Eu o w are the zero morphisms. A morphism of candidate triangles is a commutative diagram X

tl

~y ~z ~EX

gl

hl

~~1

v'

u' Y' -,------> Z' ~EX' X' -,------> where each row is a candidate triangle.

DEFINITION 1.1.2. A pre-triangulated category 'J is an additive category, together with an additive automorphism E, and a class of candidate triangles (with respect to E) called distinguished triangles. The following conditions must hold:

TRO: Any candidate triangle which is isomorphic to a distinguished triangle is a distinguished triangle. The candidate triangle X ~ X is distinguished. TRl: For any morphism triangle of the form

f:

X ~ Y

--:--->

0

--:--->

EX

X---+ Yin 'J there exists a distinguished --:--->

Z

--:--->

EX

TR2: Consider the two candidate triangles X~Y~Z~EX

1. ELEMENTARY PROPERTIES

30

and ~Y.

If one is a distinguished triangle, then so is the other. TR3: For any commutative diagram of the form X~Y~Z~~X

X'~Y'~Z'~~X' where the rows are distinguished triangles, there is a morphism h : Z ---+ Z', not necessarily unique, which makes the diagram X

~y

fl

gl

X' commutative.

u'

------7

~z ~~X

Efl

hl

Y' ~Z' ~~X'

REMARK 1.1.3. Parts of Definition 1.1.2 are known to be redundant. For instance, it is not necessary to assume that distinguished triangles are candidate triangles. In other words, we can assume that the distinguished triangles are sequences X~Y~Z~~X

without necessarily postulating that the composites v o u, w o v and ~u ow vanish. It follows from the other axioms that the composites v o u, w o v and ~u o w must be zero. Just consider the diagram X __!_____. X

------+

0

------+

~X

X~Y~Z~~X

The bottom row is a distinguished triangle by hypothesis, the top by [TRO]. But by [TR3] the diagram may be completed to a commutative X __!_____. X

X~Y

------+

0 ------+ ~X

1

w

------7

~X

and we deduce that v o u = 0. The vanishing of w o v and ~u o w follows from the above and axiom [TR2]. Similarly, it is not necessary to assume that the category 'J is additive; something slightly less suffices. It suffices to assume that the category 'J is

1.1. PRE-TRIANGULATED CATEGORIES

31

pointed (there is a zero object), and that the Hom sets are abelian groups. The fact that finite coproducts and products exist and agree follows from the other axioms. NOTATION 1.1.4. Let 'J be a pre--triangulated category. If we speak of "triangles" in 'J, we mean distinguished triangles. When we mean candidate triangles, the adjective will always be explicitly used. REMARK 1.1.5. If 'J is a pre--triangulated category, then clearly so is its dual 'J0 P. For 'J0 P, the functor E gets replaced by E- 1 . PROPOSITION 1.1.6. Let 'J be a pre-triangulated category. Then the functor E preserves products and coproducts. Let us state this precisely. Suppose {X.x, A. E A} is a set of objects of 'J, and suppose the categorical coproduct il>.EA X.x exists in 'J. Then the natural map

II {EX.x}---+ E

>.EA

{II

x.x}

>.EA

is an isomorphism. In other words, the natural maps EX.x ---+ E {

II X.x}

>.EA

give E {il>.EA X.x} the structure of a coproduct in the category 'J. Similarly, if ILEA X.x exists in 'J, then the natural maps E

{IT

x.x} ---+ EX.x

>.EA

give E

{TI>.EA

X.x} the structure of a product in the category 'J.

Proof: The point is that E, being invertible, has both a right and a left adjoint, namely E- 1 . There are natural isomorphisms

and

Hom(X,EY) ~Hom (E- 1 X, Y), induced by E- 1 . A functor possessing a left adjoint respects products, a functor possessing a right adjoint respects coproducts. Thus E respects both. D Because 'J0 P is a pre-triangulated category with E- 1 playing the role of E, it follows that E- 1 also respects products and coproducts.

32

1. ELEMENTARY PROPERTIES

DEFINITION 1.1.7. Let 'J be a pre-triangulated category. Let H be a functor from 'J to some abelian category A. The functor H is called homological if, for every (distinguished} triangle

X~Y~Z~~X

the sequence H(X)

~ H(Y) ~ H(Z)

is exact in the abelian category A. REMARK 1.1.8. Because of axiom [TR2], it follows that the sequence above can be continued indefinitely in both directions. In other words, the infinite sequence

H(~- 1 Z) H(~-lw) H(X) ~ H(Y) ~ H(Z) ~ H(~X) is exact everywhere. REMARK 1.1.9. A homological functor on the pre-triangulated category 'J0 P is called a cohomological functor on 'J. Thus, a cohomological functor is a contravariant functor H : 'J--+ A such that, for any triangle X~Y~Z~~X

the sequence

H(Z)

~ H(Y) ~ H(X)

is exact in the abelian category A. LEMMA 1.1.10. Let 'J be a pre-triangulated category, U be an object of 'J. Then the representable functor H om(U, -) is homological.

Proof:

Suppose we are given a triangle w

X~Y~Z ------+ ~X.

We need to show the exactness of the sequence

H om(U, X)

------+

H om(U, Y)

------+

H om(U, Z)

We know in any case that the composite is zero. Let f E H om(U, Y) map to zero in Hom(U, Z). That is, let f: U--+ Y be such that the composite

U~Y~Z is zero. Then we have a commutative diagram

u

------+

-1

------+ ~u

1.1.

PRE-TRIANGULATED CATEGORIES

33

The bottom row is a triangle by [TR2], the top row by [TRO] and [TR2]. There is therefore, by [TR3], a map h : U ---+ X such that Eh : EU ---+ EX makes the diagram above commute. But this means in particular that the square

EU~EU

Ehl

Efl

EX~EY commutes, and hence f = uoh. Thus, we have produced an h E H om(U, X) mapping to f E Hom(U, Y). D

REMARK 1.1.11. Recall that the dual of a pre-triangulated category is pre-triangulated. It follows from Lemma 1.1.10, applied to the dual of 'J', that the functor Hom(-, U) is cohomological. DEFINITION 1.1.12. Let H : 'J' ---+ A be a homological functor. functor H is called decent if

The

1.1.12.1. The abelian category A satisfies AB4*; that is products exist, and the product of exact sequences is exact. 1.1.12.2. The functor H respects products. For any collection {X>.,>. E A} of objects X>. E 'J' whose product exists in 'J', the natural map

is an isomorphism. EXAMPLE 1.1.13. The functor Hom(U,-) : 'J'---+ Ab is a decent homological functor. It is homological by Lemma 1.1.10, the abelian category Ab of all abelian groups ~atisfies AB4*, and H om(U,-) preserves products. DEFINITION 1.1.14. Let 'J' be a pre-triangulated category. A candidate triangle v

----+ z~

EX

is called a pre-triangle if, for every decent homological functor H : 'J'---+ A, the long sequence

is exact.

34

1.

ELEMENTARY PROPERTIES

EXAMPLE 1.1.15. Every (distinguished) triangle is a pre-triangle. A direct summand of a pre-triangle is a pre-triangle. The next little lemma will show that an arbitrary product of pre-triangles is a pre-triangle. CAUTION 1.1.16. There are pre-triangles which are not distinguished. See for example the discussion of Case 2, pages 232-234 of [22]. An example of a pre-triangle which is not a triangle is the mapping cone on the map of triangles in the middle of page 234, loc. cit. LEMMA 1.1.17. Let A be an index set, and suppose that for every >. E A we are given a pre-triangle X>-.

Y>-.

---*

---*

Z>-.

~

L;X>-..

Suppose further that the three products

rrx)o., >-.EA

>-.EA

exist in 'J. The sequence

II X>-.

---*

>-.EA

II Y>-.

II Z>-.

---*

>-.EA

~

>-.EA

II {L;X>-.}

is identified, using Proposition 1.1. 6, with

II X>-.

---*

>-.EA

II Y>-.

---*

>-.EA

II Z>-. ~ r;

>-.EA

{II

x>-.}

>-.EA

.

We assert that this candidate triangle is a pre-triangle. Thus, the product of pre-triangles is a pre-triangle. Proof: Let H : 'J - A be a decent homological functor. Because for each >. E A the sequence. X>-.

---*

Y>-.

---*

Z>-.

~

L;X>-.

is a pre-triangle, applying H we get a long exact sequence in A H (L;- 1 Z>-.)

---*

H(X>-.)

---*

H(Y>-.)

---*

H(Z>-.)

---*

H(L;X>-.)

and because A is assumed to satisfy AB4*, the product of these sequences is exact. But we are assuming H decent, and in particular by 1.1.12.2, the maps H

(rr

X>-.)

~

H(ITY>-.)

~

>-.EA

(II

>-.EA

z>-.)

II H(Y>-.)

>-.EA

>-.EA

H

II H(X>-.)

>-.EA

---*

II H(Z>-.)

>-.EA

1.1. PRE-TRIANGULATED CATEGORIES

35

are all isomorphisms. This means that the functor H, applied to the sequence

gives a long exact sequence. This being true for all decent H, we deduce that the sequence is a pre-triangle. 0 LEMMA

diagram

1.1.18. Let H be a decent homological functor 'J'-+ A. Let the

X

u ----+

y

v ----+

g1

!1

z

w

--------7

:EX

E/1

h1

' ~Z' ~:EX' X' ~Y' be a morphism of pre-triangles. Suppose that for every n E Z, H(:En f) and H(:Eng) are isomorphisms. Then H(:Enh) are all isomorphisms.

Proof: Without loss, we are reduced to proving H(h) an isomorphism. But then the diagram

H(f)

1

H(X')

H(g)

~

1

H(Y')

H(h)

1

H(Ef)

1

~ H(Z') ~ H(:EX')

H(Eg) H(Eu')

1

H(:EY')

is a commutative diagram in the abelian category A with exact rows. By the 5-lemma, we deduce that H(h) is an isomorphism. 0 LEMMA

1.1.19. In the morphism of pre-triangles

X

~y

!1

g1

~z ~:EX

h1

E/1

' ' X' ~Y' ~Z' ~:EX' if f and g are isomorphisms, then for any decent homological functor H, H(h) is an isomorphism.

Proof: If f and g are isomorphisms, so are r;n f and :Eng for any n. Hence Lemma 1.1.18 allows us to deduce that H(h) is an isomorphism. 0

36

1.

ELEMENTARY PROPERTIES

PROPOSITION 1.1.20. If in the morphism of pre-triangles X

~y ~z ~~X

g1

f1

EJ1

h1

X' ~Y' ~Z'

w'

---t

~X'

both f and g are isomorphisms, then so is h. Proof: By Lemma 1.1.19, we already know that for any decent homological functor H : 'J ~A, H(h) is an isomorphism. By Example 1.1.13, all representable functors H om(U, -) are decent, for U E 'J. We know therefore that the natural map

H om(U, h) : H om(U, Z)

---t

H om(U, Z')

is an isomorphism for every U. But then the map

Hom(-, h): Hom(-,Z)

---t

Hom(-,Z')

is an isomorphism. It follows from Yoneda's Lemma that his an isomorphism. D REMARK 1.1.21. Let u : X pleted to a triangle. Let X~

y

~

Y be given. By [TR1] it may be comv

---t

z

w

- - - t ~X

and w'

v'

u

X ----+ y - - - t Z' - - - t ~X be two distinguished triangles "completing" u. We have a diagram X

11

~y ~z~ ~X

11

11 w'

v'

Z' - - - t ~X which by [TR3] may be completed to a morphism of triangles X

~y - - - t

X

w ~y ~z - - - t ~X

11

h1

11 I

11 w'

X ~y ~Z' - - - t ~X and since 1 : X ~ X and 1 : Y ~ Y are clearly isomorphisms, Proposition 1.1.20 says that his an isomorphism. It follows that Z is well defined up to isomorphism. In fact, the entire triangle is well defined up to isomorphism. But this isomorphism is not in general canonical.

1.2. COROLLARIES OF PROPOSITION 1.1.20

37

1.2. Corollaries of Proposition 1.1.20 In this section, we will group together some corollaries of Proposition 1.1.20, which have in common that they concern products and coproducts. PROPOSITION 1.2.1. Let 'J be a pre-triangulated category and A any index set. Suppose for every A E A we are given a (distinguished} triangle

X>-. ------> Y>-. ------> Z>-. ------>

~X>-.

in 'J. Suppose the three products

exist in 'J. We know by Lemma 1.1.17 that the product is a pre-triangle

l!

X>-. ------>

l!

Y>-. ------>

l!

Z>-. ------>

~

{

l!

X>-.}

We assert that it is a distinguished triangle. Proof:

By [TRl], the map

II X)..

------>

>..EA

II Y>-.

can be completed to a triangle

II X)..

------>

>-.EA

II y)..

------>

Q

------>

>-.EA

~ { II X)..}. >-.EA

For each A E A, we get a diagram where the rows are triangles

IIx)..

------>

>-.EA

1

X>-.

IIY>-. ------> Q ------>

>-.EA

1

------>

y)..

------>

z)..

~{II X)..} >-.EA

1

------>

~X>-.

By [TR3] we may complete this to a morphism of triangles

38

1.

ELEMENTARY PROPERTIES

Taking the product of all these maps, we get a morphism

AEA

AEA

rrxA ~ IIYA ~ rrzA

AEA

AEA

AEA

Both rows are pre-triangles. The top row because it is a triangle, the bottom row by Lemma 1.1.17. It follows from Proposition 1.1.20 that this map is an isomorphism of the top row (a distinguished triangle) with the bottom, which is therefore a triangle. D REMARK 1.2.2. Dually, the coproduct of distinguished triangles is distinguished. PROPOSITION 1.2.3. Let 'J' be a pre-triangulated category. Let x~Y~z~Ex

X'~Y'~z'~EX'

be candidate triangles. Suppose the direct sum XEBX'

~

YEBY'

~

ZE!1Z'

~

EXEBEX'

is a distinguished triangle. Then so are the summands. Proof:

The situation being symmetric, it suffices to prove that x~Y~z~Ex

is a triangle. Since it is the direct summand of a pre-triangle, it is in any case a pre-triangle, by Example 1.1.15. Let x~Y~Q~Ex

be a distinguished triangle. The diagram

Q

X

( ~ )1

XEBX' ~ YEBY' ~ ZE!1Z' may be completed to a morphism of triangles

X

( ~ )1

~

y

( ~ )1

Q

1

~

EXEBEX'

~

EX

( ~ )1

XEBX' ~ YEBY' ~ ZEBZ' ~ EXEBEX'

1.2. COROLLARIES OF PROPOSITION 1.1.20

39

If we compose this with the projection

X EB X'

--t

Y EB Y'

--t

Z EB Z'

--t

:EX EB :EX'

:EX we get a morphism of pre-triangles

X --tY

X

--t

Y

--t

Z

--t

:EX

where the bottom is a pre-triangle because it is the direct summand of a triangle, and the top is a triangle. Once again, Proposition 1.1.20 implies that h is an isomorphism. Thus the bottom row is isomorphic to the top row, which is a distinguished triangle. By [TRO], the bottom row is also a triangle. D There is one special case of the above which we will have occasion to use later, especially in Section 1.4. Observe first LEMMA

1.2.4. Suppose we are given a candidate triangle

X

AEBY

--t

AEBZ

--t

:EX

Then this candidate triangle is isomorphic to a direct sum of candidate triangles

0---tA---tA---t X

--t

y

--t

z

--t

0

:EX

Proof: Consider the diagram

A

(1 0

X

--t

)j

AEBY

(

1 a (3 'Y

)

AEB Z

1( 1 A

--t

0 )

:EX

40

1. ELEMENTARY PROPERTIES

Since the composite

AEBZ ( 1 O) A AEBY is the map 1 A -----+ A, we deduce that the composite of the maps of candidate triangles

A

0

------+

1

X

1

A

(n1

------+

1

A

------+

(~ ) a 1'

AEBY

1

AEBZ

1(

1

1

A ------+ 0 ------+ is the identity. Thus the triangle

0

------+

I: X

1 0 )

A

0 ------+ A~ A is a direct summand of

------+

1 0

------+

-----7

D~)

1

0

~EX

AE!lZ X ------+ A EB Y The other direct summand may be computed, for example, as the kernel of the map

X

------+

1

(~ ~ )

AEBY

1(1

1

1

A 0 ------+ It is a candidate triangle X

------+

A EB Z

A

------+

Y

------+

Z

------+

------+

0 )

I:X

1

-----7

0

I:X

and the maps are all very explicitly computable. REMARK

X

1.2.5. If we start with a distinguished triangle

------+

A EB Y

O~)AE!lZ~Ex

0

1.2. COROLLARIES OF PROPOSITION 1.1.20

41

then, by Proposition 1.2.3, the direct summand X

-t

Y

-t

Z ----+ :EX

is also distinguished. Next we give two rather trivial corollaries, which nevertheless are useful. COROLLARY 1.2.6. The map f: X ---r Y is an isomorphism if and only if, for some Z (necessarily isomorphic to zero}, the candidate triangle

X _1____, Y ~ Z ~:EX is distinguished. Proof: If f is an isomorphism, then the diagram below

X ~X

11

!1

-t

0 ----+ :EX

11

1

X _1____, y - t 0 ----+ :EX defines an isomorphism of candidate triangles. The top is distinguished, hence so is the bottom. Thus, we can take Z = 0. Conversely, assume X _1____, Y ~ Z ~:EX is a distinguished triangle. It is the sum of the two candidate triangles X _1____, Y

-t

0

-t

:EX

and 0 - t O - t Z - - - - + 0, which by Proposition 1.2.3 must both be triangles. But then the diagram

X ~X

11

!1

-t

0

-t

:EX

11

1

X _1____, y - t 0 ----+ :EX gives a morphism of triangles. We know that 1 : X ---r X and 1 : 0 ---r 0 are isomorphisms. Proposition 1.1.20 implies that the morphism f : X ---r Y also is. Now in the morphism of triangles

x~x

0----+ :EX

0

----+ :EX

1.

42

ELEMENTARY PROPERTIES

we know that 1 : X -7 X and f : X -7 Y are isomorphisms, hence so is 0: 0 -7 Z. Thus, Z is isomorphic to zero. 0 COROLLARY 1.2.7. Any triangle of the form

X

Y

------+

------+

Z ~ IjX

is isomorphic to X

------+

that is, if the map Z

-7

X E9 Z

0

Z

------+

------+

IjX;

ljX vanishes, then the triangle splits.

Proof: By Corollary 1.2.6, for any isomorphism triangle

Z ~ Z In particular, we may take

f

------+

f :Z

--+

Z, there is a

0 ------+ IjZ.

= -1. By [TR2] it then follows that

0 ------+

z

z

~

------+

0

is a distinguished triangle. By [TRO], so is

X ~ X

------+

0 ------+ IjX.

From Proposition 1.2.1, we learn that so is the direct sum

X

------+

X E9 Z

------+

Z

------+

0

IjX.

XE9Z

------+

z

------+

0

IjX

z

0

But now the diagram

X

------+

11 X

11 ------+

y

------+

11 ------+

IjX

can be completed to a morphism of triangles, which must be an isomorphism by Proposition 1.1.20. 0

It is often necessary to know not only that Y is isomorphic to X E9 Z, but also to give an explicit isomorphism. We observe LEMMA 1.2.8. Let us be given a triangle

If v' : Z

--+

Y is a map such that

1.2. COROLLARIES OF PROPOSITION 1.1.20

composes to the identity on Z, then the map of triangles X

11 X

~XEBZ -------+

( u v' u

------+

11

)1 y

~

z

:EX

11

w ~z ------+

:EX

is an isomorphism.

Proof: First let us establish that there is a map of triangles X

11 X

------+

X EB Z

( u v' u

~

-------+

0

~

11

)1 y

z

~z

:EX

11 w

~

:EX.

We need to show the squares commutative, and perhaps the square

Z~:EX

needs a little reflection; we must prove that the composite

z Z~:EX

vanishes. We are given that the identity on Z may be factored as

Z~Y~Z, from which we conclude that the composite

z Z~:EX

may be rewritten

z w

y~z -------+

:EX.

43

44

1. ELEMENTARY PROPERTIES

Since wv

= 0, we do have a map of triangles 0

X ----+ X EB Z ----+ Z ----+ EX

X~ Y ~Z~EX. It is a map of triangles, where two of the vertical maps are isomorphisms; by Proposition 1.1.20, so is the third. P REMARK

1.2.9. Dually, given a triangle X~Y~Z~EX

and a map u' : Y

----+

X so that u'

----+X is the identity, then the map of triangles

X ----+ X EB Z ----+ Z ----+ EX is an isomorphism. In other words, given a triangle X~Y~Z~EX,

we get a canonical isomorphism whenever we give either a factoring of the identity on Z as

Z~Y~Z or a factoring of the identity on X as u

u'

X ----+ Y ----+ X. LEMMA

1.2.10. Suppose we have a triangle X~Y~Z~EX,

a factoring of the identity on Z as

z

v1

----+

y

v

----+

z,

and a factoring of the identity on X as

X~Y~X. Then we have two isomorphisms Y

~XEBZ,

1.3. MAPPING CONES

45

one from each factoring. These two isomorphisms will agree if and only if the composite v' u' Z ______. Y ______. X

vanishes. Proof: By Lemma 1.2.8, the map v': Z----+ Y gives an isomorphism

XEBZ

( u

v' )

Y,

and by the dual of Lemma 1.2.8, the map u' : Y phism

y

----+

X gives an isomor-

X EEl Z.

We need to decide whether the isomorphisms agree, meaning whether they are inverse to each other. To do this, it suffices to check whether

XEBZ

( u v' )

Y

XEBZ

composes to the identity on X EEl Z. But the composite is clearly

XEBZ

( u'u u'v' ) vu vv'

X EEl Z.

On the other hand, we know that u'u = 1 and vv' = 1, and vu = 0 since it is the composite of two maps in a triangle. This makes the matrix

X EEl Z

( 1 u'v' ) O 1

X EEl Z,

and it will be the identity precisely if u' v' vanishes.

0

1.3. Mapping cones, and the definition of triangulated categories DEFINITION 1.3.1. Let 'J be a pre-triangulated category. Suppose that we are given a morphism of candidate triangles

EX

X' ~ Y' ~ Z' ~EX'

1. ELEMENTARY PROPERTIES

46

There is a way to form a new candidate triangle out of this data. It is the diagram ( -v g

0

u'

( -w h

)

Z ffi Y'

Y ffi X'

( -Eu

0 ) v'

Ef

EX ffi Z'

0

w'

)

EY EEl EX'.

This new candidate triangle is called the mapping cone on a map of candidate triangles. DEFINITION

1.3.2. Two maps of candidate triangles

X

!1

~y ~z ~EX

h1

g1 I

~~l

X' ___::_____.. y I ~Z' ~EX'

and

X

f'

1

u

y

---t

g'

z

v

1

---t

h'

w

1

---t

EX

~f'l

X' ~Y' ~Z' ~EX' are called homotopic if they differ by a homotopy; that is, if there exist

Z ---->

~X,

then the long sequence H(~- 1 Z) - - >

H(X)

H(Y)

-->

---->

H(Z)

----> H(~X)

is exact. In fact, this is true not only for decent homological functors H, but for any additive functor. The point is that the identity on Cis homotopic to the zero map. There are 8,

H(X)

-->

H(Y)

---->

H(Z)

----> H(~X)

is chain homotopic to the zero map; hence the sequence is exact.

D

PROPOSITION 1.3.8. Let C be a contractible candidate triangle. Then C is a distinguished triangle.

Proof:

If C is the sequence w

X~Y ~z ----> ~X

then we can complete u : X

---+

Y to a triangle

1.3. MAPPING CONES

49

Because C is contractible, there are the three maps

X

X

u

y

u

y

v

z

w

v

z

w

1:X

1:X

giving the homotopy of 1c to the zero map. Consider the map ww. Clearly, 1:u o {ww} = 0, since 1:u o w = 0. But because H om(1:X, -) is a homological functor, when we apply it to the triangle

X~Y~Q~1:X we get an exact sequence. Since 1:u kills 1:X ---+ Q with w'w' = ww. Now form the map of pre-triangles

there must be a map

I]!' :

z~ 1:X

X~Y

11

ww,

w'w+v'il>

1

X~Y~Q

w'

-----+

1:X

with I]!' as above. The reader will easily show that it is a map of pretriangles; the diagram commutes. But as 1 : X ---+ X and 1 : Y ---+ Y are isomorphisms, so is w'w + v' by Proposition 1.1.20. We deduce that the top candidate triangle is isomorphic to the bottom, and the bottom is a distinguished triangle. D From now on, when we speak of contractible candidate triangles, we will call them contractible triangles. Since we know by Proposition 1.3.8 that they are all distinguished, there is no risk of confusion. Remember that, in this book, the word "triangle", with no adjective preceding it, means distinguished triangle. LEMMA

1.3.9. Let the diagram

X

!1

v ~y -----+

91

z

h1

w -----+

1:X

Ef1

X' ~Y' ~Z' ~1:X' be a map of pre-triangles in the pre-triangulated category 'J. mapping cone is a pre-triangle.

Then the

50

1. ELEMENTARY PROPERTIES

Proof: Let H be a decent homological functor. We need to show that H takes the mapping cone to an exact sequence. Because each row is a pre-triangle, we have two exact sequences

H(E- 1 Z)

H(X)

------+

H(Y)

--+

H(Z)

H(X')

------+

H(Y')

------+

H(Z')

------+

H(EX)

------+

and

H(E- 1 Z')

------+

------+

H(EX')

and a map between them. The mapping cone on this map of exact sequences is exact. But it agrees with what we get if we apply H to the candidate triangle ( -w h YEBX 1

( -L:u L:f

0 ) v1

0 wl

) L:Y EB L:X 1 •

L:X EB Z 1

ZEBY 1

Hence the candidate triangle is a pre-triangle.

D

Now suppose we are given a morphism of triangles

X

~y

~z ~EX

gl

fl I

~~1

hl I

X' ~Y' ~Z' ~EX' Then the mapping cone is a pre-triangle by Lemma 1.3.9. It turns out that it need not always be a triangle. Let us first analyse the trivial cases. LEMMA

is a triangle. Proof:

1.3.10. The mapping cone on the zero map between triangles

Consider the zero map

X

v ~y ------+

ol

ol I

z

ol v'

X' ~Y' ----+ Z' The mapping cone is the sequence

( -~ YEBX 1

0 ul

( -~

) ZEBY 1

w ------+

~I

EX

ol w' ------+

EX'

(

-L:~

)

L:X EB Z 1

This is nothing other than the direct sum of the sequences

X'~Y'~Z'~EX' and

0 ) w' L:Y EB L:X 1 •

1.3.

MAPPING CONES

51

The first row is a triangle by hypothesis. The second row is a triangle by [TR2], and because · X~Y~Z~2:X

is. The direct sum is therefore a triangle, by Proposition 1.2.1.

D

Because the mapping cone does not change, up to isomorphism, if we replace a map by a homotopic one, we immediately deduce CoROLLARY 1.3.11. Given a map of triangles

X

w ~y ~z ----+

/1

91

r:x

E/1

h1

X' ~Y' ~Z' ~2:X' if the map is homotopic to zero, then the mapping cone is a triangle.

D

COROLLARY 1.3.12. If either w

----+

r:x

or u'

v'

w'

X'

----+

X

~y ~z ~

Y' ----+ Z' ----+ 2:X' is a contractible triangle, and the other is a triangle, then the mapping cone on any map

/1

91

r:x

Ef1

hl

X' ~Y' ~Z' ~2:X' is a triangle. Proof: By Lemma 1.3.6, the map is homotopic to the zero map. Then by Corollary 1.3.11, the mapping cone is a triangle. D This is as far as one gets without further assumptions. Now we come to the main definition of this section: DEFINITION 1.3.13. Let 'Y be a pre-triangulated category. Then 'Y is triangulated if it satisfies the further hypothesis TR4': Given any diagram w

X~Y~Z ----+

r:x

X'~ Y' ~ Z' ~ 2:X'

52

1.

ELEMENTARY PROPERTIES

where the rows are triangles, there is, by [TR3j, a way to choose an h : Z -+ Z' to make the diagram commutative. This h may be chosen so that the mapping cone

y EfJ X'

~X EfJ

Z'

(

-~u

~~

0 w'

) ~y

EfJ ~X'

is a triangle. DEFINITION 1.3.14. A morphism of triangles will be called good if its mapping cone is a triangle. REMARK 1.3.15. [TR4'] can be restated as saying that any diagram

X'~Y'~Z'~EX' where the rows are distinguished triangles, may be completed to a good morphism of triangles. The author does not know an example of a pre-triangulated category which is not triangulated.

1.4. Elementary properties of triangulated categories We begin with the definition of homotopy cartesian squares. DEFINITION 1.4.1. Let 'J' be a triangulated category. Then a commu-

tative square

Y~Z

Y'

--------+

f'

Z'

is called homotopy cartesian if there is a distinguished triangle

y for some 8: Z'-+ EY.

Y' EB Z

(

f'

g' ) Z'

~

EY

1.4. PROPERTIES OF TRIANGULATED CATEGORIES NOTATION

1.4.2. If

Y~Z Y'

-------+

f'

Z'

is a homotopy cartesian square, we call Y the homotopy pullback of

z Y'

-------+

f'

Z'

and Z' the homotopy pushout of

Y~Z

Y' It follows from [TRl] that any diagram

Y~Z

Y' has a homotopy pushout; the morphism

Y

-------+

Y' EB Z

can be completed to a triangle

Y

( -f)

Y' EB Z

-------+

Z'

-------+ ~y

and this triangle defines a homotopy cartesian square y~z

y

I

-------+

f'

Z'.

53

54

1. ELEMENTARY PROPERTIES

By Remark 1.1.21, the homotopy pushout is unique up to non-canonical isomorphism. Also, any commutative square

Y' _____.... p corresponds to a map Y' EB Z

~

P so that the composite

y vanishes. But Hom(-, P) is a cohomological functor, and in particular takes the triangle

y

( _,)

Y' EB Z

------+

Z'

------+

L:Y

to a long exact sequence. We are given a map in H om(Y' EB Z, P) whose image in H om(Y, P) vanishes. It must therefore come from Hom( Z', P). There is a map Z' ~ P (non-unique) which maps the square

y ~

g1

z

1g'

------+

Z'

y ~

z

Y'

f'

to the square

g1 Y'

1

------+

P.

Dually, homotopy pullbacks always exist and are unique up to noncanonical isomorphism. And given a commutative square p ------+

1g'

1

Y'

z

------+

f'

Z'

1.4. PROPERTIES OF TRIANGULATED CATEGORIES

there is always a map P back square

-+

Y mapping this square to the homotopy pull-

Y' LEMMA

gles for rows

55

-----*

f'

Z'.

1.4.3. Let the following be a commutative diagram with trianX ~ Y

-----*

Z

-----+ }:;X

X ~ Y' -----* Z' -----+ 1;X. It may be completed to a morphism of triangles X ~ Y

g1

11 X

so that

-----*

gf

-----+

Y' y

Z

1

-----+ -----*

1

Z'

-----+ }:;X

11 w' -----+

}:;X,

z

1

Y' -----* Z' is homotopy cartesian. In fact, the differential 8 : Z' to be the composite Z' ~}:;X

-+

1;Y may be chosen

1;Y.

Proof: By [TR4'] we may complete

X~Y

11

Z

g1

X ~y

-----*

g1

X ~Y'

}:;X

11

X ~Y' -----* Z' to a good morphism of triangles

11

-----+

z

1 -----*

Z'

-----+

}:;X.

-----+

}:;X

11 w' -----+

1;X.

1. ELEMENTARY PROPERTIES

56

Then the mapping cone is a triangle

X EB Y - + Y' EB Z - + :EX EB Z' ------> :EX EB :EY. An elementary computation [see Lemma 1.2.4] allows us to show that this triangle is isomorphic to the direct sum of the two candidate triangles x-o-:Ex-:Ex

and

Y - + Y' EB Z ------> Z' - + :EY. By Proposition 1.2.3, each direct summand of a triangle is a triangle. In particular, Y - + Y'EBZ-+ Z ' - + :EY must be a distinguished triangle. It is also easy to compute that the differential is as in the statement of the Lemma. D LEMMA

1.4.4. Let

Y-+Z

Y'-+Z' be a homotopy cartesian square. If Y ~ Y' - + Y'' is a triangle, then there is a triangle

-+

:EY

Z ~ Z ' - + Y 11 -+:EZ which completes the homotopy cartesian square to a map of triangles :EY

Y~Y'-+Y''

1

1

1

Z ~ Z' -+Y''-+:EZ. That is, the differential Z' ---+ :EY is the composite Z'

-+

Y''

-+

:EY.

Proof: We know that the square

Y-z

Y'-+Z' is homotopy cartesian, in other words we have a triangle Y - + Y'EBZ-+ Z ' - + :EY.

1.4. PROPERTIES OF TRIANGULATED CATEGORIES

57

But then the diagram

Y

------+

Y'

EE1

Z

Z'

------+

------+

:EY

1

Y ------+ Y' ------+ Y'' ------+ :EY may be completed to a good morphism of triangles Y

------+

Y'

EE1

Z

Z'

------+

------+

:EY

1

1

Y ------+ Y' ------+ Y'' ------+ :EY and in particular the mapping cone is a triangle. The mapping cone can easily be written as a direct sum of the three candidate triangles

1

Y'

------+

Y'

------+

0

------+

:EY'

y

------+

0

------+

:EY

------+

:EY

1

z ------+ Z' ------+ Y" ------+ :EZ which must therefore all be triangles; in particular, Z

------+

Z'

------+

Y''

------+

:EZ

is a distinguished triangle. Computing the various maps, we deduce the map of triangles

y

1 z

g

------+

h

------+

Y'

------+

1

Z'

Y''

------+

:EY

------+

:EZ.

1

11 ------+

Y''

of the Lemma.

0

REMARK 1.4.5. Combining Lemmas 1.4.3 and 1.4.4, we have that good maps of triangles

X ~ Y

------+

Z

------+

:EX

X ~ Y' ------+ Z' ------+ :EX are closely related to homotopy cartesian squares

y

------+

z

Y'

------+

Z'.

1. ELEMENTARY PROPERTIES

58

One can pass back and forth from one to the other, of course not uniquely. By Lemma 1.4.3 a good map gives a homotopy cartesian square, and by Lemma 1.4.4, a homotopy cartesian square

y

-----+

z

Y'-----+ Z' and a triangle

give a good morphism of triangles X ___.!___, Y

X ~ Y'

-----+

Z

----+

EX

----+

EX.

1

---+

Z'

Putting together Lemmas 1.4.3 and 1.4.4 slightly differently, we deduce 1.4.6. Let 'J be a triangulated category. Let f : X --t Y Y' be two composable morphisms. Let us be given triangles

PROPOSITION

and g : Y

--t

X

---+

f

y

-----+

z

----+

EX

X

gf

-----+

Y'

----+

Z'

---+

EX

y

g

Y'

-----+

Y"

-----+

EY.

---+

Then we can complete this to a commutative diagram f

y

X

---+

X

gf

---+

Y'

---+

Y"

11 1 0

1

EX

'f:.j

---+

g1

---+

z

EX

----+

EX

1 ---+

Z'

1 1 1 Y" 1 1

EY

---+

----+

----+

EZ

11

1 ----+

0

1

---+

E2 X

where the first and second row and second column are our given three triangles, and every row and column in the diagram is a distinguished triangle.

1.4. PROPERTIES OF TRIANGULATED CATEGORIES

59

Furthermore, the square

y

---------+

1

1

Y

---------+

1

z

Z1

is homotopy cartesian, with differential being given by the equal composites Z 1 ---------+

~X ---------+ ~Y,

Z 1 ---------+ Y 11 ---------+

~Y.

Proof: By Lemma 1.4.3, the diagram

z-

X ~y ---------+

g1

11

11

X ____!!}____, y I ---------+ z~---------+ may be completed to a morphism of triangles

z

X ~y ---------+

g1

11

~X

1

X ____!!}____, y I ---------+

z~

so that ---------+

z

Yl ---------+

z~

y

1

-

~X

~X

11 ~X,

1

is homotopy cartesian. By Lemma 1.4.4, the homotopy cartesian square y ---------+

z

1

yl ---------+

1 z~

and the triangle Y ---------+ Y 1 ---------+ Y 11

may be completed to a map of triangles

-~Y

11

g y ---------+ Yl --------+YII

1 z

h

---------+

1 z~

--------+YII ---------+

~y

1 ~z,

60

1. ELEMENTARY PROPERTIES

and the diagram

~ Y

X

~ Y'

X

-----+

Z

-----+

:EX

-----+

:EX

1

-----+

1

1 0

-----+

Z'

1

1

Y'' ~Y''

1

1

0

1

:EX ~ :EY -----+ :EZ just assembles all this information together.

1 -----+

:E 2 X 0

REMARK 1.4.7. Proposition 1.4.6 is generally known as [TR4], or the Octahedral Axiom. The diagram whose existence the Proposition asserts is known as an octahedron. The reason for this name is that we do in fact have 8 triangles which can be assembled to an octahedron. The rows and the columns of the diagram give 4 distinguished triangles. But there are also 4 commutative triangles X ~ Y

and

-----+

Y''

-----+

:EX

1

X~Y' Y'

Z

Z' and

-----+~X

Y''-----+

1

~y

1

Z' ~ Y'' Y'' -----+ ~z. These 8 triangles make an octahedron, and the reader is referred elsewhere for the study of its symmetries. It may be shown that a pre-triangulated category satisfies [TR4] if and only if it satisfies [TR4 ']. We have shown the "if". The proof that any pre-triangulated category satisfying [TR4] also satisfies [TR4'] may be found in [22], Theorem 1.8. 1.5. Triangulated subcategories DEFINITION 1.5.1. Let 'J be a triangulated category. A full additive subcategory S in 'J is called a triangulated subcategory if every object isomorphic to an object of S is in S, if ~S = S, and if for any distinguished

1.5.

TRIANGULATED SUBCATEGORIES

61

triangle X

----t

Y

----t

Z

----t

EX

such that the objects X andY are inS, the object Z is also inS. REMARK 1.5.2. From [TR2] we easily deduce that if Sis a triangulated subcategory of 'J and X

----t

Y

----t

Z

----t

EX

is a triangle in 'J, then if any two of the objects X, Y or Z are inS, so is the third. DEFINITION 1.5.3. Let 'J be a triangulated category, S a triangulated subcategory. We define a collection of morphisms M ors C 'J by the following rule. A morphism f : X --+ Y belongs to M ors if and only if, in some triangle X __.!____, Y

----t

Z

----t

EX,

the object Z lies inS. REMARK 1.5.4. Note that it is irrelevant which particular triangle X __.!____, Y

----t

Z

----t

EX

we take. By Remark 1.1.21 the object Z is unique up to isomorphism, and by Definition 1.5.1, S contains all objects in 'J isomorphic to its objects. LEMMA 1.5.5. Every isomorphism f : X

Proof: Let f : X

--+

--+

Y is in M ors.

Y be an isomorphism. By Corollary 1.2.6, the

diagram X __.!____, Y

----t

0

----t

EX

is a triangle in 'J. But since S is, among other things, an additive subcategory of 'J, 0 must be in S. Thus f is in M ors. D LEMMA 1.5.6. Let f: X--+ Y and g: Y--+ Y' be two morphisms in 'J. If any two off: X--+ Y, g: Y--+ Y' and gf: X--+ Y' lie in Mors, then so does the third.

62

1.

ELEMENTARY PROPERTIES

Proof: By Proposition 1.4.6, there is a diagram of triangles f

X

~

11

y

~

X

~

1 0

~

1

"Ef

~

Y'

~

Z'

----t

1 1 1 1 1

Y"

----t

Y"

EX

11

1

91 gf

z

EX

1 ----t

0

1

EY - - - - t EZ - - - - t E 2 X EX Now f lies in M ors if and only if Z lies in S, gf lies in M ors if and only if Z' lies in S, and g lies in M ors if and only if Y" lies in S. From the triangle ~

Z

~

Z'

~

Y''

----t

EZ

we learn that if any two of Z, Z' andY" lie inS, then so does the third.D LEMMA 1.5.7. If S is a triangulated subcategory of 'Y, then there is a subcategory of'Y whose objects are all the objects of'Y, and whose morphisms are the ones in M ors.

Proof: Let X be an object of 'J. From Lemma 1.5.5 we learn that the identity morphisms 1 : X -+ X, being isomorphisms, lie in M ors. But by Lemma 1.5.6 we know that the composite of two morphisms in M ors is again in M ors. Thus M ors is a subcategory of 'J. D LEMMA 1.5.8. Let the square

y

f

----t

1g'

91 Y'

z

----t

f'

Z'

be a homotopy cartesian square. Then f is in M ors if and only if f' is, and g is in M ors if and only if g' is. Another way to phrase this is that homotopy pushout and homotopy pullback of morphisms in M ors give morphisms in M ors. Proof: The two statement being transposes of each other, we will prove only that g is in M ors if and only if g' is. By Lemma 1.4.4 the homotopy

1.6.

HOMOTOPY LIMITS

63

cartesian square above may be completed to a morphism of triangles g

y

Y'

------+

fl

------+

Z'

------+ g'

------+ ~y

11

lf'

z

Y"

------+

Efl

Y"

------+ ~z.

Now Y" will lie in S precisely if both maps g and g' are in M ors. This 0 proves that g is in M ors if and only if g' is.

1.6. Direct sums and products, and homotopy limits and co limits DEFINITION 1.6.1. Let a be an infinite cardinal. A triangulated category 'J is said to satisfy [TR5(a)] if, in addition to the other axioms, the following holds.

TR5(a): For any set A of cardinality< a, and any collection {X;.,>. E A} of objects of 'J, the coproduct X;. exists in 'J. If 'J satisfies

II

.AEA

[TR5(a)] for all infinite cardinals a, we say 'J satisfies [TR5]. DEFINITION 1.6.2. If the dual triangulated category 'J0 P satisfies the condition [TR5(a)], we say 'J satisfies [TR5*(a)J. If the dual satisfies [TR5], we say that 'J satisfies [TR5*]. REMARK 1.6.3. It follows from Proposition 1.2.1 and Remark 1.2.2 that coproducts and products of triangles are triangles. More precisely, if 'J satisfies [TR5*], then given any set of triangles, the product exists, and is a triangle by Proposition 1.2.1. Dually, if 'J satisfies [TR5], the coproduct of any collection of triangles exists and is a triangle. DEFINITION 1.6.4. Suppose 'J is a triangulated category, and assume it satisfies [TR5(N 1 )]. That is, countable coproducts exist in 'J. Let

X0

h ------+

X1

h ------+

X2

i3 ------+ ...

be a sequence of objects and morphisms in 'J. The homotopy colimit of the sequence, denoted Hocolim Xi, is by definition given, up to non-canonical isomorphism, by the triangle

gxi 00

1 - shift

Here, the shift map of Ji+1

gxi oo

II xi oo

i=O

:

------+

shift

------+

.

~xi

------+

~

{

gxi 00

}

.

II xi is understood to be the direct sum oo

i=O

Xi ---) Xi+l· In other words, the map {1- shift} is the infinite

64

1. ELEMENTARY PROPERTIES

matrix lx0 -jl

0 0

LEMMA

0 0 lx2

0 lx1 -j2 0

-j3

0 0 0 lx3

1.6.5. If we have two s.equences

Xo

-----7

X1

-----7

X2

Yo

-----7

Y1

-----7

Y2

-----7

• • •

and then,

-----7

· · •

non~canonically,

Hocolim {Xi EB Yi} =

{~Xi} EB {~ Yi}.

Proof: Because the direct sum of two triangles is a triangle by Proposition 1.1.20, there is a triangle

{gxi} EB {gYi} 1

{HocolimXi} EB {~ Yi} and this triangle identifies

{~xi}EB{~Yi} D LEMMA

1.6.6. Let X be an object of 'J, and let

X~X~·X~

be the sequence where all the maps are identities on X. Then ~X=X,

even canonically. Proof: The point is that the map

II X 00

i=O

1 - shift

II X 00

i=O

1.6. HOMOTOPY LIMITS

65

is split. Perhaps a simpler way to say this is that the map

( io

{1-shijt} )

ll

X

i=O

is an isomorphism, where i 0 : X -----+ U:o X is the inclusion into the zeroth summand. In other words, the candidate triangle

gx

1-shift

gx ~ x~ ~{gx}

where pr : ll:o X -----+ X is the map which is 1 on every summand, is isomorphic to the sum of the two triangles

and

0-------+ X~ X

Hence X is identified as LEMMA

----t

0. 0

~X.

1.6.7. If in the sequence

Xo

-------+ X1 -------+ X2 0

all the maps are zero, then

0

~Xi=

0

----t

0.

Proof: The point is that then the shift map in

fi xi

1 - shift

i=O

fi xi

i=O

vanishes. But by [TRO] there is a triangle

~xi~ gxi-------+ o

----t

~{gxi} 0

PROPOSITION 1.6.8. Suppose 'J is a triangulated category satisfying {TR5('N 1 )]. Let X be an object of 'J, and suppose e : X ---> X is idempotent; that is, e2 = e. Then e splits in 'J. There are morphisms f and g below

X ~ Y __f!___, X with gf

=e

and fg

= 1y.

66

1. ELEMENTARY PROPERTIES

Proof: Cosider the two sequences x~x~x

e

-------7

and

x~x

1-e

-------7

X~

Let Y be the homotopy colimit of the first, and Z the homotopy colimit of the second. We will denote this by writing Y = ~ (e) and Z = Hocolim (1- e). By Lemma 1.6.5, Y EB Z is the homotopy colimit of the direct sum of the two sequences, that is of

XEBX

XEBX

But the following is a map of sequences ( e

0

(

0

)

1-e

e

1-e

and in fact, the vertical maps are isomorphisms. The map (

e 1-e

1-e ) e

:

X EB X

----+

X EB X

is its own inverse; its square is easily computed to be the identity. It follows that the homotopy limits of the two sequences are the same. Thus Y EB Z is the homotopy limit of the bottom row, and the bottom row decomposes as the direct sum of the two sequences

X~X~X~··· and

x~x~x

0

-------7

By Lemma 1.6.6, the homotopy colimit of the first sequence is X, while by Lemma 1.6.7, the homotopy colimit of the second sequence is 0. The homotopy colimit of the sum, which is Y EB Z, is therefore isomorphic to XEBO =X.

1.6. HOMOTOPY LIMITS

67

More concretely, consider the maps of sequences X ~X ~X

el

el

el

X ~X ~X

and X

1-el

1-e

-----+

X

1-e

-----+

1-el

X

1-el

X ~X ~X

e -------+ ...

1

-------+

1-e

-------+

1

-------+

What we have shown is that the induced maps on homotopy colimits, that is g : Y - t X and g' : Z - t X can be chosen so that the sum Y EB Z - t X is an isomorphism. In the sequence x~x~x~

defining Y as the homotopy colimit, we get a map f map from a finite term to the colimit. In the sequence

:X

-t

Y, just the

X~X~X~··· the map from the finite terms to the homotopy colimit is the identity. We deduce a commutative square

X~Y x~x. Similarly, from the other sequence we deduce a commutative square X

!'

-----+

Z

x~x. In other words, we conclude in total that e = gf and 1 - e = g' f'. The composite

X

Y EB Z

( g

gl

)

--'------'-t

X

is e+(l-e) = 1. Since we know that the map YEBZ - t X is an isomorphism, it follows that the map X - t Y EB Z above is its (two-sided) inverse. The

68

1.

ELEMENTARY PROPERTIES

composite in the other order is also the identity. In particular, fg and f'g' = lz.

=

ly D

REMARK 1.6.9. Dually, if 'J' satisfies [TR5* (N1)], then idempotents also split. REMARK 1.6.10. In this entire section, we have used nothing more than countable coproducts and products. 1. 7. Some weak "functoriality" for homotopy limits and colimits In the last section we saw the definition and elementary properties of the homotopy colimit of a sequence. We also saw how homotopy colimits can be useful; for example; they prove that idempotents split. See Proposition 1.6.8. Since homotopy colimits are defined by a triangle, they are not in any reasonable sense functorial. But they do have some good properties. Let us mention one here. LEMMA 1.7.1. Let 'J' be a triangulated category satisfying {TR5(N1)]. Suppose we are given a sequence of objects and morphisms in 'J' Xo

-----+

X1

-----+

X2

------t

Suppose we take any increasing sequence of integers

0 ::; i 0 < i 1 < i 2 < i 3 < · · · Then we can form the subsequence xio -----+ xil -----+ xi2 ------t . . .

The two sequences have isomorphic homotopy colimits. Proof: For each integer n, define in to be the smallest im such that n::; im. Then there is a map of sequences Xo

-----+

X1

-----+

1

1

X2

------t · · ·

1

Xjo -----+ Xjl -----+ Xj2 ------t ...

This seems much more complicated than it actually is. If we start with the sequence

0, and the functor Funiv is the identity on objects. Now we turn to defining the morphisms in D je. Recall Definition 1.5.3. For the triangulated subcategory e c D, we defined a category More c D. A morphism f : X - t Y lies in More if and only if in the triangle X ~ Y -------+ Z -------+ :EX the object Z lies in e. We refer the reader to Section 1.5 for the elementary properties of More. Note that a triangulated functor F takes an object Z to zero if and only if it takes the distinguished triangle X ~ Y -------+ Z to a triangle isomorphic to the image of

X

1

-------+

X

-------+

0

-------+

----+

:EX :EX;

in other words F(f): F(X) - t F(Y) must be an isomorphism. Therefore, in the category Dje all the morphisms in More will become invertible. It is natural therefore to define DEFINITION 2.1.11. For any two objects X, Y in class of diagrams of the form

X

J/

z

y

1)

let o:(X, Y) be the

76

2.

VERDIER LOCALISATIONS

such that f belongs to More. Consider a relation R(X, Y) on n(X, Y) such that [(Z, f, g), (Z', f', g')) belongs to R( X, Y) if and only if there is an element (Z", J",g") in n(X, Y) and morphisms u: Z"-+ Z v: Z"-+ Z' which make the diagram !'

X

,/

vj

+--

Z"

!" f

"""

commute.

Z'

u!

z

g'

"\.

g" ----+ g

y

/'

LEMMA 2.1.12. With the notation as in Definition 2.1.11, u and v must be in More. Proof: The two cases being symmetric, it suffices to consider u. But fu = J", and f and f" are in More. By Lemma 1.5.6, it follows that u is in More. D REMARK 2.1.13. Elements of n(X, Y) should be thought of as maps gf- 1 in 'Dje, which would be well defined because in 'Dje any f E More is invertible. The existence of a diagram as in R(X, Y), that is

Z' !'

X

,/

vj

!"

Z"

+--

f

"""

says in particular

gf-1

u!

z

g'

"\.

g" ----+ g

y

/'

guu-1 f-1

g"{f"}-1 g'vv-1{!'} -1 g'{f'} -1 and it is therefore natural to identify (Z,f,g) with (Z',j',g'). LEMMA 2.1.14. The relation R(X, Y) is an equivalence relation.

2.1. VERDIER LOCALIZATION AND THICK SUBCATEGORIES

77

Proof: The only thing we need to check is that R is transitive. Let (Zi, li, gi), i = 1, 2, 3 be three elements in o:(X, Y) such that

[(Zz, Jz, gz), (Z3, /3, 93)]

and

both belong to R(X,Y). Then there are elements (Z,p,q), (Z',p',q') in o:(X, Y) and morphisms v: Z---+ Zz

u': Z'---+ Zz

v': Z'---+ Z3

which make the corresponding diagrams h

X

z1

,/

uj

+--

z

h ~

vl Zz

91

""

---->

y

92

/

and h

,/ X

+--

is

~

Zz

u' i Z'

v' l z3

92

""

---->

y

93

/

commutative. Consider a homotopy pullback diagram as in Notation 1.4.2 Z''~Z

u'

Z' -----. Zz

Lemma 2.1.12 says that v and u' are in More, and from Lemma 1.5.8 we deduce that so are w and w'. One can easily see now that ( Z", fz o v o w,g2 ovow) is an element of o:(X, Y), and that the morphisms morphisms

Z" ---+ Z' ---+ Z3

2. VERDIER LOCALISATIONS

78

make the diagram

z1

h

X

91

i

,/

""

zll

+---

h

---+

l

""'

y

93

/

z3

commutative. Therefore the pair [( Z1, h, gl), (Z3, fa, 93)] must belong to

R(X,Y).

D

2.1.15. We denote by 11 Hom';;;e (X, Y) the class of equivalence classes in a(X, Y) with respect to R(X, Y). DEFINITION

For an element (W1,f1,g1) in a(X,Y) and an element (W2,h,g2) in a(Y, Z), consider the diagram

w3

u

vl

92 --t

w2

--t

z

hl

W1 ___!!!___, y

hl X where the square

w3

u

--t

vl

w2

hl

w1 ___!!!___, y

is obtained by homotopy pullback. Since his in More, by Lemma 1.5.8 so is its homotopy pullback v. Since his also in More, so is the composite Av. We deduce therefore LEMMA

2.1.16. There is a map

a(X, Y) x a(Y, Z)

--->

11

H om';;;e (X, Z),

sending an element (W1,f1,g1) in a(X, Y) and an element (W2,f2,g2) in a(Y,Z) to (W3,f1 ov,g2 ou) in 11 Hom';;;e(X,Z).

2.1. VERDIER LOCALIZATION AND THICK SUBCATEGORIES

79

Proof: Since h ov is in More, the triple (W3,!I ov,g2 ou) is indeed in Hom'.JJ;e (X, Z). And since the homotopy pullback square

11

w3~W2

w1~ Y

is unique up to isomorphism in 1>, and all isomorphisms are in More, the image in 11 H om'.JJ; e (X, Z) is well-defined. 0 Next comes a little lemma which is useful in identifying these products. LEMMA

2.1.17. Given an element (W1 , h,gl) in a(X, Y), an element

(W2, h, g2) in a(Y, Z), and a commutative diagram p

u'

--7

hl

v'l

w1

Ill

w2 ~z

~y

X with v' in more, the elements (W3,j1 v,g2u) and (P,f1 v',g2u') agree in 11 Hom'.JJ;e(X,Z), where (W3,j v,g u) is obtained from a diagram 1 2

w3 ~w2 ~z

vl w1

Ill

hl ~y

X where the square

w3

vl w1

u

--7

w2

hl ~y

is a homotopy pullback.

Proof: The point is that, as in Notation 1.4.2, there is a map from the commutative square to the homotopy pullback square; there is a morphism

2. VERDIER LOCALISATIONS

80

p -

w3 in 1) which maps the square

v'

1

/2

1

to the square

This map gives the desired diagram, which establishes the equivalence of (W3, / 1 v, g2 u) and (P,/1 v', g2 u') modulo the relation R(X, Z). D We leave it to the reader to check the next Lemma LEMMA

2.1.18. The map

a(X, Y) x a(Y, Z)

-+

11

H om~;e (X, Z)

is consistent with equivalence relations R(X, Y) and R(Y, Z), and thus gives a pairing 11 Hom~;e(X,

Y) x 11 Hom~;e (Y, Z)-+ 11 Hom~;e (X, Z).

It is easy to check that the element (X, 1, 1) in 11 Hom~;e(X,X) is a two-sided identity for this composition. It is also easy to check that the composition satisfies the associative law. Let us actually do this check, by way of illusatration. LEMMA

11

2.1.19. The composition map

H om~;e (X, Y) x 11 H om~;e (Y, Z)

satisfies the associative law.

-+

11

H om~;e (X, Z)

2.1. VERDIER LOCALIZATION AND THICK SUBCATEGORIES

81

Proof: Let us be given (W, f,g) in a(X,X'), (W', J',g') in a(X',X") and (W", f", g") in a(X", X"'). Let

u

---+

r1

7"1

v

i'

---+

1

w

!1

V'

---+

W" ---+X'''

!" 1

W' ---+X''

!'1

---+X'

X be a diagram in which all the small squares are homotopy pullback squares. That is, the three squares

U---+ V'

V'---+ W''

V---+ W'

W'---+ X''

W---+ X'

7''1 V---+ W'

are homotopy cartesian. Then since f' is in More, so is its homotopy pullback f'. Since f" is in More, so is also its homotopy pullback f", and so is ]'', being the homotopy pullback of f". We deduce that all the vertical maps in the diagram are in More. But now just from the commutativity of the diagram, the fact that all the vertical maps are in More, and from Lemma 2.1.17, we deduce that by reading parts of the diagram we get the composites in the two orders of (W,J,g) in a(X,X'), (W',f',g') in a(X',X") and (W",J",g") in a(X",X"'). Since both come out to be

U---+ X'''

1

X the associative law follows.

D

2.1.20. With 11 H om~;e (X, Y) for the morphisms from X to Y, 'D je is therefore a category. From now on, we write H om'D;e (X, Y) instead of 11 H om~;e (X, Y) for the morphisms in this category. Define a functor Funiv : 'D --+ 'Dje to be the identity on objects, and to take DEFINITION

82

2. VERDIER LOCALISATIONS

f:X~Yto

X LEMMA 2.1.21. Let f: gory 'Dje, the morphisms

Y be a morphism in More. In thecate-

X~

X~Y

X

are inverse to each other. Proof: Consider the diagram

X

1

-----T

11

X ~X

!1

X ~y

11 X It shows that the composite X identity. But the diagram

X

~

1

-----T

11

Y ~ X, in the category 'Dje, is the

X ~y

11

X ~X

!1 y computes the composite Y

~X~

Yin 'Dje to be

2.1.

and the map

VERDIER LOCALIZATION AND THICK SUBCATEGORIES

f :X

---t

83

Y shows that this is equivalent to the identity

y 0

LEMMA 2.1.22. For any map in 'Dje of the form

one can write it as the composite of the two maps

w

w~w

__!!___., y

w Proof: The diagram

w

1

-----7

w

11

11

w

~w

!1

__!!___., y

X amounts to the proof; it computes the composite.

0

REMARK 2.1.23. Lemma 2.1.21 asserts that iff: X ---t Y is in More, then Funiv(f) is invertible in 'Dje. Lemma 2.1.22 states that every morphism in 'Dje is of the form Funiv(g)Funiv(f)-\ with f E More and g E 'D. Finally, Lemma 2.1.17 asserts that to compose two morphisms Funiv(91)Funiv(/I)- 1 :X - 7 Y and Funiv(92)Funiv(h)- 1 : Y ~ Z, we

84

2. VERDIER LOCALISATIONS

find a commutative diagram in 'D where all the vertical maps are in More p

v'l

w1

u'

-----7

w2

hl

~z

~y

frl

X and then

Funiv(92)Funiv(f2)- 1Funiv(91)Funiv(ft)- 1 = Funiv (gl)Funiv (u')Funiv (v') - 1 Funiv (it) - 1 . In other words, to give two composable morphisms X ---+ Y and Y ---+ Z in 'D je and their composite, is nothing other than to give X' --+ Y' and Y' --+ Z' in 'D and maps X' ---+ X, Y' ---+ Y and Z' ---+ Z in More expressing the isomorphism of the composable pair in 'D je with a composable pair in 'D. We may choose Z' ---+ Z to be the identity. PROPOSITION

all functors F : 'D morphisms.

2.1.24. The functor Funiv : 'D --+ 'Dje is universal for --+ ':J which take all morphisms in More to invertible

Proof: This is really immediate from the construction. If ':J is any category, and F : 'D --+ ':J is a functor sending More to invertible maps, then F extends to any diagram in a(X, Y), and obviously sends two diagrams equivalent modulo R(X, Y) to the same. D REMARK 2.1.25. The universal property of Funiv is self-dual. Thus the same construction on the dual category 'D 0 P will lead to {'DfetP. We deduce that morphisms in H omwe (X, Y) can also be described as diagrams

y

f'l X~W' where f' is in More. The equivalence relation on such diagrams is the dual of R(X, Y). Such a diagram is to be thought of as Funiv(f')- 1Funiv(g'). LEMMA 2.1.26. Let f and g be two morphisms X the following are equivalent

2.1.26.1. Funiv(f) = Funiv(g).

---+

Y in 'D. Then

2.1. VERDIER LOCALIZATION AND THICK SUBCATEGORIES

2.1.26.2. There exists a map o: : W

--7

85

X in More with fo: =

go:. 2.1.26.3. The map f- g: X

--7

Y factors as

x--7c-Y with C E

e.

Proof: Let us first prove the equivalence of 2.1.26.1 and 2.1.26.2. The morphisms Funiv(f) and Funiv(g) will agree in 'D je if and only if the diagrams

X

X

are equivalent. This will happen if and only if there is an object WE 'D, and maps o: 1 : W --7 X and o: 2 : W --7 X in More, rendering commutative the squares w~x

x~x

X _!1___, Y.

But the commutativity of the first square implies o: 1 = o: 2 = o:, and form the second square we learn that fo: =go:. Now let us prove the equivalence of 2.1.26.2 and 2.1.26.3. 2.1.26.2 will hold if and only if for some o: : W --7 X in More, (! - g )o: = 0. Consider now the triangle W ~ X

-------*

C

-------*

L;W

Because Hom(-, Y) is a cohomological functor, (!- g )o: = 0 if and only if f- g factors through C. But o: E More if and only if C E e. Consequently there exists an o: E More with (! - g )o: = 0 if and only if f - g factors through C E e. D LEMMA 2.1.27. Any commutative square in 'D je isomorphic to the image of a commutative square in 'D. More precisely, if

W-------* X

1

y

1

------7

z

86

2. VERDIER LOCALISATIONS

is a commutative square in 'D je, there is a commutative square in 'D

W'

-------+

X'

1

1

Y' -------+ Z' and maps in More W'---+ W, X'---+ X, Y'---+ Y and Z'---+ Z which, being isomorphisms in 'D ;e, express the isomorphism of the two diagrams. Proof: From Remark 2.1.23 we know that we can lift the composites W ---+ X ---+ Z and W ---+ Y ---+ Z to W1 ---+ X' ---+ Z and W2 ---+ Y' ---+ Z. Replacing W1 and W2 by the homotopy pullback

w3

-------+

w2

-------+

1

we may aSSUllle W1 diagram

w2.

w1

1

w

But now the commutativity in 'D je of the

w3

-------+

Y'

-------+

1

X'

1 z

means, by Lemma 2.1.26, that there is a map W' ---+ W 3 in More which equalises the two composites in 'D. Hence a commutative diagram in 'D

W'

-------+

1

X'

1

Y' -------+ Z. Thus we may even take Z' ---+ Z to be the identity.

D

Next we want to prove that 'D je is an additive category. LEMMA

2.1.28. The object 0 E 'D is a terminal and initial object in

'Dfe.

Proof: The two statements being dual, we may restrict ourselves to proving 0 terminal. Let X be an object of 'Djf!., that is an object of 'D. The diagram X

X

-------+

0

2.1. VERDIER LOCALIZATION AND THICK SUBCATEGORIES

exhibits a morphism

X~

87

0 in 'JJje. Given any other, p---+ 0

fl X then the map

f :P

~

X shows that

p---+ 0

X---+ 0

X are equivalent. Hence there is only one map X

~

0 in 1J je.

0

LEMMA 2.1.29. Let X and Y be two objects of 1J je, that is objects of ']). Then the direct sum X EB y in ']) is a biproduct in '])I e. It satisfies the universal properties both of product and coproduct.

Proof: There are maps in 1J

XEBY X

y

X

P1 ,/

y

XEBY

which give X EB Y the structure of a product (respectively coproduct) in 'JJ. We will show that these maps work also in 'JJje. The two statements being dual, we may restrict ourselves to showing the statement about coproducts. We need to show that X EB Y is the coproduct of X andY in 'JJje. Giving two morphisms X~ Q andY~ Q in 1J je is the same as giving equivalence classes of diagrams P'~Q

y

X

Since a and a' lie in More, they fit into triangles

P ~ X

---+

Z

---+

'EP

c/ ---+

---+

Z'

---+

'EP'

P'

Y

where Z and Z' are in e. The direct sum of these triangles is a triangle by Proposition 1.2.1. That is, we have a triangle

P EB P' ~ X EB Y

---+

Z EB Z'

---+

'EP EB 'EP'.

88

2 .. VERDIER LOCALISATIONS

But Z EB Z' is in

e,

and hence a EB a' is in More. This makes

PEBP'

1

XEBY a well-defined representative for a morphism in 1> je. The composite of it with

X~ XEBY is computed by the commutative diagram

P---+ PEBP'

to be just

X

and similarly for the composite with i 2 : Y ----+ X EB Y. Hence a pair of maps X--+ Q andY--+ Q in 1>je does factor through the object X EB Y. Now we need to show the uniqueness of the factorisation. For the uniqueness, it is handier to work with the dual description of morphisms in 1> je, as in Remark 2.1.25. Suppose therefore that we are given two morphisms in 1> je

Q

Q

1

1

XEBY ~ P

so that the composites with i1 and i2 agree. First, we may assume that P = P' and the vertical maps Q --+ P agree, by replacing P and P' by the homotopy pushout Q

1

---7

p

1

P'---+ N

2.1. VERDIER LOCALIZATION AND THICK SUBCATEGORIES

89

Thus, we may assume we have two diagrams

so that the composites with i 1 and i 2 induce the same morphism in 'Dje; that is

and Funiv(a)- 1Funiv(fi2)

= Funiv(a)- 1Funiv(9i2)·

Multiplying by Funiv(a), this means Funiv(fi1) = Funiv(9i1)

and

Funiv(fi2) = Funiv(9i2)·

By Lemma 2.1.26, this means that (!- g)h factors through C E e and (!- g)i2 factors through C' E e. This means that f- g factors through C EB C', and again by Lemma 2.1.26 we deduce that Funiv(f) = Funiv(g). But then Funiv(a)- 1Funiv(f)

= Funiv(a)- 1Funiv(g),

giving the uniqueness. LEMMA 2.1.30.

functor Funiv : 'D

0

The category 'D je is an additive category, and the 1> je is an additive functor.

--+

Proof: The category 'Dje is a pointed category by Lemma 2.1.28 (there is an object which is simultaneously initial and terminal). Also, 'D je has finite biproducts, by Lemma 2.1.29. Furthermore, the functor Funiv: 'D--+ 'Dje respects biproducts and the 0 object. There is therefore a well-defined addition on 'D je, and the functor Funiv : 'D --+ 1> je respects the addition. Given f : X --+ Y and g : X --+ Y, then f + g : X --+ Y is given by either of the composites

X

(; )

which agree. It remains only to show that this addition gives a group law, rather than just a commutative monoid; we need to show that a map X --+ Y in 'D je has an additive inverse. Let us be given a map in 'D je, it can be expressed as Funiv(a)- 1Funiv(f).

90

2. VERDIER LOCALISATIONS

But Funiv(a)- 1 Funiv(f)

+ Funiv(a)- 1 Funiv(- f)

=

Funiv(a)- 1 Funiv

(f + (-f))

Funiv(a)- 1 Funiv(O) 0 D

Next we need to discuss the triangulated structure on 'D je. There is a slight subtlety introduced by (TRO]. Recall that (TRO] asserts, among other things, that any object isomorphic to a distinguished triangle is a distinguished triangle. Before we continue the proof of Theorem 2.1.8, it will be helpful to study isomorphisms in 'Dje. LEMMA

2.1.31. If a morphism in 'Dje of the form

P~X

X is in the equivalence class of the identity 1 : X

-+

X, then f E More.

Proof: An equivalence of the above morphism with

x~x

11

X will consist of two commutative squares

w u'

1

~p

a1

w u'

1

u

p

~

/1

X ~X X ~X where in the first square, all the morphisms are in More; that is, u and u' are morphisms in More. But the second square tells us that u' = fu. From Lemma 1.5.6 we deduce that f must also be in More. D LEMMA

2.1.32. A morphism in 'D je of the form p~y

X

2.1. VERDIER LOCALIZATION AND THICK SUBCATEGORIES

91

will be invertible if and only if there exist morphisms f and h in 'D so that gf and hg are both in More.

Proof: The sufficiency is obvious. If there exist f and h so that gf and hg are both in More, then Funiv(hg) and Funiv(gf) are both invertible. This forces Funiv (g) to have both a right and a left inverse in '])I e' hence be invertible. But then the diagram above, which stands for the morphism Funiv(g)Funiv(a)- 1 , is also invertible. Next we wish to prove the necessity. Let us suppose, therefore, that Funiv(g)Funiv(a)- 1 is invertible in 'Die. Then Funiv(9) must also be invertible. We wish to show that there exist f and h with hg and gf in More. The statements being dual, it suffices to produce f. Let the diagram

Q~P

{31 y be a right inverse in 'Die to Funiv(9) : Funiv(P) composite

---+

Funiv(Y). Then the

Q ~ p __!!___.. y

{31 y

is in the equivalence class of the identity. From Lemma 2.1.31 we learn that gf must therefore be in More. D LEMMA

in

2.1.33. The morphism X---> 0 in']) becomes an isomorphism E ']) with X EB y E e.

'])I e if and only if there exists a y

Proof: Suppose the zero map g : X ---> 0 is a morphism in 'D such that Funiv takes it to an invertible map. By Lemma 2.1.32, there exists h : 0---> EY so that the composite X__!!___.. 0 ~ EY is in More. But then we have a triangle X ~ EY

----t

E(X EB Y)

---+

EX,

and since 0 : X ---> EY is in More, E(X EB Y) must be in e. Since e is triangulated, X EB Y is also in e. Conversely, suppose there exists a Y E 'D with X EB Y in e. Define h : 0 ---> EY and f : 0 ---> X to be the zero maps (the only possibilities).

92

2. VERDIER LOCALISATIONS

Then g f : 0 ·-+ 0 is an isomorphism, while hg : X and fits in a triangle X ~ ~y

with ~(X EB Y) E invertible in 'D ;e.

e.

------+

~(X EB Y)

-+ ~y

----+

is the· zero map,

~X,

Hence both hg and gf are in More, forcing g to be D

REMARK 2.1.34. In Corollary 4.5.12 we will see that not only does there exist some Y with X EB Y E e, but that Y may always be chosen to be ~X. It is always true that X EB ~X lies in e, if X is isomorphic to 0 in 'Dfe. PROPOSITION 2.1.35. Let g : Y -+ Y' be a morphism in 'D. Funiv(g) is an isomorphism if and only if in any triangle

Y ~ Y'

------+

Z

Then

----+ ~y

the object Z E 'D is a direct summand of an object of e; there is an object Z' E 'D so that Z EB Z' E e. Proof: Suppose there is an object Z' with Z EB Z' E e. To prove one implication, we want to show that Funiv(g) is an isomorphism. By Lemma 2.1.32, this is equivalent to producing hand f with gf and hg in More. Starting with the two triangles

Y ~ Y'

------+

Z

----+ ~y

0 ------+ Z' ------+ Z' ----+ 0 we can form the direct sum, which is a triangle by Proposition 1.2.1,

y

Y' EB Z'

Since Z EB Z' is in

------+

Z EB Z'

e, the map Y--+ Y' EB Z'

----+ ~Y.

is in More. But it factors as

Y ~ Y' ~ Y'EBZ'. This exhibits h with hg E More. But by the dual argument, there is also an f with gf in More. Thus Funiv(g) is invertible. Now we need the converse. Suppose therefore that Funiv(g) is invertible. Then there exists an h: Y'-+ Y" with hg E More. Now consider the morphism of triangles y~

Y'

(: )1 Y''

------+

Y'' EB Z

----+

Z

0

----+ ~Y"

2.1. VERDIER LOCALIZATION AND THICK SUBCATEGORIES

93

The bottom triangle is contractible, and by Corollary 1.3.12 all maps into contractible triangles are good. Hence the above map is a good map of triangles, and from the proof of Lemma 1.4.3 we learn that the square y~

Y'

Y" - - - t Y" EB Z is homotopy cartesian. But hg lies in More, and Lemma 1.5.8 allows us to conclude that the map

Y'

(: )

Y" EB Z

lies in More. Among other things, it follows that the map is an isomorphism in 'D je. But since g : Y -+ Y' also is, so is the composite

Y ~ Y'

(: )

Y"EBZ,

and since ag = 0 (the composite of two maps in a triangle), we have that

y

Y" EB Z

induces an isomorphism in the additive category 'D je. Then the composite

( : ) Y"

0

---t

z

are inverse to each other. In particular, the map Z - t 0 must be an isomorphism in 'Dje. By Lemma 2.1.33 we conclude that there is a Z' E 'D with Z E9 Z' E e. D LEMMA 2.1.36. Let us be given a commutative diagmm in 1J whose rows are triangles X

f

-------->

y

-------->

z

:EX

---t

91

11

X ~Y'

11 _______. Z'

------>

:EX.

Suppose Funiv(g) is invertible. Then it is possible to extend the diagmm to X ~y

11

-------->

91

X ~Y'

z

---t

h1 -------->

Z'

:EX

11 ---t

:EX

where Funiv(h) is also invertible.

Proof: By Proposition 1.4.6, it is possible to extend to a diagram X

f

-------->

y

11

91

X

~ Y'

1 0

1

-------->

1 :EX

~~

-------->

-------->

z

h1 -------->

Z'

Y"

1 1 --------> Y"

1

1

:EY

---t

-------->

:EZ

:EX

11 ---t

:EX

1 ---t

0

---t

:E2X

1

2.1. VERDIER LOCALIZATION AND THICK SUBCATEGORIES

95

and since Funiv(g) is invertible, Proposition 2.1.35 for the triangle

Y ~ Y' - - + Y'' - - + ~y implies that Y" is a direct summand of an object in e, which by Proposition 2.1.35 for the triangle Z ~ Z' - - + Y'' - - + ~z implies that Funiv(h) is also an isomorphism.

0

REMARK 2.1.37. The dual statement is that given a diagram with g deleted, that is

X ___.!___. y

--+

11

z

---t

11

h1 Y'

X ~

~X

Z' - - - t ~X, and if his an isomorphism in 1Jje, then g: Y ---t Y' can be chosen to also be an isomorphism in 1Jje. LEMMA

----t

2.1.38. Given two triangles in 1J X --+ Y --+ Z ---t

~X

X' - - + Y' - - + Z' - - - t ~X' and a commutative square in 1Jje with vertical isomorphisms X--+ y

1

1

X'--+ Y' it may be extended to an isomorphism in 1Jje of the triangles. Proof: The isomorphism Y represented as a diagram

---t

Y' is a map in 1J je, that is it may be

Y'' ___.!___. y Y'

where a E More and Funiv(f) is invertible. In other words, both a and f are invertible in 'D ;e. But then the diagrams in 1J whose rows are triangles

X''

--+

Y''

--+

Z'

---t

~X''

X'

--+

Y'

--+

Z'

---t

~X'

96

2. VERDIER LOCALISATIONS

and

x"

----+

Y"

----+

Z ---+ EX"

X

----+

Y

----+

Z ---+ EX

can be completed to isomorphisms in 1J je by the dual of Lemma 2.1.36 (see Remark 2.1.37). We may therefore assume that Y = Y' and that in the commutative square

X---+Y

X'

----+

y

the vertical map Y --+ Y is the identity. But now the isomorphism X in 1J1e may be represented by

--+

X'

X''~X

X' and we deduce a square in 'D, which commutes in 1J je,

X''~X X'

----+

Y.

By Lemma 2.1.26, replacing X" by W with a map W--+ X" in More, we may assume the square commutes in 1J. But now the diagrams

X"

----+

----+

11

!1 X

y

----+

y

----+

Z'' ---+EX''

z

~!1 ---+ EX

and

X" ----+Y

a1

----+

11

X' ----+Y

Z'' ---+EX''

~a1

Z' ---+EX' can be extended to isomorphisms in 1J je, completing the proof. ----+

D

2.1. VERDIER LOCALIZATION AND THICK SUBCATEGORIES

97

Proof of Theorem 2.1.8 Now it only remains to show that 2:>/"e is a triangulated category, that Funiv : 2J --+ 1Jje is a triangulated functor, and that Funiv is universal. Define the suspension functor on 2J je to be just the suspension functor of 2J on objects, and the suspension functor on diagrams defining morphisms for morphisms. Let ~univ : ~Funiv --+ Funiv~ be the identity. Define the distinguished triangles in 2) I e to be all candidate triangles isomorphic to Funiv(X)

---7

Funiv(Y)

---7

Funiv(Z) ------+ ~Funiv(X)

where X ------+ Y - - + Z ------+

~X

is a distinguished triangle in 2:>. Then [TRO] and [TR2] are obvious for 2J ;e. To prove [TRl], note first that any morphism in 2Jje may be written in the form Funiv(u)Funiv(f)-1, where f: P--+ X is in More and u: P--+ Y is any morphism in 2:>. By [TRl] for 2:>, complete u to a triangle in 2J

Then Funiv(X)

Funiv(u)Funiv(f)-1

Funiv(Y)

Funiv(v)

Funiv(Z)

Funiv(Ef)Funiv(w)

1

~Funiv(X)

is isomorphic in 2:>1e to Funiv(P)

Funiv(u)

Funiv(Y)

Funiv(v)

Funiv(Z)

Funiv(w)

~Funiv(X)

and hence is a distinguished triangle. It remains only to prove [TR3] and [TR4']. Since [TR4'] is stronger, we will prove only it. We need to show that, given a diagram in 2J je where the rows are triangles X ------+ Y ------+ Z ------+

1

X'

1

------+

Y'

------+

Z'

~X

1

------+ ~X'

then there is a way to choose Z --+ Z' making the above a good map of triangles; that is, the mapping cone is a triangle. We remind the reader that [TR3] is weaker, asserting only that the diagram can be made commutative.

98

2. VERDIER LOCALISATIONS

Observe first that by Lemma 2.1.27, the commutative square X--- y

1

1

X' - - - - t Y' can be lifted from 'DIe to 'D. Choose a square in 'D isomorphic to it as in Lemma 2.1.27. Then in that square, which we will denote X--- y

1

1

x'--- y'

the rows can be extended to triangles in 'D, and the diagram X --- Y --- Z

1

----t

1

~X

1

x' - - - y' --- z' --- ~x'

can be extended to a good morphism of triangles in 'D, hence also in 'D 1 e. But by Lemma 2.1.38 the commutative square with vertical isomorphisms

X ----tY

1

1

X ----tY extends to an isomorphism of candidate triangles in 'D 1e y X ~X

z ------1 1 1 1 --- --- ---

X

z

y

and similarly, the commutative square

x'

_,

----tY

1

X'

X

---

1

Y' extends to an isomorphism of candidate triangles in 'D je

x' --- y' ---

z' ---

X' - - - Y'

Z'

1

1

----t

1

----t

~x'

1

X'

2.2. SETS AND CLASSES

99

and hence we have defined a good morphism, as required, X ------+ Y ------+ Z ------+ X

1

1

X'

------+

Y'

1

------+

Z'

1

------+

X'.

Thus 'D 1e is triangulated. The fact that Funiv is a triangulated functor is obvious; by the definition of triangles in 'DIe' F univ takes triangles to triangles. The fact that e is contained in the kernel of Funiv follows, for instance, from Lemma 2.1.33. The universality is also obvious, since a triangulated functor taking e to zero must take all maps in More to isomorphisms, and by Proposition 2.1.24 Funiv is the universal functor with this property. D

REMARK 2.1.39. Let 'D be a triangulated category, e a triangulated subcategory. By Remark 2.1.7, the kernel of Funiv : 'D---+ 'Die is a thick subcategory of 'D, that is a triangulated category containing all direct summands of its objects. From Lemma 2.1.33 we learn that the kernel contains e, and can be described as the full subcategory whose objects are the direct summands of objects in e. We will call this category the thick closure of e, and denote it Because the kernel of a triangulated functor is always triangulated, we deduce that for any triangulated subcategory e c 'D, the subcategory is triangulated. The triangulated subcategory e c 'D is thick if and only if

e.

e

e=e.

2.2. Sets and classes In Section 2.1, we gave a proof of Verdier's localisation theorem. Given a triangulated category 'D and a triangulated subcategory e, one can form a triangulated quotient 'D 1e, with a universal property. But there is here a technical point which deserves mention. Suppose 'D is a category with small Hom-sets. That is, for any objects x and y in 'D, we require that H om'D ( x, y) should be a set. Then of course e, being a subcategory, also has small Hom-sets. However, the quotient category 'D 1e in general does not! Recall that the class H om'D; e (x, y) is defined by taking the class of all diagrams p

X

Q

------+

y

with f E More, modulo a suitable equivalence. There is no reason in general to expect there to only be a small set of diagrams as above; usually

100

2. VERDIER LOCALISATIONS

the collection of such diagrams forms a class. It may happen that even after identifying equivalent diagrams, we end up with a class of equivalence classes. See, for example, Exercise 1, Chapter 6, pp 131-132 in [12]. Note that since f : p - x is assumed in More, there is a triangle p

____.!___..X~ Z ~ ~p

with z E e. If it so happened that e is a small category (that is, a small set of objects), then we would deduce that there is only a small set of choices for z, and for each z only a small set of choices for g: x- z. This gives a small set of choices (up to isomorphism) of f : p - x, and for each of these a small set of choices for a : p - y. Thus we end up with a small set of equivalence classes. We conclude PROPOSITION 2.2.1. Let']) be a triangulated category with small Homsets. Let e be a small triangulated subcategory of']). Then '])I e' the Verdier quotient of Section 2.1, is a category with small Hom-sets. D

This is true, but not very useful. In fact, one mostly cares about subcategories e c ']) which are not small, and frequently one needs criteria which guarantee that ']) je has small Hom-sets. In the corning Chapters, we will, among other things, develop criteria on ']) and e which guarantee the smallness of the Hom-sets in ']);e.

2.3. History of the results in Chapter 2 The main result of the Chapter, Theorem 2.1.8, is due to Verdier. Verdier, in [35], constructed the quotient .'s, .A EA. Now each j>. < /3, and there are fewer than f3 of them (the set A has cardinality less than /3). Therefore the sum of the j>.'s is a sum of fewer than f3 cardinals, each smaller than /3. But f3 is regular. It follows that the cardinality of"'( must be strictly less than /3. In other words, 'Y < /3. We deduce that all the X>.'s lie in 8,. But then 8"1+1 C 8;3 contains an object isomorphic to the coproduct. Hence 8 ;3 contains all coproducts of fewer than f3 of its objects. In other words, 8;3 satisfies 3.2.1.1 and 3.2.1.2. Since 8;3 satisfies 3.2.1.1 and 3.2.1.2, it contains (8)/3. But since 8;3 is equivalent to the small category 8;3, it is essentially small. It follows that the smaller subcategory (8)/3 is essentially small. 0 DEFINITION 3.2.6. Let f3 be an infinite cardinal. Let 'J be a triangulated category satisfying [TR5}. A subcategory S C 'J is called /3-localising if it is thick and closed with respect to the formation of coproducts of fewer than f3 of its objects. That means that the coproduct of fewer than f3 objects of S exists, as a coproduct in 'J, and is an object of S. A triangulated subcategory S C 'J is called localising if it is /3-localising for all f3. Equivalently, S is localising if it is closed under the formation of all small 'J-coproducts of its objects. That is, if {X>., .A E A} is a family of objects in S, then the 'J-coproduct

is an object of S.

REMARK 3.2.7. If f3 > No, then any triangulated subcategory S C 'J closed under the formation of coproducts of fewer than f3 of its objects is automatically thick, hence localising. It is redundant to assume S thick. The point is that if f3 > No, then S contains all countable coproducts of its objects. By Proposition 1.6.8 every idempotent inS splits, and hence S contains all direct summands of its objects.

3.2.

GENERATED SUBCATEGORIES

107

EXAMPLE 3.2.8. Let 'J be a triangulated category satisfying [TR5]. Let 8 be a class of objects of 'J. Then (8)(3 is ,6-localising, whereas

(8) =

U(8)!3 (3

is localising, as in Definition 3.2.6. The statement for (8)(3 is really part of its definition, as the smallest ,6-localising subcategory with certain properties. The statement for (S) requires a little proof. Let us give the proof. Let {X>., A E A} be any collection of objects in (8). Let the cardinality of A be ,6, and suppose that for each A E A, the object X>. lies in (8)(3>-, for some cardinal ,6>.. Let 'Y be a cardinal greater than the sum of ,6 and the ,6>. 's. The X>. 's are. all in (8)', and since 'Y > ,6, the coproduct of the X>.'s, that is of ,6 objects of (8)', must lie in (8)'. Hence in (8), which is therefore localising. In fact, (8) is the smallest localising subcategory containing 8. Any localising subcategory containing 8 will satisfy 3.2.1.1, 3.2.1.2 and 3.2.1.3 for all infinite ,6. Therefore it must contain (8)(3 for all ,6, and hence (8), the union. DEFINITION 3.2.9. Let 'J be a triangulated category satisfying [TRS}. Let 8 be a class of objects of 'J. Then the union of all the (8)!3 's, that is

u(8)(3 {3

will be denoted (8). Note that even when 8 is a small set and 'J has small Hom-sets, in which case Proposition 3.2.5 tells us that the categories (8)(3 are all essentially small, the category (8) will usually be gigantic. It is called the localising subcategory generated by 8.

LEMMA 3.2.10. Let ,6 be an infinite cardinal. Let 'J be a triangulated category closed under the formation of coproducts of fewer than ,6 of its objects. Let S be a ,6-localising subcategory of 'J. Then 'J/S is closed with respect to the formation of coproducts of fewer than ,6 of its objects, and the universal functor F : 'J----. 'J/S preserves coproducts.

Proof: (cf. proof of Lemma 2.1.29). Since the objects of 'J and 'J/S are the same, it suffices to show that the coproduct in 'J of fewer than ,6 objects is also the coproduct in 'J/S. Let A be a set of cardinality less than ,6, {X>., A E A} a collection of objects of 'J. Form their coproduct in 'J, which we know exists,

II X>..

>.EA

We need to show this is also a coproduct in 'J/S.

108

3. PERFECTION OF CLASSES

Let Y be an arbitrary object of 'J'. We need to show that any collection of morphisms in 'J'/S {XA---+

Y,.X E A}

factors uniquely through llAEA X A. That is, we need to show the existence and uniqueness of a factorisation. The morphisms X A ---+ Y in 'J'/S can be represented by diagrams

PA

h,l

y

------+

where fA: PA---+ XA are in Mors. That is, in the triangle

PA

~ XA

------+

ZA

----+

EPA

the object ZA must be inS. But now the coproduct is a triangle

and llAEA ZA, being the coproduct of fewer than (3 objects inS, must be inS. The map

therefore lies in More. The diagram

II PA

------+

y

AEA

II ~Al II XA

AEA

AEA

represents a morphism in 'J'/S of the form with the inclusions

llAEA

XA

---+

Y. Its composites

3.2. GENERATED SUBCATEGORIES

109

are computed by the commutative diagrams

p).. f>.

1

X>-.

--+

-

II >-.EA

IIP>-. >-.EA

~

y

~>-.1 IIx>-. >-.EA

to be the given maps

X>-.

We have therefore factored X>-. --+ Y through li>-.EA X>-.. Now for the uniqueness of the factorisation. Suppose we are given a map in -.EA

so that the composites¢ o i>-. with every i)..: X)..--+ II X).. >-.EA

vanish. We need to show the map ¢ vanishes. Represent ¢ as a diagram

y

gl IIx).. --+Q >-.EA

with g : Y --+ Q in Mors. Then for each )., the composite with i>-. is represented by y

x)..-Q and must vanish in -. E S so that X>-. --+ Q factors as

110

3. PERFECTION OF CLASSES

But then

factors as

II X

A -----+

AEA

II z

Q

A -----+

AEA

and lhEA ZA, being a coproduct of fewer than S. Lemma 2.1.26 tells us that the map y

f3 objects of S, must lie in

must then vanish in 'J/S.

D

The following is an immediate corollary. COROLLARY 3.2.11. Let 'J be a triangulated category satisfying [TR5). Let S be a localising subcategory. Then 'J/S satisfies [TR5}, and the universal functor 'J -----+ 'J/S preserves coproducts.

3.3. Perfect classes DEFINITION 3.3.1. Let 'J be a triangulated category satisfying [TR5). Let f3 be an infinite cardinal. A class of objects S c 'J is called /3-perfect if the following hold

3.3.1.1. S contains 0. 3.3.1.2. Suppose {XA, A E A} is a collection of objects in 'J. Suppose the cardinality of A is less than /3. Let k be an object of S. Then any map

factors as k

-------*

II kA

AEA

~

II XA

AEA

with kA E S. More precisely there is, for each A E A, an object kA E S and a map fA : kA -----+ XA, so that the map factors as

3.3. PERFECT CLASSES

111

3.3.1.3. Suppose again that A is a set of cardinality < (3. Suppose k and the k>.. 's, .X E A, are objects of S and the X>.. 's are any objects of 'Y, and the composite

vanishes. Then it is possible to factor each!>.. : k>.. ---+X>.. as

k >..

9>.

-----7

l

h)..

>.

-----7

X

>.

so that l >.. E S, and the composite

already vanishes. We begin with a trivial lemma. LEMMA 3.3.2. Let 'Y be a triangulated category satisfying [TR5j. Suppose (3 is an infinite cardinal, and T is a (3-perfect class in 'J. Suppose that S C T is an equivalent class; that is, any object ofT is isomorphic to some object of S. Then S is also (3-perfect.

Proof: Trivial.

D

LEMMA 3.3.3. Let 'Y be a triangulated category satisfying [TR5]. Let (3 be an infinite cardinal. LetS be a (3-perfect class of 'J. LetT be the class of all objects in 'Y which are direct summands of objects of S. Then T is also (3-perfect.

Proof: Suppose we are given an object k E T, a set A of cardinality< (3, a family of X>. 's in 'Y and a map k

II X>..

-----7

>.EA

Because k E T, there must exist some k' E 'Y so that k EB k' E S; T is defined to be the class of all direct summands of objects of S. Consider the composite k tB k'

( 1 0 ) k

-----7

II X>.. >..EA

112

3. PERFECTION OF CLASSES

Since k EB k' E S and S is ,8-perfect, this factors

with k>.. ESC T. But then

is a factorisation of the original map k

-'-----t

II X>...

>..EA

Suppose now that we are given a vanishing composite

with k, k>.. E T. Choose k' so that k EB k' E S, and for each A choose k~ with k>.. EB k~ E S. Then we have a vanishing composite

k EB k'

( 1

0 ) k

-'-----t

II k>..

l!l (

>..EA

f>..

0)

IJx>..

>..EA

Because S is ,8-perfect, for each A the map ( f>..

0 ) : k>.. EB k~ ~X>..

must factor as k>.. EBk~ ~ l>.. ~x>.. with l>.. E S, where the composite

k EB k'

( 1

0 ) k

-'-----t

II k>.. >..EA

II k>.. EB k~

>..EA

--+

IJl>.. >..EA

3.3. PERFECT CLASSES

113

already vanishes. But then

k

(~)

kffik'

!l ( ~)

(l o)

IJkA ---~ >-.EA

also vanishes, completing the proof of the ,8-perfection ofT.

0

DEFINITION 3.3.4. Let 'J be a triangulated category satisfying (TR5]. Let S C 'J be a triangulated subcategory. Let ,B be an infinite cardinal. An object k E 'J is called ,8-good if the following holds.

3.3.4.1. Let {X>-., .A E A} be a family of objects of 'J. Suppose the cardinality of A is less than ,B. Then any map k

----t

II X)..

>-.EA

has a factorisation

with kA E S. More intuitively, 3.3.1.2 sort of holds for the single object k; it holds where the k >-. may be taken in S. LEMMA 3.3.5. Let 'J be a triangulated category satisfying (TR5}, and let S be a triangulated subcategory. If k is a ,B -good object of 'J, then the following automatically holds.

3.3.5.1. Suppose A is an index set of cardinality < ,B. Given a vanishing composite

with k >-. E S, then it is possible to factor each fA : k >-.

k >-.

9>.

--->

l

A

h>.

----t

X

A

---+

X>-. as

114

3. PERFECTION OF CLASSES

so that l>-. E S, and the composite

also vanishes.

Proof: Assume that A is a set whose cardinality is less than (3. Suppose we are given a vanishing composite

with k>-. E S. For each .X, consider the triangle Y>-. ----+ k>-.

~

X>.

~ ~Y>-..

Summing these triangles, we obtain a triangle

>-.EA

Since the composite

vanishes, the map

must factor as

But then the hypothesis that k is good guarantees that the map

k----+

I1 y)..

>-.EA

factors as

3.3.

PERFECT CLASSES

115

with j>. E S. The composite

clearly vanishes, and if we define l >. by forming the triangle

.

)).. - - - +

then clearly k >.

---+

k )..

9 >.

---+

l)..

~. LJ])..,

-

X>. factors as

Since j>. and k>. are inS and Sis triangulated, l>. must also be inS. And the composite k---+

II k)..

---+

>.EA

must vanish, since it is k---+ IIj).. - - - + >.EA

II z)..

>.EA

II k)..

AEA

-

II z)..

>.EA

and 9>.

j).. - - - + k).. - - - + l)..

are two morphisms in a triangle.

D

REMARK 3.3.6. From Lemma 3.3.5 we learn that, if S is a triangulated subcategory of 'J all of whose objects are ,6-good, then S is a ,@-perfect class; more precisely, the collection of all objects of S is a ,@-perfect class. For any k E S, if it satisfies 3.2.1.2 then 3.2.1.3 is automatic. The next couple of Lemmas allow us to cut the work even shorter. One need not check that every object of Sis ,6-good. It is enough to check all the objects of a large enough generating class. Before the next Lemmas, we should perhaps remind the reader of the notation of Section 3.2. Let S be a class of objects of 'J. Then T(S) is the smallest triangulated subcategory containing S. For an infinite cardinal a, (8)01. is the smallest a-localising subcategory containing S. More explicitly, (S) 01. is the minimal thick subcategory S C 'J such that 3.2.1.1: S contains S. 3.2.1.2: Sis closed under the formation of coproducts of fewer than a of its objects. Recall that by Remark 3.2.7, if a > N0 it is redundant to assumeS is thick; 3.2.1.2 already tells us that idempotents split in S.

116

3. PERFECTION OF CLASSES

LEMMA 3.3. 7. Let a and {J be infinite cardinals. Let 'J be a triangulated category satisfying [TR5]. Let S be a class of objects of 'J. Suppose that every object k E S is (3-good, as an object of the triangulated subcategory (S)a C 'J. Then the objects of (St form a {J-perfect class.

A similar Lemma, whose proof is nearly identical, is LEMMA 3.3.8. Let {J be an infinite cardinal. Let 'J be a triangulated category satisfying [TR5]. Let S be a class of objects of 'J. Suppose that every object k E S is {J-good as an object of the triangulated subcategory T(S) C 'J. Then the objects ofT(S) form a (3-perfect class.

Proofs of Lemmas 3.3. 7 and 3.3.8. Before we start the proof, let us make one observation. Assume Lemma 3.3.8; that is assume T(S) is {Jperfect. By Lemma 3.3.3, the collection of all direct summands of objects of T(S) is also (3-perfect; but this is precisely (S)No, the thick closure of T(S). Without loss, we may therefore assume a> ~ 0 in Lemma 3.3.7; the case a = ~o is an immediate consequence of Lemma 3.3.8. The proofs of the two Lemmas are so nearly the same, that we will give them together. Consider the following two full subcategories ~ C 'J and S c 'J, given by S = {k E (S)alk is (3-good}, ~

=

{k E T(S)Ik is (3-good}.

It suffices to show that ~contains T(S) and S contains (St. To show that ~ contains T(S), it suffices to prove that ~ is triangulated and contains S. To show that S contains (S) a, it suffices to prove that S is a thick subcategory of 'J satisfying 3.2.1.1 and 3.2.1.2. Since the argument for~ is similar and slightly simpler, we leave it to the reader. We assume therefore that a > ~ 0 , and we will prove that S is triangulated, and satisfies 3.2.1.1 and 3.2.1.2. Since a> ~ 0 , S would then be closed under the formation of countable coproducts, hence idempotents would split inS. It is redundant to prove the thickness. We need to prove three things, of which 3.2.1.1 is the hypothesis of the Lemma. By hypothesis we know that all the objects inS are (3-good as objects of (S)a. Hence S C S, S being the subcategory of all (3-good objects. Proof that S satisfies 3.2.1.2. We need to show that, if {kJL, Ji, EM} is a collection of fewer than a objects of S, then the coproduct is in S. Let {X>., .A E A} be a collection of objects of 'J, with A of cardinality < (3. Suppose that we are given a map

II kJL

-

II X)..

3.3. PERFECT CLASSES

117

That is, for each p, E M, we have a map

IIx;...

k~-' ~

>.EA

Because kJ.' E S, this map factors as

kJ.' ____.

II

II

I{>.,J.'}

>.EA

k{>.,J.t}

>.EA

with

k{>.,J.t} E

(8) 0 • It follows that our map

factors as

J.tEM

>.EA J..EA

and as (S) a. is closed under the formation of coproducts of fewer than a of its objects,

II

k{>.,J.'} E

(8) 0

.

J.tEM

D Proof that S is triangulated. Note that the category (B) a. is triangulated, in particular closed under suspension. It follows that factoring k ~

IIx;..

>.EA

as a composite

>.EA

with k;.. E (8)

0

>.EA

is the same as factoring Ek ~

IIEX;..

>.EA

as a composite

IIEX>.

>.EA

118

3. PERFECTION OF CLASSES

with Ek.x E (8)01.. Thus S satisfies ES = S. Now suppose f: k ---7 lis a morphism inS. Complete it to a triangle

k ~ l ----+ m ----+ Ek. We know that k and l are inS. We must prove that so ism. Let A E A} be a set of objects of 'J, with A of cardinality < (3. We deduce an exact sequence

{X.x;

Hom

(m, MX.x)

----+

Hom (l,

MX.x)

----+

Hom ( k,

MX.x)

Suppose now that we are given a map

m~ IlX>. >.EA

By the exact sequence, this gives a map

so that the composite k ----+ l

II X.x

----7

>.EA

vanishes. But l is in S, meaning it is (3-good; hence there exists a factorisation

l

----+

Ill>.

>.EA

with l.x E (8)01.. Furthermore, the composite

k

----+

l

----+

Il l>.

Ilt>. >.EA

>.EA

II X>. >.EA

vanishes, and hence since k E S, we deduce by Lemma 3.3.5 that

l>. ~X>. factors as so that the composite

k ----+ l ----+ Ill>. >.EA

3.3. PERFECT CLASSES

119

vanishes. But then the map ----> I J mA AEA

factors through m; there is a map

m~

IJmA AEA

giving it. Now the composite

m ~

IJmA AEA

need not agree with the given map

m~

IJxA. AEA

However, by construction, the composites with l therefore factors as a map

m _____. Ek

-+

----> I J

m agree. The difference

XA.

AEA

But Ek E S, and hence the map

Ek _____.

II XA AEA

factors as

Ek---->

II m~

II h~

AEA

AEA

with m~ E (S) a, and the map h factors as

m

----> I J {mA AEA

with {mA E9 m~} E (St.

E9 m~} D

THEOREM 3.3.9. Let a and f3 be infinite cardinals. Let 'J be a triangulated category satisfying [TR5}. Let {Si, i E J} be a family of /3-perfect classes of 'J. Recall that if uBi is the union

120

3. PERFECTION OF CLASSES

then T(USi) is the smallest triangulated subcategory of 'J containing USi, and (USi)"' is the smallest thick, a-localising subcategory S C 'J containing usi. Our theorem asserts that for any (3-perfect Si 's as above, the collection of objects of T(USi) is a (3-perfect class. Furthermore, for any infinite a, the objects of (UBi)"' also form a (3-perfect class. Proof: Since the proofs are virtually identical, we will prove the statement about (USi)"'. By Lemma 3.3.7, it suffices to show that any k E UiEISi is (3-good. More explicitly, let {XA., A. E A} be a collection of objects of 'J', where A has cardinality < (3. Given any map k

----t

II XA

A.EA

we must show there is a factorisation

k

----t

II kA

A.EA

with kA. E (USi)"'. So take k E UiEISi. For some i E I, k must lie in Si. But then a map k

----t

II XA

A.EA

as above must factor as

k

----t

II kA

A.EA

with kA. E Si, hence kA. E (USi)"'.

D

CoROLLARY 3.3.10. Let (3 be an infinite cardinal. Let 'J' be a triangulated category satisfying [TR5j. Let S be a triangulated subcategory. The collection of (3-perfect classes Si C S of 'J has a unique maximal member S. Furthermore, any (3-perfect class R of 'J, whose objects lie in S, is contained in the maximal class S.

Proof: Take the collection of {Si, i E I} to be the class of all (3-perfect classes in 'J', all of whose objects lie in S. Then by Theorem 3.3.9, the objects of the category T(USi) form an (3-perfect class. On the other hand, sis triangulated and contains usi, hence also T(USi)· Thus T(USi) is contained in S, is (3-perfect, and contains all the (3-perfect Si 's. Putting S = T(USi), we clearly have that Sis maximal. D

3.3. PERFECT CLASSES

121

DEFINITION 3.3.11. Let (3 be an infinite cardinal. Let 'J be a triangulated category satisfying [TRS}. Let 3 be a triangulated subcategory. Then the full subcategory whose object class is the maximal (3-perfect class S of Corollary 3.3.10 will be called 3,a. CoROLLARY 3.3.12. Let 3 be a triangulated subcategory of a triangulated category 'J. Suppose 'J satisfies [TRS}. Let (3 be an infinite cardinal. Then 3,a is triangulated.

Proof: 3 is a triangulated subcategory containing 3,a, and hence 3 contains the minimal triangulated subcategory containing S,a. That is,

s{3 c T(3{3) c

3.

But S,a is a (3-perfect class. By Theorem 3.3.9, so is T(S,a). Because S,a is the maximal (3-perfect class, it must contain T(Sf3), hence is equal to it, 0 hence is triangulated. COROLLARY 3.3.13. LetS be a thick subcategory of a triangulated category 'J satisfying [TRS}. Let (3 be an infinite cardinal. Then the category S,a is thick.

Proof: Let T be the class of all objects isomorphic to direct summands of objects of S,a. Since S contains S,a and is thick, it must contain T. We have

S,a

c

T

c 3.

But T is (3-perfect by Lemma 3.3.3, and by the maximality of S,a we must have T c S,a. Hence the two are equal, and S,a is thick. 0 CoROLLARY 3.3.14. Let a and (3 be infinite cardinals. Suppose S is an a-localising subcategory of a triangulated category 'J satisfying [TRS}. That is, S is a thick subcategory closed under the formation of coproducts of fewer than a of its objects. Then the subcategory S,a is also a-localising.

Proof: S is a-localising and contains S{3, and hence contains (S,a) a, the smallest a-localising subcategory containing S,a. We get an inclusion

Sf3

c (S,a)a c S.

On the other hand, S,a is a (3-perfect class. By Theorem 3.3.9, so is (S,a) 0 By the maximality of S,a we must have

•

(S,a) a c S,a. Hence the two are equal and Sf3 is a-localising.

0

REMARK 3.3.15. If Sis ')'-perfect, and 'Y > (3, then Sis also (3-perfect. We deduce that for any 3 C 'J, S1 , being (3-perfect, must be contained in

122

3. PERFECTION OF CLASSES

the rriaximal ,8-perfect class s 13 . If ::R C S C 'J', then ::R/3 is a ,8-perfect class ins, hence contained in the maximal s/3. EXAMPLE 3.3.16. Let 'J' be a triangulated category. Let k be any object of 'J'. Then the class S = {0, k} of only two objects is l-t 0 perfect. Given any set A of cardinality < l-t 0 (that is, a finite set), and a map

k

-----t

II X).

>-.EA

we can factor it as >-.EA

>-.EA

where ~ is the diagonal map from k to kn. If the composite vanishes, then in fact each f >-. must vanish, and the map factors as k

~

II k

>-.EA

-----t

II 0

>-.EA

-----t

II X)..

>-.EA

It follows that if s is any triangulated subcategory of 'J', then The case ,8 = l-t 0 is the trivial case for perfection.

s~o =

s.

3.4. History of the results in Chapter 3 The results of Section 3.2 are very standard. The definition of localising subcategories (Definition 3.2.6) is probably due to Bousfield, [4], [6] and [5]. Let 'J' be a triangulated category satisfying [TR5]. As in Definition 3.2.1, let (S) 13 be the smallest ,8-localising subcategory of 'J' containing the set S of objects. The fact that (S) 13 is essentially small (Proposition 3.2.5) is obvious. The fact that quotients by localising subcategories respect coproducts (Corollary 3.2.11) may be found in Bokstedt-Neeman [3]. Section 3.3 is completely new. This book introduces the notion of perfect classes to imitate some standard constructions involving transfinite induction on the number of cells of a complex. Whatever the motivation, the definition is new, and in the rest of the book, we will attempt to explain what one can do with it.

CHAPTER 4

Small objects, and Thomason's localisation theorem 4.1. Small objects DEFINITION 4.1.1. Let 'J be a triangulated category satisfying [TR5} (that is, coproducts exist}. Let a be an infinite cardinal. An object k E 'J is called a-small if, for any collection {X.x; A E A} of objects of 'J, any map

k

-------+

II X>.

>-EA

factors through some coproduct of cardinality strictly less than a. In other words, there exists a subset A' C A, where the cardinality of A' is strictly less than a, and the map above factors as k

-------+

II X

A -------+

>.EA'

II X>..

>-EA

EXAMPLE 4.1.2. The special case where a = ~ 0 is of great interest. An object k E 'J is called ~ 0 -small if for any infinite coproduct in 'J, say the coproduct of a family {X>.; A E A} of objects of 'J, any map

k

-------+

II X>.

>.EA

factors through a finite coproduct. That is, there is a finite subset

{X1,X2, ... ,Xn}

c

{X>.; A E A}

and a factorisation

>-EA

i=l

Expressing this still another way, the natural map

II H om(k, X>.)

>-EA

is an isomorphism.

-------+

Hom

(k, II x>.) >-EA

124

4. THOMASON'S LOCALISATION

DEFINITION 4.1.3. Let a be an infinite cardinal. Let '1 be a triangulated category satisfying [TR5}. The full subcategory whose objects are all the asmall objects of '1 will be denoted 'J(a). LEMMA 4.1.4. Let a be an infinite cardinal. Let '1 be a triangulated category satisfying [TR5}. Then the subcategory 'J(a) C '1 is triangulated.

Proof: To begin with, observe that comes about from the identity

Hom

kE

(Ek, II xA)

'J(a)

if and only if

Hom (k,E-

AEA

1

Ek E

'J(a).

This

{y XA})

Hom (k, II E- x>-.) 1

AEA

where the second equality is the fact that the suspension functor respects coproducts, i.e. Proposition 1.1.6. Let k -+ l be a morphism in 'J(a). It may be completed to a triangle in '1. There is a triangle in '1

k l m Ek. -+

-+

-+

We know that k and l are a-small. To show that 'J(a) is triangulated, we need to show that so is m. Let A A} be a set of objects of '1. Because

{XA; E

Hom(-, IJ xA) is a cohomological functor on '1, we deduce an exact sequence

Hom(m,IJxA) ~ Hom(l,yxA) ~ Hom(k,yxA) Suppose now that we are given a map

m ____!!:___,

II X AEA

By the exact sequence, this gives a map

so that the composite

A

4.1. SMALL OBJECTS

125

vanishes. But l is a-small; hence there exists a subset A' C A of cardinality < a, so that the map

z - - 7 IIx>. >.EA

factors as

Of course, we know that the composite k

--7

l

II X>.

--7

--7

>.EA'

vanishes. On the other hand, the map

II X>.

--7

II X>.

>.EA

II X>.

>.EA

is the inclusion of a direct summand, hence a monomorphism in 'J'. We deduce that the composite k

--7

--7

II X>.

>.EA'

is already zero, and therefore that

factors through m. We therefore have produced a map

m~

IIx>.

>.EA'

so that the composite

m ~

II X>.

--t

II X>.

>.EA

differs from the map

m ~

IIx>.

>.EA

by a map vanishing on l. On the other hand, the exact sequence

Hom (

~k, Jl X>.)

--t

Hom ( m,

Jl X>.)

tells us that the difference factors as

m

-----+

~k

-----+

--t

IJ X>..

>.EA

Hom

(z,

Jl X>.)

126

4. THOMASON'S LOCALISATION

But k E 'J(o:) implies 'Ek E 'J(o:), and we deduce that there is a subset A" C A, of cardinality < a, so that the map

factors as 'Ek ~

II X>,

II X>,.

------>

AEA"

AEA

Putting this all together, one easily deduces that the given map

m~

IIx>-

>-EA

factors as

II

m ~

X>.

------>

AEA'UA"

II X>. AEA

and because a is infinite and A' and A" are of cardinality < a, the cardiD nality of A' U A" is also < a. Therefore 'J(o:) is triangulated. LEMMA 4.1.5. Suppose a is a regular cardinal. That is, a is not the sum of fewer than a cardinals, each of which is less than a. Then 'J(o:) is a-localising. That is, the coproduct of fewer than a objects of 'J(o:) is an object of 'J(o:).

Proof: Let {kJ.L, J.L E M} be a collection of objects in 'J(o:), where the cardinality of M is less than a. Let {X>.,>. E A} be an arbitrary collection of objects of 'J. Suppose we are given a map J.LEM

AEA

This means that, for every J.L E M, we are given a map kJ.L ~

IIx>-.

AEA

Because kJ.L is a-small, for each J.L there exists an AJ.L < a, so that the map

factors as

c

A, with cardinality

4.1. SMALL OBJECTS

127

Thus the map

factors as

II k~

II

~

X;.~

J.EU,.eMA,.

~EM

IIx;.,

AEA

and the cardinality of U~EMA~ is bounded by the sum of the cardinalities of A~ over all p, EM, which is a sum of fewer than a: cardinals, each less than a:. Because a: is regular, this sum is less than a:. D LEMMA

4.1.6. Let a: be an infinite cardinal. The category 'J(o:) is thick.

Proof: By Lemma 4.1.4 we know that 'J(o:) is a triangulated subcategory. To prove it thick, we need to show that any direct summand of an object in 'J(o:) is in 'J(). Let k, l be objects of 'J and assume that the direct sum ktBl is a-small. We wish to show that k is a-small. Take any map k

~

IIx>-

>-EA

and consider the map k ffi m

( h

II X;.

0 )

J.EA

Because k tB m is small, there is a subset A' C A of cardinality the above factors as kffim ~

II X;.

~

< a:, so that

II X;.

J.EA

and hence the given map k

~

IIx>-

>-EA

factors as

k

kffim--.

II X;.~ II X;.

AEA'

>-EA

and in particular, it factors through a coproduct of fewer than a: terms.

D

128

4. THOMASON'S LOCALISATION

REMARK 4.1.7. If a: is a regular cardinal greater than ~o, Lemma 4.1.6 is redundant. By Lemma 4.1.5, 'J(a) is o:-localising. But since a: > ~ 0 , Remark 3.2.7 tells us that idempotents split in 'J(a), and 'J(a) must be thick.

4.2. Compact objects Let 'J be a triangulated category satisfying [TR5]. In Section 4.1 we learned how to construct, for each infinite cardinal a:, a triangulated subcategory 'J(a) of a-small objects in 'J. In Section 3.3, we learned that given any triangulated subcategory S c 'J and an infinite cardinal (3, there is a way to construct a triangulated subcategory S13 c S. In this section, the idea will be to combine the constructions and study {'J(a)}f3' LEMMA 4.2.1. Let 'J be a triangulated category satisfying [TR5j. Let

a: be an infinite cardinal. Let S be an a-perfect class of a-small objects.

Then S is also (3-perfect for all infinite (3.

Proof: Suppose k is an object in S, and {X>.,).. E A} a family of fewer than (3 objects of 'J. Because k is a--small, any map

factors as

with the cardinality of A' being ::; a:. Since Sis a-perfect, the map k

------+

Il X>.

.\EA'

factors as

k

------+

Il k>.

>.EA'

with k>. E S. For)..

f/. A',

Il f>.

.\EA'

Il X>.

.\EA'

define k>. = 0. Then we deduce a factorisation

where of course most of the k>.'s vanish, but in any case they are all inS.

4.2. COMPACT OBJECTS

129

Suppose now that we have a vanishing composite

with k and k>. all inS and A of cardinality< map

/3. Because k is a-small, the

must factor as

where the cardinality of A' is < a. The composite

vanishes, and since S is a-perfect, we deduce that for each A E A', the map f>.: k>. ~X>. factors as

k)..

9>.

------+

l)..

h>.

------+

X>.

so that l>. E S and the composite

k

------+

I1 k)..

I1 g)..

>.EA'

Ill)..

>.EA'

already vanishes. For A tj. A', define g>. : k>. we still have the vanishing of

>.EA'

~

l>. to be the identity. Then

0 DEFINITION 4.2.2. Let 'J be a triangulated category satisfying [TR5]. Let a be an infinite cardinal. Define a triangulated subcategory 'J"' by

LEMMA

4.2.3. If a

< /3 are infinite cardinals, then 'J"'

C 'Jf3.

130

4.

THOMASON'S LOCALISATION

Proof: 'J'"' is an a-perfect class, whose objects are a-small. We know, from Lemma 4.2.1, that 'J'"' must be ,8-perfect. Since its objects are asmall, they are also ,8--small. Thus 'J'a is a f)-perfect class in 'J'(.B), hence it must be contained in the maximal one, { 'J'(,B)} ,B. 0 LEMMA 4.2.4. For every infinite a, 'J'a is thick. Proof: By Lemma 4.1.6, 'J'(a) is thick. By Corollary 3.3.13, for every infinite cardinal ,8, { 'J'( a) } ,B is also thick. In particular, letting ,8 = a, 'J'a is thick. 0 LEMMA 4.2.5. Let a be a regular cardinal. Then 'J'a is a-localising. Proof: a is regular, and hence Lemma 4.1.5 says that 'J'(a) is a-localising. But then Corollary 3.3.14 asserts that, for any infinite (3, {'J'(a)},a is also a-localising. Letting ,8 =a, we deduce that 'J'a is a-localising. 0 REMARK 4.2.6. In the special case a = No, all classes are a-perfect; see Example 3.3.16. Thus

In other words,

DEFINITION 4.2. 7. The objects of 'J'a will be called the a-compact objects ofT. In the case a= N0 , the objects of'J'~ 0 will be called the compact objects. They are ,8-compact for any infinite ,8. We will permit ourselves to write 'J'c for 'J'~ 0 ; the superscript c stands for compact. 4.3. Maps factor through (S),B REMINDER 4.3.1. Let 'J' be a triangulated category satisfying [TR5]. Let S be a class of objects of 'J'. We remind the reader of Definition 3.2.9; (S) stands for the localising subcategory generated by S, that is (S) =

U(B) ,a. ,B

(S),B is the smallest thick subcategory containing Sand closed with respect to forming the coproducts offewer than ,8 ofits objects; See Definition 3.2.1. LEMMA 4.3.2. Let 'J' be a triangulated category satisfying [TR5]. Let ,8 be a regular cardinal. Suppose S is a class of objects of 'J'.B. That is, S C 'J'.B. Then the subcategory (S),B is also contained in 'J'.B.

4.3. MAPS FACTOR THROUGH (S)f3

131

Proof: If {3 is regular then, by Lemma 4.2.5, 'Jf3 is {3-localising. By hypothesis, S is contained in 'Jf3. But (8) 13 is the minimal (3-localising subcategory containing S, hence (B) f3 c 'Jf3. D The main theorem of this section is the following. THEOREM 4.3.3. Let 'J be a triangulated category satisfying [TR5j. Let {3 be a regular cardinal. Let S be some class of objects in 'Jf3. Let x be a {3-compact object of 'J. Let z be an object of (S). Suppose f : x ---+ z is a morphism in 'J. Then there exists an object y E (8) 13 so that f factors as

X---+ y ---+

Z.

Proof: We define a full subcategory S of 'J as follows. If Ob(S) is the class of objects of S, then

Ob(S)

= {

z E Ob('J) I Vx E. Ob.('Jf3), \:If: x---+ z, 3y E (8) 13 } • and a factonsat10n of f as x ---+ y ---+ z

It suffices to prove that (B) C S. To do this, we will show that S contains S, is triangulated, and contains all coproducts of its objects. Since (S) is minimal with these properties (see Example 3.2.8), it will follow that

(B) c S.

The fact that S contains S is obvious. Take any objects z E S and x E 'Jf3, and any morphism x ---+ z. Since we know that z E S, clearly z E (8) 13 , the smallest thick category containing S and closed with respect to coproducts of fewer than {3 of its objects. Put y = z, and factor f : x ---+ z as 1

~z.

Equally clearly, z E S if and only if Ez E S. After all, x factored as

x if and only if

E- 1 x ---+

~

y

---+

Ez can be

·-----+ Ez

z can be factored as

E- 1 x

~

E- 1 y

~

z

and x E 'Jf3 iff E- 1 x E 'Jf3, y E (8) 13 iff E- 1 y E (8) 13 . Suppose now that ¢ : z ---+ z' is a morphism in S. Complete it to a triangle z ~ z' ~ z" ~ Ez.

We know that z and z' are in S. To show that S is triangulated, we must establish that z" is also in S. Choose any x E 'Jf3, and any map f : x ---+ z". We need to factor it as

x

---+

y

---+

z",

132

4. THOMASON'S LOCALISATION

withy E (8)(3. First of all, the composite x

---+

z"

---+

'Ez

gives a map from x E 'J'f3 to 'Ez E S, which must factor as

x

---+

y

---+

'Ez,

with y E (8)(3. Now the composite

x

---+

z"

---+

'Ez---+ 'Ez'

clearly vanishes, and is equal to the composite

x Complete x

---+

---+

y

---+

'Ez

---+

'Ez'.

y to a triangle x ---+ y ---+ C ---+ 'Ex;

then the map y---+ 'Ez' must factor through C. We deduce a commutative square y---+

c

1

1

'Ez Now, in the triangle hypothesis, x E 'J'f3. (8) 13 c 'J'f3. Because deduce that the map

---+

'Ez'.

defining C, the other two objects are x and y. By By construction, y E (8)(3, and by Lemma 4.3.2, x andy are both in 'J'f3, so is C. Since 'Ez' E S, we C ---+ 'Ez' factors as C---+ y'---+ 'Ez'

with y' E (8)(3. Our commutative square above gets replaced by another, y

---+

1

'Ez

y'

1

---+

'Ez',

where the top row involves only objects in (8) 13 . Note also that since the composite x--+y--+C vanishes, so does the longer composite X---+ y---+ C---+ y 1 •

4.3. MAPS FACTOR THROUGH (S)f3

133

Now complete the commutative square y

y'

~

1 ~z

1 ~z'

~

to a map of triangles ~-1y"

y

~

1

1

z"

y'

~

______...

1

y"

1

~ ~z ~ ~z' -------> ~z"

and note that, because y and y' are in (S)fJ, so is y", and because the composite X ---+

vanishes, the map x

y

---+

y'

y factors as

---+

X ---+ ~-1y" ---+

y.

We deduce a commutative diagram y

X~ ~-1y'' ~

1

1

z"

~ ~z.

The composite x is not the map

f :x

---+

~ - 1 y" ---+ z"

z" we began with. But when we compose

---+

x

---+

~- 1 y" ---+

z"

---+

~z

we do get the given map

x ___.!___, z'' ~ ~z. In other words, the difference between the composite x

---+ ~-ly" ---+

and f : x ---+ z" factors as x hence x ---+ z' must factor as x withy E (S)fJ. But then

x and

f:

x

~

z'

---+

---+

---+

---+

z"

z". We know that z'

y ---+ z'

y factors as

{y EB ~- 1 y"} ------... z",

y EB ~- 1 y" is in (S)fJ. Thus z"

E S, as required.

E

S, and

134

4. THOMASON'S LOCALISATION

It remains to prove that S is closed with respect to the formation of coproducts of its objects. Let {z.A, A E A} be a set of objects of S. We wish to show that

II

ZA

.AEA

is an object of S. Rephrasing this again, we wish to show that if x E 'Jf3 and

f :X

----+

II

Z_A

.AEA

is any map, then there is a factorisation

x ~y~

IIz.A

.AEA

with y E (8)(3. Take therefore any map

f:

X----+

II

ZA .

.AEA

Now recall that x E 'Jf3 C 'J(f3) . In particular, x is .8-small. Therefore there is a subset A' C A, where the cardinality of A' is less than .8, so that f factors as X

~

II

Z_A

~

.AEA'

II

ZA .

.AEA

Now x is in fact not only .8-small, but also .8-compact. In other words, x belongs to the .8-perfect class 'Jf3. Therefore the map g factors as

X~

II

II

h.A

XA

.AEA'

for some collection of h.A : x.A ----+ z.A, with x.A in 'Jf3. For each A E A', we have x.A E 'Jf3, Z>. E Sand a map h.A: x.A ----+ z.A. It follows that, for each A, we may choose a Y>. E (S)f3 and a factorisation of h.A : X.A ----+ Z.A as XA ----+ Y>. ----+ Z_A,

withY>. E (8)(3. But then X

f factorises as

~

II

Y>.

~

.AEA'

and ILEA' Y>., being a coproduct of fewer than in (S)f3.

II

Z_A,

.AEA

.B objects of (8)f3,

must lie 0

4.4. MAPS IN THE QUOTIENT

135

4.4. Maps in the quotient As we have said in Section 2.2, one of the problems with Verdier's construction of the quotient is that one ends up with a category in which the Hom-sets need not be small. Suppose 'J is a triangulated category with small Hom-sets, and S is a triangulated subcategory. Then the quotient 'JIS of Theorem 2.1.8 is a category, which in general need not have small Hom-sets. Nevertheless, we can already give one criterion that guarantees the smallness of the Hom-sets. The criterion is Corollary 4.4.3. Then we will further explore some of the consequences of the machinery that has been developed so far. We lead up to Thomason's localisation theorem (Theorem 4.4.9), which will give a summary of the results in this Section. PROPOSITION 4.4.1. Let 'J be a triangulated category. Let (3 be a regular cardinal. Let 8 be a subclass of the objects of 'Jf3. Let y E 'J be an arbitrary object, x E 'J a (3-compact object (i.e. x E 'Jf3 ). Then

{'JI(8)}(x,y) = {'JI(8)f3} (x,y). In other words, the maps x ---+ y are the same in the Verdier quotient categories 'JI (8) and 'JI (8)(3. Proof: There is a natural map

and we want to prove it an isomorphism. We need to show it injective and surjective. Let us begin by proving it surjective. Proof that ¢ is surjective. Let x ---+ y be a morphism in 'JI (8). That is, an equivalence class of diagrams p

"'

-------+

y

X

with

f

in M or (S). In the triangle

p

---+ X ---+

z

---+

Ep

we must have z E (8). On the other hand, from the hypothesis of the Proposition, x E 'Jf3, 8 c 'Jf3 and z E (S). By Theorem 4.3.3, we know that there is a z' E (8)(3 so that x---+ z factors as

x

---+

z'

---+

z.

136

4. THOMASON'S LOCALISATION

We deduce a map of triangles p' ~X

g1

--t

11

p ~X

z'

--t

1

1 --t

z

'Ep'

--t

'Ep

The morphism fg: p'--+ x lies in Mor(S) 13 since z' E (S)fJ. The diagram PI

ag

--t

y

X

is therefore a morphism in -EA

By Corollary 3.2.11, the coproduct Il>.EA X;. exists in 'JI (8); in fact, it agrees with the coproduct in 'J. We are given a map in 'JI (8) k

------:-+

Il X>..

>-EA

It is a map in 'JI (8) from an object in k E 'Jf3 to an object and by Proposition 4.4.1,

{'JI(8)}

Il>.EA

X;. E 'J,

(k, II X;.)= {'JI(8) 13 } (k, Il X;.). AEA

>-EA

That is, the map k

------:-+

Il X>.

>.EA

comes from a morphism in 'JI (8) 13 . It therefore is represented by a diagram p

------:-+

Il X>.

>.EA

4. THOMASON'S LOCALISATION

140

where f

E

M or (S)f3. This means that there is a triangle p

-----+

k

-----+

z

-----+

"Ep

with z E (8){3 C 'J'f3. Since k is also assumed in 'J'f3, it follows that p E 'J'f3. But then the ,6...,-smallness of p E 'J' guarantees that the map in 'J' p

Il X

-----+

A

AEA

factors as p

-----+

Il X

A -----+

AEA'

Il X

A

AEA

where A' c A has cardinality less than ,6. The ,6-compactness of p in 'J' says that the map in 'J'

factors as P -----+

Il k

A -----+

AEA'

Il X

A

AEA'

with kA E 'J'f3. In other words, in 'J'/(8) we factored the map k

-----+

Il X

A

AEA

as

0

Even the seemingly stupid case, where (8) further.

= 'J', is worth considering

LEMMA 4.4.5. Let 'J' be a triangulated category satisfying [TR5]. Let 8 be a class of objects in 'J'o:, for some infinite a. Suppose (8) = 'J'. Let ,6 be a regular cardinal~ a. Then the inclusion (8){3 C 'J'f3 is an equality. In other words, every object of 'J'f3 is in (8){3.

Proof: Let x be an object of 'J'f3. We need to prove that x is in (8){3. The identity map 1 : x -----+ x is a morphism from x E 'J'f3 to x E (8) (we are assuming (8) = 'J'). By Theorem 4.3.3, it factors through some object y E (8){3. Thus x is a direct summand of y E (8)f3. But (8){3 is thick, 0 hence x E (8)f3.

4.4. MAPS IN THE QUOTIENT

141

PROPOSITION 4.4.6. Let 'J' be a triangulated category satisfying [TR5j. Let T C 'J'"' and S C 'J'"' be two classes of a-compact objects, a an infinite cardinal. Suppose that (T) = 'J'. Suppose that f3 2: a is a regular cardinal. Then the inclusion 'J'13 I (S) 13 c {'J'I (S) } 13

is almost an equivalence; every object of {'J'I (S) } 13 is isomorphic to a direct summand of something in the image. That is, the (3-compact objects of 'J'I (S) are, up to splitting idempotents, the images of (3-compact objects of 'J'. Proof: Note that the map 'J' ---t 'J'I (S) takes 'J'i3 to (3-compact objects of 'J'I (S), by Lemma 4.4.4. Hence there is a well defined map

'J'13 I (S) 13

---t

{'J'I (S) } 13

0

The fact that this map is fully faithful is a consequence of Corollary 4.4.2. We may therefore view

'J'13 I (S) 13

c

{'J'I (S) } 13

------,

as a fully faithful embedding of categories. Let 'J'/3 I (S) 13 be the thick closure of 'J'i3 I (S) 13 . Since {'J'I (S)} 13 is a thick subcategory of 'J'I (S) containing 'J'i3 I (S) 13 , we deduce We need to prove the opposite inclusion, ']'!3

I (S) 13 :J {'J'I (S) } 13 .

Now 'J'i3 is a triangulated subcategory of 'J', and contains all coproducts of fewer that (3 of its objects. By Lemma 3.2.10, coproducts in 'J' and 'J'I (S) agree. Therefore 'J'i3 I (S) 13 , and hence also 'J'/3 I (S) 13 , are closed under the formation of coproducts in 'J'I (S) of fewer than (3 of their objects. Then 'J'/3 I (S) 13 is thick, and contains the coproducts of fewer than (3 of its objects. Since T c 'J'"' c 'J'i3 = 'J'i3 I (8) 13 ,

'J'/3 I (S) 13 :J (T) 13 where (T) 13 C 'J'I (S) is the smallest (3-localising, thick subcategory containing T. On the other hand, we assume that (T) = 'J'; that is, 'J' is the smallest localising subcategory of 'J' containing T. We deduce that 'J'I (S) is the smallest localising subcategory of 'J'I (S) containing T. If we view T as a subclass of 'J'I (S), we still get (T) = 'J'I (S). By Lemma 4.4.5, applied to the class T in the category 'J'I (S), we get (T) 13

= {'J'I (S) } 13

0

142

4. THOMASON'S LOCALISATION

Hence

Thus the two subcategories are equal, as stated.

D

REMARK 4.4.7. In Proposition 4.4.6, the statement can be improved if f3 is not only regular but also f3 > N0 . In this case, 'Jf3 I (S)f3 is closed under the formation of coproducts of countably many objects, and idempotents in 'Jf3 I(S)f3 must split. Therefore, if f3 >No then every object of {'JI(S)}f3 is isomorphic in 'JI (S) to an object in 'Jf3 I (S)f3. There is no need to split idempotents; the categories {'JI (S) }{3 and 'Jf3 I (S)f3 agree, up to extending 'Jf3 I (S)f3 to include every object in 'JI (S) isomorphic to an object of 'Jf3 I (S)f3. LEMMA 4.4.8. Let 'J be a triangulated category satisfying [TR5]. Let S c 'J"' be a class of a-compact objects, a an infinite cardinal. LetS = (S) be the localising subcategory generated by S. Suppose that f3 ~ a is a regular cardinal. Then there is an inclusion

Proof: We will show that S n 'Jf3 is a /3-perfect class of objects in S(f3); hence it must be contained in the maximal such, Sf3. Let k E S n 'Jf3 be any object. Let {X>., .A E A} be a family of objects in S = (S). Suppose we are given a map

Because k E 'Jf3, it is /3-small in 'J, and hence there must be a subset A' C A of cardinality < f3, so that the map factors as · k

-------t

II X>.

>.EA'

-------t

II X>..

>.EA

This proves that k is /3-small in S. In other words, we have established that S n 'Jf3 c SCf3) . Next we want to show that S n 'Jf3 is a perfect class. By Remark 3.3.6, it suffices to show that every object k E Sn'Jf3 is /3-good. Assume therefore that A is a set of cardinality < /3, and we have a map

4.4. MAPS IN THE QUOTIENT

143

Because k E 'Jf3, this map must factor as

with k>. E 'Jf3. On the other hand, f>.: k>. ~X>. is a map from an object of 'Jf3 to an object of (S). We know from Theorem 4.3.3 that it must factor as k>. ~ k~ ~X>.

where k~ E (8) 13 C S n 'Jf3. Thus we have factored

II X>.

k -----+

>.EA

as

II k~

k -----+

-----+

>.EA

II X>..

>.EA

and k is ,8-good.

0

Summarising the work of this Section, we get THEOREM 4.4.9. LetS be a triangulated category satisfying [TR5], ~ C S a localising subcategory. Write 'J for the Verdier quotient SI~Suppose there is an infinite cardinal o:, a class of objects S C S"' and another class of objects R c ~ n S"', so that

~=

Then for any regular

(R)

and

S = (S).

.8 2::: o:, (R)/3 = ~!3 = ~ n sf3, (s)f3 = sf3.

The natural map factors as and the functor sf3 1~{3 ~ 'Jf3

is fully faithful. If .8

>

~o,

the functor sf3 1~{3 ~ 'Jf3

144

4. THOMASON'S LOCALISATION

is an equivalence of categories. If f3 = N0 , then every object of 'JP is a direct summand of an object in s.e j':RP.

Proof: From Lemma 4.4.8 we deduce an inclusion ::RnSP c ::RP. Trivially, we have an inclusion (R),e c ::R n s.e. Combining, we have inclusions

(R),e C ::R n sP c ::R,e. By Lemma 4.4.5, applied to R

c ::R n s.e c ::RP, we have (R),e

=

::R,e.

Hence equality must hold throughout, and we have

(R),e = ::R n s.e = By Lemma 4.4.5, applied to S C

:n.e.

sa, we have

(S).e = s.e. That the natural map

factors as

s.e I::R.e

---+

'J.e c 'J

is the statement that the image of a /3-compact object of S is /3--compact in 'J, that is Lemma 4.4.4. That the functor

s.e I:n.e

---+

'J.e

is fully faithful is Corollary 4.4.2. That the functor

s.e I:n.e

---+

'J.e

is an equivalence if f3 > N0 is Remark 4.4.7. The statement that, for f3 =No, every object of 'JP is a direct summand of an object of the full subcategory SP I::RP follows from Proposition 4.4.6. D

4.5. A refinement in the countable case The classical, most useful case of Thomason's localisation theorem is the case f3 = N0 . As stated, the theorem says that gNo I::RNo is embedded fully faithfully in 'JNo, and that the embedding is an equivalence up to splitting idempotents. But one gets a refinement, which we will discuss in this section. We begin with some definitions. DEFINITION 4.5.1. Let 'J be an essentially small category. Define Z('J) to be the free abelian group on isomorphism classes of the objects of'J. The object X of 'J, viewed as an element in Z('J), will be denoted [X].

4.5. A REFINEMENT IN THE COUNTABLE CASE

145

DEFINITION 4.5.2. Let 'J' be an essentially small additive category. One defines a group A('J) C Z('J) to be the subgroup generated by all

[XEBY]- [X]- [Y] where X, Y are objects of 'J'. DEFINITION 4.5.3. Let

:EX

establishing that

(XEBY]- (X]- (YJ is in T('J).

0

DEFINITION 4.5.8. Let 'J be an essentially small triangulated category. One defines a group K 0 ('J), the Grothendieck group of'J, to be Z('J) Ko('J) = T('J). REMARK 4.5.9. In particular, from the triangle X

-----+

0

-----+

:EX

-->

:EX

in 'J, we learn that (X]+ (:EX] vanishes in K 0 ('J). LEMMA 4.5.10. Suppose 'J is an essentially small triangulated category, S c 'J a triangulated subcategory. Suppose 'J is the thick closure of S. That is, every object of 'J is a direct summand of an object of S. Then in Z('J), which of course contains Z(S), one gets T('J) = A('J) + T(S). Proof: Clearly, A('J) and T(S) are both subgroups of T('J), and hence T('J) :J A('J) + T(S). The problem is to show the reverse inclusion. Choose any of the generators of T('J), that is

(YJ - (X] - [ZJ where X

-----+

Y

-----+

Z --> :EX

4.5. A REFINEMENT IN THE COUNTABLE CASE

147

is a triangle in 'J. By hypothesis, 'J is the thick closure of S; there is an object A E 'J so that X EB A E S, and there is an object B E 'J so that Z EBB E S. We deduce a triangle in 'J X EB A----+ Y EB A EBB----+ Z EBB----+ I": {X EB A},

and as X EB A E S and Z EB B E S and S is triangulated, the entire triangle lies in S. Hence

[YEBAEBB]- [XEBA]- [ZEBB] is an element of T(S). As

[YEBAEBB]- [Y]- [A]- [B],

[X EB A]- [X] - [A],

[ZEBB]- [Z]- [B] all lie in A('J), the identity

[Y]- (X]- [Z]

=

{[Y EB A EBB]- [X EB A]- [Z EBB]} -{[YEBAEBB]- [Y]- [A]- [B]} +{[X EB A] - [X] - [A]} + {[Z EBB] - [Z] - [B]}

shows that an arbitrary generator [Y]- [X]- [Z] ofT('J) lies in A('J)+T(S). Hence T('J)

c

A('J)

+ T(S),

and we are done.

D

PROPOSITION 4.5.11. Let 'J be an essentially small triangulated category. Let S C 'J be a triangulated subcategory. Suppose the thick closure of Sis all of'J. Then the natural map K 0 (S)----+ K 0 ('J) is a monomorphism. Furthermore, if X E 'J is an object so that [X] E K 0 ('J) lies in the image of Ko(S) ----+ Ko('J), then XES.

Proof: We need to show that the map fo : Ko(S) ----+ K 0 ('J) is injective, and analyse when an element [X] E Ko('J) lies in the image of f 0 . But fo is identified as the map Z(S) T(S)

Z('J) T('J)

=

We are using the fact that T('J) = A('J) therefore suffices to show that the map

Z('J) + T(S).

A('J)

+ T(S),

that is Lemma 4.5.10. It

148

4. THOMASON'S LOCALISATION

is injective. The map fo is obtained from fA by further dividing by T(S)/A(S). In other words, we are reduced to showing the injectivity of

and analysing its cokernel, which is isomorphic to the cokernel of fo Ko(S) ---t Ko('J). Let [X) - [Y) be an element of the kernel of fA; that is, X and Y are objects of S, and in 'J there exists an object P and an isomorphism

But since 'J is the thick closure of S, there must exist an object P' E 'J so that P E9 P' E S. Then the fact that there is an isomorphism

XE9PE9P' ~YE9PE9P' says that [X)- [Y) vanishes already in KA(S); the kernel of fA vanishes. Finally, assume that X is an object of 'J so that [X) lies in the image of the map fo : K 0 (S) ---t K 0 ('J), or equivalently in the image of the map fA: KA(S) ---t KA('J). Then there exist Band C inS and an identity in KA('J)

[X) = [B)- [C). This means that there is an object P E 'J and an isomorphism XE9CE9P~BE9P.

Find an object P' E 'J so that P E9 P' lies in S. We have an isomorphism

X E9 C E9 P E9 P'

~

B E9 P E9 P'.

Replacing C E S by CE9PE9P' E Sand BE S by BE9PE9P' E S, we may say that there are objects C and BinS and an isomorphism XE9C~

B.

Now consider the triangle

C ---t B ---tX---t

~C.

It is a triangle in 'J, but since C and B are in S and S is triangulated, the triangle lies inS. Hence XES. 0 A very useful special case of this is CoROLLARY 4.5.12. LetS be a triangulated subcategory of a triangulated category 'J. Suppose 'J is the thick closure of S. Then for any object X E 'J, the object X E9 ~X lies inS.

4.5. A REFINEMENT IN THE COUNTABLE CASE

149

Proof: If 'J' is small, this is an immediate corollary of Lemma 4.5.13, once one observes that X EEl EX vanishes in K 0 ('J'). One can reduce the general case to the essentially small case; but since a direct proof is very simple, let us give one. Since X E 'J' and 'J' is the thick closure of S, there exists Y in 'J' with {XEBY} E S. The suspension of {XEBY} is also inS; that is {EXEBEY} E S. There are three distinguished triangles in 'J' Y

----+

0

----+

EY

----+

EY

X

----+

X

----+

0

----+

EX

0

----+

EX

----+

EX

----+

0

The direct sum is, by Proposition 1.2.1, a triangle in 'J' X EEl Y

----+

X EEl EX

----+

EX EEl EY

----+

EX EEl EY.

But two of the terms, namely X EEl Y and EX EEl EY, lie in S. Hence so does the third term X EEl EX. D One more Lemma before we get to the main point. LEMMA 4.5.13. As in Theorem 4.4.9, let S be a triangulated category satisfying [TR5j. But unlike Theorem 4.4.9, we now insist that S have small Hom-sets. Let ~ C S be a localising subcategory. Write 'J' for the Verdier quotient S/~. Let (3 be a regular cardinal. Suppose there is a set (not just a class, as in Theorem 4-4.9} of objects 8 C g.B and another set of objects R c ~ns.B, so that

~=

(R)

and

s=

(8).

Then the categories ~.8, g.B and 'J'.B are all essentially small. Proof: By Theorem 4.4.9, we know that ~.8 = (R).B, and g.B = (8).8. Since we are assuming 8 and R are sets and S has small Hom-sets, it follows from Proposition 3.2.5 that (R).B and (8).8 are essentially small; in other words, ~.8 and g.B are essentially small. But then by Proposition 2.2.1, the Verdier quotient g.B j~.B has small Hom-sets, and since there is clearly only a set of isomorphism classes of objects, the quotient is essentially small. By Theorem 4.4.9 we know that g.B j~.B is a full subcategory of 'J'.B, and up to splitting idempotents, the two categories agree. That means that 'J'.B is obtained from g.B j~.B by splitting some idempotents. Since g.B j~.B is essentially small, it follows that so is 'J'.B. D Now we are ready for the refinement of Theorem 4.4.9 in the countable case.

150

4. THOMASON'S LOCALISATION

COROLLARY 4.5.14. LetS be a triangulated category with small Homsets, satisfying [TR5]. Let ~ C S be a localising subcategory. Write 'J for the Verdier quotient Sf~. Suppose there is a set of objects S c s~o and another set of objects R c ~ n s~o, so that

~= (R)

and

S

= (S).

Then not only is it true that 'J~ 0 is the thick closure of the full subcategory s~o ~~~ 0 ' but in fact every object in X E 'J~O which lies in the image of Ko (s~o ~~~o)

-------+

Ko (T~o)

is isomorphic to an Object in s~o ~~~O C 'J~O.

Proof: 'J~o is a triangulated category, the category containing all objects of 'J~O isomorphic tO ObjeCtS Of s~o ~~~O iS a triangulated SUbcategory WhOSe thick closure is 'J~ 0 • Both categories are essentially small by Lemma 4.5.13. Proposition 4.5.11 applies, and in particular we learn that X E 'J~o will be isomorphic to an object in s~o ~~~o if and only if [X] lie in the image of the map

0

REMARK 4.5.15. The most useful case turns out to be the object X EB :EX. See Corollary 4.5.12 above. Thus for anyX E 'J~o, there is Y E s~o so that in 'J there is an isomorphism X EB

:EX~

Y.

4.6. History of the results in Chapter 4 The short way to describe the history is to say that everything is classical if a = ~0 . An N0 -small object, usually known as a compact object, is such that any map from it into a coproduct factors through a finite coproduct. In Remark 4.2.6 we saw that any No-small object is N0 -compact. Thomason proved his localisation theorem when a = No and 'J is the derived category of the category of quasi-coherent sheaves on a semiseparated scheme X. This proof may be found in [34]. The theorem is what Thomason calls his "key lemma" . For an arbitrary 'J, but still with a = N0 , there is a proof based on Bousfield localisation in the author's [23]. The proof here, not appealing to Bousfield localisation, is new. And, of course, the statement for all a, that is Theorem 4.4.9, is entirely new.

4.6. HISTORY OF THE RESULTS IN CHAPTER 4

151

Let a be a regular cardinal. Suppose 'J = ('JNo ). That is, 'J is the smallest localising subcategory containing 'JNo. Then by Theorem 4.4.9 'J"'

= ('JNo )"'.

If 'J is the homotopy category of spectra, then 'JNo are the finite CWcomplexes. By the above, 'J"' becomes identified with ('JNo)"', that is all the spectra with fewer than o: cells. This perhaps explains how the definitions of a-perfection and a-smallness were motivated by the attempt to copy classical arguments, which work by induction over the cardinality of the set of cells in a complex.

CHAPTER 5

The category A(S) 5.1. The abelian category A(g) Let g be an additive category. We do not assume that g is essentially small. We define DEFINITION

5.1.1. The categoryeat(goP,Ab) hasforits objects all the

additive functors F : gop ------+ Ab. The morphisms in eat(goP,Ab) are the natural transformations. It is well-known that eat(gDP,Ab) is an abelian category. We remind the reader what sequences are exact in eat(goP,Ab). Suppose we are given a sequence 0 ------+ F' (-) ------+ F (-) ------+ F" (-) ------+ 0

of objects and morphisms in eat(goP,Ab), that is functors and natural transformations gop~ Ab. This sequence is exact in eat(goP,Ab) if and only if, for every s E g, the sequence of abelian groups 0 ------+ F' (s) ------+ F (s) ------+ F" (s) ------+ 0

is exact. LEMMA 5.1.2. Let g be an additive category. Let eat(goP,Ab) be as in Definition 5.1.1. Then the representable functor g( -, s) is a projective object in the category eat(goP,Ab).

Proof: The functor Y;,(-) = g(-,s) is additive, hence an object of eat(goP,Ab). Let F be any object of eat(goP,Ab). Yoneda's lemma tells us that morphisms in eat(goP,Ab), that is natural transformations Ys(-)=g(-,s)------+ F(-)

are in one-to-one correspondence with elements of F(s). Suppose we are given an exact sequence in eat(goP,Ab) 0 ------+ F' (-) ------+ F (-) ------+ F" (-) ------+ 0.

154

5.

THE CATEGORY

A(S)

Applying the functor

to the exact sequence gives

0

~

F'(s)

~

F(s)

~

F"(s)

----7

=

But this sequence of abelian groups is exact. Hence Ys projective object.

0. g( -, s) is a D

DEFINITION 5.1.3. Let g be a triangulated category. Recall that we do not assume g essentially small. The category eat (gop, Ab), as in Definition 5.1.1, is the category of all additive functors gop--+ Ab. We define

c

A(g)

to be the full subcategory of all objects F which admit presentations

g(-,s)

~

g(-,t)

----7

F(-)

----7

0.

REMARK 5.1.4. In other words, the objects of A( g) are the objects of eat (gop, Ab) admitting a presentation by nice projective objects, namely the representable ones. LEMMA 5.1.5. The functors in A(g) take coproducts of objects in g to products of abelian groups.

Proof: Let F be an object of A(g); that is, it admits a presentation

g(-,s)

~

g(-,t)

----7

F(-)

----7

0.

It is clear that the reprel'?entable functors g( -, s) and g( -, t) take coproducts to products. Let {x>., .A E A} be a family of objects of g whose coproduct exists in g_ We have a commutative diagram with exact rows

~ g (y x>.,t)

----7

II g (x>., t)

(y

x>.)

----7

0

----7

0

1

llz

~

F

----7

II F (x>.)

and we immediately deduce that F sends coproducts to products.

D

LEMMA 5.1.6. Suppose F --+ G is a morphism in A(g). Then the cokernel is an object of A(g).

5.1. THE ABELIAN CATEGORY A(S)

155

Proof: We have an exact sequence of functors F (-) -------+ G (-) -------+ H (-) -------+ 0,

and we are given that F and G are in A(S); we would like to show that so is H. But F and G admit presentations S (-, s')

-------+

S (-, t')

-------+

F (-) -------+ 0

S (-, s)

-------+

S (-, t)

-------+

G (-) -------+ 0

and we have a map F--+ G, hence a composite

S( -, t')

-------+

F(-)

1

G(-). By Lemma 5.1.2, the representable functor S( -, t') is a projective object, and hence the map S( -, t') --+ G(-) factors through the surjection S( -, t) --+ G(-). In other words the composite

S( -, t')

-------+

F(-)

1

G(-) factors to render commutative the square

S( -, t')

-------+

F(-)

-------+

G(- ).

1

S( -, t)

1

We deduce a commutative diagram with exact rows S (-, t')

S(-,s)

-------+

F (-)

----+

G(-)

1

-------+

S(-,t)

and the cokernel H of the map F

S(-, s EB t')

----+

0

-------+

0

1

--+

S( -, t)

-------+

G has a presentation ----+

H(-)

-------+

0. 0

LEMMA5.1.7. SupposeF is an object ofA(S), and¢>: S(-,x)--+ F(-) is an epimorphism. Then the kernel of¢> is an object of A(S).

156

5. THE CATEGORY A(S)

Proof: Because F is an object of A(S), it has a presentation

S(-,s)

S(-,t)

------+

F(-)

--+

We are also given a map¢: S( -, x)

---t

--+

0.

F(- ); that is we have a diagram S(-,x)

~1 S(-,s)

-----+

S(-,t)

--+

F(-)

--+

0

where the row is exact. The object S(-,x) is projective in eat(S 0 P,Ab) by Lemma 5.1.2, and hence the map¢: S(-,x) ---t F(-) factors as S(-,x)

S(-,t)

------+

--+

F(-)

--+

0.

--+

F(-)

--+

0

We have a diagram

S( -, x)

1

S(-,s)

S(-,t)

------+

where the bottom row is exact and the composite

S(-,x)

1

S(-,t)

--+

F(-)

is surjective. We deduce that S( -, x) EB S( -, s)

--+

S( -, t)

is surjective. But by Yoneda this natural transformation of representable functors is induced by a morphism inS X

EB s -----+ t.

To say that this is surjective implies that

S(t,xEBs)-----+ S(t,t) is epi, and hence 1 : t

---t

t Complete x EB s

---t

r

t lies in the image. The identity on t factors as ------+ X

EB s -----+ t.

t to a triangle ------+

x EB s -----+ t

~

Er.

157

5.1. THE ABELIAN CATEGORY A(S)

This triangle is split. We must have an isomorphism x EB s have a commutative square in 'J

~

r EB t. We

r ----> x

1

1

s ____. t

which is bicartesian for the split exact structure in 'J. Anyway, we have a bicartesian diagram of functors

S(-,r) ----> S(-,x)

1

1

S(-,s) ----> S(-,t) in the abelian category eat(S 0 P,Ab), and the cokernels of the rows are isomorphic; there is a commutative diagram with exact rows

S(-,r) ----> S(-,x) ----> F(-)

1

1

------>

0

------>

0

1

1

1

S(-,s) ----> S(-,t) ----> F(-)

The surjection S(-,x)----+ F(-) has been completed to a presentation

S(-,r) ----> S(-,x) ----> F(-)

------>

0.

Now complete r ----+ x to a triangle q ----> r ----> x

------>

:Eq.

We have an exact sequence

S(-,q) ____. S(-,r) ___r:___.. S(-,x) _____. S( -, :Eq). (}

The functor F (-) is identified as the image of (}, and the kernel K (-) of S(-,x)----+ F(-) as the image of p. We have an exact sequence

S(-,q) ----> S(-,r) ----> K(-)

------>

0,

which establishes that K is an object of A(S). LEMMA

0

5.1.8. Suppose

0----> F(-)

-----+

G(-) ----> H(-)

------>

0

158

5. THE CATEGORY A(S)

is an exact sequence of functors in eat(S 0 P,Ab). Suppose F and H lie in A(S). Then there is a commutative diagram .with exact rows 0

----7

S( -,f)

0

----7

F(-)

-1

----7

S(-,g)

----7

G(-)

1

where all the vertical maps are surjective.

--

S(-, h)

----7

0

----7

0

1

H(-)

Proof: Since F and H lie in A(S), we may certainly choose surjections S( -,f)

0

S(-, h)

1

F(-)

----7

----7

G(-)

1

---+

H(-)

---+

0

By Lemma 5.1.2, the objectS(-, h) is projective in eat(S 0 P,Ab). The map S( -,h) ---+ H(-) must factor through the surjection G ---+ H. Letting g = f EB h, we have a commutative diagram with exact rows 0

----7

S( -,f)

0

----7

F(-)

----7

S(-,g)

----7

G(-)

1

---+

S(-,h)

---+

H(-)

1

----7

0

----7

0

1

and the fact that the two outside vertical maps are surjective forces the middle to also be. D LEMMA

0

5.1.9. Suppose

----7

F(-)

----7

G(-)

---+

H(-)

---+

0

is an exact sequence of functors in eat(S 0 P,Ab). If any two ofF, G and H lie in A(S), then so does the third. Proof: IfF and G lie in A(S), then so does H by Lemma 5.1.6. It remains to consider the other two cases. Suppose F and H lie in A(S). By Lemma 5.1.8, we may complete to a diagram with exact rows and surjective vertical maps

0

----7

S(-,f)

0

----7

F(-)

----7

S(-,g)

----7

G(-)

---+

S(-,h)

---+

H(-)

1

1

1

-o ----7

0

5.1. THE ABELIAN CATEGORY A(S)

159

Taking the kernels in the columns, we have a 3 x 3 diagram with exact rows and columns

0

1

0 -------> F'(-)

1

1

G'(-) -------> H'(-)

--+

S(-,g) -------> S(-,h)

1

1

0 -------> F(-)

0

--+

1

0 -------> S(-,f)

0

1

------+

0

------+

0

1

G(-) -------> H(-)

1

0

0

1

1

--+

------+

1 0

0

and Lemma 5.1.7, applied to the exact sequences

0 -------> F'(-)

--+

S(-,f) -------> F(-) -------> 0

0 -------> H'(-)

--+

S(-,h) -------> H(-)

------+

0,

allows us to deduce that F' and H' are in A(S). Applying Lemma 5.1.8 to the exact sequence

0 -------> F' (-)

--+

G' (-) -------> H' (-) -------> 0

we deduce a commutative diagram with exact rows and surjective columns

0-------> S(-,f')

--+

1

0 -------> F'(-)

S(-,g') -------> S(-, h')

1

--+

------+

0

------+

0.

1

G'(-) -------> H'(-)

Putting this all together we have a commutative diagram with exact rows and columns 0 -------> S( -, f') - - + S( -, g') -------> S( -, h') ------+ 0

1

0 -------> S( -,f)

0 ------->

1

--+

1

F(-)

1 0

1

--+

1

S( -,g) -------> S(-, h)

0

------+

0

1

G(-)

-------> H(-)

1

1

0

------+

0

160

5. THE CATEGORY A(S)

and in particular the middle column is exact

S(-,g') -----+ S(-,g) -----+ G(-)

-----+

0,

meaning that G is in A(S). It remains to show that if in the exact sequence

0 -----+ F(-) -----+ G(-)

-----+

H(-) -----+ 0

the functors G and Hare in A(S), then so is F. Since G E A(S), we may choose a surjection S(-,g) --+ G(-). Consider the diagram with exact rows 0 -----+ 0 -----+ S(-,g) -----+ S(-,g) -----+ 0

1

1

0 -----+ F(-) -----+ G(-)

1

-----+

H(-) -----+ 0

We wish to apply the snake lemma to it. The kernels and cokernels fit in a six-term exact sequence. The kernels and cokernels are computed by the diagram with exact rows and columns 0 0

0

0-----+

1 0

1

-----+ G'(-)

H'(-)

1

1

1

1

-----+ S(-,g) -----+ S(-,g) -----+ 0

1

0 -----+ F(-) -----+ G(-)

1

1

-----+

F(-) -----+

1 0

-----+

-----+

H(-) -----+ 0

1 0

1

0 Lemma 5.1.7 applied to the exact sequences 0 -----+ G'(-) -----+ S(-,g)

-----+

G(-) -----+ 0

0-----+ H'(-) -----+ S(-,g) -----+ H(-) -----+ 0 allows us to deduce that G' and H' are in A(S). The snake lemma gives an exact sequence

0 -----+ G' (-) -----+ H' (-)

-----+

F(-) -----+ 0,

5.1. THE ABELIAN CATEGORY A(S)

161

and hence Lemma 5.1.6 establishes that F is also in A(S).

D

PROPOSITION 5.1.10. The full subcategory A(S) C eat(S 0 P,Ab) is an abelian subcategory closed under extensions. That is, if F ----+ G is a morphism of objects in A(S), then the kernel, image and cokernel, computed in the abelian category eat(S 0 P,Ab), lie in A(S). Also, any extension of objects of A(S) is in A(S).

Proof: Let f : F ----+ G be a morphism in A(S). By Lemma 5.1.6, its cokernellies in A(S). The image fits in an exact sequence

0

--t

Im(f)

-------->

G

--t

Coker(!)

--t

0,

and since G and Coker(!) lie in A(S), Lemma 5.1.9 tells us that so does Im(f). The kernel fits in the exact sequence 0

--t

Ker(f)

-------->

F --------> Im(f)

--t

0,

and since Im(f) and F lie in A(S), Lemma 5.1.9 tells us that so does Ker(f). Finally, if we have an exact sequence 0

--t

F

-------->

G

-------->

H

--t

0,

with F and H in A(S), Lemma 5.1.9 tells us that G is also in A(S). The subcategory A(S) c eat(S 0 P,Ab) is closed under extensions. D LEMMA 5.1.11. The representable functors S(-,x) lie in A(S), and are projective objects in it. Furthermore, every projective object in A(S) is a direct summand of a representable S (-, x). If all idempotents in S split, for example if S is N1 -localising, then the projective objects in A(S) are precisely the representables.

Proof: The objects S( -, s) have a presentation

S(-,0) -

S(-,s)

-------->

S(-,s) -

0,

which shows that they lie in A(S). By Lemma 5.1.2, they are projective objects already in the larger category eat (gop, Ab). Hence they must clearly be projective in A(S). Suppose F is a projective object in A(S). Being an object of A(S), it has a presentation S(-,s)

-----+

S(-,t)

-----+

F(-)

--t

0.

But F is projective, and the identity 1 : F ----+ F must factor through the epimorphism S( -, t) ----+ F(- ). That is, F must be a direct summand of S( -, t). If the category S is closed under the splitting of idempotents, then clearly F is representable. D

5. THE CATEGORY A(S)

162

REMARK 5.1.12. We remind the reader: the objects F E A(S) can be identified with the objects of the larger category eat(S 0 P,Ab) which admit projective presentations S(-,s)

~

S(-,t)

F(-)

~

-----+

0.

By Lemma 5.1.5, we know that such functors must take coproducts to products. The reader should be warned that not every functor taking coproducts to products need lie in A(S). Not even for very goodS. Of course in any abelian category, morphisms of objects give rise to morphisms of projective presentations. This leads to another description of the category A(S). DEFINITION 5.1.13. Let B(S) be the additive category whose objects are morphisms { s --+ t} E S. M orphisms

{s

--+

t}

-----+

{s'

--+

t'}

in B(S) are equivalence classes of commutative squares s

----t

t

1

1

s

t

s' -----+ t' The equivalence relation on morphisms is additive, and a morphism is equivalent to zero if in the diagram

the map ¢ : t

-7

----t

s' -----+ t' t' factors as t - 7 s' - 7 t'.

PROPOSITION 5.1.14. There is a functor B(S) object { s --+ t} E B(S) to the cokernel of S(-,s)

-----+

-7

A(S) sending the

S(-,t).

This functor is an equivalence of categories.

Proof: Clearly, every object of A(S) is in the image of the functor. And given projective presentations S(-,s)

~

S(-,t)

-----+

F(-)

-----+

0

S(-,s')

~

S(-,t')

-----+

F'(-)

-----+

0

then maps F(-) - 7 F'(-) are in one-to-one correspondence with homotopy equivalence classes of maps of projective presentations. Precisely,

THE ABELIAN CATEGORY

5.1.

A(S)

163

maps F (-) ---+ F' (-) correspond one-to-one to homotopy equivalence classes of chain maps

S(-,s)

S(-,t)

-----+

1

1

S( -, s') -----+ S( -, t')

that is with chain maps, where we identify two if the difference of the maps S( -, t) ---+ S( -, t') factors through S( -, s'). Thus the functor B(S) ---+ A(S) is fully faithful. D COROLLARY 5.1.15. Suppose S is a triangulated category with small Hom-sets. Then the abelian category A(S) has small Hom-sets.

Proof: In its description as B(S), this is clear; there is only a set of equivalence classes of diagrams s -----+ t

1

s'

1

t'.

-----+

D

REMARK 5.1.16. Note that the category eat(S 0 P,Ab) need not have small Hom-sets, even if S does. LEMMA 5.1.17. The Yoiwda mapS---+ A(S), sending s to the representable functor S (-, s), is a homological functor.

Proof: A(S) is an exact subcategory of eat(S 0 P,Ab). Exact sequences in the two categories agree. It therefore suffices to show that the functor S ---+ eat (gop, Ab) is homological. But this is clear: given a triangle in S

r

-----+

s

-----+

t

:Er,

----+

the sequence

S(-,r)

----+

S(-,s)

-----+

S(-,t)

is exact by Lemma 1.1.10.

D

THEOREM 5.1.18. LetS be a triangulated category. The functorS ---+ A(S) is a universal homological functor. Suppose we are given a homological functor H : S ---+ A, where A is some abelian category. There is,

up to canonical isomorphism, a unique exact functor of abelian categories

A(S) ---+A so that the composite

S ----+ A(S) ~ A

5. THE CATEGORY A(S)

164

is H. Furthermore, any natural transformation of homological functors S ---+ A factors uniquely through a natural transformation of the asociated exact functors A(S) ---+ A. Proof: Given an abelian category B with enough projectives, any additive functor on a generating subcategory of projectives extends uniquely, up to canonical isomorphism, to a right exact functor on B. This is standard. In our case, we haveS c A(S) is the full subcategory of representable functors. The objects of S are projective objects when viewed in A(S), and we are given an additive functor H : S ---+ A. Therefore H extends uniquely to a right exact functor

A(S)

ii

~A.

Given an object F of A(S) and a projective presentation

S(-,s)

~

S(-,t)

~

F(-)

~

0,

ii (F) is defined to be the cokernel of H(s)

~

H(t).

If H is assumed not only additive but homological, we need to prove that ii is left exact. Note that it is obvious that the composite of ii with S ---+ A(S) is H. Suppose first that we are given an exact sequence in A(S) of the special form 0

~

F'(-)

~

S(-,f)

~

F(-)

~

0.

We want to begin by showing that ii takes these to exact (as opposed to only right exact) sequences. Since F' E A(S), we may choose a surjection S( -, f') ---+ F'(-). In other words, we have a resolution ofF S(-,f') ~ S(-,f) ~ F(-) ~ 0, and F' is identified as the image of ¢. Complete

f''

~

f'

~

f

f'

~

---+

f to a triangle

Ef''.

The sequence S(-,f") ~ S(-,f') ~ S(-,f) ~ F(-) ~ 0 is exact in A(S), the functor F is the cokernel of¢ and the functor F' the cokernel of p. But the functor H is homological; it takes the triangle

f" ~ f' ~ f ~ Ef" to an exact sequence

H(f") ~ H(f') ~ H(f)

5.1. THE ABELIAN CATEGORY A(S)

and H(F) = coker(H(¢)) while H(F') deduce an exact sequence

=

165

coker(H(p)). We immediately

0 ----+ H(F') ~ H(f) ----+ ii(F) ----+ 0 as desired. Suppose now that we are given a general exact sequence in A(S)

0 ----+ F(-)

~

G(-) ----+ K(-) ----+ 0.

By Lemma 5.1.8 we can produce in A(S) a 3 x 3 diagram with exact rows and columns 0 0 0

0 ----+

1

F'(-)

~

1

0----+ S(-,f)

0----+

~

1

1

F(-)

1

1

1

1

G'(-) ----+ K'(-) ----+ 0

S(-,g) ----+ S(-,k) ----+ 0

1

G(-)

1

0 0 Note that for the sequence of projectives 0----+ S(-,f)

-t

1

----+ K(-) ----+ 0

1 0

S(-,g) ----+ S(-,k) ----+ 0

to be exact it must actually be split exact. Applying the functor H to this diagram we get a diagram with exact rows and columns

0

0

0

1

1

1

1

1

1

1

1

1

ii(F') -

1

0----+ H(f)

1

-t

H(F) -

1 0

ii(G') ----+ H(K') ----+ 0

H(g) ----+ H(k) ----+ 0

H(G) ----+ H(K) ----+ 0

0

0

166

5. THE CATEGORY A(S)

The exactness of the columns is the case just discussed. The middle row is exact because it ~s split exact. The other two rows are a priori only right exact. But the two maps

H(F')

1

H(f)

------+

H(g)

are both injective, hence so is their composite. The commutativity of

H(F')

------+

ii (G')

------+

H(g)

1

H(f)

1

tells us that the map H(F') ----t H(G') must be injective. Therefore we deduce the exactness of the rows and columns in the commutative diagram 0

0

1

1

0

------+

H(F')

0

------+

H(f)

------+

H( G')

------+

H(g)

------+

H(G)

1

1

1

------+

H(K')

------+

H(k)

------+

H(K)

1

1

H(F)

0

0

------+

0

------+

0

1

1 1

------+

1

1

0 0 0 and the 3 x 3 lemma tells us the bottom row must also be exact. So far, we have proved that a homological functor H : S ----t A extends uniquely, up to canonical isomorphism, to an exact functor ii : A(S) ----t A. The statement about the extensions of natural tranformations is easy, and we leave it to the reader. D REMARK 5.1.19. The universal property of the homological functor S ----t A(S) is clearly self-dual. In other words, the dual gop ----t { A(S)} op must satisfy the same property. Despite appearances, our construction must be self-dual. In the next few lemmas we elaborate on this point.

5.1. THE ABELIAN CATEGORY A(S)

167

DEFINITION 5.1.20. Define the category C(S) as follows. The objects are triangles in S

r

----+

s

----+

t

------+

:Er

and the morphisms are equivalence classes of morphisms of triangles r ----+ s ------+ t ------+ :Er

1

1

1

1

r' ----+ s' ------+ t' ------+ :Er' The equivalence relation is additive. A morphism is equivalent to zero if in the commutative square s ---7 t

1

s'

1 t'

------+

the equal composites s

s

1

s'

---7

------+

t

1

t'

t'

both vanish. LEMMA

5.1.21. Define a functor C(S)

r

----+

s

----+

t

r

------+

s.

---+

B(S) by taking the triangle

------+

:Er

to the pair This functor is an equivalence of categories. Proof: First let us note that the functor is well-defined. If a morphism in C(S) is equivalent to zero, we must show that its image in B(S) is also equivalent to zero. Take therefore a morphism in C(S) equivalent to zero

r

----+

1

s

------+

1

t

------+

1

:Er

1

r' ----+ s' ------+ t' ------+ :Er' The fact that it is equivalent to zero means that the composite s

1

s'

------+

t'

168

5. THE CATEGORY A(S)

vanishes. But

r' - - - > s' - - - > t' - - - > Er' is a triangle, and hence the map s ___. s' must factor as s ___. r' ___. s'. In other words, the square r ---> s

1

1

r'---> s'

defines a morphism in B(S) equivalent to zero. Every object in B(S), that is every {r ~ s} E S, can be completed to a triangle

r

--->

s

--->

t

----+

Er

and therefore lies in the image of the functor C(S) ___. B(S). Any commutative square r ----+ s

1

1

r'----+ s'

can be completed to a morphism of triangles r - - - > s ----+ t ----+ Er

1

1

1

1

1

1

r' - - - > s' ----+ t' ----+ Er' so the functor C(S) ___. B(S) is full. But also, if r ----+ s

r'----+ s' is equivalent to zero, then the map s ___. s' factors through r' ___. s', and in any completion to a morphism of triangles r - - - > s - - - > t ----+ Er

1

r'

--->

1

s'

1

---t

t'

the map s

1

s'

---t

t'

---t

1

Er'

5.1.

THE ABELIAN CATEGORY

A(S)

169

will have to vanish. The functor C(S) ------+ B(S) is fully faithful, and surjective on objects. Hence it is an equivalence of categories. D REMARK 5.1.22. The dual of the equivalence C(S) ------+ B(S) gives a functor C(S 0 P) ------+ B(S 0 P), which must also be an equivalence. Clearly C(S 0 P) = {C(S)} 0 P, the construction being self-dual. We deduce that the functor taking a triangle r

----+

s

----+

t

----+

Er

to the morphism

t ----+ Er is an equivalence of C(S) with {B(S 0 P)} op. The objects of C(S) of the form 0 ----+ s ~ s map in B(S) to the projective objects

0

----+

----+

0

s.

Dually, they must map to the projective objects s

----+

0

in B(S P). Since C(S) is naturally the dual of B(S 0 P), the objects 0

0 ----+ s ~ s ----+ 0 are not only projective objects in C(S), but they are also injective objects( =projective objects in the dual). We conclude

COROLLARY 5.1.23. The representable functors S( -, s) are not only projective objects in A(S), but also injective objects. Any projective or injective object is a direct summand of someS(-, s). If every idempotent in the category S is split, then the projective(=injective) objects of A(S) are precisely the representable functors S( -, s). Any object FE A(S) has both a projective presentation

S(-,s)

----+

S(-,t)

----+

F(-)

----+

0

and an injective copresentation

S( -, s')

+---

S( -, t')

+---

F(-)

+---

0

In other words, the category A(S) = B(S) = C(S) is a Jilrobenius abelian category. It is an abelian category with enough projectives and enough injectives, where it so happens that projectives and injectives coincide. LEMMA 5.1.24. Let (3 be an infinite cardinal. Suppose the category S satisfies [TR5((3)]. That is, coproducts of fewer than (3 objects exist in S. Then the category A(S) satisfies [AB3((3)]; that is, it is closed with respect to coproducts of < (3 objects. The universal homological functor

170

5. THE CATEGORY A(S)

S - - t A(S) respects coproducts of< /3 objects. Furthermore, a homological functor S - - t A respects coproducts of of < /3 objects if and only if the induced exact functor A(S) - - t A of Theorem 5.1.18 does.

Proof: This is easier to see in the description B(S) or C(S) of the category. Let us do it for C(S). Suppose that we have a set of < /3 objects in C(S). That is, we have a set A, of cardinality < /3, and for every A E A an object of C(S), i.e. a triangle in S r >. ----+ s >. ----+ t >. ----+ Er >..

Since the category S satisfies [TR5(/3)], we can form the coproduct of these triangles

II r>.

>.EA

----+

II s>.

----+

>.EA

II t>.

----+

E

>.EA

{II

r>.}.

>.EA

By Proposition 1.2.1 this is a triangle, that is an object of C(S). The reader will easily see that this object satisfies the universal property of a coproduct in the category C(S). In the special case of triangles of the form 1

0 ----+ s>. ----+ s >.

----7

0,

that is objects in the image of the universal homological functorS their coproduct is

o ----+

II s>.

~

>.EA

II s>.

----7

--t

C(S),

o

>.EA

and it immediately follows that the functorS - - t C(S) respects coproducts of < /3 objects. Finally, we need to show that a homological functor S - - t A, which by Theorem 5.1.18 factorises uniquely as S

----+

A(S)

3!

----7

A,

preserves coproducts of< /3 objects if and only if A(S) - - t A does. The "if" part is trivial. We know by the above that S - - t A(S) preserves coproducts of < /3 objects. If A(S) - - t A also does, then so does the composite S

----+

A(S)

----7

A.

Suppose therefore that H : S - - t A is a homological functor preserving coproducts of < /3 objects: We need to show that so does the hiduced functor ii : A(S) - - t A. In its realisation ii : B(S) - - t A, the induced

5.1. THE ABELIAN CATEGORY A(S)

171

exact functor takes the object {r---> s} E B(S) to the cokernel of the map H(r) ----+ H(s). That is, we have an exact sequence

H(r) ---- H(s)

-----+

H({r----+ s}) ---- 0.

Suppose A is a set of cardinality < /3, and for A E A, we have objects {rA B(S). The coproduct in B(S) is the object

--->sA} E

II rA

AEA

----

II sA.

AEA

The functor B(S) ----+A takes it to the cokernel of H

(y

rA) ---- H

(y sA) ;

we have an exact sequence H

(y rA)

-----+

H

(M sA) ---- H (y{rA --->sA})

------->

0.

On the other hand, the functor H respects coproducts; in the commutative square below, the vertical maps are isomorphisms

We deduce a commutative diagram with exact rows AEA

AEA

AEA

-----t

0.

The exactness of the top row is because coproducts are right exact, in any abelian category A. The 5-lemma now tells us that the map h must be an isomorphism, that is H preserves coproducts of < f3 objects. D REMARK 5.1.25. Dually, assumeS is a triangulated category satisfying [TR5*(/3)]. Then the category A(S) satisfies [AB3*(/3)], and the homological functor S ----> A(S) respects products of < f3 objects. A homological

5. THE CATEGORY A(S)

172

functor S ---+ A preserves products of < (3 objects if and only if the exact functor A(S) ~ A does.

5.2. Subobjects and quotient objects in A(S) In any abelian category, it is customary to study the behavior of subobjects and quotient objects. An abelian category is called well-powered if any object has a small set of subobjects, or equivalently a small set of quotient objects. Well-powered abelian categories are reasonable. In this sense, the categories A(S) are very unreasonable. In this section, we propose to study the elementary properties of subobjects and quotient objects in the abelian category A(S), where S is a triangulated category. In Appendix C we will show by example that even the simplest S may have an A(S) which is not well-powered. For the purpose of our study, it is convenient to introduce yet another model for the category A(S). DEFINITION 5.2.1. Let S be a triangulated category with small Homsets. The category D(S) has for its objects the morphisms { s --> t} in S. A morphism {s --> t} ---+ {s' --> t'} in D(S) is an equivalence class of commutative squares in S

s---+ t

1

1

s'---+ t'.

The equivalence relation on such squares is additive, and a square is defined equivalent to zero if the equal composites s---+ t

s

1

1

s' - - - + t'

t'

both vanish. LEMMA 5.2.2. LetS be a triangulated category with small Hom-sets. There is a natural functor C(S) ---+ D(S). It takes an object of C(S), that is a triangle in S

r

---+

s

---+

t

to the morphism s---+ t,

---+

Er

5.2. SUBOBJECTS AND QUOTIENT OBJECTS IN A(S)

173

which defines an object in D(S). It takes a morphism in C(S), that is an equivalence class of morphisms of triangles r

s

-------+

1

r'

-------+

1

-------+

s'

t

L:r

-------+

1

1

-------+

t'

L:r'

-------+

to the square s

t

-------+

1

1

s' -------+ t'. This functor is an equivalence of categories.

Proof: The functor is clearly well-defined; equivalent morphisms map to equivalent morphisms. In fact, the definition of the equivalence is the same in both C(S) and D(S). In both cases, a morphism is equivalent to zero if the equal composites s

s

1

s'

-------+

1

t'

-------+

t

t'

both vanish. Therefore the functor C(S) - - t D(S) is clearly faithful. By [TRl], any morphism {s ---+ t} in S may be completed to a triangle r

-------+

s

-------+

t

-------+

L:r;

hence the functor C(S) - - t D(S) is surjective on objects. By [TR3], any commutative diagram with triangles for rows r

-------+

r'

-------+

r

-------+

s

-------+

t

-------+

1

1

s

t

L:r

s' -------+ t' -------+ L:r' may be completed to a morphism of triangles

1

r'

-------+

-------+

1

s'

-------+

-------+

1 t'

-------+

L:r

1

L:r'.

Thus the functor C(S) - - t D(S) is surjective also on morphisms. It is a fully faithful functor, surjective on objects; therefore it gives an equivalence D of categories.

5.

174

THE CATEGORY

A(S)

REMARK 5.2.3. By Lemma 5.1.21, there is an equivalence of categories C(S) --+ B(S). We remind the reader what this functor does. It takes an object of C(S), that is a triangle in S

r

---+

s

---+ t ---+

:Er,

to an object of B(S), which was explicitly the morphism inS r---+ s. By Proposition 5.1.14, there is an equivalence of categories B(S) sending the object

--+

A(S),

r---+ s to the cokernel of the map S(-,r)

---+

S(-,s).

s

---+

t

Since for the triangle

r

---+

---+

:Er

the sequence S(-,r) ~ S(-,s) ~ S(-,t) is exact, the composite C(S) that is a triangle

r

B(S)

--+

---+

t

--+

---+

s

A(S) sends an object of C(S),

---+

:Er,

to the cokernel of a, which is the same as the image of S(-,s)

(3 ---+

S(-,t).

In other words, the equivalence C(S) --+ A(S) of Lemma 5.1.21 and Proposition 5.1.14 factors through the equivalence C(S) --+ D(S) of Lemma 5.2.2. We deduce an equivalence D(S) --+ A(S), taking an object { s---+ t} of D(S) to the image of the map S(-,s)

---+

S(-,t).

In Remark 5.1.12 and Proposition 5.1.14, we saw that objects in the the category B(S), that is morphisms {r ---+ s} inS, may be viewed as a projective presentation for the object F(-) of A(S). We have an exact sequence S(-,r)

---+

S(-,s)

---+

F(-)

---+

0.

The objects in the category D(S) may be viewed as an object F(-) of A(S), together with an embedding in an injective and a projective mapping onto it. We may think of an object { s---+ t} in D(S) as S(-,s) ~ F(-) ~ S(-,t)

5.2. SUBOBJECTS AND QUOTIENT OBJECTS IN A(S)

175

where ¢ is surjective and() injective. The natural functor D(8) ---+ A(8) sends { s--+ t} in D(8) to the image of

S(-,s)

-----4

S(-,t).

LEMMA 5.2.4. Let 8 be a triangulated category with small Hom -sets. Lets be an object of 8. Via the universal homological functor 8 ---+ A(S) = D(8), we view s as an object of D(8). Any quotient object of s can be presented as an object {s--+ t} of D(8), for some t and some morphism s---+ t in 8.

Proof: Suppose we are given a quotient object of s, that is a surjective maps--+ F. Choose any embedding ofF in an injective object t E 8; then F is the image of {s--+ t}, in D(S). 0 LEMMA 5.2.5. Let 8 be a triangulated category with small Hom-sets. Lets be an object of 8. Let {s--+ t} and { s--+ t'} be two quotient objects of s in D(8). These quotients are isomorphic if and only if there exist maps t ---+ t' and t' ---+ t rendering commutative the diagrams

s

s

-----4

t

s

1

-----4

t'

s

-----4

t'

1 -----4

t

Proof: Suppose we are given isomorphic quotients { s --+ t} and { s --+ t'} in D(S). That is, we have a quotient s ---+Fin D(8), and two embeddings of it in injectives t and t'. Because t' is injective, the map F ---+ t' factors through the injection F ---+ t; we deduce a commutative square s

-----4

t

s

-----4

t'.

By symmetry, we also have a commutative square s

-----4

t'

s

-----4

t.

Now suppose we have two commutative squares s

s

-----4

t

1

-----4

t'

s

s

-----4

t'

1 -----4

t

176

5. THE CATEGORY A(S)

we need to show that the two quotients of s agree. But these commutative squares define morphisms in D(S) of the quotient objects. The composites of the two morphisms, in both orders, give diagrams s - - - - 7 t' s--+ t

11

11+p

ll+r

11

s - - - - 7 t' s --+ t and these differ from the identity morphisms s - - - - 7 t' s--+ t

11

11

11

11

s--+ t

s

----7

t'

s--+ t

s

----7

t'

p1

ol

s--+ t

s

s

s

ol

ol

by

o1 Since it is clear that

s--+ t vanish, it follows that the composites s--+ t

11

11+p

Tl ----7

t'

s

----7

t'

s

----7

t'

ll+r

11

s --+ t s - - - - 7 t' are equivalent to the identities. Thus the objects { s --t t} and { s --t t'} of D(S) are isomorphic to each other in D(S), and the isomorphisms respect the quotient map from s. D PROPOSITION 5.2.6. LetS be a triangulated category with small Homsets. Let s be an object of S. The quotient objects of s in the category A(S) = D(S) can be represented as {s --t t} E D(S), and two pairs {s --t t} and {s --t t'} in D(S) give isomorphic quotients if there are commutative diagrams in S s--+ t s - - - - 7 t'

1

s - - + t'

s

1 ----7

t.

5.3.

THE FUNCTORIALITY OF

A(S)

177

The subobjects of s may be represented as pairs {t ---t s} E D(S), and two pairs {t ---t s} and {t' ---t s} in D(S) give isomorphic subobjects if there are commutative diagrams in S

t

-------+

1 t'

-------+

s

s

t'

-------+

1 t

-------+

s

s.

Proof: For quotients, the existence of a representation { s ---t t} for a quotient of sis Lemma 5.2.4. The characterisation of isomorphic quotients is Lemma 5.2.5. The statements about subobjects are simply the duals of the statements about quotients. D REMARK 5.2.7. In Proposition 5.2.6, we described the quotients and subobjects of s E S C A(S). Since every projective object of A(S) is a direct summand of s E S, we have described the quotient objects of any projective object of S. Similarly, since every injective object of A(S) is a direct summand of s E S, we have described the subobjects of injective objects of A(S). Since any object of A(S) may be embedded in an injective and is a quotient of a projective, we have in some sense described all quotients and subobjects of any object of A(S). CAUTION 5.2.8. An abelian category A is called well-powered if, for every object a E A, the class of isomorphism classes of subobjects of a is a small set. Equivalently, the class of isomorphism classes of quotient objects of a is a small set. Subobjects and quotient objects are in 1-to-1 correspondence, with a subobject b C a corresponding to the quotient afb. If S is a small category, then so is A(S), and the category is obviously well-powered. But for almost all non-trivial large categories S, the category A(S) is decidedly not well-powered. In Appendix C, more precisely in Proposition C.3.2, Corollary C.3.3 and Remark C.3.4, we will show that if D(Z) is the derived category of the category of abelian groups, then A(D(Z)) is not well-powered.

5.3. The functoriality of A(S) Given a triangulated category 'J, we learned in Section 5.1 how to associate to it an abelian category A('J). LEMMA 5.3.1. LetS and 'J be triangulated categories with small Homsets, F : S ---+ 'J a triangulated functor. Up to canonical isomorphism, there is a unique natural exact functor A(F) : A(S) ---+ A('J) making

178

5. THE CATEGORY A(S)

commutative the diagram

s

1

1

A(S) ~ A('J). Let (3 be an infinite cardinal. If S and 'J satisfy [TR5((3)}, and the functor F : S - - - t 'J preserves coproducts of < (3 objects, then the induced functor A(F) : A(S) - - - t A('J) also preserves coproducts of< (3 objects.

Proof: We have a diagram

1

1

A('J)

A(S)

and the composite S - - - t 'J - - - t A('J) is a homological functor. By Theorem 5.1.18, any homological functorS - - - t A factors uniquely through the universal homological functor S - - - t A(S). There is a unique exact functor A(F) : A(S) - - - t A('J) making commutative the diagram

s

F ------+

1

1

A(S) ~ A('J). It remains to prove the statement about coproducts: if the categories S and 'J both satisfy [TR5((3)] and the functor F preserves coproducts of < (3 objects, we need to show that the functor A(F) : A(S) - - - t A('J) also preserves coproducts of < (3 objects. But in the composite

s~

'J

1

A('J)

the functor F preserves coproducts of < (3 objects, by hypothesis. The functor 'J - - - t A('J) preserves coproducts because 'J satisfies [TR5(f3)], and by Lemma 5.1.24. Hence the composite preserves coproducts of< (3 objects. But this composite is equal to ·

s

1 A(S)

~ A('J),

5.3.

THE FUNCTORIALITY OF

A(S)

179

and by Lemma 5.1.24 and the facts that S satisfies [TR5(,8)], the composite preserves coproducts of < ,8 objects if and only if A(F) : A(S) ~ A('J) does. Hence A(F) : A(S) ~ A('J) preserves coproducts of < ,8 objects. 0 REMARK 5.3.2. Dually, if Sand 'J satisfy [TR5*(,8)] and F preserves products of < ,8 objects, then A(F) : A(S) ~ A('J) preserves products of < ,8 objects. REMARK 5.3.3. Let F : S ~ 'J be a triangulated functor of triangulated categories with small Hom-sets. In Lemma 5.3.1 we defined the functor A(F) : A(S) ~ A('J). By Proposition 5.1.14 and Lemma 5.1.21 we know that A(S) c:=- B(S) c:=- C(S). In terms of B and C, the functors B(F) : B(S) ~ B('J) and C(F) : C(S) ~ C('J) are very simple to describe explicitly. An object in B(S) (respectively C(S)) is a map { s ~ s'} in S (respectively a triangle { s ~ s' ~ s" ~ ~s} in S). The functor B(F) (respectively C(F)) takes this to the morphism {Fs ~ Fs'} in 'J (respectively to the triangle { F s ~ F s' ~ F s" ~ ~F s} in 'J). LEMMA 5.3.4. The assignment A(-) is a lail functor from the 2category of triangulated categories and triangulated functors, to the 2category of abelian categories and exact functors. It takes a triangulated category S to the abelian category A(S), takes a triangulated functor F to the exact functor A(F), and takes a natural trasformation ¢ to a natural transformation A(¢). It respects composition (in the lax sense} and identities.

Proof: Obvious.

0

REMARK 5.3.5. The functors B(-) and C(-) are strict functors of 2categories, not just lax functors. This follows from the explicit description of B(F) and C(F) in Remark 5.3.3. Next we wish to say something about the relation between adjoint functors between triangulated categories S and 'J, and adjoint functors between A(S) and A('J). But first a lemma. LEMMA 5.3.6. Let S and 'J be triangulated categories, F : S ~ 'J a triangulated functor. Suppose F has an adjoint (left or right} G: 'J---"' S. Then G is also a triangulated functor.

Proof: We may assume G is a right adjoint, the case of left adjunction being dual. We are given that F commutes with ~' up to natural isomorphism. Thatis F~

~F.

1 A lax functor between 2--categories only respects composition up to 2isomorphisms.

180

5. THE CATEGORY A(S)

Taking right adjoints of this identity and recalling that the right adjoint of E is E- 1 (see the proof of Proposition 1.1.6), we deduce

E- 1 G

GE- 1 .

=

Hence G commutes with E- 1 , and therefore also with E. Let X ----7 Y ----7 Z ----7 EX be a triangle in 'J; we need to show that GX ----7 GY ----7 GZ ----7 GEX is a triangle in S. Complete GX ----7 GY to a triangle GX ----7 GY ----7 C ----7 EGX in S. Because F is triangulated, FGX ----7 FGY ----7 FC ----7 EFGX is a triangle in 'J. Let c:x : FGX ----7 X and C:y : FGY ----7 Y be the counit of adjunction; it is the map corresponding to the identity 1 : G X ----7 G X under the natural isomorphism

S(GX,GX)

'J(FGX,X).

By the naturality of c: we have a commutative square in 'J FGX ----+ FGY

X ----+ y which we may complete to a morphism of triangles FGX - - - - + FGY ----+ FC----+ EFGX

X ----+ Y Let R be an object of S. Then

S(R, C)

----+

Z

----+

e defines a map eoF(-)

EX.

'J(FR, Z),

and we deduce a commutative diagram with exact rows: S(R, GX) ----7 S(R, GY) ----7 S(R, C) ----7 S(R, EGX) ----7 S(R, EGY)

111

111

1

111

111

'J(F R, X)----7 'J(FR, Y)----7 'J(FR, Z)----7 'J(F R, EX)----7 'J(F R, EY) The 5-Lemma implies that S(R, C) ----7 'J(F R, Z) is an isomorphism, i.e. that e : FC ----7 Z is precisely the counit of adjunction. That. is C = G Z, and the result follows. D REMARK 5.3.7. It should be noted that the situation here is better than with abelian categories. If F is an exact functor of abelian categories with a right adjoint G, then G is in general only left exact. If G is the left adjoint of F, then it is in general only right exact. If we think of triangulated functors between triangulated categories as the natural analog of exact functors between abelian categories, then Lemma 5.3.6 is surprising.

5.3. THE FUNCTORIALITY OF A(S)

181

LEMMA 5.3.8. Suppose F : S -----+ 'J is a triangulated functor of triangulated categories. Suppose F has a right adjoint G : 'J -----+ S. By Lemma 5.3.6, G is triangulated. By Lemma 5.3.1, both F and G induce exact functors between A(S) and A('J). We assert that these induced functors on abelian categories are also adjoint to each other. We have exact functors

A(S)

~ A('J)

A('J)

~

A(S)

and A(F) is left adjoint to A( G).

Proof: Let us use the equivalent categories B(S) ~ A(S), B('J) ~ A('J). An object in B(S) is a morphism s -----+ s' in S. An object in B('J) is a morphism t -----+ t' in 'J. The object B(F)[ { s ----> s'}] is just F applied to {s----> s'}. A morphism

B(F)[{s----> s'}]

-------+

{t----> t'}

is an equivalence class of commutative diagrams in 'J F s -------+ F s'

1

1 t

-------+

t'

where two diagrams are equivalent if the difference of the maps F s' -----+ -----+ t'. But by adjunction this is the same as an equivalence class of diagrams s -------+ s'

t' factors through t

1

Gt

1

-------+

Gt'.

There is therefore a natural 1-to-1 correspondence of maps

B(F)[{s ~ s'}]

-------+

{t----> t'}

{s----> s'}

-------+

B(G) [{t----> t'}]. D

PROPOSITION 5.3.9. LetS and 'J be triangulated categories, closed under splitting idempotents. The triangulated functor F : S -----+ 'J will have a right adjoint G : 'J-----+ S if and only if the exact functor A( F) : A(S) -----+ A('J) has a right adjoint A( G) : A('J) -----+ A(S).

Proof: If F : S -----+ 'J has a right adjoint G : 'J -----+ S, we know from Lemma 5.3.8 that A(F) : A(S) -----+ A('J) has a right adjoint A(G) A('J) -----+ A(S). We need the converse. Suppose therefore that A(F) A(S) -----+ A('J) has a right adjoint G: A('J) -----+ A(S).

182

5. THE CATEGORY A(S)

The functor G : A(':T) ----> A(S) has an exact left adjoint. It therefore takes injectives to injectives. But we are assuming that idempotents split both inS and in ':T. By Corollary 5.1.23, the injectives in A(S) (resp. A(':T)) are S C A(S) (resp. ':T c A(':T)). Therefore G : A(':T) ---t A(S) restricts to a functor G : ':T ----> S. This functor is clearly right adjoint to F. By Lemma 5.3.6, this adjoint G must be a triangulated functor of triangulated categories. 0 REMARK 5.3.10. The dual of Proposition 5.3.9 says that a triangulated functor G : ':T ----> S will have a left adjoint if and only if the functor A(G) : A(':T) ----> A(S) does. Thus, looking for adjoints to triangulated functors reduces to looking for adjoints to the associated exact functors. The problem with this proposition is that it is nearly impossible to apply. Existence theorems for adjoints usually depend on the categories being well-powered. The reader is referred to Caution 5.2.8 and Appendix C, for a discussion of just how far the categories A(':T) are from being well-powered.

5.4. History of the results in Chapter 5 The definition of the category A(S), its universal property for homological functors, and the fact that it is a Frobenius abelian category may all be found in Freyd's [13], more precisely in Section 3 on pages 127-133. Verdier's thesis, unpublished until very recently, also contains the results. See Sections 3.1 and 3.2, pages 135-144 of [36). I think Freyd had the result first; although one can not be sure. Verdier submitted his thesis in 1967, Freyd's result was already in print by 1966. The treatment we give is more similar to Verdier's. The only result in the Chapter with a claim to originality is Proposition 5.3.9. We remind the reader: Proposition 5.3.9 asserts that a triangulated functor F : S ---t ':T has an adjoint· if and only if the induced functor A(F) : A(S) ----> A(':T) does. The analysis of subobjects and quotient objects of an object in A(':T), given in Section 5.3, was certainly known to the experts. I have no doubt that either Freyd or Verdier could easily have provided the same treatment. More recently, Strickland was certainly aware of the results, as was Grandis. Part of the purpose of Section 5.3 is to prepare the ground for Appendix C, in which we show that the objects of A(':T) may well have classes, not sets, of subobjects.

CHAPTER 6

The category 6.1.

t:x(S P,Ab) 0

ex (sop, Ab)

is an abelian category satisfying [AB3] and [AB3*]

Let a be a regular cardinal. Throughout this Chapter, we fix a choice of such a cardinal a. Let S be a category, satisfying the following hypotheses HYPOTHESIS

if

6.1.1. The category S is said to satisfy hypothesis 6.1.1

6.1.1.1. S is an essentially small additive category. 6.1.1.2. The coproduct of fewer than a objects of S exists inS. 6.1.1.3. Homotopy pullback squares exist in S. That is, given a diagram inS X

1

y it may be completed to a commutative square x'-----+

p

-----+X

1

1

x'-----+

y

so that any commutative square S

-----+X

1 x'-----+

1 y

is induced by a (non-unique) map s ------7 p. The object p is called the homotopy pullback of the diagram X

1 x'-----+

y

184

6. THE CATEGORY ex(S 0 P,Ab)

and the homotopy pullback square p

--------7

1

x'

X

1 y

--------7

is unique up to (non-canonical) isomorphism. 6.1.1.4. Coproducts of fewer than o: homotopy pullback squares are homotopy pullback squares. In other words, let A be a set of cardinality< o:. If for every .A E A we are given a homotopy pullback square P>.

--------7

XA

1

1

then the coproduct

liP>-

--------7

.AEA

1

Il X~

.AEA

Il X>.

.AEA

1

--------7

Il Y>.

.AEA

is also a homotopy pullback square. EXAMPLE 6.1.2. The example of most interest is when S is an essentially small triangulated category. Being essentially small, it satisfies 6.1.1.1. If it is closed under the formation of coproducts of fewer than o: objects, then it satisfies .also 6.1.1.2. The existence of homotopy pullback squares is automatic; see Definition 1.4.1 and Notation 1.4.2. Hence we automatically have 6.1.1.3 whenever S is triangulated. The fact that the coproducts of fewer than o: homotopy pullback squares is a homotopy pullback square is also automatic for triangulated categories S, following from Proposition 1.2.1. Hence 6.1.1.4 also comes for free, when Sis triangulated. As I said, this is the important example for us. We treat the slightly more general case only for the purpose of constructing certain counterexamples; see Section A.5. The reader not interested in counterexamples may safely restrict his attention to only essentially small triangulated categories S, satisfying [TR5(o:)]. Recall that S satisfies [TR5(o:)] if it is closed under the formation of coproducts of fewer than o: objects. The categories satisfying Hypothesis 6.1.1 are more general, but mostly this will play no role.

6.1. [AB3] AND [AB3*]

185

We remind the reader of Definition 5.1.1. Given an additive category g, the category eat (gop, Ab) is the category whose objects are all additive functors gop -----* Ab, where Ab is the category of all abelian groups. The morphisms in eat (gop, Ab) are the natural transformations. The category eat(goP,Ab), being a functor category into the abelian category Ab, inherits the abelian structure of Ab. A short exact sequence in eat(goP,Ab) is three additive functors F', F and F" from gop to Ab, and two natural transformations F' ===:::;. F,

F

===?

F"

so that for every object x E g, the sequence

0

F'(x)

~

~

F(x)

~

F"(x)

~

0

is exact in Ab. In this Chapter, we will be assuming that g satisfies Hypothesis 6.1.1, in particular is essentially small. It follows that the category eat (gop, Ab) has small Hom-sets. DEFINITION 6.1.3. Suppose g satisfies Hypothesis 6.1.1. The category £x (gop, Ab) is defined to be the full subcategory of eat (gop, Ab), whose objects are the functors gop -----* Ab which take coproducts of fewer than a objects in g to products in Ab. In other words, let F be an additive functor gop-----* Ab. Then F will lie in the subcategory £x(goP,Ab) c eat(goP,Ab) if, for every family {s.>.,..\ E A} of fewer than a objects of g, the natural map

is an isomorphism. LEMMA 6.1.4. Suppose g satisfies Hypothesis 6.1.1. Then the category £x(goP,Ab) is an abelian subcategory ofeat(goP,Ab). That is, £x(goP,Ab) is an abelian category, and the inclusion

£x(goP,Ab)

c

eat(gop ,Ab)

is an exact functor.

Proof: Suppose F-----* F' is a morphism in £x(goP,Ab). That is, F and F' are functors gop -----* Ab taking coproducts of fewer than a objects to products, and F -----* F' is a natural transformation. We need to show that the kernel and cokernel of the natural transformation, which are clearly objects of the big category eat (gop, Ab), actually lie in the subcategory £x(goP,Ab). Complete the map F-----+ F' to an exact sequence in eat(gop,Ab) 0

~

K ____, F ____, F'

----+

Q

----+

0.

186

6. THE CATEGORY ex(S 0 P,Ab)

Let {sA, A E A} be a set of fewer than a objects in g_ Because F and F' lie in ex(goP,Ab), the natural maps F F'

(II sA) AEA

------+

s )

------+

(II

AEA A

II F(sA)

AEA

II F'(sA)

AEA

are both isomorphisms. We deduce that in the commutative square ------+

F'

(y

sA)

111

------+

AEA

II F'(sA)

AEA

the vertical maps are both isomorphisms. But Ab satisfies [AB4*]. Hence the product of the exact sequences

0

------+

K(sA) ------+ F(sA) ------+ F'(sA) ------+ Q(sA) ------+ 0

over A E A is an exact sequence. In the comparison map K

(y

sA) ------+ F

1

II K(sA)

(y

sA) ------+ F'

1

1

------+

sA) ------+ Q

1

1

II F(sA)

(y 1

------+

AEA

sA)

1

1

II F'(sA)

(y

------+

II Q(sA)

AEA

both the top and bottom rows are exact. It easily follows that the natural maps K Q

are both isomorphisms.

(II (II

AEA

sA) ------+ s )

AEA A

------+

II K(sA)

AEA

II Q(sA)

AEA

D

LEMMA 6.1.5. Suppose g satisfies Hypothesis 6.1.1. Then the category ex(goP,Ab) satisfies [AB3* }; it contains the product of any small set of its objects. Furthermore, the inclusion functor ex (gop, Ab) c eat (gop, Ab) respects products.

6.1. [AB3] AND [AB3*]

187

Proof: Let {F"',J.t EM} be a set of objects in t:x(S 0 P,Ab). We can form the product in eat(S 0 P,Ab) by the usual definition; for any s E S,

{ IJ F"'} (s) ~LEM

What we need to show is that

fl~LEM F~'- is an object of the category £x(S 0 P,Ab). Let therefore {s>., A E A} be a family offewer than a objects of S. For each J.t, the natural map

is an isomorphism, because F"' E t:x(S 0 P,Ab). Taking the product of these isomorphisms over all J.t, we have that

is an isomorphism, proving that gory £x(S 0 P,Ab).

fl~LEM

F"' is indeed an object of thecate0

So far everything was very painless. Next we want to prove that the category C:x (sop, Ab) has coproducts, and want to have a very explicit description of coproducts in C:x (sop, Ab). 6.1.6. The existence of coproducts in C:x(S 0 P,Ab) can be proved purely formally. One notes that the inclusion of C:x (sop, Ab) into eat(S 0 P,Ab) is exact and preserves products, hence preserves all inverse limits. Then one can show it has a left adjoint L; see for example GabrielUlmer [16]. The coproduct of a family of objects in £x(S 0 P,Ab) is the functor L applied to their coproduct in eat(S 0 P,Ab). But as I said, we want an explicit description of the coproduct. REMARK

DEFINITION 6.1.7. Suppose S satisfies Hypothesis 6.1.1. Let s be an object ofS. Let {F"',J.t EM} be a set of objects in £x(S 0 P,Ab). Define the set

isomorphism classes of pairs } { s----+ ILEA s>., f3 E fl>.EA F>.(s>.) with A C M of cardinality< a

188

6. THE CATEGORY £x(S 0 P,Ab)

Note that since Sis essentially small, { VpEM Fp} (s) is indeed a small set. The idea is to construct { llpEM Fl-'} (s) by dividing { VpEM Fl-'} (s) by a suitable equivalence relation. DEFINITION 6.1.8. Suppose S satisfies Hypothesis 6.1.1. Let s be an object of S, and let {FI-', 11- EM} be a set of objects in ex(S 0 P,Ab). Let us

be given two elements of { VpEM Fl-'} (s), that is pairs

f3 E Il.>.EA F.>. (s.>.) (3' E Il.>.EA' F.>, (t.>.)

s----+ ll.>.EA s.>., s ----+ ll.>.EA' t.>.,

If>. ~ A, we adopt the convention that s.>. is defined to be zero, and similarly if>. ~ A' define t.>. = 0. Recall that S is an additive category by Hypothesis 6.1.1.1, hence contains a zero object. The two maps s----+

II

s.>., .>.EA can be combined to a single map

s

II

SA Ef7 t.>. .>.EAUA' which exists because S is assumed an additive category, so the coproduct ----->

is a biproduct, in particular a product of the first and second sum. The two pairs S----+ ll.>.EA s.>., S ----+ ll.>.EA' t .>.' are defined to be equivalent if the map s ----->

II

.>.EAUA'

SA Ef7 t.>.

factors as

s ____.

II

k.>.

.>.EAUA' so that the images of (3 and (3' in

II

.>.EAUA' agree.

F.>.(k.>.)

6.1. [AB3] AND [AB3*]

189

LEMMA 6.1.9. SupposeS satisfies Hypothesis 6.1.1. Lets be an object of S, and let {FJJ.,I-L EM} be a set of objects in £x(S 0 P,Ab). Then the equivalence defined in Definition 6.1. 8 is an equivalence relation.

Proof: The relation is clearly reflexive and symmetric. We need to show it transitive. Suppose therefore that we have pairs of equivalent elements in { VJJ.EM FJJ.} (8) S ----+ ll.>..EA r.>.., 8 ----+ ll.>..EA' 8.>..,

and 8 ----+ ll.>..EA' 8.>..' 8 ----+ ll.>..EA" t .>..'

We need to show the equivalence of 8 ----+ ll.>..EA r.>.., 8 ----+ ll.>..EA" t .>..'

The equivalence of the pairs S----+ ll.>..EA r.>.., ll.>..EA' 8),.'

8 ----+

means that the map

factors as

8

II

_______.

II

k),.

.>..EAUA'

~.>..

.>..EAUA'

so that the images of (3 and (3' in

agree. The equivalence of the pair 8 ----+ ll.>..EA' 8.>.., 8 ----+ ll.>..EA" t .>..'

means that the map 8

_______.

II .>..EA'UA"

8),.

EEl t),.

6. THE CATEGORY ex(S 0 P,Ab)

190

factors as

s

II

II

----+

AEA'UA"

AEA'UA''

so that the images of /3 1 and

/3 11

in

IT

FA(lA)

AEA'UA"

agree. For each

>. E A U A' U A" we have arrows rA lEElg;,

EEJ

lA

1

J;,ffil

kA EEJ tA ----+ rA EEJ sA EEJ t A and by Hypothesis 6.1.1.3 these may be completed to homotopy pullback squares

1

J;,ffil

kA EEJ t A ----+ rA EEJ SA EEJ t A By Hypothesis 6.1.1.4, the coproduct of these is a homotopy pullback square

II

rnA

----+

AEAUA'UA"

II

AEAUA'UA"

1

II

1

AEAUA'UA''

AEAUA'UA"

On the other hand, we have an obvious commutative square

s

II

II

----+

rA

EEJ

lA

AEAUA'UA"

1

k A EEl t A

II

----+

AEAUA'UA''

1 rA

EEJ SA EEJ

tA

AEAUA'UA"

and the defining property of homotopy pullback squares tells us there is a map s

----+

II

AEAUA'UA"

191

6.1. [AB3] AND [AB3*]

giVmg a map of the squares. But f3 and /3' have the same image in IhEAUA' F>.(k>..), and /3' and /3" have the same image in Il>.EA'UA" F>.(l>.). Hence /3, {31 and /3" all have the same image in Il>.EAUA'UA" F>. (m>.)· D DEFINITION 6.1.10. Suppose g satisfies Hypothesis 6.1.1. Lets be an object of g, and let { FJ.L, JL E M} be a set of objects in ex (gop, Ab). The set

is defined to be the quotient of { VJ.LEM FJ.L} (s) by the equivalence relation of Definition 6.1.8. LEMMA 6.1.11. Suppose g satisfies Hypothesis 6.1.1. Let s be an object of g, and let {FJ.L,JL E M} be a set of objects in ex(goP,Ab). The set

{ llJ.LEM FJ.L} (s) has a natural structure of an abelian group. Proof: Given two elements S ----+ S ----+

ll>.EA S >.' ll>.EA' t >.'

we need to define their sum. It is the pair

f3+/3 1 E

IJ

F>.(s>.EIH>.)·

AEAUA'

We leave it to the reader to check that this addition is well-defined; equivalent elements have equivalent sums. We also leave it to the reader to check that the associative and commutative law hold for this addition. The zero element of this group action is a pair

all such pairs are equivalent. We leave it to the reader to check that this is a neutral element for the addition, and that every element of { lJJ.LEM FJ.L} (s) has an additive inverse. D LEMMA 6.1.12. Suppose g satisfies Hypothesis 6.1.1. Let {FJ.L,JL EM} be a set of objects in ex (gop, Ab). The assignment which sends s E g to

the abelian group { llJ.LEM FJ.L} (s) can be naturally extended to an additive contravariant functor.

192

6. THE CATEGORY £x(S 0 P,Ab)

Proof: We have to define this functor on morphisms. Given any morphism f : s - t in S, we need a map

Suppose therefore that we are given an element of { ujtEM a pair

Composition with s -

Fl'} (t), that is

t gives the pair

which we may view as representing an element of { ujtEM our map

Fl'} (s). This is

We leave it to the reader to show that the map

is a group homomorphism with the group structure as in Lemma 6.1.11. The reader will also easily check that the assignment sending f : s - t to

respects composition and identities; it defines a functor. Finally, it needs to be checked that the functor ujtEM Fl' is additive. This is also immediate from the definition; the functor clearly respects finite biproducts, since each Fl' does. In fact, we have more, as the next Lemma shows. D

6.1. [AB3] AND [AB3*]

193

LEMMA 6.1.13. SupposeS satisfies Hypothesis 6.1.1. Let {F11 , IL EM} be a set of objects in ex (sop' Ab). Then the functor UJ1EM FJ1 lies in the category ex (sop, Ab); it sends coproducts of fewer than a objects to products.

Proof: We need to check that U11 EM F 11 is an object in the subcategory ex(SOP' Ab) of the large category eat(S 0 P, Ab). In other words, let { s')', 'Y E r} be a family of fewer than a objects in S. We need to show that

L!P•}[ys,] L!P•} [ys,]--L YL!P+•,);

In any case, there is a natural map

we need to prove that ¢ is injective and surjective. Let us prove surjectivity first. An element of n')'H {llJ1EM FJ1} (s')') is, for every 'Y Era pair s'Y

IJ tl,

---+

IJ

(3'1' E

AEA~

FA (tl)

AEA~

where A'Y C M is a subset of cardinality < a. Let us put

A= UA')',

rt

')'Er

and adopt the notation that if A A'Y we define tl to be 0. Recall that, since each A'Y is of cardinality< a and the index set r is also of cardinality< a, the union has cardinality bounded by the sum of fewer than a cardinals, each smaller than a. Since a is a regular cardinal, the cardinality of A is

.EA

to the pair s ----+

II s>. ~ II s>.,

>.EA >.EA by the definition of llJLEM FJL applied to the morphism s ----+ ll>.EA s>.; see

Lemma 6.1.12. The commutativity of

LY/"} (M·A) 1

G (ys>.)

------+

tells us that in order to compute the image of

G(s)

6. THE CATEGORY ex(S 0 P,Ab)

198

in G(s), it suffices to figure out what the composite

does to

1:

II

S;.. --+

J\EA

II

S;..,

J\EA

Any element of the product

with /3;.. E F;..(s;..)- Write composite

TI>.EA

F;.. (s;..) can be written as a product

f3 this way. Since we are assuming that the

FI-L ------+

II FI-L

~

G

11-EM

is the given map

FJJ---+

G, the pair

must map to the image of /3;.. E F;..(s;..) by¢;.. : F;..(s;..) product of these elements, that is the pair

1:

II

S;.. --+

J\EA

II

S;..,

J\EA

--+

II (3;.. E II F;..(s;..) J\EA

J\EA

must map to

II

¢>. ((3;..)

E

J\EA

This computes the image of

1:

II J\EA

under

S;..--+

II J\EA

S;..,

II fJ;.. E II F;..(s;..) J\EA

J\EA

G(s;..)· The

6.1. [AB3] AND [AB3*]

199

But then under the longer composite

the element 1:

II

AEA

sA ~

II

II {JA E II FA(sA)

sA,

AEA

AEA

AEA

must map to the image of E

under G

(II

AEA

sA) -------+ G(s).

The commutativity of

btF•} (ys,) 1

G (ysA)

-------+

allows us to deduce that the image of s ~II sA, AEA

via ¢ must be the image of E

via a

(II

AEA

sA) -------+ G(s).

G(s)

6. THE CATEGORY ex(S 0 P,Ab)

200

Thus the map ¢> is uniquely determined by the above. The reader is left to verify that this map is well-defined (takes equivalent pairs to the same element of G(s)), is a group homomorphism, is natural ins and that the composites

are the given maps Fl-' ----. G.

D

REMARK 6.1.16. We have proved in this Section that the category £x(S 0 P,Ab) is an abelian category satisfying [AB3] and [AB3*]; it is closed with respect to products and coproducts of its objects. The inclusion functor £x(S 0 P,Ab) C eat(S 0 P,Ab) is exact and respects products, but decidedly does not respect coproducts. LEMMA 6.1.17. Suppose S satisfies Hypothesis 6.1.1. The inclusion functorS----. £x(S 0 P,Ab) respects coproducts of< a objects

Proof: There is an obvious, inclusion functorS ----. £x(S 0 P,Ab), which takes an object s E S to the representable functor S(-, s). We wish to show that this functor preserves coproducts of fewer than a objects. Therefore let {t_x, .A E A} be a set of objects in S, with A of cardinality < a. By 6.1.1.2, the coproduct of these objects exists in GX (sop' Ab). We have a natural map

IJ S(-, t_x) ~ S (-, IJ t_x) -\EA

-\EA

and we wish to show the map an isomorphism. But this is an essentially immediate consequence of Definition 6.1.10. The point is that the map has an obvious inverse. The inverse

takes an element of S

(s, IJ

t_x),

-\EA

that is a map s ----.

lhEA

t_x, to the pair

6.2. THE CASE OF S = T"

201

which is an element of UAEA S( -, tA). It is obvious that ¢>'1/J is the identity. But 'ljJ¢> takes an element of

II S(s, tA),

AEA

that is an equivaelnce class of pairs

to s -----+

II sA -----+ II t AEA

A,

AEA

which is equivalent to it. 6.2. The case of S =

D

:ra

Let 'J' be a triangulated category satisfying [TR5]. Let o: be the same regular cardinal that we have fixed throughout this Chapter. LEMMA 6.2.1. Suppose that S is an essentially small category, and is an o:-localising subcategory of'J'. Then S satisfies Hyposthesis 6.1.1. Proof: We need check that S satisfies all four parts of Hypothesis 6.1.1. We are supposing Sis essentially small, and hence satisfies Hypothesis 6.1.1.1. We assume also that S is o:-localising. This means S is triangulated and closed under coproducts of fewer than o: of its objects. The closure under coproducts is Hypothesis 6.1.1.2. As in Example 6.1.2, Hypotheses 6.1.1.3 and 6.1.1.4 are automatic for a triangulated category; therefore :ra satisfies all of Hypothesis 6.1.1. D REMARK 6.2.2. In particular, the entire discussion of Section 6.1 applies, and we understand coproducts in ex (sop, Ab) quite explicitly. If we do not insist that S be essentially small, it is not in general true that eat(S 0 P,Ab) will have small Hom-sets. Here we are primarily interested in categories with small Hom-sets. In the situation S c 'J' as above, we have LEMMA 6.2.3. Let 'J' be a triangulated category satisfying [TR5], and assume S C 'J' is an essentially small o:-localising subcategory. There is a natural functor 'J' -----+ ex(S 0 P,Ab) sending t E 'J' to the representable functor 'J'(-, t), or more precisely to its restriction to S. We will denote this restriction 'J' (-, t) I8 . Proof: Clearly the restriction gives a functor to eat(S 0 P,Ab). We need to show that for t E 'J', the functor 'J' (-' t) Is lies in ex (sop, Ab), that is the

202

6. THE CATEGORY ex(S 0 P,Ab)

functor 'T ( -, t)Js sends coproducts inS offewer than a objects to products in Ab. This is true simply because the coproducts in S agree with those in 'T. The subcategory S c 'T is a-localising. D LEMMA 6.2.4. Let 'T be a triangulated category satisfying [TR5], and assume S C 'T is an essentially small a-localising subcategory. Then the functor 'T--+ t:x(S 0 P,Ab) respects products.

Proof: Let {t .x, A E A} be a set of objects in 'T whose product exists in 'J. Then, for each s E S C 'T,

and the right is just the product of 'J(-,t.x)ls applied to s, as defined in Lemma 6.1.5. D LEMMA 6.2.5. Let 'T be a triangulated category satisfying [TR5], and assume S C 'T is an essentially small a-localising subcategory. If S is aperfect, then the functor 'T --+ t:x(S 0 P,Ab) respects coproducts of fewer than a objects. If S is not only a-perfect, but every object of S is also a-small, then 'T--+ t:x(S 0 P,Ab) respects all coproducts.

Proof: Let {t .x, A E A} be a set of objects in 'J. Since 'T satisfies [TR5], the coproduct exists in 'J. We need to show that if S is a-perfect and the cardinality of A is < a, or if S is not only a-perfect but also S C 'J(a) and the cardinality of A is unrestricted, then

'T (-.

n t;..) s 1

>.EA

is the coproduct in t:x(S 0 P, Ab) of the functors 'T (-, t;..)) 3 . By the universal property of the coproduct in t:x(S 0 P,Ab), there is a natural map

We need to check that ¢ is an isomorphism. We begin by showing it surjective. Let us therefore begin with an element of

'T

(s, ll t;..) I , >.EA

S

6.2. THE CASE OF S

= 'J"'

203

that is a map in 'J from an object s of S to the coproduct. If S C 'J(o:) then s is a-small, and any map s

---t

II t;.

-\EA

factors as a map s

II t;.

---t

IIt,\

c

-\EA'

-\EA

where A' C A is a subset of cardinality < a. If we are not assuming S C 'J(o:), then we assume anyway that the cardinality of A is .EA'

already vanishes. Complete each s>. --? q>. to a triangle k>. ----+ s>. ----+ q>. - - + ~k>.

with k>. E S. By Proposition 1.2.1 the direct sum is a triangle IIk>.----+ I I s > . - - + >.EA'

>.EA'

II q>.

--+

~

>.EA'

II k>.

>.EA'

and the vanishing of the composite s ----+

II s>.

--+

>.EA'

II Q>.

>.EA'

means the map s ----+

II s>.

>.EA'

factors as s ----+

II k>.

--+

II s>..

>.EA'

>.EA'

On the other hand, the composites k>. - - - - + s>. - - + q>. - - + t>.

all vanish, since the first couple of maps are two maps of a triangle. We deduce that the pair s--?

II s>., AEA'

II f>.

E

II 'J(s>.,

t>.)ls

>.EA'

AEA'

is equivalent to the pair s--?

II s>., >.EA'

IIo

>.EA'

E

II 'J(s>.,

t>.)ls,

>.EA'

in other words to the zero map. The kernel of ¢ is trivial, and¢ is an 0 isomorphism. PROPOSITION 6.2.6. Let 'J be a triangulated category satisfying {TRS], and assume S C 'J is an essentially small a-localising subcategory. The natuml functor 'J --? ex (gop, Ab) is homological and respects products. If S is, a-perfect, then the functor 'J--? ex(S 0 P,Ab) respects coproducts of fewer than a objects. If S is not only a-perfect, but every object of S is also a-small, then 'J - - t ex(S 0 P,Ab) respects all coproducts.

6.2.

THE CASE

OF S = 'I"'

205

Proof: The fact that the functor respects products is Lemma 6.2.4, and the statements about coproducts are Lemma 6.2.5. The fact that 'J' ----+ ex (sop, Ab) is homological is easy. Given a triangle in 'J' of the form

r

--'------+

s

t

------T

'J'(-,s)is

------T

------T

:Er

we need to check that

'J'(-,r)ls

------T

'J'(-,t)is

is exact in ex(S 0 P,Ab), in other words gives an exact sequence when we evaluate it on any k E S. But this is the exactness of 'J'(k, r)

------T

'J'(k, s)

------T

'J'(k, t)

which we know from Lemma 1.1.10.

0

REMARK 6.2.7. The most interesting case of the above isS= 'J'"', as in Definition 4.2.2. Suppose 'J'"' is essentially small. Lemma 4.2.5 asserts that 'J'"' is o:-localising. By its definition, 'J'"' is contained in 'J'(a), that is consists only of o:-small objects. Furthermore, 'J'"' is o:-perfect; it is in fact the largest o:-perfect class in 'J'(a). It follows that S = 'J'"' C 'J' satisfies all the hypotheses of Proposition 6.2.6. We deduce that the natural functor

'J'

-----+

ex ( { 'J'"' YP, Ab)

is a homological functor respecting all products and coproducts. There is one more useful fact about the functors 'J' ----+ To state it, we need to introduce one important definition.

ex (sop, Ab).

DEFINITION 6.2.8. Let 'J' be a triangulated category satisfying [TR5}, and assume S C 'J' is a triangulated subcategory. We say that S generates 'J' if Hom(S,x) = 0

X=

That is, if x is an object of 'J' and for all s x is isomorphic to zero in 'J.

E S

0.

we have 'J'(s, x)

= 0, then

LEMMA 6.2.9. Let 'J' be a triangulated category satisfying [TR5], and assume S C 'J' is an essentially small o:-localising subcategory. Suppose that S generates 'J', as in Definition 6.2.8. A morphism x ----+ y in 'J' is an isomorphism if and only if its image by the functor

'J' is an isomorphism in

------T

ex (sop, Ab)

ex (sop, Ab).

Proof: One direction is obvious. If x so is its image by

----+

y is an isomorphism in 'J', then

6. THE CATEGORY ex(S0 P,Ab)

206

just because functors take isomorphisms to isomorphisms. We need to prove the converse. Suppose therefore that x ---+ y is a morphism in 'J', and that its image in ex(goP,Ab) is an isomorphism. We need to show that X ---+ y is an isomorphism in 'J'. In any case, we may complete to a triangle in 'J' X

- - + y - - + Z - - + ~X - - + ~y.

Since the functor 'J'---+ ex(goP,Ab) is homological (see Proposition 6.2.6), this maps to an exact sequence in ex(goP,Ab). We are assuming that the functor 'J' ---+ ex (gop' Ab) takes X ---+ y to an isomorphism. But then it also takes ~x ---+ ~y to an isomorphism. After all,

'J' (-, ~x)ls

--+

'J'( -, ~Y)Is

can also be written as 'J'(~- 1 -, x)ls - - + 'J'(~- 1 -, Y)ls

and ~- 1 : g ---+ g is an equivalence. From the exact sequence it follows that 'J'(-,z)ls must vanish. That is, for all s E g, 'J'(s,z) = 0. But g generates; this means that z is isomorphic to zero, and by Corollary 1.2.6, x ---+ y is an isomorphism in 'J'. 0

6.3. ex(goP,Ab) satisfies [AB4] and [AB4*], but not [AB5] or [AB5*] We return now to considering the general case of

c that is for this Section, g is an arbitrary. category satisfying Hypothesis 6.1.1. The case g = 'J'" which we considered in Section 6.2 is a spe.

-----+

Ff

-----+

0.

We need to show that the coproduct of these sequences is exact. Since right exactness is clear, we need to show that 0

II F{

-----7

~

>.EM

II F>.

>.EM

is exact. For s an object inS, pick an element in the kernel of

{II F{}

(s)

>.EM

~

{II F>.}

(s).

>.EM

It is given by a subset A C M, where the cardinality of A is < o:, and a pair (3 E

II F{ (s>.)

>.EA

and the fact that the pair lies in the kernel means that under the map

II F{(s)

~

>.EA

II F>.(s)

>.EA

(3 E TI>.EA F{ (s) goes to an element ¢((3), so that the pair

is equivalent to zero. But by the definition of the equivalence relation (see Definition 6.1.8) this means that s

-----7

II s >.

>.EA

must factor as

s

-----7

II k>.

>.EA

so that under the map

II F>.(s>.)

----+

II F>.(k>.)

>.EA

208

6. THE CATEGORY

cx(S P,Ab) 0

the element ¢({3) maps to zero. In other words, we have a commutative square of abelian groups

II Fl{8_x)

.XEA

II

------+

II F~(k.x)

.XEA

1

II

1

F_x(8_x) ------+ F.x(k.x) .XEA .XEA and we have figured out that under the composite

II Fl{8_x)

.XEA

1

II F_x(8_x)

------+

.XEA

II F.x(k.x)

the element {3 E TI.xEA Fl{8_x) maps to zero. But by the commutativity it also maps to zero under the composite

II F~(8_x)

------+

.XEA

II Fl(k.x)

.XEA

p1

II F.x(k.x)·

.XEA

Since the map pis injective (each map Fl(k_x) --+ F.x(k.x) is assumed injective), we deduce that the image of {3 via the map

II F~(8_x)

------+

.XEA

II F~(k.x)

.XEA

already vanishes. But then our pair

is such that 8 ------+

II

.XEA factors as

8 ------+

II k .x

.XEA

8 .x

6.3. [AB4] AND [AB4*]

and f3 vanishes under the map

II FHs>.)

-------+

>.EA

This means the pair s

II

----7

209

II F~(k>.)·

>.EA

S>,,

>.EA

is equivalent to zero. The kernel of

II F{} (s) ~ {II F>.} (s) {>.EM >.EM vanishes, and we have proved the left exactness.

D

REMARK 6.3.3. Before we end this Section, we should warn the reader that in general, the category C:x(S 0 P, Ab) satisfies neither [AB5] nor [AB5*]. We remind the reader that an abelian category is said to satisfy [AB5] if filtered direct limits of exact sequences are exact. It is said to satisfy [AB5*] if the dual category satisfies [AB5]. For [AB5*], this is clear; even the category Ab does not satisfy [AB5*], and we can hardly expect a category of functors into Ab to satisfy the condition. More surprising is the fact that, in general, the category C:x(S 0 P,Ab) need not satisfy [AB5]. In fact, let 'J be a triangulated category satisfying [TR5]. Let S C 'J be an ~ 1 -localising subcategory. That is, S is closed under the formation of countable coproducts (in 'J) of its objects. Suppose furthermore that Sis N1-perfect. Let a be the cardinal N1. By Proposition 6.2.6, the Yoneda map

'J -------+ C:x (gop, Ab),

that is the map sending an object t E 'J to the functor 'J(-,t)j 8 , is a homological functor respecting coproducts of fewer than a = N1 objects. Let

Xo-xl-x2 -··· be a sequence of objects and morphisms in S. As in Definition 1.6.4, we can form the homotopy colimit, which is given by the triangle

gxi 00

1 - shift

gxi 00

-------+

• ~xi

-------+

E

{

gxi 00

}

.

This sequence only involves countable coproducts, hence lies in S. The functor

6. THE CATEGORY ex(S 0 P,Ab)

210

which respects countable coproducts, takes the map

i=O

i=O

to the map

DO 'J(-,Xi)ls li

1- shift

i=O

liDO 'J(-,Xi)ls· i=O

Since the abelian category ex (gop, Ab) satisfies {AB4], the kernel and cokernel compute colim 1 and colim terms, respectively. For a discussion of derived functors of limits see Section A.3, more particularly Remark A.3.6. In our case here, for the sequence

in the abelian category kernel of

ex (gop, Ab), we have that colim1

i=O

is precisely the

i=O

But we have a vanishing composite :E- 1 Hocolim Xi

------7

DO li Xi

1- shift

i=O

00 liX•. i=O

with :E- 1 ~Xi E g_ In other words, the map :E- 1

~xi ~

00 li xi

i=O

is an element of 'J ( :E- 1 ~xi,

u:o xi) 1s, and lies in the kernel of

Since it is easy to find examples where this map fails to vanish, we see that in general, colim 1 can fail to vanish.

6.4. PROJECTIVES AND INJECTIVES

211

6.4. Projectives and injectives in the category ex(S 0 P,Ab) Lemma 5.1.2 taught us that in the category eat (sop, Ab), the representable objects S(-, s) are projective. For this we needed nothing; S was only an additive category, not necessarily essentially small. Now we leave the realm of the very general, and return to essentially smallS's. The regular cardinal a is the one fixed throughout the Chapter, and the category S satisfies Hypothesis 6.1.1. It is an immediate consequence of Lemma 5.1.2, that the category ex(S 0 P,Ab) has enough projectives. The category turns out in general not to have enough injectives. In this section, we will give a general discussion of the consequences of the existence of enough injectives; after all, some categories ex (sop, Ab) do have them. We refer the reader to an Apendix (see Section C.4) for a counterexample, showing that ex(S 0 P,Ab) can fail to have enough injectives. LEMMA 6.4.1. Let S be a category satisfying Hypothesis 6.1.1. Let s be an object of the category S. Then the representable functorS(-, s) is a projective object in the category ex(S 0 P,Ab).

Proof: Observe that the functor Y,(-) = S(-,s) is an object of the category ex (sop, Ab). This is true because it clearly carries any coproduct of objects of S to a product of abelian groups. But by Lemma 5.1.2, the functor S(-,s) is projective as an object of eat(S 0 P,Ab), hence also as an object of the exact subcategory ex(S 0 P,Ab). D LEMMA 6.4.2. The projectives { S(-, s), s E S} give a generating set of projectives.

Proof: We need to show that every non-zero object F E ex(S 0 P,Ab) admits a non-zero map S(-,s)

----+

F(-)

for some s E S. But this is clear; if F in non-zero, then for some s E S, F(s)-:/= 0. Yoneda's lemma says that F(s) is in one-to-one correspondence with maps S(-,s)

----+

F(-).

Hence there is a non-zero map S(-,s)---+ F(-).

D

REMARK 6.4.3. Let P = IlsES S( -, s). Then P is a projective generator in the category ex (sop, Ab). From standard arguments, it follows formally that the category ex(S 0 P,Ab) has enough projectives, and that any projective object is a direct summand of a coproduct of P's. We remind the reader how this goes.

212

6.

THE CATEGORY ex(S P,Ab) 0

LEMMA 6.4.4. The category ex(S 0 P,Ab) has enough projectives, and any projective object is a direct summand of a coproduct of P 's.

Proof: Let F be an object of ex(S 0 P,Ab). Consider the set of all maps P----+ F, and take the coproduct. Let Q be the cokernel; that is we have an exact sequence

lJ P

------+

F

------+

Q

-----+

0.

Take any map P----+ Q. Since Pis projective, the map factors through the epimorphism F----+ Q; it may be written as a composite P----+ F----+ Q. But any map P ----+ F factors thorough IJ P ----+ F, the coproduct of them all. Hence P ----+ F ----+ Q composes to zero. This being true for all P ----+ Q, it follows that Q vanishes. After all, P is a generator; any non-zero Q admits a non-zero map P ----+ Q. Thus Q = 0 and IJ P ----+ F is surjective. This proves that ex(S 0 P,Ab) has enough projectives. Now suppose F is projective. By the above, there is a surjective map IJP-----+ F.

But F is projective. Hence the identity 1 : F ----+ F must factor through the surjective map lJ P ----+ F. We conclude that F is a direct summand cllJ~

0

REMARK 6.4.5. The dual statements are far more subtle. It turns out that the category ex(S 0 P,Ab) does not, in general, have enough injectives. For now, let us observe the trivial case. If Q = No, then ex(S 0 P,Ab) = eat(S 0 P,Ab) is a Grothendieck abelian category, and thus the existence of enough injectives is classical. But if a > No, we remind the reader that the category ex(S 0 P,Ab) does not in general satisfy [AB5]. See Remark 6.3.3. Even though enough injectives need not always exist, let us remind ourselves, briefly, what happens when they do. LEMMA 6.4.6. Suppose the category ex(S 0 P,Ab) has enough injectives. Then it has an injective cogenerator. If I is an injective cogenerator, then any object Fin ex(S 0 P,Ab) admits an injection F - - t ITI into a product of I 's. Any injective object is a direct summand of a product of I 's.

Proof: By Lemma 6.4.4, the category ex(S 0 P,Ab) has a projective generator, which we will call P. Then every non-zero object k of ex(S 0 P,Ab) admits a non-zero map P ----+ k. That is, k contains the image of some non-zero map P ----+ k, in other words, k contains a non-zero subobject, isomorphic to a quotient object of P. Suppose that ex(S 0 P,Ab) has enough injectives. For every quotient object of P, that is for every isomorphism class of exact sequences p

------+

q

---t

0,

6.4. PROJECTIVES AND INJECTIVES

213

choose an embedding q------+ Iq, with Iq an injective object in ex(S 0 P,Ab). Let

=

I

I assert that I is an injective cogenerator of ex (gap, Ab). For let F be any object in ex(S 0 P,Ab). Consider the set of all maps F------+ I, and take the product map, F----+

II

I.

Let k be the kernel of the map. I assert k = 0. For otherwise, k would have a non-zero subobject q, which is a quotient of P. We have a short exact 0 ----+ q ----+ k

and hence the embedding q------+ Iq must factor as a map k ------+ Iq. But k is a subobject ofF, hence the map factors further as F------+ Iq, and finally

c

II

p___,q___,O

Iq

I

and we have a map F------+ I. By construction, this map fails to vanish on q c k c F. We have a map F------+ I, which fails to vanish on k c F, and k was defined as the kernel of F----+

II

I.

This is a contradiction, proving k = 0. For every F, we have shown there is a monomorphism F----+

II

I.

F___,J

IfF is injective, this monomorphism must split, and F is a direct summand of a product of I's. D REMARK 6.4.7. The most interesting counterexample we have shows that, even when S = 'J' for 'J' a triangulated category satisfying [TR5], it may happen that ex(S 0 P,Ab) does not have a cogenerator. If there is no cogenerator, then by Lemma 6.4.6, there cannot possibly be enough injectives. In the counterexample, which may be found in Section C.4, 'J' = D(R), where R is any discrete valuation ring, and a is any regular cardinal~ ~1- But a slight modification ofthe argument allows us to find the same counterexample in 'J' = D(Z), the derived category of Z, or in 'J' the homotopy category of spectra. There is one more well-known fact about projective generators and injective cogenerators of which we want to remind the reader.

6. THE CATEGORY ex(S 0 P,Ab)

214

LEMMA 6.4.8. Suppose I is an injective cogenerator for the category ex (gop, Ab). Then the sequence

F'

-----+

F

-----+

F"

is exact in ex (gop, Ab) is and only if the sequence of abelian groups ex(goP,Ab)

{F'', I} -----+ ex(goP,Ab) { F,I} -----+ ex(goP,Ab){ F',I}

is exact.

Proof: Let the homology to the sequence

F' ~ F __!!_____. F'' be H; that is,

H =

Ker(,B) Im(o:) ·

Because I is injective, we easily show that the homology of ex(goP,Ab) { F",

I} -----+ ex(goP,Ab) { F, I}

-----+

ex(goP,Ab) { F',

I}

is precisely ex (gop, Ab) { H,I}. This will vanish precisely when H does. D

6.5. The relation between A('T) and ex ( {'T"'Yp, Ab) As in the rest of this Chapter, let a be a fixed, regular cardinal. In Chapter 5, more precisely Theorem 5.1.18, we learned about the universal homological functor. We remind the reader: given a triangulated category 'J', there is a universal homological functor 'T

-----+

A('T).

If 'j has small Hom-sets, and furthermore satisfies [TR5], there is a natural homological functor

'T

-----+

ex( {'T"'} 0 P,Ab).

This homological functor must factor through the universal homological functor. We have 'T-----+ A('T)

~ ex({'T"'YP,Ab).

We now propose to begin studying the functor A('T) ----+ ex ( {'T"'YP, Ab). To simplify the notation, we will once again put g = 'J'"', so that

6.5.

THE RELATION BETWEEN

A('J)

AND ex ( {'J"'}"P, Ab)

215

In this Section, we will prove that ex (gop, Ab) is a quotient of the category A('J), in the sense of Gabriel. The reader is assumed to have some familiarity with Gabriel's [15). For the reader's convenience, there is a condensed summary of the results we need in Appendix A, more specifically in Section A.2. In fact, this might be a good time for the reader to skim through Appendix A. In this Section, we use the results of Section A.2. In Section 7.1 we appeal to the work of Sections A.l. In the Sections 7.3 and 7.4, we depend mostly on the theory of Sections A.3 and A.4, although Sections A.l also plays a role. Thus, for the next few Sections, we will be making heavy use of the theory of abelian categories, summarised for the reader's convenience in Appendix A. LEMMA 6.5.1. Let 'Y be a triangulated category with small Hom-sets, satisfying [TRS}. The natural functor n : A('J) ---+ ex (gop, Ab) above is exact and respects coproducts.

Proof: By Proposition 6.2.6 the functor 'j---+ ex(goP,Ab) is homological and respects coproducts. From the fact that it is homological and from Theorem 5.1.18, we have that it factors 'j

---'>

A('J) ~ ex(S 0 P,Ab)

with 1r an exact functor of abelian categories. Lemma 6.2.5 asserts that the functor 'j---+ ex(S 0 P,Ab) preserves coproducts. Lemma 5.1.24 tells us that, since 'Y satisfies [TR5) and 'Y ---+ ex (gop, Ab) preserves coproducts, it follows that preserves coproducts.

0

LEMMA 6.5.2. The functor 1r: A('Y)---+ ex(goP,Ab) above is just the functor taking F : 'J0 P ---+ Ab to its restriction to S C 'j.

Proof: The homological functor H : 'j ---+ ex(goP,Ab) is the functor taking an object t E 'j to 'Y( -, t)ls· The functor A('Y) ---+ ex(goP,Ab) is obtained as the universal factorisation 'j - - - . A('Y) ~ ex(goP,Ab) of Theorem 5.1.18. By the proof of Theorem 5.1.18, the functor 1r is given as follows. An object F E A('J) is a functor F : 'J0 P ---+ Ab, admitting a presentation 'Y(-,s) - - - . 'J(-,t) ---. F(-) ---. 0 with s, t E 'J. The object n(F) was obtained as the third term in the exact sequence

H(s) ---. H(t) ---. n(F) ---. 0.

6. THE CATEGORY ex(S 0 P,Ab)

216

In our case, we have an exact sequence of functors on 'J

'J( -, s)

-----t

'J( -, t)

-----t

'Y(-,t)ls

-----t

---->

0.

c 'J, we get an exact sequence

If we restrict this sequence to S = 'J"' functors on S

'J(-,s)ls

F(-)

---->

F(-)ls

---->

0.

Since we know that 'J(-,s)ls = H(s) and 'J(-,t)ls = H(t), we deduce

7r(F)=F(-)Is·

o

PROPOSITION 6.5.3. Let 'J be a triangulated category with small Homsets, satisfying [TR5j. PutS= 'J"'. The category t:x(S 0 P,Ab) is the quotient, in the sense of Gabriel, of the category A('J) by a colocalizant subcategory we will denote 13, or 13(a) when we wish to remind ourselves of the dependence on the regular cardinal a. Precisely, 1r: A('J) ~ t:x(S 0 P,Ab) is the quotient map, and it has a left adjoint L.

Proof: We want to prove that the functor 1r : A('J) ~ ex (gop, Ab) identifies t:x(S 0 P,Ab) as the Gabriel quotient A('J)/13, with 13 a colocalizant subcategory. It suffices, by the dual of Proposition A.2.12, to produce a left adjoint L: t:x(S 0 P,Ab) ~ A('J) to the functor 1r, so that the unit of adjunction (dual to counit) is an isomorphism rJ: 1 ~ 1rL. By Lemma 6.4.2, the representable functors on S, that is the functors S( -, s), form a set of projective generators for C:x(S 0 P,Ab). Every object FE t:x(S 0 P,Ab) admits a presentation

II S(-, s;.)

.\EA

-----t

II S( -, t J 1

---->

F(-)

---->

0.

f.' EM

Put Ys>. for the representable functor; that is

Ys>.

S( -, s;.).

Lemmas A.2.13 and A.2.15 give us that we need only prove the existence of the adjoints and the fact that rJ is an isomorphism on objects Il>.EA Ys>of t:x(S 0 P,Ab). More precisely, to prove the existence of the left adjoint L it suffices to show that the functor

C:x(S 0 P,Ab)

{II

Y 8 >.,1r(-)}

.\EA

is representable in A('J). To show that the unit of adjunction is an isomorphism, observe that the functor 1r is exact by Lemma 6.5.1. Therefore Lemma A.2.15 applies, and we need only check that the natural transformation rJ: 1 ~ 1rL is an isomorphism on objects Il>.EA Ys>.. Let G be an arbitrary object in A('J); that is, G is a functor G: 'J0 P ~ Ab, and by Lemma 5.1.5, G takes coproducts in 'J to products of abelian

6.5. THE RELATION BETWEEN A('J) AND

ex ({'J"} op, Ah)

groups. By Lemma 6.5.2, 7r( G) is just the restriction of G to S give a natural transformation

II S(-,s>.)

---t

217

c 'J'. To

G(-)ls

>.EA

is to give, for each .X E A, a natural transformation

S(-,s,x)

---t

G(-)ls·

By Yoneda, this is the same as giving, for each .X E A, an element of G(s>.)· There is a 1-to-1 correspondence between natural transformations

II S(-,s,x)

---t

G(-)ls

>.EA

and elements of

IJ G(s,x).

>.EA

Now by Lemma 5.1.5, G takes coproducts in 'J' to products of abelian groups. That is, G

(II

AEA

s,x)

By Yoneda, elements of G (ll>.EA s,x) correspond 1-to-1 with natural tranformations

'J' (-,

II s,x)

-->

G(- ).

AEA

Summarising, there is a 1-to-1 natural correspondence between natural transformations .\EA

This exactly asserts that the functor

ex(S

0

P,Ab)

{II

YsA, 7r(- )}

>-EA

is representable in A('J'); in fact, it is represented by the object

'J' (-,

II s>.) ·

.\EA

218

6.

THE CATEGORY t:x(S P,Ab) 0

Thus the left adjoint L of 1r exists, and more concretely L takes to

'J (-,

II s>..) ·

>..EA

But then 1r takes this to its restriction to S c 'J, again by Lemma 6.5.2. But by Lemma 6.2.5 the restriction functor 'J ---+ ex(S 0 P,Ab) respects coproducts. That is,

'J (-,

II s>..) I

>..EA

Thus 1r L takes

S

il>..EA Y 8 >-

to itself, proving that TJ is an isomorphism.

D

CoROLLARY 6.5.4. Let 'J be a triangulated category with small Homsets, satisfying [TR5j. The natural functor 1r : A('J) ---+ ex (gop, Ab) respects products. Proof: The functor

1r

has a left adjoint, and therefore must be left exact. D

REMARK 6.5.5. As a formal consequence of Proposition 6.5.3, (see also Proposition A.2.12) we have that the category ex(S 0 P,Ab) is the quotient of A('J) by a full subcategory 13 c A('J). The objects of 13 c A('J) are the objects F so that 1r(F) = 0. By Lemma 6.5.2, 1r(F) is the restriction ofF to S C 'J. Then 13 is the full subcategory of A('J) consisting of the functors vanishing on S. The functor

respects products and coproducts. It respects coproducts by Lemma 6.5.1, products by Corollary 6.5.4. It follows that products and coproducts of objects in 13 lie in 13. The subcategory 13 is closed under subquotients, extensions, limits and colimits. LEMMA 6.5.6. We have an equivalence of categories D('J) ~ A('J); see Definition 5.2.1 and Lemma 5.2.2. An object of D('J) is a map {x-+ y} E 'J. The object {x -+ y} E D('J) lies in the kernel of

D('J) = A('J) if and only if the image of {x 'J is zero.

-+

ex(S P,Ab)

----+

0

y} via the map

----+

ex(S P,Ab) 0

6.5. THE RELATION BETWEEN A('J) AND

Proof: By Remark 5.2.3, the object { x as a pair of maps in A('J)

--->

ex ({'J"'} op, Ab)

219

y} E D('J) may be thought of

'J(-,x) ~ F(-) ____L_, 'J(-,y), with a epi and f3 mono. The equivalence D('J) --+ A('J) is simply the map that takes { x ---> y} E D('J) to F(-), that is to the image of the induced map in A('J). By Lemma 6.5.2, the map A('J)--+ ex(S 0 P,Ab) is just the restriction to S C 'J. It takes F(-) to F(-)ls· That is, it takes {x---> y} E D('J) to the image of 'J(-,x)\s

-----+

'J(-,y)\s·

This image vanishes precisely when the image of { x

'J

-----+

--->

y} under the functor

ex (gop, Ab)

vanishes.

D

This makes the next definition very natural. DEFINITION 6.5.7. A morphism if, under the natural map

'J

------+

f :x

--->

y in 'J is called a-phantom

ex( {'JayP ,Ab),

it maps to zero. REMARK 6.5.8. With Definition 6.5.7, we can restate Lemma 6.5.6, to say that the kernel of the functor

D('J) = A('J)

-----+

ex ( {'Ja YP, Ab)

consists precisely of the a-phantom maps in 'J, viewed as objects of D('J). As above, we denote this kernel by 13 C D('J). When we want to emphasize the dependence on a, we will denote the kernel by 13 (a). We know, from Proposition 6.5.3, that ex( {'JayP,Ab) is the Gabriel quotient of D('J) by 13(a) (see Remark 6.5.5). For many 'J's, we know that 'Ja is essentially small for all a; for example, this is true when 'J = D(7!..), the derived category of 7!... When this is the case, the categories

ex( {'JayP,Ab) are well-powered for every a. This means every object of ex ( {'Ja} op, Ab) has only a small set of quotients. We see in Corollary C.3.3 and Remark C.3.4 that the category D('J) is not in general well-powered. Specifically, it is not well-powered for 'J = D(7!..). It follows that, when 'J = D(7!..), the category D('J) can never agree with D('J) B(a)

220

6. THE CATEGORY C:x(S0 P,Ab)

We conclude that, for every regular cardinal a, B(a) zero a-phantom maps for every a.

"1- 0. There are non-

6.6. History of the results of Chapter 6 The results of this Chapter are essentially all new. Only the trivial case, that is a = No, has been studied at all.

CHAPTER 7

Homological properties of

ex (sop, Ab)

We have learned, in the previous Chapter, some of the basic properties of the categories ex (gop, Ab). In Appendix C, more specifically in Section C.4, we can see that in general the categories ex(goP,Ab) need not have enough injectives; in fact, they can fail to have cogenerators. See also Lemma 6.4.6 for the fact that, if ex (gop, Ab) fails to have a cogenerator, it certainly cannot have enough injectives. But nevertheless, something positive is true. This Chapter will be devoted to proving the positive results we have. These positive results are fragmented and inconclusive. They are included for the benefit of future workers on the subject, who will hopefully be able to push them further. The casual reader is advised to skip this Chapter; the results do not affect the development later in the book. One of the first questions to ask, is whether the categories ex (gop, Ab) satisfy [AB4.5]. In Proposition A.5.12, we saw that this need not be the case, for a general g_ In order to have a hope, we must at the very least assume g to be triangulated. Even if g is triangulated, I have not been able to prove that ex(goP,Ab) satisfies [AB4.5]. In this Chapter, I include what I could prove in that direction. Also included are some other amusing homological facts about the categories ex (gop, Ab) . Although for these it is not essential to assume that g is triangulated, we will make our life simpler by treating the restricted case. In other words, as was the case in the previous Chapter, a is a fixed regular cardinal. But from now on, g will be an essentially small triangulated category satisfying [TR5 (a)]. 7.1. ex(goP,Ab) as a locally presentable category

In this Section, we will be appealing to the material of Section A.l. Recall that Section A.l deals with locally presentable categories. Lemma A.l.3 tells us that a-filtered colimits agree in ex(goP,Ab) and eat(goP,Ab), and as an immediate consequence a-filtered colimits in ex(goP,Ab) are exact. Proposition A.l.9 tells us that the category ex(goP,Ab) is locally presentable. That is, for every object a E ex (gop, Ab), there exists a cardinal

222

7.

HOMOLOGICAL PROPERTIES OF

ex(S0 P,Ab)

/3, in general depending on a, so that Hom( a,-) commutes with /3-filtered colimits. As a very special case, we have LEMMA 7.1.1. Lets____. ex(S 0 P,Ab) be the Yoneda inclusion; it is the functor sending s E s to S( -, s). For every s E s, its image in ex(S 0 P,Ab) is an a-presentable object. That is, Hom(s,-) commutes with a-filtered colimits in ex (sov, Ab).

Proof: By Yoneda's Lemma, for s E Sand FE ex(S 0 P,Ab), Hom(s,F)

=

ex(S 0 P,Ab){S(-,s),F(-)}

=

F(s).

By Lemma A.l.3, a-filtered colimits agree in ex(S 0 P,Ab) and eat(S 0 P,Ab). For an a-filtered colimit,

{c~mF} (s)

colim{F(s)}. ____.

But Yoneda's Lemma identifies this as Hom

c~{ Hom(s, F)}·

(s,c~mF)

0

CoROLLARY 7.1.2. In particular, any map S(-,s)

----+

II F.>.(-)

>-EA

factors as

II

F.>.(-) c >-EM where M c A, and the cardinality of M is< a. S(-,s)

----+

Proof: The object ll.>.EA F.>. (-) is the direct limit of all the coproducts over subsets of cardinality < a. This is an a-filtered colimit. Hence by Lemma 7.1.1, a map from S( -, s) into this a-filtered colimit must factor through one of the terms. 0 So far, we have not used the fact that the category S is triangulated. Now we will start. LEMMA 7.1.3. LetS be an essentially small triangulated category, satisfying [TR5(a)j. Then the abelian category A(S) satisfies [AB4.5(a)J, and there is a natural full embedding A(S)

----+

ex(S 0 P,Ab)

which is exact, and respects coproducts of fewer than a objects.

7.1.

ex(S 0 P,Ab) AS A LOCALLY PRESENTABLE CATEGORY

223

Proof: By Lemma 5.1.24, if Sis a triangulated category satisfying [TR5(a)], then A(S) is an abelian category which satisfies [AB3(a)]; coproducts of fewer than a objects exist in S. From Corollary 5.1.23, we know that the category A(S) has enough injectives. It therefore follows that A(S) satisfies [AB4.5(a)]; see Lemma A.3.15, Definition A.3.16 and Remark A.3.17. The other statements of the Lemma, which still remain to be proved, are that the natural functor A(S)

-----+

ex(S 0 P,Ab)

is a full embedding, is exact, and preserves coproducts of fewer than a objects. First we should remind the reader what the natural functor is. Recall that we always have a map

S

-----+

ex(S 0 P,Ab)

takings E S to the representable functorS(-, s). This functor is homological, and therefore factors uniquely (up to canonical equivalence) through the universal homological functor of Theorem 5.1.18. It factors as S

-----+

A(S) ~ ex(S 0 P,Ab)

where the canonical functor A(S) ----+ ex(S 0 P,Ab) is exact. We need to show that this unique functor is a full embedding(=fully faithful), and respects coproducts of fewer than a objects. Observe that by Lemma 6.1.17, the composite functor

s

-----+

A(S)

-----+

ex(S 0 P,Ab)

respects coproducts of fewer than a objects. From the fact that S satisfies [TR5(a)], coupled with Lemma 5.1.24, we deduce that the functor A(S) ----+ ex(S 0 P,Ab) also respects coproducts of fewer than a objects. It remains only to show the functor fully faithful. Consider now the longer composite

c It is clear that the canonical functor A(S)----+ eat(S 0 P,Ab) is a full embedding. The category A(S) is, by definition, afullsubcategoryofeat(SaP,Ab); see Definition 5.1.3. But then the inclusion of A(S) in ex(S 0 P,Ab) must also be fully faithful. D REMARK 7.1.4. In Lemma 7.1.1, we proved that every object in S

c

ex(S 0 P,Ab) is a-presentable. In Lemma 7.1.3, we factored the inclusion asS c A(S) c ex(S 0 P,Ab). It is natural to ask if the objects in the larger A(S) ::::> S are also a-presentable. The answer is yes, and now we wish to prove it. LEMMA 7.1.5. Every object FE A(S) C ex(S 0 P,Ab) is a-presentable.

7. HOMOLOGICAL PROPERTIES OF ex(S 0 P,Ab)

224

Proof: Let F be an object in A(S). By Definition 5.1.3, it admits a presentation S(-,s)

~

S(-,t) ----+ F(-) ----+ 0,

and S(-, s), S(-, t) are object inS c £x(S 0 P,Ab), hence a-presentable by Lemma 7.1.1. Taking maps in £x(S 0 P,Ab) into an a-filtered colimit, we deduce an exact sequence 0 ----+ Hom(F,colim¢) ----+ Hom(t,colim¢) ----+

----+

1

Hom( s, colim¢) ----+

Since s and t are a-presentable, this identifies as 0

~ Hom(F,c~¢)----+ c~{Hom(t,¢)}

1

c~{ Hom(s, ¢) }· But the category Ab of abelian groups satisfies [AB5]; filtered colimits are exact. Hence this identifies as 0----+

c~{Hom(F,¢)}----+ c~{Hom(t,¢)}

1

c~{ Hom(s, ¢) }, which allows us to deduce that

c~{ Hom(F, ¢)}

Hom(F, colim¢). ----+

0

7.2. Homological objects in £x(S 0 P,Ab) In keeping with our conventions for this Chapter, the regular cardinal a is fixed, and S is an essentially small triangulated category satisfying [TR5(a)], that is closed under the formation of coproducts of < a of its objects. REMARK 7.2.1. The next few lemmas concern homological functors. Recall: an object FE £x(S 0 P,Ab), that is a functor F:

gop

----+ Ab

is called homological if it takes triangles to long exact sequences.

7.2.

HOMOLOGICAL OBJECTS

IN ex(S 0 P,Ab)

225

In the remainder of the Chapter, we will be considering many functors into the abelian category cx(goP,Ab), and some of these will be homological. There exist interesting homological functors into the abelian category ex (gop, Ab). But the abelian category is itself a category of functors, whose objects may be homological. Because this could lead to nightmarish confusions, an object of ex(goP,Ab) which is homological, as a functor F : gop

Ab,

------+

will be called a homological object of ex(goP,Ab). LEMMA 7.2.2. Let g be an essentially small triangulated category, satisfying [TR5(a)J. An a-filtered colimit of homological objects in ex(gop, Ab) is a homological object in ex (gop, Ab).

Proof: By Lemma A.l.3, a-filtered colimits agree in ex(goP,Ab) and eat(goP,Ab). Suppose {Fi,i E :J} is an a-directed family of homological objects. Let x

y

------+

------+

z

I;x

------+

be a triangle in g_ For each i E :J, the sequence Fi ( z)

------+

Fi (y)

------+

Fi (X)

is an exact sequence of abelian groups. The category Ab of abelian groups satisfies [AB5], and hence the sequence of abelian groups colimFi(z) iE~

colimFi(Y) iE3

------+

colimFi(x)

------+

iE~

is exact. But since these colimits agree in cx(goP,Ab) and eat(goP,Ab), this is precisely the colimit of the functors Fi, taken in ex(gop' Ab), applied to x

------+

y

------+

z

------+

Since the sequence is exact, it follows that

I;x.

colimiE~

Fi is homological.

D

EXAMPLE 7.2.3. Any coproduct in ex(goP,Ab) of representable functors is homological. For let {s>., A E A} be a set of objects in g_ The object of ex(goP,Ab)

II g(-,s>.)

>.EA

is, as in Corollary 7.1.2, the colimit of all coproducts over subsets of A of cardinality .)

>.EA

226

7. HOMOLOGICAL PROPERTIES OF cx(sov,Ab)

is homological when the cardinality of A is< a. But from Lemma 6.1.17, we learn that

Il S( -, s>.)

S (-,

Il s>.) ·

AEA

AEA

Being representable, it is homological. It turns out that there is a converse to Lemma 7.2.2. Lemma 7.2.2 asserts that every a-filtered colimit of homological objects is homological. But it turns out that every homological object is an a-filtered colimit of representables. LEMMA 7.2.4. Let S be an essentially small triangulated category, satisfying {TR5(a)j. Any homological object FE cx(S 0 P,Ab) is an a-filtered colimit of representable objects.

Proof: Since a-filtered colimits are the same in the categories cx(S 0 P,Ab) and eat(S 0 P,Ab), it suffices to prove that F is an a-filtered colimit, in eat (sop, Ab), of representables. Clearly, F is the colimit of all the representables mapping to it. It needs to be shown that this colimit is a-filtered. Suppose we are given any collection of < a representables mapping to F. I assert that, in the category of representables mapping to F, there is a coproduct. We have

S( -, s>.)

---+

F(-)

for A E A, with the cardinality of A being < a. By Yoneda, this gives, for every A E A, an element off>. E F(s>.)· That is, we.have an element ILEA f>. E ILEA F(s>.)· But Flies in cx(S 0 P,Ab), and therefore

We have constructed an element corresponds to a morphism

S (-,

Il>.EA

Il s>.)

f>. E F

---+

(ILEA

F(-).

AEA

What we have proved is that the morphisms

all factor through

S (-,

Il s>.)

AEA

---+

F(-).

s>.)· By Yoneda, it

7.2. HOMOLOGICAL OBJECTS IN

ex(S P,Ab)

227

0

We next want to show that

ll sA)

S (-,

F(-)

-----t

AEA

is the coproduct, in the category of representables mapping to F, of the objects S(-,sA)

-----t

F(-).

Suppose therefore that we are given aS(-, t) A E A a commutative diagram

S( -,sA)

1

F(-)

-----t

F(- ), and for each

~

S( -, t)

1

__!_____. F(- ).

To prove that

S (-,

ll sA)

F(-)

-----t

AEA

is the coproduct of

we must produce a unique factorisation through S( -,sA)

1

F(-)

-----t

S ( -,

---+

S(-,t)

1

F(-).

1

1

F(-)

-----t

But our morphism S( -, t) commutative diagrams

ll SA)

AEA

-+

F(-) corresponds to

S( -,sA)

1

F(-)

-----t

1

---+

f

E F(t). The given

S( -, t)

1

__!_____. F(-)

tells us that, for every A E A, the the natural map F(t) f E F(t) to fA E F(sA). But then the natural map

ll sA

AEA

-----t

t

~

F(sA) takes

7. HOMOLOGICAL PROPERTIES OF ex(S0 P,Ab)

228

induces a map F(t) ----+ F(lJ.>.EA s.x), which takes f E F(t) to fhEA f.x E F(lJ.>.EA s.x)· This establishes the factorisation S( -, s.x)

~

1

F(-)

s

(-,II

1

~

.AEA

S.x)

--->

1

F(-)

1

--->

S( -, t)

1

F(-).

The uniqueness is obvious. To prove that the category of natural transformations

S(-,s)

--->

F(-)

is a-filtered, it remains to show that any two morphisms are coequalised. Given two objects

S(-,s)

~

F(-)

and

S(-,t) ___. F(-)

and two morphisms between them, we need to show they can be coequalised. Two objects as above, and two morphisms between them, amount to a diagram

S(-,s)

f

F(-)

(;}

S( -, t)

g

with () f = Og. By Yoneda's Lemma, the natural transformation

S(-,t) ~ F(-) corresponds to an element x E F(t). The fact that Of= Bg corresponds, via Yoneda, to the statement that the two maps F(t)

F(f) F(g)

F(s)

take x E F(t) to the same element of F(s). In other words, f- g: s----+ t induces a map F(t) ----+ F(s), taking x E F(t) to zero. Now complete f - g : s ----+ t to a triangle f-g r~s~t~Er.

Because F is homological, the sequence F(Er) ~ F(t)

F(f-g)

F(s)

is exact. Now x E F(t) lies in the kernel of the map F(f- g), and hence there is a y E F(Er) mapping to x E F(t). By Yoneda, this means the natural transformation

S(-,t)

(;}

~

F(-)

7.2. HOMOLOGICAL OBJECTS IN C:x(S 0 P,Ab)

229

factors as S(-,t)

--+

S(-,1ir)

---t

F(-)

and the two composites f

S(-,s)

S( -, t)

g

S( -, 1ir)

are equal. The given two maps f

S(-,s)

F(-)

()

S( -, t)

g

are coequalised by a map into S(-,1ir)

--+

F(-). D

LEMMA 7.2.5. Let S be an essentially small triangulated category, satisfying [TR5(a)j. Suppose

0

---t

F --+ G

---t

H

---t

0

is a short exact sequence in £x(S 0 P,Ab). If any two ofF, G or H are homological objects, then so is the third. Proof: Let x

---t

y

---t

z

---t

1ix

be a triangle inS. Consider the short exact sequence of chain complexes 0

0

1

F {1ix}

1

---t

Fz

---t

Gz

1

G {1ix}

1

Fy

--+

Gy

Hz

1

Fx

---t

Gx

Hy

---t

Hx

1

1

---t

F {1i- 1 z}

---t

G{1i- 1 z}

---t

H {1i- 1 z}

1

1

--+

0

1

---t

1

1

---t

0

1

--+

1

1

H {1ix}

0

1

1 1

1

1

0 0 0 0 0 From the long exact sequence of the homology of these chain complexes we D learn that, if any two of the rows are exact, then so is the third.

7. HOMOLOGICAL PROPERTIES OF £x(S 0 P,Ab)

230

7.3. A technical lemma and some consequences

The regular cardinal a is fixed, as was assumed throughout the Chapter. The category S is an essentially small triangulated category satisfying [TR5(a)], that is closed under the formation of coproducts of < a of its objects. We warn the reader that this section and the next rely heavily on the material developed in Appendix A. We will make frequent use of the notions of Sections A.3 and A.4. The reader should be familiar with the definition of a sequence of length 'Y (Definition A.3.5), a Mittag-Leffler sequence (Definition A.3.10), and the lemma about the vanishing of limn of Mittag-

-

Leffler sequences in the presence of enough projectives (Lemma A.3.15). Definition A.3.16 encapsulates it concisely, if somewhat mysteriously: an abelian category satisfies [AB4.5*] if it satisfies the vanishing of limn for Mittag-Leffler sequences. This summarises the main facts we appeal to in Section A.3. From Section A.4, we need the fact that the derived functors of the limit of a sequence agree with those of a cofinal subsequence (Proposition A.4.8), and the fact that we can compute the derived functor of the limit of a Mittag-Leffler sequence very concretely by injective resolutions, as in Construction A.4.10.

-

REMARK 7.3.1. Actually, we will be applying mostly the duals of the results in Appendix A. We will be studying sequences of length 'Y which are covariant; that is F: :J('Y) ~ £x(S 0 P,Ab) (see Definition A.3.5). We consider, in this Section, sequences :J ~ A, not sequences :rop ~ A. We will be considering their colimits rather than limits, and the interesting sequences will be co-Mittag-Leffler, which is the dual of Mittag-Leffler. We will freely apply the duals of facts proved in Appendix A. Suppose 'Y is an ordinal, and 'Y < a. Suppose we are given a sequence F: :J('Y) ~ £x(S 0 P,Ab). If it so happens that F is co-Mittag-Leffler and factors through

:J('Y) ~ A(S) ~ £x(S 0 P,Ab), then the functor G : :J('Y) ~ A(S) must also be co-Mittag-Leffler, since ¢: A(S) ~ £x(S 0 P,Ab) is an exact embedding respecting coproducts of

- 1. ~

Now A(S) and £x(S 0 P,Ab) are abelian categories satisfying [AB4(a)]; for A(S) we see this because, by Lemma 7.1.3, A(S) even satisfies the stronger [AB4.5(a)]. For £x(S0 P,Ab) we have that, by Lemma 6.3.2, it satisfies [AB4], and hence the weaker [AB4(a)]. The inclusion A(S) ~ £x(S 0 P,Ab) is exact and respects coproducts of< a objects. Lemma A.3.2

7.3.

A TECHNICAL LEMMA AND SOME CONSEQUENCES

231

now tells us that, for colimits over partially ordered sets of cardinality < a,

c~n{¢G}

¢{c~nc}.

Our index set :J('y) has cardinality colimn{¢G} _____.

if>{c~nc}

¢{0}.

That is, for n;:::: 1, F = ¢G has vanishing colimn. _____. The essence of this Section is to reduce the case of a general co-MittagLefRer sequence in ex(S 0 P,Ab) to the above. The unfortunate technical detail is that this does not quite work. We end up proving the following. Suppose for all co-Mittag-LefRer sequence F of length < 'Y and all n ;:::: 1, colimnF = 0. Then any co-Mittag-LefRer sequence F of length 'Y satisfies _____. colimn _____. F = 0 for n

> - 2. If we wish to do a transfinite induction, this

is not quite enough. In Section 7.4, we will see how the author tried (unsuccessfully so far) to worm his way around this difficulty. Because we will need some variants of the statement given above, the main lemma of this Section is quite awkward to state. REMARK 7.3.2. From now until the end of Section 7.4, we will assume that our fixed regular cardinal a is >No. When a= No, the inclusion

ex(S 0 P,Ab)

C

eat(S 0 P,Ab)

is an equality. The functors in eat(S 0 P,Ab) are the additive functors gop -----> Ab, and hence they must respect finite biproducts. Therefore they send coproducts of fewer than a = N0 objects in S to products of abelian groups. In the case a = No, the category ex(S 0 P,Ab) = eat(S 0 P,Ab) is a Grothendieck abelian category. It satisfies [AB5], hence [AB4.5]. In this Section and the next, we are out to prove that [AB4.5] is true even for a > N0 . The case a = N0 being trivial, in the remainder of the Section we assume a > No. One consequence is that the category S is closed under countable coproducts. After all, S satisfies [TR5(a)] and a> N0 . By Proposition 1.6.8, the category S is closed under splitting idempotents. By Corollary 5.1.23, the projective-injective objects of A(S) are precisely the representable functors. Recall that for a general S, the projective-injectives are direct summands of representables. The assumption that a > N0 guarantees that a direct summand of a representable functor is representable. REMARK 7.3.3. We wish to remind ourselves about projective objects. The objects S( -, s) are projective in ex(S 0 P,Ab) by Lemma 6.4.1. They are projective in A(S) by Lemma 5.1.11. Any projective object in ex(S 0 P,Ab) which happens to lie in A(S) c ex (gop, Ab) is clearly projective in the

232

7. HOMOLOGICAL PROPERTIES OF ex(S 0 P,Ab)

exact subcategory A(S). By the last paragraph of Remark 7.3.2, it must be a representable S (-, s). The representable functors S (-, s) are precisely the projective objects of A(S), and they can also be characterised as the projective objects in cx(S 0 P,Ab) which happen to lie in A(S). Recall that the category cx(S 0 P,Ab) has enough projectives. The objects S( -, s) are projective, as are all their coproducts. Lemma 6.4.2 shows that the projectives S( -, s) generate. More precisely, in Lemma 6.4.4 we see that any object of cx(S 0 P, Ab) is a quotient of a coproduct of S( -, s)'s, and any projective object is a direct summand of such a coproduct. Let P be the class of all coproducts of representable functors S(-, s) in cx(S 0 P,Ab). By the above, every object of £x(S 0 P,Ab) is a quotient of an object in P. By the dual of Lemma A.4.3, every object in the category of sequences

:J('Y) ~ ex(S 0 P,Ab) is a quotient of a projective object. The projective objects may be chosen to be coproducts of the duals of F~i 's. That is, define G~:

:J('Y) ~ cx(&0 P,Ab)

by the formula G~(j)

=

{ 0 ifi>j a ifi$:j

Then for any (covariant) sequence F oflength "(,there is a surjection

II

G~i ~ F,

iE:J.(1')

where the ai's are projective in cx(S 0 P,Ab). By Remark A.4.4, we may even choose the a/s to be objects in the class P. That is the a/s are coproducts of representables. In other words, given any functor

we may find a surjection G ~ F, where G is a particularly nice projective object as above. In Construction A.4.10, we saw that concretely, such G's satisfy

Gi colimGj j.EA;'

This defines a summand hf C gf, containing uf, so that ... ----+

II

h2

II

97 ----+

iE:J(I')

•

----+

II

hl

II

g}

•

iE:J(I')

II

ho

II

9i .

----+

iE:J(I')

•

is a subcomplex of ... ----+

----+

iE:J(I')

iE:J(I')

0

iE:J(I')

0

7.3.5.3. Suppose we are in the situation of the proof of Lemma 7.3.5. To remind the reader: suppose the complex · · · ----+

II

97 ----+

iE:J(I')

is obtained from a slightly special projective resolution of a co-MittagLeffier sequence F in ex (gop, Ab). And suppose further that F admits a short exact sequence of co-Mittag-Leffiers

0

----+

F

----+

G

----+

H

----+

0

and that G is of the special form, so that

Gi colimGj j.)

>.EA?+ 1

factors through a coproduct of fewer than a terms, which we call uf+ 1 C gf+l. We deduce the commutative diagram S(-,p)

------7

1

ur:+1

•

------7

hi

t1

8

------7

h'?-1

•

1

gf ---------+ gin-1

and this completes the proof of 7.3.5.3.1; the kernel of hi ----+ hf- 1 is the image of S( -,p), and it maps under f : hi ---+ gf into the image of Un+1 ---t gn .

.

.

242

7. HOMOLOGICAL PROPERTIES OF ex(sov,Ah)

Now for the proof of 7.3.5.3.2. Consider the diagram

h? g[

1

gp

~

-----+

9i·

Recall that we are assuming about gi that it is a coproduct; more precisely

II Gi,

=

>.EA

with Gi E A(S). In any case, since h? is projective in A(S), it is isomorphic to S( -, t) for some t E S. The map

II Gi

S( -, t)

AEA

must factor, by Corollary 7.1.2, through a coproduct of fewer than a objects. This coproduct lies in A(S). Call it h-; 1 . We deduce a commutative diagram

g[

~

1

1

gp

-----+

9i

where the bottom row is exact, and the top row lies in A(S). As in the proof of 7.3.5.3.1, we may find a direct summand u} C g}, u} E A(S), and a commutative diagram

S(-,p) ~

1

u[

h?-----+

1

1

~ 9? -----+ 9i

where the top row is exact. This says precisely that the image of h? hi 1 C gi agrees with the image of h? in the cokernel of u[ - gp. It remains to prove 7.3.5.3.3. We assume therefore that we are in the situation of 7.3.5.1. We have exact sequences ... ~ 9? ~

g[

-----+

gp

-----+

fi ----+ 0

and 0 ~ fi ~ 9i

-----+

hi -----+ 0,

and Ji, 9i and hi are homological objects in ex(S 0 P,Ab). We are given a direct summand h? c gp. The kernel of the composite h? ~

gp

-----+

fi

----+

9i

7.3. A TECHNICAL LEMMA AND SOME CONSEQUENCES

243

is the same as the kernel of

h?

~9? ~ fi,

since the map fi ---+ 9i is mono. On the other hand, in the proof of 7.3.5.3.2 above we saw that the image of the natural map h? ---+ 9i is the cokernel of S(-,p)---+ h?. We have a vanishing composite S(-,p) ~

h?

~

k

Now recall that h? is projective; it must be S( -, q) for some q E S. Complete to a triangle p~q~r~~p.

Our vanishing composite becomes S(-,p)

~

S(-,q)

~

k

By Yoneda, the map S(-, q) ---+ fi corresponds to an element () E fi (q). The vanishing of the composite S(-,p)

~

S(-,q)

~

fi

means that the image of () under

fi(q) ~ fi(P) is zero. But fi is a homological object of

cx(S P,Ab). 0

The sequence

fi(r) ~ fi(q) ~ fi(P) is exact. This means that ()lies in the image of Again, by Yoneda's lemma, this means the map

S( -, q)

----+

fi

----+

S( -, r)

----+

must factor as

S( -, q)

Recalling that

h? = S( -, q), we have a factoring h? ----+ S(-,r) ~

fi·

fi·

On the other hand, S( -, r) is projective, and the mapS(-, r) ---+ fi must factor through the epimorphism 9? ---+ fi. It factors as S(-,r)

----+

9?

~ fi·

But 9? is a coproduct ofrepresentables, and by Corollary 7.1.2, a map from S( -, r) into a coproduct factors through a coproduct offewer than a terms. There exists u? C 9?, u? E A(S), and a factorisation

h?

----+

S(-, r)

----+

u? C 9?

----+

fi.

244

ex(S P,Ab)

7. HOMOLOGICAL PROPERTIES OF

0

Put another way, we have a commutative diagram where the top row is exact S( -,p)

------+

h9

•

a

-----+

1

u9

•

Yi

11

a

-----+

Yi·

The inclusion of the cokernel of S( -,p) -----+ h? into gi factors through h? -----+ S( -, r) -----+ u? as above. In other words, the assertion of 7.3.5.3.3 is true. There exists a commutative square

1

h?E9u? ~ Yi out choice of u?, the h?

as asserted. In fact, for of this square is superfluous.

in the bottom left corner D

Next we use the above two steps, to conclude the proof of Lemma 7.3.5. We want to show the exactness of the sequence ... ------+

We showed that any homology class is really contained in lliE~ ui, with uf c gf a direct summand, and ui E A(S). In 7.3.5.2 we showed that it is possible to find a subcomplex . . . ------+

II

h~

------+

II

ht

II

------+

h?

iE~(r)

contained in A(S), so that ui c hi. Call this complex proceed to construct a sequence of subcomplexes of .. . ------+

II

g~

------+

iE~(r)

Our first subcomplex is · · · ------+

II

II

g}

------+

iE~(r)

oh, which we denote also

oh~

------+

iE~(r)

Suppose we have constructed mh, that is a complex .. · ------+

II iE~(r)

mh~

oh.

We will now

II Y?. iE~(r)

7.3. A TECHNICAL LEMMA AND SOME CONSEQUENCES

245

Then, as in 7.3.5.3, we can construct objects ui. Recall: these objects come with commutative diagrams

8(-,p)

---+

{)

mhi

---+

m

!1

1

un+l

•

Yi

---+

---+

hn-1 i

1 n-1 9i

where the top row is exact. Here, if n = 0, we interpret g; 1 to be 9i· These diagrams define objects u~+ 1 for n 2:: 0. In the situation where the hypothesis of 7.3.5.3.3 holds, that is 9i and hi are homological, then the conclusion of 7.3.5.3.3 allows us to also choose u? C gp. By 7.3.5.2, there is a subcomplex · · · ----+

Il

m+lh~

Il

---+

iE:I(I')

m+lh}

Il m+1 h?

----+

iE:I(I')

iE:I("Y)

of the complex

iE:I(I')

iE:I('y)

iE:I(7)

m+1h·

so that m+lhi C gi contains mhi and ui. This defines the complex What we have done is constructed a sequence of complexes mh, and the inclusion mh ~ 1 h is, for each i E :J, a map of complexes

m+

... ----+

mhr

mh}

----+

1 ... ----+

m+lhr

mh?

----+

1 ----+

m+lh}

1 ----+

m+lh?.

The fact that ui C m+lhi means that we have commutative diagrams

8( -,p)

---+

mhi

---+

m+lhi

1

u~+1

•

1 m+1 hn+1 i

{)

---+

1 ---+

11 ---+

m+lhi

---+

hn-1

ffii

1 m+1 hn'--1 11 m+1 hn-1 i

i

The exactness of the top row means that the kernel of[) : mhi ~ mh~- 1 maps to zero in the homology of the bottom row. In other words, the map

246

7. HOMOLOGICAL PROPERTIES OF ex(S 0 P,Ab)

of chain complexes ' · · ----+

m h i2

---t

m hoi

1

1

1 · · · ----+

-~

m hli

-----+

1 m+10Q 0 -----+ m+lhi -----+ m+l hi

m+lhT

induces the zero map in homology, except when n = 0. The statement for n = 0 is that the image of the cokernel of mOo in the cokernel of m+ 1 8 0 injects into 9i· If we further assume the hypothesis of 7.3.5.3.3 holds, that is gi and hi are homological, then the construction of mh? E9 u? C m+lh?

guarantees that the map of images

Im {mh? ----+ 9i}

Im {m+lh?----+ 9i}

---t

factors through some representable S( -, r m)· Now let the subcomplex h be the colimit of the mh. That is, we have a complex

ll

h}

---t

iE:J(r)

It clearly lies in A(S), since A(S) is closed under colimits of fewer than a objects, and N0 < a. I assert that this complex has the property that for each i E 1, the sequence . . . ----+

hr

---t

hi

---t

h?

is exact. The sequence is the colimit of sequences and there is a spectral sequence for computing its homology. The easiest way to see the existence of this spectral sequence is to consider the double complex given by the map of complexes

oo ll

m

h. t

1- shift

m=O

lloo

m

h.

2·

m=O

There are two spectral sequences that compute the homology of this double complex. One quickly degenerates to the colimit sequence . . . -----+

hr

---t

hf

---t

h?.

The other can be used to compute its homology. The terms in the spectral sequence involve colimH-n(mhi) ----+

and

colim 1 H-n(mh,·). ----+

7.3. A TECHNICAL LEMMA AND SOME CONSEQUENCES

247

If n 2: 1, the map

H-n(mhi)

-----t

H-n(m+lhi)

vanishes. In other words, the sequence of maps

H-n(ohi)

-----t

H-n(1hi)

-----t

H-n(2hi)

H-n(1hi)

~

~

···

is a cofinal subsequence of

H-n(ohi)

~

0

-----t

0

~

···

as is the sequence 0

-----t

0

-----t

0

~

...

By Proposition A.4.8, all three sequences have the same colim and colim 1 ; ~

~

but for the zero sequence this clearly vanishes. If n = 0, we have that the image of

H 0 (mhi)

-----t

H 0 (m+1hi)

injects into gi. In other words, it stabilises. The sequence

H 0 (ohi) ~ H 0 (1hi) ~ H 0 (2hi) is a cofinal subsequence of

H 0 (ohi)

-----t

Im(¢0 )

-----t

H 0 (1hi) ~ Im(¢ 1 ) ~

Im(¢0 )

-----t

Im(¢ 1 )

-----t

···

as is Im(¢ 2 ) ~

· · ·

By Proposition A.4.8, all three sequences have the same colim and colim 1 . ~

~

On the other hand, the sequence Im(¢ 0 )

-----t

Im(¢ 1 )

-----t

Im(¢ 2 ) ~

·· ·

is co-Mittag-Leffier. Being a co-Mittag-Leffier sequence in A(S), it must have vanishing colim 1 . ~

If we furthermore assume the hypothesis of 7.3.5.1, that is that gi and hi are homological, then we know a little more. We know that in the sequence Im(¢ 0 )

------+

Im(¢ 1 )

-----t

Im(¢ 2 )

------+ · · ·

above, the maps Im( 4>n) ~ Im( 4>n+l) factor through representable objects S(-, r n)· The sequence above is cofinal in Im(¢ 0 ) as is

-----t

S(-,r0 )

-----t

Im(¢ 1 )

------+

S(-,r1 )

------+ ·· ·

7. HOMOLOGICAL PROPERTIES OF ex(S 0 P,Ab)

248

Once again, Proposition A.4.8 allows us to conclude that all three sequences have the same colim and colim 1 . We already know that colim 1 vanishes. ~

~

~

It follows that we have an exact sequence in A(S) 1-shijt

00

0----+

us(-,ro)

c ex(S P,Ab)

00

0

•

us(-,ro)----+ c~----+

n=O

0.

n=O

Representables are projective-injective in the Frobenius abelian category A(S). Their coproduct is projective, hence injective. And since the first two terms in this exact sequence are injective, the sequence splits and the third term is also projective-injective, hence representable. Under the hypothesis of 7.3.5.1, the only non-vanishing term in the spectral sequence is a representable object in A(S). In the spectral sequence, there is only one non-zero term. And that term is the homology of the complex . . . ----+

hf ----+ h} ----+ h?.

The homology is all concentrated in degree 0. If we furthermore assume the hypothesis of 7.3.5.1, that is that 9i and hi are homological, then we know further that this unique non-zero term is a representable S(-, s). But now it follows that the complex . . . ----+

Il

Il

h: ----+

iE~('y)

h} ----+

iE~('y)

Il

h?

iE~("Y)

is induced by a projective resolution of a co-Mittag-Leffi.er sequence in A(S). Define F'(j) to be the cokernel of

Il h}

----+

Il h?. i-:!:_j

i-:!,_j

This gives a co-Mittag-Leffier sequence

F': :7(1')

----+

A(S).

This sequence comes with a projective resolution, and h is a chain complex computing its colimn. Since co-Mittag-Leffler sequences in A(S) have ~

vanishing colimn, the chain complex ~

is acyclic except at n = 0. Since any homology class in the complex . . . ----+

Il

iE~("Y)

gf

----+

Il

iE~('y)

gf

----+

Il

gp

iE~('y)

is supported on some subcomplex h as above, it follows that the complex g is acyclic.

7.3. A TECHNICAL LEMMA AND SOME CONSEQUENCES

Let the co-Mittag-Leffier sequence F' : 1('y) We have an exact sequence ···~h?~h~~h9~ '

'

'

--+

249

A(S) be as above.

{'i

CO lmF'j

~o

j

colimG. --+

But colimF = Gi, and hence the map --+

Gi

---->

colim Gj j

£x(S 0 P,Ab).

Suppose it satisfies 7.4.3.1. For j 2: 0, the map

Gj

---->

G(j + 1)

7.4. THE DERIVED FUNCTORS OF COLIMITS IN

ex(S P,Ab)

257

0

is mono, and the quotient G(j + 1)/G(j) is a coproduct of objects in A(8) c ex(8°P,Ab). 7.4.3.2. For j a limit ordinal, Gj

=

colimGi. i. E A, we denote the object in A(S) that corresponds to it by t>.. For each>. E A, choose an exact sequence 0

r>. -----+ S( -, s>.)

-----+

---t

t>.

---t

0

where S( -, s>.) is a projective object. Now we define a map which sends an object x of ex(S 0 P,Ab) to a morphism x---+ x. The construction is as follows. Let M be the set of all maps r>. ---+ x; that is M We have maps

=

II

r>.

j >.

{r>. -+ x

A}.

ll

---t

{r>.-+x}EM

E

S(-,s>.)

{r>.-+x}EM

1 X

and we define x

x to be given by the pushout

---+

ll

r>.

II

---t

{r>. -+x}EM

S(-,s>.)

{r>. -+x}EM

1

1

X

Now recall that the sequences

0

-----+

are exact, and that quence 0

-----+

r>. -----+

S(-,s>.)

---t

t>.

---t

0

ex(S P,Ab) satisfies [AB4] by Lemma 6.3.2. These-

ll r>.

0

-----+

l1 S( -, s>.)

---t

l1 t>.

---t

0

is therefore also exact, and the pushout gives a commutative diagram with exact rows 0

0

-----+

llr>. -----+

1

-----+

X

-----+

l1 S( -, s>.)

---t

1 x

---t

ll t)..

11

ll t)..

---t

0

---t

0.

7.4. THE DERIVED FUNCTORS OF COLIMITS IN C:x(S 0 P,Ab)

261

We deduce that x ~ xis mono, and xjx is a coproduct of objects in A(S). Now recall; we are given an object F E ex (gop, Ab), and want to enibed it in a homological object G. We propose to define a sequence of length a+ 1 in ex(S 0 P,Ab). We define this sequence, 8: :J(a + 1) ~

ex(S P,Ab) 0

inductively, as follows.

ex(S P,Ab)

7.4.6.1. 8(0) E

0

7.4.6.2. The map S(i) S(i)

is defined to be

F.

S(i + 1) is defined to be

~

S(i)

~

as above. 7.4.6.3. If i is a limit ordinal, then S(i)

col. ~ G

J. E A(S) is a-presentable, and therefore its map into the a-filtered colimit must factor as r>. ~

S(j)

By the construction of S(j + 1) r>.

~

J.)

S(j) ~ S(j

This says that the map r >.

climS(j).

--+ G

1

+ 1).

must factor as

r>. ~ S(-,s>.) ~G.

262

7. HOMOLOGICAL PROPERTIES OF

cx(S P,Ab) 0

But now we have the exact sequence

0

----+

r >.

----+

S(-, s >.)

-----+

t >.

-----+

0.

Applying Hom(-, G) to it, we deduce an exact sequence Hom{ S( -, s;.), G}

----+

Hom(r>., G) -----+

Ext 1 (t;., G)

1 Ext 1 { S( -, s;.), G }· We have just proved that any map r>. r;. ----+

---*

S( -, s>.)

G factors as ----+

G,

that is the map Hom{ S( -, s;.), G}

----+

Hom(r>., G)

is surjective. Since S(-, s ;.) is projective in the category must have Ext 1 {S(-,s;.),G}

ex (sov, Ab), we

0.

The exact sequence now permits us to conclude that

O· '

this is true for all >. E A, meaning for all isomorphism classes oft;. E A(S). By Lemma 7.4.5, G must be homological. D LEMMA 7.4.7. LetS be an essentially small triangulated category, satisfying {TR5(a)J. Let"( be an ordinal, "(

F ----+ G ----+ H ----+ 0

Suppose fori E :J(j), Gi E cx(S 0 P,Ab) is homological. We need to define the extension to J(j + 1). In any case, by the hypothesis of the Lemma, the co-Mittag-LefH.er sequence H on :J(j) has vanishing colim 1 . The exact sequence ----+

colim 1 H ----+ colim F ----+ colim G i.

-----+

o.

AEA 0

By Proposition 7.5.5, for n 2: 1 we have Lnj{iF} = 0. For n = 0, observe L 0 j{iF} = jiF, and since i is fully faithful with a left adjoint j, Lemma A.2.9 tells us that the counit of adjunction jiF --+ F is an isomorphism. We deduce an exact sequence in ex (gop, Ab) -----+

II ~>.

AEA1

--

II ~>.

AEA 0

-----?

F

-----?

0.

7.6. HISTORY OF THE RESULTS IN CHAPTER 7

271

This is, of course, nothing more nor less than a projective resolution, in ex(S 0 P,Ab), for the functor F. Mapping the complex . . . ---+

II Y >. 8

---+

AEA2

II Y,>.

II Y >.

---+

AEA1

8

---+

0

---+

O

AEAo

into G and taking homology, we obtain the groups Extn (

ex

Mapping the complex . . . ---+

EB Y,A

---+

AEA2

S 0 P,Ab

) (F, G).

EB Y >. 8

EB Y,>.

---+

AEA1

AEAo

into iG and taking homology, we obtain the groups Extn (

eat S 0 P,Ab

) (iF,iG).

But

It immediately follows that the complexes of abelian groups whose homology is

Extn (

ex

S 0 P,Ab

) (F, G)

respectively

Extn (

eat S 0 P,Ab

coincide.

) (iF, iG) D

REMARK 7.5.7. In this Section, we assumed that the category S is essentially small, triangulated and satisfies[TR5(a)]. However, the statements remain true even if it only satisfies Hypothesis 6.1.1. The reader will note that in the proof of Proposition 7.5.5, we only apply Lemma 7.5.4 to an object P which is a coporduct of representables. Such objects are a-filtered colimits of representables even when S is not triangulated.

7.6. History of the results in Chapter 7 The results in this Chapter are all new.

CHAPTER 8

Brown representability 8.1. Preliminaries In this Chapter, all categories are assumed to have small Hom sets. Sometimes we will explicitly remind the reader of this; even when we do not, it is assumed. Let us make some definitions about possible sets of generators for 'J. DEFINITION 8.1.1. (cf Definition 6.2.8). Let 'J be a triangulated category satisfying [TR5j. A set T of objects ofT is called a generating set if 8.1.1.1. {Hom(T, x)

= 0} ==} {x = 0}; that is,

if x E 'J satisfies

Vt E T,Hom(t,x) = 0 then x is isomorphic in 'J to 0. 8.1.1.2. Up to isomorphisms, Tis closed under suspension and desuspension; that means that given an object t E T and an integer n, there is an object in T isomorphic to r;nt; DEFINITION 8.1.2. Let 'J be a triangulated category satisfying [TR5]. A set T of objects of 'J is called a ,8-perfect generating set for 'J if it is a generating set as in Definition 8.1.1, and T U {0} is ,8-perfect as in Definition 3.3.1. Note that we do not insist that 0 E T; this is mostly for convenience of notation. REMARK 8.1.3. Let a and ,B be any infinite cardinals. Let 'J be a triangulated category satisfying [TR5]. Let T be a ,6-perfect generating set for 'J. By Proposition 3.2.5 (T)a is an essentially small category. Let S be a set of objects of (T)a containing T, and containing at least one representative in each isomorphism class of objects. We assert that S is also a ,8-perfect generating set for 'J. After all, it is a set closed under suspension (up to ismorphism). Since it contains T, it generates. The fact that it is ,8-perfect is Theorem 3.3.9 and Lemma 3.3.2.

274

8.

BROWN REPRESENTABILITY

DEFINITION 8.1.4. Let f3 be an infinite cardinal. A triangulated category 'J satisfying [TR5] is called /3-perfectly generated if it contains some /3-perfect generating set T. REMARK 8.1.5. Replacing T by a setS equivalent to (T)a, we may assume that S is, up to equivalence of subcategories, the set of objects of an a-localising triangulated subcategory. An even better situation is when the perfect generating set may be chosen so that its objects are also /3-small. This gives DEFINITION 8.1.6. Let 'J be a triangulated category satisfying [TR5}. Let /3 be a regular cardinal. A set T of objects of 'J is called a /3--compact generating set for 'J if it is a /3-perfect generating set as in Definition 8.1.2, and all the objects ofT are /3-small. That is, T C 'J(,6). REMARK 8.1.7. Let T be a /3-compact generating set for 'J. Then T is a /3-perfect subset of 'J(,6), hence contained in the largest such, which is 'J,a. That is, T is contained in the subcategory 'J,a of /3--compact objects in 'J. Hence the name /3--compact generating set. The categories possessing /3--compact generating sets for some regular cardinal /3 are particularly nice. We will call them well generated. We cannot call them compactly generated, since the term already exists in the literature, referring to categories which possess an N0 --compact generating set. Before we leave this introductory section, which focused mostly on the definitions of various types of generating sets, let us remind the reader of the (unrelated) definition of homotopy colimits of countable sequences. DEFINITION 1.6.4 Let 'J be a triangulated category satisfying [TR5}. {This means that 'J is closed under small coproducts). Let

X0

- i1 +

X1

- h+

X2

ia

-------7

...

be a sequence of objects and morphisms in 'J. The homotopy colimit of the sequence, denoted ~Xi, is by definition given, up to non-canonical isomorphism, by the triangle

where the shift map shift

-+

00

BROWN REPRESENTABILITY

8.2.

is the direct sum of ii+l is the infinite matrix

:

275

Xi ---+ Xi+l· In other words, the map {1 - shift}

1x0 -jl

0 0

0 1xl

-h 0

0 0 1x2

-h

0 0 0 1x3

8.2. Brown representability

We will be studying representable functors from triangulated categories 'J0 P to the category Ab of abelian groups. Clearly, the functor 'J(-, h) is homological and takes coproducts to products. We define DEFINITION 8.2.1. A triangulated category 'J is said to satisfy the representability theorem if 8.2.1.1. The category 'J satisfies (TR5]. 8.2.1.2. Any functor H : 'J0 P ----+ Ab, which is homological, and sends coproducts in 'J to products in Ab, is representable; it is naturally isomorphic to 'J( -,h) for some hE 'J. REMARK 8.2.2. The main theorems of the Chapter will show that, if 'J or 'J0 P have sufficiently nice generating sets (see Section 8.1), then the representability theorem holds in 'J. The first theorem of this type was proved by Brown [7]. For this reason, theorems of this type are usually referred to as Brown representability. The key to all the representability theorems we will prove is the following Lemma. LEMMA 8.2.3. Let 'J be a triangulated category with small Hom-sets satisfying (TR5}, T a set of objects in 'J. Suppose T is essentially closed under suspension; that means it contains objects isomorphic to all suspensions of its objects. Let H : 'J0 P ----+ Ab be a homological functor. That is, H is contravariant and takes triangles to long exact sequences. Suppose the natural map

H

(II

>.EA

t>.)

------+

II H(t>.)

>.EA

is an isomorphism for all small coproducts in 'J. Then it is possible to construct a sequence of objects and morphisms in 'J 12 i1 X 1 ------+ X 2 ------+ ia X 0 ------+ ...

so that

8. BROWN REPRESENTABILITY

276

8.2.3.1. For every i, the objects Xi lie in (T}, the smallest localising subcategory of 'J" containing T. 8.2.3.2. For each i, there is a natural transformation of functors on 'J"

These are compatible in that the diagram

/ H(-) commutes for every i. 8.2.3.3. Let X= ~Xi. There is a natural transformation 'J"( -,X) ~ H(-) rendering commutative the triangles

/ H(-)

'J"( -,X) for every i.

8.2.3.4. For every object t E T, the image of the map

:r(t, xi)

~

:r(t, xi+l)

maps isomorphically to H(t) via :r(t, xi+l)

~

H(t).

Proof: The rest of this Section will be devoted to the proof of this Lemma. Since the proof is somewhat technical, on first reading the reader might do well to skip this proof and pass on to the applications. We have fixed a set T of objects in 'J", where up to isomorphism T = 'ET. Let Uo be defined as Uo =

UH(t).

tET

8.2. BROWN REPRESENTABILITY

277

Elements of U0 can be thought of as pairs (a, t) with a E H(t). Put

ll

Xo =

t.

(a,t)EUo

Clearly, X 0 is an object of (T). Also, by the hypothesis that H takes coproducts to products,

IT

H(Xo) =

H(t),

(a,t)EUo

and there is an obvious element in H(Xo), namely the element which is a E H(t) for (a, t) E Uo. Call this element a 0 E H(Xo). The construction is t corresponding such that if t ---4 Xo is the inclusion oft into Xo =

l1

(a,t)EUo

to (a, t) E Uo, then the induced map H(Xo) ---4 H(t) takes a 0 E H(Xo) to a E H(t). To give an object Xo and an element a 0 E H(Xo) is by Yoneda's lemma the same as giving a natural transformation . E A} be a set of objects in 'J. For each >. E A, the functor 'J(-,X,x) is a cohomological functor by Remark 1.1.11. Being representable, it sends coproducts in 'J to products of abelian groups. But then the functor

H(-)

=

II'J(-,X,x) AEA

is also a cohomological functor sending coproducts in 'J to products of abelian groups. Because the representability theorem holds for 'J, the functor H is representable. There is an object X E 'J so that 'J( -,X)

=

II 'J( -, X,x). -XEA

But then X satisfies the universal property of a product; it is the product of the X,x's. D 8.5. Applications in the presence of injectives In this Section, we look a little more closely at what happens, if the category C.x ( {'J} op, has enough injectives: We remind the reader that enough injectives need not exist. See the counterexample of Section C.4.

Ab)

8.5.1. Before proceeding to Proposition 8.5.2, let us remind the reader of Proposition 8.4.2 and Theorem 8.3.3. Let 'J be a wellgenerated triangulated category. By Proposition 8.4.2, more precisely by 8.4.2.2, we have that for every infinite o:, 'J is essentially small. In Theorem 8.3.3, we learned that 'J satisfies the representability theorem. The next Proposition therefore applies to well-generated categories. REMARK

PROPOSITION 8.5.2. Let 'J be a triangulated category with small Homsets, satisfying [TR5]. Let o: be a regular cardinal. Suppose the category 'J

8.5. IN THE PRESENCE OF INJECTIVES

289

is essentially small. By the results of Chapter 6, more specifically Proposition 6.2.6 and Remark 6.2. 7, the natural homological functor

'J

----T

ex({'Ja}op,Ab)

preserves coproducts and products. To simplify the notation, write g = 'Ja. We remind the reader that ex(goP,Ab) is the category of functors gop---tAb, sending coproducts of fewer than a objects to products. Suppose 'J satisfies the representability theorem. Let I be an injective object in ex(goP,Ab). There exists an object GI E ' j so that:

8.5.2.1. For any object x 'J(x,GI)

----T

E

'J, there is a natural isomorphism

ex(gov,Ab)['J(-,x)J 3 ,I(-)].

Proof: Let I be an injective object in the category ex(goP,Ab). Then we can construct a functor 'J0 P ---+ Ab, denoted

ex(S 0 P,Ab) [ 'J(-,x)Js, I(-)] which takes

X

E

'J to the group of maps in ex(goP,Ab) 'J(-,x)Js

----T

I(-).

By Remark 6.2. 7, the functor taking x to 'J (-, x) Is is a homological functor respecting coproducts 'J

----T

ex(goP,Ab).

In ex (gov, Ab), the object I is injective. Taking maps into I preserves exact sequences, and clearly takes coproducts to products. Thus the functor taking X E 'J to the group of maps in ex(goP,Ab)

ex(S 0 P,Ab) [ 'J( -, x)Js, I(-)] is a homological functor 'J0 P ---+ Ab, taking coproducts to products. We are assuming the representability theorem holds for 'J. It follows that there exists an object GI E 'J, so that maps x ---+ GI are in one-to-one correspondence with maps

'J(-,x)Js

----T

I(-). D

COROLLARY 8.5.3. Let 'J be a triangulated category with small Homsets, satisfying [TR5]. Let a be a regular cardinal. Suppose the category 'Ja is essentially small. Suppose 'J satisfies the representability theorem. Finally, suppose that the category ex( {'Jayv,Ab) has enough injectives.

290

BROWN REPRESENTABILITY

8.

Then the natural functor of Section 6.5 1r:

A('J)

------+

£x(S 0 P,Ab)

has a right adjoint G.

Proof: In Proposition 8.5.2, we proved that for x E 'J and I an injective object in £x(S 0 P,Ab), the functor sending x E 'J to £x(S 0 P,Ab) [1rx,l] is representable. In other words, there is an isomorphism, natural in x E 'J,

'J(x,GI)

~

£x(S 0 P,Ab) [1rx,I(-)].

Both functors, viewed as functors in x, are homological functors 'J0 P Ab. They are the restrictions to x E 'J c A('J) of, respectively, A('J)(x,GI)

and

---+

£x(S 0 P,Ab) [1rx,I(-)].

The restrictions of these functors to 'J C A('J) are isomorphic. From the canonical nature of the factorisation of a homological functor through the universal one (see Theorem 5.1.18), it follows that the functors are isomorphic on all of A('J). We want to show that the functor 1r has a right adjoint. We have already shown that the adjoint is well-defined on injective objects I E £x(S 0 P,Ab). But by hypothesis, the category £x(S 0 P,Ab) has enough injectives. Every object has an injective copresentation. From Lemma A.2.13, it follows that the functor G extends to all of £x (sop, Ab), to define a right adjoint for 1r. D REMARK 8.5.4. From Proposition 6.5.3, we already know that the category £x(S 0 P,Ab) is a Gabriel quotient A('J)/13. In Proposition 6.5.3, we proved that the quotient is colocalizant. In Corollary 8.5.3, we saw that if 'J satisfies the representability theorem and if £x ( {T"'YP, Ab) has enough injectives, then the quotient is also localizant. The map to the quotient has a right adjoint as well as a left adjoint. From Proposition A.2.10, we know that the counit of adjuction c 1rG---+ 1 is an isomorphism. Concretely, for any IE £x(S 0 P,Ab), 'J ( -, GI)Is

I(-).

LEMMA 8.5.5. Let 'J be a triangulated category satisfying [TR5]. Let a be a regular cardinal. Suppose that S = 'J"' is essentially small. If furthermore the natural projection 1r:

A('J)

----+

£x(S 0 P,Ab)

8.5. IN THE PRESENCE OF INJECTIVES

admits a right adjoint G: ex(S 0 P,Ab) ex(S 0 P,Ab) has a cogenerator.

----+

291

A('J), then the abelian category

Proof: By Lemma 6.4.4, the category ex(S 0 P,Ab) has a projective generator P. Let {q>., .A E A} be a set of representatives for all the quotients of P. Every non-zero object of ex(SoP,Ab) contains a a non-zero subobject of the form q>.. Let I be an injective object in A('J), which admits an embedding

II GqA

I.

----+

AEA

We assert that 1r I is a cogenerator for the abelian category ex (sop, Ab). Take any X E ex(S 0 P,Ab), and consider the map X

II

----+

1ri.

X----+11" I

Let k be the kernel of the map, We wish to show that k = 0. Let q>. be a quotient of P embedding ink. Then we have monomorphisms

q>.

----+

k

----+ X.

But the functor G has a left adjoint, and therefore is left exact. We deduce monomorphisms in A('J)

Gq>.

----+

Gk

Gx.

----+

Now I is an injective object of A('J), containing Gq>.. The monomorphism Gq>. ----+I extends to a map Gx----+ I, and applying the functor 1r we have a commutative diagram 0 ----+ 1rGq>.

0 ----+

q>.

----+

1rGx

----+

X

----+ 1r I

The map 1rGq>. ----+ 1r I is a monomorphism since Gq>. ----+ I is, and 1r is exact. By Proposition A.2.10, we know that the vertical maps c: 1rG----+ 1 are isomorphisms. We have therefore found a map x ----+ 1ri, which is a monomorphism on q>. C k, and k is the kernel of all maps x ----+ 1ri. It follows that q>. = 0. The only subobject of k of the form q>. is the zero D object. Hence k = 0, and x embeds into a product of 1rI's. REMARK

8.5.6. We therefore know the implications

ex(S 0 P,Ab) { has enough injectives

}={

A('J)----+ t:x(S 0 P,Ab) has a right adjoint

}={

ex(S 0 P,Ab) } has a cogenerator

292

8. BROWN REPRESENTABILITY

In Section C.4, we see an example where £x(S 0 P,Ab) does not have a cogenerator. This example therefore teaches us not only that there need not be enough injectives, but also that there need not be a right adjoint to 1r: A('J)----+ £x(S 0 P,Ab). It might be instructive to reformulate some of the statements about injectives. Given an injective in £x(S 0 P,Ab), we saw in Proposition 8.5.2 that there is an object GI in 'J with certain properties. We next want to identify the objects GI E 'J. Recall Proposition 6.5.3: £x (gop, Ab) is the Gabriel quotient of A('J), where we send to zero the class of objects vanishing under rr. If we use the identification A('J) = D('J) of Definition 5.2.1 and Lemma 5.2.2, the class of objects mapping to zero under 1r is easier to understand; see Lemma 6.5.6 and Definition 6.5.7. Let us remind the reader. The category D('J) has for its objects the morphisms {x ----+ y} E 'J. A morphism in D('J) is an equivalence class of commutative squares x~y

1

1

x'~y'.

A square as above is equivalent to zero if the two equal composites X

1

1

y'

x'~y'

both vanish. The functor 1r: D('J) ----+ £x(S P,Ab) takes the object {x----+ y} E D('J) to the image of 0

'J(-,x)ls ~ 'J(-,y)is· The objects {x ----+ y} E D('J) map to zero under 1r precisely if the induced map 'J(-,x)ls ~ 'J(-,y)Js vanishes; that is, x ----+ y is an a-:-phantom map. LEMMA 8.5. 7. Let 'J be a triangulated category satisfying the representability theorem, and let a be a regular cardinal. Put S = 'J"', and assume S is essentially small. Let I be an injective object in £x(S 0 P,Ab). Then any a-phantom map x ----+ G I vanishes. Furthermore, rrG I is naturally isomorphic to I.

Proof: By Proposition 8.5.2, thereis a natural isomorphism 'J(x, GI)

~ £x(S 0 P,Ab) [ 'J(-,x)ls,J(-)].

8.5. IN THE PRESENCE OF INJECTIVES

Let x

----+

293

y be an a-phantom map; it induces the zero map

'J(-,x)ls

~

'J(-,y)fs,

hence the zero map

£x(S 0 P,Ab) [ 'J( -, y)fs, I(-)] ~ £x(S 0 P,Ab) [ 'J (-, x)ls, I(-)]. By the naturality of the isomorphism of Proposition 8.5.2, this is also the zero map

'J(y,GI)

~

'J(x,GI).

Now if y = GI, we discover that any a-phantom map f: x----+ GI induces the zero map

'J(GI,GI)

~

'J(x,GI).

In particular, f = l{GI} of= 0. Thus any a-phantom map f: x----+ GI must vanish. It remains to show that 1rGI is naturally isomorphic to I. In Remark 8.5.4, we observed that when there are enough injectives and G extends to a right adjoint to 1r, then the counit of adjunction c1 : 7rGI----+ I is an isomorphism for all I, not only for I injective. But even when there are not enough injectives, for any injective I we have a natural isomorphism 1rGI ~I. This is what we now want to prove. By the definition of G I, there is a natural isomorphism

'J[x,GI]

=

£x(S 0 P,Ab){1rx,f}.

Now put x = s E S c 'J. Then

1rGI(s)

=

'J[s, GI]

£x(S0 P,Ab){ s,

I}

=

I(s). 0

LEMMA 8.5.8. Let 'J be a triangulated category satisfying the representability theorem, and let a be a regular cardinal. Put S = 'Jo:, and assume S is essentially small. Suppose t is an object of 'J, and suppose further that any a-phantom map x ----+ t in 'J vanishes. Let 1r:

A('J)

--+

t.x(Sop,Ab)

be the usual projection. If a is an object of A('J) with 1ra = 0, then A('J) [a, t}

= 0.

Proof: It is convenient to work with D('J) ~ A('J). An object a E D('J') is a morphism{!: x----+ y} in 'J. The object t E 'J maps to {1 : t----+ t} E

8. BROWN REPRESENTABILITY

294

D('J), under the universal homological functor 'J----> D('J). A morphism a ---> t is an equivalence class of commutative squares

1

1

t~t.

If 1ra = 0, then {! : x composite x

L

--->

y} is an a-phantom map in 'J. Hence so is the

y ----> t. By the hypothesis of the lemma, any a-phantom

map into t is zero. In particular, the map x other words, the two equal composites

L

y ----> t must vanish. In

X

1

1 t

vanish, making the commutative square

1

1

equivalent to zero. That is, any map a----> t, in D('J), must vanish.

D

LEMMA 8.5.9. Let 'J be a triangulated category satisfying the representability theorem, and let a be a regular cardinal. Put S = 'J"', and assume S is essentially small. Suppose t is an object of 'J, and,suppose further that any a-phantom map x ---> t in 'J vanishes. Let

be the usual projection. Then 7rt is an injective object of ex(S 0 P,Ab), and t

= G7rt.

Proof: Let X -----4 y be a monomorphism in ex (gop' Ab). The functor 7r has a left adjoint L, which is not left exact. But we can form an exact sequence 0

-------+

k

-------+

Lx

---t

Ly.

8.5. IN THE PRESENCE OF INJECTIVES

295

The functor 1r is exact; hence we have a commutative diagram, where the bottom row is an exact sequence

0

--+

1rk

--------+

X

--+

y

1rLx

--+

1rLy.

By Proposition 6.5.3, the maps 'f/x, 'fJy are both isomorphisms. Since x ~ y is a monomorphism in ex(S 0 P,Ab), it follows that 1fk = 0. By Lemma 8.5.8 we conclude that A('J')[k, t] = 0. But as t E 'J' C A('J'), Corollary 5.1.23 tells us that tis injective, as an object of the abelian category A('J'). The exact sequence 0

---+

k

Lx

---+

--+

Ly

gives rise to an exact sequence

A('J')[Ly, t]

--+

A('J')[Lx, t]

--+

A('J')[k, t]

--+

0.

Since we know that A('J')[k, t] = 0, we deduce a surjective map

A('J')[Ly, t] But L is left adjoint to

1r,

---+

A('J')[Lx, t].

hence this surjective map identifies as

Since this is surjective for any monomorphism X ~ y in ex (gop' Ab) ' it follows that 1rt is an injective object. Proposition 8.5.2 now tells us that we can form an object G1rt E 'J', and I assert that it is canonically isomorphic to t. Recall that for any injective object IE ex(S 0 P,Ab), GI E 'J' is defined by representing the functor

In the case where I = 1rt, with t E 'J' as above, we have

A('J') [L1rx, t] just because Lis left adjoint to 1r. But we also have the counit of adjunction C:x :

L1rx

---+

x.

Complete it to an exact sequence in A('J')

0

--------+

k

---+

L1rx ~ x

--------+ q - - +

0.

296

8.

BROWN REPRESENTABILITY

Now 7r : A('J) --+ ex(S 0 P,Ab) is an exact functor; in the commutative diagram below, the bottom row is exact 1rX

1

----+ 1rX

0 ----+ nk ----+ nLnx ~ nx ----+ nq ----+ 0.

Since 'f/: 1 us that

===}

nL is an isomorphism, so is

7rcx·

The exact sequence tells

nk = 0 = nq. From Lemma 8.5.8, it now follows that

A('J)[k, t]

A('J)[q, t].

0

But t E 'J C A('J) is an injective object. The exact sequence 0

----t

k

----t

L 1rX

E:.,

----t X

---->

q

---->

0

gives an exact sequence

0 = A('J)[q, t]

----+

A('J)[x, t]

----+

A('J)[Lnx, t]

---+

A('J)[k, t]

= 0.

Hence 'J(x, t) = A('J)[x, t] is naturally isomorphic to A('J)[Lnx, t], which in turn is naturally isomorphic to

ex(SOP' Ab) { nx, nt }· This means Gnt = t.

D

REMARK 8.5.10. Lemmas 8.5.7 and 8.5.9 allow us to identify injective objects IE ex(S 0 P,Ab) with objects t E 'J so that all a-phantom maps x --+ t vanish. If t E 'J admits no non-zero a-phantom maps x --+ t, then ntis an injective object of ex(S 0 P,Ab). If I is an injective object of ex(S 0 P,Ab), then GI E 'J admits no non-zero a-phantom maps X--+ GI. Furthermore, nGI ~I, and Gnt ~ t. We will say that an object t E 'J is orthogonal to the a-phantom maps if all a-phantom maps X--+ t vanish. Thus injective objects in ex(S 0 P,Ab) are in 1-to--1 correspondence with objects t E 'J, orthogonal to the aphantom maps. It might be helpful to state this in more generality. DEFINITION 8.5.11. Let 'J be an additive category. An ideal :J of morphisms in the category 'J is a class of morphisms, closed under addition, and so that if g E :J and f and h are arbitrary morphisms, then fgh E :J, whenever the composite exists. EXAMPLE 8.5.12. If 'J is a triangulated category satisfying [TR5] and a is a regular cardinal, the class :J of all a-phantom maps is an ideal.

8.5.

IN THE PRESENCE OF INJECTIVES

297

DEFINITION 8.5.13. Given an ideal :J of morphisms, its orthogonal, denoted :J..L, is the collection of all objects t E 'J', so that if {x ---t t} E :J, then x ---t t is the zero map. If T is any class of objects in 'J', the ideal I {T} is the ideal of all morphisms {! : x ---t y} E 'J', so that for any t E T, all composites x ...!_. y ---t t vanish. REMARK 8.5.14. Clearly for any ideal :J, :J C I{:J..L}. It may happen that equality holds. That is, under favorable conditions, one has

The next lemma treats the case where :J is the ideal of a-phantom maps, in a triangulated category 'J'. LEMMA 8.5.15. Let 'J' be a triangulated category satisfying the representability theorem, and let a be a regular cardinal. Put g = 'J'"', and assume g is essentially small. Let :J be the ideal of a-phantom maps. The category ex (gop, Ab) has enough injectives if and only if

Proof: Recall that 7r:

D('J')

=

A('J')

------+

ex(goP,Ab)

is a Gabriel quotient map. Every object in ex(goP,Ab) is of the form 1ra, with a an object of D('J'). The category ex(goP,Ab) will have enough injectives if and only if every non-zero object 1ra in ex (gop, Ab) admits a non-zero map to an injective object. By Remark 8.5.10, injective objects in ex (gop, Ab) are all of the form 1rt, with t E :J..L. In other words, the category ex (gop, Ab) has enough injectives iff, for any object a E D('J') not in the kernel of 1r, there is a non-zero morphism 1ra ---t 1rt, with t E :J..L. By Lemma 8.5.9, whenever t E :J..L,

D('J') [a, t]

cx(goP,Ab) { 1ra, 1rt }.

Therefore ex (gop, Ab) will have enough injectives if and only if, for any object a E D('J') with 1ra non-zero, there is a non-zero map (in D('J')) a ---t t, with t E :Jl.. Now remember that an object a E D('J'), with 1ra non-zero, is a morphism {x ---t y} E 'J' which is not a-phantom. The category cx(goP,Ab) has enough injectives precisely if, for every non-a-phantom { x ---t y} E 'J', there is a non-zero map, in D('J'), into t E :J..L.

298

8. BROWN REPRESENTABILITY

But a map from {f : x tive squares

y} to t is an equivalence class of commuta-

~

1

1

t~t. To say that the square is not equivalent to zero, is to say that the equal composites X

1

t

do not vanish. In other words, a non-zero map from {f : x if and only if there is a non-zero composite x ~ y

~

y} tot exists

t. There will be enough injectives, if and only if a non-zero composite x ~ y ~ t exists for every non-a-phantom {! : x ~ y} E 'J. Equivalently, if x ~ y ~ t vanishes for every t E :J.l, then f : x ---+ y should be an a-phantom map. ---+

That is, D

REMARK 8.5.16. From the counterexample of Section C.4, we learn that in general, the inclusion :J C I {:J.l} can be proper. Lemma 8.5.15 gives an equivalence between the existence of enough injectives in ex(S 0 P,Ab), and a statement purely about a-phantom maps in 'J. Let us find another, equivalent formulation. LEMMA 8.5.17. Let 'J be a triangulated category satisfying the representability theorem, and let a be a regular cardinal. Put S = 'Ja, and assume S is essentially small. Let :J be the ideal of a-phantom maps. The category ex (gop, Ab) has enough injectives if and only if, for every object z E 'J, there exists a triangle y ~ z

----7

t

----7

"Ey

where t,E :J.l and f: y ~ z is an a-phantom map (that is, f E :J). Proof: The category ex (sop, Ab) has enough injectives if and only if every object a E ex(S 0 P,Ab) has an embedding into an injective object. Now recall that the functor

8.5. IN THE PRESENCE OF INJECTIVES

299

has a left adjoint L. The category A('J) always has enough injectives; they are the objects z E 'J. Given an object a E ex(goP,Ab), there is an embedding La-----+ z, z E 'J. Since 7f is exact, this gives an embedding a= 7f La

-----+

1fZ.

Therefore, the category ex (gop, Ab) has enough injectives if and only if all objects of the form {1rz, z E 'J} can be embedded in injective objects IE ex(goP,Ab). By Remark 8.5.10, the injective objects in ex(goP,Ab) canbe identified with objects t E :Jj_. The category ex(goP,Ab) has enough injectives if and only if, for every object z E 'J, there is an embedding in ex (gop, Ab) 7f

z

---+

1ft,

with t E :Jj_. By Lemma 8.5.9, whenever t E :Jj_, 'J(z,t)

ex(goP,Ab){1fz,1ft}.

In other words, there are enough injectives in ex (gop, Ab) if and only if, for every object z E 'J, there is a morphism {z--+ t, t E :Jj_} in 'J, so that 1r z --+ 1rt is a monomorphism. Consider now the triangle y ~ z - - - + t - - - + L:y.

The functor

7f

takes it to an exact sequence in ex(goP,Ab). That is, nf

1ry - - - + 1rz - - - + 1rt

is a exact in ex(goP,Ab). The map 1rz-----+ 1rt will be a monomorphism if and only if 7rf = 0, that is f E :J. Summarising, the category ex(goP,Ab) has enough injectives if and only if, for every object z E 'J, there is a triangle y ~ z - - - + t - - - + L:y

with t E :Jj_ and

f

E :J.

0

Combining Lemmas 8.5.15 and 8.5.17, we have PROPOSITION 8.5.18. Let 'J be a triangulated category satisfying the representability theorem, and let a be a regular cardinal. Put g = 'J"', and assume g is essentially small. Let :J be the ideal of a-phantom maps. The following are equivalent:

8.5.18.1. The category ex(goP,Ab) has enough injectives. 8.5.18.2. The ideal :J satisfies :J = J{:Jj_}

8. BROWN REPRESENTABILITY

300

8.5.18.3. For every object z E 'J, there is a triangle y

____!.___., z ~ t ~ ~y

with t E 'Jj_ and f E 'J.

Proof: By Lemma 8.5.15, we have the equivalence 8.5.18.1

~

8.5.18.2.

By Lemma 8.5.17, we also have the equivalence

8.5.18.1

~

8.5.18.3. D

COROLLARY 8.5.19. Let 'J be a triangulated category satisfying the representability theorem, and let a be a regular cardinal. Put S = 'J, and assume S is essentially small. Suppose ex (gop, Ab) has enough injectives. Then every object z E 'J admits a maximal a-phantom map y ---+ z. That is, y---+ z is an a-phantom map, and every a-phantom map x---+ z factors as X ---+

y

---+ Z.

Proof: Let 'J be the ideal of a-phantom maps. We are assuming that ex(S 0 P,Ab) has enough injectives, and hence by Lemma 8.5.17, every object z E 'J admits a triangle y

____!.___., z ~ t ~ ~y,

with t E 'Jj_ and f E 'J. I assert that the map f : y ---+ z is a maximal a-phantom map. The fact that f : y ---+ z is a-phantom is given; f E 'J is granted to us. What remains to prove is its maximality. Suppose g : x ---+ z is any a-phantom map. The composite x~z~t

is an a-phantom map x the triangle y

---+

t, with t

E

'Jj_. All such maps vanish. From

____!.___., z ~ t ~ ~y,

we have that x---+ z must factor as x---+ y---+ z.

D

Recall that in Remark 8.5.6 we saw the implications

ex(S 0 P,Ab) { has enough injectives

}={

A('J)---+ ex(S 0 P,Ab) has a right adjoint

}={

ex(S 0 P,Ab) } has a cogenerator

8.5.

IN THE PRESENCE

OF

INJECTIVES

301

In the last few Lemmas, summarised in Proposition 8.5.18, we have explained concretely, in terms of the ideal of a-phantom maps and its orthogonal, what it means to have enough injectives in C:x(S 0 P,Ab). In other words, the reader can now express somewhat more concretely, what is implied by the fact that enough injectives need not exist, as in the counterexample of Section C.4. But the counterexample of Section C.4 says more than just that there are not enough injectives. It says there is, in general, no cogenerator. In the spirit of reformulating everything in terms of a-phantom maps, let us see what it means to say that there is no right adjoint to A('.T) ---+ C:x (gop, Ab). LEMMA 8.5.20. Let 'J be a triangulated category satisfying [TR5}. Let a be a regular cardinal. Put S = 'J"', and assume S is essentially small. The functor 1r:

A('J)

-----+

C:x(S 0 P,Ab)

will have a right adjoint if and only if, for every object z E 'J, there is a maximal a-phantom map y ---+ z. That is, there is an a-phantom map y ---+ z, and given any other a-phantom map x ---+ z, there is a (nonunique) factorisation X ---+

y

---+ Z.

Proof: The map 1r:

A('J)

-----+

C:x(SoP,Ab)

is a quotient map by Proposition 6.5.3. The abelian category A('.T) has enough injectives. Therefore Proposition A.2.20 applies: a right adjoint will exist if and only if every injective object z E 'J C A('.T) has a maximal subobject, among those in the kernel of 1r. It is convenient to use the description D('J) ~ A('.T). Recall Proposition 5.2.6. The subobjects, in D('J), of an object z E 'J, may be represented by morphisms in 'J of the form y ---+ z. A subobject y ---+ z contains x ---+ z if the map x ---+ z factors as X ---+

y---+

Z.

The kernel of 1r is identified with the a-phantom maps x ---+ z. Therefore the existence of a maximal subobject of z, among those belonging to the kernel of 1r, is equivalent to the existence of a maximal a-phantom map y---+ z.

D

REMARK 8.5.21. Proposition 8.5.18 rephrases for us the existence of enough injectives, in terms of a-phantom maps. Lemma 8.5.20 rephrases the existence of an adjoint G : C:x (Sop, Ab) ---+ A ('.T), also in terms of a-phantom maps. From Proposition 8.5.2 we know, that the existence of

302

8. BROWN REPRESENTABILITY

enough injectivesimplies the existence of a right adjoint. Corollary 8.5.19 rephrases this in terms of a-phantom maps. That is, in Proposition 8.5.18 we saw that ex(goP,Ab) has enough injectives if and only if, for every object z E 'J there is a triangle

y~z~t~~y, with t E :J..L and f E :J. In Corollary 8.5.19, we learned that in any such triangle, the map y ---+ z is a maximal a-phantom map into z. Finally, Lemma 8.5.20 teaches us that the existence of maximal a-phantom maps y ---+ z is equivalent to the existence of a right adjoint G to the natural functor 7r:

A('J) ~ ex(goP,Ab).

In other words, in terms of a-phantom maps, we recover Proposition 8.5.2. We learn the implication

ex(goP,Ab) } { has enough . . t' mJec 1ves

==?

{ A('J)---+ ex(goP,Ab) } h

. ht d' . t as a ng a JOlll

.

From Lemma 8.5.20, we know that there is a right adjoint

G: ex(goP,Ab) ~ A('J), if and only if every z in 'J admits a maximal a-phantom map y---+ z. If we furthermore know that there are enough injectives in ex(goP,Ab), then the maximal a-phantom map y ---+ z may be so chosen, that in the triangle y~z~t~~y,

the object tis orthogonal to the a-phantom maps. From Section C.4, we know by example that ex(goP,Ab) need have neither enough injectives, nor a right adjoint to 1r : A('J) ---+ ex (gop, Ab). There need not be enough objects t E :J..L, but even worse than that, there may be objects z E 'J, admitting no maximal a-phantom map y---+ z. REMARK 8.5.22. If I is an injective cogenerator of ex (gop' Ab)' we will denote GI by llllC. The notation stands for Brown-Comenetz objects of 'J. When we wish to remind ourselves of the dependence on the choice of a and on 'J, we will write them llll y - - y" - - 0 where the rows are exact and x", y' E ~.

---->

328

A. ABELIAN CATEGORIES

The natural functor A -----+ A/13 is the identity on objects, and sends a morphism to its equivalence class. The following lemma may be found in [15], more precisely Corollaire 2, in Paragraphe 1 of Chapitre III; see page 368. LEMMA A.2.3. The category A/13 is an abelian category. The functor F : A -----+ A/13 is exact, and takes the objects of 13 to objects in A/13 isomorphic to zero. Furthermore, F is universal with this property. The subcategory 13 C A is the full subcategory of all objects b E A so that Fb is isomorphic to zero. REMARK A.2.4. Unlike in the case of triangulated categories, which we studied in Chapter 2, there is usually no set theoretic problem in forming A/13. If A is a well-powered abelian category with small Hom-sets, then the objects x andy have sets of subobjects and quotient objects, and hence the colimit in Definition A.2.2.2 is a colimit of sets indexed over a set; it is a set. Thus A/13 will have small Hom-sets.

As with triangulated categories, it is very interesting to study the situation in which the quotient map A -----+ A/13 has an adjoint, right or left. We will for now restrict attention to right adjoints; the case of a left adjoint is dual. The Serre subcategory 13 C A is called localizant if the functor A -----+ A/13 has a right adjoint. LEMMA A.2.5. Let F: A-----+ 'J be an exact functor of abelian categories, G : 'J -----+ A its right adjoint. Let 13 C A be the full subcategory of all objects b E A so that Fb is isomorphic to zero. If an object y E A is isomorphic to Gt for some t E 'J, then

A.2.5.1. For every object bE 13, A(b, y) = 0. A.2.5.2. For every object bE 13, Ext,1(b,y)

= 0.

Proof: Suppose t E 'J, y = Gt. Then by the adjunction, for all x E A

A(x, y)

A(x, Gt)

'J(Fx, t).

In particular, if x =bE 13, then Fb = 0 and we deduce the isomorphisms A(b, y) = 'J(Fb, t) = 0. Suppose z is an extension of y = Gt by some object bE 13. That is, we have an exact sequence in A

0

----+

y

----+

z

----+

b

----+

0.

Applying F to it, we get an exact sequence in 'J

0

----+

Fy

----+

Fz

----+

and since Fb = 0 we must have that Fy then

'J(Fz, t)

----+

-----+

Fb

-----t

0,

Fz is an isomorphism. But

'J(Fy, t)

A.2. FORMAL PROPERTIES OF QUOTIENTS

329

is also an isomorphism. Consider the commutative square below

'J'(Fz, t) ~ 'J'(Fy, t).

The vertical maps are isomorphisms by adjunction. The bottom row is an isomorphism by the above. Hence so is the top row. Recalling that Gt = y, we have that the identity 1: y ---t y must be in the image of A(z, y); there is a map z ---t y so that the composite y ---t z ---t y is the identity. The short exact sequence 0

-----+ y -----+

z

-----+

b

-----+

0

splits, and Ext1(b,y) = 0.

D

DEFINITION A.2.6. Let A be an abelian category, ~ C A a Serre subcategory. An object y E A will be called ~-local if the following two conditions hold, as in Lemma A.2.5. A.2.5.1: For every object bE ::B, A(b, y) = 0. A.2.5.2: For every object b E ::B, Ext1 (b, y) = 0. Lemma A.2.5 asserts that, ifF: A - 'J' is exact and has a right adjoint G, then every y = Gt is :B-local, where ::B is the full subcategory of all bE A with Fb ~ 0. LEMMA A.2.7. Let A be an abelian category, ~ C A a Serre subcategory. If y E A is a :B-local object, then for every x E A

A/:B(x, y).

A(x, y)

Proof: Let y be a ::B-local object in A. There is a natural map

A(x, y)

-----+ A/~(x,

y);

we need to prove it an isomorphism. Before we start, let us make one helpful observation. Recall that a map in A/:B(x, y) is an equivalence class of diagrams in A

0

-----+

y'

-----+

0

-----+

x'

y

-----+

y"

-----+

x

-----+

0

1

-----+

x"

-----+

0

where the rows are exact and x", y' E ~. Since y is ~-local, by A.2.5.1 any map b - y , bE ::B must vanish. But we are given y' E ::B, and hence the monomorphism y' ---t y vanishes. Therefore y' = 0, andy= y". In other

A. ABELIAN CATEGORIES

330

words, if y is :B-local, morphisrns x -... y in A/:B are equivalence classes of diagrams

x' ____. Y

1 X

In the larger diagrams y

1

x' ____. y"

1 X

the map y --> y" must be an isomorphism. Next we want to prove the map

A(x, y) ----. Aj:B(x, y) an isomorphism. Let us first prove surjectivity. Take any map in Aj:B(x, y), that is a diagram

x' ____. y

1 X

as above. Now the map x' bicartesian square

-->

x is mono, so we may push out to get a

x' ____. y

1 X

1

____. z

and this extends to a map of exact sequences

0 ____. x'

____.

X

____. x"

1____. 1

0 ____. y

0 ----. y ----. z

0

---+

0

111

z ____. x"

But x" E :B, and by A.2.5.2, Ext_1(x",y) sequence

---+

=

---+

0. In other words, the exact

x''

---+

0

A.2. FORMAL PROPERTIES OF QUOTIENTS

331

splits, and using any splitting we have a commutative diagram x'~y

1

1

x~z~y

which gives us a map x

--->

y in A, in the equivalence class of x'~y

1 X

The map A(x, y) ~ Aj13(x, y)

is surjective. Next we want to prove it injective. Suppose we have a map x---> yin A, which becomes zero in A/13. This means that in the equivalence class there is a diagram x'~y

1 X

where x' ---> y vanishes. More precisely, this means that for x' C x with xjx' E 13, the composite x' C x ---> y vanishes. But then the exact sequence o~x'~x~x"~o

tells us that x ----. y must factor through x ---> x" ---> y. By hypothesis, x / x' = x" E 13 and y is 13-local; hence x" ---> y vanishes, and therefore so does the composite x ---> x" ---> y. D LEMMA A.2.8. Let A be an abelian category, 13 C A a localizant subcategory. Let F : A ---> A/13 be the projection, G its right adjoint. The counit of adjunction FG =? 1 is an isomorphism.

Proof: Let x E A, y E A/13. Then by adjunction Aj13(Fx,y)

A(x,Gy).

By Lemma A.2.5, Gy is 13-local. By Lemma A.2.7, A(x,Gy)

A/13(Fx, FGy).

Hence Aj13(Fx,y)

A(x,Gy)

Aj13(Fx, FGy).

332

A. ABELIAN CATEGORIES

In other words, the natural map Cy :

FGy

y

------+

induces, for every x E A, an isomorphism

Aj13(Fx, FGy)

------+

Aj13(Fx, y).

Since every object of A/13 is of the form Fx, it follows that the map Cy :

FGy

------+

y

must be an isomorphism in A/13.

0

LEMMA A.2.9. Let A and 'J' be any categories. Let F : A - - t 'J' be a functor, G : 'J' - - t A its right adjoint. The counit of adjunction

c: : FG

------+ 1

is an isomorphism if and only if G is fully faithful. Proof: For any x, y E 'J', composition with the counit gives a map

'J'(x, y)

'J'(e.,,y)

'J'(FGx, y),

and by adjunction

'J'(FGx, y)

A(Gx,Gy).

The composite is the natural map

'J'(x, y)

------+

A(Gx, Gy);

it is an isomorphism if and only if

'J'(x, y)

'J'(e.,,y)

'J'(FGx, y)

~

A(Gx,Gy)

is an isomorphism. To say that

'J'(x,y)

is an isomorphism for every x and y is to say that G is fully faithful, while

'J'(x, y)

'J'(e.,,y)

'J'(FGx, y)

will be an isomorphism for every x and y if and only if

c: : FG

------+

1

is an isomorphism.

0

PROPOSITION A.2.10. Let A be an abelian category, 13 c A a localizant subcategory. Let F : A - - t Aj'B be the projection to the quotient, G : Aj'B - - t A its right adjoint. The functor G : Aj'B - - t A is fully faithful, the counit of adjunction

c: : FG

------+

1

A.2. FORMAL PROPERTIES OF QUOTIENTS

333

is an isomorphism, and the objects isomorphic to Gt, t E Aj'B are exactly the 'B-local objects. More precisely, if y is a 'B-local object of A, then the unit of adjunction 7Jy :y

~

GFy

is an isomorphism.

Proof: By Lemma A.2.8, the counit of adjunction is an isomorphism

c:FG

~

1.

By Lemma A.2.9 this means that G is fully faithful. It only remains to show that if y E A is 'B-local, then the unit of adjunction

7Jy :y

~

GFy

is an isomorphism. In any case, we have that the composite

Fy ~ FGFy ~ Fy is the identity, while the second map is an isomorphism by Lemma A.2.8. Therefore the first must be its two-sided inverse; F7Jy : Fy

~

FGFy

is an isomorphism. Because y is 'B-local, Lemma A.2. 7 tells us that for any x E 'J,

A(x, y)

Aj'B(Fx, Fy).

In particular, if we let x = GFy,

A(GFy,y) ~ Aj'B(FGFy,Fy) is an isomorphism. There exists a unique map p-lCFy:

GFy

~

y

lifting

CFy : FGFy

~

Fy.

The fact that y is 'B-local means also that the map

A(y,y) ~ Aj'B(Fy,Fy) is an isomorphism. The composite

is a lifting of FrJy

c:Fy

Fy ~ FGFy ~ Fy, and the latter is 1 : Fy----. Fy. We deduce that

334

A. ABELIAN CATEGORIES

must be the identity. Finally, the object GFy is and so by Lemma A.2.7 the map

'B~local

by Lemma A.2.5,

A(GFy, GFy) ----) A/'B(FGFy, FGFy) is an isomorphism. The composite 'f/y

F-lEFy

GFy y ----) GFy maps to the identity under F, hence must be the identity. This establishes that 'fly has a two-sided inverse, namely p~leFy· D Proposition A.2.10 can also be turned into a construction of the adjoint. We have PROPOSITION A.2.11. Let A be an abelian category, 'B C A a Serre subcategory. Let F : A -----+ A/'B be the projection to the quotient. A right adjoint G : A/'B -----+ A will exist if and only if for every object t E A there is a morphism t -----+ y in A such that

A.2.11.1. y is

'B~local.

A.2.11.2. Ft-----+ Fy is an isomorphism in Aj'B.

Proof: If there is a right adjoint G : A/'B -----+A, then consider the unit of adjunction 'Tlt : t-----+ GFt. By Lemma A.2.5 the object GFt is 'B~local, while Lemma A.2.7 tells us that eFt : FGFt -----+ Ft is an isomorphism. The composite

Ft ~ FGFt ~Ft is the identity, making Fryt the .two-sided inverse of the invertible eFt. Now we need to prove the converse. Suppose every object t E A admits a map t -----+ y satisfying A.2.11.1 and A.2.11.2. We need to show that the functor F : A -----+ A/'B has a right adjoint G. Let t be any object of A/'B, which is the same as an object of A. We wish to show that the functor X

A/'B(Fx, Ft)

is representable, as a functor on A. But we can choose a map t -----+ y satisfying A.2.11.1 and A.2.11.2. By A.2.11.2 the map .Ft -----+ Fy is an isomorphism. Hence

A/'B(Fx, Ft)

A/'B(Fx, Fy).

But by A.2.11.1 the object yEA is 'B-local. Hence Lemma A.2.7 tells us that

A/'B(Fx,Fy) This proves the representability.

=

A(x,y). D

A.2. FORMAL PROPERTIES OF QUOTIENTS

335

PROPOSITION A.2.12. Let F: A ---? '.T be an exact functor of abelian categories, and G : '.T ---? A its right adjoint. If the counit of adjunction

c:: FG

---t

1

is an isomorphism, then '.T = Aj'B for a localizant subcategory 'B C A, and F is the projection. In fact, 'B is the full subcategory of all b E A with Fb isomorphic to zero. Proof: Define 'B to be the full subcategory of all b E A with Fb isomorphic to zero. Observe that every object isomorphic to an object in 'B is in 'B. Furthermore, if 0

---t

x'

---t

x

---t

x"

---t

0

is an exact sequence in A, the fact that the functor F is exact guarantees that 0

---t

Fx'

---t

Fx

---t

Fx"

~

0

is exact in '.T. But then Fx is isomorphic to zero if and only if both Fx' and Fx" are. In other words, x E 'B if and only if x', x" E 'B. The category 'B is a Serre subcategory of A; see Definition A.2.1. One can form the universal quotient functor A ---? Aj'B. The exact functor F : A ---? '.T factors (uniquely) as a composite of exact functors A

---t

A/'B

---t

'.T.

We wish to show that the functor Aj'B ---?'.Tis an equivalence. For any object t E '.T, the object Gt is 'B-local, by Lemma A.2.5. Let a be an object of A. The object GFa is 'B-local, and the unit of adjunction

"'a :a

---t

GFa

satisfies the property that the composite

Fa~ FGFa ~Fa is the identity. Since we are assuming that

FGFa ~Fa is an isomorphism, it follows that

FGFa is its two--sided inverse, in particular is invertible in '.T. But now consider the exact sequence 0

---t

k

---t

a~ GFa ~ q ~ 0.

The functor F is exact; hence we get an exact sequence 0

~ Fk

---t

Fa~ FGFa ~ Fq ~ 0.

336

A. ABELIAN CATEGORIES

Since Frya is an isomorphism, it follows that Fk = 0 = Fq. Thus k, q E 1j_ But this means that "'a is an isomorphism already in Aj1j. In other words, for every object a E A we have produced a morphism

'fla:a-+ GFa so that GFa is 1j-local, and "'a becomes an isomorphism in Aj1j. It follows from Proposition A.2.11 that the projection F' : A --> A/1j has a right adjoint G' : A/1j --> A. By Lemma A.2.8 the counit of adjunction e' : F'G' ===? 1 is an isomorphism. Lemma A.2.9 now tells us that both G and G' are fully faithful. In other words, G : 'J' ---+ A and G' : A/1j ---+ A are, up to equivalence, inclusions of full subcategories. Their left adjoints F and F' will agree if and only if G and G' agree; that is, if the images of G' and G

G'(A/1j) c A

G('J')

cA

are equivalent subcategories. For every object t E 'J', Lemma A.2.5 tells us that Gt is 1j-local; by Proposition A.2.10 this implies that Gt is isomorphic to an object of the form G'(p), for some p E Aj1j. Up to equivalence, G('J') C G'(A/1j). On the other hand, we showed above that the map

'fla:a-+ GFa becomes an isomorphism in A/1j. But GFa E G('J'). This means that the composite

G('J')

c

G'(A/1j)

c

A - + A/1j

is surjective. Since the composite

G'(A/1j)

c

A ~A/1j

is an equivalence by Lemma A.2.9, we deduce that G'(A/1j)

= G('J').

D

An additive functor F : A --> 'J' has a right adjoint G : 'J' ---+ A if and only if, for every t E 'J', the functor

'J'(F(-),t) :A0 P - + Ab is representable; that is,

'J'(F(x), t)

A(x, G(t))

for some G(t) E A, and naturally in x. This much is completely standard. Now observe LEMMA A.2.13. Let F : A ---+ 'J' be an additive functor of abelian categories. Suppose we are given a class I of objects in 'J' so that

A.2. FORMAL PROPERTIES OF QUOTIENTS

337

A.2.13.1. Every object t E 'J admits a copresentation (an exact sequence) 0

----t

t

i

----t

----t

j

with i,j E I.

A.2.13.2. For each i E I, the functor 'J(F(- ), i) : A 0 P

----t

Ab

is representable. Then the functor F has a right adjoint G : 'J ---+ A. Proof: For each i E I, choose an object G(i) E A so that

'J(F(x),i)

A(x,G(i))

with the isomorphism natural in x. Let t be an object of 'J. Choose a copresentation 0

----t

t

----t

i

----t

j

with i,j E I. The map i---+ j induces a natural transformation

'J(F(-),i)

----t

'J(F(-),j)

and hence a natural transformation

A(-,G(i))

----t

A(-,G(j)).

By Yoneda's lemma, this is induced by a morphism G(i) ---+ G(j). LetT be the kernel in the exact sequence 0

----t

T

----t

G(i)

G(j).

----t

We deduce a commutative diagram with exact rows 0----+

'J(F(-),t)

----t

'J(F(-),i)

----+

111 0----+

A(-,T)

----t

'J(F(-),j)

111

A(-,G(i))

----+

A(-,G(j))

The diagram allows us to identify the kernels; we get a natural isomorphism

'J(F(- ), t)

A(-,T) 0

REMARK A.2.14. For example, let F: A---+ 'J be an additive functor of abelian categories, and suppose the category 'J has enough injectives. Every object t E 'J admits an injective copresentation; that is, an exact sequence

0 ---+ t

----t

i

----t

j

A. ABELIAN CATEGORIES

338

with i and j injective. To prove the existence of the right adjoint G : 'J' ---+ A it suffices to show that, for every injective object i E 'J', the functor 'J'(F(- ), i) is representable. What is more, ifF is exact, then it is possible to check on the i's that the counit of adjunction is an isomorphism. LEMMA A.2.15. Let F :A ---+ 'J' be an exact functor of abelian categories. Suppose a right adjoint G : 'J' --+ A exists. Suppose I is a class of objects in 'J' satisfying A.2.13.1. We remind the reader: A.2.13.1: Every object t E 'J' admits a copresentation o~t~i---+j

with i,j E I. Suppose that for every i

E

I, the counit of adjunction c:i : FGi

----+ i

is an isomorphism. Then for every t E 'J', the counit c:t : FGt isomorphism.

--+

t is an

Proof: Let t be an object of 'J', and choose a copresentation o~t---+i---+j

with i, j E I. The functor G, being a right adjoint, is left exact. It takes the exact sequence above to an exact sequence 0

~

Gt

----+

Gi

----+

Gj.

By hypothesis, the functor F is exact. Hence the sequence 0

~

FGt

~

FGi

----+

FGj

is also exact. Now the naturality of c:t gives a commutative diagram 0 ~ FGt----+ FGi----+ FGj

o~

t

~

i----+

j

from which we immediately deduce that the counit c:t : FGt isomorphism.

---+

t is an D

These lemmas give us a practical criterion to identify when a functor F : A --+ 'J' is the map to a quotient by a localizant subcategory. PROPOSITION A.2.16. Let F: A ---+ 'J' be an exact functor of abelian categories. Suppose we are given a class I of objects in 'J' so that A.2.13.1: Every object t E 'J' admits a copresentation o~t~i---+j

with i,j

E

I.

A.2. FORMAL PROPERTIES OF QUOTIENTS

339

A.2.13.2: For each i E J, the functor 'J(F(- ), i) : A 0 P

--t

Ab

is representable. By A.2.13.2, whenever i E I we have an isomorphism 'J(F(- ), i)

'J( -, G(i)).

We have a map from a representable functor 'J (-, G (i)) to the functor 'J(F(-),i), and by Yoneda this correspond to an element of'J(FGi,i), in other words a map

c:i : FGi

--t

i.

Suppose further that all the maps c:i are isomorphisms. Then the projection F : A ---> 'J is equivalent to A """'-----+ A/13, where 13 C A is the full subcategory on which F : A ---> 'J vanishes.

Proof: By Lemma A.2.13, conditions A.2.13.1 and A.2.13.2 already imply that the functor F has a right adjoint G. We are supposing further that for i E J, the maps c:i: FGi

--t

i

are isomorphisms. By Lemma A.2.15 the counit of adjunction c: is an isomorphism. But then Proposition A.2.12 tells us that 'J = A/13. D Until now, we have mostly presented results permitting us to prove that some given map F :A ---> 'J can be identified as A ---> A/13. That is, if it turns out that F has a right adjoint G and that c: : FG ---> l is an isomorphism, we know that 'J = A/13. Proposition A.2.16 tells us how to check that the adjoint G exists, and that c: : FG ---> 1 is an isomorphism. We might find ourselves in the situation where we are given 'J = A/13. Can one give practical hints on how to produce the adjoint G? The next few lemmas should help. LEMMA A.2.17. Suppose F : A ---> A/13 is the projection of an abelian category A, to the quotient by a Serre subcategory 13. Suppose F has a right adjoint G. Then in every object a E A, there is a maximal subobject belonging to 13.

Proof: Let k be the kernel of the unit 'TJa : a ---> GFa. I assert that k is maximal, among subobjects of a belonging to 13. Because F is exact, Fk is the kernel of the isomorphism Frya :Fa---> FGFa. Hence Fk -::::: 0, that is k E 13. Next we need to show that k is maximal with this property.

A. ABELIAN CATEGORIES

340

Let b C a be a subobject of a, belonging to the category 13. We form the composite

b ----+ a ~ GFa. It is a morphism from bE 13 to the 13-local object GFa; by A.2.5.1, it must vanish. Hence b

c

k

Ker{1Ja: a----+ GFa}.

Therefore k is maximal.

D

NOTATION A.2.18. Suppose F : A ----+ A/13 is the projection of an abelian category A, to the quotient by a Serre subcategory 13. Let a be an object of A. If a has a maximal subobject belonging to 13, we will denote this maximal13-subobject am C a. Note that if b C a is another subobject betonging to 13, then the union buam also belongs to 13. By the maximality of am, we deduce =

that is b C am. In other words, am contains all other 13-subobjects of a, and is therefore unique. LEMMA A.2.19. Suppose F : A ----+ A/13 is the projection of an abelian category A, to the quotient by a Serre subcategory 13. Suppose the category A has enough injectives. If every injective object i E A contains a maximal 13-subobject im C i, then so does every object a EA.

Proof: Let a be an object of A. Since A has enough injectives, we may embed a in an injective i. By hypothesis, i E A contains a maximal 13subobject im c i. If b c a C i is a subobject contained in 13, then by the argument of Notation A.2.18, b C im. But then b

c

and a n im is a maximal 13-subobject of a.

c

a, D

PROPOSITION A.2.20. Suppose F : A ----+ A/13 is the projection of an abelian category A, to the quotient by a Serre subcategory 13. Suppose the category A has enough injectives. There is a right adjoint G: A/13 ----+A if and only if every injective object i E A has a maximal13-subobject im C i.

Proof: The necessity is clear. Lemma A.2.17 established that if the right adjoint G exists, then every object a E A (not just the injectives) has a maximal13-subobject am Ca. We need to prove the sufficiency. Suppose therefore that every injective object i E A has a maximal 13-subobject im C i. We must prove the existence of a right adjoint G : A/13 ----+ A.

341

A.2. FORMAL PROPERTIES OF QUOTIENTS

The objects of A and Aj'B are the same. Pick an object a E Ob(A) = Ob(Aj'B). By Lemma A.2.19, it has a maximal 'B-subobject am C a (even though a need not be injective). We can find an exact sequence in A 0

-----+

aj am

-----+

i

-----+

j,

with i and j injective in A. We can complete this to a commutative diagram with exact rows 0 -----+ a/ am -----+ i -----+ j

1

1

1

0 -----+ Ga -----+ i / im -----+ j / im and I assert that this definition of Ga works; that is,

Aj'B(Fx,a)

A(x,Ga).

Now we must prove this fact. First of all, we are given a map, in A,

a - a f a m - Ga. It clearly is an isomorphism in A/'B. The map a -----+ afam is division by am E 'B, hence an isomorphism in Aj'B. And if we apply the exact functor F to the diagram 0

-----+

afam

1

0

------+

Ga

we get a diagram

oo-

F{afam}

1

FGa

-----+

-

i

j

------+

1 i/im

1 -----+

Fi

jfjm

------+

111 F{i/im}

Fj

111 ------+

F{j/jm}

where the rows are exact, and two of the columns are clearly isomorphisms. Hence so is the third; the map a/am -----+ Ga is an isomorphism in A/'B. Therefore the composite

a - a f a m - Ga is also an isomorphism in A/'B. This gives us a natural map

A(x, Ga) -

Aj'B(Fx, a).

We take a map in A(x, Ga), view it as a map in A/'B(Fx, FGa), and compose with the inverse of the isomorphism Fa -----+ FGa. We need to prove that this natural map is an isomorphism A(x, Ga) -----+ A/'B(Fx, a).

A. ABELIAN CATEGORIES

342

First let us prove surjectivity. Suppose therefore that we are given a map in A/'B(Fx, a); we need to show that it is the image of something in A(x, Ga). But now recall that a map in A/'B(Fx, a) is an equivalence class of diagrams in A 0

----->

x'

----->

x

----->

0

1

0 -----> a' -----> a -----> a''

----->

x''

-----+

0

with a', x" in 'B. Because a' is a 'B-subobject of a, it is contained in the maximal one, am. The above is therefore equivalent to a diagram 0 ----->

x'

----->

x

----->

x"

----->

0

1

0 -----> am -----> a -----> a/am -----> 0

Now recall the exact sequence 0 -----> a/ am ----->

----->

j

that we chose above. The composite x' ~a/am~ i is a map from x' to an injective object i, and therefore factors through the inclusion x' ~ x. We deduce a commutative square with exact rows 0

0

x'

----7

----7

1

----7

a/am

X ----->

x"

1

1

i

----7

----7

----->

0

j

which we can further extend to a diagram 0

----7

x'

0

----7

a/am

----7

i

0

----7

Ga

----7

i/im

----7

X

----7

1

1 1

x"

----7

0

1 ----7

j

----->

j/Jm

1

1

Now x" ~ j is a map from an object x" E 'B to j. Its image is therefore a 'B-subobject of j, hence contained in the maximal Jm c j. Hence the composite x" ~ j ~ j /Jm must vanish. From the commutative diagram, the map x ~ i/im ~ j /Jm must also vanish, and hence x ~ i/im must factor through Ga C i/im. This produces a map x ~ Ga, in the

A.2. FORMAL PROPERTIES OF QUOTIENTS

equivalence class of the composite of the map x

0

0 - - - - t a' and the map

------+

a

a

------+

----t

x'

----t

a given by the diagram x - - - - t x" - - - - t 0

----t

0

1

----t

a"

a''

----t

343

afam

------+

------+

Ga.

In other words, we have shown the surjectivity of A(x, Ga)

----t

Aj'B(Fx, a).

It only remains to prove the injectivity. Let f : x ------+ Ga be a map whose image in Aj'B(Fx, a) vanishes. That is, the composite ofF f with the isomorphism FGa ------+ Fa vanishes. Hence F f must vanish. It follows that the image of f : x ------+ Ga is a 'B-subobject of Ga. By consrtuction, Ga embeds in i/im. Therefore Im(f) is a 'B-subobject of i/im. But then Im(f) = 0, and so f = 0. This proves that the kernel of the map

A(x, Ga)

----t

Aj'B(Fx, a)

is trivial.

0

After this extensive discussion of how to construct the adjoint, maybe we should briefly comment on the implications of its existence. If the functor F : A Aj'B has a right adjoint G, then F must preserve coproducts. We note LEMMA A.2.21. Let A be an abelian category satisfying [AB4]; coproducts exist in A, and coproducts of exact sequences are exact. Let 'B be a Serre subcategory. The functor F: A ------+ Aj'B preserves coproducts if and only if 'B is closed under the formation of A-coproducts of its objects.

Proof: Suppose F preserves coproducts. Let {b>., A E A} be a set of objects in 'B. For each b>. we have F(b>.) ~ 0. But F preserves coproducts. Hence

F

(M

b>.)

M

F(b>.)

0.

Therefore ll>.EA b>. lies in the subcategory 'B on which the functor F vanishes. Coversely, suppose the category 'B is closed under coproducts. We wish to show that the functor F :A ------+ Aj'B preserves coproducts. Let {a>., A E A} be a set of objects in A. We can form the coproduct in A. We wish to show that it also has the universal property of a coproduct in Aj'B.

344

a>.

A. ABELIAN CATEGORIES

Suppose therefore that, for every >. E A, we are given a morphism yin Aj'B. That means an equivalence class of diagrams in A

---t

0

o ----+

----+ a~ -------+ a>. -------+ a~ ----+

1

y~ ----+ y ----+ yr ----+

0

o

where the rows are exact and a~, y~ E 'B. By hypothesis, the coproduct of the objects y~ lies in 'B. Therefore for each >. E A the diagram above is equivalent to

II y~

1

----+

y ----+ y" ----+

o

.AEA

and these assemble to 0

----+

II a~

----+

.AEA

II y~

----+

y ----+

1

y"

II a>.

-------+

.AEA

----+

II a~

----+

0

.AEA

0

.AEA

which is a map il>.EA a>. ---t y in Aj'B. To give a map il>.EA a>. ---t y in Aj'B is to give a diagram 0

0

----+

A'

----+

1

----+

II a>.

-------+

A''

----+

0

.AEA

y' ----+ y ----+ y" ----+

0

where the rows are exact and A", y' E 'B. For each >. E A we can extend to a larger diagram

0

0

----+ a'>. -------+

0

----+

1

A'

1

----+

----+

y' ----+ y ----+ y" ----+

a>.

1

II

.AEA

0

-------+ a" >. -------+

0

A"

0

a>. -------+

1

-------+

A.3. DERIVED FUNCTORS OF LIMITS

where the rows are exact and a~, A", y' E to be the kernel of the composite a.>..

1

ll a.>..

~.

345

We can for example take

a~

--+A" .

.>..EA

To say that in

A/~

the composite

a.>..

--+

ll a.>..

Y

--+

.>..EA

vanishes is to say that, in the diagram above, the image of a~ ---+ y" lies in~. Replacing a~ by the kernel of a~---+ y", we may assume that in the diagram above, the composite a~ - - +

A'

--+

y"

vanishes. But now the diagram 0--+

ll a~--+ ll a.>..~ ll a~~o

.>..EA

.>..EA

1

0--+

0

--t

y'

--t

y

A'

--+

y"

ll a.>..~

1

A"

~

0

.>..EA

1

----t

.>..EA

0

--t

immediately implies the vanishing, in

A/~,

of

ll a.>..--+ y .

.>..EA

D

A.3. Derived functors of limits Let A be an abelian category satisfying [AB3*]; products exist in A. Then of course all small limits exist in A. Let 1 be a small category. The category eat (J0 p' A) is the category of all functors :JDP - - - - t

A.

This is of course an abelian category. There is a functor ~: eat(1°P,A)

----t

A,

A. ABELIAN CATEGORIES

346

taking an object F E eat(:J 0 P,A) to its limit. The limit exists since we are assuming A satisfies [AB3*). The functor lim is trivially left exact. It +---

is natural to wonder if it has right derived functors. There are two wellknown sufficient conditions which guarantee the existence of right derived functors. The first is that the category eat (J0 p' A) have enough injectives. We will discuss this further in the next Section. The second is that the category A satisfy [AB4 *). In this Section we remind the reader how this goes. But we wish to treat a slightly more general case. DEFINITION A.3.1. Let A be an abelian category. Let a be an infinite cardinal. We say that A satisfies [AB3*(a)j if A is closed under products of fewer than a of its objects. We say that A satisfies [AB4*(a)j if it satisfies [AB3*(a)j, and products of fewer than a exact sequences are exact.

LEMMA A.3.2. Let a be an infinite cardinal. Let A be an abelian category satisfying [AB4*(a)J. Suppose :J is a small category of cardinality< a. This means there are fewer than a morphisms in :J. Then the functor

~: eat(:J 0 P,A)

-------t

A

has right derived functors, denoted limn. Furthermore, the limn are June+---

+---

torial in exact functors preserving products. Precisely, suppose

A.3.2.1. A, 'J' are abelian categories satisfying [AB4*(a)j. A.3.2.2. ¢ : A Then, for any F : :J0 P

----7

----7

'J' is an exact functor preserving products.

A,

limn{¢F}

+---

Proof: Recall that the nerve of the category :J is defined as the simplicial set :N.(:J), where :Nk(:J) (the k-simplices) are sequences of k composable morphisms If the cardinality of :J is < a, so is the cardinality of each :Nk(:J). Now A satisfies [AB3*(a)). Given a functor F : J0 P ----7 A, we form a chain complex

where

II

F(i0 ).

{i 0 -->i 1 -->···-->ik}E:Nkp)

The products exist, being products of fewer than a objects in A. The differential ak : N k (F) ----7 N k+ 1 (F) is given by the usual alternating sum.

A.3. DERIVED FUNCTORS OF LIMITS

347

That is,

ak

k+l

I)-lFaL j=O

and 8~ is the map induced by deleting the lh term in the sequence

We define Tn(F) to be the nth cohomology of the complex

No(F)

~ N1(F) ~ N2(F) ~ · · ·

and I assert that, if A satisfies [AB4*(ao)], then the collection of functors Tn are the right derived functor of lim. +--

Suppose we have a short exact sequence in eat(:JoP, A), that is an exact sequence of functors :J 0 P __, A

0

--------+

F'

--------+

F

--------+

F''

--------+

0.

By [AB4*(a)], products of fewer than a exact sequences are exact in A, and so the sequence

is exact. This being true for all k, we have a short exact sequence of chain complexes, hence a long exact sequence in cohomology. It follows that T is a J-functor. It is also clear that Tn commute with products. That is,

To prove that the Tn give the derived functor of lim , it suffices to show +--

two things.

A.3.2.3. There is a natural isomorphism lim = T 0 . +-----

A.3.2.4. Every object F E eat(:J 0 P,A) can be embedded in an object F', with Tn(F') = 0 for all n > 0. The proof that T 0 is naturally isomorphic to lim is obvious. It therefore +--

remains to prove A.3.2.4. We must show that any object can be embedded in a T -acyclic. For every i E :J and every object a E A, we define a functor F~ : :J __,A by the rule:

IT a.

J(i,j)

A. ABELIAN CATEGORIES

348

That is, F~ (j) is the product over all morphisms i complex

----+

j of a. The chain

is obviously homotopy equivalent to the chain complex a~o~O------+···

Hence the functor F~ is T -acyclic. Since the functor T commutes with products, the product of any F~ 's is T -acyclic. But any functor G can be embedded in the product over all i of Fbi· This proves that, in an abelian category satisfying [AB4*(a)], the functors rn compute limn. But rn (F) is defined as the cohomology of the chain +--

complex

No(F) ~ N1(F) ~ N2(F) ~ · · · and any functor ¢;:A complex to

----+

'J' preserving products will send the above chain

No(r/JF) If the functor ¢; is also exact, ¢; of the cohomology of a chain complex is the cohomology of ¢; of the chain complex; In our specific case above,

limn{ cf;F}

+--

¢limn F. 'J'(-,~- 1 Y1)Is ~ ':T(-,Yo)ls

1

':T(-,X)Is

1

0 is exact; it is, by construction, the sequence computing colimn of an a---+

filtered colimit in £x(S 0 P,Ab). In particular

'J'(-,~- 1 Y1)Is - - > ':T(-,Xo)ls - - > 'J'(-,X)Is ~ 0

is exact. On the other hand, from the triangle ~- 1 y1-->

Xo--> X1 ~ Y1

we deduce an exact sequence 'J'(-,~-IY1)Is --> ':T(-,Xo)ls ~ ':T(-,X1)1s·

We have a commutative diagram with exact rows ':T(-,~- 1 YI)Is - - > ':T(-,Xo)ls - - > ':T(-,X)Is

111 ':T (-,~-I Y1) Is

~o

111 ~

':T(-,Xo)ls - - > ':T(-,X1)1s

which may be completed, uniquely, to ':T ( -, ~- 1 Y1) Is

~

':T ( -, Xo)ls - - > ':T (-,X) Is

111

~o

fl

111

':T ( -, ~- 1 YI) Is - - > ':T(-,Xo)ls - - > ':T(-,XI)Is The composite

':T(-,Xo)ls - - > ':T(-,X)Is __!____. ':T(-,Xl)ls

'J(-,g)ls

':T(-,X)Is

is the natural surjection

':T(-,Xo)ls - - > ':T(-,X)Is· In other words, the two composites

{'J'(-,g)Js} 0 f ':T(-,Xo)ls - - > ':T(-,X)Is are equal, and we deduce that

1

':T (-,X) Is

374

B. HOMOLOGICAL FUNCTORS INTO [AB5"'] CATEGORIES

'J(-,X)Is ~ 'J(-,X1)1s

'J(-,X)Is

'J'(-,gJis

must compose to the identity. Hence 'J (-, X 1) Is contains 'J (-, X) Is as a direct summand. The other direct summand has no choice. From the exact sequence

'J ( -, E-1 Y1) Is

1

'J( -, Xo)ls----+ 'J (-, X1)ls

----t

'J( -, Y1)1s

----t

'J (-, EXo)ls

it clearly has to be the kernel of

which is also the image of

Suppose therefore, by induction, that up to some integer n, we have constructed a sequence Xo

----t

X1

-----+ . . .

----t

Xn-1

.----t

Xn

-----+

X.

We already know that there is a map 'J(-,X)Is---+ 'J(-,X1)Is, so that the composite

'J(-,X)Is

-----+

'J(-,Xdls

----t

'J(-,X)Is

is the identity. So 'J (-,X) Is must be a direct summand of each 'J( -,Xi) Is, 1 :::; i :::; n. Suppose further that

B.l.3.5. Fori < n, the other direct summand is the kernel of 'J(-,Xi)ls---+ 'J(-,Xi+l)ls· B.l.3.6. For each i :::; n, we have a triangle

B.l.3.7. The images of

'J(-,Xi)ls

----t

'J(-, ~)Is

'J (-,~)Is

----t

'J (-, EXi-1) Is

agree, respectively, with

Im{ 'J(-,E- 1 ~+1)1s---+ 'J(-,~)Is} Im{ 'J(-,~)Is---+ 'J(-,E~-1)1s}·

B.l. A FILTRATION

375

B.1.3.8. Fori< n, the isomorphisms of the image

Im{ 'J(-,E- Yi+1)1s----+ 'J(-,Yi)ls} 1

with

Im{ 'J(-,E- 1Yi+l)ls----+ 'J(-,Xi)is} Im{ 'J(-,Xi)ls----+ 'J(-, Yi)ls}

are compatible. That is, the composite

'J(-,E- 1 Yi+I)Is------* 'J(-,Xi)ls-----+ 'J(-,Yi)ls is the given map. After all, the sequence of the Y 's is given. By B.1.3.4, this has to mean that the composite

E- 1Yi+l ------* xi -----+ Yi is also correct in 'J. Maps Y ----+ Z in 'J are in 1-to-1 correspondence with their images in £x (gop, Ab). We wish to show that the sequence of X's may be continued to n By the induction hypothesis B.1.3.6, we have a triangle

+ 1.

Xn-1 ------* Xn ------* Yn -----+ EXn-l· Hence an exact sequence

'J(-,Xn-1)1s------* 'J(-,Xn)ls-----+ 'J(-,Yn)ls· By hypothesis B.1.3.5, the image of 'J (-, xn-l) Is ----+ 'J (-, Xn)ls is the direct summand 'J (-,X) Is C 'J (-, Xn_ 1) Is. After all,

'J(-,Xn-I)Is

KtJJ 'J(-,X)Is,

where K is the kernel of the map. From the exact sequence, 'J(-,Xn)ls is a direct sum of 'J (-,X) Is and the image of

'J(-,Xn)ls ------* 'J(-, Yn)is· Again by induction hypothesis, this image is also the image of

'J (-, E- 1yn+1) Is ------* 'J( -, Yn)is· This image is a direct summand of 'J(-,Xn)ls· It is the kernel of the map

'J(-,Xn)ls ------* 'J(-,X)Is induced by xn ----+X. It follows that the map from 'J ( -, E- 1 yn+1) Is to this image also gives a map to the kernel above. We have an exact sequence

'J(-,E- 1Yn+l)ls ------* 'J(-,Xn)ls -----+ 'J(-,X)Is, and by construction, the composite

'J(-,E- 1 Yn+l)ls------* 'J(-,Xn)ls-----+ 'J(-,Yn)ls

376

B. HOMOLOGICAL FUNCTORS INTO [AB5"'] CATEGORIES

is right. Hence B.l.3.8 is proved for n+ 1, modulo the fact that we have yet to define a map :E- 1 yn+1 ---+ xn in 'J lifting the map constructed above in cx(S 0 P,Ab). But now B.l.3.4 applies. The map

'J{-,:E- 1Yn+l)ls

-->

'J(-,Xn)ls

is a morphism from a 'J (-, Y) Is, and must come from a morphism in 'J

:E- 1yn+1

--t

xn.

The vanishing of the composite

'J(-,:E- 1Yn+1)1s

---->

'J(-,Xn)ls

-->

'J(-,X)Is

implies the vanishing of

:E-1 yn+1

---->

Xn ------+ X.

Form the triangle

:E- 1yn+1

---t

xn ------+ xn+1 ------+ yn+1·

The vanishing of

:E-1 yn+1 ------+ Xn ------+ X means that Xn

---+

X must factor as Xn

---->

Xn+1 _ _ . X.

Finally, the map 'J(-,:E- 1Yn+ 1)1s ---+ 'J(-,Xn)ls was constructed so that its image is precisely the image of 'J (-, :E- 1Yn+ 1) Is ---+ 'J (-, Yn) Is· The triangle

:E- 1yn+1

---t

xn ------+ xn+1 ------+ yn+1

gives an exact sequence

'J(-,Xn+1)1s------+ 'J(-,Yn+l)ls------+ 'J(-,:EXn)ls· This means that the image of 'J ( -, Xn+l) Is ---+ 'J (-, Yn+ 1) Is must be indentified with the kernel of 'J (-, Yn+ 1) Is ---+ 'J (-, :EYn)ls, which is also the image of 'J (-, :E- 1Yn+2) Is ---+ 'J ( -, Yn+l) Is· Inductively, this allows us to construct the sequence

x1

---->

x2 ------+ x3 __. ...

which maps to X. There is therefore a map from the homotopy colimit to X. Applying the homological functor

'J ------+ cx(S 0 P,Ab) to this sequence, we get

'J(-,X1)1s

---->

'J(-,X2)1s ------+ 'J(-,X3)1s

- - > .. ·

B.l. A FILTRATION

377

and by construction, this is the direct sum of the two sequences ----+

~

----+

K1(-)

1

1

'J(-,X)Is ~ 'J(-,X)Is

---+ 0

0

K2(-)

---+

It follows that the colimit of this sequence is 'J (-, X) I3 , while its colim1 ----7

vanishes. We get an exact sequence

II 'J (-,Xi) Is 00

0

----+

II 'J (-, Xi)ls 00

1- shift

i=O

i=O

1

'J(-,X)Is Now we remind ourselves that the functor 'J----+ products; the above can be written as 1- shift

'J (-,

ex(S

:fi •=0

0

---+

o.

P,Ab) respects co-

xi)

1

Is

'J(-,X)Is

---+

0

But the triangle

i=O

~X

i=O

gives another exact sequence 'J (-,

:fi

•=0

xi)

11s

shift 'J (-,

:fi

xi)

•=0

1---. s

'J ( -,

~mX) Is

and the two must clearly agree; the map 'J( -, Hocolimxi)ls is an isomorphism. Complete Hocolim Xi

---+

X

----+

'J(-,X)Is

~

Xi ----+ X to a triangle

----+

Z

---+

~ {~Xi} .

378

B. HOMOLOGICAL FUNCTORS INTO [AB5"] CATEGORIES

The homological functor to ex(S 0 P,Ab) gives an exact sequence, from which we easily deduce that 'J' (-, Z) I3 = 0. Since S generates, Z = 0 D and X is isomorphic to Hocolim Xi.

B.2. Abelian categories satisfying [AB5"'] Let 'J' be a triangulated category satisfying [TR5]. In this Section, we will be studying functors into triangulated categories satisfying weak versions of [AB5]. We first need to define the abelian categories we are dealing with. DEFINITION B.2.1. An abelian category A is said to satisfy [AB5"' J if it satisfies [AB4}, and a-filtered colimits in A are exact. LEMMA B.2.2. Let a be a regular cardinal. Let 'J' be an a-compactly generated triangulated category. Let X be an object in 'J', and suppose we have a sequence

Xo

~

xt

~

x2

- - - + ...

as in Lemma B.1.3. We remind the reader: B.l.3.1: X ~Xi. B.l.3.2: Put Y 0 = X 0 . For any i ~ 1, let Yi come from the triangle

xi-1 ~

xi

~

Yi

---+

Exi-t·

B.l.3.3: The composites

Yi

~ EXi-t - - - + EYi-t

give a chain complex

...

~

E-2y2

~

E-tyt

---+

Yo.

The entire collection of data can be so chosen, so that there exists an a-filtered functor f : :J ---+ 'J'"' and the sequence of Y 's is exactly N(f), the realisation of the nerve of the functor f. Let H : 'J' ---+ A be a homological functor, preserving coproducts. Suppose the category A satisfies [AB5"']. Then the sequence

H (E- 1 Y1 ) ~ H (Y0 )

---+

H(X)

---+

0

is exact in A.

Proof: Let H : 'J' ---+ A be a functor satisfying the hypothesis of the Lemma; that is, the category A satisfies [AB5"'] and H is homological and respects coproducts. Observe first that for every integer n E Z, the functor HEn also satisfies the hypothesis; after all, E : 'J ---+ 'J respects coproducts, and up to sign also respects triangles. We may take the product over all n E Z of these. We deduce a functor from 'J to the category of graded objects in A, also satisfying the hypothesis of the Lemma. Replacing

B.2. ABELIAN CATEGORIES SATISFYING [AB5"]

379

A by the category of graded objects in A, and H by the product of all {H~n,n E Z}, we may assume that the abelian category A admits an automorphism ~, and that we have a natural isomorphism H~~~H,

that is, H commutes with ~- From now until the end of the proof, we assume this is the situation. Recall that the chain complex ... ~ ~-2y2 -------+ ~-1y1 ~

Yo

was chosen isomorphic to N(f), for some a-filtered functor The homological functor H respects coproducts. Hence · · · ~ H (~- 2 Y2 )

-------+

f : :J

----+

'J'01 •

H (~- 1 Y1 ) ~ H (Y0 )

is just N(H f). But by Lemma A.3.2, N(H f) is a chain complex computing colimn, for the a-filtered functor H f : :J ----+ A. Since A satisfies [AB5 01 ), ----+

colimn vanishes if n ----+

> 1. It follows that the sequence -

-------+ H (~- 2 Y2 ) -------+ H (~- 1 Y1 ) -------+ H

(Y0 ) ~ Z ~ 0

is exact, where Z is defined to be the cokernel of H (~- 1 Y1 )

Recall that X 0

H (Yo) ·

-------+

= Y0 , and we have, in 'J', a triangle

Xo

-------+

X1 -------+ y1

~ ~Xo.

This gives an exact sequence H (~- 1 Y1)

-------+

H (X0 ) ~ H (X1) .

We have a commutative diagram with exact rows

H (~- 1 Y1 )

H (X0 )

-------+

-------+

which means we can extend uniquely to H (~- 1 Y1 )

H (Y0 )

-------+

111 H (~- 1 Y1 )

-------+

~o

H (X1)

z

~o

1

111 -------+

z

H (X0 ) ~ H (X1 )

This gives a monomorphism Z

----+

H (X 1 ), and the exact sequence

H (~- 1 Y1 ) ~H (X0 ) - - - H (X1 ) ~H (Y1 ) ~H (~X0 )

allows us to identify the cokernel of Z H (Y1 ) ----+ H (~X0 ).

----+

H (X1 ) with the kernel of

380

B. HOMOLOGICAL FUNCTORS INTO [AB5"'] CATEGORIES

Now we propose to prove, by induction on n 2::: 1, that B.2.2.1. The composite Z -------. H (X1 )

-------.

H (Xn)

is a monomorphism, and its cokernel is the kernel of the morphism H (Yn) ---+ H (EYn_ 1 ). What is more, the identification of the cokernel with a subobject of H (Yn) is from the natural map, the one induced by the triangle Xn-1 -------. Xn -------. Yn -----) EXn-1·

B.2.2.2. The composite Z -------. H (X1) -----) H (Xn)

is a split monomorphism, and where Kn is the kernel of H (Xn) -------. H (Xn+1) . Proof: The strategy of the proof will be to show that

{B.2.2.1 for n 2::: 1}

===?

{B.2.2.1 for n + 1} 1\ {B.2.2.2 for n}.

Since we already know B.2.2.1 for n = 1, this suffices. Suppose therefore that we know B.2.2.1 for n. The triangle Xn-1 -------. Xn -------. Yn -----) EXn-1

gives a map Xn---+ Yn, and we have an exact sequence 0 -----) Z -------. H (Xn) -----) H (Yn)

with the image of H(Xn) ---+ H(Yn) being the kernel of H(Yn) ---+ H(EYn_ 1). This kernel is also the image of H(E- 1Yn+l) ---+ H(Yn)· By hypothesis on the X's and Y's, the map E- 1 Yn+l ---+ Yn factors as

E-1yn+1 _______. Xn -----) Yn. Hence, in the category A, we have a factorisation H

(E- 1Yn+1)

-------. H (Xn) -----) H (Yn) .

Let E- 1 Kn+ 1 be the kernel of H(E- 1 Yn+ 1) ---+ H(Yn), Kn the image of H(E- 1Yn+ 1)---+ H(Yn), which is also the kernel of H(Yn)---+ H(EYn_ 1). We have a commutative diagram with exact rows

0 -------. E-1 Kn+1 -------. H (E-1 yn+1) -----) Kn -----) 0

0 -------.

1 z

1

1

1

381

B.2. ABELIAN CATEGORJES SATISFYING [AB5"']

From the triangle

xn

---t

xn+l

----+

yn+1

---t

~xn

we have a diagram with exact rows and columns 0 0

1

1 H

(~- 1 Xn+l)

---t

~-1Kn+1

---t

1 H (I::-1 Yn+1)

11 H (~-1Xn+1)

----+

z

----+

1 H(Xn)

1

----+

1 0 But the image of the composite H (I:-2Yn+2)

----+

H(Xn+l)

11 ---t

H(Xn+l)

1 Kn

1

Kn

---t

1 0

H (I::-1 Xn+1)

--->

H (1::- 1Yn+l)

is the kernel of

H (~- 1 Yn+1)

----+

H (Yn) '

that is I;- 1Kn+l C H (~- 1 Yn+l). The commutative diagram

H (I:-2Yn+2)

----+

H (I::-1 Xn+l) ___. 1

1

I:;-1 Kn+1

1

H (1::-1 Xn+l) ___. H (~-1 yn+l) shows that the composite

H (I:-2Yn+2) ----. H (I::-1 Xn+l)

--->

I:;-1 Kn+l

must be surjective. The map H (I::- 1Xn+ 1) -----t I:;- 1 Kn+l is onto, and hence the map I:;- 1Kn+ 1 ---> Z is zero. We deduce a short exact sequence

0

----+

Z

----+

H (Xn+l)

--->

This proves B.2.2.1 for n + 1. Finally, in the commutative square

I:;-1 Kn+l

1

----+

Kn+ 1

Z

1

--->

0.

382

B. HOMOLOGICAL FUNCTORS INTO [AB5"] CATEGORIES

we now know that the map E- 1 Kn+l equal composites

------*

Z vanishes. Hence so do the

z H

1

(E- 1 Yn+1)

----t

H (Xn)

It follows that the map H (E- 1 Yn+ 1) ------* H (Xn) must factor through the cokernel of E- 1 Kn+ 1 ------* H (E- 1 Yn+ 1), which is Kw We deduce a map Kn ------* H (Xn) splitting the exact sequence

0

-----+

Z

-----+

H (Xn)

----t

Kn

----t

0.

And the fact that H (E- 1Yn+l)

-----+

vanishes, while H (E- 1 Yn+ 1) Kn

-----+

Kn ------*

H (Xn)

-----+

----t

H (Xn+l)

Kn is surjective, tells us that

H (Xn)

----t

H (Xn+l)

vanishes. Hence we have B.2.2.2 for n. By induction, we have proved B.2.2.1 and B.2.2.2 for all n :2: 1. D Now, to end the proof of the Lemma, observe what we now know about the sequence Xo

-----+

X1

X2

-----+

We know that, for n :2: 1, H(Xn) isomorphism of the sequence

- - - - t ...

Z EEl Kn, and that this gives an

with the direct sum of the two sequences 1

z~

----t

Hence colimH (Xn) = Z, while colim 1 H (Xn) = 0. There is an exact ------* ------* sequence

n=O

From the triangle

n=O

B.2. ABELIAN CATEGORIES SATISFYING [AB5"']

383

we deduce an exact sequence

H(X)

Since H respects coproducts, the two exact sequences must agree, and we have Z = H(X). 0 COROLLARY B.2.3. Let a be a regular cardinal. Assume 'J' is an acompactly generated triangulated category. Let A be an abelian category satisfying [AB5a}. Let X be an object of 'J'. Suppose H : 'J' ----+ A is a homological functor respecting coproducts. Then there exists an object Y E 'J', with

y

=

II s>.

>.EA

where the objects s>. lie in S = 'J'O!. There is a map Y H(Y)

~

----+

X in 'J', and

H(X)

is an epi in A. Proof: By Lemma B.l.3, there is a sequence

Xo

·~

xl

~

x2

--+

satisfying some special properties we will not repeat fully. Among these properties, we have also an object :E- 1Y1 and two maps

:E-lyl

~

Xo

--+

X

which compose to zero. By Lemma B.2.2, these induce a short exact sequence in A

H(:E- 1Y1) ~ H(X0 ) ~ H(X)--+ 0. But X 0 = Y0 is a coproduct of objects inS. In other words, the assertions of the Corollary are satisfies if Y = Y0 = X 0 , and the map X 0 ----+X is as 0 in Lemma B.l.3. COROLLARY B.2.4. Let a be a regular cardinal. Assume 'J' is an acompactly generated triangulated category. Let A be an abelian category satisfying [ABsaj. Suppose H : 'J' ----+ A is a homological functor respecting coproducts.

384

B. HOMOLOGICAL FUNCTORS INTO [AB5"'] CATEGORIES

Then for any a-phantom map X inA.

----t

Z in 'J, H(X)----> H(Z) vanishes

Proof: We should perhaps remind the reader; a map a-phantom if the induced map

'J(-,X)Is

~

f :X

----t

Z is called

'J(-,Z)Is

vanishes in ex(S 0 P,Ab). See Definition 6.5.7. By Corollary B.2.3, we may choose Y a coproduct of objects in S, and a map g: Y ----t X, so that H(g) : H(Y) ----t H(X) is surjective in A. But now the composite

'J(-,Y)Is

'J"(-,g)ls

'J(-,X)Is

'J"(-,f)ls

'J(-,Z)Is

vanishes, because 'J(-,f)ls does. On the other hand, Y is a coproduct of objects inS, and by Lemma B.l.1, maps Y----> Z correspond 1-to--1 with maps 'J (-, Y) Is ----t 'J (-, Z) Is. Therefore the composite Y __!!____, X ~ Z vanishes in 'J. But then so does

H(Y) ~ H(X) ~ H(Z). By Corollary B.2.3, the map H(g) is surjective; we conclude that H(f) must vanish. D THEOREM B.2.5. Let a be a regular cardinal. Let 'J be an a-compactly generated triangulated category. Then the functor

'J ~ ex({'J"'YP,Ab) is universal among coproduct-preserving homological functors to {AB5"'j abelian categories A. Any coproduct-preserving homological functor 'J ----t A factors uniquely, up to canonical equivalence, as

'J ~

ex( {'J"'tP,Ab)

~ A.

Furthermore, any natural transformation of coproduct-preserving homological functors 'J ----t A factors through a natural transformation of the coproduct-preserving exact functors ex ( {'J"'Yp, Ab} - A . Proof: By Theorem 5.1.18, any homological functor factors as ::JI

A('J) -----=----. A. Moreover, by Lemma 5.1.24, 'J ----t A preserves coproducts if and only if A('J) ----t A does. Now put S = 'J"'. By Proposition 6.5.3, the natural functor

'J

~

B.3. HISTORY OF THE RESULTS IN APPENDIX B

is a coproduct-preserving Gabriel quotient map. A('J') ----+A factors further as A('J)

--->

t:x(goP,Ab)

385

To show that a map

----+

A

it is necessary and sufficient to prove that the map A('J) ----+ A annihilates any object in the kernel of A('J)----+ t:x(goP,Ab). But in Lemma 6.5.6 we computed the kernel of A('J) ----+ £x (gop, Ab). In the description D('J) = A('J), the objects mapping to zero are precisely the a-phantom maps. Now let H be a coproduct-preserving functor into an [AB5"'] abelian category A. By virtue of being homological, it factors through A('J) ----+ A. Corollary B.2.4 tells us that H takes a-phantom maps to zero. The map A('J) ----+ A kills the kernel of A('J) ----+ t:x(goP,Ab). Thus H factors, uniquely, through the Gabriel quotient £x (gop, Ab). The statement about natural transformations comes about because both Freyd's A('J) and Gabriel quotients behave well with respect to natural transformations. D

B.3. History of the results in Appendix B In the case a= N0 , and where 'J is the homotopy category of spectra, the result is in the recent work of Christensen and Strickland [9]. The argument they give is more down-to-earth and concrete. But it does not seem to generalise. It does not even generalise to other triangulated categories, but with a still equal N0 . I would like to thank Christensen for emphasizing the relevance of the work.

APPENDIX C

Counterexamples concerning the abelian category A('J) C.l. The submodules pi M Let R be a discrete valuation ring. That is, R is a commutative, regular, noetherian local ring of height 1. The (unique) maximal ideal of R is principal. Let p be a generator. That is, Rp C R is the maximal ideal. Let us remind the reader briefly of the injective R-modules. It is standard that, for any commutative ring R and any R-module M, the module M is injective as an R-module if and only if, for every non-zero ideal I C R,

Ext 1 (R/I,M) = 0. In a discrete valuation ring R, every non-zero ideal I C R is Rpn for some n ~ 0. There is an exact sequence of R-modules pn

0----+ R----+ R----+ R/I----+ 0.

Hence there is an exact sequence of Ext-groups 1

Pn

Hom(R, M)----+ Hom(R, M)----+ Ext (R/ I, M)----+ Ext\R, M). Now Ext 1 (R, M) = 0 since R is projective, while we have a natural isomorphism Hom(R, M) = M. Hence the sequence becomes pn

M ----+ M ----+ Ext 1 (R/I,M) ----+ 0. Thus M is injective if and only if for each n ~ 0, multiplication by pn gives a surjective map pn : M ----r M. Clearly, the case n = 1 implies the general case; M is injective if and only if p : M ----r M is surjective. In the rest of this Section, we choose and fix a discrete valuation ring R and a generator p of its maximal ideal. DEFINITION C.l.l. Let M be an R-module. By transfinite induction, we define for every ordinal i, a submodule pi M C M. The definition is:

C.l.l.l. Fori= 0, piM

= p0 M = M.

C. COUNTEREXAMPLES WITH A('J)

388

C.l.l.2. For a successor ordinal i Pi+lM

+ 1,

we define

p{piM}.

That is, pi+ 1 M is the set of all elements of pi M divisible {in pi M) by p. C.l.l.3. For a limit ordinal i, we define piM

npiM. j i we have piM = piM.

Proof: The proof is an easy transfinite induction. We remind the reader briefly how these go. For C.l.2.1, fix the ordinal j and consider the set I = {i an ordinal Ii :::: j and pi M :) pi M} To prove C.l.2.1, it suffices to show that every i :::: j lies in I. To do this, it suffices, by transfinite induction, to show that j E I, that i E I ===? i+ 1 E I, and that if k E I for all j ~ k < i for some limit ordinal i, then i E I. Observe that pi M :) pi M, and hence j E I. Furthermore, if i E I then pi M :) pi M, from which we deduce piM that is, i + 1 E I. If i is any limit ordinal > j, then the set of ordinals k < i contains j. Therefore

c

npkM k i then pi M =pi M. Let J be the set of ordinals J ={jan ordinal [j:::: i and pi M = piM}.

Once again, we use transfinite induction to show that all ordinals j :::: i lie in J. Clearly, i E J. Suppose j E J. Then pi M =pi M. Hence pi+l M

p{pi M}

p{pi M}

This means pi+l M = piM, and hence j

+ 1 E J.

pi+ 1M

pi M

C.l. THE SUBMODULES p' M

389

Suppose j is a limit ordinal, j > i. Then pJM If for all ordinals k with i::::; k pkM = piM. Hence

By C.1.2.1, for k

< j we have

< i we have pk M

k E J, then for all i :::; k


j, pk M = pi M. In particular, Thus for j < i,

390

C. COUNTEREXAMPLES WITH A('J)

The contrapositive is that for j < i,

which is the assertion of the Corollary.

pi M

D

LEMMA C.l.5. Let M be an R-module. Suppose i is an ordinal with =f. pi+ 1 M. Then the cardinality of M is at least the cardinality of i.

Proof: Assume pi M =f. pi+ 1 M. By Corollary C.l.4, we deduce that for every j < i, we must have pi M =f. pi+l M. By C.l.2.1, we know that pi M ~ pi+ 1M. Since they are not equal, we may choose, for each j < i, an element

mj EpiM -pi+lM. If j < k < i, then mj E pi M- pi+l M, while mk E pk M C pi+l M. Since mj fl. pi+l M while mk E pi+l M, it follows that mj =f. mk. The elements m j are all distinct. The set {m j , j < i} is a set of distinct elements of M, whose cardinality is the cardinality of i. D

REMARK C.l.6. In Remark C.l.3, we saw that if piM = pi+ 1 M then pi M is an injective R-module, and contains every injective R-submodule of M. In Lemma C.l.5, we saw that if M has cardinality< o:, then for the ordinal o: we must have pa+l M =paM. For every R-module, there exists some ordinal with pi M = pi+ 1 M. Every module has a largest injective submodule. Of course, the inclusion of the injective pi M C M must split. M is the direct sum of the injective module pi M and a submodule isomorphic to MjpiM. The module MjpiM contains no injective submodule. LEMMA C.l.7. Let¢: M --+ N be a homomorphism of R-modules. Then for every ordinal i, ¢(pi M) c piN.

Proof: We prove this by induction on the ordinal i. If i obvious; the assertion is that

c

cp(M)

N

Suppose we are given a ordinal i for which ¢(pi M) C piN. Then

¢(p{piM}) pcp(piM)

c

p{piN} pi+lN,

0, this is

C.l. THE SUBMODULES pi M

391

that is the ordinal i + 1 satisfies ¢(pi+ 1 M) c pi+ 1 N. Finally, if i is a limit ordinal and for every j < i we have ¢(pi M) C piN, then

cp(piM)

~ (J: M--+ M taking [a] EM to itself. Then the image Im(¢>) has cardinality at least a. By [a] EM we mean of course the generator [i 0 ], where i 0 =a.

Proof: The idea is to apply Lemma C.l.8, to the map ¢> : M --+ M. By Lemma C.2.2, we have that [a] E paM. Since ¢([a]) = [a], we have ¢([a]) :f. 0. Also, we computed in Lemma C.2.2, that rf3 M is generated by [in, in_ 1 , · · · , i 1 , i 0 ] with in :2': (3. Since there are no such generators in M, it follows that pf3 M = 0 and M contains no injective submodule. Lemma C.l.8 therefore applies, and the image of¢> must have cardinality at least a. D

C.3. THE CATEGORY A(S) IS NOT WELL-POWERED

393

C.3. The category A(S) is not well-powered As in Section C.2, let us choose a discrete valuation ring R with maximal ideal Rp. Let D(R) be the derived category of all chain complexes of any R-modules. Let R be the chain complex which is R in dimension 0, zero elsewhere. Let A(D(R)) be the abelian category associated to the triangulated category D(R), as in Definition 5.1.3. LEMMA C.3.1. Let {R ~ x>.., A E A} be a set of quotient objects of R in A(D(R)); see Proposition 5.2.6. There exists an infinite cardinal a, so that for any representation of any one of these quotient objects, that is for any {R ~ y} isomorphic as a quotient object in A(D(R)) to one of the {R ~ x>..}, there is an endomorphism y ---+ y making commutative the square

R----+ y

R----+ y and the image of ¢ : H 0 (y)

H 0 (y) is of cardinality < a.

---+

Proof: Let a be any cardinal greater than the sum of the cardinalities of all { H 0 ( x >..), A E A}. Since we are assuming that { R ~ y} isomorphic as a quotient object in A(D(R)) to one of the {R ~ x>..}, there exists a A E A and an isomorphism of the two objects {R ~ y} and {R ~ x>..}· By Proposition 5.2.6, this means that there exist commutative squares in D(R)

R----+ y

R

1

----+

X>..

R

R

----+

X>..

1 ----+

y

Applying the functor H 0 to these commutative squares, we get a commutative square of R-modules

R----+ HO(y)

11

p1

R

----+

a1

11

R----+ H 0 (x>..)

R

H 0 (x>..)

----+

Ho(y)

Combining the squares, we have a commutative square

R----+ Ho(y)

ap1

11 R

----+

HO(y)

394

C. COUNTEREXAMPLES WITH

A('J)

where ap factors through H 0 (x>..), whose cardinality is by hypothesis . E A} is some set of quotient objects of R in A(D(R)). Then there exists a quotient object R-+ M not isomorphic in A(D(R)) to any of the {R-+ xA, A E A}. Proof: By Lemma C.3.1 we can, for our set of quotient objects {R -+ xA, >. E A}, choose a cardinal a satisfying the conclusions of Lemma C.3.1. Pick such an a. For this a, let M be the module of Construction C.2.1. Let 0 : R -+ M be the map

O(r)

= r[a].

This is of course an R-module homomorphism, which we may view as a map of complexes concentrated in degree 0, in the derived category. For any commutative diagram

R~M

R~M we have that ¢([a]) = ¢0(1) = 0(1) = [a]. From Proposition C.2.3 we deduce that the image of¢ has cardinality at least a. On the other hand, if { R -+ M} is isomorphic, as a quotient of R in A(D(R)), to one of {R-+ xA, >. E A}, then by Lemma C.3.1 there must exist a commutative square R ~ H 0 (M)=M 1

1

¢1

R ~ H 0 (M)=M

with the image of ¢ of cardinality < a. We deduce that {R -+ M} is not isomorphic, as a quotient of R in A(D(R)), to any of {R-+ xA, >. E A}. D COROLLARY C.3.3. Let R be a discrete valuation ring. The class of isomorphism classes of quotients of R in A(D(R)) is not a set. Given any set of quotients, there is a quotient not isomorphic to any of them. REMARK C.3.4. If R is the localisation of Z at a prime p, we have shown that R does not have a set of quotients in A(D(R)). But since A(D(R)) is a localisation, in the sense of Gabriel, of the category A(D(Z)), it follows that R does not have a set of quotients in A(D(Z)) either.

C.4. A CATEGORY

C.4. A category

ex(S 0 P,Ab) WITHOUT A COGENERATOR

395

ex(S P,Ab) without a cogenerator 0

As throughout this Appendix, R is a discrete valuation ring with maximal ideal Rp C R. Let K be the quotient field of R. LEMMA C.4.1. Let a be an infinite cardinal. There exists a non-trivial extension {a non-split short exact sequence) of R-modules 0

---t

M

~

M'

~

K/R

---t

0

so that, for any ¢ : M -----+ N, with N of cardinality < a, the induced extension of K / R by N splits. That is, if we push out the exact sequence to get 0

---t

M

~

¢1 0

---t

M'

---t

~

N

~

---t

0

11

1

N

K/R

N'

---t

K/R

---t

0

N'

~

K/R

---t

0

then the sequence 0

---t

is split. Proof: Choose an infinite cardinal f3 > a, and let M be the R-module of Construction C.2.1. The facts that are relevant to us here are: C.4.1.1. P01 M

=f. 0

C.4.1.2. pf3 M = 0. From Lemma C.l.2, more precisely by C.l.2.2, we deduce that p 01 M =f. p 01 +1 M. Choose an element x E p01 M - p 01 +1 M. It corresponds to a map R -----+ p 01 M C M. We define the extension of K/ R by M to be given by the pushout

0

---t

R

1

~

K

---t

1

0 -----tM~M' - - - +

K/R

---t

0

---t

0.

11

K/R

Now we need to prove two things. We must prove that the extension is non-trivial, and also that it becomes trivial after pushing out along any map¢: M-----+ N, if the cardinality of N is M ------> M'

q\1

~

~

KjR

~o

But the map R -----> M was chosen to factor through pa. M C M, and by Lemma C.l. 7, ¢(pa. M) C pa. N. Therefore the map R -----> N factors through pa. N C N. Now recall that the cardinality of N is assumed

R

1

------> K

~

~

0

11

1

0 ____. pa.N ------> N

KjR

~

KjR

~o

is split. Hence so is the bottom row of the commutative diagram 0------>

R

1

------> K

0------>

N

~

~o

KjR

~o

11

1 ------> N'

KjR

11

1

0 ------> pa.N ------> N

1

~

~

KjR

~o.

This establishes that the extension of K j R by M becomes trivial after extending by maps¢: M-----> N, with N of cardinality< a. To finish the proof, we must establish that the bottom row in the commutative diagram 0 ------> R ------> K

1

1

~

KjR

~

0

11

0 ------> M ------> M' ~ KjR ~ 0

is not a split exact sequence. Equivalently, we must show that the map R -----> M does not extend to a map R C K -----> M. The image of 1 E R

C.4. A CATEGORY cx(S 0 P,Ab) WITHOUT A COGENERATOR

under the map R X

--+

M is an element x

E

M, more precisely

c

E

397

M.

To extend to a map K --+ M would be equivalent to finding, for all n the image of p-n E K. Let Xn be the image of p-n. We must have

~

0,

C.4.1.3. x 0 = x. C.4.1.4. PXn+l = Xn. We must show that there is no such sequence {xn}· Suppose there exists a sequence {xn} of elements of M, satisfying the conditions C.4.1.3 and C.4.1.4. Consider the set S of ordinals, given by

S

{i ::::; a

+ 1 I 3n ~ 0 with Xn

~ pi M}.

The set S is non-empty; by our construction, x = x 0 (j. p"'+ 1 M, and so a+ 1 E S. Because the set of ordinals ::::; a+ 1 is well-ordered, there exists a minimal ordinal k E S. Since k is minimal, for all n ~ 0 and all j < k, Xn E pi M. Thus for all n,

and since at least one Xn does not lie in pk M, we must have

This means that k cannot be a limit ordinal. For limit ordinals k, by definition of pk M, we have

Hence k must be a successor ordinal. Put k = i + 1. Then for all £, X£ E piM. And there exists at least one n for which Xn (j. pi+ 1 M. Choose and fix n, so that Xn ~ pi+ 1 M. But C.4.1.4 asserts that pxn+l = Xn· We know that for all£, X£ E pi M; in particular Xn+l E pi M. Therefore E

and this is our contradiction; the sequence {xn} cannot exist.

0

Before we apply this to the derived category of R, let us remind the reader of well-known facts.

C. COUNTEREXAMPLES WITH A('J)

398

LEMMA C.4.2. Let R be a ring, of projective dimension ~ 1 (e.g. a discrete valuation ring). Let X be an object in the derived category D(R). Then

X

00

00

n=-oo

n=-oo

=

Proof: Choose a projective resolution for the R-module Hn(X)

0

~

P1

~

Po

Hn(X)

~

~

0.

Such a resolution exists since R is of projective dimension ~ 1. We may also assume that Po and P 1 are both free. They are coproducts of the module R. Since Hn(X) = Hom(:E-n R, X), the map Po ~ Hn(X) could be thought of as a map

Po

Il R

~ Hom(:E-nR,X)

Hn(X),

=

AEA

which may be viewed as the functor Hom(:E-n R, -) applied to a morphism in D(R)

But now we have a composite

:E-n pl

~

Since P 1 is free, this is a map

:E-n Po

X.

----t

Il :E-n R

X.

----t

p.EM

On the other hand, applying the functor Hom(:E-n R, -), we get P1 Hn(X), which vanishes by hypothesis. It follows that the composite

:E-n pl

~

:E-n Po

~

X

----t

vanishes in D(R). But the triangle

:E-npl

~

:E-npo

asserts that the map :E-n Po

:E-n Po

~

~X

~

:E-nHn(X)

----t

:E-n+lpl

factors, in D(R), as

:E-n Hn(X)

----t

X.

We have produced a map :E-n Hn(X) ~ X which is an isomorphism in Hn. Producing such a map for every n, we have a morphism

Il 00

nE-oo

:E-n Hn(X) ~ X,

C.4. A CATEGORY £x(S 0 P,Ab) WITHOUT A COGENERATOR

399

which is an Hn-isomorphism for every n. Hence it is an isomorphism in D(R). This establishes that X is a coproduct of suspensions of its cohomology modules. But now consider the natural map

II

II

00

00

E-n Hn(X) ~

nE-oo

E-n Hn(X).

nE-oo

It is also an Hn-isomorphism for every n, hence an isomorphism in D(R). Thus X is also isomorphic to the product of suspensions of its cohomology modules. D Let R be a ring of projective dimension :::; 1. Let 'J = D(R) be its derived category. By Lemma C.4.2, any object x E 'J can be written as

II

II

00

X=

00

E-nxn

=

E-nxn,

n=-oo

n=-oo

where for each n E Z, xn is just an R-module. If f : x in 'J = D(R), then we may write

II

---+

y is a morphism

00

Y=

._.-n L..-

Yn

n=-oo

and a map from a coproduct to a product is entirely determined by its components. In other words, to understand all possible maps in 'J, it is enough to understand the maps

Emxm

~

Enyn

with xm and Yn just ordinary R-modules. It is classical that these maps are in 1-to-1 correspodence with elements of Extn-m(xm, Yn)· But what we really need is to understand the maps in C.x ( {'Ja} op, Ab). To this end, we prove the Lemma LEMMA C.4.3. Let R be a ring of projective dimension :::; 1. Let 'J = D(R) be its derived category. Let a be a regular cardinal. PutS = 'Ja. Let

1r:

A('J)

-----t

£x(S 0 P,Ab)

be the quotient map of Section 6.5. Let x andy be ordinary R-modules. If n =f. 1, then 'J(x, Eny)

= £x(S P,Ab){ 1r(x), 1r(Eny) }· 0

That is, the maps in 'J of the form x ---+ Eny agree, via the natural map, with the maps in £x(S 0 P,Ab) of the form 1r(x) ---+ 1r(Eny).

400

C. COUNTEREXAMPLES WITH A(9)

Proof: We need to show that maps in C:x(S 0 P,Ab) of the form 1r(x) ---7 1r(Eny) correspond 1-to-1 with elements of Extn(x, y). Since the ring R has projective dimension:::; 1, Extn(x,y) = 0 unless n E {0, 1}. Thus we must show: C.4.3.1. lfn =/:- 0, 1, then all maps 1r(x) ---??r(Eny) vanish. C.4.3.2. Maps 1r(x) ---7 1r(y) correspond 1-to-1 with R-module homomorphisms x ---? y. The reader should note that the case n = 1 is specifically excluded. We do not know, nor care about, the maps 1r(x) ---7 1r(E 1y). It helps to recall what the functor 1r is. From the discusion at the beginning of Section 6.5, (see also Lemma 6.5.2), we know the functor 1r very concretely. The functor 1r takes an object x E 'J C A('J) to the functor 'J(-,x)ls:

sop

-----+

Ab.

What we must establish is that, for any R-modules x andy, C.4.3.3. If n =/:- 0, 1, then any natural transformation 'J(-,x)ls

-----+

'J(-,Eny)Js

vanishes. C.4.3.4. The natural transformations 'J(-,x)ls

-----+

'J(-,y)Js

correspond 1-to-1 with R-module homomorphisms x

---7

y.

Let us first treat C.4.3.4. The functor 1r takes an R-module homomorphism f : x ---7 y to a natural transformtion 'J(-,x)ls

'J(-,f)ls

'J(-,y)is·

But there is an inverse. We can evaluate a natural transformation 'J(-,x)ls ~ 'J(-,y)Js at any objects E S. In particular, we wish to evaluate it on the free module

R. We deduce a map of R-modules x

= 'J(R, x)

-----+

'J(R, y)

= y.

Call this map ¢(R). Clearly, 'J(-,f)ls(R)= f. What we need to show is that¢= 'J(-,¢(R))Is· Replacing¢ by¢- 'J(-,¢(R))Js, we need to show C.4.3.5. Suppose n =/:- 1, and suppose we are given a natural transformation 'J(-,x)ls ~ 'J(-,Eny)Js which vanishes on the object R. Then¢= 0.

C.4. A CATEGORY £x(S 0 P,Ab) WITHOUT A COGENERATOR

401

Note that C.4.3.5 combines C.4.3.3 and C.4.3.4. We have just seen that, if n = 0, C.4.3.5 implies C.4.3.4. But if n rf. {0, 1}, then any map

'J'(-,x)ls __.:!____. 'J'(-,~ny)ls vanishes on the object R, and hence C.4.3.5 implies the vanishing of all maps ¢, that is C.4.3.3. We wish to prove C.4.3.5, that is the vanishing of all maps

'J'(-,x)ls

¢

~ 'J'(-,~ny)ls

which vanish on R. Let ¢ be such a map. Of course, since ¢ vanishes on R, it also vanishes on any coproduct of R's, and on any direct summand of such coproducts. Therefore¢ vanishes on any projective R-module. Let s be any object of the category S = 'J'"'. We wish to show that ¢ vanishes on s. By Lemma C.4.2, 00

with sm ordinary R-modules. It therefore suffices to prove that, for any m E Z, ¢ vanishes on ~-msm. And since s E S = 'J'"' and ~-msm is a direct summand of s, we must have ~-msm E S. In other words, we are reduced to showing that, for any R-module s belonging to S = 'J'"', and for any integer m, the map¢ vanishes when evaluated on ~-ms. The fact that s is an R-module and s E 'J'"' means that s admits a resolution by projective R-modules o~P1~Po~s~o

with Po and P 1 of rank< o:. This becomes a triangle inS ~- 1 s ~

P1

~

Po ~ s

Both the functor 'J' (-, x) I3 and the functor 'J' (-, y) I3 take triangles in S to long exact sequences, and the naturality of ¢ gives a map of long exact sequences 'J'(s,x)

~

'J'(Po,x)

----4

'J'(Po, ~ny)

'J'(P1' X)

~

~

'J'(P1, ~ny)

~

1

1

'J'(s, ~ny)

~

1

'J'(~- 1 s,

X)

1

'J'(~-1 8 , ~ny)

Now note that in each row, all but four of the abelian groups are zero. Let us prove it for the top row; the case of the bottom row is parallel. The point is that 'J'(~-mP,x) = Extm(P,x). And if Pis projective, this vanishes unless m = 0. Thus the terms 'J'((~-m Pi, x), with i = 0, 1, must vanish unless m = 0. This gives two possible non-zero terms. And from the long exact sequence we learn that 'J'(~-ms, x) vanishes, unless mE {0, 1}.

C. COUNTEREXAMPLES WITH A('J)

402

Now recall that by the hypothesis of C.4.3.5, the map ¢vanishes on any projective object. The commutative diagram with exact rows ----t

----t

'J(P1,x)

'J(Pl.Eny)

----t

'J(E- 1s,x)

----t

1 'J(I;-ls,Eny)

----t

0

----t

1 'J(I;-lPo,Eny)

gives, in particular, a commutative square

'J(P1.x)

'J(E- 1 s,x)

----t

and the map

'J(P1.x)

----t

'J(P1,Eny)

vanishes since P 1 is projective. But then the composite

'J(P1,x)

'J(E- 1 s,x)

----t

must vanish, and the surjectivity of ----t

'J(E- 1s,x)

----t

'J(E- 1s, Eny).

'J(P1,x) implies the vanishing of

'J(E- 1s,x)

This much was painless. We also have a commutative diagram with exact rows

0

----t

'J(s,x)

----t

'J(s, Eny)

1

'J(EP1, Eny)

-----+

'J(Po, x)

-----+

'J(Po, Eny)

1

Because we are assuming n

1

i= 1,

'J(EP1, Eny)

=

'J{Pt, En- 1y)

0;

this is because P1 is projective, and n- 1 i= 0. The commutative diagram with exact rows above becomes 0

----t

'J(s,x)

----t

'J(Po,x)

C.4. A CATEGORY £x(S 0 P,Ab) WITHOUT A COGENERATOR

403

Since the map

'J(Po, x) -----+ 'J(P0 , ~ny) vanishes (because P0 is projective, and by the hypothesis of C.4.3.5), it follows that the map also vanishes on the subobjects. That is, the map

'J(s, x) -----+ 'J(s, ~ny) is zero. Note that this part of the argument depends on n-:/= 1. Summarising, we have shown that for any R-module s E S, the maps

'J(s,x)

-----+

'J(s, ~ny)

'J(~-ls,x)

-----+

'J(~-ls,~ny)

both vanish. But form¢ {0,1} the group the map 'J(~-ms,

x) -----+

'J(~-ms,x)

is zero, and hence

'J(~-ms, ~ny)

vanishes trivially. Thus the map vanishes for all m, completing the proof ~~~m D PROPOSITION C.4.4. Let R be a discrete valuation ring, K its quotient field. Put 'J = D(R), the derived category of R. Let a be regular cardinal~ ~ 1 . PutS = 'Ja. Then the category £x(S 0 P, Ab) does not have a cogenerator.

Proof: Observe first that K j R is an object in 'Ja. It has countably many generators, namely {p-n,n ~ 1}, and there are countably many relations. In other words, there is an exact sequence of R-modules 0-----+

II R-----+ II R-----+ 00

00

n=O

n=O

K/R-----+ 0.

This gives a triangle

But R E 'Ja, and hence so is the countable coproduct U:::o R. Two terms in the triangle above lie in S = 'Ja, and hence so does the third, K / R. Suppose t:x(S 0 P,Ab) had a cogenerator. There would be an object C, so that all other objects inject into products of C. Now recall Proposition 6.5.3; the category ex( {'Jayv,Ab) is a Gabriel quotient of thecategory A('J). There is an exact quotient map 1f:

A('J) -----+ t:x({'Jayv,Ab)

C. COUNTEREXAMPLES WITH A('J)

404

and it has a left adjoint L. But then L(C) is an object of A('J}, and may be embedded in an injective object I. The functor rr, being exact, takes this to a monomorphism

C

;:;=

rrL(C)

-----+

rr(I).

If Cis a cogenerator, any object may be embedded in a product of C's. But C embeds in rr(I), and hence any object may be embedded in a product of rr(I)'s. That is, rr(I) must also be a cogenerator. We may therefore assume that our cogenerator is of the form rr(I), with I an injective objeCt in A('J). By Corollary 5.1.23, the injectivity of I means that IE 'J C A('J}. Choose and fix such an I. Now let f3 be an infinite cardinal greater than the maximum cardinality of all the homology groups Hn(I), with I E 'J ;:;= D(R), and rr(I)

a cogenerator of ex ({'J"'} op, Ab), as above. Let M be an R-module as in Lemma C.4.1, for the infinite cardinal f3 chosen above. There exists a non-trivial extension

0----+ M----+ M'----+ K/R

----t

0

so that, for any ¢ : M -----+ N, with N of cardinality < /3, the induced extension of K/R by N splits. Now M may be viewed as an object of

D(R) ;:;= 'J, and rr(M) becomes an object in ex({'J"'VP,Ab). I assert that this object cannot possibly be embedded into a product of rr(I)'s. By Lemma C.4.2, the object I E 'J admits a decomposition

m=-oo

with Im all R-modules, and since Im ;:;= H-m(I), then by the choice of f3 their cardinalities are all < /3. Now the map 7f respects products, since it has a left adjoint L. Therefore 00

m=-oo

But we are supposing that rr(I) is a cogenerator. There is therefore an embedding in ex(S 0 P,Ab)

rr(M)

II rr(I)

c

A

II II 00

A m=-oo

For every integer m, this gives a map

rr(M)

c A

A

rr(~m Im)·

C.5. HISTORY OF THE RESULTS OF APPENDIX C

405

Now Lemma C.4.3 tells us that unless mE {0, 1}, all such maps vanish. We conclude that the following is already an embedding, the other components being zero

Now we wish to evaluate this natural transformation on the particular object E- 1 {K/R} E S. In the interest of making the formulas below more legible, let us abbreviate

K/R

= 1!,

E- 1 {K/R} = E- 11!.

that is

We have a monomorphism of abelian groups

'J(E- 1 1!, M)

----+ {

But 'J(E- 1 1!, E 1 I 1 )

IJ

'J(E- 11!, 10 )} EB {

= Ext 2 (1!, ! 1 ) = 0. 'J(E- 11!, M)

IJ

'J(E- 11!, E 1 I 1 )}.

Hence we have that the map

----+

II 'J(E- 1!, 1 1

0)

A

must be a monomorphism. But this map is nothing other that the natural transformation

1r(M)

----+

II 1r(I

0 ),

A

evaluated on the object E- 11! E S. By Lemma C.4.3, we know that the natural transformation above is induced by a map of modules M ------+ fiA ! 0 • But our extension 0

-----+

M

-----+

M'

-----+

K I R = 1! - - - 0

gives a non-zero map E- 11! ------+ M, and for any map M ------+ ! 0 , the extension E- 1 £ ---+ M ------+ I 0 must vanish, as the cardinality of ! 0 is < fJ. In other words, we have found a class in 'J(E- 11!, M), which maps to zero under the natural map to 'J(E- 11!, fiA ! 0 ). This contradicts the injectivity. D

C.5. History of the results of Appendix C The results of Sections C.l, C.2 and C.3 are certainly not new. As far as the author knows, the earliest version appeared in Freyd's article [14]. There were also accounts due to Grandis and to Morava, but those never appeared in print. The author learned the construction of the large module M from Boardman. Boardman uses it in his unpublished paper, on conditionally convergent spectral sequences.

406

C. COUNTEREXAMPLES WITH A('J)

The application of these constructions, in Section C.4, is entirely new. Since injectives in the category £x (gop, Ab) are new to this book, it is new that one can use these large modules to show that in general, £x(S 0 P,Ab) does not have enough injectives.

APPENDIX D

Where '1 is the homotopy category of spectra D.l. Localisation with respect to homology

In this Appendix, 'J' will be the homotopy category of spectra. For the reader unfamiar with spectra, Summary D.l.l lists the properties we will need in the present section. The properties used in Section D.2 are more difficult to summarise briefly. What we need there is basically some familiarity with the results of (24]. SUMMARY D.l.l. The homotopy category of spectra is a triangulated category 'J', closed under small coproducts. That is, it satisfies (TR5]. It has a smash product

'J' X 'J'

1\ :

and an object

S0 E

----t

'J'

'J', satisfying the following additional properties:

D.l.l.l. The object S 0 E 'J' is N0 -compact. That is, any map

so

----7

II t ,\

.AEA

factors through a finite coproduct. D.l.l.2. Put sn = :En 8°. Let t be an object of 'J'. If, for every n E Z, 'J'(Sn, t) = 0, then t ~ 0. D.l.l.3. For any n :::; No.

E

Z, the group 'J'(S 0 , sn) has cardinality

D.l.l.4. The smash product takes triangles to triangles. That is, there are natural isomorphisms

{:Ea} 1\ b

:E{a 1\ b}

and given an object t x

E

----t

a 1\ {:Eb},

'J' and a triangle y

----t

z

----t

:Ex,

then both of the following are triangles xl\t - - - - t yl\t - - - - t zl\t - - - - t {:Ex}/\t t 1\ X

----t

t

1\

y

----7

t

1\ Z

----t

t 1\ {:Ex}

:E{x/\t}, =

:E{t 1\ X}.

408

D. THE CATEGORY OF SPECTRA

D.l.l.5. The smash product respects coproducts. That is,

D.l.l.6. S 0 is a two-sided unit for the smash product; that is X

1\

S0

S 0 1\ X.

X

D.l.l.7. Any map

S 0 ~ xl\y factors as

S 0 ~ x'A.y'

fl\g

~xl\y

where x' and y' are objects in 'J~o.

LEMMA D.l.2. Let T be the set of suspensions of S 0 ; that is,

T

{SnlnEZ}.

Then T C 'J~ 0 , and T U {0} is a-perfect for all infinite a.

Proof: By D.l.l.1, the objects of sn E T are all ~ 0 -small; see Definition 4.1.1. The set T U {0} is ~ 0 --compact, since any class of objects containing 0 is. See Example 3.3.16. Thus T U {0} is an ~ 0 -perfect class of objects in 'J(~o), hence contained in the maximal one 'J~ 0 • Now T U {0} is an ~ 0 -perfect class of objects in 'J(~o). Lemma 4.2.1 applies, and we deduce that T U {0} is a-perfect for all infinite a. 0 LEMMA D.1.3. As in Lemma D.1.2, letT be the set of suspensions of S 0 • Then D.l.3.1. The category 'J satisfies the representability theorem. D.1.3.2. 'J = (T).

Proof: By D.1.1.2, the set T generates the triangulated category 'J; see Definition 8.1.1. In Lemma D.l.3, we saw that T U {0} is o:-perfect for all infinite a, in particular ~ 1 -perfect. By Definition 8.1.2, T is therefore an ~1-perfect generating set. Theorem 8.3.3 applies, and we conclude D.1.3.1: The category 'J satisfies the representability theorem. D.1.3.2: 'J = (T). 0

D.l.

LOCALISATION WITH RESPECT TO HOMOLOGY

409

PROPOSITION D.l.4. As in Lemma D.1.2, letT be the set of suspensions of 8°. Then for any regular cardinal a,

That is, the subcategory of a-compact objects in 'J agrees with the alocalising subcategory generated by T. Proof: From Lemma D.l.3, we know that for our set T, 'J = (T). From Lemma D.l.2, we also have that T C 'J~ 0 • Therefore Lemma 4.4.5 applies. We deduce that, for all regular a ?:: N0 ,

(T)a.

'J"'

D

REMARK D.l.5. Since T is a set, Proposition 3.2.5 guarantees that, for every regular a, 'J"' is essentially small. It is the smallest triangulated category containing all the spheres, and closed under triangles, as well as coproducts of fewer than a of its objects. This is usually referred to as the "category of spectra with fewer than a cells". An object in 'J"' can be constructed out of fewer than a spheres, by attaching. LEMMA D.l.6. Let a be a regular cardinal. If x andy are objects in 'J"', then so is x 1\ y.

Proof: Fix an object y E 'J"'. Define a full subcategory S C 'J by

Ob(S) We will prove that Sis an a-localising subcategory containing T, and hence must contain (T)"' = 'J"'. First observe that S contains T = {Sn InEZ}. For if n E Z,

sn 1\ y

=

{~nso} 1\

y}

~n{so 1\

=

y

~ny.

We are given y E 'J"', therefore sn 1\ y = ~ny E 'J"'. Hence sn E S. Next note that xES if and only if ~xES. This is because {~x} 1\

y

~

~{x 1\

and sox 1\ y E 'J"' if and only if {~x} 1\ y Now suppose we have a triangle in 'J

x

--+

x'

--+

y},

= ~{x 1\ y}

x"

E 'J"'.

- - - + ~x,

and suppose x, x' E S. By D.l.l.4, the following is a triangle in 'J

x 1\ y

---+

x' 1\ y

--+

x" 1\ y

- - - + {~x} 1\

y.

D. THE CATEGORY OF SPECTRA

410

Since x, x' E S, we know that x 1\y, x' 1\y E 'J. But 'J is triangulated, and we deduce that x" 1\ y E 'J, that is x" E S. The category S is triangulated. Let {x>., A E A} is a family of fewer than a objects of S. That is, for each A E A, x>.l\ y E 'J. By D.l.l.5,

{ II x>.} 1\y AEA

But Il>.EA { x >. 1\ y} is a coproduct of fewer than a objects in 'J. By Lemma 4.2.5, the category 'J is a-localising: it is closed under coproducts of fewer than a of its objects. Hence 'J

E

'

and Il>.EA x>. E S. The category Sis closed under coproducts of fewer than a of its objects. Since we are not assuming a> N0 , to verify that Sis a-localising, we must also check that it is thick. Suppose x EB x' E S. That is,

{x EB x'} 1\ y

{x 1\ y} EB {x' 1\ y}

E

'J.

Since 'J is thick, we have x 1\ y E 'J, that is x E S. The category S contains all the direct summands of its objects, that is S is thick. Thus Sis an a-localising subcategory, containing T. Therefore S contains the smallest such, (Tt. But by Proposition D.l.4, (T) = 'J. That is, 'J C S. In other words, for our fixed (but arbitrary) y E 'J, and any x E 'J, we have x 1\ y E 'J. D LEMMA D.l.7. Let a be a regular cardinal, a> No. Let x be an object of 'J. Then for any n E Z, the cardinality of 'J(Sn, x) is< a. Proof: Define a full subcategory S Ob(S)

=

c 'J by

{xE'Ji'v'nEZ,#'J(Sn,x) ~ 0 ; the fact that S is triangulated and closed under countable coproducts, guarantees that S must be thick. Thus S is an a-localising subcategory, containing T. Therefore S contains the smallest such, (T)a.. By Proposition D.l.4, (T)a. = 'Ja.. That is, 'Ja. C S. In other words, for every object x E 'Ja. and any n E Z, #'J(Sn,x)




x' 1\ E - - - . x" 1\ E

-----+

{:Ex} 1\ E.

By hypothesis, x 1\ E = 0 = x' 1\ E. From the triangle we have x" 1\ E = 0, that is x" E 'JE. The subcategory 'JE is triangulated. Let {x,\, A E A} be a set objects of 'JE. That is, for each A E A, X,\ 1\ E = 0. By D.1.1.5,

{ II x ,\EA

Therefore

u,\EA X,\

II {x,\ 1\ E}

} 1\ E ,\

0.

,\EA

is an object of 'JEl and 'JE is localising.

0

Now we come to the key lemma. LEMMA D.l.lO. As above, 'J is the homotopy category of spectra. Let o: be a regular cardinal, o: > ~o. Let E be an object of 'J. Let R be a set of representatives for all isomorphism classes of objects in 'J n 'JE· [By Remark D.1.5, 'J is essentially small, so we may always choose a set R of representatives for any class of objects in 'J .} Then any map t ~ x, with t E 'J and x E 'JE, factors as

t

--->

r

-----+

x

with r E R.

Proof: Suppose we are given a map t ~ x, with t E 'J and x E 'JE· We want to factor it. The strategy of the proof is first to show that. any map t ~ x as above admits a factorisation

t ___. t where

t E 'J,

-----+

x,

and 'J(Sn, t 1\ E) - - - . 'J(Sn, t 1\ E)

vanishes. Then we let r be the homotopy colimit of the sequence t~t~t~···

D.l. LOCALISATION WITH RESPECT TO HOMOLOGY

413

and show that this r works. We begin by constructing, for any map t--+ x with t E 'J'"' and x E 'J'E, a factorisation t--+ t--+ x. Complete t--+ x to a triangle k

---7

t

---7 X

---7

Ek.

By D.l.l.4, smashing withE gives a triangle k 1\ E

t 1\ E

---7

x 1\ E

---7

---7

{Ek} 1\ E.

= 0. Therefore the natural map is an

Since x E 'J'E, we have x 1\ E isomorphism ki\E

---7

ti\E.

Both t and E lie in 'J'"'. By Lemma D.l.6, ti\E E 'J'"'. By Lemma D.l.7, for any n E Z, the cardinality of the group

is< a. Let

U 'J'(Sn, t 1\ E). 00

A=

n=-oo

The set A is a union of countably many sets, each of cardinality < a. Since a > ~o and a is regular, #A < a. An element of the set A is a map sn--+ ki\E ~ ti\E. From D.l.l.7, we know that it is possible to choose k' ----+ k and E' --+ E, with k', E' E 'J'No, so that sn ----+ k 1\ E will factor as sn

---7

ft\g

k' 1\ E'

k /\E.

---7

For each>. E A, choose such a factorisation. That is, pick an object k' = k >.. E 'J'No , and a factorisation sn

---7

k>.l\ E

k /\E.

We have a map

II k>.

---7

k

---7

>..EA

We define the map t

II k>.

>..EA

--+

t.

t from the triangle

---7

t

---7

t

---7

Now t E 'J'"', and il>..EA k>. is the coproduct of fewer than a objects, each in 'J'No C 'J'"'. The coproduct lies in 'J'"', and from the triangle, we deduce

D. THE CATEGORY OF SPECTRA

414

E 'J'. Because the composite k -----+ t -----+ x vanishes, so does the longer composite

t

II k>.

___.. k ___.. t

x.

---7

AEA

From the triangle

II k>.

___.. t

---7

t,

AEA

we deduce that the map t-----+ x must factor as t-----+ t-----+ x. Also, from the triangle

II k>.

___.. t

---7

t

.>.EA

we obtain, by smashing with E, a triangle

II kA} 1\E ___.. ti\E {.>.EA

ti\E .

---7

But by construction, any morphism sn -----+ t 1\ E factored through k>. 1\ E for some .X E A. In other words, sn -----+ t 1\ E factors as

sn ___..

{II kA} 1\ E

and we deduce that, for every

sn

---7

t 1\ E,

AEA

sn -----+ t 1\ E, the composite

---->

t 1\ E

---7

t 1\ E

vanishes. Now we iterate the process. We define a sequence

to -----+ tl -----+ t2 -----+ ... in 'J', together with a map from the sequence to x. Define t 0 -----+ x to bet -----+ x. Suppose we have defined, for each i :=::; n, ti -----+ x and maps ti-l -----+ ti, with all the morphisms compatible. Then tn -----+ tn+l -----+xis defined to be tn -----+ tn -----+

as above. Put r =

~ tn.

t

= t0

X

Clearly, the map t -----+ x factors as

-----+ Hocolim tn

--t

x

To complete the proof of the Lemma, we need to show that Hocolim tn E 'J' n 'J'E. In other words, up to replacing r by an isomorph, we may choose

rER.

D.l. LOCALISATION WITH RESPECT TO HOMOLOGY

415

We have a triangle in 'J

By construction, ti E 'Ja. Since a > No and 'Ja is a-localising, the coproduct of countably many objects in 'Ja is in 'Ja. Thus in the triangle, the two coproducts lie in 'Ja. Since 'J"' is triangulated, Hocolim ti E 'J"'. Now we may smash the above triangle withE, and by D.l.l.4, we get a triangle

By D.l.l.5, the smash product commutes with coproducts. The triangle above naturally identifies with 00 l-shift 00 II{tiAE} II{tiAE}------+ i=O i=O But the third vertex in the triangle 00

II {ti A E} i=O

l-shift

{

~ti } AE.

00

II {ti A E} ------+ ~ {ti A E} i=O

is by definition Hocolim {tiAE}; we deduce a (non-canonical) isomorphism

~ { ti

{ Hocolim ti} A E We will prove that

~ {ti AE}

A E}.

vanishes. From the isomorphism above,

we deduce that { Hocolim ti} A E vanishes, that is { Hocolim ti}

E

'JE·

Now puts= 'J~ 0 , and let 1T: 'J----+ ex(S 0 P,Ab) be the usual homological functor, commuting with coproducts. It takes the triangle

oo {t.. A E} l-shift IIoo {t.. A E} H 1" { E} II ·'' ·'' ------+ ~tit\ i=O i=O to the long exact sequence --+

fi 1r{

i=O

ti A E}

l-shift

fi 1r{

ti A E} ------+

1r Hocolim {ti A E}

--+

i=O

Now note that S = 'J~ 0 , and ex(S 0 P,Ab) satisfies [AB5]. The map 1- shift is therefore injective, with cokernel c~ 1r{ ti A E}. From the long exact

416

D. THE CATEGORY OF SPECTRA

sequence, colim 1r{ti 1\ E}. ---+

But this is an equality of two objects in ex(S 0 P,Ab), that is two functors s ---+ Ab. We can evaluate these at sn E s = 'J~O. Recalling that ?rt = 'J(-,t)is, and so 1rt(s) = 'J(s,t), we have

'J(Sn, Hocolim {ti Now the map ti

---+

1\

colim 'J(Sn, {ti 1\ E} ).

E})

---+

ti+l was constructed so that the induced map

'J(Sn, {ti 1\ E}) -

'J(Sn, {ti+l 1\ E})

vanishes. Therefore colim 'J(Sn, {ti 1\ E}) ---+

'J(Sn, That is,

~

~{ti

= 0. But then

1\E}) = 0.

{ti 1\ E} is an object in 'J, so that every map

sn-

~{ti/\E}

vanishes. By D.l.l.2, Hocolim {ti 1\ E}

= 0.

D

LEMMA D .1.11. Let a be a regular cardinal, a > No. Let E be an object of 'J. Let R be a set of representatives for all isomorphism classes of objects in 'J n 'JE. Then R is an N1-perfect generating set for 'JE.

Proof: Let us first prove that R generates. It is a set of objects closed under suspension (up to isomorphism). If x is a non-zero object of 'JE' then by D.1.1.2, there exists a non-zero map in 'J

sn-x. But sn E 'J~O c 'J, and factor the above as

X

E

'JE. Lemma D.l.lO now tells us that we may

with r E R. Thus there is a non-zero map r ---+ x, in 'JE. Next we want to prove R to be N1-perfect. Note that 'J n 'JE is the intersection of two triangulated subcategories of 'J, hence is triangulated. And R is a set of representatives for the isomorphism classes. By Lemma 3.3.5, to show that R is N1-perfect, it suffices to prove that every element is N1-good. Suppose therefore that we are given a countable set of objects {xi E 'JE I 0 :=:; i < oo }. Let r be an object of R, and suppose we have a map 00

r-

IJxi.

i=O

D.l.

LOCALISATION WITH RESPECT TO HOMOLOGY

417

Now r E 'J'"', and the objects of 'J'"' form an ,/3-perfect class for every infinite cardinal ,13, in particular an N1-perfect class. The above map therefore factorises as 00

i=O

i=O

with ti E 'J'"'. But for each i, the map fi : ti

--+

xi is a morphism from

tiE 'J'"' to xi E 'J'E. By Lemma D.l.lO, the map must factor as

ti with ri E R. The map r

g, ~

h, ~xi

ri

--+ll:o xi

therefore factors as

00

00

r

~

II ti

IIgi i=O

~

00

00

II ri

II hi i=O

---7

i=O

i=O

00

II xi, i=O

and we conclude that r is N1 -good, and since this is true for all r E R, R is N1-perfect. D THEOREM D.l.12. As above, 'J' is the homotopy category of spectra. Let a be a regular cardinal, a > No. Let E be an object of'J'"'. Let R be a set of representatives for all isomorphism classes of objects in 'J'"' n 'J'E. Then the following hold D.l.l2.1. 'J'E satisfies the representability theorem, and 'J'E =

(R). D.l.l2.2. There is a Bousfield localisation functor for the inclusion 'J'E C 'J'. D.l.l2.3. For any regular cardinal ,13 2: a, we have

{'J'E},B

= 'J'f:l n 'J'E = (R),B,

and the category 'J'E is ,/3-compactly generated. D.l.l2.4. For any regular cardinal ,13 2: a,

{'J'/'J'E },B

'J'f:l I {'J'E },B'

and the category 'J'/'J'E is ,/3-compactly generated. D.l.l2.5. The representability theorem holds for 'J' j'J'E·

Proof: In Lemma D.l.9, we learned that the subcategory 'J'E C 'J' is localising. In particular, it is a triangulated category satisfying [TR5). In Lemma D.l.ll, we learned that the set of objects R C 'J'E is an N1 perfect generating set. Theorem 8.3.3 therefore applies; we conclude that

D. THE CATEGORY OF SPECTRA

418

':JE satisfies the representability theorem, and that ':JE = (R). That is, we conclude D.1.12.1. Now observe that 'JE is a localising subcategory of ':J, and the representability theorem holds for ':JE· The category ':J has small Hom-sets, and we can therefore use Proposition 9.1.19, to conclude that the inclusion ':JE c ':J has a right adjoint, and a Bousfield localisation functor exists for the pair ':JE c ':J. This proves D.l.12.2. In 0.1.12.1, we saw that ':JE = (R), where R C ':J"'. Therefore Theorem 4.4.9 now applies. If f3 is any regular cardinal > a, we conclude that

{':JE}.B = ':J.B n':JE

=

(R).B.

Since R is a set, the category {':JE}.B = (R).B is essentially small. Since {':JE }.6 contains R, {':JE }.6 generates ':JE. Therefore 'JE is (3-compactly generated. This establishes 0.1.12.3. But Theorem 4.4.9 goes on to tell us more. It asserts that, for all regular f3 2 a, the natural map ']'.6 j{':JE}.B

--t

{':Jj':JE}.B

is an equivalence of categories. It follows that the category {':J j':JE}.6 is essentially small; expressing it as ':J.B / {':JE }.6, we see that it is the quotient of two essentially small categories. Now consider the quotient functor

F: ':J

--t

':Jj':JE.

We have that p- 1{'1'/':JE}.B :J ':J.B. For any regular 'Y > (3,

p-1( {':J/':JE}.B)

'Y

is a 1-localising subcategory of ':J, containing T. Hence it contains (T) "~. Taking the union over all 'Y, we have ':J

c

and hence

But then {':J/':JE }.6 generates ':J j':JE; the two notions of generation coincide. See Proposition 8.4.1. This establishes that ':J j':JE is (3-compactly generated. Thus we have proved 0.1.12.4. Finally, the category ':J/':JE is well-generated by 0.1.12.4. Hence the representability theorem holds for ':J j':JE• by Theorem 8.3.3. This establishes 0.1.12.5. 0

D.l.

LOCALISATION WITH RESPECT TO HOMOLOGY

419

REMARK D.l.13. From D.l.12.2, we learn that a Bousfield localisation functor exists for the pair 'JE C 'J. The category ..L'JE is usually called the category of E-local spectra, for the homology theory E. From Theorem 9.1.16, we know that there is a natural equivalence ..L'JE

=

'Jf'Je.

In other words, in D.l.12.4 and D.l.12.5, we have proved that ..L'JE is /3compactly generated, and satisfies the representability theorem. The fact that a Bousfield localisation exists for the pair 'JE c 'J was first proved by Bousfield. What is new here is that we prove, for /3 2:: a, that both ..L'JE and 'JE are /3-compactly generated triangulated categories. Of course, since the concept did not exist before this book, this is new. But the concrete consequence is that both ..L'JE and 'JE satisfy the representability theorem. This is completely new. The category 'J is N0-compactly generated, and if S = ']"No, then cx(S 0 P,Ab) has enough injectives. Theorem 8.6.1 applies, and we learn that the representability theorem holds for the dual of 'J. It is natural to wonder whether the representability theorem holds for the dual of ..L'JE and 'JE. For 'Ie the answer is yes. Let us quickly prove this. LEMMA D.1.14. Let E be an object of'J. The representability theorem holds for the dual of the category 'JE. Proof: There must be a regular cardinal a so that E E (T) a: = 'Ja:. Theorem D.l.12, more precisely D.l.12.2, asserts that a Bousfield localisation functor exists for the pair 'JE C 'J. Let J : 'J ---+ 'JE be right adjoint to the inclusion I: 'JE C 'J, as in Proposition 9.1.18. Being a right adjoint of a triangulated functor, J is triangulated, and takes products to products. Now let H : 'JE ---+ Ab be a homological functor taking products to products. Then H J : 'J - - t Ab is a homological functor taking products to products. By Theorem 8.6.1, the representability theorem holds in the dual of 'J. Therefore,

HJ(-)

'J(h,- ).

Now, for any x E ..L'JE' we have J(x) = 0, and hence 'J(h, x) = H J(x) = 0. Thus h E {..L'Je} ..L = 'JE' and we conclude that the functor H is representable in 'JE. D REMARK D.1.15. It is very natural to ask, whether the duals of thecategories ..L'JE also satisfy the representability theorem. And the answer is that I do not know. There is no simple trick, allowing us to reduce the problem to a question about 'J. And since the categories ..L'JE are /3-compactly generated only for large /3, we cannot be sure that, if S = 'JP, there will be

420

D. THE CATEGORY OF SPECTRA

enough injectives in ex(S 0 P,Ab). In other words, Theorem 8.6.1 does not apply. We should maybe stress the nature of our estimate for (3. If (3 is a regular cardinal, and E E 'J'f3, then the categories 'J'E and .l'J'E are both (3compactly generated. This estimate is not best possible, but it illustrates that the bound depends on E.

D.2. The lack of injectives In Section C.4, we saw that if D(R) is the derived category of a discrete valuation ring R, if a is any regular cardinal > No, and if S = D(R)"', then the category ex(S 0 P,Ab) does not have enough injectives. Even worse, it has no cogenerator. In this Section, we want to transfer all of this to the homotopy category of spectra. For the remainder of this Section, 'J' is the homotopy category of spectra. Let p be a prime number p -=f. 2, and let R = Z(v)• the localisation of Z where all primes other than p are inverted. The ring R is a discrete valuation ring. Now we wish to quote some results from [24]. SuMMARY D.2.1. The proofs of the following facts may be found in

[24].

D.2.1.1. There is a functor F : D(R) ---+ 'J', taking a p-local abelian group to its Moore space. This functor is triangulated, and preserves coproducts. D.2.1.2. Note that the category D(R) satisfies the representability theorem. By Theorem 8.4.4 of the present book, the functor F : D(R) ---+ 'J' has a right adjoint P : 'J'---+ D(R). In [24], we essentially compute the restriction of this right adjoint, to the subcategory 'J'P C 'J', of p-local spectra. The functor P : 'J'P ---+ D(R) is denoted II in [24], but here the letter 1r is reserved for the projection 7r: A('J')---+ ex(S 0 P,Ab). D.2.1.3. On the category of p-local spectra, the functor P takes a spectrum to a direct sum of suspensions of its stable homotopy groups. Thus, on the subcategory 'J'P C 'J', the functor P commutes with coproducts. The functor F commutes with all coproducts by D.2.1.1, and takes any x E D(R) to Fx E 'J'P. It follows that the functor P F : D(R) ---+ D(R) commutes with coproducts. LEMMA D.2.2. Let a be a regular cardinal, a > N0 • The functor F above, which takes an abelian group A to its Moore spectrum, satisfies the hypothesis

F{D(R)"'}

c

'J'"'.

421

D.2. THE LACK OF INJECTIVES

Proof: The explicit description ofF makes this obvious. The point is that the Moore space of an abelian group with fewer than a elements is a space with fewer than a cells. It is necessary here to take a > N0 , since an Nocompact object of D(R) is a coproduct of suspensions of finitely generated Z(v)-modules, which need not be finitely generated abelian groups. The Moore spectrum on them is made up of finitely many p--local spheres, but countably many ordinary spheres. If we consider F as a functor to p--local spectra, then it is true that F { D(R)~-to}

c

It is for this reason, that the computation of the adjoint P is much easier on 'JP C 'J. D LEMMA D.2.3. Let F : D(R) ~ 'J and P : 'J of adjoint functors in D.2.1.2. For every object x adjunction induces a map

D(R) be the pair D(R), the unit of

~

E

"lx : X -----> P Fx.

This map is a split monomorphism in the triangulated category D(R). Proof: By Lemma C.4.2, the object x E D(R) is isomorphic to a coproduct of suspensions of R-modules; DO

n==-oo

Since P F respects coproducts by D.2.1.3, the unit of adjunction TJx:

X

----->

PFx

can be identified as the coproduct map DO

DO

DO

n=-oo

n=-oo

n=-oo

It suffices to show that, for each n E Z, the map 'f/z;nxn :

:Enxn

----->

PF:Enxn

is a split monomorphism. Desuspending, we may assume n = 0. Thus we are reduced to proving that TJx:

X

----->

PFx

is a split monomorphism, where x is an R-module, viewed as a complex in D(R), concentrated in degree 0. But now we know both F and P quite explicitly. Fx is the Moore spectrum on the abelian group x, and P Fx is the stable homotopy of this

D. THE CATEGORY OF SPECTRA

422

Moore spectrum. The map x = H 0 (x) - - t H 0 (PFx) is an isomorphism, since the zeroth stable homotopy group of the Moore spectrum on a group xis just x. Now Lemma C.4.2 tells us that H 0 (PFx) is a direct summand ~P~. D LEMMA D.2.4. Let 'J be the category of spectra, D(R) the category of p-local abelian groups, as in Summary D.2.1. Let a: be a regular cardinal, a:> No. Put G('J)

ex ({'J"'Yp, Ab),

G(R)

t:x({D(R)"'} 0 P,Ab).

We want to extend the adjoint functors F, P to G('J) and G(R). We will prove D.2.4.1. There is a functor F : G(R) adjoint P: G('J) - - t G(R).

G('J), with a right

--t

D.2.4.2. Let x be an object of G(R). The unit of adjunction

is a monomorphism in the abelian category G(R). Proof: We are given a pair of adjoint functors F : D(R)

--t

P: 'J - - t D(R).

'J,

By Lemma 5.3.8, these extend to give a pair of adjoint functors on Freyd's universal abelian category A('J). That is, we have a pair of exact functors of abelian categories A(F) : A(D(R))

--t

A('J),

A(P): A('J)

--t

A(D(R)),

and A(F) is left adjoint to A(P). Given any object x E A(D(R), we may embed it by a map f : x - - t i, with ian injective object of A(D(R). Corollary 5.1.23 asserts that injective objects i E A(D(R) lie in the subcategory D(R) C A(D(R). By the naturality of the unit of adjunction ry, there is a commutative square f

X

------+

i

A(P)A(F)x

A(P)A(F)f

A(P)A(F)i

D.2. THE LACK OF INJECTIVES

423

Lemma D.2.3 tells us that, for objects i E D(R), the map 'T/i A(P)A(F)i = P Fi is a split monomorphism. The composite

~

X

--+

i

1ry; A(P)A(F)i is theorefore the composite of two monomorphisms, and is mono. It is equal to the composite X

ryx

1

A(P)A(F)x

A(P)A(F)f

A(P)A(F)i,

"'x :

and we deduce that the map x --+ A(P)A(F)x must be mono, for every x E A(D(R)). Now we have to pass from A('J) and A(D(R)) to the quotient categories

G('J) =

ex ( {'Ja} op, Ab),

G(R)

ex({D(RttP,Ab).

=

Recall that, by Proposition 6.5.3, for every triangulated category 'J, there is a functor 1r:

A('J) = D('J) ~ ex( {'JatP,Ab),

which is a Gabriel quoient map. In Lemma 6.5.6 and Remark 6.5.8, we even identified the kernel of 7rj viewed as a subcategory of D('J), it is the a-phantom maps. We have diagrams

A(D(R)) ~ A('J)

1

1n

G(R)

G('J)

n

and

A('J) ~ A(D(R))

To prove that these diagrams can be completed, up to canonical equivalence, to

A(D(R)) ~ A('J)

n1 G(R)

~ A(D(R))

and

1n ~ G('J)

1n G('J)

amounts to showing that F : D(R) --+ 'J and G : 'J a-phantom maps to a-phantom maps.

G(R) --+

D(R) take

424

D. THE CATEGORY OF SPECTRA

For F, one reasons as follows. The functor F : D(R) ---. 'J' respects coproducts. By Proposition 6.2.6, so does the functor 7r :

'J'

G ( 'J')

----+

Hence the composite

D(R)

F

----+

'J'

is a homological functor respecting coproducts. But then the abelian category G('J') = t:x({T"'YP,Ab) is an abelian category satisfying [AB5('-")], and from Corollary B.2.4, the composite 1rF above must take a-phantom maps in D(R) to zero in G('J'). So F takes a-phantom maps in D(R) to the kernel of 1r : D('J') ---. G('J'), that is to a-phantom maps in 'J'. Now we want to show that G also takes a-phantom maps to a-phantom maps. Let f : x ---. y be an a-phantom map in 'J'. Let s be an object of D(R)"'. We are assuming a > N0 , and so by Lemma D.2.2,

F{D(R)"'} Therefore, Fs E 'J'"', and since composites

c 'J'"'.

f : x ---.

y is an a-phantom map, all

f Fs----+ x--+ y

must vanish. But under the adjunction, this corresponds to

s

----+

Gx

Gf

--+

Gy.

Since all such composites vanish, for any object s E D(R)"', it follows that the map G f : Gx ---. Gy is a-phantom. We therefore have commutative squares

A(D(R)) ~ A('J')

nl

and

ln

G(R)

A('J') ~ A(D(R))

nl

a G('J') - - +

~ G('J')

ln G(R)

and by the universality of the factorisation through the Gabriel quotient, natural transformations descend. In particular, we have natural transformations

r;:

1 ----+ GF

and

€: FG--+ 1,

D.2. THE LACK OF INJECTIVES

425

and the composites

F

Fri

FGF ~ F

G

fiG

GFG~G

are both identities. In fact, the natural transformations are given by the formulas E:

= 7rc.

It follows formally that F is left adjoint to G. Finally, we know that for every x E A(D(R)), the unit of adjunction ryx :x ~ A(P)A(F)x

is mono. But the functor fix = 1rryx is also mono.

1r :

A(D(R))

G(R) is exact, and hence

--->

D

D.2.5. Let a be a regular cardinal, a> No. Let 'J be the homotopy category of spectra. Then the category G('J) = {'J"'VP,Ab) has no cogenerator. PROPOSITION

ex(

Proof: Let the notation be as in Lemma D.2.4. Suppose k is a cogenerator of the category G('J). I assert that Pk must be a cogenerator of G(R). We know, from Section C.4, that G(R) has no cogenerator. Hence our contradiction. It remains therefore to prove our assertion, that if k is a cogenerator of the category G('J), then Pk must be a cogenerator of G(R). Let x be an object of G(R). Then Fx is an object of G('J), and since k is a cogenerator, there is an embedding Fx ~

rrk. A

Since P : G('J) monomorphism

--->

G(R) is exact and respects products, we have a

PFx ~ I1Pk. A

But we know that the unit of adjunction x Hence so is the composite

--->

P Fx is a monomorphism.

x ~ PFx ~ ITPk; A

we deduce a monomorphism from x to a product of Pk's. Since xis arbitrary, P k is a cogenerator. D

426

D. THE CATEGORY OF SPECTRA

D.3. History of the results in Appendix D The existence of Bousfield localisation with respect to homology was proved by Bousfield in [6). Up through Lemma D.l.lO, the treatment given here is, at least in spirit, lifted directly from Bousfield. There are minor changes in detail, mostly because we do not assume the reader has any familiarity with spectra. But starting with our Lemma D.l.ll, our argument is entirely different. Bousfield does not appeal to a representability theorem. He directly constructs the localisation functor, as a colimit of a long sequence. One problem with the construction, is that it depends on lifting to models, and does not generalise well to triangulated categories other that spectra. Margolis [21) gave an argument based on a representability theorem, but he used the representability theorem for 'J. To apply it, one must show that the category 'Jf'JE has small Hom-sets. See [21) for details. What we do here, is appeal to the representability theorem for 'JE· Since the fact that such a theorem holds is new to this book, it is inevitable that our proof is, at this point, quite different from the older arguments. And along the way, we also prove the representability theorem for 'jE, and ..L'JE. The results of Section D.2, about the absence of injective objects, are completely new to this book. The reader is encouraged to read Section 8.5, in particular Proposition 8.5.18 and Lemma 8.5.20, to see what consequences one can draw, about a-phantom maps. All these facts, about a-phantom maps between spectra, are very new. The theory developed in this book has new and surprising applications, even to very old problems.

cr;

APPENDIX E

Examples of non-perfectly-generated categories E.l. If 'J is N0 -compactly generated, 'J0 P is not even well-generated Assume 'J is a well-generated triangulated category, as in Remark 8.4.3. We remind the reader what this means. First, 'J must satisfy [TR5). Choose a a large enough regular cardinal, and let S = 'Ja. By Remark 8.4.3, S is essentially small and, since a is chosen big, S generates 'J. Recall also what it means to generate 'J. REMINDER E.l.l. By Proposition 8.4.1, S following equivalent conditions hold:

= 'Ja c 'J generates 'J if the

E.l.l.l. If x is an object of 'J and, for all objects s E S,

Hom(s,x) = 0, then x = 0. E.l.l.2.

'J = (S). We remind the reader that E.l.l.l can be slightly rewritten as E.l.l.3. Suppose x is an object in 'J'. If

'J(-,x)ls = 0

E

then x is isomorphic to zero. In other words, if x under the functor

E

'J maps to zero

then x = 0. Suppose 'J is a-compactly generated. Suppose further that the category ex(S 0 P,Ab) has an injective cogenerator. If a = No, the existence of an injective cogenerator is automatic, but not for larger a. By Remark 8.5.22,

428

E. EXAMPLES OF NON-PERFECTLY-GENERATED CATEGORJES

this injective cogenerator is 'T (-, lBlC)/ 3 , for some object lBlC E 'J. Furthermore, for any object x E 'J, by Proposition 8.5.2, there is a natural isomorphism

'J(x,lBlC)

---+

£x(S 0 P,Ab) [ 'J(-,x)/ 3 , ':T(-,lBlC)/ 3

J.

Now we are ready to state the main result of this Section. PROPOSITION E.l.2. Let 'T be an a-compactly generated triangulated category, and let S = 'To:. Suppose there exists an injective cogenerator in £x(S 0 P,Ab), and letlBlC E 'T be chosen so that 'J(-,lBlC)/ 3 is an injective cogenerator. Then for any infinite cardinal (3, the object lBlC is not (J-small in the category 'J0 P.

Proof: Choose a non-zero object x E 'J, and choose a regular cardinal 'Y > max( a, (3). Let :J('Y) be the set of ordinals < 'Y· Here, we do not care about the order; we consider just the set. Let P be the set of all subsets of :J('Y) of cardinality < 'Y· The set P is ordered by inclusion, and since 'Y is regular and a < "(, the union of < a subsets of :J('Y) of cardinality < 'Y has cardinality < 'Y· That is, Pis a-filtered. We define a functor F: P----; £x(S 0 P,Ab) by the formula

IT 'T( -, x)/

F(p)

3.

iEp

Here, pEP is a subset of :J('Y), and F(p) is the product of 'T (-, x)/ 3 over the index set p. For each pEP, we have a monomorphism

F(p)

---+

IT

'T(-,x)/ 3 .

iE:J(r)

Therefore there is a map colimF(p) pEP

---+

IT

'T(-,x)/ 3 .

iE:J(r)

Because P is a-filtered, Lemma A.l.3 says that the colimit is the same, whether taken in £x(S 0 P,Ab) or eat(S 0 P,Ab). The map above clearly is not surjective. After all, S generates 'J, so there is a non-zero element f in 'T(s,x), for somes E S. Now

is not in the image of colimF(p) pEP

---+

IT

'T(-,x)ls·

iE:J(r)

This is clear, since the colimit may be formed in eat(S 0 P,Ab), where it is constructed pointwise.

E.l. NOT BOTH 'J' AND 'J"'P

429

We deduce an exact sequence colimF(p) with Q E

II

----t

pEP

'J(-,x)ls

----t

Q

----t

0,

iE~(r)

t:x(SoP,Ab)

'J(-,~C)Is,

non-zero. Mapping into the injective cogenerator we deduce that the kernel of the map

t:x(S P,Ab) 0

[II

'J(-,x)ls,

iE~(r)

t:x(S P,Ab) 0

1

'J(-,~C)Is]

'J(~,lffi No, this sum is < /3. In other words, we are given a cardinal /3, which is determined by the generating set T. We want to show that T cannot be N1-perfect. Given /3, we want to choose a cardinal '"Yanda vector space W of cardinality "f, so that not all maps ()()

IJw

~Q

i=l

can factor as ()()

()()

IJw

IT!i i=l

~

i=l

()()

IJv

~Q

i=l

with V of cardinality /3. Of course, we can increase /3. Replace j3 by 2!3. It will suffice to show that there is a cardinal '"Y and a vector space W of cardinality '"Y, so that not all maps ()()

IJw~Q i=l

434

E. EXAMPLES OF NON-PERFECTLY-GENERATED CATEGORIES

can factor as 00

IT!i

00

IJw

i=l

00

--t

i=l

IJv

-----t

Q

i=l

with V of cardinality 2!3. We propose to estimate the number of maps

IIW----->Q 00

i=l

and the number of maps 00

II v

00

00

IIw

i=l

-----t

Q.

i=l

For the purpose of the estimate, we will assume 'Y is very large, to be chosen later. E.2.1.1. Upper bound for the number of maps

II v

00

00

IIw

i=l

-----t

Q.

i=l

Let us estimate the number of maps of sets with this factorisation; clearly, the linear maps are fewer. For each fi : w

-----t

v

there are {2!3} "~ choices. This is the number of maps from a set W of cardinality 'Y to a set V of cardinality 2!3. But we are assuming 'Y large, in particular 'Y > (3. Hence

The number of maps 00

00

IIw

i=l

II1i

i=l

--t

00

IIv

i=l

is the number of countable sequences of fi, that is { 2"~} No

= 2-y·l-to = 2"~.

E.2. NON N 1 -PERFECTLY GENERATED 'J

435

The number of maps 00

rrv~Q i=l

is independent of 'Y· The set

IJ: V has cardinality 1

{2,a}No

= 2,B·No = 2,13.

The number of maps 00

IJv~Q i=l

is therefore The number of pairs of maps 00

00

IJw

IT!i

i=l

-----+

i=l

00

IJv-----+

00

IJv

i=l

Q

i=l

is therefore bounded by 21 22 /l.

If 'Y is large, then this is 2'. Some pairs of maps will, of course, give the same composite. But we have that, for 'Y large, the number of composites 00 00

IJw

IT!i

i=l

-----+

i=l

00

IJv

-----+

Q

i=l

is bounded above by 2'. E.2.1.2. Lower bound for the number of maps

IJw-----+ Q. 00

i=l

IJ:

The number of elements in the set 1 W is "(No. But this is also the dimension of 1 W, as a vector space over Q. This means we may choose a basis of cardinality "(No. Let 1 E Q be a basis for the !-dimensional vector space Q. I do not want to count all the linear maps

IJ:

00

IJw----+ Q.

i=l

436

E. EXAMPLES OF NON-PERFECTLY-GENERATED CATEGORIES

Instead, let us only count the ones with a matrix of zeros and ones. In other words, every basis element in 1 W goes either to 0 E Q or 1 E Q. The number of such linear maps, all of which are distinct, is the number of ways to divide the basis into two sets, the elements mapping to 1 and the elements mapping to 0. There are 2·/'o choices. Since this is only some of the linear maps, we have that the set of all linear maps is of cardinality bounded below by 27No . To complete the proof, we need to show that for a suitable choice of a very large 'Y,

TI:

27No > 27. This will mean that there are more maps

rrw---+ Q 00

i=l

than maps that admit factorisations 00

rrv---+

rrw 00

00

i=l

Q.

i=l

Now we proceed to show this. More precisely, we will show

E.2.1.3. Let (3 be any cardinal. There exists a 'Y

> (3 with

27No > 27. The author would like to thank Shelah for pointing out this argument. Define a sequence of cardinals ci, i E N by

E.2.1.3.1. c0 = (3. E.2.1.3.2. ci+l = 2c;. Now let 'Y = L::o ci. We get the estimate that 00

i=l 00

i=l

=

Therefore

27.

E.3. T AND 2'Y. 0

E.3. For 'J = K(Z), neither 'J nor 'J0 P is well-generated. Recall that if Z is the ring of integers, the triangulated category K(Z) is defined as follows. The objects are chain complexes of abelian groups. The morphisms are the homotopy equivalence classes of chain maps. The derived category D(Z) is the Verdier quotient of K(Z), where we divide by the subcategory of acyclic complexes. It is well-known that D(Z) is No-compactly generated. In fact, the set

T =

p~nz

In

E

Z}

is a generating set, and is contained in {D(Z)f~o = {D(Z)}(~o)_ In Corollary E.l.3, we saw that the category D(ztP cannot be well-generated. Now we will deduce LEMMA

E.3.1. The category K(ztP is not well-generated.

Proof: We assume K(ztP is well-generated, and deduce a contradiction. Consider the abelian group QjZ. It is an injective cogenerator in the category of abelian groups. For any object x E K(Z), we have

The right hand side vanishes only if H-n(x) = 0. In other words,

for all n E Z if and only if x is acyclic. Let R be the set of all suspensions of Q/Z. That is,

{En{Q/Z} I n

R

E

Z}.

In the notation of Definition 9.1.11, Rj_ is the class of acyclic complexes in

K(Z).

We are assuming S

= K(ztP is well-generated. By Proposition 8.4.2,

s

us~. ~

Since R is a set of objects in 'J, it must be contained in some S~. Let :R be the category generated by R. That is,

:R

= (R).

438

E. EXAMPLES OF NON-PERFECTLY-GENERATED CATEGORIES

We now find ourselves in the situation of Thomason's localisation theorem 4.4.9. Put 'J = Sj'R. Then for any "Y ~ (3, 'R~' =

(R)'Y,

and 'J'Y

= S~' j'JP.

In particular, 'J has small Hom-sets. We are assuming the category S = K(z)op is well-generated. By Theorem 8.3.3, S satisfies the representability theorem. The hypotheses of Example 8.4.5 are satisfied, and we conclude that the map

S

---->

S/'R = 'J

has a right adjoint. A Bousfield localisation functor exists for the pair = (R) c S. By Corollary 9.1.14, we deduce that 'R = S/'Rj_. On the other hand, 'Rj_ = (R)j_ = Rj_, and we computed that this is the category of acyclic complexes. This identifies 'R0 P as the quotient of gop = K(Z) by the subcategory of acyclics; that is,

'R

'](DP

= D(Z).

But 'R = (R), with R c '](/3. This makes 'R = D(ztP a well-generated triangulated category, contradicting Corollary E.l.3. D Now we want to prove LEMMA

E.3.2. The category K(Z) does not have a generating set.

Proof: Once again, we suppose the category K(Z) has a generating set, and prove a contradiction. Choose a generating set S. The objects are chain complexes of abelian groups. Since there is only a set of them, we may choose a regular cardinal a > ~o exceeding the maximum size of these abelian groups. The objects of S are chain complexes of abelian groups, whose cardinality is ~o, we may choose G = Z. But to say that

Z

---+

C

---+

A

vanishes for every Z ---+ C is to say that the map C ---+ A takes every element of C to 0 E A. We deduce that there exists a non-zero o:-phantom map

C---+ r.A.

440

E. EXAMPLES OF NON-PERFECTLY-GENERATED CATEGORIES

That is, the map is non-zero, but for any object x E D(Zt', the composite X

----t

c

---->

:EA

vanishes. In particular, if x is an abelian group of cardinality < a, the composite vanishes. Of course, a morphism in D(Z)

C

:EA

----t

is an extension of C by A; it corresponds to an exact sequence

0

----t

A

B

----t

C

---->

---->

0.

To say that the composite X

----t

c

---->

:EA

vanishes, is equivalent to asserting that when we pull back the extension via x--+ C, we get

0

----t

A

----t

1

0 ----t A and the extension 0

----t

A

B'

----> X

1

0

---->

1

----t

B

---->

C

---->

0

----t

B'

---->

X

---->

0

is split. And this is true for all abelian groups x of cardinality < a. Now let s be any object of the generating set S for K(Z). Writes as a complex - - - - * Xo - - - - * xl ----> x2 -------.

By the choice of a, each of the groups xi has cardinality< a. Suppose we are given a map of chain complexes ----t

x_l

----t

1

xo

----t

xl

-------.

x2

----t

B ............_._

c

1

0 ----t A The map factors as ----t

1

-------.

xo

----t

xl ............_._ x2 ............_._

----t

A

----t

B' ............_._ x2 ............_._

----t

A

----t

B ............_._

1

1

1

-------.

1............_._ 1............_._

----t

1

x3

1 1

c -------.

0

E.3. 'J AND ex({'J"'VP,Ab), 218-219 definition, 219 existence for every a, 219-220 from coproducts of compacts vanish, 369-370 in D(Z), 438-440 pretriangle definition, 33 examples, 34 new out of old, 34, 49 pretriangulated category, 29, 70 definition, 29 products existence of, 288 of triangles are triangles, 37

quotient categories, 74-99, 309 commutative squares, 85-86 compact objects, 143-144 embedding via Bousfield localisation, 316-317 equality of morphisms, 84~85 existence, 74-75, 84 isomorphisms, 90, 92 preservation of products, 107, 110 size of Hom-sets, 99-100, 137, 318 zero objects, 91 regular cardinal, 103 representability of functors, 275 for l-h -perfectly generated categories, 282-284 for duals of well-generated categories, 303-306 for well-generated categories, 285-286 Serre subcategories, 327 singular cardinal, 103 six functors, 318-319

small categories, 99-100, 137 small Hom-sets, 99-100, 137, 318 small object definition, 123 subcategory of, 124 is localising, 126 is triangulated, 124 spectra E-acyclics, 411-412 are well-generated, 417-418 Brown representability, 417-418 Brown representability for dual, 419420 E-local, 419 are well-generated, 417-418 Brown representability, 417-418 ex( {'J"'YP,Ab) does not have a cogenerator, 425 Bousfield localisation for homology, 417-418 Brown representability, 408 cardinal of 'J(Sn, x), 41Q-411 elementary properties, 407-408 functor to D(R), 420 descends to ex( {'J"} 0 P, Ab), 422425 respects a-compacts, 420-421 splitting idempotent, 65 triangle with 0, 42-45 subcategory generated by a set, 103-104, 106-107 localising, 106-107 of compact objects, 129, 130 inclusion relations, 129 is localising, 130 of small objects, 124 is localising, 126 is triangulated, 124 thick, 74, 99 thick closure, 75, 99, 147-149 triangulated, 60 Thomason localisation, 143-144 applied to finding 'J"', 409 TRO, 29 TRl, 29 TR2, 29 TR3, 30 TR4 equivalent formulations, 51, 60

INDEX

TR5 dual, 63 statement, 63 triangles contractible, 47, 48 distinguished, 29 products of, 37 summands of, 38 triangulated subcategories which are perfect classes, 115 universal homological functors, 163-166, 384-385 Verdier localisation, 74-99, 309 existence theorem, 74-75 size of Hom-sets, 99-100, 137, 318 well generated categories, 274 are unions of 'Jf3, 285-286 duals satisfy Brown representability, 303-306 neither K(Z) nor K(Z) 0 P, 437-441 not both 'J and 'J0 P, 427-431 satisfy Brown representability, 285286

449

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