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Categories T. S. Blyth University of St Andrews, Scotland

.....

Longman: London and New York

Longman Group Limited

Longman House. Burnt Mill. Harlow. Essex CM20 2JE. England Associated companies throughout the world

©T

S Blyth 1986

All nghts reserved, no part of thiS publication may be reproduced, stored in a retneval system, or transmitted in any for or by any means, electronic, mechanical, photocopymg, recordmg, or othefWlSe, wtthout the pnor wntten permission of the Publishers First published 1986 British Library Cataloguing In Publication Data Blyth. T S Categories I Categories (Mathematics) I Title 512' 55 QAI69 ISBN G-582-98804-7

Pnnted and bound in Great Bntain by Biddies Ltd, Gwldford and King's Lynn

Contents

Chapter 1 - Looking at the woods rather than the trees Exercises Chapter 2 - Particular objects and morphisms Exercises Chapter 3 - Universal constructions Exercises Chapter 4 - Factorisation of morphisms Exercises Chapter 5 - Structuring the morphism sets Exercises Chapter 6 -

Functors Exercises

Chapter 7 -

Equivalent categories Exercises

Chapter 8 -

Representable and adjoint functors Exercises

1 7

10 20 24

38 42

57 59 70

73 88 93 102 104 118

Solutions to the Exercises

120

Index

151

Preface

It was in the early 1940s that Eilenberg and MacLane, the founding fathers of category theory, introduced the notions of a functor and a natural transformation in order to explain what should be meant by the 'natural' nature of certain isomorphisms, e.g. that between a finitedimensional vector space and its bidual. The first authoritative account of the enormous development that followed was published by Mitchell in 1965. In the Preface to his book, Mitchell says that 'a number of sophisticated people tend to disparage category theory as consistently as others disparage certain kinds of classical music'. Whereas the latter assertion is as true as before, the fonner no longer obtains. However, it is true to say that many shy away from categories, being put off by what they regard as the 'abstraction of all abstractions'. Indeed, though there are several texts available on the subject they tend to be immensely encyclopaedic, demand a huge amount of knowledge on the part of the reader, and all too often 'get to the big guns' too quickly. The result is that the beginner, meeting with the idea of a functor on page two, cannot help but feel that living on the first floor provides a nice view but to get anything done he has, often with some difficulty, to reach right down into the basement. The purpose of the present text is therefore quite simple : to provide, at an advanced undergraduate or beginning graduate level, an introductory course that can be covered in a couple of dozen lectures with enough material to provide a good idea of what categories are about and to whet the appetite of those who wish to proceed to the texts that are written by the experts in the field.

1 Looking at the woods rather than the trees

During the last century, mathematics has been revolutionised by the development of the theory of sets. Abstract algebra, topology, probability theory, the theory of functions (to name but a few) are wholly dependent on this foundational basis. In studying topics such as these, one has of necessity to adopt the working compromise of an intuitive or 'naive' approach to set theory, a rigorous axiomatic development usually being out of the question. The reader will be aware that there are weaknesses in too naive an approach and that a more careful examination of the foundations is required in order to avoid paradoxical situations. For example, if one were to consider as a set the 'collection' U of all sets then the subset V ={X E U; X 1:- X} would have the property V E V if and only if V 1:- V (Russell's paradox). Now in what follows we shall have occasion to consider such 'large collections' as 'all sets', 'all groups', 'all morphisms of groups', etc., these being entities which cannot be regarded as sets. Not wishing to become engrossed here in foundational considerations, we shall therefore agree to extend our naive approach to set theory by referring to all 'collections' of objects as classes, and think intuitively of a set as a 'small' class. At the axiomatic level, this is the approach that is formally adopted in the von Neumann-Bernays-GOdel system. Here a distinction is made between sets and classes; sets may be elements of classes but not conversely, and there is postulated a universal class to which all sets belong (though this class is not a set, so that Russell's paradox is avoided).

Chapter 1

2

With this extension to our vocabulary, we can proceed to 'take the plunge at the deep end' and explain precisely what is meant by a category. To the novice this will come as a rather involved concept, but we shall quickly provide many examples that not only will be familiar but also will consolidate the notion. Definition. A category is a class C of objects such that (1) for every pair of objects X, Y of C there is a set Morc(X, Y), called the set of morphisms from X toY, with More( X, Y) and Morc(X', Y') disjoint unless X= X' andY = Y' in which case they coincide; (2) for any three objects X, Y, Z of C there is a mapping Morc(X, Y) x Morc(Y, Z) _, Morc(X, Z), described by (g, f) ,..... fog, with the following properties : (a) for every object X there is a morphism idx E Morc(X, X) which is a right identity under o for the elements of Morc(X, Y) and a left identity under o for the elements of Morc(Y, X); (,B) o is 'associative' in the sense that when the composites h o (go f) and (hog) of are defined they are equal. Before we proceed to consider some examples, a few comments on notation are in order. Note first that, despite the notation o used in the definition, the morphism sets More( X, Y) need not be sets of mappings. Nevertheless, we shall write the elements of Morc(X, Y) as though they were mappings; i.e. we shall use the notation f : X -+ Y. It is for this reason that what we call morphisms are often also called simply arrows. As we shall see, this notation will cause no confusion. The morphism idx : X-+ X will be called the identity morphism on the object X; note that it is unique with respect to the property (a). The above definition shows that we can define a category essentially by specifying three things : (1) the class of objects; (2) the morphism sets; (3) how to 'compose' morphisms. In practice, when the morphisms and the operation of composition of morphisms are obvious, it is common to refer to a specific category by referring only to its objects. Example 1.1 The category Set of sets. Here the class of objects consists of all sets, the morphisms are ordinary mappings between sets, and o is the usual composition of mappings. The reader will quickly verify that the various items in the definition of a category are satisfied, with the identity morphisms being the usual identity mappings. Indeed, the notation is chosen such that this is almost obvious. Likewise we can form the category FSet of finite sets.

Looking at th.e woods

3

Example 1.2 The category Grp of groups. Here the class of objects consists of all groups, the morphisms are the ordinary group morphisms, and o is the usual composition of mappings. Likewise we can form the category FGrp of finite groups. Example 1.3 The category Ab of abelian groups. Example 1.4 The category Ring of rings. Here the class of objects consists of all rings, the morphisms are the ordinary ring morphisms and o is the usual composition of mappings. Example 1.5 The category Ring 1 of rings with a 1. Here the class of objects consists of all rings with a 1, the morphisms are the !-preserving ring morphisms, and o is the usual composition of mappings. Example 1.6 The category Top of topological spaces. Here the class of objects consists of all topological spaces, the morphisms are the continuous maps (i.e. those for which the pre-image of an open set is open), and o is the usual composition of mappings. Example 1.7 The category FVect of vector spaces over a field F. Here the objects are vector spaces and the morphisms linear transformations. Example 1.8 The category RMod of (left) R-modules. This is one of the most important categories that we shall encounter. A left R-module can be thought of essentially as 'the same as a vector space over a field, except that the scalars, instead of coming from a field F, belong to a ring R with a 1'. More formally, if R is a ring with a 1 then a left R-module is an additive abelian group M together with a mapping R x M ---> M (sometimes called an action of R on Jvf), denoted by ( r, m) ~ rm, such that (1) (V'x, y E M)('v'r E R) r(x+ y) = rx + ry; (2) (V'x E M)(V'r, s E R) (r + s)x = rx + sx; (3) (V'x E M)(V'r, s E R) (rs)x = r(sx);

(4) ('v'x EM)

1RX =X.

The adjective 'left' signifies that the scalars (elements of R) are written on the left of the elements of M. The objects of RMod are then the left R-modules, the morphisms are the R-morphisms (mappings that preserve addition and the action by the scalars, which is precisely what a linear transformation does), and o is the usual composition of mappings. Example 1.9 The category Sgp of semigroups. Example 1.10 The category Mon of monoids (semigroups with a 1). Here the morphisms are !-preserving. Example 1.11 The category Ord of ordered sets. Here the objects are sets on which there is defined an ordering ~ (i.e. a relation that is reflexive, anti-sy=etric, and transitive), and the morphisms are the

Ckapter 1

4

mappings I : A -+ B that are isotone (or order-preserving) in the sense that if x::; yin A then l(x) ::; l(y) in B. Example 1.12 To illustrate the remark made above that the morphisms of a category need not be mappings, consider an ordered set (E, :S) and construct a category E as follows : take as objects of E the elements of E and for a, b E E define

if a::; b; otherwise. To define composition : if a E MorE( a, b) and {3 E MorE( b, c) then these sets are not empty, and this happens only when a ::; b and b ::; c whence, by the transitivity of::;, we have a::; c and so MorE( a, c) -F 0. We thus define {3 o a to be the single element (a, c) of MorE( a, c). The reflexivity of::; ensures the existence of identities; we have id., = (a, a). It is now easily seen that E is a category. Example 1.13 A lattice is an ordered set L such that every two-element subset {x, y} has a supremum x V y and an infimum x II y in L. The category Lat (the category of lattices) can be formed by taking as objects the lattices and as morphisms the mappings that preserve suprema and infima of finite non-empty subsets. Example 1.14 A complete lattice is an ordered set in which every nonempty subset has a supremum and an infimum. The category CLat (the category of complete lattices) can be formed by taking as objects the complete lattices and as morphisms the mappings that preserve suprema and infima of arbitrary non-empty subsets.

In a given category C, a diagram of objects and morphisms is said to be commutative if all composite morphisms from a given departure object to .a given arrival object are equal. Thus, for example, the diagrams I

·-- ---t•

.______.. {3

are commutative if and only if, respectively, h

= go I

and {3 o a

= go I.

Example 1.15 Let C be a category and let A be a fixed object of C. We can form a category A- as follows : for objects take the morphisms of

C of the form A_!___. X and for the set of morphisrns from A_!___. X

Looking at the woods

5

to A-~ Y take the morphisms h in C such that

is commutative. Similarly, we can form a category -A as follows : for objects take the morphisms of C of the form X__!_____. A and for the set of morphisms from X__!_____. A to Y ____'!____, A take the morphisms g in C such that

is commutative. These categories are often called the comma categones, the first from and the second to the object A. Example 1.16 Let C be a category. We can form a category as follows : for the objects take the morphisms of C and for the set of morphisms from A__!_--+ B to C ~ D take the pairs a, fJ of morphisms in C such that the diagram

C---->D g

is commutative. composite of

Define composition of morphisms as follows

the

A'-_f__.B' and

a'

1

lp'

C'----->D' g'

exists if and only if A' = B and C' = D, in which case the composite is

6

Chapter 1

given by

A~B'

The resulting category is often called the arrow category over C. Definition. A category is said to be small if its objects form a set. Example 1.17 If Sis a monoid then Scan be regarded as a category with exactly one object (namely S), the morphisms being the elements of S and composition of morphisms being the semigroup operation. Thus every monoid is a small category. Example 1.18 If a category has only a few morphisms it is sometimes useful to express it in terms of a diagram that shows all the objects and all the non-identity morphisms. For example,

is a category with 3 objects and 7 morphisms (it being understood that since the convention is to exhibit all the non-identity morphisms we must have fog= idA, go f = id 8 , a= b o j, b =a o g). In contrast, note that

does not represent a category. Example 1.19 For every positive integer n we can consider (as in Example 1.12) the ordered set 0 C are residuated prove that so also is go I : A -+ C with (go !)+ = I+ o g+. Deduce that a category Ord + (the category of ordered sets and residuated mappings) can be formed by takivg as objects the ordered sets and as morphisms the residuated mappings.

1.9 A lattice is said to be bounded if it has a smallest element 0 and a greatest element 1. Say whether or not each of the following is true :

(1) the category Latri1 of bounded lattices and residuated mappings is a full subcategory of Ord+; (2} the category CLat+ of complete lattices and residuated mappings is a full subcategory of Latri1 ; (3) the category Lat01 of bounded lattices and {0, !}-preserving lattice morphisms is a full subcategory of Lat. 1.10 A boolean algebra is a bounded lattice L that is distributive (in the sense that the identity x A (y v z) = (x A y) V (x A z) holds) and complemented (in the sense that for every x E L there exists a necessarily unique x' E L such that x A x' = 0 and x V x' = 1). If A, B are boolean algebras and I: A---+ B is a {0, !}-preserving lattice morphism prove that f(x') = [l(x)]' for every x E A. Show that the category BAlg of boolean algebras and {0, 1}-preserving lattice morphisms is a full subcategory of La to!·

2 Particular objects and morphisms

We have already seen examples of categories in which the objects are sets endowed with some additional structure and the morphisms are the structure-preserving mappings. Such categories are called concrete. One of the main objectives of the theory of categories is to obtain general theorems with applications in concrete categories. To see how this can be achieved, we first show how notions that arise in concrete categories can be generalised to arbitrary categories. Of course, in so doing, we have to find properties that are independent of 'element-wise' arguments. By way of illustration, we observe (as the reader will no doubt be aware) that in the category Set the following statements concerning a mapping f : A -+ B are equivalent : {I) f is injective {in the sense that x f- y ==? f(x) f- f(y)); {2) f is left cancellable {in the sense that fog= f o h ==? g =h). In fact, to show that (I) =? {2) suppose that f is injective and that g, h : C -+ A are such that f o g = f o h. Then for all x E C we have f[g(x)) = f[h(x)) whence g(x) = h(x) and so g = h. As for {2) =? {I), suppose that f : A -+ B is left cancellable and that f(x) = f(y). Take C ={a} and define g,h: C-+ A by setting g(a) = x and h(a) = y. Then from f(x) = f(y) we have f[g(a)) = f[h(a)) so that fog= f o h, whence g =hand consequently x = g(a) = h(a) = y. Thus the notion of injectivity, which is usually defined as in (I) using elements, can be expressed as in (2) in terms only of mappings.

Particular objects and morphisffl8

11

Now the property of being left cancellable can clearly be considered in an arbitrary category, and leads to the following notion. Definition. Let C be a category. A morphism to be monic if it is left cancellable.

f :A

-+

B in C is said

We have just seen that in Set a morphism is monic if and only if it is injective. This is also true, for example, in Sgp, Grp, and RMod (see the Exercises for the details). Now it is clear from the proof of (1) =:> (2) above that in a concrete category every injective morphism is monic. However, as the following example illustrates, the converse is not true. Example 2.1 An abelian group G is said to be divisible if for every g E G and every positive integer n there exists g' E G such that g = ng'. Clearly, we can form a category DivAb by taking as objects the divisible abelian groups and as morphisms the group morphisms. Now both () and 'l/lL. are objects in DivAb; this follows from the observations and

r.q + 7L. =

n

(E..nq + z) .

Consider now the natural morphism ~ : () -+ '!/lL. described by ~(q) = q + lL.. This morphism is not injective, but is monic. In fact it is clear that Qis not injective. To see that it is monic, suppose that f, g : A --+ () are morphisms in DivAb such that f f- g. Then there exists a E A such that f(a)- g(a) = ~ f- 0 and sf- ±1. Since A is divisible we can 8 find b E A with a = rb. Then r

r[f(b)- g(b)] = rf(b)- rg(b) = f(rb)- g(rb) = f(a)- g(a) =s and so f(b)-g(b) =~which gives ~[/(b)] f- ~[g(b)] and hence 8 It follows that ~ is left cancellable, so is monic.

~off- ~og.

If Cis a category then the dual (or opposite) category is the category Cd that is defined as follows. The objects of Cd are the objects of C, and for every pair of objects A, B the set Morc•(A, B) is the set Morc(B, A). Composition of morphisms involves the obvious change of order. If P is a property that may be enjoyed by an object or morphism in C then the property pd dual to P is that obtained by postulating that an object or morphism enjoys pd in C if and only if it enjoys P in Cd. It is clear that the dual of pd is P. Dualising the notion of a monic morphism we obtain the following :

Chapter I!

12

Definition. Let C be a category. A morphism f : A --+ B in C is said to be epic if it is right cancellable (go f = h of==? g =h). In the category Set the epic morphisms are precisely the surjective mappings. To see this, suppose that f : A --+ B is right cancellable. If g, h : B --+ {0, 1} are given by

g(x) = {

~

if xEim/; otherwise,

h(x) = {

~

if xEim/; otherwise,

then for every yEA we have g[f(y)] = f(y) = h[f(y)] so that gof = hof and hence, by the hypothesis, g = h. Now if Im f C B then there exists x E B with x fl. Im /, and for such an element x we have the contradiction 0 = g(x) = h(x) = 1. Thus Im/ =Band f is surjective. Conversely, suppose that f : A --+ B is surjective and that g, h : B --+ C are such that go f = h of. If b E B then b = f(a) for some a E A and so g(b) = g[f(a)] = h[f(a)] = h(b). Consequently we have g =hand f is right cancellable. Thus we see that in Set the epic morphisms are the surjections. This is also true, for example, in Grp and RMod (see the Exercises). Now it is clear from the above that in a concrete category every surjective morphism is epic. However, as the following example shows, the converse is not true. Example 2.2 In the catgeory Sgp (or Ring), consider the inclusion map £ : 7l.--+ 4:). This morphism (with respect to multiplication) is not suryective, but is epic. In fact, it is clear that £ is not surjective. To see that it is epic, however, suppose that g, h : Q --+ A are morphisms in Sgp (or Ring) with go£= h o £. Then g(n) = h(n) for all integers n. For every m/n E Q we then have

g(m/n)

= g(m · n- 1 . 1) = g(m)g(n- 1 )g(l) = h(m)g(n- 1 )h(l) = h(m)g(n- 1 )h(n)h(n- 1 ) = h(m)g(n- 1 )g(n)h(n- 1 )

= h(m)g(l)h(n- 1 ) = h(m)h(l)h(n- 1 ) = h(m/n)

so that g = h. Thus

£

is right cancellable, so is epic.

The following simple observations will be used in the sequel without reference. Assuming that f o g is defined, we have :

Particular objects and morphisms

13

(a) if/, g are monic then fog is monic; In fact I o go h = I o g o k => g o h = g o k => h = k. (,8) if I, g are epic then I o g is epic; In fact h o I o g = k o I o g => h o I = k o I => h = k. ('y) if I o g is monic then g is monic; In fact g o h = go k => I o go h = I o go k => h = k. ( 6) if I o g is epic then I is epic; In fact h o 1 = k o 1 => h o 1 o g = k o 1 o g => h = k.

In the category Set there is another characterisation of injective mappings that does not involve elements, namely that I : A --+ B is injective if and only if I has a left inverse, in the sense that there exists g : B --> A such that go I = idA. To see this, suppose first that I is injective. Then for every bE Im I there is a unique x E A such that l(x) =b. Call this element b. Let C< be a fixed element of A and define g : B --+ A by g(b) =

{!

if bE Iml; otherwise.

Then we have (\fa E A)

g[l(a)] =[{;)=a

and so g o I = idA. Conversely, if there exists g : B g o f = idA then f oh

=f

o

k ==> g o f o h

= go f

o

k ==> h

--+

= idA oh =

A such that idA ok

=k

and so f is monic. We can therefore consider the following notion in an arbitrary category. Definition. Let C be a category. A morphism f : A --> B in C is called a section if there is a morphism g : B --> A such that g o f = idA.

We have just seen that in Set the sections are simply the monic morphisms. Now it is clear from the above that every section is monic. However, as the following example illustrates, the converse is not true. Example 2.3 In zMod consider the morphism f : 7L. --+ 7L. given by f(n) = 2n. Since f is injective there certainly exists a mapping g : 7L.--> 7L. such that go I= idz. But no such ll.-morphism can exist. For, if h were such a ll.-morphism then we would have

(Vn Ell.)

2h(n)

= h(2n) =

h[f(n)]

= id;z(n) = n. = 1 which is impossible

In particular, therefore, we would have 2h(l) since 2x = 1 has no solution in ll.. Thus we see that f is not a section. It is monic, however, since it is injective.

Chapter 2

There is, of course, the notion that is dual to that of a section : Definition. Let C be a category. A morphism f : A --+ B in C is called a retraction if there is a morphism g : B --+ A such that f o g = idB. In the category Set the retractions are precisely the surjective mappings. To see this, suppose that f : A --+ B is surjective. For every b E B there exists a E A with f(a) = b. For each bE B choose and fix such an element a, say ab. (Note that here we are making use of the axiom of choice.) Now define g: B--+ A by the prescription

('rib

E

B)

g(b) = ab.

Then for every bE B we have f[g(b)] = f(ab) = b, so that fog= idB and f is a retraction. Conversely, if there exists g : B --+ A such that f o g = idB then

hof = kof

=

hof og= k of og

=

h = h o idB = k o idB = k

shows that f is epic, hence surjective. It is clear from this that every retraction is epic. However, as the following example shows, the converse is not true. Example 2.4 In zMod consider the object G:lp defined as follows. Let p be a prime and let

G:lp = {x E (); (:Jk E Z)(:ln E IN)

x = kp-n}.

Then G:lp is a subgroup of(), and Z is a subgroup of QP. Let f : G:lv/Z --+ G:lv/Z be given by f(x + Z) = px + Z. Then it is readily seen that f is a Z-morphism. Since

kp-n

+Z =

p(kp-n-l

+ Z)

we see that f is surjective and so there is certainly a mapping h : G:lv/Z --+ G:lr/Z such that f o h = id. But no such Z-morphism can exist. For, suppose that h were such a Z-morphism. Then we would have

p- 1 + Z = f[h(p- 1 + Z)] = p[h(p- 1 + Z)] = h[p(p- 1

+ Z)J

= h(O+ Z)

=O+Z, which is impossible since x + Z = 0 + Z if and only if x E Z. Thus we see that f is not a retraction. It is, however, epic since it is surjective.

Particular objects and morphisms

Let us now combine some of the above notions : Definitions. A bimorphism is a morphism that is both monic and epic. An isomorphism is a morphism that is both a section and a retraction. Objects X, Y of a category C are said to be isomorphic if Morc(X, Y) contains an isomorphism. In this case we often write X ~ Y.

Note that f: A-+ B is an isomorphism if and only if there exists (a necessarily unique) g : B -+ A such that fog = idB and go f = idA. For, the existence of ht, h2 : B-+ A such that f o h 1 = idB and h2 of = idA gives

We write g as f- 1 • Clearly, it too is an isomorphism. Note also that in a concrete category every isomorphism is a bijection, hence is both monic and epic, hence is a bimorphism. In Set the converse is also true, but this is not so in general : Example 2.5 In DivAb the morphism Q: Cl:)-+ Cl:)/71. is monic (Example 2.1). Being surjective, it is also epic. Hence it is a bimorphism. Clearly, it is not an isomorphism.

For concrete categories the following diagram (in which an increasing line segment is taken to mean ==?) summarises the above discussion : epic

monic

./~b.1morpli/ Ism

suryectsve

~. lnJecttve

lt~b··'t· IJec sve ~sec tl.ton

re t rae wn

~ isomJphism~ Definition. A category is said to be balanced if every bimorphism is an isomorphism.

Clearly, Set is balanced. So also, for example, are Grp and RMod (see the Exercises). In a balanced category, however, not every epic need be a retraction, nor every monic a section (see Example 2.5). We now turn our attention to a particularly important type of object that may exist in a category.

16

Chapter 2

Definitions. An object U of a category C is said to be an initial object (or universally repelling) if, for every object X of C, the set More( U, X) is a singleton. Dually, U is said to be a terminal object (or universally attracting) if, for every object X of C, the set More(X, U) is a singleton.

When no confusion can arise, we shall often use the term universal object to mean an object that is initial or terminal. THEOREM 2.1

Universal objects of the same type are isomorphic.

Proof. We have to show that if U1 , U2 are universal objects of the same type then More( U1 , U2 ) contains (indeed consists of) an isomorphism. The proof is the same for U1 , U2 initial or U1 , U2 terminal. Suppose that Morc(U1 ,U2 ) = {o:} and More(U2,UI) = {,8}. Then o:o,B E Morc(U2 , U2 ) and so we must have o: o ,B = idu,- Similarly we have ,Boo:= idu,- It follows that o: is an isomorphism with ,B = o:- 1 • A is a morphism in Ring 1 then for every n E 7L. we have /(n) = n/(1) = n1A, which shows that f is uniquely determined. There is no terminal object in Ring 1 ; by convention, we assume that 0 i' 1 in a ring with a 1. Example 2.10 Let E be a non-empty set and let R be an equivalence relation on E. Consider the category C whose objects are pairs (X, f) where X is a non-empty set and f : E ---> X is a mapping such that x R y => f(x) = f(y) (in other words, such that R implies R 1 where RJ is the equivalence relation on E associated with f). Define More ((X,!), (Y, g)) to be the set of all mappings h: X ..... Y such that hoJ~-g:

We claim that the set E / R of R-dasses together with the natural map E ..... E/R, described by ~n(x) = [x]n, is an initial object in C. To see this, let g : E--> Y be such that x R y => g(x) = g(y). This means ~R:

Particular objects and morphisms that we have and so we can define a mapping h : E / R

--->

Y by setting

h([x]R) = g(x). Clearly, we have h o QR = g. That h is unique with respect to this property follows from the fact that if k is also such that k o QR = g then, since QR is surjective, hence epic, we obtain k = h by right cancellation. Example 2.11 Let S be a non-empty set and let C be a concrete category. Consider the category K whose objects are pairs (X,a) consisting of an object X of C and a mapping a : S ---+ X. Define MorK((X,a),(Y,,B)) to be the set of all morphisms h: X---+ Yin C such that h o a = .B :

An initial object (F, f) inK (when such exists) is called a free C-object on the setS. By way of illustration, consider the case where C is the category Sgp of semigroups. Let F denote the set of all finite sequences of elements from the given setS (with repetitions allowed). Clearly, F is a semigroup under the law of composition of 'concatenation', described by setting

Let f : S ---> F be given by f(x) = {x) where {x) denotes the sequence consisting of the single element x. Then we claim that ( F, f) is a free semigroup on the set S. To establish this, we must show that if Y is any semigroup and .B : S -+ Y is any mapping then there is a unique semigroup morphism h : F -+ Y such that h of= {3. Now an obvious way to define a mapping h : F -+ Y is by setting

That this mapping is a semigroup morphism is an immediate consequence of the fact that the multiplication in Y is associative. Moreover, for every x E S we have

h[f(x)j

= h[{x)] = f3(x)

Chapter f

18

so that hoI = {3. In order to show that h is unique with respect to this property, suppose that k : F ---+ Y is also a. semigroup morphism such that k o I= {3. Then we have

k(x 1 , ••• ,xm) = k[(xt) · · · (xm)] = k[(xt)]· · ·k[(xm)] = k[f(xt)]· · · k[l(xm)] = {3(xt) · · {3(xm) =h(xt,···•Xm) which shows that k = h a.s required. Thus we see that the semigroup F (consisting of the finite sequences of elements of S under concatenation) together with the above mapping I: S---+ F is a. free semigroup on the set S. By Theorem 2.1 we know that a.ll free semigroups on S are isomorphic. We can therefore choose to call the above semigroup F the free semigroup on S. Note that in the particular ca.se where Sis a. singleton, sayS= {a}, we can regard F a.s the multiplicative semigroup {an; n = 1, 2,3, ... }. The existence of other free C-objects on a. set S can a.lso be established, notably when C is Grp and RMod; see the Exercises. The notion of a. free semigroup is useful in proving that in Sgp every monic morphism is injective. To see this, suppose that t : X ---+ Y is a. morphism in Sgp that is not injective. Then there exist x, y E X with x f- y and t(x) = t(y). LetS= {a} be a. singleton and let cx,{3: S---+ X be given by cx(a) = x, f3(a) = y. If (F,!) is the free semigroup on S then there exist semigroup morphisms F---+ X such that o I = ex and j3 0 I= {3. Now clearly f- f3 but, since F can be taken to be the semigroup {an ; n = 1, 2, 3, ... }, we have t o = t o Thus t is not monic. Consequently, if t is monic in Sgp then it must be injective.

a

a, /3:

a

a

/3.

Definition. An object Z of a. category C is said to be a. zero object if it is both initial and terminal in C.

Suppose that C has a. zero object Z. Given objects A, B of C there exist, by universality, unique morphisms OAz : A ---+ Z and Oze : Z ---+ B. Consider the composite morphism Oze o OAz : A---+ B. We claim that this depends only on A, B and not on Z. To see this, suppose that Z' is also a. zero object of C. Then there exist unique morphisms f'J : Z---+ Z' and rJ 1 : Z'---+ Z. Since rJ 1 o f'J : Z---+ Z we have necessarily rJ 1 o f'J = idz. Now, by universality, we observe that

Particul11.r objects and morphisms and similarly

Z' __i____. Z ~~ B = Z' ~B. Consequently we have

Because of this observation we shall agree to write OzB oOAz as OAB. We call OAB the zero morphism from A to B and, when no confusion arises, denote it simply by 0. The following result shows that zero morphisms behave in a nice way. THEOREM 2. 2 If C has a zero object then for every morphism a of C

we have 0 o a

= 0 and a o 0 = 0.

Proof. For example, if Z is a zero object, we have

X~A~B=X~A~Z~B =X~Z~B

by universality

=X~B. Writing all zero morphisms as 0, it follows that 0 o a equality is established similarly. 0

= 0. The second

A category that has an initial object and a terminal object need not have a zero object. For example, Set has no zero object. Likewise, in the category depicted in Example 1.18, the object A is initial and the object C is terminal, but there is no zero object. Thus in the presence of an initial object and a terminal object a further condition is required in order that there exist a zero object.

Definition. A category C is said to be connected if Morc(A, B) for all objects A, B of C.

f.

0

The category depicted in the diagram g

A.....-!-- B====:c h

is a simple example of a category that is not connected; for example we have Mor(A, C)= 0.

Chapter 2

20

THEOREM 2. 3 Let C be a category with an initial object and a terminal object. Then C has a zero object if and only if C is connected.

Proof. If C has a zero object then it is clear from the discussion preceding Theorem 2.2 that C is connected. Conversely, suppose that C is connected. Let X be an initial object and Y a terminal object. We observe that if Morc(X, Y) ={a} then a is monic. In fact, since Cis connected, there exists f3 : Y -+ X and necessarily f3 o a = idx since X is initial. Thus a is a section and hence is monic. Suppose now that f,g E Morc(A,X). Then aof,aog E Morc(A,Y) and so, since Y is terminal, a o f = a o g. The fact that a is monic now gives f = g. Consequently we see that X is also a terminal object, and hence is a zero object. ¢

IExercises I 2.1 Show that a group can be considered as a category with a single object and every morphism an isomorphism. 2.2 Consider the additive semigroup IN as a category (as in Example 1.17). Show that every morphism in this category is a bimorphism, and that 0 is the only isomorphism. 2.3 Consider the following 'cube' of objects and morphisms in a given category C :

Suppose that all the faces except the top face are given to be commutative. Prove that if A4 -+ A 8 is monic then the top face is also commutative. 2.4

An additive abelian semigroup S is cancellative if (Va, b,c E S)

a + b = a + c ==* b = c.

Prove that in the category of cancellative abelian semigroups the canonical injection t : IN -+ 7L. is epic but not surjective.

Particular objects and morphisms

2.5

21

Show that in Grp the morphism

f(n) =

g

f : 7L. -+ 71. 2 given by

if n is even; if n is odd,

is surjective but not a retraction. 2.6 Prove that in a category C the following statements concerning a morphism f are equivalent : (1) f is an isomorphism; (2) f is epic and a section; (3) f is monic and a retraction. Deduce that if every epic in C is a retraction (dually, every monic is a section) then C is balanced. 2.7 Prove that the free semigroup on a singleton is isomorphic to the additive semigroup IN\ {0}. 2.8 Given a set S letS= S x {1, -1}. Associate with each xES the elements x 1 = (x, 1) and x- 1 = (x, -1) of S. Let E be the free semigroup on S. By Example 2.11 we can consider the elements of E as finite products of elements of S. Call these words and say that a word is reduced if x1 and x- 1 never stand in juxtaposition. Let F be the set of all reduced words in E, together with a symbol e that represents the empty word. Define a law of composition on F by reduced juxtaposition. Show that F is a group with identity element e. Now define f: S -+ F by setting f(x) = x 1 for every x E S. Show that (F, f) is a free group on the set S. [Hint. If G is a group and g: S-+ G is any mapping define h: F--> G as follows : if w is the empty word define h( w) to be lc; otherwise w is of the fonn x~'x; 2 • • x~" where e; = ±1, in which case define h(w) = [g(xt)]''[g(x 2 )]' 2 • • • [g(x,.)]'".] 2.9 Prove that the free group on a singleton is isomorphic to the additive group 71.. 2.10

Prove that in Grp every monic morphism is injective.

Let S be a non-empty set and let R be a ring with a 1. Let F be the set of all mappings {} : S -+ R such that ,J(x) = 0 for 'almost all' x E S (i.e. all but finitely many x E S). Given {}, ~ E F and >. E R define{}+~ and ).{}by the prescriptions({}+ ~)(x) = ,J(x) + ~(x) and (>.,J)(x) = H(x). Show that in this way F becomes an R-module. Now define f : S -+ F by assigning to each x E S the mapping f( x) : S --> R given by ift = x; [f(x)](t) = { if t oF X. 2.11

~

Chapter!!

22

Prove that (F, f) is a free R-module on S. [Hint. If M is an R-module and g : S -+ M is any mapping define h: F-+ M by h(6) = L:,Es 6(x)g(x).J 2.12 Prove that the free R-module on a singleton is isomorphic to the ring R regarded as an R-module. 2.13

Prove that in RMod every monic morphism is injective.

2.14 Suppose that in Grp the morphism f: G-+ His not surjective. If Im f has only two cosets in H then Im f is a normal subgroup of H. In this case define a,~: G-+ H/Imf by a(g) = [g] and ~(g)= [lH[· If Im f has more than two cosets let K be the group of permutations on H and, writing Im f = J, choose three different cosets J, Jh 1 ,Jh2 of J in H and define u E K by ify=xh1; if y = xh2; otherwise.

In this case define a,~ : G -+ K by a(g) = Ag (left composition by g) and ~(g) = u- 1 o a( g) o u. Prove that in either case a of =~of but a#~' so that f is not epic. Conclude that in Grp every epic morphism is surjective. 2.15

Prove that in RMod every epic morphism is surjective. f: M-+ N be epic and let Q: M-+ M/Imf be the natural R-morphism. Consider the composite R-morphism Qo f.)

[Hint. Let

2.16 If f is a morphism in Sgp, Grp or RMod that is a bijection prove that f- 1 is also a morphism. 2.17

Show that Grp and RMod are balanced but that Sgp is not.

2.18 Consider the ordered sets E, F depicted by the following Hasse diagrams (the interpretation of which is that an increasing line segment indicates P given by cp,(O) = 0 and 'Px(1) = x is residuated. Suppose now that f: P---> Q is monic in Ord(j1 • Show that if f(x) = f(y) then f o 'Px = f o 'Py· Deduce that f is injective. Prove dually that every epic in Ord(j1 is surjective. Show that for every residuated mapping f we have f o J+ o f = f. Deduce that Ord(j1 is balanced. 2.21 Construct a category K as follows. For the objects of K take triples (X, e.J) where X is a set, e E X and f : X ---> X. For the morphisms from (X,e,J) to (X',e',J') take the mappings 6: X--+ X' such that 6( e) = e' and the diagram

is commutative. If Ct : IN ---> IN is the 'successor function' defined by a(n) = n+ 1 prove that (IN,O,a) is an initial object inK. 2.22 Prove that if a category has a zero object Z then every initial object and every terminal object is isomorphic to Z.

3 Universal constructions

In Chapter 2 we saw how 'element free' characterisations of injective and surjective mappings could be used to fonnulate notions in an arbitrary category. This idea of 'turning theorems into definitions' will now be exploited in some detail using the concept of a universal object. In so doing, we shall produce several pieces of a jigsaw puzzle that we shall assemble in three different ways. I.-Products and coproducts We begin by considering the simple notion of the cartesian product of two sets E 1 and E 2 • This is the set E 1 x E 2 of pairs (x, y) where x E E 1 andy E E 2 • Associated with E 1 x E 2 are the two (surjective) mappings 1rt : Et X Ez -+ E1 and 1rz : E 1 x Ez -+ Ez given by 1rt(x,y) = x and 1r 2 (x, y) = y. Suppose now that X is a set and that ft : X -+ E 1 and fz : X -+ E 2 are mappings. Then we claim that there is a unique mapping .J: X-+ E 1 x E 2 such that the diagram

Universal constructions

is commutative. In fact, define{} :X--+ E 1 x E2 by setting (Vx EX)

Then clearly we have 1r1 o {} = !I and 1r2 o {} = h, so the diagram is commutative. To establish the uniqueness of such a mapping{}, suppose that '1 : X-+ E 1 x E 2 is also such that 1r1 o '1 = f 1 and 1r2 o I] = /2. Given x EX let IJ(x) = (e 1 ,e 2 ). Then

fi(x) h(x)

= 1ri[IJ(x)] = 1r1(el,e2) = e1; = 1r2[1J(x)] = 1r2(e1,e2) = e2,

whence IJ(x) = (e 1 , e2 ) = (! 1 (x), f 2 (x)) = {}(x) and so I]={}. More generally, we can consider the cartesian product X E; of an iEI

arbitrary family of sets (i.e. the set of all families (e;)iEI> or mappings f: I---> U E;, such that e; = f(i) E E; for every i E I) together with iEI

the family (1r3 )iEI of (surjective) mappings 1rj:

X E;-+

Ej described

iEI

by 1rj((e;);EI) = ei. The corresponding result is that if X is a set and (f;);Er is a family of mappings J; : X --+ E; then there is a unique mapping {} : X--+ XE; such that the diagram iEI

is commutative for every i E I. We can turn this theorem into a definition in the following way. Let C be a category and let (A;);EI be a family of objects of C. Construct a category K as follows : for the objects of K take the pairs ( E, (f;);EI) consisting of an object E of C and a family (J;)iEI of morphisms J; : E-+ A;, and for the set MorK((E, (h)iEI ), (E', (f:JiEI )) take the morphisms g : E ---> E' in C such that the diagram

is commutative for every i E I.

Chapter 9 Definition. A terminal object (P,(p;);ei), when such exists, in the above category K is called a product of the family (A;);ei· By abuse of language, we shall often refer simply to the object P of C as a product of the family. It follows by Theorem 2.1 that 'products are unique up to composition by an isomorphism' in the sense that if (P, (p;);e!) and (Q, (q;);ei) are products of (A;);ei then there is a unique isomorphism f: P-+ Q such that each of the diagrams

is commutative. A category C is said to have finite products if every finite set of objects in C has a product; and to have arbitrary products if every family of objects in C has a product. We can, of course, consider the notion of a product of an empty family of objects of C. This can best be visualised by covering up A; and the morphisms f;, f: in the diagram that precedes the above definition of product. What we obbin is that P is a product of an empty family if and only if P is a terminal object of C. A simple inductive argument now gives the following result. THEOREM 3. 1 A category C has finite products if and only if it has a terminal object and every pa1r of objects in C has a product. THEOREM 3. 2 If C 1s a connected category and if ( P, (p; );e 1 ) is a product of (A;);ei then each p; is epic.

Proof. In the above definition, take E = A; and f = idA,. Then the existence of 1'J such that p; o 1'J = idA, means that each p; is a retraction and so is epic. Remark. It should be noted that the above result is no longer true if the condition that C be connected is dropped. In fact, if C is not connected the the constructed category K may not be connected, and the above proof no longer works. For example, the 4-element 9-morphism category depicted by the diagram

Uniuersal constructions

is not connected. In this category we have a o c = b o c with a that cis not epic; but (C, c, d) is clearly a product of B, D.

27

#

b, so

Example 3.1 Set, Sgp, Grp, Ab, Ring, Ring 1 , FVect, RMod, Top, Ord, Lat, CLat all have arbitrary products. Models of these are sim· ply the cartesian product sets (with the necessary structure) and the projection mappings Pi: X A;---+ A; given by P;((a;)iEI) =a;. iEI Example 3.2 Let (E, ~) be an ordered set. Considering (E, ~) as a category (Example 1.12) we see that a product of x, y E E is none other than inf{x, y}, when such exists. An 1\-semilattice is an ordered set L in which x 1\ y = inf{ x, y} exists for all x, y. So, let L be an /\semilattice with a greatest element. Regarding this as a category, we see by Theorem 3.1 that L has finite products. If L fails to be /\-complete (i.e. to have infima of arbitrary subsets) then L does not have arbitrary products. Example 3.3 In the category Set 2 of two-element sets no two dis· tinct objects have a product. In fact, suppose that B = {b1 , b2 } and G = {c1 , c2 } are distinct objects of Set 2 and that they have a prod· uct (D,6 1 ,62 ) where D = {d 1 ,d2 }. Then given A = {a 1 ,a 2 } and f : A ---+ B, g : A ---+ G there is a unique h : A ---+ D such that the diagram

is commutative. By Theorem 3.2, 61 and 62 are epic, hence surjective. Suppose, without loss of generality, that 61 (d;) = b; and 62 (d;) = c; fori= 1,2. Choosing J,g such that f(ai) = f(a 2 ) = b1 and g(ai) = g( a 2 ) = c2 we see from 61 o h = f that d2 ¢:. Im h (for otherwise d2 = h(ai) or d2 = h(a 2 ) and these give b2 = 6i(di) = 6i[h(ai)] = f(ai) or b2 = 61 (d2 ) = 6i[h(a 2 )] = f(a 2 ), a contradiction). Likewise we can show that d 1 ¢:. Im h. This contradiction shows that no such h can exist, whence a product does not exist. Dual to the notion of a product is that of a coproduct. This we obtain essentially by 'reversing all the arrows'. Let C be a category and let (A;)iEI be a family of objects of C. Construct a category K as follows: for the objects of K take the pairs (E, (J;)iEI) consisting of an object E of C and a family (f;)iEI of morphisms f; : A; ---+ E, and for the set

Chapter 9

MorK((E,(f;);EI),(E',(f:J;EI)) take those morphisms '19: E' C such that the diagram

is commutative for every

t E

---t

E in

I.

Definition. An initial object (Q,(q;)iEI), when such exists, in the above category K is called a coproduct of the family (A;);EI· By abuse of language we shall often refer simply to the object Q as a coproduct of the family (A;)iEI·

By Theorem 2.1 we have that 'coproducts are unique up to composition by an isomorphism'. It is clear that we also have the following results which are dual to those given above. THEOREM 3. 3 A category C has fintte co products if and only if it has an m1tial object and every pa1r of objects m C has a coproduct. THEOREM 3.4 If C ts a connected category and if (Q,(q;);EI) is a copoduct of (A;);EI then each q; IS monic.

Example 3.4 Set has (arbitrary) coproducts. In fact, if (A;);EI is a family of sets then their disjoint union Q = U (A; x { i}) together with iEI the injections q; : A; ---> A; x {i}, described by q;(a;) = (a;,i), is a coproduct of (A;);EI· In the case where I= {1, 2} the reader may care to dualise the proof given at the beginning of this chapter : the unique mapping '19 such that

is commutative is given by if x = (a 1 , 1); if x = (a 2 , 2).

Universal constructions

29

Example 3.5 Coproducts exist in RMod. In fact, if (M;)iEI is a family of R-modules let EEl M; be the submodule of X M; consisting of those iEI iEI families ( m;);e 1 which are such that m; = 0 for 'almost all' (i.e. all but finitely many) i E J. For every j E I define the j-th injection inj : Mj --> EEl M; by inj(x) = (x;);ei where x; = 0 if i -1- j and iEI Xj = x. Then (EEl M;, (in;)iei) is a coproduct of (M;);EI· To see iEI this, let g; : M; ---+ M be an R-morphism for every i E I and define h:(ElM;---+Mby iEI h((m;);EI)

= L: g;(m;). iEI

Note that his well defined since for every family (m;);e 1 all but a finite number of m; are zero. Then h is an R-morphism and

(VxEM;)

(h o in;)(x)

= h[in;(x)] = g;(x),

so h o in; = g; for every i E J. To see that h is unique with respect to this property, suppose that k : EEl M; --> M is also an R-morphism such iEI that k o in; = g; for every i E J. Then

k((m;);ei)

= k(L:

in;(m;J)

•El

=

l:(k o in;)(m;) iEI

= z=g;(m;) iEI

= h((m;)iEI) gives k =h. Example 3.6 A V-semilattice is an ordered set L in which x V y = sup{ x, y} exists for all x, y E L. If L is a V-semilattice with a greatest element then, regarding this as a category, we see by Theorem 3.3 that L has finite products. If L fails to be V-complete (i.e. to have suprema of arbitrary subsets) then L does not have arbitrary coproducts. In what follows we shall denote products and coproducts, when they exist, in an arbitrary category C by the notation X A; and EEl A;. Other iEI iEI notation commonly used is IT A; and l1 A;. iEI iEI A useful property of the morphisms associated with a product is that they are 'left cancellable en famille'. More precisely :

Chapter 9

30 THEOREM 3. 5 If (

X A;, (p;)iei)

is a product of (A;);ei and for every

iEI

i E I the diagram

as commutative then a =

f3.

Proof. This is immediate from the universal property of

X A;.

0

iE!

There is, of course, a dual result for coproducts. 2.-Equalisers and coequalisers Let C be a category and /, g : A -> B a pair of morphisms that belong to the same given morphism set. Consider the category K whose objects are diagrams of the fonn I X~A===:B

in which f o a = g o a, and whose morphisms are the morphisms {} : X -> Y in C such that the diagram

is commutative. If

. ! T __:__. A==::: B

is terminal in K then we say that i : T -> A is an equaliser of/, g. The notion of a coequaliser j : B -> I of /, g : A -> B is defined dually. By Theorem 2.1, equalisers and coequalisers are unique up to composition by an isomorphism. Example 3.7 Given /, g: A-> B in Set, Grp or RMod let T

= {x E A;

f(x)

= g(x)},

considered appropriately as a subset, a subgroup, or a submodule. Then the canonical inclusion i : T ____, A is a equaliser of /,g. To see this,

Universal constructions

31

observe first from the definition of T that f o i = g o i. Suppose now that a : X ---+ A is such that f o a = go a. Then for every x E X we have f[a(x)] = g[a(x)], from which we see that a(x) E T for every x EX, so that Ima ~ T. Let £ 1 : Ima---+ A and £2 : Ima---+ T be the canonical inclusions. If ii: X---+ Ima, described by X>--> a(x) = a(x), is the surjective morphism induced by a then we have the commutative diagram

and so i o {} = a where {} morphism such that

= L2 o ii.

Since i is monic, {} is then the unique

is commutative. Consequently i : T ---+ A is an equa.liser of J, g. Example 3.8 Given J, g : A ---+ B in Set, Grp or RMod let Q be the smallest equivalence relation (respectively, congruence) on B such that

('ta E A)

f(a)

=g(a) (Q).

Consider B/Q endowed with the appropriate structure. Then the na.tu· ral morphism QQ : B---+ BjQ is a. coequa.liser of J, g. To see this, observe first from the definition of Q that Qq o f = QQ o g. Suppose now that 1 : B ---+ X is suclt that j of =jog. If Rj is the associated equivalence relation (respectively, congruence) we have that

('ta E A) so that, by definition, Q implies R1 . It now follows from the univer· sal property of B/Q described in Example 2.10 that there is a. unique morphism k: B/Q---+ X such that the diagram

is commutative. Consequently Qq: B---+ B/Q is a coequa.liser of J,g.

Chapter 9

32

THEOREM 3. 6 Equalisers are monic and coequalisers are epic.

Proof. Let i : E ~ A be an equaliser of I, g : A ~ B. Suppose that a, {3 : X ~ E are such that i o a= i o {3. Then we have the commutative diagram

where t = i o a = i o {3. But I o i = g o i gives I o t = go t and so, since i : E ~ A is an equaliser, there is a unique h : X~ E such that i o h = t. Clearly then we have a = {3 = h and hence i is monic. A dual proof shows that coequalisers are epic. Definition. We say that a category C has equalisers ( coequalisers) if every pair of morphisms I, g: A--> B has an equaliser (coequaliser).

It is clear from Examples 3.7 and 3.8 that Set, Grp and RMod have equalisers and coequalisers. Example 3.9 In the category Set 2 of two-element sets no two distinct morphisms have an equaliser or a coequaliser. In fact, suppose that we had the equaliser situation

By Theorem 3.6 we have that i is monic, hence surjective, hence bijective (i.e. an isomorphism in Set 2 ), and so is right cancellable. This gives the contradiction f = g. Similarly, if f # g then no coequaliser exists. 3.-Pullbacks and pushouts Let C be a category and let

A--c I be a given diagram of objects and morphisms in C. Construct a category K as follows : for the objects of K take the commutative squares

Universal constructions

33

[A, B, C; D] of the form

A-->C f

and for the set of morphisms from [A,B,C; D] to [A, B, C; XJ take the morphisms 1 : D -+ X of C such that

is commutative. A terminal object in K is called a pullback for /,g. Other names are a cartesian square, or a fibred product of J, g. The dual notion is called a p!L8hout. We say that C has pullbacks if a pullback exists for every pair of morphisms with a common arrival object. Before giving examples of pullbacks, we first establish the following useful result. THEOREM 3. 7 lfC has finite products and equalisers then C has pullbacks.

Proof. Consider the following diagram in which (A x B, 1rA, 7rB) is a product of A, Band e: T-+ Ax B is an equaliser off o 1rA and go 1rB :

f

By its construction, the outer rectangle is commutative. We show as follows that it is a pullback for /,g. Suppose that qA : Q -+ A and q8 : Q -+ B are such that f o qA = go q8 . By the universality of A x B

Chapter 9

34

there is a unique h: Q-+ Ax B such that 1rA oh = qA and 1rB oh = qB. Then f o 1rA o h = f o qA =go qB = go 1rB o h. By the universality of the equaliser, there is then a unique morphism k : Q -+ T such that e o k = h.

Consequently we have and

1rB

o

eo k = 1rB

o

h = qB.

Suppose now that 8 : Q-+ Tis such that 1rA o eo 8 = qA and 1rB o eo 8 = qB. Then since 1r A, 1rB are left cancellable en fa mille (Theorem 3.5) we have e o 8 = e o k. Since e is monic (Theorem 3.6) it then follows that 8 = k and the pullback criteria are satisfied. ~ Example 3.10 Let us apply the above result to Set, which has finite products (Example 3.1) and equalisers (Example 3.7). Iff: A-+ C and g : B -+ C are morphisms in Set let T ={(a, b) E Ax B; f(a) = g(b)}.

Then with PA : T-+ A and PB : T-+ B given by PA(a,b) =a and PB(a, b)= b we see that

A------+C f

is a pullback of J, g. Note that the same construction works, for example, in Grp and RMod.

Universal constructions

35

Example 3.11 Since universal objects are unique to within composition by an isomorphism, we have the following particular cases of the above : (a) Take A, B ~ C and let /, g be the canonical inclusions. Then

AnB~B

A

inc

C

is a pullback. (b) Take A~ C and for g: B-> C let g-(A) = {x E B; g(x) E A}. Then

A ---->C inc

is a pullback, where

fis

the restriction off to g-(A).

4.-Intersections Definition. If B is an object of a category C then by a sv.bobject of B we shall mean a pair (A, f) consisting of an object A of C and a morphism f : A -> B that is monic. Suppose now that (A;);EJ is a family of subobjects of a given object B. Construct a category K as follows : for the objects of K take those subobjects (D, d) of B for which there exists a commutative triangle

D~A;

~ll and for the set of morphisms from (D, d) to (E, e) take those morphisms a : D -> E in C such that

is commutative. Then a terminal object in K is called an intersection of the family (A;, /;);EI of subobjects of B.

Chapter 3

We say that C has (finite) intersections if every (finite) family of subobjects of C has an intersection. Note that if (D, d) is an intersection of (A;, l;)iEl then the existence for every i E I of a commutative triangle

D-d-'-~A;

~11; B means that (D, d;) is a subobject of A; for every i E I (for monic implies that d; is monic). The following result is immediate from the definitions.

I; o d;

being

THEOREM 3. 8 If (A 1 ,/i) and (A 2 , fz) are subobjects of B then a subobject (D, d) of B is an mtersection of (A 1 , 11 ) and (A 2 , h) if and only If

A2-----+ B j,

is a pullback.

Our objective now is to show how all the above notions interact. We have already seen part of this in Theorem 3.7 above. THEOREM 3. 9

The following statemonts are equivalent :

(1) C has finite products and equalisers; (2) C has finite products and fimte intersections; (3) C has pullbacks and a terminal object. Proof. (1) => (3) : If C has finite products and equalisers then, by Theorem 3.7, C has pullbacks. That C also has a terminal object follows from the fact that such an object is a product of an empty family. (3) => (2) : If T is a terminal object of C then Morc(X, T) is a singleton, {ax} say, for every object X of C. Given morphisms I : C --+ A and g : C --+ B in C we then have the commutative diagram c__!___.A

1 1

g

B-----+T

aA

Universal constructions

37

Now form a pullback of a A, an via PA : P-> A and Pn : P-> B. Then there is a unique h: C--. P such that PB o h = g and PA o h = f : P---~------tA

·{:>/tl B----------tT "B

Consequently, (P, PA,PB) is a product of A, B. (2) =:> (1) : Suppose that we have morphisms J, g : A -> B. Let (idA,!) and (idn,g) be the morphisms obtained from the universality ofAxB:

Note that (idA,!) and (idA, g) are monic since each is a section. Consider the intersection of the subobjects (A, (idA,!)) and (A, (idA, g)) of Ax B, namely (by Theorem 3.8) the pullback

D

d,

~)

A

I

)'"'I AxE

A (idA,g)

Composing each side of (idA, g) o d2 = (idA ,J) o d 1 on the left with PA we have that d2 = d 1 = d say. We now show that (D, d) is an equaliser of J, g. Composing each side of the above equality on the right with PB we obtain f o d = go d. Consider now the diagram

Chapter 9 in which f o h = go h. We have to produce a unique t'J : H that do t'J = h. For this purpose, consider the diagram

-+

D such

In this diagram we have

PB

0

(idA, g)

0

h

=g

0

h

=f

0

h

= PB

0

(idA, f)

0

h.

Now it is clear that we also have

PA o (idA, g) o h =idA oh = PA o (idA, f) o h. Since PA, PB are left cancellable en fa mille (Theorem 3.5) it follows that

and so the above diagram is commutative. Since the rectangle is a pullback, there is then a unique t'J : H -+ D such that d o t'J = h, as required. We shall return to this result in another light later.

3.1 Prove that the category Set. of pointed sets (Exercise 1.5) has finite products and finite coproducts. 3.2 Let C be a connected category in which every morphism is monic. Suppose that there exist A, B with IMorc(A, B) I ~ 2. Prove that, for every object X of C, a product of X, A does not exist. Deduce that the category Field of fields does not have finite products. 3.3 Prove that in the category FAb of finite abelian groups a co· product does not exist for the family of cyclic groups of order pk where pis a prime and k = 1,2,3,· ·

Universal constructions

39

3.4 Prove that the category of chain complexes (Exercise 1.7) has products and coproducts. 3.6 A category C is said to be regular if every monic morphism in C is an equaliser. Prove that Set is regular and that Ring is not regular. [Hint. For Set let I : A --+ B be monic. Let a 1 , a 2 ¢'. B be such that a 1 f a 2 and let 11 , /2 : B --+ B U {a 1 , a 2 } be given by

/;(b)= {

~.

ifbElml; otherwise.

For Ring consider the embedding of 7L in Q.] 3.6

Prove that every section is an equaliser.

3.7 If e : E --+ A is an equaliser of following statements are equivalent :

I, g

:

A

--+

B prove that the

(I) I= g; (2) e is an isomorphism; (3) e is epic.

3.8

In the diagram

it is given that (I) (i = I, 2, 3) w; is a coequaliser of u;, v;; (2) (i =I, 2) h; is a coequaliser of/;, g;; (3) the following 'commutativities' hold : u 2 !J

= /2u 1 , v2g1 = g2v 1 , h 2 u2 = uah 1 , h2v2 = vah 1 , Jaw! = w2/2, gawl = w2g2, haw2 = wah2.

Prove that h 3 is a coequaliser of

fa, ga.

Chapter 3

40

3.9

If in the diagram ·~•----+•

1 1 1 • ----+•----+. each of the squares is a pullback, prove that so also is the composite rectangle. 3.10

Prove that f : A

-+

B is monic if and only if

A-----.B f

is a pullback. 3.11

Suppose that

A--->C f

is commutative and let a = f o p1 = g o p2 • Prove that this square is a pullback of J,g if and only if (a: P-+ C,p 1 ,p2 ) is a product of f: A-+ C and g: B-+ C in the comma category -c (Example 1.15). 3.12

Suppose that the diagram

A-----.C I

is a pullback. Prove that if f is monic then so is

f3.

Let G be a group considered as a category (Example 1.17). Prove that G does not have finite products. Show, however, that G has pullbacks. Conclude that in Theorem 3.9(3) the existence of a terminal object cannot be omitted. 3.13

Uniuersal constructions 3.14

If

(X

A;, (pf )iei) and (

iEl

XB;, (pf);ei)

are products of the

iEl

families (A;);ei and (B;);er respectively and if/; : A; ---> B; for every i E !, denote by X /; : X A; ---> X B; the unique morphism t'J which, iEl

iEI

iEI

by universality, is such that pf o t'J = /; o pf for every i E !. If (X x Z, 11"x,1rz) and (Y x Z,py,pz) are products of X, Z and Y,Z respectively, prove that the following diagram is a pullback : XxZ~X /Xidz

Y

1 X

1/

Z------> Y .-y

3.15 If /;, g; E Mor(A;, B;) for i E I and if e; : E; equaliser of/;, g; prove that X e; is an equaliser of X /;, ~I

~I

--->

A; is an

X g;. ~I

4 Factorisation ofmorphisms

Given a morphism f : E-+ F in Set let R1 be the associated equivalence relation. Then (see Example 2.10) there is a unique ,J: E/ Rj -+ F such that

E____!___.F

~1/ E/Ri is commutative. In fact, ,J is given by .J[x] = f(x) where [x] denotes the Rrc!ass of x E E. Now Q is surjective; and ,J is injective since f(x) = f(y) implies [x] = [y]. Thus we see that every morphism in Set can be 'factorised' as a composite of the form monic o epic. Moreover, this factorisation is 'unique to within isomorphism' in the sense that if also f = a o f3 where a : A -+ F is monic and f3 : E --+ A is epic then there is a unique isomorphism (i.e. bijection) i: E/RJ-> A such that

Factorisation of morphisms

43

is commutative. In fact, by the universal property of E / R1 there is a unique i : Ef RJ -+A such that ioQ = {3. Then ao1oq = aof3 = f = 11oq and so, since q is right cancellable, a o 1 = {}. Thus the above diagram is commutative. To see that i is epic, suppose that p o 1 = q o i· Then we have pof3=poioQ=qoloQ=qo{3

whence, since f3 is epic, p = q. To see that i is also monic, suppose that i o a = i o b. Then we have 11oa=aoioa=aoiob=l1ob

whence, since 11 is monic, a = b. Thus i is a bimorphism and hence, since Set is balanced, is an isomorphism. These observations lead to the following notion. Definition. A category C will be called factonsable if every morphism moe where m is epic and e is monic; and uniquely factorisable if every such decomposition is unique to within isomorphism, in the sense that if also f = m' o e' where m' is epic and e' is monic then there is a unique isomorphism i such that the diagram

f in C can be expressed in the form

e'

·~·

is commutative. We have seen above that Set is uniquely factorisable. As we shall see later, so also is, for example, RMod. In contrast : Example 4.1 The category Ring 1 is not uniquely factorisable. To see this, consider the embedding i : 71. -+ it). We have the commutative diagram

in which every morphism is a bimorphism (see Example 2.2), and clearly there can be no diagonal filling in by an isomorphism.

Chapter

44

4

There are several types of category that enjoy the property of being factorisable. We shall concentrate on the 'mainstream' varieties, all of which contain a zero object. In this situation there are special kinds of equaliser and coequaliser that are important and which we shall now consider. Definitions. Let C be a category with a zero object. If f : A ---+ B is a morphism in C then by a kernel off we shall mean an equaliser of the pair /, OAB. By a cokernel of f we shall mean a coequaliser of the pair J, OAB. We say that C has kernels ( cokernels) if every morphism in C has a kernel (co kernel). Thus we have that, given f : A ---+ B in C, a morphism k : K -> A is a kernel of f if and only if

(1) fok=OKBi (2) if a : X --> A is such that f o a = Oxs then there is a unique morphism {} : X --> K such that the diagram

is commutative. Reversing the arrows, we obtain dually a characterisation of cokernels. Clearly, the following result is a special case of Theorem 3.6 : THEOREM 4.1 Kernels are monic and cokernels are epic.

Example 4.2 In Grp and RMod the usual notion of kernel yields the categorical equivalent. Indeed, let f : A ---+ B be a group morphism (or an R-morphism). Then the (algebraic) kernel off is the subgroup (or submodule)

Kerf={xEA; f(x)=Os}. The canonical embedding i : Kerf -> A is then a kernel of f in the categorical sense (see Example 3.7). Thus Grp and RMod have kernels. Example 4.3 Taking g = 0 in Example 3.8, we see that if f : A -> B is a group morphism and if N is the smallest normal subgroup of B containing Im f then the natural morphism Q : B --> B / N is a cokernel of f. In the case of an R-morphism f : A -> B we replace 'normal subgroup' by 'submodule' and obtain N = Imf. (The cokernel in the algebraic sense is then the quotient module Coker f = B/Imf.) Thus we see that Grp and RMod also have cokernels.

Factorisation of morphisms

45

In an arbitrary category with a zero object, morphisms need not have kernels or cokernels. Example 4.4 If R is a commutative ring with a 1 then an R -algebra is a R-module that is also a ring with a 1. For example, the set Matnxn(Z) of n x n matrices over Z is a Z-algebra. Now in the category nAlg of R-algebras the only candidate for the (algebraic) kernel of a morphism is a submodule that is also an ideal. But this in general fails to have an identity element and therefore fails to be an object in the category. Hence not every morphism in nAlg has a kernel in the categorical sense. Likewise, in certain types of topological spaces the quotient spaces (the candidates for cokernels) do not in general inherit the properties of the parent space, so fail to be objects of the category under consideration. Consequently not every morphism has a cokernel. Although kernels and cokernels need not exist in general, certain mor· phisms always have kernels and cokernels : THEOREM 4.2 If C is a category with a zero object then every zero

morphism has a kernel and a cokernel (each being an isomorphism}; and every isomorphism has a kernel and a coke mel (each being a zero morphism). Proof. In the diagram

we have, by Theorem 2.2, DAB oa = DxB and DAB oidA =DAB· Clearly, there is a unique !J : X --+ A such that idA o!J = a, namely !J = a. Thus idA is a kernel of OAB· Since kernels are unique to within composition by an isomorphism, it follows that k : K--+ A is a kernel of DAB if and only if k is an isomorphism. Suppose now that a : A -+ B is an isomorphism and consider the diagram

Z-------+ A ---->B Oz..t

a

in which Z is a zero object. Then clearly a o OzA = DzB· If now aof3 = Dxs = OzsoDxz then we have f3 = a- 1 oOzsoOxz = OzA oDxz,

Chapter.+

so that Oxz completes the diagram in a commutative manner. Since OzA is monic (see the proof of Theorem 2.3) we deduce that it is a kernel of a. It follows that the kernels of an isomorphism a : A --+ B a.re precisely the zero morphisms from arbitrary objects to A. The statements for cokernels are established similarly. THEOREM

4. 3 If C is a category with. zero object Z then

(1) every monic morphism f: A--+ B h.as a kernel, namely OzA; (2) every epic morphism f : A --+ B has a cokernel, namely OB z. Proof. (a.) H in the diagram

we have fog = OxB = OzB o Oxz = f o OzA o Oxz then, since f is monic, we have g = OzA oOxz so Oxz makes the diagram co=uta.tive. Its uniqueness follows from the fact that OzA is monic (see the proof of Theorem 2.3). The proof of (b) is similar. Kernels and cokernels are related. We have the following useful result. THEOREM

4.4 Let C be a category with. a zero object. If tt: K--+ A is --+ B and if 'I : A --+ C is a cokernel of tt then tt is a

a kernel of f : A kernel of 'I·

Proof. Since 'I is a. cokernel of tt and f ott = 0 there is a. unique v : C--+ B such that v o 'I = f. Also, 'I ott= 0.

K~A~B

~ h~ X

Now let u : X

--+

C

/v

A be such that 'I o u = 0. Then fou=vo,ou=vo0=0

and so, since tt is a. kernel of f, there is a unique tt o t = u. Thus tt is a kernel of 'I·

t:X

--+

K such that

In what follows we shall find it convenient to denote any kernel of a morphism f by simply kerf, and any cokernel by coker f. Theorem 4.4 above can then be expressed in the useful form ker coker ker f

- ker f.

Factorisation of morphisms

47

There is, of course, the dual statement cokerkercoker f- coker f. Definition. A category C will be called normal if (1) C has a zero object; (2) every morphism in C has a kernel and a cokernel; (3) every monic morphism in C is a kernel. Example 4.5 The category RMod is normal. In fact, (1) and (2) are clearly satisfied. As for (3), if f : A ~ B is monic then f is a kernel of q : A ~ B jim f. To see this, observe by Example 4.2 that the embedding i : Im f ~ B is a kernel of Q. Since f is monic, it is A ~ Im f given by J( a) = f( a) is an isomorphism. injective, so Then, using the fact that kernels are unique to within composition by an isomorphism, we deduce from f = i o that f is a kernel of Q. Example 4.6 The category Ring is not normal. For example, consider the embedding i : 7L. ~ 4;1. If this were the kernel of a ring morphism f : (I -+ A then in every diagram

1:

1

ll.---+4;1---+A i

J

there would exist a unique ring morphism {} : B -+ 7L. such that i o {} = = (I and g = idQ·· Then i would be a retraction (hence surjective), which is not the case (Example 2.2). In a normal category every monic morphism f : A -+ B is a kernel, say a kernel of h : B ~ C. Then from ker h - ker coker ker h we obtain f - kercoker f. Thus we see that f is monic if and only iff- kercoker f. Similarly, f is epic if and only iff- cokerker f. We shall make use frequently of these observations in establishing the following results concerning normal categories. g. Take B

THEOREM 4. 5 A normal category is balanced.

Proof. Let C be normal with zero object Z. Let f : A -+ B be a bimorphism in C. Since f is epic we have, by Theorem 4.3, that OzB is a cokernel of f. Since f is also monic we then have

f- kercoker fand so, by Theorem 4.2,

kerOzB

f is an isomorphism.

~

Chapter 4 THEOREM 4.6 A normal category has pullbacks.

Proof. Given

It : A ---+ C

and

h :B

k2.....,ker a.o/2

p

A / 1

C consider the diagram

---+

B

lt.c

......,kera

a-coker /1

D

We have a o h o k2 = 0 and so, since a is a cokernel of f 1 , there is a unique k1 : P ---+ A such that h o k1 = h o k2. Suppose now that g1 : X ---+ A and 92 : X ---+ B are such that /1 o g1 = /2 o 92 :

x~~B k,

1 ,,

A ------.C ---+D J. a

Since a o h o g2 = a o / 1 o g1 = 0 o 91 = 0 and k2 is a kernel of a o 12 there exists a unique "/ : X---+ P such that ~ o "/ = g2. Now

and /1 is monic, so we also have k1 o "f = 91· 0 COROLLARY

A normal category has finite products, finite intersections

and equalisers. Proof. Since every zero object is terminal, the result is an immediate consequence of Theorem 3.9. 0 THEOREM 4. 7 If C is a normal category then the following statements

concerning f : A

---+

B are equivalent :

{1) f is epic; {2) all cokernels off are zero morphisms.

'*

Proof. (1) {2) : Let g : B ---+ C be a cokernel of f. Then we have 9 o I= OAc = Osc of so that, f being epic, 9 = Osc(2) (1) : Let p,q : B ---+ C be such that p of = q of. By the Corollary to Theorem 4.6, C has equalisers, so let e : E ---+ B be an

'*

Factorisation of morphisms

49

equaliser of p, q. Then there is a. unique g : A

-+

E such tha.t e o g =

f :

A-~B___:____. C

~~·

q

E Now by hypothesis every cokernel of f is a. zero morphism. It follows tha.t so a.lso is every cokernel of e; for if in the diagram

we ha.ve xoe = 0 then xof = xoeog = Oog = 0 a.nd so, since 0: B-+ T is a. cokernel of/, there exists a. unique h: X-+ T such tha.t h ox= 0. It now follows from the fa.ct tha.t e is monic tha.t e ~ kercokere

~

kerO

a.nd so, by Theorem 4.2, e is a.n isomorphism. Thus e is a.lso epic. Consequently, we ha.ve p = q and f is epic. 9 THEOREM 4. 8 A normal category is factorisable. Proof. Given a morphism fit follows by the universal property of kernels that there is a unique morphism q such that

f = kercoker f o q. Our objective is to prove that q is epic, whence the result follows. For this purpose, note that we can apply the above observation also to q to obtain the existence of a unique morphism p such that q = kercokerq o p.

Consider now the diagram p

''""n·•C~?j, 1 •

cokerq

•

kercoker I

·-------+• coker f

so

Chapter 4

Lett= kercoker I o kercokerq. Then we have coker I~ eckert. [To see this, suppose that g is such that got = 0. Then go I = got o p = 0 o p = 0 and so there is a unique(} such that (} o coker I= g.] Now t is clearly monic, so t ~ kercokert ~ kercoker I from which it follows that kercokerq must be an isomorphism. It now follows from coker q ~ coker ker coker q that every cokernel of q is a zero morphism (being a cokernel of an isomorphism). We conclude by Theorem 4.7 that q is epic, as required.

9 The above results show that the seemingly modest axioms for a category to be normal have quite deep consequences. There is, of course, the dual notion of a conormal category (in which every epic is a cokernel), and results that are dual to Theorems 4.5, 4.6, 4.7 and 4.8. Example 4.7 The category RMod is cononnal. To see this, it suffices to prove that every epic morphism in RMod is a cokernel of any of its kernels. Now a kernel of I : A -+ B is the inclusion i : Ker I -+ A. If now g : A -+ X is such that g o i = 0 then clearly Ker I ~ Ker g, so that l(a) = l(b) implies g(a) = g(b). Since I is epic, hence surjective (see Exercise 2.15), we can therefore define a morphism (} : B-+ X by fJ[f(a)] = g(a) for every a E A :

Clearly, {}of = g and so I is a cokernel of i. Example 4.8 The category Grp is conormal but not normal. In fact, the same argument used in Example 4.7 shows that Grp is conormal. However, it is not normal. For, we know (Example 4.3) that a cokernel of I: A-+ B is Q: B---+ B/N where N is the smallest normal subgroup containing Im/, and a kernel of Q is the embedding i: N-+ B. Iff is monic then I need not be a kernel of its cokernel Q, for the embedding j : Im I ---+ N need not be an isomorphism. Definition. A category C is said to be binormal if it is both normal and conormal.

Factorisation of morphisms

51

THEOREM 4. 9 A binormal category is uniquely factonsable.

Proof. If C is binormal then from the proof of Theorem 4.8 and its dual we have, for any morphism f of C, a commutative diagram •----=-ke~r.:..f_ ___. • _ __:.c.:.:ok.:.:e.:_r.:.:ke~r-=-1--+•

coker f

ker coker f

in which q is epic and k is monic. Now ker coker f o q o kerf =

f

o kerf = 0

and the fact that ker coker f is left cancellable gives q o kerf = 0. Consequently there is a unique f* such that f* o coker kerf = q, and since the right hand side is epic so is f*. Dually, we have coker f o k o cokerker f = coker f o f = 0 and the fact that coker kerf is right cancellable gives coker f o k = 0. Consequently there is a unique 7 such that kercoker f o 7 = k, and since the right hand side is monic so is f. We now have ker coker f

0

r

0

coker ker f = ker coker f

0

7

0

coker kerf

and so, again by cancellation, f* = 7 and is a bimorphism, hence an isomorphism by Theorem 4.5. To complete the proof, suppose that f = m o e = m 1 o e' where m, m' are monic and e, e' are epic. Consider the commutative diagram a

e'

e---t•~•

Since e is epic we have bof=O

(2) : From ker coker f

coker f

- coker ker coker f

~

coker ker g = coim g.

(2) => (3) : If the sequence is coexa.ct then coim g - coker f so go f

~

im g o coim g o f

~

im g o coker f o f = im g o 0 = 0,

coker f o kerg- coimg o kerg

~

cokerkerg o kerg = 0.

(3) => (1) : go f = 0 gives g o imf o coimf = 0 a.nd so, since coima.ges a.re epic, g o im f = 0. By the universality of ker g there is a. unique morphism h such that im f = ker g o h. Similarly, from coker f o ker g = 0 we have, by the universality of kercoker /, a. unique morphism k such that kerg = kercoker f o k

~

imf o k.

It follows from these observations that kerg~kerg

o h o k

whence hok is a.n isomorphism, so that his epic a.nd k is monic. Similarly, we have imf ~ imf o k o h whence k o h is a.n isomorphism, so that k is epic a.nd h is monic. Thus h a.nd k a.re bimorphisms, hence isomorphisms, a.nd so im f ~ ker g. ¢ We shall henceforth agree to denote by 0 a.ny zero object in a. binormal category. THEOREM 4. 11 In a binormal category

( 1) 0 -~ • ____!____., • is exact if and only iff is monic;

(2) • ____!____., •---+ 0 is exact if and only iff is epic; (3) 0---+ • ____!____., • ~ • is exact if and only iff~ ker g; (4) • ____!____., • ~ • ---+ 0 is exact if and only if g ~ coker f; (5) 0----. . ____!____., •---+ 0 is exact if and only if f is an isomorphism.

Chapter

54

4

Proof. (1) The sequence is exact if and only if the zero morphism is a kernel of f. By the dual of Theorem 4.7, this is equivalent to f being monic. (3) The sequence is exact if and only if f is monic and im f ~ ker g. But when f is monic we have f ~ kercoker f = imf. (2),(4) are the duals of (1),(3). (5) follows from (1),(2) and the fact that a binormal category is balanced. ~ COROLLARY

For every morphism f in a binormal category C the se-

quence

is exact.

~

Example 4.9 In the binormal category RMod the canonical sequence

o---+ Kerf __i___. A __!__. B ____!__. B / Im f---+ o is exact. An 'image o coimage' factorisation off is

where fis given by !{a)= f(a) for every a E A, and j is the embedding of Im f into B. An exact sequence of the form 0----+. __!__.. ~. ~.

is called a short exact sequence. There are several results concerning short exact sequences that can be established in a binormal category. The proofs can be quite intricate in nature. By way of illustration, we prove the following result (often called the 3 x 3 lemma).

Factorisation of morphisms

55

THEOREM 4.12 Suppose that in a binormal category the diagram

0

0

0

is commutative and has exact rows and columns. Then there are unique morphisms 'Yl : cl ~ c2 and '12 : c2 ~ c3 such that the completed diagram is commutative. Moreover,

is also exact. Proof. The existence of 'YI results from the fact that 92

° f3t 0 It = g2 0 h

0

O!t

=0

0

O!t

=0

and g1 is a cokernel of / 1 • Similarly, the existence of 1 2 is assured. That 12 is epic follows from the fact that g3 and /3 2 are epic, whence so is 93 o/32 = '13 ° 92· Now let a = 92 o {3 1 • We claim that 72 is a cokernel of a. To see this, refer the following argument to the diagram below. Let p : C 2 ~ D be such that p o a = 0. Then p o 92 o /3 1 = o and so, since /32 is a cokernel of /3 1 , there is a unique q: B 3 ~ D such that q 0 /32 = p 0 92· Now q o /a o a2 = q o !32 o h = p o 92 o h = p o 0 = 0 and a2 is epic, so q o fa = 0. Since 93 is a co kernel of fa there is then a unique r : C3 ~ D

Chapter

4

such that ro 93 = q:

A2·----="c.::•c...__ __, A3

B,

~l,_____::f3_:__z___.l

"L/D([. C2

C3

'Yz

Since r o "12 o 92 = r o 93 o f32 = q o {3 2 = p o 92 and 92 is epic we deduce that r o 1 2 = p. Consequently we have that 12 ~ cokero.

In a dual manner (do it!) we can show that

/J

~

kero.

Consider now an 'image o coimage' factorisation of o, say

B 1 _ _ _co:..cim=-.:"~--+ F ____:i=m:..:ac__

___,

C2 •

Using the above observations, we have that 91

~

coker /J

~

coker ker o = cairn o

and so there is a unique isomorphism IJ : F 91· Since 1 1 o IJ o cairn o

--+

C1 such that IJ o cairn o =

= 1 1 o 91 = o = im o o cairn o

and cairn o is epic, we deduce that 1 1 o IJ = im o and so, since IJ is an isomorphism, 1 1 ~ imo = kercokero ~ ker"(2 • Thus, by Theorem 4.11, the sequence

is exact. ¢

Factorisation of morphisms

57

IExercises I 4.1 Prove that the catgeory Set* of pointed sets is normal but not conormal. 4.2

Let C be a normal category. Suppose that in the diagram

At_l___,A

lh it is given that g is a kernel of{}. Prove that the following are equivalent : (1) there is a unique morphism 1J : A1 -> B 1 such that the square is a pullback; (2) f is a. kernel of{} o h. Use this result to prove directly that a. normal category has finite intersections. 4.3

Let C be a. uniquely fa.ctorisa.ble category. If

·-----· ·-----· e

is commutative with m monic and e epic, prove that there is a. unique morphism"( such that"( o e = f and m o "f =g. 4.4

If, in a. binormal category C, the diagram

R----+8 is a pullback prove that

' {} o kert

~

Deduce that if t is monic then so is m.

kerm.

58

Chapter 4.&

4.

Consider the diagram 0------+

A ------+ B

1 1

0------+ A'------+ B' -----+C'

in a binormal category C. If this diagram is commutative and if 0 ------+

A' ------+ B' -----+ C'

is exact, prove that the square is a pullback if and only if 0 ------+

A ------+ B

------+ C'

is also exact. Deduce that in the completed 3 x 3 array of Theorem 4.12 the northwest square is a pullback and the south-east square is a pushout. 4.6 [The short five lemma] Suppose that, in a binormal category, the diagram 0--+•~·~•--+0

0--+.

______.....------+. ---+0 m' e'

is commutative and has exact rows. Prove that

(1) if J, hare monic then so is g; (2) if J, h are epic then so is g.

5 Structuring the morphism sets

In the category RMod the morphism sets MorR(A, B) possess a 'natural structure' in the sense that if[, g E MorR(A, B) then we can define f + g E MorR(A, B) by setting(!+ g)(a) = f(a) + g(a) for every a EA. It is readily seen that this operation of addition makes Mar R (A, B) into an abelian group. In this section we shall investigate the idea of defining a structure on the morphism sets in an arbitrary category with a zero object. The keystone in this connection is the following notion. Definition. Let C be a category with a zero object and let (Ai);EI be a family of objects of C. If B is an object of C and there exist morphisms A;~B~A; for every i E I then we say that (J.!;,B,7r;);EI is a biproduct of (A;)iEI if (1) (B, 7r;)iEI is a product of (A;)iEii (2) (J.!;, B)iEI is a coproduct of (A;)iEii (3) each of the diagrams

is commutative, where "d 6;; = { 10

if i = j; if i # j.

Chapter 5

6o

It is clear that biproducts, when they exist, are unique up to isomorphism. We shall denote the object part of a biproduct of (A;)iEI by

II A;.

n

II

In the case where I= {1, ... ,n} we shall write this as

~I

as A 1

A; or

~I

•

A2

•

· · •

A,..

Example 5.1 If (A;);EJ is a family of R-modules and I= {1, ... ,n}

then we have that

n

n

i=l

i=l

X A;= EI) A;, together with the canonical injections

and surjections, is a biproduct of At, ... , A,.. We now make the very important observation that morphisms between finite biproducts can be represented by matrices. In fact, as can be seen from the diagram 1rf A;~

A;

···!Bi~TI :: l

Bi

~>f

II

i=l

n

m

every morphism a:

m

II

II Bi

A;->

induces morphisms aii: A;-> Bf,

i=l

i=l

namely those given by Ol.fi

= 'Iff

0

01. 0

p.f.

It is immediate from Theorem 3.5 and its dual that if i,j then

01.

= (3. We can therefore associate with

01. :

Ot.fi

=

TI A;

!3ii ->

i=l

for all

TI Bi

i=l

then x m matrix [ai;J. Conversely, given ann x m matrix [ai;J where A; -> Bi for all i, j there is (by the universal property of the

Ot.fi :

n

product

II

n

m

B;) a unique

J=l

01. :

II i=l

A;

->

II

B; such that

1r7 oa =

01. ;; o1rf.

j=l

It follows that

Thus every such matrix determines such a morphism. Example 5.2 Take m = 1 in the above. By the universal property of products, there is a unique morphism (!,g) : A1 -> B 1 • B 2 such that

Structuring the morphism sets

61

the diagram

is commutative. This morphism (!,g) is represented by the matrix [

~].

Example 5.3 Take n = 1 in the above. By the universal property of coproducts, there is a unique morphism [f, g] : At • A 2 ---+ Bt such that the diagram

is commutative. The matrix that represents [f, g] is [ f Example 5.4 id :

g ].

TI A; ---+ TI A; is represented by the identity matrix i=l

i=l

[6;j]nxn· This follows from the fact that

1r;

o id op.j =

1r;

o P.i =

Oij·

With the above notation, we have the following useful result. THEOREM 5.1 lf a;; :A;---+ B; for i,j E {1, 2} then

Proof. It is readily seen that each side of the equation represents a mor· phism from At • A 2 to Bt • B 2 • Since each is represented by the matrix

the result follows. Let us now consider the idea of defining a law of composition on the morphism sets and show how this depends on the existence of biproducts.

Chapter 5 Definition. A category C with a zero object is said to be semi-additive if, for every pair of objects A, B of C there is a law of composition *on Morc(A, B) such that

(1) (More( A, B), *) is a commutative semigroup with identity 0; (2) o is bilinear, in the sense that the following identities hold : (f*g) o k = (fo k) *(go k). Example 5.5 RMod is semi-additive. Example 5.6 Let G be an additive abelian group regarded as a category G. Then G is semi-additive (the addition on the morphism sets being the group operation). Example 5.7 The additive monoid IN considered as a category is semiadditive. Example 5.8 Let L.at 0 be the subcategory of L.at (Example 1.13) consisting of those lattices that have a smallest element 0, the morphisms being the a-preserving lattice morphisrns. Then L.at 0 is semi-additive. In fact, {0} is a zero object; and if J, g E Mor(L, M) then with f * g defined by ('v'xE L) (! * g)(x) = f(x) V g(x), properties (1) and (2) above are readily seen to hold. THEOREM 5. 2

The following statements are equivalent :

( 1) C has finite biproducts; (2) C is semi-addztive and has finite products; (3) C ts semi-additive and has finite coproducts. Proof. (1) => (3) : If (1) holds then C has a zero object, arising from the biproduct of an empty family; and clearly C has finite coproducts. To prove that C is semi-additive, let J, g E Morc(A, B) and consider the following diagrams that involve the biproducts (Jl 1 , }l 2, A • A, 1r1, 1r2) and (Jli, Jl~, B • B, 1rj, 1r~) :

Z Al~·d ~lfnl 1~ 1/g/ (id,i0 is exact and splits on the right. Let {3': C---+ B be such that {3'{3 =ide. We know by Theorem 4.11(3) that a is a kernel of {3. Consider then the diagram B lidn-/3'/3

A---+B-----+C a

We have /3(ids

-!3'{3)

{3

-

= {3- {3 !3' {3 = {3- {3 = 0. id

Consequently there is a unique fJ: B---+ A such that a!'J =idE -{3'{3. It follows by Theorem 4.10 that ai'Ja

= (id 8 -{3' f3)a =

a-

-

/3' /3 a

=a

0

and so, since a is monic, !'Ja =idA. Thus the sequence splits on the left. The converse follows dually. Because of the above result, we shall say simply that a short exact sequence splits if it splits on the right or on the left. The connection between splitting and biproducts is explained in the following result. THEOREM 5. 7 Let C be an abelian category. If (J.lJ, P.2, At • A2, 1rt, 1r2) is a hiproduct of At, A2 then the top and bottom sequences tn P.l

7r'2

O~At~At •A2~A2~0

...

are each split exact. Proof. By Theorems 3.2 and 3.4 the morphisms p. 1 , p. 2 are monic and 1rt, 1r2 are epic. By Theorem 4.11, the sequences 0---+ At ~At • A2

O---+A2~At•A2

At • A2 ~At - - - + 0

At•A2~A2---+0

Chapter 5 are exact. We now observe that p. 1 is a kernel of 1r2 • In fact, we know from the definition of biproduct that 7r2 JA. 1 = 0; and if 'IJ is such that 1r2 '1J = 0 then we have ., = (JI.J71"J

+ Jl.27r2)11 =

JI.J7rJ'IJ + Jl.27r?.'IJ = JI.J71"J'IJ

so that 'IJ = p. 1 1; where 1; = 1r1 11. It now follows by Theorem 4.11(3) that the top sequence is exact. Moreover, it splits since 7r1 J1. 1 = id. Similarly, the bottom sequence is split exact. ~

IExercises I 5.1 Let A _J____, B ____!:___, D and A_!!___, C ~ D be morphisms in a semi-additive category. Prove that the composite morphism

A~B•C~D is none other than hf +kg. 5.2 Let A be a semi-additive category and let A 1 , •.• , A., be objects of A. Given an object B of A and morphisms A;~ B ~A;, prove that the following statements are equivalent : (1) (JI.J 1 ••• 1 p.n,B 1 7rt 1 ••• ,7r.,) is a biproduct of AJ, ... ,An; (2) (B,7r1 , . . . ,7rn) is a product of A 1 , . . . ,An and (Vi,j) 7r;p.; = 6;;; (3) (p. 1 , ... ,Ji.n,B) is a coproduct of A 1 , ... ,A., and (Vi,j) 1r;p.; = 8;;; n

( 4)

L: p.;1r; =

id 8 and (Vi, j) rr;p.; = 8;;.

i=l

5.3 Let A be an additive category. If B, C are objects of A and C ---> B is such that (1) 7rB has a kernel; (2) there exists f.i.B : B---> C with 7rBf.i.B =ids, prove that there is an object A and a morphism f.i.A : A ---> C such that (f.!. A, f.i.B, C) is a coproduct of A, B. 'lrB :

5.4 Let A be an additive category in which every morphism has a kernel. Prove that A has equalisers. 5.5

Let E be a binormal category and suppose that the diagram 0------+ A__!_______. B __!!___.c ------+0

wAl

la

lwc

0------+ A------+ B ------+C ------+0 I

g

Structuring the morphism sets is commutative and has exact rows. Prove that a is an isomorphism. Suppose now that E is also additive. Prove that in the above diagram there is a unique p. : C -+ B such that a- idB = p.g. Show that gJ.L = 0 and deduce that there is a unique {}: C -+ A such that a = id 8 + f{}g.

5.6

Let C be an abelian category and consider a split exact sequence 0----+ A~ B

___!_____. C ----+ 0.

Let a', {3 1 be associated splitting morphisms. Prove that the following are equivalent :

(1) a 1{3 1 =0; (2) aa 1 + {3 1{3 = idB j (3) (a, {3', A, a', {3) is a biproduct of A, C. 5.7 Prove that the category Latci1 of bounded lattices and residua.ted mappings (Example 1.9) is semi-additive and has finite products. Deduce that a. balanced semi-additive category with biproducts need not be additive. 5.8

In a.n abelian category, consider the (non-commutative) diagram

,/pl~ .

b-.~~~ A--=:;A• B _ B

1

iJ.B /

[a.-(j]/

c/(j

in which (!,g) and [a, -/3] are the morphisms induced by the universal properties of the biproduct (J.IA, J.IB, A • B, 1l'A, 1l'B) relative to J, g and to a, -{3. Suppose that the outer boundary is commutative. Prove that

and deduce that

[a, -,8] 0 (!,g) = 0. Prove that the outer boundary is a pullback for a, ,8 if and only if the sequence 0---+ P

is exact.

..J!.:!l.. A • B

[a,-f1] C

Chapter 5

72

6.9

If, in an abelian category,

A~c

a

is a pullback and a is epic, prove that g is epic. [Hint. If h o g = 0 consider [0, h] o (!,g). Now use the fact that, by Exercise 5.8, [a, -,8] is a cokernel of(!, g).] 6.10

Given an exact sequence 0 _____. A_____. B _____.C-. 0

in an abelian category, prove that for every morphism ,8 : Q --+ C there is a commutative diagram

in which the top row is also exact and the right hand square is a pullback. [Hint. Use Exercise 5.9.] 6.11 Tn a category C a morphism f : A --+ A is said to split if there is an object B and morphisms a : B --+ A, ,8 : A --+ B such that f = a o ,B and ,8 o a = ide. Prove that the following are equivalent :

(1) f: A--+ A splits; (2) f of = f and an equaliser of/, idA exists. Deduce that, in an abelian category, if idA -f.

f :A

--+

A splits then so does

6 Functors

As we have seen in the investigation of categories given in the previous sections, what is important is not so much what the objects are as how the morphisms behave. The time has now come for us to take a wider view of this. In order to do so, we shall begin by introducing the idea of a 'morphism from one category to another'. For typographical convenience, we shall henceforth denote identity morphisms idx as lx.

Definition. A covariant functor F from a category C to a category D is a prescription that assigns to every object A of C an object FA of D, and to every morphism a : A -+ B of C a morphism Fa : FA -+ F B of D, such that

(1) FlA = lFA for every object A of C; (2) if {3 o a is defined in C then Ff3 o Fa is defined in D and Ff3 o Fa = F(f3 a a). A contravariant functor from C to D is a covariant functor from C to Dd. For such a functor the only change in the above definition is that in this case we have F(f3 o a)= Fa o Ff3. We shall make the convention that whenever we speak simply of a functor we shall mean a covariant functor. Also, we shall denote a functor F from C to D by F: C-+ D. Example 6.1 If C is a category consider the assignment lc : C --+ C described by setting leX= X for every object X of C and lcf = f for every morphism of C. It is clear that this defines a functor, called the identity functor on C.

Chapter 6

74

Example 6.2 If C is a subcategory of a category D, consider the assignment I: C-+ D described by setting IX= X for every object X of C and If= f for every morphism of C. Clearly, this defines a functor, called the inclusion functor from C to D. Example 6.3 Consider the assignment U : Grp -+ Set described as follows. For every group G let UG denote the underlying set G, and for every group morphism ex : G -+ H let U ex denote the corresponding set map Ct. Clearly, U is a functor. We call it the forgetful functor from Grp to Set, essentially since it forgets that what it is looking at has any structure. Another name used is a stripping functor. Example 6.4 Let R be a ring with a 1 and for every set S let ( F S, f s) be the freeR-module on S (refer to Example 2.11 and to Exercise 2.11). H ex : A -+ B is a mapping then by the universal property of FA there is a unique R-morphism Fet : FA -+ F B such that the diagram

is commutative. Consider the assignment F : Set -+ RMod described as follows. For every set A let FA be the (object part of the) freeR-module on A, and for every mapping Ct :A-+ B let Fet :FA-+ F B be the above induced R-morphism. Then it is readily seen that FlA = lFA and, by pasting commutative diagrams together, that F(f3oet) = Ff3o Fet. Thus F is a functor, called the free R-module functor. Example 6.5 For every group G let 6(G) be the derived group of G (i.e. the subgroup generated by the elements of G of the form xyx-ly- 1). For every group morphism ex: G-+ H let 6(cx) : 6(G) -+ 6(H) be the group morphism induced by the restriction of ex to 6( G). Then the diagram G~H

i~r

rnc

6(G)-----+6(H) 6(a)

is commutative. Consider the assignment F : Grp -+ Grp described by FG = 6(G) and Fet = 6(et). Clearly, FIG = lFG and, by pasting together commutative diagrams, F(f3 o ex) = F{3 o Fet. Thus F is a functor, called the derived group functor.

Functors

75

Ezample 6.6 For every group G let FG = G/6(G). Since Q : G -+ G/6(G) is a cokernel of the embedding i: o(G) -+ G, given any group morphism a : G -+ H there is a unique group morphism Fa such that the diagram G~H

·1

1'

G/6(G) =FG-----+FH= H/6(H) Fa

is commutative. Since o(G) is the smallest normal subgroup of G for which the quotient is abelian, this construction clearly yields a functor F : Grp -+ Ab, called the abelianisation functor.

Given categories C 1 , ••• , Cn we can form a category by taking for the objects then-tuples (At, ... ,An) where A; is an object of C; for every i, for the morphisms from (A 1 , ••• , An) to (B1 , .•• , Bn) the n-tuples {11 , ••• , In) of morphisms with I; : A; -+ B;, and for composition of morphisms that given by

The resulting category is called the cartesian product category and is denoted by

X C; or by C 1 x · · · X Cn.

This notion allows us to define

i=l

more involved functors. Definition. A bifunctor is a functor of the form F : A x B -+ C. Trifunctors and more general multifunctors are defined similarly. Ezample 6.7 Given mappings I : A -+ C and g : B -+ D, we know from the universal property of products that there is a unique mapping I x g : A x B -+ C x D such that the diagram A~AxB~B

C C that is described by

F(A, _)(B) = F(A, B),

F(A,_)(f) = F(lA,f);

and for every object B of B there is the left associated functor F(_, B) : A ---> C that is described by

F(_, B)(A) = F(A, B),

F(_,B)(f) = F(f,lB)·

Each of these functors is covariant. Taking in particular F to be the above morphism bifunctor, we have for every object C of C the right associated functor More(C,_): C---> Set that is described by More(C, -)(X)= More(C,X), More(C, _)(f)= More(lc,/): {} ,_. f

o

r'J;

and the left associated functor More(_, C): Cd---> Set that is described by More(-, C)(X) =More( X, C); More(-,C)(g) = More(g, lc): {} ,_. RMod be the free R-module functor (Example 6.4) and let 1 : RMod -> RMod be the identity functor (Example 6.1). For every R-module M let (FM,iM) be the freeRmodule on the set M. Given an R-morphism f : M -+ N, we have that F f : F M -+ F N is the unique R-morphism that makes the following diagram commutative :

M 1

M=1M~FM

1

N

1Ff

/=l/1

N = 1N ------+FN iN

This situation clearly defines a natural transformation i : 1 -+ F. Example 6.10 For any set A let A x _ : Set -+ Set and _ x A Set -+ Set be the associated functors. If, for any set X, the mapping 'f/x: Ax X-+ X x A is given by 'f/x(a,x) = (x,a) then, relative to a given mapping f : E -> F we have the commutative diagram

E

AxE~ExA

/1

lAX/1

F

A

X

1/XIA F------+F

X

A

~F

This situation defines a natural transformation 'f/ : A x _

-+ _

x A.

Chapter 6 Example 6.11 Let A : Grp -+ Ab be the abelianisation functor (Example 6.6) and let I : Ab -+ Grp be the inclusion functor (Example 6.2). For every group G let q6(G) : G-+ Gjo(G) be the natural morphism. Then, relative to a given group morphism f : G -+ H we have the commutative diagram

G~AG I

1

1A/

H-------+AH -6(H)

This clearly defines a natural transformation

q6 : lcrp

-+

I o A.

Example 6.12 Let V be a vector space over a field F, let V 4 Mor(V, F) be the dual space, and V 44 = (V 4 ) 4 the bidual. The linear mapping 11 : V-+ ydd described by x >--+ x 11 where x 11 (f) = f(x) has the property that if rp : v ..... w is linear and rp11 : ydd ..... wdd is its bitranspose then there is a commutative diagram

v~vdd

v

~1

~1

w

1~.

w-------+ wdd "

This situation describes a natural transformation from the identity functor to the bidual functor. It is clear how we can define the composite of two natural transformations 'I : F-+ G and J.l.: G-+ H; we simply paste together commutative diagrams: A FA~GA~HA I

1 B

Fl

1 1 1 Gl

Hf

AB----+GB-------+HB 'IB

I'B

The commutative outer rectangle then yields a natural transformation which we shall denote by J.l. o 'I: F-+ H with (J.I. o 'I)A = J.I.A o 'lA· All this suggests, therefore, that we can define a category by taking as objects all functors F : C -+ D, as Mor(F, G) the set of natural transformations J.l. : F -+ G, and as composition that described above.

Functors

79

There is, however, a difficulty-how can we be sure that the natural transformations from F to G constitute a set? In fact, in general they do not. However, all is not lost : we simply have to restrict our sights a little. Suppose that I is a small category (i.e. its objects form a set) and that F, G : I -+ D are functors. Then clearly X Moro (FA, GA) is a AEI

set, whence so is the class of natural transformations from F to G; for if 'I: F-+ G is a natural transformation then each 1'/A E Moro(FA,GA) so 'I is determined by a subset of X Moro(F A, GA). Thus we see AEI

that when I is small we do obtain a category, which we can denote by Fnnct(I, D). More succinctly, we shall use the notation D 1 (which is suggested by the notation yx that is often used to denote the set of all mappings I: X-+ Y). We shall now show how the above observations can be used to link the ideas in Chapter 3 with that of a functor. Given a small category I, let its set of objects be denoted by I. If C is any category and A is an object of C then we can clearly define a functor KA :I-+ C by setting KAX =A for every X E I and KAI = 1A for every morphism I in I. We call this the constant functor associated with the object A. Definition. If I is a small category then by a source for a functor F : I -+ C we shall mean a pair (A, 'I) consisting of an object A of C and a natural transformation 'I : KA -+ F. The dual notion is that of a sin/c. In keeping with the diagram illustrating a natural transformation, we can represent a source by the set of all commutative diagrams

It is clear that we can form a category whose objects are the sources for F : I -+ C and whose morphisms from (A, 'I) to (A', '1') are those morphisms {} : A -+ A' in C such that o {} = '}; for every i E I. A terminal object in this category is called a limit of the functor F. The dual notion is that of a colimit. Example 6.13 If I is the small category

'1:

a

1•

==:::::: •2 fJ

So

Chapter 6

then a functor F : I --+ C is described by specifying morphisms Fa

F1====!F2 F/3

in C. A source (A, 'I) of F is then represented by the diagram

Thus we see that a limit exists for F if and only if an equaliser exists in C for Fa,F{J. Example 6.14 A category C is said to be discrete if all the morphisms in C are identities. Consider the small discrete category I described by 1•

• 2

A functor F : I --+ C is then described by specifying objects F1, F2 of C. A source (A, 'I) for F is then represented by a diagram qyF1

A

/ ~F2

Thus we see that a limit exists for F if and only if a product exists in C for F1,F2. More generally, when I is small and discrete a functor F : I--+ C is specified by a family of objects of C, and F has a limit if and only if a product of this family exists in C. Example 6.15 Consider the small category described by

•

.___... I

Functors A functor F : I ---> C is specified by morphisms F f, F g having a common terminus in C. A source (A, 17) for F is then represented by the diagram A~•

.__,. F/

Thus we see that a limit exists for F if and only if a pullback exists for Ff,Fg in C. It is clear that the situation in Example 6.15 can be generalised. For this, we require to generalise the notion of a pullback (for two morphisms with a common terminus) to that of a multiple pullback (for a family of morphisms with a common terminus). Beginning with a small category I described by the diagram

•

:~ .__,.

:/

•

we can show that a limit exists for a functor F : I ---> C if and only if a multiple pullback exists inC for (Ff;);ei· Definition. A category C is said to be finitely complete if every functor F :I---> C from a finite category I has a limit. THEOREM 6.1

The following statements are equivalent: (1) C is finitely complete; (2) C has finite products and equalisers; (3) C has finite products and finite intersections; (4) C has pullbacks and a terminal object. Proof. The equivalence of (2),(3),(4) was established in Theorem 3.9 so it suffices, for example, to show that (1) ~ (2) and (3) ~ (1). (1) ~ (2) :This is immediate from Examples 6.13 and 6.14. (3) ~ (1) : Let I be small with set I of objects and let F: I---> C be a functor. Form the product (X Fi, (7r;);ei) of the family (Fi);ei and iEI

for every morphism

f:

i---> j in I let

(Ef, e{,) be an equaliser ofFf o 7r;

Chapter 6 and

1r;.

Now let (

n E:, d)

be an intersection of the family (E:);er·

JEI

Then we obtain (1) by applying to the following diagram the subsequent argument.

nE~7l\· r ~

,er

3

d

h

"'

..\

X Fi----> F"~ - - -Ff- - > F j K;

iEI

Kj

Since Ff

o 1r; o

we see that (

d = Ff

o 1r; o

e;

d;

o

= 1r; o e; o d: = 1r; o d

n E:, 1) is a source for F

where ); =

1r;

o d. We claim

jEI

that this source is in fact a limit for F. To see this, suppose that (A, p) is a source for F. By the universality of products there is a unique h : A --+ X Fi such that 1r; o h = TJ; for iEI

every i E J. Now, by commutativity, FJ o

1r; o

h = Ff

o

TJ;

= TJ; = 1r; o h

and so, by the definition of equaliser, there exists kf : A --+ E{ such that o kf = h. By the universality of intersections, there is then a morphism {} : A -+ E: such that d o {} = h. That {} is unique with

e;.

n

jE!

respect to this property follows from the observations that, for every i, 'lr;

0

d0

{}

=

'lr;

0

h = TJ;,

that the 1r; are left cancellable en famille, and that d, being monic, is also left cancellable. ~ Definition. A category C is complete if every functor F : I --+ C from a small category has a limit. We invite the reader to review the proofs of Theorems 3.7, 3.9 and 6.1 with the objective of ignoring the finiteness conditions that were imposed. Bearing in mind the remarks made above concerning arbitrary products and multiple pullbacks, we can gather the resulting conclusions together in the following result.

Fu.nctors THEOREM 6. 2 The following statements are equivalent :

(1) (2) {3) (4)

C C C C

is complete; has products and equalisers; has products and intersections; has multiple pullbacks and a terminal object. 0

Definition. A downward directed set is an ordered set (I,:::;) in which every pair of objects has a lower bound (i.e. for all x, y E I there exists z E I with z:::; x and z:::; y). As in Example 1.12, we can consider a downward directed set I as a small category I.

Definition. We say that a category C has inverse limits if every functor from a downward directed set (considered as a category) to C has a limit. THEOREM 6.3

The following statements are equivalent:

(1) C is complete; (2) C is finitely complete and has inverse limits. Proof. It is clear that (1) => (2). (2) => (1) : By the equivalence of statements (1),(2) of Theorem 6.2, it suffices to prove that C has products. Suppose then that (A;);er is a family of objects of C. Observe that the finite subsets of I form a downward directed set (under ~) which we can consider as a category Dr. Let F : Dr ......, C be the functor described by F(J) = X A; for iEJ

every object J of Dr, and F(J,K) = PKJ for every morphism (J,K) of Dr where J ~ K and PKJ : X A;......, X A; is the morphism induced iEK

by the universality of the product

iEJ

X A;.

Suppose that (L, p) is a limit

iEJ

for F. Then we have a commutative diagram

We show that (L, (1rj o P;);er) is a product of (A;);er· This we achieve

Chapter 6 by applying to the following diagram the subsequent argument.

By the universality of the products there exist unique morphisms such that

etK, CtJ

Consequently,

By Theorem 3.5 we deduce that PKJOCtJ( =CtJ.

Consequently, (B, a) is a source for F and so, by the definition of (L, p), there is a unique (}: B---+ L such that, for every J, PJ o (}

=

CtJ.

It now follows that, for every J,

Moreover, (} is unique with respect to this property. For, if (}' : B also satisfies it then by Theorem 3.5 we have PJ o (}'

= PJ

The uniqueness of (} such that pJ o (}

---+

L

o (}.

= et J

now shows that (}1 = (}.

We now consider some 'preservation properties' of functors. Definition. If A, B are additive categories then a functor F : A ---+ B is said to be additive if, for all morphisms a,/3: A---+ A' in A, we have

F(et + {3) =Fa+ F/3.

Bs

Functors

Example 6.16 If C is an abelian category and M is an object of C we have the associated 'set valued' functors More{M, _) and More{-, M). We can also consider the associated 'group valued' functors hM : C -• Ab, given by hMX = More(M,X);

hM/=More(lM,/) and hM : C

-+

t'J>->fot'J,

Ab, given by hM X= More(X,M), hM f =More{!, 1M) : t'J

>->

t'Jo f.

If U : Ab-+ Set is the forgetful functor (Example 6.3) then clearly we have U ohM = More{M,-) and U ohM = More{-,M). The group valued functors hM and hM are additive. For example, hM(a+f3) sends

t'J ,___, (a+ {3) o t'J = (a o t'J)

+ ({3 o t'J)

and so is the same as hMa + hMf3· THEOREM 6.4 Additive functors preserve zero objects. Proof. Let F : A -> B be additive and let Z be a zero object of A. Since MorA{Z, Z) = {Ozz} we have lz = Ozz and hence lrz = Flz = FOzz = F(Ozz- Ozz) = FOzz- FOzz = Orz,rz.

For every f: FZ-+ B, we have f = f o lrz = f o Orz,Fz = Orz,B and so F Z is initial in B. Similarly, for every f : B -+ F Z we have f = lrz of= Orz,Fz of= Oe,Fz and so FZ is also terminal in B. Definition. Let A, B be binormal categories. A functor F : A -> B is said to be left exact if 0 ---+ A ---+ A' ---+ A" exact ==> 0 ---> FA ---+ FA'

--->

FA" exact;

and right exact if A ---+ A' ---+ A" ---+ 0 exact ==> FA ---+ FA' ---. FA" ---. 0 exact. If F is contravariant then we say that F is left exact if

0---+ A---. A' --+A" exact==> 0---+ FA" ---+FA' ---+FA exact; and that it is right exact if A

--+

A' ---+ A" ---+ 0 exact ==> FA" ---+ FA'

--+

FA ---+ 0 exact.

By an exact functor we shall mean a functor that is both left exact and right exact. Note that an exact covariant functor 'preserves' short exact sequences, and that an exact contravariant functor 'reverses' short exact sequences.

86

Chapter 6

THEOREM 6. 5 Let A be an abelian category. For every object A of A the covariant functor hA :A -+ Ab is left exact, and the contravariant functor hA :A-+ Ab is also left exact.

Proof. Suppose that we have an exact A-sequence 0---+ B

~ C __!!__. D

and consider the associated Ab--sequence

0---+hAB~hAC~hAD in which, we recall, hAa is simply composition on the left by a. If f E Ker hAa then a of= 0 and so, a being monic, f = 0. Thus hAa is injective, hence monic, in Ab. If now g E Ker hAf3 then f3 o g = 0 and so, a being a kernel of (3, there exists k such that a o k = g, i.e. hAa(k) = g. Thus we have KerhAf3 ~ ImhAa. But f3 o a= 0 so, by Theorem 6.4, hAf3 o hAa = 0 and consequently Im hA a ~ Ker hAf3· The resulting equality shows that the associated Ab-sequence is exact. Hence hA is left exact. A similar proof shows that the contravariant functor hA is also left exact. In general, the functors hA and hA fail to be exact, as the following examples show. Example 6.17 In zMod consider the short exact sequence

Let A = 71./271. and consider the functor hA : zMod -+ Ab. This is left exact (by Theorem 6.5), but is not right exact since the induced morphism Morz(ll./271., ()) ~ Morz(ll./271., G:l/71.) cannot be epic(= surjective). In fact, the group on the left collapses to {0} whereas that on the right does not. To see this, let {} : 71./271.-+ () be a group morphism and let x = {}(1 + 271.). We have 2x = 2{}(1

+ 271.)

= {}(2 + 271.) = {}(O

+ 271.)

= 0

whence x = 0 and consequently {} = 0. On the other hand, 0 + 271.

>---+

0 + 71.,

1 + 271.

>---+

~

describes a non-zero element of Morz(ll./271., G:l/71.).

+ 71.

Functors Example 6.18 In zMod consider the short exact sequence 0 ~ 71. ___j_____, Q _____!____. Q/71. ~ 0

and the left exact contravariant functor hz : zMod ---> Ab. This functor is not right exact since the induced morphism Morz(Q, 71.)

--+

Morz(ll., 71.)

cannot be epic (=surjective). In fact, the group on the left collapses to {0} whereas that on the right does not. To see this, let 11 : Q -+ 71. be a group morphism and suppose that 11(1) I 0. Then for every non-zero integer r we have 11(1) = rl1(1/r), whence r divides 11(1). But, by the fundamental theorem of arithmetic, 1?( 1) has only finitely many divisors. Hence we must have 1?( 1) = 0. For all non-zero integers p, q we then have 0 = pl1(1) = pl?(qfq) = pql?(1/q) = ql?(p/q) whence 11(pfq) = 0 and so 1? = 0. On the other hand, 1z is clearly a non-zero element of Morz(ll., 71.). In view of these observations, it is natural to investigate those objects A of an abelian category A for which the functors hA and hA are exact. We can in fact do this at a more general level. Definition. An object P of a category C is said to be projective if the right associated functor Morc(P, -): C---> Set preserves epics, and injective if the left associated functor Morc(-,P): C-+ Set preserves monies. THEOREM 6.6 An object P of a category Cis projective if and only if, for every epic morphism f : B -+ C and every morphism g : P -+ C there is a morphism 'IJ : P -+ B such that the diagram

p

/lg B~c

f epic

is commutative. Proof. hpf: Morc(P, B)-+ Morc(P, C) is surjective if and only if for every g : P -+ C there exists 11 : P -+ B such that hp f( 11) = g. Since the action of hpf is simply composition on the left by J, the result follows. ~

There is, of course, a dual characterisation of injectives.

88

Chapter 6

Example 6.19 In Set every P diagram

# 0 is projective. In fact, given a

B----+C f epir

we have that, f being a retraction, there exists 1 : C f o 1= 1c. Then tJ = 1o g is such that f o tJ =g.

---> B such that

Example 6.20 In Set every Q # 0 is injective. The proof is dual to that in Example 6.19.

In general, determining the projectives and injectives, when they exist, in a given category is not an easy task. It can be shown, for example, that in RMod the projectives are those R-modules P for which there exists a free R-module F and an R-module Q such that F = P EB Q. Also, in Ab the injectives are the divisible abelian groups.

IExercises I 6.1 For every set X let 2X denote the power set of X. For every mapping f: X---> Y let f- : 2x---> 2Y be given by f-(A) ={!(a) ; a E A}, and let f-: 2Y---> 2x be given by f-(B) = {x EX; f(x) E B}. Prove that the assignment

Ft=r defines a functor F: Set ---> Set (called the covariant power set functor). Prove also that the assignment

Ft=r defines a functor F : Set ---> Set (called the contravanant power set functor). 6.2 For every set E let E 2 = Ex E and for every mapping f : E ---> F let P: ~---> F 2 be given by the prescription f 2 (x,y) = (/(x),J(y)) .. Show that the assignment

describes a functor F : Set ---> Set (called the squaring functor).

Functors

8g

6.3 Show that if an abelian group G is considered as a category G then the addition in G is a bifunctor + : G x G --+ G. 6.4 Let A,B be categories with A small. Consider the definition of a bifunctor H(_, _) : BA x A --+ B as follows : (1) for every functor F: A-+ B and every object A of A let

H(F,A) =FA; (2) for every natural transformation 'I : F --+ G in BA and every morphism f: A--+ A' in A let H('l, f): FA-+ GA' be given by

H('l,/)=? Determine what should be chosen for? in order that H(_,_) becomes a bifunctor, and verify that your choice does produce a bifunctor. For every functor F : A -+ B describe the right associated functor H(F, _),and for every object A of A describe the left associated functor H(_,A). 6.5 Let F : A -+ B be a functor. Consider the definition of a bifunctor Mor(F _,_):Ad x B-+ Set as follows : Mor(F _,_)(A, B)= Mor(A, B);

Mor(F_,_)(f,g) =? Determine what should be chosen for?. Now let G : A -+ B be a functor. Consider the definition of a bifunctor Mor(_, G_): Ad x B-+ Set as follows :

Mor(_, G_)(A, B) = Mor(A, GB); Mor(_, G_)(f, g) =??

Determine what should be chosen for ?? . 6.6 If C, D are binormal categories and F : C prove that the following statements are equivalent : (1) F preserves exact sequences; (2) F preserves short exact sequences; (3) F preserves kernels and epics; (4) F preserves cokernels and monies; (5) F preserves kernels and images; (6) F preserves cokemels and coimages.

-+

D is a functor,

Chapter 6

go

6. 7 Let A, B be additive categories. If F : A prove that the following statements are equivalent : (1) (2) (3) (4)

F F F F

->

B is a functor,

is additive; preserves products; preserves coproducts; preserves biproducts.

6.8 Prove that if A, B are abelian categories and F : A left exact additive functor then F preserves pullbacks. [Hint. Use Exercises 6.7 and 5.8.]

->

B is a

6.9 Let A be a category aud I a small category. Let A have au initial object X and let Kx : I -> A be the constant functor. Prove that for every object Fin A 1 there is a unique natural transformation '7: Kx-> F. Deduce that Kx is initial in A 1 . Prove similarly that if A has a terminal object then so does A 1. Conclude that if A has a zero object then so does A 1 .

6.10

If A, B are binormal categories and 0------> hx A~ hxB ~ hxC

is exact for every object X of A, prove that

is exact. 6.11 If C, D are binormal categories then a functor F : C said to be semi-exact if for every short exact sequence

-+

D is

in C the induced sequence

FX~FY~FZ is exact in D. Prove that if A, B are abelian categories and F : A exact functor then F is additive.

-+

B is a semi-

6.12 If Cis au abelian category and (P;);er is a family of projectives in C prove that fi P; is also projective. iEI

Functors

91

Let C be an abelian category. Given objects A, B in C let B) be the set of morphisms f : A --> B that 'factor through projectives' in the sense that there is a projective object P and morphisms a,/3 such that 6.13

Mor~(A,

A __!_____. B = A~ P _.!!___. B. Prove that Mor~(A, B) is a subgroup of Morc(A, B). Define

Q(A, B)= Morc(A, B)/Mor~(A, B). Prove that for every morphism f : B' --> B in C there is a unique morphism Q(A, f) in Ab such that the diagram

hAB1 ~ hAB I'

1

1~

Q(A,B')---+Q(A,B) Q(A,!)

is commutative. Prove that the functor

Q(A,_): C ...... Ab so defined is semi-exact. 6.14 Let C be a concrete category in which every epic is surjective. Suppose that for every setS a free C-object on Sexists (Example 2.11). Prove that (1) every free C-object is projective; (2) an object P of C is projective if and only if there is a free C-object F and a retraction {}: F--> P.

6.15

In an abelian category C suppose that the diagram

P'

P"

a1

1p

O---+E1--->E---+E11 ---+0 J g is such that the row is exact and P', P" are projective. Prove that this diagram can be extended to a commutative diagram

0---+P' __!___.p~P"-----+0 a

1 1 7

lp

0---+E'---+E---+E11 --->0 J g in which the top row is exact and P is also projective.

Chapter 6

6.16 Let A be a non-empty set and let KA : Set ~ Set be the constant functor. Prove that the projection maps 7rB : Ax B ~ A given by 7rB(a,b) =a define a natural transformation 1r: Ax_.- KA. Show also that there is no natural transformation ~ : KA .- A x _.

Let F, G : A -+ B and H, K : B .- C be given functors. If G and 6 : H -+ K are natural transformations, establish for every object A of A the commutative diagram 6.17

'7 : F

-->

HFA~KFA

H~A

1

1K~A

HGA-----KGA 0GA.

Prove that p. : H F

-+

KG described by p. ~ (P.A) where

is a natural transformation. Writing p. = 6 * 1'}, prove that the diagram

commutes.

7 Equivalent categories

In any study of algebraic structures it is important to know when two structures of the same type are 'essentially the same' or isomorphic. We shall now explore what it means to say this about categories. We first introduce what is a fairly natural definition of an 'isomorphism' between categories. Definition. We say that categories C and D are isomorphic, and write C ~ D, if there exist functors F : C -+ D and G : D -+ C such that FoG= 10 a.nd Go F = 1c. Example 7.1 Associate with every ring R the dual ring Rd (in which the multiplication is given by x · y = yx ). This clearly yields a functor F: Ring-+ Ring such that F oF= 1Ring· We obtain in this way an isomorphism from Ring to itself. Example 7.2 Let U : zMod -+ Ab be the forgetful functor and let G : Ab -+ zMod be the functor that regards an abelian group as a Z-module. Then U o G = 1 and Go U = 1, so that zMod ~A b. Example 7.3 Given categories A,B let F: AxB--> BxA be described by F(A, B)= (B, A), F(cr, P) = (.P, cr) and let G : B x A -+ A x B be described by G(B, A) =(A, B),

Clearly, FoG= 1uxA and Go F

G(j3,cr) = (cr,P). = 1AxB so that Ax B ~ B

x A.

Chapter 7

94

A less trivial example is provided by the following result. THEOREM 7.1 If A, B are small categories then, for every category C,

Proof. Since A X B ~ B x A by Example 7.3, it clearly suffices to establish the first isomorphism. We begin by defining a functor t.: cAxB-+ (C 8 )A. Every object of cAx 8 is a bifunctor F : A x B -+ C. For every object A of A the right associated functor F(A,_): B-+ Cis an object of C 8 . Define, for every object A of A,

t.FA = F(A, _). For every morphism a : A -+ A' of A define f.Fa to be the natural transformation (i.e. morphism) in C 8 such that, for every object B of B, (t.Fa)s = F(a, ls). This then defines a functor t.F: A-+ C 8 , i.e. an object of (CB)A. To complete our definition of a functor f. we have to associate with each natural transformation '7 : F -+ G a natural transformation f>f'} : t.F -+ t.G. Now for such an f'J we have the commutative diagram F(A,B) ~ G(A,B)

(A, B) (a,/1)

1

F(a,/1)

f>f'}

1

G(a,/1)

F(A', B')------+G(A', B')

(A',B') and for such a

1

'I(A.'.B')

we have the commutative diagram F(A,_)

= t.FA~t.GA = G(A,_)

~Fal

1~Ga

F(A' -) = t.F A'-----+t.GA' = G(A' -)

'

So we define

(~.,)~

'

Equi11alent categories

95

That D. is a functor now follows from the equalities ((fllF)A)B = (lF)(A.B) = lF(A.B) = lf(A,_)(B) = lt.FABi

((fl(17 ° ~))A)B =('I o

d(A,B) = 'I(A,B) o

~(A,B) = ((Ill]

0

fl~)A) B'

We now define a functor r: The objects of (C 8 )A are functors H : A --+ C 8 • For every morphism a : A --+ A' in A we have that Ha: HA--+ HA' is a morphism in C 8 , so we have the commutative diagram (C 8 )A--+ cAxB.

HAB~HA'B

1

1

HAP

HA'P

H AB' ---->H A' B' (Ha)s•

Define rH(A,B) = HAB;

r H( a, {3)

= the diagonal of the above commutative diagram.

Then clearly we obtain a functor r H : A x B --+ C, i.e. an object of cAxB. To complete our definition of a functor r we have to associate with every natural transformation 1 : H --+ K a natural transformation r~: rH--+ rK. For such a 1 we have HA~KA

A a

1

Ha

A'

1

1

Ka

HA'----->KA' (A'

and for such a (A, B) (a,P)

1

(A',B') So we define

(**)

r~

we have HAB= rH(A,B)

~

l'H(a,p)1

rK(A,B) =KA(B) 11'K(a.P)

HA'B' =rH(A',B')-----> rK(A',B') = KA'B' (I'(I(A',B')

Clw.pter 7

g6

That

r

is a functor follows from the equalities (flH)(A,B) = ((!H)A)B = (!HA)B = !HAB = lrH(A,B);

(r('l o I))(A,Bl = (('7 o I)A) B = ('IA)B o (IA)B = (r'7 o ri)(A,Bl·

Now for every morphism 1 in (C 8 )A we have, by(*) and(**),

so that l:t.(fl) = 1 and hence l:t. or= I. Also, for every morphism '7 in cAxB,

(r(l:t.q))(A,B) = ((l:t.I'/)A)B = 'l(A,B)

so that f(l:t.'7) = '7 and hence r 0 l:t. =I. 0 Isomorphisms of categories are relatively rare. What is more common is the situation that we shall now describe. Let F, G : A --> B be functors and let 'I : F transformation. Recall as usual the diagram

-->

G be a natural

H each 'lA is an isomorphism then we say that '7 is a natural isomorphism. Definition. We say that functors F, G : A --> B are naturally equivalent, and write F R$ G, if there is a natural isomorphism 'I : F --> G. THEOREM 7. 2 A natural transformation '7 : F ---+ G is a natural isomorphism if and only if there is a natural transformation J.l : G ---+ F such that J.l o '7 = IF and 'I o J.l = I a.

Proof. => : If '7 : F --> G is a natural isomorphism where F, G : A --> B, assign to every object A of A the morphism J.IA = '7::1 1 : GA -+ FA. This clearly defines a natural transformation 1.1 : G --+ F. Since

(J.I 0 'I)A = /.lA (I'/

0

0

'lA = '7::1 1

0

I'/ A = IF A,

J.I)A = 1'/A o J.IA = 1'/A o '7::1 1 = laA

we deduce that J.1 o '7 = IF and '7 o J.1 = la.

Equitmlent categories

97

B is monic and that F I o g = F I o h. Since F is representative there exists an isomorphism k : FC --> B for some object C; and since F is full there exist morphisms p, q such that Fp = go k and Fq = h o k. Thus we have that

Flog= Flo h ==}Flo go k =Flo ho k

=* F(! o p) =Flo Fp

= Ff o Fq = F(f o q)

since F is faithful =*lop=loq =}p=q since f is monic

=* g o k = Fp = Fq = h o k =* g = h since k is epic. Thus F I is monic. Similarly, F preserves epics. The importance of these ideas is highlighted in the following result. THEOREM 7. 6 Categories A and B are equivalent if and only if there is a functor F :A --> B that is full, faithful, and representative.

Proof. => : If A and B are equivalent then there are functors F : A --> B and G: B-. A such that GoF Rl lA and FoG Rl lB. Let 'I : lA - t GoF be a natural isomorphism. To show that F is faithful, suppose that /, g : A --> A' are such that F I= Fg. Then we have the commutative diagram

in which a= Fl = Fg. It follows that 'f/A' of= 'f/A' o g whence, 'f/A' being an isomorphism and hence monic, I = g. Thus F is faithful. Note that a. similar proof shows that G is a.lso faithful. To show that F is representative we observe that, for every object B of B, we have FGB ~ 18 B = B, i.e. A= GB is such that FA~ B.

Chapter 1

100

f

Finally, to show that F is full consider g : FA --+ FA'. H we define = 'l::i) o Gg o 'lA then we have the commutative diagram GFA~A~GFA Gg

1 1/

1

GF/

GF A'+---A----+GFA '1A 1

'lA'

Consequently Gg o 'lA = GFJ o 'lA and so, since 'lA is an isomorphism, hence epic, we have Gg = GFJ. Now, as observed above, G is also faithful. Thus we obtain g = F f and so F is full. B. For every morphism g: B--> B' of B we then have {}8! o go {}B : FGB--+ FGB'. Since F is full and faithful there is then a unique morphism Gg: GB--+ GB' such that FGg = {}8! o go {}B· We now observe that G : B

--+

A so defined is a functor. In fact,

= {}8 1 o ls o {}B = lFGB = FlGs faithful, Gls = 1GB. Also, if g

FGls

and so, since F is h : B' --> B" then

FG(ho g)=

{}Ii!, o hogo

B

--+

B' and

"B

= {}B~ o h o {}B' o

{}8! o go {}B

= FGho FGg = F(GhoGg)

and so, again since F is faithful, G(h o g) = Gh o Gg. Thus G is indeed a functor. It now follows from the commutative diagram

Equivalent categories

10.1

and the fact that each I'Js is an isomorphism that FoG~ 18 • We now establish a natural isomorphism 71 : Go F ~ lA. For this purpose we observe that, for every object A of A, we have an isomorphism

{}FA: FGFA----+ FA. Then t'Jj;:~ :FA~ FGF A and so, since F is full and faithful, there is a unique morphism 'lA : A ~ GF A such that F'IA = t'Jj;:~. Moreover, by Theorem 7.4, 'lA is an isomorphism. We thus have, for every morphism f: A~ A', the commutative diagram

FA~FGFA

F/1

lFGF/

FA'---->FGFA' FqA,

Since, by Theorem 7.5, F reflects commutative diagrams we deduce that the diagram A~GFA

J

1 1

GF/

A'---->GFA' qA'

commutes. Since each 'lA is an isomorphism, Go F ~ lA. ¢ Definition. A functor that is full, faithful, and representative is said to be an equivalence.

Ezample 7.10 Let FDVF denote the category of finite-dimensional vector spaces over a field F and let MatF be the category of matrices over F (here the objects are positive integers, Mor( m, n) is the set of n x m matrices over F, and composition is matrix multiplication; see Exercise 1.2). Define a functor G: FDVF ---+ MatF by setting GV = dim V for every object V of FDVF and, for every morphism f of FDVF, by defining G f to be the matrix that represents f relative to fixed bases. Then G is full and faithful, since every matrix Anxm : m---+ n represents (relative to the fixed bases) a unique linear transformation. Also, G is representative since every positive integer m is the dimension of the vector space Fm. Thus G is an equivalence and, by Theorem 7.6, we have that the categories FDVF and MatF are equivalent. Ezample 7.11 IfF is a field then the functor Mor(_, F) : FDVffi ---+ FDVF that sends each finite-dimensional vector space to its dual is full, faithful, and representative. Thus Mor( _,F) is an equivalence.

Chapter 1

102

Functors that are full, faithful, and exact a.re also of considerable importance. However, we shall not develop the details here, other than give the following indications. It can be shown that if I is a small category then for the most part the category D 1 inherits the properties of D (e.g. binormal, additive). Thus, for example, Ab 1 is abelian. An important subcategory L of Ab 1 is that consisting of the left exact functors. It can be shown that the functor H : I ---+ L given by HI = h1 (recall Theorem 6.5) is full, faithful, and exact. This leads to the following important result. THEOREM 7. 7 [Mitchell] For every small abelian category I there is a full faithful exact functor F : I ---+ RMod for a suitable ring R. From this result there can be derived a 'metatheorem' (i.e. a theorem about theorems) which asserts that every statement of the form 'P implies Q', where P is a categorical statement about a finite commutative diagran1 and Q is a statement about the existence of finitely many additional morphisms which makes some categorical statement true of the extended diagram, is true in an abelian category A if it is true in RMod for every ring R. Thus, for example, to prove that certain results concerning commutative diagrams are true in an abelian category it suffices to prove that they hold in RMod. This means that certain otherwise difficult manipulations with morphisms in an abelian category can be made easier on occasions by element-wise diagram chasing with modules.

I Exercises I 7.1

If 2 denotes the category

• --------+. prove that for every category A the category A 2 is isomorphic to the arrow category over A (Example 1.16). 7.2

If 3 denotes the category

•--------+•

~1•

prove that for every category A the category A 3 is isomorphic to the triangle category over A (Exercise 1.6).

Equivalent categories

103

7.3 If A is an initial object of C prove that the comma category A(Example 1.15) is isomorphic to C. 7.4

If R is a ring with a 1 prove that the functor

MorR(R,_): RMod-+ Ab is naturally equivalent to the forgetful functor U : RMod -+ Ab. 7.5 Prove that the forgetful functor U : Grp -+ Set is naturally equivalent to hz = Mor(Z, -)and that Nat(U, U) is equipotent to Z. 7.6 If 2 denotes the two-element set {0, 1} prove that the squaring functor (Exercise 6.2) is naturally equivalent to Mor(2, _). 7.7 Let C and D be categories of modules and suppose that D has arbitrary products and coproducts. If (F;)iEI is a family of additive functors F; : C -+ D, define X F; and 9 F; from C to Din the obvious iEI

iEI

way; for example, for every morphism

XF;)Ul = XFd, (iEI iEI

f

of C define

(9 F;)(!J = 9 Fd iEI

iEI

where the right hand sides are the morphisms induced by the universality of products and coproducts in D. With this notation, given a family (M;);EI of R-modules establish the following natural equivalences of functors from RMod to zMod : (1)

MorR ( ffi M;, _)

R>

iEI

(2)

MorR(-,

XMorR(M;, _); iEI

X M;)

iEI

R>

X MorR(-,M;).

iEI

7.8 Determine if the cartesian product bifunctor (Example 6.7) is full, faithful, or representative. 7.9 Determine if the abelianisation functor (Example 6.6) is full, faithful, or representative. 7.10 Let K be the category whose objects are those small categories C satisfying the properties (1) for all objects A, B of C, IMorc(A, B) I ::; 1; (2) if IMorc(A, B) I= IMorc(B,A)I = 1 then A= B. Prove that K is equivalent to Ord. 7.11 Let Cat 1 be the category whose objects are those categories consisting of a single object. Prove that Cat 1 is equivalent to Mon. 7.12 jects.

Prove that an equivalence preserves initial and terminal ob-

8 Representable and adjoint functors

In this section we shall be particularly interested in set valued functors. Recall that if C is any category then associated with every object A of C there is the set valued functor hA = Morc(A, _). H F, G : C -+ Set are given functors then, as we have observed, the class of natural transformations from F toG may fail to be a set. There is, however, an important situation where this class does form a set. THEOREM 8. 1 [Yoneda] Let C be a category and F : C ---> Set a set valued functor. Then for every object A of C the class Nat(hA, F) of natural transformations from hA to F constitutes a set that is equipotent to FA.

Proof. Let 'fJ : hA ---> F be a natural transformation. Then for every morphism f : A ---> X we have the commutative diagram A

Morc(A, A) = hAA~F A

fl

hAJl

X

Morc(A,X) =hAX--FX

lFJ ~X

By the commutativity of this diagram we see that

Representable and adjoint functors

105

Consequently, for every object X of C the mapping 'IX is completely determined by the element 'IA(lA) ofFA. Since FA is a set, it follows that so also is Nat(hA,F). For every natural transformation 'I : hA -> F let

(t) Clearly, this defines a mapping t1: Nat(hA,F) ->FA. Our objective now is to produce a mapping A: FA---> Nat(hA, F) that is an inverse for 11. Given f : A -> X we have F f : FA -+ F X and so if a E FA we have Ff(a) E FX. We can therefore define a mapping (A(a))x: hAX-+ FX by the prescription

(ttl

(A(a))xUl = Ff(a).

This gives rise to a natural transformation A(a) : hA seen from the fact that the diagram (>.ia))x

-+

F, as can be

FX

lFg (>.(a))y

FY

co=utes; for, via the north-east route,

!

>--->

FJ(a)

>----->

Fg[Ff(a)],

via the south-west route,

f

>--->go f

>----->

F(g of)( a),

and the outcomes are the same. We can therefore define A : FA Nat(hA, F) by A: a>---> A(a) for every a E FA. Now since, by (tt), (A(fJA(l_.t)))x(f)

-+

= Ff['IA(lA)] = 'lx(/)

we have that A(fJA(lA)) = 'I so that, by (t), A[t1('1)] = fJ and hence A o t1 = lNat(hA,F)' Also, for every a E FA we have, by (t) and (ttl, 11[A(a)J and hence tJ o A=

= (A(a))A(1A) = FlA(a) = 1FA(a)

1FA·

~

Chapter 8

106

If A,B are objects of a category C then Nat(hA,hs) is a set that i.! equipotent to Morc(B,A).

COROLLARY

Proof. Take F = hs in the above. ¢

In the course of establishing Theorem 8.1 we produced, for every a E FA, a natural transformation

.A( a): hA -+F. We shall now investigate the situation where, for some such element a, the natural transformation .A( a) is a natural isomorphism, in which case we have F ~ hA· When this situation obtains, we say that F is a representable functor, and that (A, a) represents F. A useful equivalent way of looking at this is provided by the following result. THEOREM 8. 2 The set valued functor F : C -+ Set is represented by the pair (A, a) zf and only if, for every object X of C and every x E F X, there is a unique morphism f : A -+ X in C such that F I( a) = x. Moreover, a representing pair is unique up to isomorphism, in the sense that if (B, b) also represents F then there is a unique isomorphism g: A-> B such that Fg(a) =b.

Proof. Recalling how the functors .A(a) were defined, we see that (A, a) represents F if and only if, for every X, the mapping (.A(a))x: hAX-> F X is a bijection. This is equivalent to saying that, for every X and every x E FX, there is a unique I E hAX = Morc(A, X) such that

= (.A(a))xUl = x. H now (B, b) also represents F then there are unique morphisms I :

Fl(a)

A-+ Band g: B-+ A such that Fl(a) =band Fg(b) =a. Then F(g of)( a)= Fg[Ff(a)]

=

Fg(x) =a= !FAa= FlA(a)

and so, by the uniqueness, we obtain go I= lA. Similarly I o g = ls and consequently I is an isomorphism. ¢ It is Theorem 8.2 that lies behind all universal constructions. In fact, given a functor F : C -> Set construct a category Cp as follows. For the objects, take the pairs (A, a) where A is an object of C and a E FA, and for the set of morphisms from (A, a) to (B, b) take those morphisms I: A-+ Bin C such that Fl(a) =b. Then by Theorem 8.2 we see that (A, a) represents F if and only if (A, a) is an initial object in the category Cp. We can do the same with contravariant functors: we form a category pC in which the morphisms from (A, a) to (B, b) are the morphisms I : B -> A of C such that F I( a) = b.

Representable and adjoint functors

107

Example 8.1 For every object A of a category C the functor hA is represented by the pair (A, 1A)· In fact, for every object X of C and every g E hAX = Morc(A, X) there is a unique morphism f: A -+ X such that hAf(lA) = g, namely I = g. Similarly, the contravariant functor hA is represented by the pair (A, 1A)· Example 8.2 The forgetful functor U : Grp -+ Set is represented by the pair (Z, 1). In fact, for every group G and every x E UG = G there is a unique group morphism f: Z-+ G such that U 1(1) = x (recall that every group morphism I: Z-+ G is completely determined by f(l)). Example 8.3 Let ( C;);ei be a family of objects of a category C. Define a contravariant functor F : C -+ Set as follows : for every object A of C let FA= XMorc(A,C;) iEI

and for every morphism g: Y-+ X let Fg: FX-+ FY be given by

Then, by Theorem 8.2, (P, (p;);er) represents F if and only if, for every object X of C and every (x;);er E FX there is a unique morphism g : X -+ P such that Fg(p;) = x;; in other words, if for every object X of C and every family (x;);er of morphisms x; : X-+ C; there is a unique morphism g : X----> P such that p; o g = x;.

Thus we see that (P, (p;);er) represents F if and only if it is a product of (C;)iei in C. Given categories A, B and a functor G : A -+ B, we can compose G on the left by the (representable) functor hB : B -+ Set to obtain, for every object B of B, a set valued functor GB = hB o G = Mor(B, G_): A-+ Set. We now wish to answer the question : under what conditions is GB representable for every object B of B?

Chapter 8

108

THEOREM 8.3 Given a functor G: A-+ B, the following statements

are equivalent :

(1) for every object B of B the functor GB = hB o G : A -+ Set is representable; (2) there is a functor F : B --+ A such that the morphism bifunctors Mora(-, G_): Ad x B-+ Set, MorA(F _,-):Ad x B--+ Set

are equivalent.

*

Proof. (1) (2) :For evry BE B let (AB, "s) represent Gs = hB o G. Recall that AB is an object of A and "BE GBAs = Mora(B, GAs). By Theorem 8.2, for every object A' of A and every "' E GBA' = Mora(B, GA') there is a unique '7 :As --+A' such that

i.e. such that the diagram

is co=utative. In particular, if g : B --+ B' is a morphism in B and (AB',"B·) represents Gs• then there is a unique '1B: As--+ As• such that

is commutative. Now define, for every object B ofB and every morphism g : B -+ B' of B, By pasting together two of the above co=utative rectangles we see, by the uniqueness of the morphism rJB, that F(H o g) = Fh o Fg. Since clearly FiB = 1A 8 = 1Fs it follows that we have thus defined a functor

F: B--+ A.

&pre.sentable and adjoint functors

109

Observe now that if F B' ~ A' in A then in B we have

Consider therefore the diagram t~-+GtotJ B'

MorA(F B', A')-----':...._--+Mora(B', GA') MorA(Fg,!)

1

1

Mora(g,Gf)

MorA(FB,A)

Mora(B,GA)

t~-+Gto{J B

This diagram is commutative. In fact, via the north-east route

t >--+ Gt o {}B'

>--+

Gf o Gt o {}B' o g,

1--t

G(f 0 t

via the south-west route, t

>--+I

0

t

0

Fg

0

Fg)

{}B,

0

and these outcomes are the same since, using the previous commutative diagram, G(f o to Fg) o {} 8 = Gf o Gt o GFg o {}B

= G(f

0

t)

0

GTJB

= G(f

0

t)

0

{}B'

0

0

{}B

g.

In order to show that Mor A(F _, -) ~ Mor 8 (_, G_) it therefore suffices to prove that each of the mappings t ,.._. Gt o {}B is a bijection. But since (FB,I'JB) represents G 8 we have, by Theorem 8.2, that for every I'JEGBA=Mora(B,GA) there is a unique TJ :FB-+ A such that

Thus each mapping t ,__. Gt o {} 8 is indeed a bijection and the above bifunctors are equivalent. (2)

=?

(1) :Suppose that there is a functor F: B-+ A such that

Our task is, given any B E B, to find a representing pair (AB, {}B) for GB. Recall that AB is to be an object of A and {}B E GBAB =

Chapter 8

110

MorB(B, GAs). Now because of the above equivalence of bifunctors there is, for all objects A of A and B of B, a bijection

MorA(F B, A) ___qc......:..•r_s__-+Mora(B, GA). Taking A= FB we see that 'IFs,Fs(lFs) E MorB(B,GFB). Thus, consider the pair (As, t?s) where As= FB,

t?s

= 'IFB,Fs(lFs).

We claim that this is a representaing pair for G s. To prove this, it suffices by Theorem 8.2 to show that for every object X of A and every t1 E GsX = MorB(B,GX) there is a unique a: FB-+ X such that Gsa(t?s) = 11, i.e. such that the triangle

B~GFB

~lGa GX

is commutative. Now since 'IX.FS is a bijection there is a unique a : FB-+ X such that 'lx.Fs(a) = t'J. Consider now the following commutative diagram :

MorA(F B, FB)---qr_s...;,_r_s_--+Mora(B, GF B) MorAilrs,a) 1

1Mors(ls,Ga)

Mora(B,GX)

MorA(FB,X) 'IX,FB

We have, by the north-east route, lFs,........ t'Js >---+Gao t?s

and, by the south-west route, 1FS ,...._.a>---+ t'J.

Since the destinations coincide, the above triangle is therefore commutative. That a is unique with respect to this property follows from the fact that if {3 : F B -+ X is such that G/3 o {} s = {} then there is a similar diagram to the above for {3, and necessarily {3 = 'lx~Fs(t'J) =a. 0 Definitions. If G : A -+ B and F : B -+ A are functors such that MorA(F _,_) r::. Mora(-, G_) then we say that F is a left adjoint of G, that G is a right adjoint ofF, and that (F, G) form an adjoint pair.

Representable and adjoint functors

111

Using this terminology, we can rephrase Theorem 8.3 in the form : the functor G : A -+ B has a left adjoint if and only if, for every object B of B, the functor GB = hB o G: A ...... Set is representable. There is, of course, a dual condition concerning the existence of right adjoints, namely that the contravariant functors GB = hB o G : A --+ Set be representable.

Example 8.4 The abelianisation functor t. : Grp --+ Ab is a left adjoint of the inclusion functor J : Ab ...... Grp. To see this, consider the diagram Morcrp(A,JB) - - -1"-'""'-fi:..:'---+ MorAb(~A, B) Mor(!,Jg)

1

1

Mor(£>./,g)

Morcrp( A', J B')

t't-+~1'

Mor Ab( t.A', B')

in which the horizontal mappings are given as follows. H t : A group morphism with B abelian then

-+

B is a

~ Kert where i: 6(A)--+ A is the inclusion morphism. Since A/6(A) is a cokernel of i there exists a unique abelian group morphism 11t such that

gives Imi Q:

A

--+

A______!_____. B = J B

·Al / t.A = Af6(A) is commutative. The mapping described by t ,..... 111 is clearly injective. It is also surjective since if g : t.A --+ B is an abelian group morphism then go QA : A -+ B and so, by the uniqueness, g = 1J go lA. The assertion

now follows from the fact that the above rectangle is commutative. To see this, travel via the north-east route to obtain

and via the south-west route to obtain

t ,__..... Jgoto f

~ !JJgotoj

Chapter 8

112

and observe that, by the following commutative diagram, the destinations are the same :

~t ~r.::.B 6

tlA=Af8(A)~ •

T~ vgotof

!>./I ; · .---

B'=JB'

'J••••I

tlA' =A'/8(A') We shall now discuss under what conditions a given functor has a left or right adjoint. For this purpose, suppose that G : A --+ B is a functor, that B is a fixed object of B, and consider the category G 8 in which the objects are pairs (A, f) where A is an object of A and f: B--+ GA, and the set of morphisms from (A, f) to (A', f') is the set of morphisms " : A --+ A' in A such that the diagram

/GA

B

lG.t ~GA'

commutes. An initial object in GB, when such exists, is said to be G-free ouer B. Note that this is a more general situation than that described in Example 2.11 where we defined free C-objects; in fact the situation there can be obtained from the present one by taking A to be a concrete category and G : A --+ Set to be the forgetful functor. It is clear by the universal property of initial objects that G-free objects, when they exist, are unique up to isomorphism. THEOREM 8.4 A functor G : A --+ B has a left adjoint for euery object B of B, a G-free object ouer B exists.

if and only if,

Proof. Given B, consider the set valued functor h 8 o G. By Theorem 8.2, this is represented by the pair (A, a) if and only if for every object X of A and every x E hBGX = MorB(B,GX) there is a morphism f : A --+ X such that Gf o a= hBGf(a) = x;

Representable and adjoint functors

113

i.e. such that the diagram

is commutative. Thus h 8 o G is representable if and only if a G-free object on B exists. The result therefore follows from Theorem 8.3. ¢ Note from the above results that if G : A --+ B has a left adjoint F : B --+ A then, for every object B of B, a G-free object over B is (FB, l?s) where l?s is defined as in the proof of Theorem 8.3. There is, of course, a dual result concerning the existence of right adjoints; this involves the notion of a G-cofree object over B, which is defined dually. Our next (and final) goal will be to determine necessary and sufficient conditions under wiLich a given functor G : A --+ B from a complete category A has a left adjoint, a result that is often referred to as the adjoint functor theorem. For this purpose, we require the following notion, which is more general than that of a G-free object. Definition. Let G : A --+ B be a functor. If B is an object of B then by a. G-family over B we shall mean a family (A;,/;);er, where A; is a.n object of A and /; : B --+ GA; for each i E I, such that for every object X of A and every morphism g: B--+ GX there is for some i E I a morphism {} : A; --+ X such that the diagram

is commutative. It is clear that if (A, f) is a G-free object over B then the singleton set {(A, f)} is a G-family over B. THEOREM 8. 5 Let A be a complete category and G : A

Then G has a left adjoint if and only if

(1) for every object B ofB a G-family over B exists; (2) G preserves products and equalisers.

-+

B a functor.

Chapter 8 Proof. => : If G has a left adjoint then, by Theorem 8.4, for every object

B of BaG-free object (F B, 118) exists over B. As we have just observed, {(FB,I18)} is a G-family over B. To see that G preserves products, apply to the following diagram the subsequent argument :

GFB~G(X

A;)

··]~}, B

GA;

g;

Let (F B, 118 ) be G-free over B. Then there is a unique morphism tp; : F B ----+ A; such that G1r; o 118 = g; for every i E I. Also, by the universality of the product, there is a unique morphism 'P : F B ----+ X A; iEI

such that 1r; o 'P = tp; for every i E I. Consequently, (ViE I)

Now suppose that 'r/1 : B

----+

G(1r;

o tp) o

c( X A;)

118 = g;.

is such that G'/1";

0

'r/1 = g; for

iEI

every i E I. Then since (F B, 118) is G-free over B there exists a unique morphism t: F B----+ XA; such that Gt o 118 = 'r/J. Consequently, iEI

(ViE I)

G(1r;

o

t)

o

118 = g;.

By the uniqueness in the definition of a G-free object, it now follows that (ViE I)

1r; 0 'P =

'II";

0

t

and hence, since the 1r; are left cancellable en famille, we deduce that 'P = t. Thus we have that

is a product of (GA;);EI· To see that G preserves equalisers, apply to the following diagrams

Representable and adjoint functors

115

the subsequent argument :

Let e : E

I

-->

At be an equaliser of At=====:A 2 • Then certainly we g

have

Gfo Ge = G(Jo e)= G(go e)= Ggo Ge. Suppose now that Gf o h = Gg o h. Let !'J : F B morphism such that

-->

At be the unique

Then we have

G(f o !'J)

o

I'Js

= GJ o G!'J o I'Js = GJ o h = Ggoh = Gg o G!'J o I'Js = G(g o !'J) o 6s

and so, by the uniqueness property concerning I'Js, we deduce that foi'J = go !'J. Since e : E --> At is an equaliser of/, g there is a unique morphism 'P: F B--> E such that eo 'P = !'J. Then

Suppose now that a : B -> GE is such that Ge o a = h. There is a unique morphism t : F B -> E such that Gt o I'Js = a and so G(e o t)

o

I'Js

= Ge o Gt o I'Js = Ge o a= h.

Again by the uniqueness property concerning I'Js, we deduce that eo t = e o 'P and hence that

Thus we see that Ge: GE-> GAt is an equaliser of GJ, Gg.

Chapter 8

116

¢::To establish the sufficiency of the conditions (1) and (2) it suffices to produce a free G-object over B for every object B of Band then apply Theorem 8.4. Suppose first that (A;, f;);er is a G-family over B. Given any morphism f: B---+ GX in B, consider the diagram

in which some fJ : A; --> X is such that GiJ o f; = f (definition of Gfamily) and et is the morphism induced through the preservation of the product, so that Gtr; oCt = k The commutativity of this diagram gives G(iJ o tr;) o et =f.

For convenience of notation in what follows, let

A= X A;. iEI

What we have shown so far is that the singleton {(A, et)} is a G-family over B. However, in general (A, et) is not G-free; some more and rather intricate arguments are required in order to produce G-free objects. Let S be the subset of MorA( A, A) consisting of those morphisms 'P : A--> A such that Gip o et = et, and consider the product

where AI'= A for every 'PES. For every 'PES the universality of the product A induces morphisms f3, "! : A--> A such that in the diagram

.Y-il~

A+--A--+A p'P ptp

Representable and adjoint functors

Let e : E

-+

A be an equaliser of f3, 1.

Then we have the diagram

in which, by hypothesis, Ge is an equaliser of G/3, G"f. Now, by the definition of S,

Gp'P o G/3 o cr = 1GAo cr = cr = Gtp ocr = Gp'P o G1 o cr. Since Ge is an equaliser of G/3, G"f there is then a unique 'I : B --+ GE such that Ge o 17 = cr. We now show that in fact (GE, 17) is a G-free object over B. For this purpose, we observe from the above diagram that iff : B -+ GX is a given morphism then G (tJ o 1r; o e) : G E --+ G X is such that

B____.!!______.GE

~

1

G("o>r;oe)

GX

is commutative; for

G(tJ o 1r; o e) o 'I= G(tJ o 1r;) o Ge o 'I= G(tJ o 1r;) ocr= f. Suppose now that f3 : E--+ X is also such that Gf3 o 17 = f. We show as follows that f3 = tJ o 1!"; o e whence (GE, 'I) will beG-free and the proof complete. Write J = ,Jo 1r; oe and let j : J -+ E be an equaliser of f3 and J. Then Gj : GJ -+ GE is an equaliser of Gf3 and GJ. Since, by the hypothesis, Gf3o17 = f = GJo'l there is a unique g: B--+ GE such that Gjog = "' and since (as observed above) {(A, cr)} is a G-family over B there exists

u8 t:

Chapter 8

A--+ J

with Gt o a = g, so that we have the commutative diagram

GX

Gf3na~ B __!!_______. G E

aL~ jGj GA----+GJ Gt

Let ijJ = e o j o t : A -+

A and observe that Ge o Gj o Gt o a = Ge o Gj o g = Ge o 'I = a,

Gi{J o a = so that ijJ E S. Consequently, eo

j

o

t

o

e = ijJ o e =

P;; o "' o e = p; o {3 o e = 1A o e = e,

and the fact that e is monic gives j otoe = we have

lE.

Finally, since {3o j = :fo j,

{3 = {3 olE = {3 o jot o e = ~ o jato and the objective is achieved. 0

e

= J olE = ~

IExercises I 8.1

Given a diagram

•

.____.. a

in a category C define a functor F : C -+ Set as follows. For every object X of C let F X be the set of morphism pairs (!,g) such that x~

•

._____... a

is commutative; and if h : Y --+ X is a. morphism inC let Fh : F X-+ FY be given by Fh(f, g)=(! o h, go h). Show that F is a contravariant functor which is representable if and only if a pullback exists for a, {3.

Representable and adjoint functors

119

8.2 Let A be a singleton set. Prove that the covariant/contravariant constant functor KA : C --+ Set is representable if and only if C has an initial/terminal object. 8.3 Prove that the contravariant power set functor (Exercise 6.1) is represented by the pair ({0, 1},{1}), and that the covariant power set functor is not representable. 8.4 Let U : Grp --+ Set be the forgetful functor. Prove that the free group functor F : Set --+ Grp is a left adjoint of U. 8.5 Let CG be the category of groups that coincide with their commutator subgroup. Prove that the commutator subgroup functor C : Grp --+ CG is a right adjoint of the inclusion functor I : CG --+ Grp. 8.6 Let Set* be the category of pointed sets and P : Set* --+ Set the puzzled functor (i.e. it doesn't see the point!). Show that P has a left adjoint. 8. 7 If G : A --+ B and F : B --+ C have left adjoints prove that so also does FoG. 8.8 Let A, B be abelian categories. If G : A --+ B and F : B --+ A are such that (F, G) is an adjoint pair, prove that G is left exact and F is right exact. [Hint. Use Exercise 6.10.] 8.9 Prove that the squaring functor (Exercise 6.2) has as a left adjoint the functor F: Set--+ Set described by FA= Ali A (the disjoint union of A with itself). 8.10 Consider an ordered set X as a small category X. Show that an isotone mapping f : A --+ B can be regarded as a functor f : A --+ B. Show also that to say that f is residuated (Exercise 1.8) is equivalent to saying that there is a functor r+ : B --+A such that (f, f+) forms an adjoint pair.

8.11 Given a set X consider the power set P(X), ordered by set inclusion, as a category. H A, B are sets and f : A --+ B is a mapping show that the induced mappings J- : P(A) --+ P(B) and f- : P(B) --+ P( A) are functors. Show also that (J-, j-) is an adjoint pair.

Solutions to the exercises

1.1--+1.7: Trivial verification ofthe axioms. 1.8 If I: A--+ B is residuated Jet I+ : B--+ A be given by l+(y) = max{x E A; l(x) ::=; y}. Then I+ is isotone. Moreover, for every x E A, J+[l(x)] ~ x; and, for every y E B, ![J+(y)] ::=; y. Thus J+ol ~idA and I o 1+ ::=; id 8 • Conversely, suppose that h : B--+ A is isotone and such that hoI~ idA and I o h ::=; idB. Given y E B we have l[h(y)] ::=; y and so {:r; E A; l(x) ::=; y} f. 0. For every x E A such that l(x) ::=; y we have, since his isotone, :r; ::=; h[l(x)] ::=; h(y). Thus max{x E A; l(x) ::=; y} exists and is h(y). This shows that I is residuated, and that such a mapping h is unique. If I : A --+ B and g : B --+ C are re.~iduated then go I is isotone and

r

r

1+ o g+ o go 1 ~ o idB o1 = o 1 ~ idA, g o I o 1+ o g+ ::=; go idB og+ = g o g+ ::=; ide

and so go I is residuated with (go!)+ = I+ o g+. A routine verification of the axioms (observing that identity maps are residuated) shows that we can form the category Ord+. 1.9

( 1) Yes; (2) Yes; (3) No.

1.10 Since l(x) A l(x') = l(x Ax') = I(O) = 0 and l(x) V l(x') l(x V x 1) = 1(1) = 1, it follows that l(x') = [l(x)]'.

=

Solution&

121

2.1 Consider the category G formed as follows : for the only object of G take the group G, for the set of morphisms from G toG take the set G, and define composition of morphisms to be the group operation. Then, since every element of G has an inverse, every morphism is an isomorphism. 2.2 Since + is commutative and the cancellation laws hold every morphism in IN is both monic and epic, hence is a bimorphism. The only morphism that has an inverse under + is the identity morphism 0; so this is the only isomorphism. 2.3 Let [i,j] denote the morphism from A; to Ai. We have to prove that [3, 4] o [1, 3] = [2, 4] o [1, 2]. Now [4, 8] is given to be monic and (writing o as juxtaposition)

[4, 8][3, 4][1, 3]

= [7, 8][3, 7][1, 3] = [7, 8][5, 7][1, 5] = [6, 8][5, 6][1, 5] = [6, 8][2, 6][1, 2] = [4, 8][2, 4][1, 2]

so the result follows by left cancellation of [4, 8]. 2.4

£

Then f(n) f(lm])

J

Consider IN ------+ z===::::::x. Suppose that f o ' = g o '·

= g(n) for all n E IN.

+ f( -]m]) =

g

For every mE Z we then have

/(]m]-]m])

= f(O)

= g(O) = g(]m])

+ g( -]m]).

Since /(]m]) = g(]m]) it follows that /( -]m]) = g( -]m]). Hence f = g and so ' is right cancellable, so epic. Clearly, ' is not surjective. 2.5 Clearly, the given f is surjective. If it were a retraction, there would exist a group morphism g : Z 2 -> Z such that fog = idz 2 whence f[g(1)] = 1 forces g(1) to be odd. But g(1) + g(1) = g(O) = 0 and so we have the contradiction g(1) = 0.

r

1 exists. This implies that f is 2.6 (1) =? (2) : If (1) holds then right cancellable, hence epic, and is a section since f- 1 of= id. (2) => (1) : Suppose that (2) holds and let g be such that go f = id. Then f o go f = f and so, since f is right cancellable, f o g = id. Hence f is an isomorphism with f- 1 =g. The equivalence of ( 1) and (3) is similar. Finally, if every epic is a retraction then every bimorphism is both monic and a retraction whence (by the above) is an isomorphism. Thus C is balanced.

2.7 The free semigroup on {a} is isomorphic to the multiplicative semigroup {an ; n = 1,2,3, ... },which is isomorphic to the additive semigroup IN \ {0}.

Solutions

122

2.8

Follow the hint.

2.D

The free group on {a} is isomorphic to the multiplicative group

{an ; n = 0, ±1, ±2, ... }, which is isomorphic to the additive group 71.. 2.10 Let f : A -+ B be an additive group morphism and let K = Kerf. The natural embedding iK : K-+ A is a group morphism and f o iK = 0. Since f is monic by hypothesis, we deduce that iK = 0 whence K = {0} and so f is injective. 2.11

Follow the hint.

2.12 Free R-modules on a given set, being initial objects in a suitable category, are isomorphic. It suffices to show that the ring R regarded as an R-module is free on {1R}· Let f: {1R}-+ R be given by /(1R) = 1R and let g : {1R} -+ X be any mapping into an R-module X. Define(}: R-+ X by (}(r) = r g(1R)· Then(} is an R-morphism and (}of = g. If now h : R -+X is an R-morphism such that h of= g then for every rE Rwe have h(r) = rh(1R) = r[(hof)(1R)] = rg(1R) = (}(r) and soh= 6. 2.13 In the solution to Exercise 2.10 replace additive group morphisms by R-morphisms. 2.14

The proof is outlined in the question.

= 0 so Q = 0 so Im f = M.

2.15

Follow the hint; Qo f

2.16

Since f is a morphism that is a bijection we have

so f- 1 (x) 2.17

+ /- 1(y) =

r 1(x + y), so rl

is also a morphism.

In Grp and in RMod it follows by Exercises 2.10, 2.14 and

Exercises 2.13, 2.15 that every bimorphism is an isomorphism, so these categories are balanced. In Sgp this is not so; for example (see Exercise 2.4), ' : IN -+ 71. is a bimorphism that is not an isomorphism. 2.18 It is clear that f is an isotone bijection. But f- 1 is not isotone since (3 -+ xP is a monomorphism different from idF. 3.3 If (A, (qk)k;:>::l) were a coproduct of (7l./pk7l.)~e:::.: 1 then each qk would be monic. Thus IAI would be divisible by pk for every k, which is not possible since A is finite.

Solutions

3.4

125

To show the existence of products, consider the diagram

By the universal property of products there exists a sequence (t.,.),.ez of morphisms, with t.,. : A,. x A~ --+ A,._ 1 x A~_ 1 , that completes the diagram in a commutative manner. Since then 11",._ 2 6.,._ 1 6.,. = 0 and 11"',._ 2 6.,._ 1 6.,. = 0, cancellation en famille gives 6.,._ 1 6.,. = 0 so this middle row is also a complex. To prove that the middle row is a product of the complexes, consider the diagram _.-"

A..-1

~~~-1 A,.

~n-1 X~~

Tn-1

··9,::-Y/~:-?~- 1

A.x•\.'Y~-• ~

n

For each n there is a unique morphism in : T,. 11"nln =a,. and ~in= a~. Then

--+

A,. x

A~

such that

and similarly ~- 1 in-1'~"n = 11"~_ 1 6.,. In· So, again by cancellation en famille, we deduce that in- 1 1",. = 6.,. In whence the entire diagram is commutative. The result now follows from the uniqueness of the sequence (i,.),.~t· A dual proof establishes the existence of coproducts.

Solutiona

126

3.5 Follow the hint. An equaliser of /I, h is given by the canonical inclusion of {bE B; / 1 (b) = h(b)} in B, which is simply the inclusion of Im f into B. Since f is injective the mapping j+ : A -+ 1m f given by x ~--+ f(x) is a. bijection, so is an isomorphism in Set. Since equalisers are unique to within composition by a.n isomorphism it follows that f is an equaliser of / 1 , /2. H now L : 7l -+ Q were an equa.liser (say of J, g : Q -+ A) then since we know (Example 2.2) that L is epic we must then have f = g. The condition for equaliser then reduces, given h : B -+ Q, to showing the existence of a unique 6 : B -+ 7l sucli that L o 6 = h. Choosing an h with Im h \!1: Im L = 7l we obtain a counter-example. 3.6 If f : A -+ B is a section then there exists g : B -+ A such that go f =idA. Consider the morphisms f o g,idB : B-+ B. We have (fog) of= f =idE of. Also, if h: X-+ B is such that (fog) o h = h then taking 6 = g o h we have that f o 6 = h; and 6 is unique since if also f o t = h then, f being monic, we obtain t = g o h. Thus we see that f is an equaliser of the pair f o g,idB.

3.7 (1) => (2) : Consider idA :A-+ A. Since e is an equaliser there exists a unique 6 : A -+ E sucli that eo 6 = idA. Thus e is a. retraction. Since e is monic, it follows by Exercise 2.6 that e is an isomorphism. (2) => (3) is trivial. (3) => (1) : e is right cancellable. 3.8

Note first that

hafawi = haw2!2 = wah2h = wah2g2 = haw2g2

= hagawi

and w1 is epic, so hafa = haga. Suppose now that w: Ba-+ Dis sucli that w/3 = wg3 • Then

ww2h = wfawi = wgawi = ww2g2 and so, since h 2 is a co-equaliser, there exists i : C2 ih2 = ww2. Now

-+

D such that

)uahl = )h2u2 = ww2u2 = ww2v2 = ww2 g2 and so, since h1 is epic, )u3 = )v3 • Since w3 is a co-equaliser there then exists 6: C3 -+ D such that 6w 3 = i· Now 6haw2 = 6wah2 = )h2

= ww2

and so, since w2 is epic, fJh 3 = w. Suppose finally that {}' : C3 also sucli that fJ' ha = w. Then fJ'wah2 = fJ'haw2 = fJhaw2 whence {}'={}since w 3 h 2 is epic.

-+

D is

Solutiom

127

3.9 In the following diagram suppose that 'I : X --. C, are such that '1'1 = 91I'

e.

e: X

--+

D

X~], D----+E----+F !'

g'

e

We have to prove that there exists a unique {} : X --+ A such that et{} = and 91{} = 'I· Now since the right hand square is a pullback there is a unique 1 : X --+ B such that 91 = 'I and /3i = I' and since the left hand square is a pullback there is a unique {} : X -> A such that et{} = and I{}= I· It follows that 91{} = 91 = 'I and so a{} with the required properties exists. Suppose now that {}' : X -+ A is also such that et{}' = and 91{}' = 'I· Then f31,Y = l'et{} = and so, by the uniqueness of 1, we have I{}'= I· It now follows from the uniqueness of {} (with respect to et{} = and f{} = 1) that {}' = {}.

e;

e

e

f'e

e

Let 0t : X-+ A and {3 :X-+ B be such that let = 1/3. Then the square is a pullback if and only if there is a unique {} : X -+ A such that idA{}= Ot and idA{}= {3, i.e. if and only if Ot = {3, i.e. if and only if I is monic. 3.10

3.11

Compare the commutative diagrams

From the first we see that (et: P-+ C,p 1 ,p 2 ) is a product of I: A-> C and 9 : B -+ C in the comma category if and only if there is a unique 1: X-> P such that Pll = {}1 and P2i = {}2· From the second diagram (which is the same as the first) we see that this is precisely the condition that the square be a pullback in C. 3.12 Suppose that {3t 1 = {3t2. Then since the square is commutative we have lett1 = g{3t1 = 9f3t2 = lett2 so that, since I is monic, 0tt1 = ett2. Since the square is a pullback we deduce from {3t 1 = {3t 2 and 0tt1 = ett2 that t 1 = t 2 • Hence {3 is monic.

Solutions

128

3.13

Consider the commutative diagram

Suppose that :r; f. ab- 1 • If there existed y : G --> G such that ay = :r; and by = 1 then we would have the contradiction :r; = ab- 1 • So finite products do not exist. But pullbacks do exist : consider the diagram

G

q

~]. G-----+G a

in which the square is commutative. Suppose that ap = bq and ac = bd. Then pq- 1 = a- 1 b = cd- 1 and so p- 1 c = q- 1 d =a, say, and we have qa

= d,pa =c.

3.14

Consider the diagram

z

~.:x~~:/. xl'

~/. YxZ------+Y py

The rectangle commutes by the definition off x idz. Let u1 : P -+X and u 2 : P --> Y x Z be such that fu 1 = pyua. By the universal property of X x Z there is a unique .U : P -+ X x Z such that 7rx.U = u 1 and 7rzfl = pzu2. Now py(f x idz)fl = J1rxfl = ful = pyu2 and pz(f x idz )fl = 1rzfJ = pzu2 • By cancellation en famille we obtain (! x idz )fl = u2 , so the entire diagram is commutative. If rp : P --> X x Z is also such that 7rx'P = u 1 and (f x idz)'P = u2 then the latter gives pzu2 = pz(f x idz)rp = 7rzrp whence rp = fJ by the uniqueness of fl.

Solution. 3.15

129

Consider the diagram

X Using /; o e; = g; o e; and commutativity we have

XJ; o Xe; =

pf o

(ViE I)

iEI

and so, by cancellation en famille, now that

X /; o h = X g; o h. iEI

Xg; o Xe;

pf o

iEI

iEI

iEI

X/; o Xe; = Xg; o Xe;. iEI

iEI

iEI

Suppose

iEI

Composing on the left with

pf we obtain

iEI

/;opfoh=g;opfoh and so, since e; is an equaliser of /;, g; there exists {}; : X .... E; such that e; o t'J; = pf o h. By the universality of XE; there then exists a unique i: X ....

XE; ~uch that pf o i =

iEI

{};. Finally

iEI

pf o

X e; o i = e; o pf o i = e; o t'J; = pf o h iEI

and cancellation en famille gives

X e; o i

= h. That i is unique with

iEI

respect to this property follows from the fact that if 'I : X such that

Xe; o 'I =

-+

X E;

is

iEI

h then

iEI

e; o pf

o

17

= pf o X e; o 17 = pf o h = e; o pf o i iEI

so, since e; is monic, pf o 17 = pf o i and cancellation en famille gives , = i·

Solutio'flll

4.1 Every singleton set yields a zero object ({x},x). Given f : (A, a) ---+ (B, b) let /-(b) = {x E A ; f(x) consider

=

b} and

We have Im f oi = ( {b}, b) so f oi = 0. H now g: (X, x)---+ (A, a) is such that f o g = 0 then (Vy E X) f[g(y)J = b and so Im g ~ f- (b). There is then a unique h : X ---+ f-(b) such that i o h = g and in particular h(x) =a. Thus i is a kernel of f. To identify cokernels, consider

(A,a) ____!___.(B,b) ___J_.(BfR,[b]) where R is the equivalence relation whose classes are Imf [y] = { {y}

ifyElmf; otherwise.

Since b = f(a) E lmf we have [b] = lmf. Clearly, qof= 0. If now h: (B,b)---+ (X,x) is such that hof=O then lmf ~ h-(x) so we have that

whence there is a unique g : B f R---+ X such that goq = h, and g([b]) = x. Thus q is a cokernel of f. To show that Set. is normal, it suffices to show that if f is monic (=injective) then f is a kernel of any of its cokernels, in particular a kernel of q above. Suppose then that g : (X, x) ---+ (B, b) is such that qog = 0. Then (Vx EX) g(x) = [b] = lmf and so lmg ~ Imf. There is therefore a unique 11 :X---+ A such that /11 =g. Also, f[t9(x)] = g(x) = b = f(a) whence, f being injective, 11(x) =a. Thus f is a kernel of q. To see that Set. is not conormal, consider .

I

({1},1) ___!___._,({1,2,3},1) ----+({2,3},2) where i is the canonical inclusion and /(1) = 2, /(2) = f(3) = 3. It is clear that f is epic (=surjective) and that i is a kernel of f. Hid is the identity map on {1, 2, 3} then we also have id oi = 0. But there is no 11: ({2,3},2)---+ ({1,2,3}, 1) such that 11/ = id since f is not surjective. Hence f is not a co kernel of its kernel i.

Solutions 4.2

131

Refer to the diagram

(1) =? (2) : If I is a kernel of 6h then 6hl = 0. Since g is a kernel of 6 there is a unique 'I : At -+ Bt such that 9'1 = hi. Suppose that p : X -+ A and q : X -+ B are such that hp = gq. Then 6hp = 6gq = Oq = 0 so there is a unique t : X -+ At such that It = p. Now g'f/t = hit = gq so, g being monic, 'It = q. Thus we have a pullback. (2) =?(I) : Suppose now that we have a pullback. Then 6hl = 6g'f/ = O'f/ = 0, and if p : X -+ A is such that 6hp = 0 then, since g is a kernel of 6, there exists q : X-+ Bt such that gq = hp. The pullback morphism t : X -+ At is now the required morphism that ensures that I is a kernel of 6h. For the last part, recall Theorem 3.8. It suffices to observe that if, in the above, g and h are monic and I is a kernel of 6h then 'I is also monic.

4.3

Consider the diagram

in which I= m1e1 and g = mgeg are image o coimage factorisations of I, g. Since e is epic we have m.(ege) is a monic o epic factorisation of ge; and since m is monic (mm1 )e1 is a factorisation of mi. Now it is given that ge = ml and so m.(ege) = (mm,)e,. Since Cis uniquely factorisable there is an isomorphism 1 : F -+ E such that the completed diagram is co=utative. Then k = m11eg is such that ke = I and mk =g. Moreover, k is unique since m is monic (or since e is epic).

Solutions

132

4.4 ofm.

Let a be a. kernel oft. Then we ha.ve to show tha.t !Ja is a. kernel

First we observe tha.t m!Ja = )ta = 10 = 0. If now h : Y -+ q is such tha.t mh = 0 then the pullback produces k : Y -+ P such tha.t !Jk = h a.nd tk = 0. Since a is a. kernel of t there exists j : Y -+ X such tha.t aj = k. Consequently !Jaj = l?k = h a.nd so l?a is a. kernel of m, i.e. 11 o kert - kerm. It follows tha.t if t is monic then 0 - kert whence 0 - ker m a.nd m is monic. 4.5

Refer to the dia.gra.m

in which g = g1{3, f'a = {3/, a.nd 0-+ A'-+ B'-+ C' is exact. Suppose first tha.t we ha.ve a pullback. We ha.ve gf = g'f3f = g' f'a = Oa = 0. If 'IJ : X -+ B is such tha.t gl? = 0 then g' {311 = 0 and so, since f' is a. kernel of g', there exists p.: X-+ A' such tha.t f'p. = {311. There then exists 1 : X -+ A such tha.t fi = 11. Thus f is a. kernel of g a.nd 0 ..... A ..... B ..... C' is exact. Conversely, suppose that 0 ..... A -+ B ..... C' is exact. Since gl? = g1f' p. = Op. = 0 a.nd f is a kernel of g there exists a unique t : X -+ A such tha.t ft = 'IJ. But f'at ={3ft= {3{} = f'p. a.nd !'is monic, so at = p. a.nd we have a pullback. In the 3 x 3 a.rra.y of Theorem 4.12 we ha.ve a 1 - ker faa 2 . In fact faa 2a 1 = faO = 0; a.nd if h is such tha.t faa2h = 0 then, since fa is monic, a2h = 0 whence, at being a. kernel of a2, there exists 11 such tha.t a 1 11 = h. Consequently 0 -+ At -+ A2 -+ Ba is exact. Thus the north-west square is a. pullback. Dually, the south-east square is a. pushout.

Solution.a

133

4.6 (1) Let T- kerg. Then gt = 0 gives 0 = e'gt = het whence, since h is monic, 0 = et. But m - kere so there exists fJ such that mi'J = t. Then m' fi'J = gm!'J = gt = 0 and so, since m' is monic, fi'J = 0. But f is also monic, so t = 0 and hence g is monic. The proof of (2) is dual. 5.1

[h, k] o {!,g) is represented by [ h

5.2 (1) biproduct. (3)

=?

=? ( 4)

(2) and {1)

: (

f: ~;1r;)

=?

J1.t =

s=l

k

J[ ~] = [ hf +kg].

{3) are i=ediate from the definition of

t ~;7r;JI.t t ~;5;t =

s=l

= Jl.t. It follows

s=l

from the universality of the coproduct (11 1 , ••• , ~n, B) that

n

L: J1;7r;

=

i=l

ids. (4)

=?

n

L: J~;/;.

(2) : Let /;: C--+ A; and define f =

Then

i=l

1rcf

= 1rt

(.E ~d•) = t, 1rt~d• = f: 5td; = It. s=l

s=l

s=l

Moreover, if g is such that 7rtg = ft for every t then g

=ids og = (

E

J1;7r;)g

=

i=l

E~;1r;g = E!ldi

i=l

=f.

i=l

Thus (B, 1r11 ••• , 7rn) is a product. Dually, we have ( 4) =? (3). Finally, since (2) and (3) together imply ( 1), we see that all four are equivalent. 5.3 Let JIA : A --+ C be a kernel of 1rs. Then 7rsJIA = 0 and, by Exercise 5.2, it suffices to show the existence of 7rA : C --+ A such that 1rAJIA = idA,7rA~s = 0 and JLA7rA +J~s7rs =ide. Consider the diagram A~C~B

~ida

-P.B1rB

c Since 7rs(idc -~s7rs) = 1rs- 7rsJ~s7rs = 1rs - 1rs = 0 and since JIA is a kernel of 1rs, there exists 'IrA: C--+ A such that JIA7rA =ide -J~s'lrs. Furthermore, JIA 1rAJIS =(ide -J~s7rs )~s = JIA -J~s1rsJIA = JLA -0 =~A and so, since JIA is monic, 7rA~S = 0; and

~A 1rAJLA =(ide -J~s'lrs )JIA = JIA -J~A 1rSJIA = JIA- 0 =~A so, since JIA is monic, 7rA~A =idA.

Solutio'M 6.4 Given f,g: A--+ B let k: K-+ A be a kernel off- g. Then (! - g)k = 0 gives fk = gk. Suppose that h : X -+ A is such that fh = gh. Then (!- g)h = 0 and so, since k is a. kernel off- g, there is a unique {} : X -+ K such that kt'J = h. Thus k is an equaliser of f, g. 6.5 Let i : K -+ B be a kernel of a. Then gai = gO = 0 and the co=utativity of the right hand square gives gi = 0. Now f is a kernel of g, so there exists t : K --+ A such that ft = i. The co=utativity of the left hand square now gives i = ft = aft = ai = 0. Thus all kernels of a are zero, so a is monic. A dua.l proof using the exactness of the bottom row shows that a is a.lso epic. Since E is balanced, a is then an isomorphism. If E is additive then the left hand square gives (a - idB )I = 0. But g is a. cokernel of/, so there is a. unique p.: C-+ B such tha.t a-idB = p.g. From the right ha.nd square we have ga = g, so g( a- idB) = 0 and hence gp.g = 0. Since g is epic, we deduce that gp. = 0. Since f is a kernel of g we now infer the existence of {} : C -+ A such that ft'J = p.. Consequently a- idB = ft'Jg; a.nd such a{} is unique since f is monic and g is epic. 5.6 (1) =? (2) : If a'/3' = 0 let f = idA -aa'- {3'{3. Then, since a is a. kernel of {3, we have /3/ = {3- {3aa1 - {3{3'{3 = {3- 0- {3 = 0 and so there exists a. unique {} : A -+ A' such that at'J = f. But a' f = a' - a' a a' - a'{3' {3 = a' - a' - 0 = 0, so 0 = a' f = a' at'J = {} a.nd consequently f = 0. (2) =? (3) is immediate from Exercise 5.2, a.nd (3) =? (1) follows from Theorem 5. 7. 5. 7 Given /, g : L --+ M in Lat(j1 define f V g : L -+ M by setting (! V g)(x) = f(x) V g(x) for all x E L. Then f V g is residuated with (! V g)+(y) = j+(y) 1\ g+(y); in fa.ct, we ha.ve

(! V g)(x) S y f(x), g(x) S y x S J+(y), g+(y). Tha.t (! V g)h = fh v gh is clear, and h(f V g) = hf V hg follows from the fa.ct tha.t residua.ted ma.ps preserve suprema.. [In fa.ct, it is clear tha.t /( V x;) ;::: f(x;) for every i E I. If now y ;::: f(x;) for iE!

every i E I then j+(y) :::: y ;::: /[J+(y)) ;::: /(

V x;). ~I

r [f(x;)) :::: x; so J+(y) :::: v Thus we have /( V ~I

X;

and hence

iEI

x;) = V /(x;).]

This

~I

then defines a semi-additive structure on Lat(j1 . Moreover, Lat(j1 is balanced, by Exercises 2.1B and 1.9. Now (PA,PB,A x B) is a. product of A,B in Lat(j1 • For PA,PB are residuated (p~, for example, is given by p~(a, b) = (a, 1)), and if f : L -+ A, g : L -+ B are residuated

Solutions

135

maps then there is a unique residuated map 11 such that PA 11 =

I and

Psl1 = g. In fact, 11 is given by !J(x) = (f(x), g(x)), with !J+ given by u+(a,b) = f+(a) 1\ g+(b). Thus Lat;j1 has finite products. The final statement follows from Theorem 5.2 and the fact that V is not a group operation. 6.8 (ali"A -f31rs)J.LA = ali"AJ.LA -f311"BJ.LA =a and (all" A -{31rs)J.LB = a'lrAJ.LB -/31rsJ.LB = -{3 so, by the uniqueness property, [a, -/31 = a1rA{31rs. It follows that

[a, -/3] o (!,g) =all" A(!, g) - /31rs (!,g) = al- {3g = 0. Suppose now that the outer boundary is a pullback. It suffices to show that (!,g) is a kernel of [a, -/31· Let h: H--+ A • B be such that [a, -/31 o h = 0. Then ali"Ah = /31rsh and so we have li"Ah : D --+ A and ll"sh : D --+ B with a o li"Ah = {3 o ll"sh. The pullback property produces a unique e : D --+ P such that le = li"Ah and ge = 1rsh. Thus li"A(f,g)e = 1rAh and ll"s(/,g}e = 1rsh whence, by cancellation en famille, (!, g)e = h. Conversely, if the sequence is exact then (!,g) is a kernel of [a, -/31· To show that the outer boundary is a pullback, let p : D --+ A and q : D --+ B be such that ap = {3q. By the universal property of A • B there exists h : D --+ A • B such that 1rAh = p and 1rsh = q. Then [a, -/31 o h = a1rAh- {31rsh = ap- {3q = 0. Since (!,g) is a kernel of [a, -/31 there is then a unique e : D --+ P such that (!, g)e = h. Then le = li"A (!, g)e = 1rAh = p and likewise ge = q. Finally, e is unique with respect to these properties since if also I e' = p, ge' = q then 1rA{!,g)e' = 1rAh and 1rs{!,g)e' = 1rsh whence (!,g)e' =hand e' =e. 6.9

If h : B

--+

X is such that hg = 0 then

[O,h.lo(f,g)~[o h.J[;]=[O+hg]~o. But, by Exercise 5.8, [a, -/31 is a cokernel of {!,g), so there exists a unique k : C --+ X such that k o [a, -/31 = [0, hi. Consequently ka = 0 and -k/3 = h. Since a is epic it follows that k = 0, whence h = 0. Thus every cokernel of g is zero, so g is epic. 6.10

In the diagrant

Solutions

let the square be a pullback of {J, g (see Theorem 3. 7; e is an equaliser of g1rs and {3-trq). Since g is epic, so is 1rqe by Exercise 5.9. Lett: A-+ P be the morphism induced by the morphisms f : A -+ B and 0 : A -+ Q (so that 11'qe o t = 0 and 1rse o t = !). It suffices to prove that t is a kernel of 1rqe. Let h : X-+ P be such that 1rqeh = 0. Then g1r8 eh = {31rqeh = 0 and so, since f is a kernel of g, there exists k : X-+ A such that fk = 1rseh, i.e. 1rsetk = 1rseh. Since also 1rqetk = 0 = 1rqeh it follows by cancellation en famille that tk = h. 6.11 (I)=? {2) : f = af3 and {3a =ids so f of= afJafJ = afJ =f. Also, a is an equaliser of /,idA since on the one hand fa = af3a = a = idA a and on the other, if fh = idA h = h then afJh = h, and if tJ : X -+ B is such that atJ = h then tJ = {3atJ = fJh. {2) =? {1): Suppose that P = f and that e is an equaliser of /,idA. Then in the diagram

there is a unique tJ : A -+ E such that etJ = f. But then etJe = fe = idA e = e and so, since e is monic, tJe =idE. Thus f splits. For the last part observe that equalisers exist in an abelian category so f splits if and only if P = f. Now in this case {idA-!){idA -!) = idA -/ - f + P = idA -/, so idA - f also splits.

n- =

8.1 1:4 is the identity map on 2A so F1A = 1FA. Since (g 0 g-o 1- it follows that F is a functor. The second part is similar.

8.2 1~ is the identity map on A 2 soFIA = 1FA· Since (go /) 2 = g2 o p it follows that F is a functor. 8.3 We have +(g, h)= g+ h so +(1, 1) = 1. It now follows from the equality (g + g') + (h + h') = (g +h)+ (g' + h') that +is a bifunctor. For every morphism '1 in BA, i.e. every natural transformation G, we have associated with every morphism f : A -+ A' the commutative diagram 8.4

'7 : F

-+

FA~GA

F/1

1

G/

FA1 ---+GA' 'lA'

Solution'

137

We require H('l,/) to be a morphism from FA toGA', so let H('l,/) be the diagonal of the above commutative diagram : H('I,Jl = GF o 'lA = 'lA• oFf. We have H(lF, lA) = FIA o IFA = IFA o IFA = IFA by considering the commutative diagram A

/1

= IH(F,A)

and,

FA~GA~KA

lF/

lG/

lK/

·1 r·~T~~T~· A"

F A"---+GA"----.KA" "A" eA" we see that H((€, g) o ('I, Jl) = H(€'1, gf) = the diagonal of the outer rectangle= H(€,g) o H('J,/). Thus H(_,_) : BA x A -+ A is a bifunctor. For every functor F : A -+ B we have H(F, _)(A) = H(F, A) = FA and H(F,_)(f) = H(IF,Jl = Fj o IFA = Ff. Thus H(F,_) =F. For every object A of A we have (H(_, A)(F) = H(A, F) = FA and H(_,A)('I) = H('l, lA) = IGA 0 'lA = 'lA·

6.5

Consider the diagram FA...,_!..!...._FA'

al B---+ B' g

Mor(FJ, g) has to be an arrow from Mor(FA, B) to Mor(FA', B'), so define it to be g o a o F f. Then it is readily seen that Mor( F _, _) is a bifunctor. Consider now the diagram A..,___!__ A'

pl GB---+GB' Gg

Mor(f,Gg) has to be an arrow from Mor(A,GB) to Mor(A',GB'), so define it to be Gg o {3 o f. Then it is readily seen that Mor( _, G _) is a bifunctor.

Solutions 6.6

By Theorem 4.ll, the following statements are equivalent :

(a) 0---+ A____!__. B ____!___, C---+ 0 is exact; (b) I is monic and g is a cokernel of I; (c) g is epic and I is a kernel of g. (1) => (2) is clear; (2) ¢} (3) is immediate from (a)¢} (c); (2) ¢} (4) is immediate from (a) ¢}(b). Since im- kercoker we then have (2) => (5). That (5) => (1) follows from the definition of exactness. Dually, we have (2) => (6) and (6) => (1). 6.7 If (B, 11"1> ••• , 1r,.) is a product of A1 , ••• , A,. in A then for every j there is (by the universal property of products) a unique morphism

Pi : A;

-+

B such that

commutes. In other words, every product in a semi-additive category can be completed to a biproduct. A similar statement holds for coproducts. The equivalence of (2),(3),(4) now follows by Exercise 5.2. That (1) => (4) follows from the equivalence of (1},(4) in Exercise 5.2 and the fact that an aditive functor preserves zero morphisms. Finally (4) => (1) follows from the fact that the existence of biproducts implies that of a unique semi-additive structure. Recall that I + g is given by [1, g] o (1, 1). Applying F to the diagram

t/Al(~l

~

!,1)~

A====:; A • A~ A

~lr.a!/ I~ [l/11 B we deduce from (4) that F I+ Fg is given by

F[l,g] oF{1,1) = F([l,g] o (1,1)) = F(f +g). 6.8

A left exact functor preserves exact sequences of the form 0---+ X'---+ X---+ X"

and, by Exercise 6.7, an additive functor preserves biproducts. The result follows by applying these observations in the light of Exercise 5.8.

Solutions

139

6.9 Kx is given by Kx A = X for every A in I. For every A in I define 'lA to be the unique element of Mor(X, FA). Then for every morphism f : A -+ B we have the diagram KxA=X~FA

A

/1

Kx/=lxl

B

lF/

KxB = X-------+FB liB

which is co=utative since X is initial. This defines a natural transfer· mation 'I : Kx -+ F. The uniqueness of 'I is clear, so Kx is initial in A I. Thus if A has an initial object then so does A I. Argue dually for terminal objects; and combine the results for zero objects. 6.10 U 0-------+ hx A~ hx B ~ hxC is exact for every object X then in particular it is exact for X = A, in which case we have 0 = hA/3 o hAa = hA(/3 o a) and so, for any t : A--. A, {3at = 0. Taking t = lA we obtain {3a = 0. Also, if 0 = hxf3(k) = {3 o k then k = hxa(e) =a o e for some e (which is unique since hxa is monic). Thus we see that a is a kernel of {3 and so 0 ---. A ~ B __!___. C is exact.

6.11 In view of Exercise 6.7 it suffices to prove that F preserves biproducts. Now if (~t ,~2, At • A 2 , 1r 1 , 1r2 ) is a biproduct then by Theorem 5.7 we have the split exact sequences

...

.

,..

Now F7rt oF~t = F(7rt o~t) = FlA, = lFA, and similarly F1r2oFp.2 lFA,· Since F is semi-exact, the sequences

=

are then split exact. Since F1r1 o Fp. 2 = F(1r 1 o p.2 ) = FO = 0 it follows by Exercise 5.6 that (Fp. 1 , F~2 ,F(A 1 • A2 ), F1r1 , F1r2 ) is a biproduct of FA1,FA2.

Solution~~

140

6.12

Consider the diagram

in which g is epic. Since P; is projective there exists h; : P; --+ A with gh; = fp.;. By the universal property of coproducts there is a unique fJ : fl P; --+ A such that fJp.; = h; for every i E /. Then iEI

gfJp.;

= gh; =

fp.; and so, by cancellation en famille, gfJ

TI P; is projective.

=

f. Thus

iEI

6.13 Given f,g E Mor~(A,B) we have fact, consider the diagram

f-

g E Mor~{A,B). In

in which f factors through the projective P 1 and g factors through the projective P2 • By Exercise 6.12, P 1 • P2 is also projective so there exists {)A: A-+ P1 • P2 with 1r1fJA = k1 and 1r2fJA = k2. Then h11r1fJA = h1k1 =I and h21r2fJA = h2k2 = g whence I- g = {h11r1- h21r2)fJA. Thus f- g factors through the projective P1 • P2. Given the exact sequence 0-------> B' ~ B __l___. B"----+ 0 we know, since hA is left exact, that hAB' ~ hAB ~ hAB" is exact. Now if a E Mor~(A, B') then hAf(a) = fa E Mor~{A, B). Thus we see that hAl maps Mor~{A, B') into Mor~{A, B) and hence we have the induced sequence of quotient groups Q(A, B') ___!:___, Q(A, B)~ Q(A, B").

Solutiona

141

Now the diagram

is co=utative and hence g./.~' = ~" o hAg o hAl= ~ 11 o 0 = 0. Since~· is epic we then have g./.= 0 so Jmf. ~ Kerg•. To obtain the reverse inclusion, note that Ker g.= {,8 + Mor(;(A, B) ; g,B E Mor(;(A, B")}. But if g,B E Mor(;(A, B") then there is a co=utative diagram

in which P is projective. Since g is epic there exists {} : P -+ B with g!'J = k. Then g(,B- !'Jh) = 0 and so, since f is a kernel of g, there exists 1: A-+ B' such that /i = ,8- !'Jh. Clearly, !'Jh E Mor(;(A, B) so

,8 + Mor(;(A, B)= /i + Mor(;(A, B) Elm/•. Thus Ker g. semi-exact. 6.14

~

Ker /. and the resulting equality shows that Q(A, _) is

(1) Consider the diagram

A-------+B g

in which F is free on S and g is epic. Since, by hypothesis, g is surjective it is a retraction so there exists g' : B -+ A with ggl = lB. Since F is free on S there exists '1 : F-+ A with l]i = g'hi whence gl]i = gg'hi =hi,

Solution& and then gf/ = h by the uniqueness in the definition of a free ~bject. Thus F is projective. (2) Suppose that P is projective and let (Fp, fp) be a free C-object on P. Then there exists 6: Fp--+ P such that 6fp = lp :

Since 6 is epic and Pis projective there exists a morphism 1: P--+ Fp such that t11 = lp. Thus 6 is a retraction. Conversely, let Fp be free on P and let tJ : Fp --+ P be a retraction. Then there exists a morphism 1 with t11 = lp. Consider the diagram

in which g is epic. Since Fp is projective there exists a : Fp --+ A such that ga = ftJ. Then a1: P--+ A is such that ga) = ftJ) = f and hence P is projective. Consider the biproduct (p.', p.", P' • P", 1r1 , 7r11 ). By Exercise P' • P" is projective. Let i = p.' and 7f = 1r11 • If 7J is a projective lifting of {3 (in the sense that gfj = {3) let 1 = f a1f1 + /J7f. We have "ti = ja1r1 p.' + /J1f 11 p.' = falp' + fjo =fa;

6.16 6.12, P

=

g"t

= gfa7r 1 + gfj1r" = 0a1f 1 + {31f11 = {31f

so the completed diagram is commutative. Clearly, the top row is exact. 6.16

The diagram

AxC-->A is commutative. So mation.

1r -

( 1fB)

:Ax _

--+

K defines a natural transfor-

Solutions

143

To see that there is no natural transformation IJ : KA -+ A x _, suppose that for every f : B -+ C we had a commutative diagram

A~AxB

1A1

11AX/

A---->A

X

tJo

c

Then for every a E A we have IJ 8 (a) =(a', b) and !Jc(a) = (a",c}, and the commutativity requires f(b) = c for every /, which is impossible when ICI > 2. 6.17 The commutativity of the initial diagram is immediate from the fact that 'I and 5 are natural transformations. For every morphism f : A -+ A' we have, since 'I : F -+ G is a natural transformation, the commutative diagran1 FA~GA

1

F/1

G/

F A'---->GA' 'lA'

Appling H to this and using the fact that 5 : H -+ K is a natural transformation, we obtain the commutative diagram

HFA~HGA~KGA

1

HF/1

1

HG/

KG/

HFA ---->HGA ---->KGA' 1

1

HrrA.'

6aA'

in which the composite morphism in the top row is the diagonal ~A of the initial diagram. Thus ~ is a natural transformation from H F to KG. Taking G = F and 'I= IF we obtain 5FA = (5 * lF)Ai and taking K =Hand 5 = ls we obtain H'IA = (le * 'I)A· It now follows from the initial diagran1 that

HF~KF

lH*'I 1

11K*'I

HG---->KG 6*1G

commutes.

Solutioflll

7.1 The objects of A 1 are functors F: 2 ~A, which can be described by specifying a morphism • ~ • in A, i.e. an object in the arrow category. The morphisms of A 1 are the natural transformations 'I between functors, which can be described by specifying a commutative diagram

·-· "'

i.e. a morphism in the arrow category. 7.2

Proceed as in Exercise 7.1.

7.3 With each object X of C associate the object A____!__. X of the arrow category (! being unique since A is initial), and with each morphism h : X ~ Y of C associate the obvious morphism in the arrow category:

7.4

Define 'lA by 'IA(t1)

= t1(1) for every t1: R ~A.

Then

MorR(R,A)~A=UA

Mor,.(R,IJ

1

1/=U/

MorR(R, A1 )---+A1 = U A' fl.--+ g(l) >--+ l[g(l)J a.nd the southwest route takes g >--+I o g >--+ (! o g)(l). Now define p. - (p.a) : U --+ h.z as follows : for every x E UG let p.a(x) = g where g: Z-+ G is the group morphism such that g(l) = x. Then p. is a natural transformation since UG=G~h.zG

lhzl

11

UH=H--+hzH I'H

commutes; for the north-east route takes x >-+ g >--+I o g where l[g(l)] = x, and the south-west route leads to the same destination. Since clearly f/GP.G = 1 and P.G'IG = 1 we see that hz and U are equivalent. '1.6 Let f/E : Mor(2,E)-+ W be given by 'IE(a) = (a(O),a(l)). Then the diagram Mor(2,E)~W Mor(2,!)

1

11• ,,

Mor(2,F)--+F2 commutes and we have a natural transformation '1 ~ ('IE)· Now define P.E : W-+ Mor(2,E) by P.E(x,y) = 6:r;, 11 where 6:r;, 11 (0) = x and 6z,11 (1) = y. Then the diagram W~Mor(2,E) j 2

1

1

Mor(2,!)

,.,

F 2 ---+Mor(2, F) commutes and we have a natural transformation p. ~ (P.E)· A be a left adjoint of G : A -> B and let B be a left adjoint of F : B -> C. It suffices to observe that the composite of the isomorphisms F* : C

->

(C,FGA) ----+(F*C,GA) ----+(G*F*C,A) is an isomorphism. Thus G* F* is a left adjoint of FG. 8.8

Suppose that 0 - - - + B' __!___. B __!!_____. B"

----+ 0

is exact. Since hFA = Mors(FA,_) is left exact we have the co=utative diagram (where, for considerations of space, we have written the morphism sets as [ , ]) [A,GB'J~[A,GBJ

tA,B'

1

hAGg

11A,B

[A,GB"j

1tA,B"

0----+[FA,B']----+[FA,B]----+[FA,B"] hpAf

hpAfl

Solutions

149

in which the bottom row is exa.ct and the vertical maps are isomorphisms. It follows that the top row is also exact with hAGf monic. Thus 0 - - + MorA(A, GB') ~ Mor A(A, GB) ~ MorA(A, GB") is exa.ct. Using Exercise 6.10 we deduce that 0 - - + G B' ....!!..L..... G B ___!!!__. GB" is also exa.ct, whence G is left exact. A similar proof shows that F is right exa.ct. 8.9

A mapping f : A

-+

B x B can be described by f(a) =

(!1(a),/2 (a)) and so determines mappings /J, 12 : A -+ B. The universal property of the coproduct (JL 1 , Jl2 , A II A) then produces a unique fJ1 :A II A-+ B such that A __!_!___. B

l'•

~·1

AIIA+---A ~·

is commutative. The mapping f ...... fJ1 is injective since ig fJg = fJ1 then 91 = fJgJLl = fJJJLt = /J and similarly 92 = h and so 9 = f. It is also surjective since if h : A II A -+ B then h induces mappings ht, h2 :A-+ Band if h* :A-+ B x B is given by h*(a) = (ht(a),h2(a)) then {)h. = h. The result now follows on noting that the diagram

Mor( A, B x B) - - - '1-.......:_".:.'- - - > Mor( A II A, B) Mor(l,g 2 )

1

1

Mor(!Il/,g)

Mor(A',B' x B')

Mor(A' II A', B') t't-+fJt'

is commutative. In fa.ct, by the north-east route,

t ...._, fJ 1 ...._, 9 o fJ 1 o (! II/) and, by the south-west route,

Solutions and these destinations are the same by virtue of the commutativity of the diagram

8.10

Regarding an ordered set X as a small category X, we have Morx(a,b) = { {(a0b)}

if a~ b; otherwise.

Given an isotone map f: A-+ B, form a functor f: A-+ B as follows: associate with each object a of A the object fa = f(a) of B, and with each morphism (a, b) of A the morphism f(a, b) = (!(a), f(b)) of B (this is possible since f is isotone). Then the properties of a functor follow easily. To say that f: A-+ B is residuated is equivalent to saying that there is an isotone map ·J+ : B -+ A such that

f(:c) ~B Y

X

~A J+(y).

In terms of functors, this says that

IMors(f:c,y)l

= 1

IMorA(:c,f+y)l

=1

i.e. that (f, r+) forms an adjoint pair. 8.11

This is a special case of Exercise 8.10 since the induced map =

!- : P(A) -+ P(B) is isotone. In fact it is residuated with(!-)+

r-.

Index

abelian category, 67 abelianisation functor, 75 additive category, 66

additive functor, 84 adjoint functor theorem, 112 adjoint pair, 110 balanced category, 15 bidual functor, 78 bifunctor, 75 bimorphism, 15 binormal category, 50 biproduct, 59

concrete category, 10 connected category, 19 conormal category. 50 constant functor, 79 contravariant functor, 73 coproduct, 28 covariant functor, 73

derived group functor, 74 discrete category, 80 downward directed set, 83

dual category, 11 epic, 12

cancellation en famille, 29 cartesian product bifunctor, 75 cartesian square, 33 category, 2 coexact sequence, 52 cofree object, 112

equaliser, 30 equivalence, 101 equivalent categories, 77 exact functor, 85 exact sequence, 52

cokernel, 44 colimit, 79 commutative diagram, 4 complete category, 82

factorisable, 43 faithful functor, 97 fibred product, 33 finitely complete, 81

Index

forgetful functor, 14 free group, 21 free R-module, 21 free R-module functor, 14 free object, 112 free semigroup, 18 full functor, 97 full subcategory, 1 functor, 73 identity functor, 73 identity morphism, 2 inclusion functor, 74 initial object, 16 injective, 87 intersection, 35 inverse limit, 83 iSomorphic categories, 98 isomorphism, 15

kernel, 44 left &