Solutions to "Basic Category Theory" by Tom Leinster [1 ed.]

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Solutions to Basic Category Theory Tom Leinster Solutions by positrón0802 https://positron0802.wordpress.com

https://positron0802.wordpress.com/basic-category-theory-leinster/ 1 January 2021

Contents

Contents 0 Introduction

2

1 Categories, functors and natural transformations 1.1 Categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Natural Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4 4 4 10

2 Adjoints 2.1 Definition and examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Adjunctions via units and counits . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Adjunctions via initial objects . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

16 16 22 27

3 Interlude on sets 3.1 Constructions with sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Small and large categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Historical remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

31 31 33 36

4 Representables 4.1 Definitions and examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 The Yoneda lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Consequences of the Yoneda lemma . . . . . . . . . . . . . . . . . . . . . . . . . .

36 36 39 40

5 Limits 5.1 Limits: definition and examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Colimits: definition and example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Interactions between functors and limits . . . . . . . . . . . . . . . . . . . . . . .

42 42 46 50

6 Adjoints, representables and limits 6.1 Limits in terms of representables and adjoints . . . . . . . . . . . . . . . . . . . . 6.2 Limits and colimits of presheaves . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Interactions between adjoint functors and limits . . . . . . . . . . . . . . . . . . .

53 53 53 60

Appendix: Proof of the general adjoint functor theorem

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1

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0. Introduction

0

Introduction

Exercise 0.10. Let 𝑖 : 𝑆 → 𝐼 (𝑆) to the identity function of sets. Then the topological space 𝐼 (𝑆) together with the function 𝑖 satisfy the universal property illustrated in the following diagram: 𝑖

𝐼 (𝑆)

𝑆

∃! continuous 𝑓

∀ functions 𝑓

∀𝑋 . Explicitly, given any topological space 𝑋 and any function 𝑓 : 𝑋 → 𝑆, there exists a unique (continuous) map 𝑓 : 𝑋 → 𝐼 (𝑆) such that 𝑓 = 𝑖 ◦ 𝑓 . Indeed, there is only one way to define 𝑓 , namely to be given by 𝑓 (𝑥) = 𝑓 (𝑥) for all 𝑥 ∈ 𝑋, and this function is continuous since 𝐼 (𝑆) has the indiscrete topology. Briefly, this property says that any function into an indiscrete space is continuous. Exercise 0.11. Let 𝜀 : 𝐺 → 𝐻 denote the trivial homomorphism. The pair (ker 𝜃, 𝜄) satisfies the following universal property: 𝜀◦𝑖

ker 𝜃 ∃!𝑗 0

𝜃

𝑖

𝐺

𝐻

∀𝑗

.

𝜀◦𝑗

∀𝐾

That is, we have 𝜃 ◦𝑖 = 𝜀 ◦𝑖, and if (𝐾, 𝑗) is any other pair of a group 𝐾 and a morphism 𝑗 : 𝐾 → 𝐺 such that 𝜃 ◦ 𝑗 = 𝜀 ◦ 𝑗, then there exists a unique homomorphism 𝑗 0 : 𝐾 → ker 𝜃 such that 𝑖 ◦ 𝑗 0 = 𝑗 . Indeed, such 𝑗 0 must be defined by 𝑗 0 (𝑘) = 𝑗 (𝑘) for all 𝑘 ∈ 𝐾, and this is well-defined since 𝑗 (𝑘) ∈ ker 𝜃 for all 𝑘 ∈ 𝐾 . Exercise 0.12. We are given the following diagram: 𝑖

𝑈 ∩𝑉

𝑈 𝑗0

𝑗

∀𝑓

𝑖0

𝑉

𝑋 ∃!ℎ

∀𝑌 .

∀𝑔

Given a space 𝑌 and maps 𝑓 : 𝑈 → 𝑌 , 𝑔 : 𝑉 → 𝑌 such that 𝑔 ◦ 𝑗 = 𝑓 ◦ 𝑖, there exists a unique map ℎ : 𝑋 → 𝑌 such that the diagram commutes. The condition 𝑔◦ 𝑗 = 𝑓 ◦𝑖 is saying that 𝑔|𝑈 ∩𝑉 = 𝑓𝑈 ∩𝑉 , so there is a unique well-defined function ℎ : 𝑋 → 𝑌 making the diagram commute, namely the one given by ℎ(𝑥) = 𝑔(𝑥) if 𝑥 ∈ 𝑉 and ℎ(𝑥) = 𝑓 (𝑥) if 𝑥 ∈ 𝑈 , and this function ℎ is continuous by the gluing lemma as 𝑈 and 𝑉 are open in 𝑋 . 2

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0. Introduction Exercise 0.13. (a) Let 𝑅 be a ring and 𝑟 ∈ 𝑅. Any ring homomorphism 𝜙 : Z[𝑥] → 𝑅 such that 𝜙 (𝑥) = 𝑟 must be given by ! 𝑛 𝑛 𝑛 Õ Õ Õ 𝜙 𝑎𝑖 𝑥 𝑖 = 𝜙 (𝑎𝑖 )𝜙 (𝑥)𝑖 = 𝑎𝑖 𝑟 𝑖 , 𝑎𝑖 ∈ Z, 𝑖=0

𝑖=0

𝑖=0

and it is clear that this is indeed a homomorphism. (b) Since 𝐴 is a ring and 𝑎 ∈ 𝐴, it follows from part (a) that there is a unique ring homomorphism 𝜄 : Z[𝑥] → 𝐴 such that 𝜄 (𝑥) = 𝑎. By assumption, there is also a unique ring homomorphism 𝜙 : 𝐴 → Z[𝑥] such that 𝜙 (𝑎) = 𝑥 . Then 𝜙 ◦ 𝜄 is a ring homomorphism Z[𝑥] → Z[𝑥] such that 𝑥 ↦→ 𝑥, so by uniqueness we have 𝜙 ◦ 𝜄 = 1Z[𝑥 ] . Similarly 𝜄 ◦ 𝜙 = 1𝐴, so 𝜄 is an isomorphism. Exercise 0.14. (a) We seek for (𝑃, 𝑝 1, 𝑝 2 ) with the universal property illustrated below: ∀𝑓1

∀𝑉 ∃!𝑓

𝑝1

𝑃 ∀𝑓2

𝑋

𝑝2

.

𝑌

Consider 𝑃 = 𝑋 ⊕ 𝑌 with 𝑝 1 : 𝑃 → 𝑋 and 𝑝 2 : 𝑃 → 𝑌 the canonical projections. Then, given a cone (𝑉 , 𝑓1, 𝑓2 ), there is only one way to define 𝑓 : 𝑉 → 𝑃 so that 𝑝 1 ◦ 𝑓 = 𝑓1 and 𝑝 2 ◦ 𝑓 = 𝑓2, namely to be given by 𝑓 (𝑣) = (𝑓1 (𝑣), 𝑓2 (𝑣)), and this is linear since both 𝑓1 and 𝑓2 are linear. (b) Let (𝑃, 𝑝 1, 𝑝 2 ) and (𝑃 0, 𝑝 10 , 𝑝 20 ) be cones having the property stated in (a). Then, since 0 (𝑃 , 𝑝 10 , 𝑝 20 ) has this property, there is a unique linear map 𝑖 : 𝑃 → 𝑃 0 such that 𝑝 10 ◦ 𝑖 = 𝑝 1 and 𝑝 20 ◦ 𝑖 = 𝑝 2 . Similarly there exists a unique linear map 𝑗 : 𝑃 0 → 𝑃 satisfying 𝑝 1 ◦ 𝑗 = 𝑝 10 and 𝑝 2 ◦ 𝑗 = 𝑝 20 , and the equations 𝑖 ◦ 𝑗 = 1𝑃 0 and 𝑗 ◦ 𝑖 = 1𝑃 hold by uniqueness. (c) We seek for (𝑄, 𝑞 1, 𝑞 2 ) with the universal property illustrated below: ∀𝑓1

∀𝑉 ∃!𝑓

𝑄 ∀𝑓2

𝑞1

𝑋

𝑞2

𝑌

.

Consider 𝑄 = 𝑋 ⊕ 𝑌 with 𝑞 1 : 𝑋 → 𝑄 and 𝑞 2 : 𝑌 → 𝑄 the canonical inclusions, 𝑞 1 (𝑥) = (𝑥, 0) for all 𝑥 and 𝑞 2 (𝑦) = (0, 𝑦) for all 𝑦. Then, given a cocone (𝑉 , 𝑓1, 𝑓2 ) there is only one way to define a map 𝑓 : 𝑄 → 𝑉 fitting in the diagram above, namely to be given by 𝑓 (𝑥, 𝑦) = 𝑓1 (𝑥) + 𝑓2 (𝑦), and this is linear since both 𝑓1 and 𝑓2 are linear. 3

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1. Categories, functors and natural transformations (d) Given another cocone (𝑄 0, 𝑞 10 , 𝑞 20 ) satisfying the property stated in (c), there is a unique linear map 𝜙 : 𝑄 → 𝑄 0 such that 𝜙 ◦ 𝑞 1 = 𝑞 10 and 𝜙 ◦ 𝑞 2 = 𝑞 20 , and similarly a unique linear map 𝜓 : 𝑄 0 → 𝑄 such that 𝜓 ◦ 𝑞 10 = 𝑞 1 and 𝜓 ◦ 𝑞 20 = 𝑞 2, which is the inverse of 𝜙 by uniqueness.

1 1.1

Categories, functors and natural transformations Categories

Exercise 1.1.12. The category of modules over a (fixed) ring 𝑅; the category of based topological spaces and based map; the category of smooth manifolds and smooth maps; the category of ringed spaces and morphisms of ringed spaces. Exercise 1.1.13. If 𝑔, 𝑔 0 : 𝐵 → 𝐴 are both inverses of 𝑓 : 𝐴 → 𝐵, then 𝑔 = 𝑔1𝐵 = 𝑔𝑓 𝑔 0 = 1𝐴𝑔 0 = 𝑔 0 . Exercise 1.1.14. In the product category 𝒜×ℬ, identities are given by 1 (𝐴,𝐵) = (1𝐴, 1𝐵 ) : (𝐴, 𝐵) → (𝐴, 𝐵), and composition of morphisms by (𝑓 , 𝑔) ◦ (𝑓 0, 𝑔 0) = (𝑓 𝑓 0, 𝑔𝑔 0). Exercise 1.1.15. To prove that Toph is a category we need the following facts from topology: • Composition is well-defined: if 𝑓 , 𝑓 0 : 𝑋 → 𝑌 are homotopic and 𝑔, 𝑔 0 : 𝑌 → 𝑍 are homotopic then 𝑔𝑓 , 𝑔 0 𝑓 0 : 𝑋 → 𝑍 are homotopic. • Existence of identities: given a topological space 𝑋 there exists a map 𝑐 : 𝑋 → 𝑋 such that for all maps 𝑓 : 𝑋 → 𝑌 , 𝑓 𝑐 is homotopic to 𝑓 , and for all maps 𝑔 : 𝑍 → 𝑋, 𝑐𝑔 is homotopic to 𝑔. We can take 𝑐 = 1𝑋 . • Composition is associative: given maps 𝑓 : 𝑋 → 𝑌 , 𝑔 : 𝑌 → 𝑍 and ℎ : 𝑍 → 𝑊 , then ℎ◦(𝑔◦𝑓 ) is homotopic to (ℎ ◦ 𝑔) ◦ 𝑓 . This is clear (they are equal). Two objects 𝑋, 𝑌 ∈ Toph are isomorphic if and only if they are homotopy equivalent, i.e. whenever there exist maps 𝑓 : 𝑋 → 𝑌 and 𝑔 : 𝑌 → 𝑋 such that 𝑔𝑓 and 𝑓 𝑔 are homotopic to the respective identity maps.

1.2

Functors

Exercise 1.2.20. • Generalising the functor 𝜋1, for all 𝑛 ≥ 2 there are functors 𝜋𝑛 : Top∗ → Ab assigning to a based space its 𝑛th homotopy group (which happens to be abelian). • The functor Grp → Ab sending a group to its abelianisation. • The functor Ab → Ab sending an abelian group to its torsion subgroup. • For 𝑛 = 0, 1, 2, . . . , the bifunctors Tor𝑛𝑅 (−, −), Ext𝑛𝑅 (−, −) : Mod𝑅 × 𝑅 Mod → Ab for a fixed ring 𝑅. 4

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1.2

Functors

Exercise 1.2.21. Let 𝑓 : 𝐴 → 𝐴 0 be an isomorphism with inverse 𝑔 : 𝐴 0 → 𝐴. Then 1𝐹 (𝐴0) = 𝐹 (1𝐴0 ) = 𝐹 (𝑓 𝑔) = 𝐹 (𝑓 )𝐹 (𝑔) and 1𝐹 (𝐴) = 𝐹 (1𝐴 ) = 𝐹 (𝑔𝑓 ) = 𝐹 (𝑔)𝐹 (𝑓 ), so 𝐹 (𝑓 ) : 𝐹 (𝐴) → 𝐹 (𝐴 0) is an isomorphism with inverse 𝐹 (𝑔) : 𝐹 (𝐴 0) → 𝐹 (𝐴). Exercise 1.2.22. Given an order-preserving map 𝑓 : 𝐴 → 𝐵 we obtain a functor 𝐹 : 𝒜 → ℬ sending 𝑎 ∈ 𝐴 to 𝐹 (𝑎) = 𝑓 (𝑎), and an arrow 𝑎 → 𝑎 0 to 𝑓 (𝑎) → 𝑓 (𝑎 0). This is well-defined since 𝑎 ≤ 𝑎 0 in 𝐴 implies 𝑓 (𝑎) ≤ 𝑓 (𝑎 0) in 𝐵. Given arrows 𝑎 → 𝑎 0, 𝑎 0 → 𝑎 00, their composition has to be the unique arrow 𝑎 → 𝑎 00, and we have unique arrows 𝑓 (𝑎) → 𝑓 (𝑎 0), 𝑓 (𝑎 0) → 𝑓 (𝑎 00) whose composition is the unique arrow 𝑓 (𝑎) → 𝑓 (𝑎 00), so 𝐹 ((𝑎 → 𝑎 0) ◦ (𝑎 0 → 𝑎 00)) = 𝐹 (𝑎 → 𝑎 00) = 𝑓 (𝑎) → 𝑓 (𝑎 00) = (𝑓 (𝑎) → 𝑓 (𝑎 0)) ◦ (𝑓 (𝑎 0) → 𝑓 (𝑎 00)). Thus 𝐹 is a functor. Conversely, if 𝐹 : 𝒜 → ℬ is a functor then 𝑓 : 𝐴 → 𝐵 given by 𝑓 (𝑎) = 𝐹 (𝑎) is orderpreserving. Indeed, if 𝑎 ≤ 𝑎 0 then there is an arrow 𝑎 → 𝑎 0, which is sent by 𝐹 to an arrow 𝑓 (𝑎) → 𝑓 (𝑎 0), which means that 𝑓 (𝑎) ≤ 𝑓 (𝑎 0). 𝑔op

ℎ op

Exercise 1.2.23. (a) Given 𝑔, ℎ ∈ 𝐺 and ∗ −−→ ∗ −−→ ∗ in 𝐺 op, the composition ℎ op ◦op 𝑔op is the opposite arrow of ∗ −−→ ∗ in 𝐺, that is, ℎ op ◦op 𝑔op = (𝑔ℎ) op . Thus 𝐺 op is the group with the same underlying set as 𝐺 with multiplication : 𝐺 op × 𝐺 op → 𝐺 op given by ℎ 𝑔 = 𝑔ℎ. Define 𝜙 : 𝐺 → 𝐺 op by 𝜙 (𝑔) = 𝑔−1 . Then 𝜙 is clearly bijective. Furthermore, 𝜙 is a group homomorphism: if 𝑔, ℎ ∈ 𝐺 then 𝜙 (𝑔ℎ) = (𝑔ℎ) −1 = ℎ −1𝑔−1 = 𝑔−1 ℎ −1 = 𝜙 (𝑔) 𝜙 (ℎ). Thus 𝜙 is an isomorphism of groups 𝑔ℎ

(b) Let 𝑋 be any set with |𝑋 | ≥ 2 and consider 𝑀 = End(𝑋 ). Consider a constant function 𝑐 ∈ 𝑀. Then 𝑐 satisfies 𝑐 𝑓 = 𝑐 for all 𝑓 ∈ 𝑀. Since there is no element 𝑐 0 in 𝑀 such that 𝑓 𝑐 0 = 𝑐 0 for all 𝑓 ∈ 𝑀, it follows that 𝑀 is not isomorphic to 𝑀 op . Exercise 1.2.24. (This is also Mac Lane’s Exercise I.3.4.) We will show there is no such 𝑍 . Consider the symmetric groups 𝑆 2  Z/2 and 𝑆 3 . The inclusion 𝑖 : 𝑆 2 → 𝑆 3 and the projection 𝜋 : 𝑆 3 → 𝑆 3 /𝐴3  𝑆 2 (where 𝐴3 denotes the alternating group of degree 3) compose to the identity 1𝑆2 : 𝑆 2 → 𝑆 2 . As 𝑆 3 has trivial centre, such functor 𝑍 : Grp → Ab would send 1𝑆 2 to 1𝑆 2 = 1𝑍 (𝑆2 ) = 𝑍 (1𝑆 2 ) = 𝑍 (𝜋)𝑍 (𝑖) = 0, which is a contraction. Exercise 1.2.25. (a) Let 𝐴 ∈ 𝒜. For all 𝐵 ∈ ℬ we have 𝐹 𝐴 (1𝐵 ) = 𝐹 (1𝐴, 1𝐵 ) = 1𝐹 (𝐴,𝐵) , and if 𝑓 : 𝐵 → 𝐵 0 and 𝑔 : 𝐵 0 → 𝐵 00 are morphisms in ℬ then 𝐹 𝐴 (𝑔𝑓 ) = 𝐹 (1𝐴, 𝑔𝑓 ) = 𝐹 ((1𝐴, 𝑔) ◦ (1𝐴, 𝑓 )) = 𝐹 (1𝐴, 𝑔) ◦ 𝐹 (1𝐴, 𝑓 ) = 𝐹 𝐴 (𝑔) ◦ 𝐹 𝐴 (𝑓 ). Thus 𝐹 𝐴 is a functor. Similarly, given 𝐵 ∈ ℬ, 𝐹𝐵 : 𝒜 → 𝒞 defined by 𝐹𝐵 (𝐴) = 𝐹 (𝐴, 𝐵) on objects and 𝐹𝐵 (𝑓 ) = (𝑓 , 1𝐵 ) on morphisms is a functor.

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1.2

Functors

(b) The equation 𝐹 𝐴 (𝐵) = 𝐹 (𝐴, 𝐵) = 𝐹𝐵 (𝐴) is clear. Given 𝑓 : 𝐴 → 𝐴 0 in 𝒜 and 𝑔 : 𝐵 → 𝐵 0 in ℬ we have 0

𝐹 𝐴 (𝑔) ◦ 𝐹𝐵 (𝑓 ) = 𝐹 (1𝐴0 , 𝑔) ◦ 𝐹 (𝑓 , 1𝐵 ) = 𝐹 (𝑓 , 𝑔) = 𝐹 (𝑓 , 1𝐵0 ) ◦ 𝐹 (1𝐴, 𝑔) = 𝐹𝐵0 (𝑓 ) ◦ 𝐹 𝐴 (𝑔) by functoriality of 𝐹 . (c) We must have 𝐹 (𝐴, 𝐵) = 𝐹 𝐴 (𝐵) = 𝐹𝐵 (𝐴) on objects. If 𝑓 : 𝐴 → 𝐴 0 and 𝑔 : 𝐵 → 𝐵 0 are morphisms in 𝒜 and ℬ respectively, then 𝐹 (𝑓 , 𝑔) must be given by 𝐹 (𝑓 , 𝑔) = 𝐹 (𝑓 , 1𝐵0 ) ◦ 𝐹 (1𝐴, 𝑔) = 𝐹𝐵0 (𝑓 ) ◦ 𝐹 𝐴 (𝑔), so we define 𝐹 (𝑓 , 𝑔) by this formula. At the same time, 𝐹 (𝑓 , 𝑔) must be given by 0

𝐹 (𝑓 , 𝑔) = 𝐹 (1𝐴0 , 𝑔) ◦ 𝐹 (𝑓 , 1𝐵 ) = 𝐹 𝐴 (𝑔) ◦ 𝐹𝐵 (𝑓 ). By the conditions in (b), these two definitions agree. It remains to show that 𝐹 is indeed a functor when defined in this way. First note that 𝐹 (1𝐴, 1𝐵 ) = 𝐹𝐵 (1𝐴 ) ◦ 𝐹 𝐴 (1𝐵 ) = 1𝐹𝐵 (𝐴) ◦ 1𝐹 𝐴 (𝐵) = 1𝐹 (𝐴,𝐵) for all 𝐴 ∈ 𝒜 and 𝐵 ∈ ℬ. If (𝑓 , 𝑔) : (𝐴, 𝐵) → (𝐴 0, 𝐵 0) and (𝑓 0, 𝑔 0) : (𝐴 0, 𝐵 0) → (𝐴 00, 𝐵 00) are morphisms in 𝒜 × ℬ then 𝐹 ((𝑓 0, 𝑔 0) ◦ (𝑓 , 𝑔)) = 𝐹 (𝑓 0 𝑓 , 𝑔 0𝑔) = 𝐹𝐵00 (𝑓 0 𝑓 ) ◦ 𝐹 𝐴 (𝑔 0𝑔) = 𝐹𝐵00 (𝑓 0) ◦ 𝐹𝐵00 (𝑓 ) ◦ 𝐹 𝐴 (𝑔 0) ◦ 𝐹 𝐴 (𝑔) 0

= 𝐹𝐵00 (𝑓 0) ◦ 𝐹 𝐴 (𝑔 0) ◦ 𝐹𝐵0 (𝑓 ) ◦ 𝐹 𝐴 (𝑔) = 𝐹 (𝑓 0, 𝑔 0) ◦ 𝐹 (𝑓 , 𝑔). Hence 𝐹 : 𝒜 × ℬ → 𝒞 is indeed a functor, the unique one satisfying the conditions in (b). Exercise 1.2.26. (This is also Mac Lane’s Exercise II.3.5.) Given a topological space 𝑋 let 𝐶 (𝑋 ) be the ring of continuous functions 𝑋 → R, and given a map 𝑓 : 𝑋 → 𝑌 of spaces let 𝐶 (𝑓 ) : 𝐶 (𝑌 ) → 𝐶 (𝑋 ) be defined by ℎ ↦→ ℎ ◦ 𝑓 for all maps 𝑔 : 𝑌 → R. This is well-defined since composition of continuous functions is continuous. Moreover, this gives a contravariant functor from Top to Ring: given 𝑓 : 𝑋 → 𝑌 and 𝑔 : 𝑌 → 𝑍, we have 𝐶 (𝑓 𝑔) = 𝐶 (𝑔)𝐶 (𝑓 ); furthermore, clearly 𝐶 (1𝑋 ) = 1𝐶 (𝑋 ) for all spaces 𝑋 . Exercise 1.2.27. Let 𝒜 be a category with precisely three objects 𝑎, 𝑏, 𝑐 and only two non-identity morphisms 𝑓 : 𝑎 → 𝑐, 𝑔 : 𝑎 → 𝑐, and ℬ be a category with exactly two objects 𝑑, 𝑒 and only one non-identity morphism ℎ : 𝑑 → 𝑒. Let 𝐹 : 𝒜 → ℬ be given by 𝐹 (𝑎) = 𝑑 and 𝐹 (𝑏) = 𝐹 (𝑐) = 𝑒 on objects, and by 𝐹 (𝑓 ) = 𝐹 (𝑔) = ℎ on morphisms. An easier example would be letting 𝒜 have two elements and only identity morphisms, ℬ have one element and only the identity, and letting 𝐹 : 𝐴 → 𝐵 be the only possible functor. Also, any non-injective function of sets 𝒜 → ℬ, considered as discrete categories, works. Exercise 1.2.28. (a) We list them here: 6

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Functors • Forgetful functors that forget “structure”, as Grp → Set, Ring → Set, Vect𝑘 → Set, Ring → Ab and Ring → Mon. All of these are faithful; for instance, given two groups 𝐺, 𝐻 ∈ Grp, if 𝑓 : 𝐺 → 𝐻 is a function between the underlying sets then there is at most one group homomorphism 𝐺 → 𝐻 whose underlying set function is 𝑓 , namely 𝑓 itself if it is a homomorphism. Thus Grp → Set is faithful, and the same argument works for the others. However, none of these is full, as forgetting some structure gives rise to more morphisms. For instance, we can find two groups 𝐺, 𝐻 and a function 𝐺 → 𝐻 between the underlying sets which is not a group homomorphism, so Grp → Set is not full, and the same argument works for the other functors. • The forgetful functor Ab → Grp (which forgets a “property”) is full and faithful. It is faithful by the same reason as in the previous example, but now it is also full since any group homomorphism between abelian groups is an abelian group homomorphism. • Free functors: Set → Grp, Set → CRing, Set → Vect𝑘 , etc. All of these satisfy the same universal property in their corresponding categories. For instance, if 𝑆 is a set then the free group 𝐹 (𝑆) on 𝑆 have the following universal property: if 𝑖 : 𝑆 → 𝐹 (𝑆) is the natural inclusion, 𝐻 is a group and 𝑓 : 𝑆 → 𝐻 a (set) function, there is a unique group homomorphism 𝑓 : 𝐹 (𝑆) → 𝐻 such that 𝑓 𝑖 = 𝑓 : 𝑖

𝐹 (𝑆)

𝑆

∃! homomorphism 𝑓

∀ functions 𝑓

∀𝐻 Thus, if 𝑈 : Grp → Set denote the forgetful functor, for any group 𝐻 we have a bijection Set(𝑆, 𝑈 (𝐻 ))  Grp(𝐹 (𝑆), 𝐻 ). In particular, for two sets 𝑆,𝑇 we have a bijection Set(𝑆, 𝑈 𝐹 (𝑇 ))  Grp(𝐹 (𝑆), 𝐹 (𝑇 )). We have an inclusion 𝑇 ⊂ 𝑈 𝐹 (𝑇 ) of sets, so if 𝑓 : 𝑆 → 𝑇 is a function inducing a homomorphism 𝐹 (𝑆) → 𝐹 (𝑇 ), then 𝑓 is unique for 𝑓 can be seen as a function 𝑆 → 𝑈 𝐹 (𝑇 ). Thus the free functor Set → Grp is faithful, and the same argument works for Set → CRing and Set → Vect𝑘 . However neither of these is full. For instance, if a homomorphism 𝜑 : 𝐹 (𝑆) → 𝐹 (𝑇 ) corresponds via the bijection to a function 𝑆 → 𝑈 𝐹 (𝑇 ) with image not in 𝑇 , then there is no function 𝑆 → 𝑇 inducing 𝜑. • The fundamental group functor 𝜋1 : Top∗ → Grp. This functor is not faithful since two different based maps 𝑋 → 𝑌 can induce the same morphism in 𝜋1 . For example, any two (based) maps 𝑆 2 → 𝑆 2 will induce the trivial map on 𝜋 1 since 𝜋 1 (𝑆 2 ) = 1. (Moreover, since 𝜋2 (𝑆 2 ) = Z ≠ 0, then 𝜋 1 as a functor Toph∗ → Grp is not faithful either.) To show that 𝜋1 : Top∗ → Grp is not full consider the spaces R𝑃 3 and R𝑃 2, whose fundamental group is Z/2. Recall that 𝐻 ∗ (R𝑃 𝑛 ; Z/2Z) = Z/2Z[𝑥]/(𝑥 𝑛+1 ) for all 𝑛 ≥ 1. Thus any map 𝑓 : R𝑃 3 → R𝑃 2 induces the zero map on 𝐻 1 (−; Z/2Z), for if 𝑥 ∈ 𝐻 1 (R𝑃 2 ; Z/2Z) is a generator then 0 = 𝑓 ∗ (𝑥 3 ) = 𝑓 ∗ (𝑥) 3, so 𝑓 ∗ (𝑥) = 0 ∈ 𝐻 1 (R𝑃 3 ; Z/2Z). Since there are 7

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Functors natural isomorphisms 𝐻 1 (R𝑃 𝑛 ; Z/2Z)  𝐻 1 (R𝑃 𝑛 ; Z)  𝐻 1 (R𝑃 𝑛 ; Z)  𝜋 1 (R𝑃 𝑛 ), then any map R𝑃 3 → R𝑃 2 induces the trivial map on 𝜋 1, so 𝜋 1 : Top∗ → Grp is not a full functor. (The above example could also show that 𝜋1 : Toph∗ → Grp is not full as long as we could show that there is a map R𝑃 3 → R𝑃 2 which is not nullhomotopic. Is there?) • The 𝑛th homology group functor 𝐻𝑛 : Top → Ab. As in the previous examples, any map 𝑆 𝑛+1 → 𝑆 𝑛+1 will induce the trivial map on 𝐻𝑛 , so none of the functors 𝐻𝑛 : Toph → Ab is faithful. The same example as before shows that 𝐻 1 : Toph → Ab is not full. Now assume 𝑛 ≥ 2 and let 𝑝, 𝑞 > 0 be such that 𝑝 + 𝑞 = 𝑛. We show that any map 𝑓 : 𝑆 𝑛 → 𝑆 𝑝 × 𝑆 𝑞 induces the trivial map on 𝐻𝑛 . Let 𝛼 ∈ 𝐻 𝑝 (𝑆 𝑝 ) and 𝛽 ∈ 𝐻 𝑞 (𝑆 𝑞 ) be generators, and 𝛼 0 = 𝛼 × 1 ∈ 𝐻 𝑝 (𝑆 𝑝 × 𝑆 𝑞 ), 𝛽 0 = 1 × 𝛽 ∈ 𝐻 𝑞 (𝑆 𝑝 × 𝑆 𝑞 ). Then 𝛾 = 𝛼 0 ∪ 𝛽 0 ∈ 𝐻 𝑛 (𝑆 𝑝 × 𝑆 𝑞 ) is a generator by Künneth theorem, so 𝑓 ∗ : 𝐻 𝑛 (𝑆 𝑝 ×𝑆 𝑞 ) → 𝐻 𝑛 (𝑆 𝑛 ) sends 𝛾 to 𝑓 ∗ (𝛼 0)∪𝑓 ∗ (𝛽 0) = 0. Thus 𝑓 is trivial on 𝐻 𝑛 , hence also on 𝐻𝑛 (by the universal coefficient theorem).This shows that 𝐻𝑛 : Top → Ab is not full. (And neither is 𝐻 𝑛 .) (Moreover, consider now 𝐻𝑛 as a functor Toph → Ab. For 𝑛 ≥ 3 consider 𝑝 = 𝑛 − 1, 𝑞 = 1. Since 𝜋𝑚+1 (𝑆 𝑚 ) ≠ 0 for all 𝑚 ≥ 2 then 𝜋𝑛 (𝑆 𝑛−1 × 𝑆 1 ) ≠ 0, so there are maps 𝑆 𝑛 → 𝑆 𝑛−1 × 𝑆 1 which are not nullhomotopic, and hence 𝐻𝑛 : Toph → Ab is not full for 𝑛 ≥ 3. For 𝑛 = 2 this does not work since all maps 𝑆 2 → 𝑆 1 × 𝑆 1 are nullhomotopic. In this case we should find another example. Or maybe it is full?) • A monoid (group) homomorphism 𝐹 : 𝐺 → 𝐻, considered as a functor 𝐹 : 𝒢 → ℋ between the corresponding categories. 𝐹 is faithful if and only if it is injective as a map 𝐺 → 𝐻, and full if and only if it is surjective as a map 𝐺 → 𝐻 . • Let 𝐺 be a monoid, regarded as a category 𝒢 with one element. Consider a functor 𝐹 : 𝒢 → Set, i.e. a left 𝐺-set. Let ∗ ∈ 𝒢 be the unique object and 𝑆 = 𝐹 (∗). 𝐹 is faithful if and only if the following condition holds: given 𝑔, 𝑔 0 ∈ 𝐺, if 𝑔 · 𝑠 = 𝑔 0 · 𝑠 for all 𝑠 ∈ 𝑆 then 𝑔 = 𝑔 0 . This is the same as saying that the action of 𝐺 in 𝑆 is faithful (in the sense of the definition in this context). Now, 𝐹 is full if and only if for all functions 𝑓 : 𝑆 → 𝑆 there exists some 𝑔 ∈ 𝐺 such that 𝑓 (𝑠) = 𝑔 · 𝑠, but this is only possible for bijections 𝑓 . Thus 𝐹 is full if and only if |𝑆 | ≤ 1. The same conclusions hold for a functor 𝐺 : 𝒢 → Vect𝑘 , a 𝑘-linear representation of 𝐺 . • 𝐴, 𝐵 (pre)ordered sets, 𝒜, ℬ the corresponding categories. Let 𝐹 : 𝒜 → ℬ be a functor, that is, an order-preserving map 𝐹 : 𝐴 → 𝐵. Given 𝑎, 𝑎 0 ∈ 𝐴 the set 𝒜(𝑎, 𝑎 0) is empty unless 𝑎 ≤ 𝑎 0, and in this case there is a unique arrow 𝑎 → 𝑎 0 . Thus 𝐹 is faithful since there is at most one arrow between two given elements of 𝒜. But 𝐹 is not necessarily full, for we can have an arrow 𝑓 (𝑎) → 𝑓 (𝑎 0) in ℬ such that 𝑎, 𝑎 0 ∈ 𝐴 are not related, i.e. there is no arrow 𝑎 → 𝑎 0. • The functor 𝐶 : Topop → Ring sending a space to its ring of continuous real-valued functions. Let 𝑋, 𝑌 be topological spaces, and assume that 𝑌 has the indiscrete topology. If 8

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Functors 𝑞 : 𝑌 → R is a map and 𝑦 ∈ 𝑌 , then 𝑞(𝑦) ∈ R is non-empty closed, so 𝑞 −1 ({𝑞(𝑦)}) = 𝑌 . It follows that any map 𝑌 → R is constant. Thus, if 𝑓 , 𝑔 : 𝑋 → 𝑌 are two arbitrary maps then 𝑓 ∗ = 𝑔∗ : 𝐶 (𝑌 ) → 𝐶 (𝑋 ). By letting 𝑌 have at least two elements we see that 𝐶 is not faithful. Moreover, in the same example we see that the induced morphism 𝐶 (𝑌 ) → 𝐶 (𝑋 ) of a map 𝑋 → 𝑌 have image precisely the constant functions. Hence 𝐶 is not full either. op

• The contravariant Hom functor Hom(−,𝑊 ) : Vect𝑘 → Vect𝑘 for a fixed 𝑘-vector space 𝑊 . If 𝑊 = 0 this functor is clearly full but not faithful, so assume 𝑊 ≠ 0. First we show that Hom(−,𝑊 ) is faithful. Let 𝑈 , 𝑉 ∈ Vect𝑘 and 𝑓 , 𝑔 : 𝑈 → 𝑉 be linear maps. Suppose that 𝑓 ≠ 𝑔, so that there exists 𝑢 ∈ 𝑈 such that 𝑣 B 𝑓 (𝑢) ≠ 𝑔(𝑢) B 𝑣 0 . Extend {𝑣 } to a basis for 𝑉 , such that if 𝑣 and 𝑣 0 are linear independent then 𝑣 0 belongs to this basis. Let 𝐿 : 𝑉 → 𝑊 be a linear map sending 𝑣 to some 𝑤 ≠ 0 and all other basis elements to 0. If 𝑣, 𝑣 0 are linear dependent there exists 𝛼 ∈ 𝑘, 𝛼 ≠ 0, 1 such that 𝛼𝑣 = 𝑣 0, so that (𝑓 ∗ 𝐿) (𝑢) = 𝐿(𝑣) = 𝑤 and (𝑔∗ 𝐿) (𝑢) = 𝐿(𝑣 0) = 𝛼𝑤 ≠ 𝑤 . If 𝑣, 𝑣 0 are linear independent then (𝑓 ∗ 𝐿) (𝑢) = 𝐿(𝑣) = 𝑤 ≠ 0 and (𝑔∗ 𝐿) (𝑢) = 𝐿(𝑣 0) = 0. Thus 𝑓 ∗ ≠ 𝑔∗ . It follows that the functor Hom(−,𝑊 ) is faithful. Now we show that Hom(−,𝑊 ) is not full. It is enough to consider the case 𝑊 = 𝑘. (For then it will hold for all 𝑊 ≠ 0 as every 𝑊 contains 𝑘 as a subspace.) Recall that if 𝑈 is a vector space with countable infinite dimension then 𝑈 ∗ has uncountable dimension. Thus not all linear maps 𝑈 ∗ → 𝑘 ∗  𝑘 are duals of a linear map 𝑘 → 𝑈 , so Hom(−, 𝑘) is not full. • The functors 𝐻 𝑛 : Topop → Ab assigning a space to its 𝑛th cohomology group. None of these is faithful, for any map 𝑆 𝑛+1 → 𝑆 𝑛+1 induces the trivial map on 𝐻 𝑛 . This shows moreover that 𝐻 𝑛 is not faithful as a functor Tophop → Ab. The same examples as for 𝐻𝑛 above show that 𝐻 𝑛 : Topop → Ab is not full for 𝑛 ≠ 2. (And 𝐻 𝑛 : Tophop → Ab is not full for 𝑛 ≥ 3. Is it for 𝑛 = 2?) • A functor 𝐹 : 𝒢op → Set, i.e. a right 𝐺-set. This is analogous to example above of functors 𝒢 → Set, i.e. left 𝐺-sets. • A presheaf on a category 𝒜, i.e. a functor 𝒜 op → Set. One does not expect this to be faithful or full unless we have an specific example. Consider a presheaf on a space 𝑋 ∈ Top, i.e. a functor 𝐹 : 𝒪(𝑋 ) op → Set. Then 𝐹 is faithful since for any two elements 𝑈 , 𝑉 ∈ 𝒪(𝑋 ), there is at most one morphism 𝑉 → 𝑈 . In general one does not expect 𝐹 to be full, for instance if 𝑈 ⊄ 𝑉 then there is no morphism 𝑉 → 𝑈 in 𝒪(𝑋 ) op, but there could be functions 𝐹 (𝑉 ) → 𝐹 (𝑈 ), e.g. if they are both non-empty. In the particular case in which 𝐹 is the presheaf of real continuous functions, 𝐹 is also not full in general for the same reason. (b) We can use one the previous examples: Let 𝐺 and 𝐻 be monoids; let 𝒢 and ℋ denote the corresponding categories. The a functor 𝒢 → ℋ is full (resp. faithful) if and only if the 1Z

corresponding homomorphism 𝐺 → 𝐻 is surjective (resp. injective). Thus Z −→ Z is full and 9

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Natural Transformations 2

(2 0)

1

faithful, Z → − Z is faithful but not full, Z → − Z/2 is full but not faithful, and Z × Z −−−→ Z is neither full nor faithful. Exercise 1.2.29. (a) Let (𝑃, ≤𝑃 ) be a poset, i.e. a (partially) ordered set. Then a subcategory of 𝑃 is a poset (𝑄, ≤𝑄 ) such that 𝑄 ⊂ 𝑃 and if 𝑥 ≤𝑄 𝑦 for 𝑥, 𝑦 ∈ 𝑄, then 𝑥 ≤𝑃 𝑦. A subcategory 𝑄 ⊂ 𝑃 is full if and only if 𝑄 is a subposet of 𝑃, i.e. if 𝑥 ≤𝑄 𝑦 ⇔ 𝑥 ≤𝑃 𝑦 for all 𝑥, 𝑦 ∈ 𝑄. (b) As we do not require closure under inverses, the subcategories of a group are precisely its submonoids (and the empty category). Only one subcategory is full, namely the whole group itself.

1.3

Natural Transformations

Exercise 1.3.25. 𝑅 Mod.

• Given a ring 𝑅, there is a natural isomorphism 1𝑅 Mod  Hom𝑅 (𝑅, −) : 𝑅 Mod →

• Given a ring 𝑅, there is a natural isomorphism 1𝑅 Mod  𝑅 ⊗𝑅 − : 𝑅 Mod → 𝑅 Mod. • Let 𝑛 ≥ 2 and consider the functors 𝜋𝑛 , 𝐻𝑛 : Top∗ → Ab. Then there is a natural transformation ℎ : 𝜋𝑛 ⇒ 𝐻𝑛 such that for all based spaces (𝑋, 𝑥 0 ) ∈ Top∗, ℎ (𝑋 ,𝑥 0 ) : 𝜋𝑛 (𝑋, 𝑥 0 ) → 𝐻𝑛 (𝑋 ) is the Hurewicz homomorphism. Exercise 1.3.26. First assume that 𝛼 is an isomorphism and let 𝛽 : 𝐺 ⇒ 𝐹 be an inverse of 𝛼 . Then 𝛼𝐴 ◦ 𝛽𝐴 = 1𝐺 (𝐴) and 𝛽𝐴 ◦ 𝛼𝐴 = 1𝐹 (𝐴) for all 𝐴 ∈ 𝒜, so 𝛼𝐴 : 𝐹 (𝐴) → 𝐺 (𝐴) is an isomorphism for all 𝐴 ∈ 𝒜. Conversely, assume that 𝛼𝐴 : 𝐹 (𝐴) → 𝐺 (𝐴) is an isomorphism for all 𝐴 ∈ 𝒜. Then for each 𝐴 ∈ 𝒜 there exists an inverse 𝛽𝐴 : 𝐺 (𝐴) → 𝐹 (𝐴) of 𝛼𝐴 . If we prove that 𝛽 defines a natural transformation 𝐺 ⇒ 𝐹 then 𝛽 is an inverse of 𝛼 . For this purpose, we need to show that for each 𝐴, 𝐴 0 ∈ 𝒜 and 𝑓 : 𝐴 → 𝐴 0, the diagram 𝐺 (𝐴)

𝐺 (𝑓 )

𝛽𝐴0

𝛽𝐴

𝐹 (𝐴)

𝐺 (𝐴 0)

𝐹 (𝑓 )

𝐹 (𝐴 0)

commutes. By naturality of 𝛼 we have 𝐺 (𝑓 ) ◦ 𝛼𝐴 = 𝛼𝐴0 ◦ 𝐹 (𝑓 ), so that 𝐹 (𝑓 ) ◦ 𝛽𝐴 = 𝛽𝐴0 ◦ 𝛼𝐴0 ◦ 𝐹 (𝑓 ) ◦ 𝛽𝐴 = 𝛽𝐴0 ◦ 𝐺 (𝑓 ) ◦ 𝛼𝐴 ◦ 𝛽𝐴 = 𝛽𝐴0 ◦ 𝐺 (𝑓 ). It follows that 𝛽 : 𝐺 ⇒ 𝐹 is a natural transformation which is inverse to 𝛼 . Exercise 1.3.27. Define a functor 𝐹e: [𝒜 op, ℬop ] → [𝒜, ℬ] op as follows. Given a functor 𝐻 : 𝒜 op → ℬop, let 𝐹e(𝐻 ) : 𝒜 → ℬ be the functor given on objects 𝐴 ∈ 𝒜 by 𝐹e(𝐻 ) (𝐴) = 𝐻 (𝐴)

10

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1.3

Natural Transformations

and on morphisms 𝑓 ∈ 𝒜(𝐴, 𝐴 0) = 𝒜 op (𝐴 0, 𝐴) by 𝐹e(𝐻 ) (𝑓 ) = 𝐻 (𝑓 ) ∈ ℬop (𝐻 (𝐴 0), 𝐻 (𝐴)) = ℬ(𝐻 (𝐴), 𝐻 (𝐴 0)). Given a natural transformation 𝐻

𝒜 op

𝛼

ℬop

𝐾

we need 𝐹e(𝛼) ∈ [𝒜, ℬ] op ( 𝐹e(𝐻 ), 𝐹e(𝐾)) = [𝒜, ℬ] ( 𝐹e(𝐾), 𝐹e(𝐻 )), i.e. a natural transformation 𝐹e(𝛼) : 𝐹e(𝐾) ⇒ 𝐹e(𝐻 ). For each 𝐴 ∈ 𝒜 we let 𝐹e(𝛼)𝐴 = 𝛼𝐴 ∈ ℬop (𝐻 (𝐴), 𝐾 (𝐴)) = ℬ(𝐾 (𝐴), 𝐻 (𝐴)). Given 𝑓 ∈ 𝒜(𝐴, 𝐴 0) = 𝒜 op (𝐴 0, 𝐴) the diagram 𝐹e(𝐾) (𝐴) = 𝐾 (𝐴)

𝐹e(𝐾) (𝑓 )=𝐾 (𝑓 )

𝐹e(𝐾) (𝐴 0) = 𝐾 (𝐴 0) 𝐹e(𝛼)𝐴0 =𝛼𝐴0

𝐹e(𝛼)𝐴 =𝛼𝐴

𝐹e(𝐻 ) (𝐴) = 𝐻 (𝐴)

𝐹e(𝐻 ) (𝑓 )=𝐻 (𝑓 )

𝐹e(𝐻 ) (𝐴 0) = 𝐻 (𝐴 0)

commutes by naturality of 𝛼, so 𝐹e(𝐾)

𝒜

𝐹e(𝛼)

ℬ

𝐹e(𝐻 )

is a natural transformation. If 𝛼 : 𝐻 ⇒ 𝐾 and 𝛽 : 𝐾 → 𝐿 are natural transformations of functors 𝐻, 𝐾, 𝐿 : 𝒜 op ⇒ ℬop then 𝐹e(𝛼 ◦ 𝛽)𝐴 = 𝛼𝐴 ◦ 𝛽𝐴 = 𝐹e(𝛼)𝐴 ◦ 𝐹e(𝛽)𝐴, so 𝐹e(𝛼 ◦ 𝛽) = 𝐹e(𝛼) ◦ 𝐹e(𝛽). It follows that 𝐹e is a functor [𝒜 op, ℬop ] → [𝒜, ℬ] op . e : [𝒜, ℬ] op → [𝒜 op, ℬop ] as follows. Given a functor 𝐻 : 𝒜 → ℬ let Now define a functor 𝐺 op op e(𝐻 ) : 𝒜 → ℬ be the functor given on objects 𝐴 ∈ 𝒜 op by 𝐺 e(𝐻 ) (𝐴) = 𝐻 (𝐴) and on morph𝐺 op 0 0 e isms 𝑓 ∈ 𝒜 (𝐴 , 𝐴) = 𝒜(𝐴, 𝐴 ) by 𝐺 (𝐻 ) (𝑓 ) = 𝐻 (𝑓 ) ∈ ℬ(𝐻 (𝐴), 𝐻 (𝐴 0)) = ℬop (𝐻 (𝐴 0), 𝐻 (𝐴)). e(𝛼) ∈ [𝒜 op, ℬop ] (𝐺 e(𝐾), 𝐺 e(𝐻 )) be given by Given 𝛼 ∈ [𝒜, ℬ] op (𝐾, 𝐻 ) = [𝒜, ℬ] (𝐻, 𝐾) let 𝐺 op e 𝐺 (𝛼)𝐴 = 𝛼𝐴 ∈ ℬ(𝐻 (𝐴), 𝐾 (𝐴)) = ℬ (𝐾 (𝐴), 𝐻 (𝐴)) for all 𝐴 ∈ 𝒜. Thus, if 𝑓 ∈ 𝒜 op (𝐴 0, 𝐴) = 𝒜(𝐴, 𝐴 0) then e(𝐻 ) (𝐴) = 𝐻 (𝐴) 𝐺

e(𝐻 ) (𝑓 )=𝐻 (𝑓 ) 𝐺

e(𝐻 ) (𝐴 0) = 𝐻 (𝐴 0) 𝐺 e(𝛼)𝐴0 =𝛼𝐴0 𝐺

e(𝛼)𝐴 =𝛼𝐴 𝐺

e(𝐾) (𝐴) = 𝐾 (𝐴) 𝐺

e(𝐾) (𝑓 )=𝐾 ( 𝑓 ) 𝐺

e(𝐾) (𝐴 0) = 𝐾 (𝐴 0) 𝐺

commutes by naturality of 𝛼, so e(𝐾) 𝐺

𝒜 op

e(𝛼) 𝐺

ℬop

e(𝐻 ) 𝐺

11

Solutions by positrón0802

1.3

Natural Transformations

e : [𝒜, ℬ] op → [𝒜 op, ℬop ]. is a natural transformation. This defines a functor 𝐺 It is clear that 𝐹𝐺 = 1 [𝒜,ℬ] op and 𝐺𝐹 = 1 [𝒜 op,ℬop ] , so 𝐹 is an isomorphism [𝒜 op, ℬop ]  [𝒜, ℬ] op . Exercise 1.3.28. (a) Let 𝑓 : 𝐴 × 𝐵𝐴 → 𝐵 be given by 𝑓 (𝑎, 𝑔) = 𝑔(𝑎) for all 𝑎 ∈ 𝐴 and 𝑔 ∈ 𝐵𝐴 . (b) Let ℎ : 𝐴 → 𝐵 (𝐵

𝐴)

be given by ℎ(𝑎) (𝑓 ) = 𝑓 (𝑎) for all 𝑎 ∈ 𝐴 and 𝑓 ∈ 𝐵𝐴 .

Exercise 1.3.29. First assume that the family (𝛼𝐴,𝐵 : 𝐹 (𝐴, 𝐵) → 𝐺 (𝐴, 𝐵))𝐴∈𝒜,𝐵 ∈ℬ is a natural transformation 𝐹 ⇒ 𝐺 . Fix 𝐴 ∈ ℬ and consider the family (𝛼𝐴,𝐵 : 𝐹 𝐴 (𝐵) → 𝐺 𝐴 (𝐵))𝐵 ∈ℬ . To show this defines a natural transformation 𝐹 𝐴 ⇒ 𝐺 𝐴 we need to prove that for all 𝐵, 𝐵 0 ∈ ℬ and 𝑓 ∈ ℬ(𝐵, 𝐵 0), the diagram 𝐹 𝐴 (𝐵) = 𝐹 (𝐴, 𝐵)

𝐹 𝐴 (𝑓 )=𝐹 (1𝐴 ,𝑓 )

𝐹 𝐴 (𝐵 0) = 𝐹 (𝐴, 𝐵 0) 𝛼𝐴,𝐵0

𝛼𝐴,𝐵

𝐺 𝐴 (𝐵) = 𝐺 (𝐴, 𝐵)

𝐺 𝐴 (𝑓 )=𝐺 (1𝐴 ,𝑓 )

𝐺 𝐴 (𝐵 0) = 𝐺 (𝐴, 𝐵 0)

commutes, and it indeed does since 𝛼𝐴,𝐵 : 𝐹 ⇒ 𝐺 is natural. Similarly, the family (𝛼𝐴,𝐵 : 𝐹𝐵 (𝐴) → 𝐺 𝐵 (𝐴))𝐴∈𝒜 is a natural transformation 𝐹𝐵 ⇒ 𝐺 𝐵 . Conversely, assume that the above families define natural transformations 𝐹 𝐴 ⇒ 𝐺 𝐴 and 𝐹𝐵 ⇒ 𝐺 𝐵 for all 𝐴 ∈ 𝒜, 𝐵 ∈ ℬ. Let (𝐴, 𝐵), (𝐴 0, 𝐵 0) ∈ 𝒜 × ℬ and (𝑓 , 𝑔) ∈ 𝒜 × ℬ((𝐴, 𝐵), (𝐴 0, 𝐵 0)). We want to show that the diagram 𝐹 ( 𝑓 ,𝑔)

𝐹 (𝐴, 𝐵)

𝐹 (𝐴 0, 𝐵 0) 𝛼𝐴0,𝐵0

𝛼𝐴,𝐵

𝐺 (𝐴, 𝐵)

𝐺 (𝑓 ,𝑔)

𝐺 (𝐴 0, 𝐵 0)

commutes. Since 𝐹 (𝑓 , 𝑔) = 𝐹 (𝑓 , 1𝐵 ) ◦ 𝐹 (1𝐴, 𝑔), 𝐺 (𝑓 , 𝑔) = 𝐺 (𝑓 , 1𝐵 ) ◦ 𝐺 (1𝐴, 𝑔), and the diagram 𝐹 (𝐴, 𝐵)

𝐹 (1𝐴 ,𝑔)

𝐹 (𝐴, 𝐵 0) 𝛼𝐴,𝐵0

𝛼𝐴,𝐵

𝐺 (𝐴, 𝐵)

𝐹 (𝑓 ,1𝐵 )

𝐺 (1𝐴 ,𝑔)

𝐺 (𝐴, 𝐵 0)

𝐹 (𝐴 0, 𝐵 0) 𝛼𝐴0,𝐵0

𝐺 (𝑓 ,1𝐵 )

𝐺 (𝐴 0, 𝐵 0)

commutes by hypothesis, the result follows. Exercise 1.3.30. Let ∗Z denote the unique object in Z and ∗𝐺 the unique object in 𝐺 . The oneto-one correspondence described assigns to an element 𝑔 ∈ 𝐺 the functor Z → 𝐺 sending a morphism 𝑛 ∈ Z(∗Z, ∗Z ) to 𝑔𝑛 ∈ 𝐺 (∗𝐺 , ∗𝐺 ). Note that every natural transformation between functors Z → 𝐺 is an isomorphism by Exercise 1.3.26, since every element in a group is invertible. 12

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Natural Transformations

If ℎ, 𝑔 are functors Z → 𝐺, a natural isomorphism 𝛼 : ℎ → 𝑔 amounts to a morphism 𝛼 ∈ 𝐺 (∗𝐺 , ∗𝐺 ) such that the diagram ∗𝐺

ℎ (𝑚)

∗𝐺

𝛼

∗𝐺

𝛼

∗𝐺

𝑔 (𝑚)

commutes for all 𝑚 ∈ Z, i.e. 𝛼 ∈ 𝐺 is such that 𝛼ℎ𝑚 = 𝑔𝑚 𝛼 for all 𝑚 ∈ Z. This holds if and only if 𝛼ℎ𝛼 −1 = 𝑔. Thus, the equivalence relation on 𝐺 is that of conjugacy. Exercise 1.3.31. (a) Given 𝑋, 𝑋 0 ∈ ℬ and 𝑓 ∈ ℬ(𝑋, 𝑋 0), define Sym(𝑓 ) ∈ Set(Sym(𝑋 ), Sym(𝑋 0)) to be given by ℎ ↦→ 𝑓 ℎ𝑓 −1 for ℎ ∈ Sym(𝑋 ). Now define Ord(𝑓 ) ∈ Set(Ord(𝑋 ), Ord(𝑋 0)) as follows. Given a total order ≤ on 𝑋 let Ord(𝑓 )(≤) be the order  on 𝑋 0 given by 𝑦  𝑧 if and only if 𝑓 −1 (𝑦) ≤ 𝑓 −1 (𝑧). It is clear that both these definitions give functors ℬ → Set. (b) Suppose that 𝛼 : Sym ⇒ Ord is a natural transformation. Let 𝑋, 𝑋 0 ∈ ℬ be finite groups of the same cardinality and 𝑓 : 𝑋 → 𝑋 0 be a bijection. Then there is a commutative diagram Sym(𝑋 )

Sym(𝑓 )

Sym(𝑋 0) 𝛼𝑋 0

𝛼𝑋

Ord(𝑋 )

Ord(𝑓 )

Ord(𝑋 0) .

Consider the elements 1𝑋 ∈ Sym(𝑋 ), 1𝑋 0 ∈ Sym(𝑋 0) and denote 𝛼𝑋 (1𝑋 ) by ≤𝑋 and 𝛼𝑋 0 (1𝑋 0 ) by ≤𝑋 0 . By definition, we have Sym(𝑓 )(1𝑋 ) = 1𝑋 0 . Hence Ord(𝑓 ) (≤𝑋 ) = ≤𝑋 0 . This means that no matter what 𝑓 is, Ord(𝑓 ) sends ≤𝑋 to ≤𝑋 0 . This situation is clearly impossible by considering, e.g., 𝑋 = 𝑋 0 and |𝑋 | ≥ 2 since there is at least two different bijections giving different total orders in 𝑋 . (c) It is clear that |Sym(𝑋 )| = |Ord(𝑋 )| = 𝑛!. Since Sym(𝑋 ) and Ord(𝑋 ) have the same cardinal, we have Sym(𝑋 )  Ord(𝑋 ) for all 𝑋 ∈ ℬ. But this isomorphism is not natural in 𝑋 as there is no natural transformation Sym ⇒ Ord. Exercise 1.3.32. (a) By assumption there exists a functor 𝐺 : ℬ → 𝒜 and natural isomorphisms 𝜂 : 𝐺𝐹 ⇒ 1𝒜 and 𝜀 : 𝐹𝐺 ⇒ 1ℬ . If 𝐵 ∈ ℬ, then 𝐺 (𝐵) ∈ 𝒜 and 𝐹𝐺 (𝐵)  𝐵 via 𝜀. Hence 𝐹 is essentially surjective on objects. Now let 𝐴, 𝐴 0 ∈ 𝒜 and consider the function 𝒜(𝐴, 𝐴 0) → ℬ(𝐹 (𝐴), 𝐹 (𝐴 0)) 𝑓 ↦→ 𝐹 (𝑓 ).

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1.3

Natural Transformations

To show it is injective assume that 𝐹 (𝑓 ) = 𝐹 (𝑔) for 𝑓 , 𝑔 ∈ 𝒜(𝐴, 𝐴 0). Then 𝐺𝐹 (𝑓 ) = 𝐺𝐹 (𝑔). By naturality of 𝜂 the diagram 𝐺𝐹 (𝐴)

𝐺𝐹 (𝑓 )=𝐺𝐹 (𝑔)

𝐺𝐹 (𝐴 0) 𝜂𝐴0

𝜂𝐴

𝐴0

𝐴 𝑓

commutes. Since 𝜂𝐴 is an isomorphism we have 𝑓 = 𝜂𝐴0 ◦ 𝐺𝐹 (𝑓 ) ◦ 𝜂𝐴−1, and the same can be said about 𝑔, so 𝑓 = 𝑔. Thus 𝐹 is faithful. Now we show that the above function is surjective for all 𝐴, 𝐴 0 ∈ 𝒜. Let ℎ ∈ ℬ(𝐹 (𝐴), 𝐹 (𝐴 0)) be arbitrary. There exists a unique 𝑓 ∈ 𝒜(𝐴, 𝐴 0) making the diagram 𝐺𝐹 (𝐴)

𝐺 (ℎ)

𝐺𝐹 (𝐴 0) 𝜂𝐴0

𝜂𝐴

𝐴0

𝐴 𝑓

commute, namely 𝑓 = 𝜂𝐴0 ◦ 𝐺 (ℎ) ◦ 𝜂𝐴−1 . Since 𝐺 is faithful (by the previous proof since it is an equivalence), then for proving 𝐹 (𝑓 ) = ℎ it suffices to prove 𝐺𝐹 (𝑓 ) = 𝐺 (ℎ). But both 𝐺𝐹 (𝑓 ) and 𝐺 (ℎ) fill the dashed arrow in the commutative diagram 𝐺𝐹 (𝐴 0)

𝐺𝐹 (𝐴)

𝜂𝐴0

𝜂𝐴

𝐴 𝑓

𝐴0 ,

so both equal 𝜂𝐴−10 ◦ 𝑓 ◦ 𝜂𝐴 . Thus 𝐹 (𝑓 ) = ℎ and it follows that 𝐹 is full. (b) Define 𝐺 : ℬ → 𝒜 as follows. Since 𝐹 is essentially surjective, given 𝐵 ∈ ℬ there exists 𝐺 (𝐵) ∈ 𝒜 and an isomorphism 𝜀𝐵 : 𝐹𝐺 (𝐵) → 𝐵. We want 𝜀 to define a natural isomorphism 𝐹𝐺 ⇒ 1𝐵 , i.e. that for all 𝐵, 𝐵 0 ∈ ℬ and 𝑓 ∈ ℬ(𝐵, 𝐵 0) the diagram 𝐹𝐺 (𝐵)

𝐹𝐺 (𝑓 )

𝐹𝐺 (𝐵 0) 𝜀𝐵0

𝜀𝐵

𝐵0

𝐵 𝑓

commutes. Since 𝜀𝐵0 is an isomorphism this amounts to having 𝐹𝐺 (𝑓 ) = 𝜀𝐵−10 ◦ 𝑓 ◦ 𝜀𝐵 . Thus, given 𝑓 ∈ ℬ(𝐵, 𝐵 0) we define 𝐺 (𝑓 ) to be the unique morphism 𝐺 (𝐵) → 𝐺 (𝐵 0) corresponding to 𝜀𝐵−10 ◦ 𝑓 ◦ 𝜀𝐵 : 𝐹𝐺 (𝐵) → 𝐹𝐺 (𝐵 0) under the function 𝒜(𝐺 (𝐵), 𝐺 (𝐵 0)) → ℬ(𝐹𝐺 (𝐵), 𝐹𝐺 (𝐵 0)) ℎ ↦→ 𝐹 (ℎ), 14

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1.3

Natural Transformations

which is a bijection since 𝐹 is full and faithful. If 𝑓 ∈ ℬ(𝐵, 𝐵 0) and 𝑔 ∈ ℬ(𝐵 0, 𝐵 00) then the equation 𝐹𝐺 (𝑔) ◦ 𝐹𝐺 (𝑓 ) = 𝐹𝐺 (𝑔 ◦ 𝑓 ) guarantees 𝐺 (𝑔) ◦ 𝐺 (𝑓 ) = 𝐺 (𝑔 ◦ 𝑓 ), so 𝐺 is a well-defined functor ℬ → 𝒜. By construction, 𝜀 : 𝐹𝐺 ⇒ 1𝐵 is a natural isomorphism. It remains to construct a natural isomorphism 𝜂 : 𝐺𝐹 ⇒ 1𝒜 . Given 𝐴 ∈ 𝒜, let 𝜂𝐴 : 𝐺𝐹 (𝐴) → 𝐴 be the unique morphism corresponding to 𝜀 𝐹 (𝐴) : 𝐹𝐺𝐹 (𝐴) → 𝐹 (𝐴) under the bijection 𝒜(𝐺𝐹 (𝐴), 𝐴) → ℬ(𝐹𝐺𝐹 (𝐴), 𝐹 (𝐴)) ℎ ↦→ 𝐹 (ℎ). Since 𝜀 𝐹 (𝐴) is an isomorphism, so is 𝜂𝐴 . We thus have a family of isomorphisms (𝜂𝐴 : 𝐴 → 𝐺𝐹 (𝐴))𝐴∈𝒜 . To show that 𝜂 is natural, let 𝐴, 𝐴 0 ∈ 𝒜, 𝑓 ∈ 𝒜(𝐴, 𝐴 0) and consider the diagram 𝐺𝐹 (𝐴)

𝐺𝐹 (𝑓 )

𝐺𝐹 (𝐴 0) 𝜂𝐴0

𝜂𝐴

𝐴0 .

𝐴 𝑓

We want to show that it commutes. But applying 𝐹 we obtain the commutative diagram 𝐹𝐺𝐹 (𝑓 )

𝐹𝐺𝐹 (𝐴)

𝐹𝐺𝐹 (𝐴 0) 𝐹 (𝜂𝐴0 )=𝜀 𝐹 (𝐴0 )

𝐹 (𝜂𝐴 )=𝜀 𝐹 (𝐴)

𝐹 (𝐴)

𝐹 (𝐴 0) ,

𝐹 (𝑓 )

so the first diagram commutes since the functor 𝐹 is faithful. Thus 𝜂 is a natural isomorphism. We conclude that 𝐹 is an equivalence. Exercise 1.3.33. (This is also Mac Lane’s Exercise I.4.6.) Composition in Mat is given by multiplication of matrices. For each finite-dimensional 𝑘-vector space 𝑉 fix an ordered basis {𝑒𝑖𝑉 }1≤𝑖 ≤dim 𝑉 . If 𝑘 = 𝐹 𝑛 for some 𝑛, we take the (ordered) standard basis. Define a functor 𝑇 : FDVect → Mat as follows. For 𝑉 ∈ FDVect let 𝑇 (𝑉 ) = dim 𝑉 . If 𝐿 ∈ FDVect(𝑉 ,𝑊 ) let Mat(dim 𝑉 , dim𝑊 ) be the dim𝑊 ×dim 𝑉 matrix of 𝐿 with respect to the bases {𝑒𝑖𝑉 }𝑖 , {𝑒𝑊 𝑗 } 𝑗 . Then 𝑇 is a functor FDVect → Mat. On the other hand, for each 𝑛 ∈ Mat let 𝐺 (𝑛) = 𝑘 𝑛 , and given 𝐴 ∈ Mat(𝑛, 𝑚) let 𝐺 (𝐴) : 𝑘 𝑛 → 𝑘𝑚 have matrix 𝐴 with respect to the standard bases for 𝑘 𝑛 and 𝑘𝑚 . Then 𝐺 : Mat → FDVect is a functor such that 𝑇𝐺 = 1Mat . Furthermore, for all 𝑉 ∈ FDVect there is an isomorphism 𝜂𝑉 : 𝑉 → 𝑘 𝑛 , 𝑛 = dim 𝑉 , sending the ordered basis {𝑒𝑖𝑉 }1≤𝑖 ≤𝑛 onto the (ordered) standard basis for 𝑘 𝑛 . Then, if 𝑓 : 𝑉 → 𝑊 is a linear map, the square 𝜂𝑉

𝑘𝑛

𝑉

𝐺𝑇 (𝑓 )

𝑓

𝑊

𝑘𝑚

𝜂𝑊

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2. Adjoints Í 𝑘 𝑊 commutes, where 𝑛 = dim 𝑉 , 𝑚 = dim𝑊 . Indeed, if 𝑓 (𝑒𝑖𝑉 ) = 𝑚 𝑗=1 𝐴𝑖 𝑒 𝑗 for each 𝑖 = 1, . . . , 𝑛, Í 𝑘 𝑘𝑚 then both 𝐺𝑇 (𝑓 ) ◦𝜂𝑉 and 𝜂𝑊 ◦ 𝑓 send 𝑒𝑖𝑉 to 𝑚 𝑗=1 𝐴𝑖 𝑒 𝑗 for all 𝑖. Then 𝜂 is a natural isomorphism 1FDVect  𝐺𝑇 and we conclude that FDVect is equivalent to Mat. Our equivalence does not involve a canonical functor FDVect → Mat since we fixed bases for all finite-dimensional vector spaces. However, the functor 𝐺 : Mat → FDVect is canonical. Exercise 1.3.34. For any category 𝒞 we have 𝒞 ' 𝒞 via the identity functor 1𝒞 . It is also clear that this relation is symmetric; it remains to prove it is transitive. So assume 𝒞, 𝒟, ℰ are categories such that 𝒞 ' 𝒟 and 𝒟 ' ℰ. Then there exist functors 𝐹 : 𝒞 → 𝒟, 𝐺 : 𝒟 → 𝒞 and 𝐻 : 𝒟 → ℰ, 𝐾 : ℰ → 𝒟 such that 𝐺𝐹  1𝒞 , 𝐹𝐺  1𝒟, 𝐾𝐻  1𝒟 and 𝐻𝐾  1ℰ . Consider the functors 𝐻 𝐹 : 𝒞 → ℰ and 𝐺𝐾 : ℰ → 𝒞; we show there are isomorphisms 𝐺𝐾𝐻 𝐹  1𝒞 and 𝐻 𝐹𝐺𝐾  1ℰ . Let 𝛼 : 𝐾𝐻 ⇒ 1𝒟 be a natural isomorphism. Then 𝛼 𝐷 : 𝐾𝐻 (𝐷) → 𝐷 is an isomorphism for all 𝐷 ∈ 𝒟 (Exercise 1.3.26). In particular, given 𝐶 ∈ 𝒞, 𝛼 𝐹 (𝐶) : 𝐾𝐻 𝐹 (𝐶) → 𝐹 (𝐶) is an isomorphism, so 𝐺 (𝛼 𝐹 (𝐶) ) : 𝐺𝐾𝐻 𝐹 (𝐶) → 𝐺𝐹 (𝐶) is an isomorphism by functoriality of 𝐺 . Let 𝛽 : 𝐺𝐹 ⇒ 1𝒞 be a natural isomorphism. Define 𝜂𝐶 = 𝛽𝐶 ◦ 𝐺 (𝛼 𝐹 (𝐶) ) : 𝐺𝐾𝐻 𝐹 (𝐶) → 𝐶. Then (𝜂𝐶 )𝐶 ∈𝒞 defines a family of isomorphisms. It remains to prove naturality: Let 𝐶, 𝐶 0 ∈ 𝒞 and 𝑓 ∈ 𝒞(𝐶, 𝐶 0). Consider the diagram 𝐺𝐾𝐻 𝐹 (𝐶)

𝐺𝐾𝐻 𝐹 (𝑓 )

𝐺 (𝛼 𝐹 (𝐶 0 ) )

𝐺 (𝛼 𝐹 (𝐶 ) )

𝐺𝐹 (𝐶)

𝐺𝐾𝐻 𝐹 (𝐶 0)

𝐺𝐹 (𝑓 )

𝐺𝐹 (𝐶 0) 𝛽𝐶 0

𝛽𝐶

𝐶0

𝐶 𝑓

The upper square commutes by naturality of 𝛼 and functoriality of 𝐺, and the lower square commutes by naturality of 𝛽. By definition of 𝜂, this shows that 𝜂 : 𝐺𝐾𝐻 𝐹 ⇒ 1𝒞 is a natural transformation, hence a natural isomorphism (Exercise 1.3.26 again). Similarly there is a natural isomorphism 𝐻 𝐹𝐺𝐾 ⇒ 1ℰ and therefore 𝒞 ' ℰ. We conclude that equivalence of categories is an equivalence relation.

2 2.1

Adjoints Definition and examples

Exercise 2.1.12. Adjoint functors: • If 𝑅, 𝑆 are rings, and 𝐴 is an (𝑆, 𝑅)-bimodule, then the functor 𝐴 ⊗𝑅 − : 𝑅 Mod → 𝑆 Mod is left adjoint to the functor 𝑆 Mod(𝐴, −) : 𝑆 Mod → 𝑅 Mod. (Tensor-hom adjunction.)

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Definition and examples • Let 𝑓 : 𝑋 → 𝑌 be a function of sets. Let 𝑓 −1 (−) : 𝒫(𝑌 ) → 𝒫(𝑋 ), 𝐵 ↦→ 𝑓 −1 (𝐵), be the inverse image functor. Then the image functor 𝑓 (−) : 𝒫(𝑋 ) → 𝒫(𝑌 ), 𝐴 ↦→ 𝑓 (𝐴), is left adjoint to 𝑓 −1 (−). Moreover, 𝑓 −1 (−) has a right adjoint given by sending 𝐴 ∈ 𝒫(𝑋 ) to the largest subset 𝐵 of 𝑌 such that 𝑓 −1 (𝐵) ⊂ 𝐴. • Let Σ : Toph∗ → Toph∗ denote the (reduced) suspension functor, and Ω : Toph∗ → Toph∗ denote the loop space functor. Then 𝑆 is left adjoint to Ω. Initial and terminal objects: • The category Top∗ of pointed topological spaces has no initial object, and any one-point space is terminal. • For a vector space 𝑘, Vect𝑘 the zero vector space 0 ∈ Vect𝑘 is both initial and terminal. • It 𝒞 is a discrete category with more than one object, then 𝒞 has no initial nor terminal object. • The category Field has no initial nor terminal object, since given 𝐹, 𝐾 ∈ Field there is no morphism 𝐹 → 𝐾 if char 𝐹 ≠ char 𝐾 . • In Exercise 0.13(a) we showed that (Z[𝑥], 𝑥) is initial in the category of pointed rings and basepoint-preserving ring homomorphisms.

Exercise 2.1.13. Let 𝐹 : 𝒜 → ℬ and 𝐺 : ℬ → 𝒜 be adjoint functors between discrete categories, so that for all 𝐴 ∈ 𝒜 and 𝐵 ∈ ℬ we have a bijection ℬ(𝐹 (𝐴), 𝐵)  𝒜(𝐴, 𝐺 (𝐵)). This means that 𝐹 (𝐴) = 𝐵 if and only if 𝐺 (𝐵) = 𝐴. So 𝐹 and 𝐺 are mutually inverse and therefore 𝒜  ℬ. Exercise 2.1.14. First assume that the equation 𝑝

𝑓

𝐹 (𝑝)

𝐺 (𝑞)

𝑓

𝑞

(𝐴 0 → − 𝐴→ − 𝐺 (𝐵) −−−−→ 𝐺 (𝐵 0)) = (𝐹 (𝐴 0) −−−→ 𝐹 (𝐴) → − 𝐵→ − 𝐵 0) holds for all 𝑝, 𝑓 and 𝑞. By considering 𝑝 = 1𝐴 : 𝐴 → 𝐴 we obtain (the transpose of) equation (2.2) for all 𝑓 and 𝑞 (with 𝑔 = 𝑓 in that equation), and by considering 𝑞 = 1𝐵 → 𝐵 we obtain equation (2.3) for all 𝑝 and 𝑓 . Conversely, assume (2.2) and (2.3) hold and let 𝑝 : 𝐴 0 → 𝐴, 𝑓 : 𝐴 → 𝐺 (𝐵) and 𝑞 : 𝐵 → 𝐵 0 . Then 𝑝

𝑓

𝐺 (𝑞)

𝐹 (𝑝)

𝐺 (𝑞)◦𝑓

𝐹 (𝑝)

𝑓

𝑞

(𝐴 0 → − 𝐵→ − 𝐵 0), − 𝐴→ − 𝐺 (𝐵) −−−−→ 𝐺 (𝐵 0)) = 𝐹 (𝐴 0) −−−→ 𝐹 (𝐴) −−−−−−→ 𝐵 0 = (𝐹 (𝐴 0) −−−→ 𝐹 (𝐴) → the first equality by (2.3) and the second one by (the transpose of) (2.2). Exercise 2.1.15. Given 𝐵 ∈ ℬ there is a bijection ℬ(𝐹 (𝐼 ), 𝐵)  𝒜(𝐼, 𝐺 (𝐵)). Since 𝐼 ∈ 𝒜 is initial there is precisely one map 𝐼 → 𝐺 (𝐵), hence precisely one map 𝐹 (𝐼 ) → 𝐵. Thus 𝐹 (𝐼 ) ∈ ℬ is initial. Similarly, if 𝑇 ∈ ℬ is terminal and 𝐴 ∈ 𝒜, the bijection 𝒜(𝐴, 𝐺 (𝑇 ))  ℬ(𝐹 (𝐴),𝑇 ) shows that there is precisely one map 𝐺 (𝑇 ) → 𝐴, so 𝐺 (𝑇 ) ∈ 𝒜 is terminal. 17

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2.1

Definition and examples

Exercise 2.1.16. (a) First consider the forgetful functor 𝑈 : [𝐺, Set] → Set, whose object function sends a 𝐺-set 𝑓 ∈ [𝐺, Set] to its underlying set 𝑓 (∗) ∈ Set. We will show that 𝑈 has both left and right adjoints. First let 𝐹 : Set → [𝐺, Set] be given on objects by sending 𝑋 ∈ Set to 𝐺 × 𝑋 ∈ [𝐺, Set], where the 𝐺-action on 𝐺 × 𝑋 is given by ℎ · (𝑔, 𝑥) = (ℎ𝑔, 𝑥) for all 𝑔, ℎ ∈ 𝐺 and 𝑥 ∈ 𝑋 . It is clear that this defines a 𝐺-action on 𝐺 × 𝑋, and a map 𝑓 : 𝑋 → 𝑌 of sets induces a map 𝐹 (𝑓 ) : 𝐺 × 𝑋 → 𝐺 × 𝑌 of 𝐺-sets given by (𝑔, 𝑥) ↦→ (𝑔, 𝑓 (𝑥)), which clearly preserves the 𝐺-action. We claim that 𝐹 a 𝑈 , i.e., for 𝑋 ∈ Set, 𝑌 ∈ [𝐺, Set] there is an isomorphism 𝜑𝑋 ,𝑌 : [𝐺, Set] (𝐺 × 𝑋, 𝑌 ) → Set(𝑋, 𝑈 (𝑌 )) natural in 𝑋 and 𝑌 . Let 𝑒 ∈ 𝐺 denote the identity element. For 𝑓 ∈ [𝐺, Set] (𝐺 × 𝑋, 𝑌 ) define 𝜑𝑋 ,𝑌 (𝑓 ) = 𝑓e ∈ Set(𝑋, 𝑈 (𝑌 )) to be given by 𝑓e(𝑥) = 𝑓 (𝑒, 𝑥) for all 𝑥 ∈ 𝑋 . On the other hand, for 𝑓e ∈ Set(𝑋, 𝑈 (𝑌 )) define 𝜓𝑋 ,𝑌 ( 𝑓e) = 𝑓 ∈ [𝐺, Set] (𝐹 (𝑋 ), 𝑌 ) by 𝑓 (𝑔, 𝑥) = 𝑔 · 𝑓e(𝑥). Note that this is well-defined since 𝑓 (ℎ · (𝑔, 𝑥)) = 𝑓 (ℎ𝑔, 𝑥) = (ℎ𝑔) · 𝑓e(𝑥) = ℎ · (𝑔 · 𝑓e(𝑥)) = ℎ · 𝑓e(𝑔, 𝑥). We claim 𝜑𝑋−1,𝑌 = 𝜓𝑋 ,𝑌 . Indeed, for 𝑓 ∈ [𝐺, Set] (𝐺 × 𝑋, 𝑌 ) and 𝑥 ∈ 𝑋 we have 𝜓𝑋 ,𝑌 ◦ 𝜑𝑋 ,𝑌 (𝑓 ) (𝑔, 𝑥) = 𝑔 · 𝜑𝑋 ,𝑌 (𝑓 ) (𝑥) = 𝑔 · 𝑓 (1, 𝑥) = 𝑓 (𝑔 · (𝑒, 𝑥)) = 𝑓 (𝑔, 𝑥), and for 𝑓e ∈ Set(𝑋, 𝑈 (𝑌 )) and (𝑔, 𝑥) ∈ 𝐺 × 𝑋 we have 𝜑𝑋 ,𝑌 ◦ 𝜓𝑋 ,𝑌 ( 𝑓e) (𝑥) = 𝜓𝑋 ,𝑌 ( 𝑓e) (𝑒, 𝑥) = 𝑒 · 𝑓e(𝑥) = 𝑓e(𝑥). Hence 𝜑𝑋 ,𝑌 is a bijection with inverse 𝜓𝑋 ,𝑌 . It remains to show naturality. This amounts to showing that for all 𝑋, 𝑋 0 ∈ Set and 𝑌 , 𝑌 0 ∈ [𝐺, Set], and morphisms 𝑓 : 𝑋 0 → 𝑋, 𝑔 : 𝑌 → 𝑌 0, the diagram [𝐺, Set] (𝐺 × 𝑋, 𝑌 )

𝜑𝑋 ,𝑌

Set(𝑓 ,𝑈 (𝑔))

[𝐺,Set] (𝐹 (𝑓 ),𝑔)

[𝐺, Set] (𝐺

Set(𝑋, 𝑈 (𝑌 ))

× 𝑋 0, 𝑌 0)

𝜑𝑋 0,𝑌 0

Set(𝑋 0, 𝑈 (𝑌 0))

commutes (c.f. Exercise 2.1.14). So let 𝑞 ∈ [𝐺, Set] (𝐺 × 𝑋, 𝑌 ) and 𝑥 0 ∈ 𝑋 0 . On the one hand we have Set(𝑓 , 𝑈 (𝑔)) (𝜑𝑋 ,𝑌 (𝑞)) (𝑥 0) = 𝑈 (𝑔) ◦ 𝜑𝑋 ,𝑌 (𝑞) ◦ 𝑓 (𝑥 0) = 𝑔(𝑞(𝑒, 𝑓 (𝑥 0))), and on the other hand 𝜑𝑋 0,𝑌 0 ([𝐺, Set] (𝐹 (𝑓 ), 𝑔) (𝑞)) (𝑥 0) = 𝜑𝑋 0,𝑌 0 (𝑔 ◦ ℎ ◦ 𝐹 (𝑓 )) (𝑥 0) = 𝑔 ◦ 𝑞 ◦ 𝐹 (𝑓 ) (𝑒, 𝑥 0) = 𝑔(𝑞(𝑒, 𝑓 (𝑥 0))). Thus the diagram indeed commutes and it follows that the isomorphism 𝜑𝑋 ,𝑌 is natural in 𝑋 and 𝑌 . We conclude that 𝐹 : Set → [𝐺, Set] is left adjoint to 𝑈 : [𝐺, Set] → Set. Now define 𝐻 : Set → [𝐺, Set] as follows. For 𝑋 ∈ Set set 𝐻 (𝑋 ) = 𝑋 𝐺 ∈ [𝐺, Set], the set of functions 𝐺 → 𝑋, with 𝐺-action given by (𝑔 · 𝑓 ) (ℎ) = 𝑓 (ℎ𝑔) for all 𝑔, ℎ ∈ 𝐺, 𝑓 ∈ 𝑋 𝐺 . It 18

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Definition and examples

is clear that 𝑒 · 𝑓 = 𝑓 for all 𝑓 ∈ 𝑋 𝐺 . Moreover, if 𝑔, 𝑔 0, ℎ ∈ 𝐺 then (𝑔 0𝑔 · 𝑓 ) (ℎ) = 𝑓 (ℎ𝑔 0𝑔) = (𝑔 · 𝑓 )(ℎ𝑔 0) = (𝑔 0 · (𝑔 · 𝑓 )) (ℎ), so this indeed defines a 𝐺-action. A map 𝑝 : 𝑋 → 𝑌 of sets induces a map 𝐻 (𝑝) : 𝑋 𝐺 → 𝑌 𝐺 given by [𝐻 (𝑝) (𝑓 )] (ℎ) = 𝑝 ◦ 𝑓 (ℎ). This is a morphism of 𝐺-sets by the equation [𝐻 (𝑝) (𝑔 · 𝑓 )] (ℎ) = 𝑝 ◦ (𝑔 · 𝑓 ) (ℎ) = 𝑝 ◦ 𝑓 (ℎ𝑔) = [𝑔 · 𝐻 (𝑝) (𝑓 )] (ℎ), so 𝐻 : Set → [𝐺, Set] is a functor. We claim 𝑈 a 𝐻, so that we seek a natural bijection 𝜑𝑌 ,𝑋 : Set(𝑈 (𝑌 ), 𝑋 ) → [𝐺, Set] (𝑌 , 𝑋 𝐺 ). For 𝑓 ∈ Set(𝑈 (𝑌 ), 𝑋 ) define 𝜑𝑌 ,𝑋 (𝑓 ) : 𝑌 → 𝑋 𝐺 by 𝜑𝑌 ,𝑋 (𝑓 ) (𝑦) (𝑔) = 𝑓 (𝑔 · 𝑦) for all 𝑦 ∈ 𝑌 , 𝑔 ∈ 𝐺 . If 𝑓 ∈ Set(𝑈 (𝑌 ), 𝑋 ), 𝑦 ∈ 𝑌 and 𝑔, ℎ ∈ 𝐺 we have [𝜑𝑌 ,𝑋 (𝑓 ) (𝑔 · 𝑦)] (ℎ) = 𝑓 (ℎ𝑔 · 𝑦) = [𝑔 · 𝜑𝑌 ,𝑋 (𝑓 ) (𝑦)] (ℎ), so 𝜑𝑌 ,𝑋 (𝑓 ) is a morphism of 𝐺-sets. Now define a map 𝜓𝑌 ,𝑋 : [𝐺, Set] (𝑌 , 𝑋 𝐺 ) → Set(𝑈 (𝑌 ), 𝑋 ) by 𝜓𝑌 ,𝑋 ( 𝑓e) (𝑦) = 𝑓e(𝑦) (𝑒) for all 𝑓e ∈ [𝐺, Set] (𝑌 , 𝑋 𝐺 ) and 𝑦 ∈ 𝑌 . It is straightforward to show that 𝜓𝑌 ,𝑋 = 𝜑𝑌−1,𝑋 , so 𝜑𝑌 ,𝑋 is a bijection. It remains to show that this bijection is natural. For 𝑋, 𝑋 0 ∈ Set and 𝑌 , 𝑌 0 ∈ [𝐺, Set], and morphisms 𝑓 : 𝑌 0 → 𝑌 , 𝑘 : 𝑋 → 𝑋 0, we need to show that the diagram Set(𝑈 (𝑌 ), 𝑋 )

𝜑𝑌 ,𝑋

[𝐺,Set] ( 𝑓 ,𝐻 (𝑘))

Set(𝑈 (𝑓 ),𝑘)

Set(𝑈 (𝑌 0), 𝑋 0)

[𝐺, Set] (𝑌 , 𝑋 𝐺 )

𝜑𝑌 0,𝑋 0

[𝐺, Set] (𝑌 0, (𝑋 0)𝐺 )

commutes. Given 𝑞 ∈ Set(𝑈 (𝑌 ), 𝑋 ), 𝑦 0 ∈ 𝑌 0 and 𝑔 ∈ 𝐺 we compute 𝜑𝑌 0,𝑋 0 (𝑘 ◦ 𝑞 ◦ 𝑈 (𝑓 )) (𝑦 0) (𝑔) = (𝑘 ◦ 𝑞 ◦ 𝑈 (𝑓 )) (𝑔 · 𝑦 0) = 𝑘 (𝑞(𝑓 (𝑔 · 𝑦 0))) and (𝐻 (𝑘) ◦𝜑𝑌 ,𝑋 (𝑞) ◦ 𝑓 ) (𝑦 0) (𝑔) = 𝐻 (𝑘) [𝜑𝑌 ,𝑋 (𝑞) (𝑓 (𝑦 0))] (𝑔) = 𝑘 (𝜑𝑌 ,𝑋 (𝑞) (𝑓 (𝑦 0)) (𝑔)) = 𝑘 (𝑞(𝑔 · 𝑓 (𝑦 0))). Both results agree since 𝑓 is a morphism of 𝐺-sets, so that 𝑓 (𝑔 · 𝑦 0) = 𝑔 · 𝑓 (𝑦 0). It follows that the bijection 𝜑𝑌 ,𝑋 is natural in 𝑋 and 𝑌 , and we conclude that 𝑈 a 𝐻 . Consider now the functor Δ : Set → [𝐺, Set] that equips a set with the trivial left 𝐺-action. That is, if 𝑋 ∈ Set then Δ𝑋 is the 𝐺-set such that 𝑔 · 𝑥 = 𝑥 for all 𝑥 ∈ 𝑋 . We have two interesting functors [𝐺, Set] → Set as follows. We have the functor (−)𝐺 : [𝐺, Set] → Set sending a 𝐺set 𝑋 to its 𝐺-fixed point subset 𝑋 𝐺 = {𝑥 ∈ 𝑋 | 𝑔 · 𝑥 = 𝑥 for all 𝑔 ∈ 𝐺 }, and the functor (−)/𝐺 : [𝐺, Set] → Set sending a 𝐺-set 𝑋 to the set 𝑋 /𝐺 = {𝐺 · 𝑥 | 𝑥 ∈ 𝑋 } of orbits of 𝑋 under 𝐺 . Then we have adjunctions (−)/𝐺 a Δ a (−)𝐺 . Similarly as in the previous adjunctions, these can be proved directly. Alternatively, we are going to deduce them on Exercise 6.1.6. (b) Definitions analogous to those in (a) give functors [𝐺, Vect𝑘 ] → Vect𝑘 left and right adjoint to the forgetful functor 𝑈 : Vect𝑘 → [𝐺, Vect𝑘 ], with the 𝑘-vector space structure on 19

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2.1

Definition and examples

𝐺 × 𝑋 given by 𝛼 (𝑔, 𝑥) = (𝑔, 𝛼𝑥), and the vector space structure on the set of functions 𝐺 → 𝑋 given by (𝛼 𝑓 ) (𝑔) = 𝛼 𝑓 (𝑔). We also have a functor Δ : Vect𝑘 → [𝐺, Vect𝑘 ] endowing a 𝑘-vector space with the trivial left 𝐺-action. For a 𝐺-vector space 𝑉 , the 𝐺-fixed point subset 𝑉 𝐺 is a 𝑘-vector subspace of 𝑉 , so we have a functor (−)𝐺 : [𝐺, Vect𝑘 ] → Vect𝑘 which is right adjoint to Δ. However, we no longer have a functor (−)/𝐺, as there is no natural way of endowing the set of orbits 𝑋 /𝐺 of a 𝐺-vector space 𝑋 under 𝐺 with a 𝑘-vector space structure. Exercise 2.1.17. More precisely, the functor Δ : Set → [𝒪(𝑋 ) op, Set] is given as follows. If 𝐴 ∈ Set then Δ𝐴(𝑈 ) = 𝐴 for all 𝑈 ∈ 𝒪(𝑋 ), and Δ𝐴(𝑓 ) = 1𝐴 for all morphisms 𝑓 : 𝑉 → 𝑈 in 𝒪(𝑋 ) op . If 𝑝 : 𝐴 → 𝐴 0 is a map of sets, Δ(𝑝)𝑉 = 𝑝 for all 𝑉 ⊂ 𝑋 . For 𝑈 ⊂ 𝑉 ∈ 𝒪(𝑋 ) let 𝑟𝑉 ,𝑈 : 𝑉 → 𝑈 denote the unique arrow 𝑉 → 𝑈 in 𝒪(𝑋 ) op . First we find a right adjoint Γ to Δ. For 𝐹 ∈ [𝒪(𝑋 ) op, Set] let Γ(𝐹 ) = 𝐹 (𝑋 ), and for a natural transformation 𝜂 : 𝐹 ⇒ 𝐺 of functors 𝒪(𝑋 ) op → Set let Γ(𝛼) = 𝜂𝑋 : 𝐹 (𝑋 ) → 𝐺 (𝑋 ). Then Γ defines a functor [𝒪(𝑋 ) op, Set] → Set. To show that Δ a Γ we need bijections 𝜑𝐴,𝐹 : [𝒪(𝑋 ) op, Set] (Δ𝐴, 𝐹 ) → Set(𝐴, 𝐹 (𝑋 )) natural in 𝐴 ∈ Set and 𝐹 ∈ [𝒪(𝑋 ) op, Set]. Given a natural transformation 𝜂 : Δ𝐴 ⇒ 𝐹, let 𝜑𝐴,𝐹 (𝜂) = 𝜂𝑋 : 𝐴 → 𝐹 (𝑋 ). Conversely, define 𝜓𝐴,𝐹 : Set(𝐴, 𝐹 (𝑋 )) → [𝒪(𝑋 ) op, Set] (Δ𝐴, 𝐹 ) as follows. If 𝑓 : 𝐴 → 𝐹 (𝑋 ) is a map of sets and 𝑈 ∈ 𝒪(𝑋 ), let 𝜓𝐴,𝐹 (𝑓 )𝑈 : 𝐴 → 𝐹 (𝑈 ) be the composition 𝐹 (𝑟 𝑋 ,𝑈 ) ◦ 𝑓 . That 𝜓𝐴,𝐹 (𝑓 ) defines a natural transformation Δ𝐴 ⇒ 𝐹 follows from the equation 𝐹 (𝑟𝑉 ,𝑈 ) ◦𝜓𝐴,𝐹 (𝑓 )𝑉 = 𝜓𝐴,𝐹 (𝑓 )𝑈 for all 𝑈 ⊂ 𝑉 , which holds by functoriality of 𝐹 . It is clear that 𝜓𝐴,𝐹 and 𝜑𝐴,𝐹 are inverses to each other. It remains to show that the bijection 𝜑𝐴,𝐹 is natural, that is, that given 𝐴, 𝐴 0 ∈ Set, 𝐹, 𝐹 0 ∈ [𝒪(𝑋 ) op, Set], and morphisms 𝑓 : 𝐴 0 → 𝐴, 𝛼 : 𝐹 ⇒ 𝐹 0, the diagram [𝒪(𝑋 ) op, Set] (Δ𝐴, 𝐹 )

𝜑𝐴,𝐹

Set(𝐴, 𝐹 (𝑋 ))

[𝒪 (𝑋 ) op ,Set] (Δ𝑓 ,𝛼)

Set(𝑓 ,Γ (𝛼))

[𝒪(𝑋 ) op, Set] (Δ𝐴 0, 𝐹 0)

𝜑𝐴0,𝐹 0

Set(𝐴 0, 𝐹 0 (𝑋 ))

commutes. This is a straightforward computation: for a natural transformation 𝜂 : Δ𝐴 ⇒ 𝐹 we have Γ(𝛼) ◦ 𝜑𝐴,𝐹 (𝜂) ◦ 𝑓 = 𝛼𝑋 ◦ 𝜂𝑋 ◦ 𝑓 = (𝛼 ◦ 𝑋 ◦ Δ𝑓 )𝑋 = 𝜑𝐴0,𝐹 0 (𝛼 ◦ 𝜂 ◦ Δ𝑓 ). It follows that the bijections 𝜑𝐴,𝐹 define an adjunction Δ a Γ. We now find a right adjoint ∇ to Γ. Given 𝐴 ∈ Set let ∇𝐴 ∈ [𝒪(𝑋 ) op, Set] be defined as follows. On objects define ∇𝐴 by ∇𝐴(𝑋 ) = 𝐴 and ∇𝐴(𝑉 ) = {∗} for all 𝑉 ≠ 𝑋 . On morphisms, let ∇𝐴(1𝑋 ) = 1𝐴, and if 𝑋 ≠ 𝑈 ⊂ 𝑉 and 𝑟𝑉 ,𝑈 is the unique morphism 𝑉 → 𝑈 on 𝒪(𝑋 ) op, let ∇𝐴(𝑟𝑉 ,𝑈 ) : ∇𝐴(𝑉 ) → {∗} be the unique possible map. Then ∇𝐴 is a functor 𝒪(𝑋 ) op → Set. For a map of sets 𝑓 : 𝐴 → 𝐴 0, let ∇𝑓 : ∇𝐴 ⇒ ∇𝐴 0 be the natural transformation given by (∇𝑓 )𝑋 = 𝑓 ,

20

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2.1

Definition and examples

and (∇𝑓 )𝑉 = 1 {∗} if 𝑉 ≠ 𝑋 . Then ∇ is a functor Set → [𝒪(𝑋 ) op, Set]. To show it is right adjoint to Γ we seek bijections 𝜑 𝐹,𝐴 : Set(𝐹 (𝑋 ), 𝐴) → [𝒪(𝑋 ) op, Set] (𝐹, ∇𝐴) natural in 𝐹 ∈ [𝒪(𝑋 ) op, Set] and 𝐴 ∈ Set. For 𝑓 : 𝐹 (𝑋 ) → 𝐴 let 𝜑 𝐹,𝐴 (𝑓 ) : 𝐹 ⇒ ∇𝐴 be given by 𝜑 𝐹,𝐴 (𝑓 )𝑋 = 𝑓 and 𝜑 𝐹,𝐴 (𝑓 )𝑉 : 𝐹 (𝑈 ) → {∗} the unique map if 𝑉 ≠ 𝑋 . Then 𝜑 𝐹,𝐴 (𝑓 ) is clearly natural. Conversely, if we are given a natural transformation 𝜂 : 𝐹 ⇒ ∇𝐴 let 𝜓 𝐹,𝐴 (𝜂) = 𝜂𝑋 : 𝐹 (𝑋 ) → 𝐴, so that we have a function 𝜓 𝐹,𝐴 : [𝒪(𝑋 ) op, Set] (𝐹, ∇𝐴) → Set(𝐹 (𝑋 ), 𝐴). Then it is clear that −1 , so 𝜑 𝜓 𝐹,𝐴 = 𝜑 𝐹,𝐴 𝐹,𝐴 is a bijection. It remains to show naturality. Given 𝐴, 𝐴 0 ∈ Set, 𝐹, 𝐹 0 ∈ [𝒪(𝑋 ) op, Set], and morphisms 𝑓 : 𝐴 → 𝐴 0, 𝛼 : 𝐹 0 ⇒ 𝐹, we need to show commutativity of diagram Set(𝐹 (𝑋 ), 𝐴)

𝜑 𝐹 ,𝐴

[𝒪(𝑋 ) op, Set] (𝐹, ∇𝐴) [𝒪 (𝑋 ) op ,Set] (𝛼,∇𝑓 )

Set(Γ (𝛼),𝑓 )

Set(𝐹 0 (𝑋 ), 𝐴 0)

𝜑 𝐹 0,𝐴0

[𝒪(𝑋 ) op, Set] (𝐹 0, ∇𝐴 0) .

For 𝑞 : 𝐹 (𝑋 ) → 𝐴, we want to show that (∇𝑓 ◦ 𝜑 𝐹,𝐴 (𝑞) ◦ 𝛼)𝑉 = [𝜑 𝐹 0,𝐴0 (𝑓 ◦ 𝑞 ◦ 𝛼𝑋 )]𝑉 for all 𝑉 ⊂ 𝑋 . For 𝑉 ≠ 𝑋 this is clear as there is only one map 𝐹 0 (𝑉 ) → {∗}, and for 𝑉 = 𝑋 we have (∇𝑓 ◦ 𝜑 𝐹,𝐴 (𝑞) ◦ 𝛼)𝑋 = (∇𝑓 )𝑋 ◦ 𝜑 𝐹,𝐴 (𝑞)𝑋 ◦ 𝛼𝑋 = 𝑓 ◦ 𝑞 ◦ 𝛼𝑋 = [𝜑 𝐹 0,𝐴0 (𝑓 ◦ 𝑞 ◦ 𝛼𝑋 )]𝑋 . It follows that 𝜑 𝐹,𝐴 is a bijection natural in 𝐴 and 𝐹 . We conclude that Γ a ∇. Now we construct a left adjoint Π to Δ. Given 𝐹 ∈ [𝒪(𝑋 ) op, Set] set Π(𝐹 ) = 𝐹 (∅), and for a natural transformation 𝜂 : 𝐹 ⇒ 𝐺 let Π(𝛼) = 𝛼 ∅ : 𝐹 (∅) → 𝐺 (∅). This defines a functor Π : [𝒪(𝑋 ) op, Set] → Set. To show that Π a Δ, define, for all 𝐴 ∈ Set and 𝐹 ∈ [𝒪(𝑋 ) op, Set], 𝜑 𝐹,𝐴 : Set(𝐹 (∅), 𝐴) → [𝒪(𝑋 ) op, Set] (𝐹, Δ𝐴) by 𝜑 𝐹,𝐴 (𝑓 )𝑉 = 𝑓 ◦ 𝐹 (𝑟𝑉 ,∅ ) : 𝐹 (𝑉 ) → 𝐴 for all 𝑓 ∈ Set(𝐹 (∅), 𝐴) and 𝑉 ⊂ 𝑋 . Then 𝜑 𝐹,𝐴 (𝑓 ) is a natural transformation 𝐹 ⇒ Δ𝐴 since given 𝑈 ⊂ 𝑉 we have 𝜑 𝐹,𝐴 (𝑓 )𝑉 = 𝜑 𝐹,𝐴 (𝑓 )𝑈 ◦ 𝐹 (𝑟𝑉 ,𝑈 ) by functoriality of 𝐹 . On the other direction define 𝜓 𝐹,𝐴 : [𝒪(𝑋 ) op, Set] (𝐹, Δ𝐴) → Set(𝐹 (∅), 𝐴) by −1 , so 𝜓 𝐹,𝐴 (𝛼) = 𝛼 ∅ : 𝐹 (∅) → 𝐴 for all 𝛼 : 𝐹 ⇒ Δ𝐴. It is straightforward to check that 𝜓 𝐹,𝐴 = 𝜑 𝐹,𝐴 𝜑 𝐹,𝐴 is a bijection. To prove naturality of this bijection let 𝐴, 𝐴 0 ∈ Set, 𝐹, 𝐹 0 ∈ [𝒪(𝑋 ) op, Set], and morphisms 𝑓 : 𝐴 → 𝐴 0, 𝛼 : 𝐹 0 ⇒ 𝐹, and consider the diagram Set(𝐹 (∅), 𝐴)

𝜑 𝐹 ,𝐴

[𝒪(𝑋 ) op, Set] (𝐹, Δ𝐴) [𝒪 (𝑋 ) op ,Set] (𝛼,Δ𝑓 )

Set(Π (𝛼),𝑓 )

Set(𝐹 0 (∅), 𝐴 0)

𝜑 𝐹 0,𝐴0

[𝒪(𝑋 ) op, Set] (𝐹 0, Δ𝐴 0) . 21

Solutions by positrón0802

2.2

Adjunctions via units and counits

To prove it is commutative let 𝑞 : 𝐹 (∅) → 𝐴 and 𝑉 ⊂ 𝑋 . Then 𝜑 𝐹 0,𝐴0 (𝑓 ◦ 𝑞 ◦ ◦𝛼 ∅ )𝑉 = 𝑓 ◦ 𝑞 ◦ 𝛼 ∅ ◦ 𝐹 0 (𝑟𝑉 ,∅ ) = 𝑓 ◦ 𝑞 ◦ 𝐹 (𝑟𝑉 ,∅ ) ◦ 𝛼𝑉 = (Δ𝑓 ◦ 𝜑 𝐹,𝐴 (𝑞) ◦ 𝛼)𝑉 , the second equality by naturality of 𝛼 . It follows that 𝜑 𝐹,𝐴 is a bijection natural in 𝐴 and 𝐹 and hence Π a Δ. At last, we find a left adjoint Λ to Π. For 𝐴 ∈ Set let Λ(𝐴) be the functor 𝒪(𝑋 ) op → Set given as follows. On objects, Λ(𝐴) (∅) = 𝐴 and Λ(𝐴) (𝑉 ) = ∅ if 𝑉 ≠ ∅; on morphisms, let Λ(𝐴)(1 ∅ ) = 1𝐴 and let Λ(𝐴) (𝑟𝑉 ,𝑈 ) be the empty function if 𝑉 ≠ ∅. If 𝑓 : 𝐴 → 𝐴 0 is a map of sets let Λ(𝑓 ) : Λ(𝐴) ⇒ Λ(𝐴 0) have ∅-component Λ(𝑓 )∅ = 𝑓 , and 𝑉 -component Λ(𝑓 )𝑉 the empty function if 𝑉 ≠ ∅. The naturality of Λ(𝑓 ) is automatic. Then Λ is a functor Set → [𝒪(𝑋 ) op, Set]. We now find natural bijections 𝜑𝐴,𝐹 : [𝒪(𝑋 ) op, Set] (Λ(𝐴), 𝐹 ) → Set(𝐴, 𝐹 (∅)). Define 𝜑𝐴,𝐹 by 𝜑𝐴,𝐹 (𝛼) = 𝛼 ∅ for all 𝛼 : Λ(𝐴) ⇒ 𝐹 . On the other hand, let 𝜓 𝐹,𝐴 : Set(𝐴, 𝐹 (∅)) → [𝒪(𝑋 ) op, Set] (Λ(𝐴), 𝐹 ) send 𝑓 : 𝐴 → 𝐹 (∅) to the transformation 𝜓 𝐹,𝐴 (𝑓 ) : Λ(𝐴) ⇒ 𝐹 with ∅component 𝜓 𝐹,𝐴 (𝑓 )∅ = 𝑓 and 𝑉 -component 𝜓 𝐹,𝐴 (𝑓 )𝑉 the empty function if 𝑉 ≠ ∅. Then 𝜓 𝐹,𝐴 (𝑓 ) −1 . is of course natural, so 𝜓 𝐹,𝐴 is well-defined, and moreover 𝜓 𝐹,𝐴 = 𝜑 𝐹,𝐴 0 0 op 0 Given 𝐴, 𝐴 ∈ Set, 𝐹, 𝐹 ∈ [𝒪(𝑋 ) , Set], and morphisms 𝑓 : 𝐴 → 𝐴, 𝛼 : 𝐹 ⇒ 𝐹 0, consider the diagram [𝒪(𝑋 ) op, Set] (Λ(𝐴), 𝐹 )

𝜑𝐴,𝐹

[𝒪 (𝑋 ) op ,Set] (Λ(𝑓 ),𝛼)

Set(𝐴, 𝐹 (∅)) Set(𝑓 ,Π (𝛼))

[𝒪(𝑋 ) op, Set] (Λ(𝐴 0), 𝐹 0)

𝜑𝐴0,𝐹 0

Set(𝐴 0, 𝐹 0 (∅)) .

If 𝜂 : Δ𝐴 ⇒ 𝐹 is a natural transformation then Π(𝛼) ◦ 𝜑𝐴,𝐹 (𝜂) ◦ 𝑓 = 𝛼 ∅ ◦ 𝜂 ∅ ◦ Λ(𝑓 )∅ = (𝛼 ◦ 𝜂 ◦ Λ(𝑓 ))∅ = 𝜑𝐴0,𝐹 0 (𝛼 ◦ 𝜂 ◦ Λ(𝑓 )), so the diagram commutes. It follows that 𝜑 is a natural isomorphism, so that Λ a Π. This completes the chain of adjoint functors Λ a Π a Δ a Γ a ∇.

2.2

Adjunctions via units and counits

Exercise 2.2.10. First assume (a) holds. Given 𝑎 ∈ 𝐴, then 𝑓 (𝑎) ∈ 𝐵 and 𝑓 (𝑎) ≤ 𝑓 (𝑎), so 𝑎 ≤ 𝑔𝑓 (𝑎) by assumption. Similarly, given 𝑏 ∈ 𝐵 then 𝑔(𝑏) ≤ 𝑔(𝑏) implies 𝑓 𝑔(𝑏) ≤ 𝑏. Now assume (b) holds. Let 𝑎 ∈ 𝐴 and 𝑏 ∈ 𝐵. First suppose that 𝑓 (𝑎) ≤ 𝑏. We have 𝑔𝑓 (𝑎) ≤ 𝑔(𝑏) as 𝑔 is order-preserving, so 𝑎 ≤ 𝑔𝑓 (𝑎) ≤ 𝑔(𝑏) by assumption. Similarly, suppose that 𝑎 ≤ 𝑔(𝑏). We have 𝑓 (𝑎) ≤ 𝑓 𝑔(𝑏) since 𝑓 is order preserving, so 𝑓 (𝑎) ≤ 𝑓 𝑔(𝑏) ≤ 𝑏 by assumption. 22

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2.2

Adjunctions via units and counits

Exercise 2.2.11. (a) Let 𝐴 ∈ Fix(𝐺𝐹 ). Then 𝜂𝐴 is an isomorphism, so 𝐹 (𝜂𝐴 ) is an isomorphism. Since 𝜀 𝐹 (𝐴) ◦ 𝐹 (𝜂𝐴 ) = 1𝐹 (𝐴) it follows that 𝜀 𝐹 (𝐴) is also an isomorphism. Thus 𝐹 restricts to a well-defined functor 𝐹 0 : Fix(𝐺𝐹 ) → Fix(𝐹𝐺). Similarly, 𝐺 restricts to a well-defined functor 𝐺 0 : Fix(𝐹𝐺) → Fix(𝐺𝐹 ). Furthermore, 𝜂 0 : 1Fix(𝐺𝐹 ) ⇒ 𝐺 0𝐹 0 given by 𝜂𝐴0 = 𝜂𝐴 and 𝜀 0 : 𝐹 0𝐺 0 ⇒ 1Fix(𝐹𝐺) given by 𝜀𝐵0 = 𝜀𝐵 are well-defined. Since by definition 𝜂 0 and 𝜀 0 are isomorphisms, it follows that (𝐹 0, 𝐺 0, 𝜂 0, 𝜀 0) is an equivalence. (b) Let 𝑈 : Top → Set be the forgetful functor. Recall (from Example 2.1.5) that 𝑈 has a left adjoint 𝐷 : Set → Top which equips a set 𝑋 with the discrete topology. The unit 𝜂 of the adjunction 𝐷 a 𝑈 is given at a set 𝑆 by the identity function 𝜂𝑆 = 1𝑆 : 𝑆 → 𝑈 𝐷 (𝑆), so that Fix(𝑈 𝐷) = Set. On the other hand, the counit 𝜀 at a topological space 𝑋 is the map 𝜀𝑋 : 𝐷𝑈 (𝑋 ) → 𝑋 whose underlying set function is the identity function. Thus 𝜀𝑋 is a homeomorphism if and only if 𝑋 is discrete, and therefore Fix(𝐷𝑈 ) is the full subcategory Topd of Top spanned by the discrete topological spaces. It follows that 𝑈 and 𝐷 induce an equivalence Topd ' Set which associates to a discrete space its underlying set, and equips a set with the discrete topology. Similarly, 𝑈 : Top → Set has a right adjoint 𝐼 : Set → Top which equips a set with the indiscrete topology. Its counit is given at a set 𝑆 by the identity function 1𝑆 = 𝑈 𝐼 (𝑆) → 𝑆, which is always a bijection. Its unit is given at a space 𝑋 by the map 𝑋 → 𝐼𝑈 (𝑋 ) whose underlying set function is the identity function; it is a homeomorphism if and only if 𝑋 is indiscrete. It follows that 𝑈 and 𝐼 induce an equivalence Topi ' Set, where Topi is the full subcategory of Top spanned by the indiscrete topological spaces. Consider now the forgetful functor 𝑈 : Ab → Grp, whose left adjoint 𝐹 : Grp → Ab sends a group to its the abelianisation (see Example 2.1.3 (c)). Since 𝐹 is the identity on Ab ⊂ Grp (or more precisely, can be taken in this way), the counit of 𝐹 a 𝑈 at an abelian group 𝐴 is the identity morphism 1𝐴 = 𝐹𝑈 (𝐴) → 𝐴, so it is always an isomorphism. On the other hand, the counit of 𝐹 a 𝑈 at a group 𝐺 is the canonical quotient 𝐺 → 𝑈 𝐹 (𝐺) onto its abelianisation, so it is an isomorphism if and only if 𝐺 is abelian. It follows that the equivalence induced by the adjunction 𝐹 a 𝑈 is just the identity Ab ' Ab. Exercise 2.2.12. (a) Let 𝐹 a 𝐺, 𝐹 : 𝒜 → ℬ, be an adjunction with counit 𝜀. Let 𝜑𝐴,𝐵 : 𝒜(𝐴, 𝐺 (𝐵)) → ℬ(𝐹 (𝐴), 𝐵) be the structural isomorphisms of the adjunction. Then 𝜀𝐵 = 𝜑𝐺 (𝐵),𝐵 (1𝐺 (𝐵) ) for all 𝐵 ∈ ℬ, by definition. By naturality of 𝜑 we have 𝑞 ◦ 𝜀𝐵 = 𝑞 ◦ 𝜑𝐺 (𝐵),𝐵 (1𝐺 (𝐵) ) = 𝜑𝐺 (𝐵),𝐵0 ◦ 𝐺 (𝑞)∗ (1𝐺 (𝐵) ) = 𝜑𝐺 (𝐵),𝐵0 ◦ 𝐺 (𝑞) for all 𝑞 : 𝐵 → 𝐵 0, that is, the composite map 𝜑𝐺 (𝐵),𝐵0

𝐺 (−)

ℬ(𝐵, 𝐵 0) −−−−→ 𝒜(𝐺 (𝐵), 𝐺 (𝐵 0)) −−−−−−→ ℬ(𝐹𝐺 (𝐵), 𝐵 0) is given by 𝑞 ↦→ 𝑞 ◦ 𝜀𝐵 . 23

Solutions by positrón0802

2.2

Adjunctions via units and counits

First suppose that 𝐺 is full and faithful. Then the above composition is bijective for all 𝐵, 𝐵 0 ∈ 𝐵. In particular, given 𝐵 ∈ ℬ then for 𝐵 0 = 𝐹𝐺 (𝐵) there exists a map ℎ : 𝐵 → 𝐹𝐺 (𝐵) such that ℎ◦𝜀𝐵 = 1𝐹𝐺 (𝐵) . Moreover, since 𝐺 is full and faithful then 𝜀𝐵 has a right inverse if and only if 𝐺 (𝜀𝐵 ) has a right inverse, and the latter holds by the triangle identity 𝐺 (𝜀𝐵 ) ◦𝜂𝐺 (𝐵) = 1𝐺 (𝐵) . Thus 𝜀𝐵 has both a left and a right inverse, which then must coincide. It follows that 𝜀𝐵 is an isomorphism. Conversely, suppose that 𝜀 is a natural isomorphism. Let 𝐵, 𝐵 0 ∈ ℬ. Then the composition written above is bijective with inverse given by 𝑝 ↦→ 𝑝 ◦ 𝜀𝐵−1 for all 𝑝 : 𝐺𝐹 (𝐵) → 𝐵 0 . Since 𝜑𝐺 (𝐵),𝐵0 is a bijection, it follows that 𝐺 (−) is a bijection. Thus 𝐺 is full and faithful. (b) We want to see which of the right adjoints given in the examples are full and faithful. • Forgetful functors that forget “structure”, as Vect𝑘 → Set, Grp → Set, Grp → Mon, Top → Set, etc. They are all right adjoint to “free” functors. These forgetful functors are not full, so these adjunctions are not reflections. • The forgetful functor 𝑈 : Ab → Grp, which forgets a “property”. Its right adjoint 𝐹 : Grp → Ab is given by the abelianisation. The functor 𝑈 is full and faithful, so the adjunction 𝐹 a 𝑈 is a reflection. • The forgetful functor 𝑈 : Top → Set has a right adjoint 𝐼 : Set → Top endowing a set with the indiscrete topology. The functor 𝐼 is full and faithful, so 𝑈 a 𝐼 is a reflection. • For 𝐵 a set, the functor (−) 𝐵 : Set → Set is right adjoint to − × 𝐵 : Set → Set. If 𝐵 = ∅, the functor (−) 𝐵 sends every set to the empty set, so is clearly not faithful. If 𝐵 has precisely one element, (−) 𝐵 is the identity functor, hence full and faithful, so that (− × 𝐵) a (−) 𝐵 is a reflection. Finally, if 𝐵 has more than one element, the functor (−) 𝐵 is not full. For instance, if 𝐵 has precisely two elements, then (−) 𝐵 is the functor 𝐶 ↦→ 𝐶 × 𝐶, and for sets 𝐶 and 𝐷, in general not all maps 𝐶 × 𝐶 → 𝐷 × 𝐷 have the form 𝑓 × 𝑓 for some 𝑓 : 𝐶 → 𝐷. • Let 𝒜 be a category. If 𝒜 has an initial object, then the unique functor 𝒜 → 1 has a left adjoint (given by the initial object). The right adjoint 𝒜 → 1 is full and faithful if and only if 𝒜 has precisely one arrow between every two objects, so this adjunction is a reflection if and only if this is the case. If 𝒜 has a terminal object, a functor 1 → 𝒜 given by a terminal object is right adjoint to 𝒜 → 1. Such a functor 1 → 𝒜 is full and faithful, so this adjunction is a reflection. Exercise 2.2.13. (a) Let ℎ : 𝒫(𝐾) → 𝒫(𝐿), 𝑇 ↦→ 𝑓 (𝑇 ), be the image functor. (It is a functor being order-preserving.) The equations 𝑇 ⊂ 𝑓 −1 (𝑓 (𝑇 )) for all 𝑇 ∈ 𝒫(𝐾) and 𝑓 (𝑓 −1 (𝑆)) ⊂ 𝑆 for all 𝑆 ∈ 𝒫(𝐿) (which hold for any set function) translate into 𝑇 ≤ 𝑓 ∗ℎ(𝑇 ) for all 𝑇 ∈ 𝒫(𝐾),

ℎ𝑓 ∗ (𝑆) ≤ 𝑆 for all 𝑆 ∈ 𝒫(𝐿).

This means precisely that ℎ a 𝑓 ∗ (see Example 2.2.7, c.f. Exercise 2.2.10).

24

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2.2

Adjunctions via units and counits

Now let us find a right adjoint 𝑔 : 𝒫(𝐾) → 𝒫(𝐿) to 𝑓 ∗ . For 𝑇 ⊂ 𝐾, let 𝑔(𝑇 ) be the largest subset 𝑆 of 𝐿 such that 𝑓 −1 (𝑆) ⊂ 𝑇 . More explicitly, 𝑔 is given as follows. Let 𝑀 = 𝐿 \ 𝑓 (𝐾). Given 𝑇 ⊂ 𝐾 let 𝑇 0 ⊂ 𝑇 be the subset consisting of points 𝑡 ∈ 𝑇 such that there exists some 𝑢 ∉ 𝑇 with 𝑓 (𝑢) = 𝑓 (𝑡). Then 𝑔(𝑇 ) = 𝑓 (𝑇 \ 𝑇 0) ∪ 𝑀. Suppose 𝑇1 ⊂ 𝑇2 ⊂ 𝐾 . If 𝑡 ∈ 𝑇1 \ 𝑇10, then there is no 𝑢 ∉ 𝑇1 such that 𝑓 (𝑢) = 𝑓 (𝑡). In particular there is no 𝑢 ∉ 𝑇2 such that 𝑓 (𝑢) = 𝑓 (𝑡), and hence 𝑡 ∈ 𝑇2 \ 𝑇20 . Thus 𝑔 is order-preserving, i.e. a functor 𝒫(𝐾) → 𝒫(𝐿). Now, proving 𝑓 ∗ a 𝑔 amounts to showing the equations 𝑆 ⊂ 𝑔(𝑓 −1 (𝑆)) for all 𝑆 ∈ 𝒫(𝐿),

𝑓 −1 (𝑔(𝑇 )) ⊂ 𝑇 for all 𝑇 ∈ 𝒫(𝐾).

Given 𝑆 ∈ 𝒫(𝐿), note that (𝑓 −1 (𝑆)) 0 = ∅, for if 𝑡 ∈ 𝑓 −1 (𝑆) and 𝑢 ∈ 𝐾 are such that 𝑓 (𝑢) = 𝑓 (𝑡), then 𝑢 ∈ 𝑓 −1 (𝑆). Thus 𝑔(𝑓 −1 (𝑆)) = 𝑓 (𝑓 −1 (𝑆)) ∪ 𝑀, and this set contains 𝑆. Furthermore, if 𝑇 ⊂ 𝐾 then 𝑓 −1 (𝑔(𝑇 )) = 𝑓 −1 (𝑓 (𝑇 \ 𝑇 0) ∩ 𝑀) = 𝑓 −1 (𝑓 (𝑇 \ 𝑇 0)) ⊂ 𝑇 \ 𝑇 0 ⊂ 𝑇 . Hence the above equations hold, and we conclude that 𝑓 ∗ a 𝑔. (b) Let us write our definitions of 𝑔 and ℎ above for the particular case 𝑝 ∗ : 𝒫(𝑋 ) → 𝒫(𝑋 ×𝑌 ). Given 𝑇 ⊂ 𝑋 × 𝑌 we have ℎ(𝑇 ) = 𝑝 (𝑇 ). This is the set of points 𝑥 ∈ 𝑋 such that (𝑥, 𝑦) ∈ 𝑇 for some 𝑦 ∈ 𝑌 , i.e., the set of points 𝑥 ∈ 𝑋 satisfying 𝑆 (𝑥) = ∃𝑦 ∈ 𝑌 such that (𝑥, 𝑦) ∈ 𝑇 . The unit of the adjunction ℎ a 𝑝 ∗ at a subset 𝑇 ⊂ 𝑋 is the inclusion 𝑇 ⊂ 𝑝 −1 (𝑝 (𝑇 )). This can be written as a logical implication as (∀𝑇 ⊂ 𝑋 × 𝑌 ) (∀(𝑥, 𝑦) ∈ 𝑋 × 𝑌 ) ((𝑥, 𝑦) ∈ 𝑇 =⇒ (∃𝑦 0 ∈ 𝑌 ) ((𝑥, 𝑦 0) ∈ 𝑇 )). Similarly, the counit at a subset 𝑆 ⊂ 𝑋 is the inclusion 𝑝 (𝑝 −1 (𝑆)) ⊂ 𝑆. This can be written as (∀𝑆 ⊂ 𝑋 ) (∀𝑥 ∈ 𝑋 ) ((∃𝑦 ∈ 𝑌 ) ((𝑥, 𝑦) ∈ 𝑝 −1 (𝑆)) =⇒ 𝑥 ∈ 𝑆). On the other hand 𝑔(𝑇 ) = 𝑝 (𝑇 \ 𝑇 0) ∪ 𝑀 with 𝑇 0 defined as in (a) and 𝑀 = 𝑋 \ 𝑝 (𝑋 × 𝑌 ) = ∅. By definition, the set 𝑇 \ 𝑇 0 consists of those pairs (𝑥, 𝑦) ∈ 𝑇 such that (𝑥, 𝑦 0) ∈ 𝑇 for all 𝑦 0 ∈ 𝑌 . Thus 𝑔(𝑇 ) is the set of points satisfying 𝑆 0 (𝑥) = ∀𝑦 ∈ 𝑌 , (𝑥, 𝑦) ∈ 𝑇 . The unit of the adjunction 𝑝 ∗ a 𝑔 can be written as a logical implication as (∀𝑆 ⊂ 𝑋 ) (∀𝑥 ∈ 𝑋 ) (𝑥 ∈ 𝑆 =⇒ (∀𝑦 ∈ 𝑌 ) ((𝑥, 𝑦) ∈ 𝑝 −1 (𝑆))), and the counit as (∀𝑇 ⊂ 𝑋 × 𝑌 ) (∀(𝑥, 𝑦) ∈ 𝑋 × 𝑌 ) ((∀𝑦 0 ∈ 𝑌 ) ((𝑥, 𝑦 0) ∈ 𝑇 ) =⇒ (𝑥, 𝑦) ∈ 𝑇 ). 25

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2.2

Adjunctions via units and counits

Exercise 2.2.14. Let 𝜂 : 1𝒜 ⇒ 𝐺𝐹 and 𝜀 : 𝐹𝐺 ⇒ 1ℬ be the unit and counit of the adjunction, respectively. We will show that they induce natural transformations 𝜂 ∗ : 1 [𝒜,𝒮] ⇒ 𝐹 ∗𝐺 ∗ and 𝜀 ∗ : 𝐺 ∗ 𝐹 ∗ ⇒ 1 [ℬ,𝒮] which will be the unit and counit, respectively, of the adjunction 𝐺 ∗ a 𝐹 ∗ . For all 𝐻 ∈ [𝒜, 𝒮] let 𝜂𝐻∗ : 𝐻 ⇒ 𝐻𝐺𝐹 be the transformation with components (𝜂𝐻∗ )𝐴 = 𝐻 (𝜂𝐴 ) : 𝐻 (𝐴) → 𝐻𝐺𝐹 (𝐴) for all 𝐴 ∈ 𝒜. Since 𝜂 is natural, so is 𝜂𝐻∗ . Moreover, given 𝐻, 𝐻 0 ∈ [𝒜, 𝒮] and 𝛼 : 𝐻 ⇒ 𝐻 0 consider the diagram ∗ 𝜂𝐻

𝐻

𝐻𝐺𝐹

𝛼

𝛼𝐺𝐹

𝐻0

∗ 𝜂𝐻 0

𝐻 0𝐺𝐹 .

For all 𝐴 ∈ 𝒜 we have (𝛼𝐺𝐹 ◦ 𝜂𝐻∗ )𝐴 = 𝛼𝐺𝐹 (𝐴) ◦ 𝐻 (𝜂𝐴 ) = 𝐻 0 (𝜂𝐴 ) ◦ 𝛼𝐴 = (𝜂𝐻∗ 0 ◦ 𝛼)𝐴 by naturality of 𝛼, so the diagram commutes. Thus 𝜂 ∗ : 1 [𝒜,𝒮] ⇒ 𝐹 ∗𝐺 ∗ is indeed natural. Similarly, for all 𝐾 ∈ [ℬ, 𝒮] let 𝜀𝐾∗ = : 𝐾𝐹𝐺 ⇒ 𝐾 have components (𝜀𝐾∗ )𝐵 = 𝐾 (𝜀𝐵 ) : 𝐾𝐹𝐺 (𝐵) → 𝐾 (𝐵) for all 𝐵 ∈ ℬ. Then 𝜀 ∗ is a well-defined natural transformation 𝐺 ∗ 𝐹 ∗ ⇒ 1 [ℬ,𝒮] . To complete the proof it suffices to show that 𝜂 ∗ and 𝜀 ∗ satisfy the triangle identities 𝐺∗

𝐺 ∗𝜂 ∗

1𝐺 ∗

𝐺 ∗ 𝐹 ∗𝐺 ∗

𝜂∗𝐹 ∗

𝐹∗

𝜀 ∗𝐺 ∗

1𝐹 ∗

𝐺∗ ,

𝐹 ∗𝐺 ∗ 𝐹 ∗ 𝐹 ∗𝜀 ∗

𝐹∗ .

First consider the left-hand side triangle. Given 𝐻 ∈ [𝒜, 𝒮], we want to show that the triangle ∗ 𝐺 𝜂𝐻

𝐻𝐺

𝐻𝐺𝐹𝐺 ∗ 𝜀𝐻𝐺

1𝐻𝐺

𝐻𝐺 commutes, and this follows from the computation ∗ (𝜀𝐻𝐺 ◦ 𝜂𝐻∗ 𝐺)𝐵 = 𝐻𝐺 (𝜀𝐵 ) ◦ 𝐻 (𝜂𝐺 (𝐵) ) = 𝐻 (𝐺 (𝜀𝐵 ) ◦ 𝜂𝐺 (𝐵) ) = 𝐻 (1𝐺 (𝐵) ) = 1𝐻𝐺 (𝐵)

for 𝐵 ∈ ℬ, where we use one triangle identity for 𝜂 and 𝜀 in the third equality. Similarly, given 𝐾 ∈ [ℬ, 𝒮] and 𝐴 ∈ 𝒜 we have ∗ (𝜀𝐾∗ 𝐹 ◦ 𝜂𝐾𝐹 )𝐴 = 𝐾 (𝜀 𝐹 (𝐴) ) ◦ 𝐾𝐹 (𝜂𝐴 ) = 𝐾 (𝜀 𝐹 (𝐴) ◦ 𝐹 (𝜂𝐴 )) = 𝐾 (1𝐹 (𝐴) ) = 1𝐾𝐹 (𝐴) ,

so that 𝜂 ∗ and 𝜀 ∗ satisfy the right-hand side triangle identity above. We conclude that 𝐺 ∗ a 𝐹 ∗ with unit 𝜂 ∗ and counit 𝜀 ∗ . 26

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2.3

2.3

Adjunctions via initial objects

Adjunctions via initial objects

Exercise 2.3.8. Let 𝐻, 𝐾 be groups, 𝐹 : 𝐻 → 𝐾, 𝐺 : 𝐾 → 𝐻 be functors (i.e. group homomorphisms), and suppose that we have an adjunction 𝐹 a 𝐺 . The unit of this adjunction amounts to an element ℎ ∈ 𝐻 such that 𝐺𝐹 (𝑥) = ℎ𝑥ℎ −1 for all 𝑥 ∈ 𝐻 (c.f. Exercise 1.3.30). Similarly, the counit is an element 𝑘 ∈ 𝐾 such that 𝑦 = 𝑘𝐹𝐺 (𝑦)𝑘 −1 for all 𝑦 ∈ 𝐾 . Moreover, the triangle identities say that 𝑘𝐹 (ℎ) = 1 and 𝐺 (𝑘)ℎ = 1, so that 𝐹 (ℎ) = 𝑘 −1 and 𝐺 (𝑘) = ℎ −1 . It follows that an adjunction 𝐹 a 𝐺 amounts to group homomorphisms 𝐹 : 𝐻 → 𝐾, 𝐺 : 𝐻 → 𝐾 such that there exist ℎ ∈ 𝐻, 𝑘 ∈ 𝐾, with 𝐹 (ℎ) = 𝑘 −1, 𝐺 (𝑘) = ℎ −1, such that 𝐺𝐹 is conjugation by ℎ and 𝐹𝐺 is conjugation by 𝑘 −1 . In particular, both 𝐺𝐹 and 𝐹𝐺 are isomorphisms, and therefore so are 𝐹 and 𝐺 . Exercise 2.3.9. Dual of Corollary 2.3.7. Let 𝐹 : 𝒜 → ℬ be a functor. Then 𝐹 has a right adjoint if and only if for each 𝐵 ∈ ℬ, the category (𝐹 ⇒ 𝐵) has an initial object. The proof goes through the dual statements of Lemma 2.3.5 and Theorem 2.3.6. Dual of Lemma 2.3.5. Take an adjunction 𝐹 a 𝐺, 𝐹 : 𝒜 → 𝐵, and an object 𝐵 ∈ ℬ. Then the counit map 𝜀𝐵 : 𝐹𝐺 (𝐵) → 𝐵 is a terminal object of (𝐹 ⇒ 𝐵). 𝑓

Proof of Dual of Lemma 2.3.5. Let (𝐴, ℎ) ∈ (𝐹 ⇒ 𝐵). We want a unique arrow (𝐴, ℎ) → − (𝐺 (𝐵), 𝜀𝐵 ), i.e. a unique 𝑓 : 𝐴 → 𝐺 (𝐵) such that ℎ = 𝜀𝐵 ◦ 𝐹 (𝑓 ). By Lemma 2.2.4, 𝑓 = ℎ is the unique such arrow.  Dual of Theorem 2.3.6. Take categories and functors 𝐹 : 𝒜 → ℬ, 𝐺 : ℬ → 𝒜. There is a one-toone correspondence between: (a) adjunctions 𝐹 a 𝐺; (b) natural transformations 𝜀 : 𝐹𝐺 ⇒ 1ℬ such that 𝜀𝐵 : 𝐹𝐺 (𝐵) → 𝐵 is terminal in (𝐹 ⇒ 𝐵) for all 𝐵 ∈ ℬ. Proof of Dual of Theorem 2.3.6. It remains to show that if 𝜀 : 𝐹𝐺 ⇒ 1ℬ satisfies (b) then there exists a unique natural transformation 𝜂 : 1𝒜 ⇒ 𝐺𝐹 such that (𝜂, 𝜀) satisfies the triangle identities. To prove uniqueness assume that 𝜂 and 𝜂 0 are natural transformations 1𝒜 ⇒ 𝐺𝐹 such that both (𝜂, 𝜀) and (𝜂 0, 𝜀) satisfy the triangle identities. For 𝐴 ∈ 𝒜, the diagram 𝐹 (𝐴)

𝐹 (𝜂𝐴 )

𝐹𝐺𝐹 (𝐴) 𝜀 𝐹 (𝐴)

1𝐹 (𝐴)

𝐹 (𝐴) shows that 𝜂𝐴 is a map (𝐴, 1𝐹 (𝐴) ) → (𝐺𝐹 (𝐴), 𝜀 𝐹 (𝐴) ) in (𝐹 ⇒ 𝐹 (𝐴)). The same can be said about 𝜂𝐴0 . Since 𝜀 𝐹 (𝐴) is terminal in (𝐹 ⇒ 𝐹 (𝐴 0)) it follows that 𝜂𝐴0 = 𝜂𝐴 . Hence 𝜂 = 𝜂 0 .

27

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2.3

Adjunctions via initial objects

To show existence define 𝜂 : 1𝒜 ⇒ 𝐺𝐹 as follows. Given 𝐴 ∈ 𝒜 let 𝜂𝐴 : 𝐴 → 𝐺𝐹 (𝐴) be the unique map (1𝐹 (𝐴) , 𝐴) → (𝐺𝐹 (𝐴), 𝜀 𝐹 (𝐴) ) in (𝐹 ⇒ 𝐹 (𝐴)). We need to show that 𝜂 is indeed natural. Given 𝐴, 𝐴 0 ∈ 𝒜 and 𝑓 : 𝐴 → 𝐴 0, the diagrams 𝐹 (𝐴)

𝐹 (𝜂𝐴 )

𝐹𝐺𝐹 (𝐴)

𝐹𝐺𝐹 (𝑓 )

𝐹𝐺𝐹 (𝐴 0)

𝐹 (𝐴)

𝐹 (𝑓 )

𝐹 (𝐴 0)

𝜀 𝐹 (𝐴0 )

𝐹 ( 𝑓 )◦𝜀 𝐹 (𝐴)

𝐹 (𝜂𝐴0 )

1𝐹 (𝐴0 )

𝐹 (𝐴 0)

𝐹𝐺𝐹 (𝐴 0) 𝜀 𝐹 (𝐴0 )

𝐹 (𝐴 0) 𝐹 (𝑓 )

𝐹 (𝑓 )

commute. Thus 𝐺𝐹 (𝑓 ) ◦ 𝜂𝐴 and 𝜂𝐴0 ◦ 𝑓 are both maps (𝐴, 𝐹 (𝑓 )) → (𝐺𝐹 (𝐴 0), 𝜀 𝐹 (𝐴0) ) in (𝐹 ⇒ 𝐹 (𝐴 0)); since 𝜀 𝐹 (𝐴0) is terminal we have 𝐺𝐹 (𝑓 ) ◦𝜂𝐴 = 𝜂𝐴0 ◦ 𝑓 , and hence 𝜂 is natural. By definition (𝜂, 𝜀) satisfies the triangle identity given in the triangle of the right-hand side diagram above, it remains to show the other one, i.e. we want to show that the diagram 𝐺 (𝐵)

𝜂𝐺 (𝐵)

𝐺𝐹𝐺 (𝐵) 𝐺 (𝜀𝐵 )

1𝐺 (𝐵)

𝐺 (𝐵) commutes for all 𝐵 ∈ ℬ. This follows by considering the commutative diagrams 𝐹𝐺 (𝐵)

𝐹 (𝜂𝐺 (𝐵) )

𝐹𝐺𝐹𝐺 (𝐵)

𝐹𝐺 (𝜀𝐵 )

𝐹𝐺 (𝐵)

𝐹𝐺 (𝐵)

𝐹 (1𝐺 (𝐵) )

𝜀𝐵 𝜀𝐵 ◦𝐹𝐺 (𝜀𝐵 )

𝐹𝐺 (𝐵) 𝜀𝐵

𝜀𝐵

𝐵,

𝜀𝐵

𝐵.

Indeed, we see that both 𝐹𝐺 (𝜀𝐵 ) ◦ 𝜂𝐺 (𝐵) and 1𝐺 (𝐵) are maps (𝐺 (𝐵), 𝜀𝐵 ) → (𝐺 (𝐵), 𝜀𝐵 ) in (𝐹 ⇒ 𝐵), so they are equal. This completes the proof.  Proof of Dual of Corollary 2.3.7. By the dual of Lemma 2.3.5 it remains to show that if for each 𝐵 ∈ ℬ, (𝐹 ⇒ 𝐵) has an initial object, then 𝐹 has a right adjoint. Given 𝐵 ∈ ℬ let (𝐺 (𝐵), 𝜀𝐵 : 𝐹𝐺 (𝐵) → 𝐵) be an initial object of (𝐹 ⇒ 𝐵). For 𝑔 : 𝐵 → 𝐵 0 let 𝐺 (𝑔) be unique map (𝐺 (𝐵), 𝑓 ◦ 𝜀𝐵 ) → (𝐺 (𝐵 0), 𝜀𝐵0 ) in (𝐹 ⇒ 𝐵 0). Then 𝐺 : ℬ → 𝒜 is a functor, and the definition ensures that 𝜀 : 𝐹𝐺 ⇒ 1ℬ is a natural transformation. By the dual of Theorem 2.3.6 it follows that 𝐹 a 𝐺 .  Exercise 2.3.10. In view of Remark 2.2.8 the tuple (𝐹, 𝐺, 𝜂, 𝜀) is not necessarily an adjunction since (𝜂, 𝜀) may not satisfy the triangle identities. However we may show that 𝐹 a 𝐺 with unit 𝜂 (and some counit 𝜀 0) using Theorem 2.3.6. More precisely, we will show that 𝜂 : 1𝒜 ⇒ 𝐺𝐹 is such that 𝜂𝐴 : 𝐴 → 𝐺𝐹 (𝐴) is initial in (𝐴 ⇒ 𝐺) for all 𝐴 ∈ 𝒜.

28

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Adjunctions via initial objects

First we define a natural transformation 𝜀 0 : 𝐹𝐺 ⇒ 1ℬ so that the triangle identity 𝐺 (𝜀𝐵0 ) ◦ 𝜂𝐺 (𝐵) = 1𝐺 (𝐵) holds. Define 𝜀𝐵0 : 𝐹𝐺 (𝐵) → 𝐵 by requiring the following diagram to be commutative: 𝜀𝐵0

𝐹𝐺 (𝐵)

𝐵

𝐹𝐺 (𝜀𝐵−1 )

𝜀𝐵

𝐹𝐺𝐹𝐺 (𝐵)

−1 ) 𝐹 (𝜂𝐺 (𝐵)

𝐹𝐺 (𝐵) .

Then 𝜀 0 : 𝐹𝐺 ⇒ 1ℬ is indeed natural. (Even so, naturality of 𝜀 0 will not be used in this proof.) To show the desired triangle identity we want to show that the composite map 𝜂𝐺 (𝐵)

−1 ) 𝐺𝐹 (𝜂𝐺 (𝐵)

𝐺𝐹𝐺 (𝜀𝐵−1 )

𝐺 (𝜀𝐵 )

𝐺 (𝐵) −−−−→ 𝐺𝐹𝐺 (𝐵) −−−−−−−→ 𝐺𝐹𝐺𝐹𝐺 (𝐵) −−−−−−−−→ 𝐺𝐹𝐺 (𝐵) −−−−→ 𝐺 (𝐵) is the identity 1𝐺 (𝐵) . First note that by naturality of 𝜂 we have 𝐺𝐹𝐺 (𝜀𝐵−1 ) ◦ 𝜂𝐺 (𝐵) = 𝜂𝐺𝐹𝐺 (𝐵) ◦ 𝐺 (𝜀𝐵−1 ), and 𝐺𝐹 (𝜂𝐺 (𝐵) ) ◦ 𝜂𝐺 (𝐵) = 𝜂𝐺𝐹𝐺 (𝐵) ◦ 𝜂𝐺 (𝐵) . Since 𝜂𝐺 (𝐵) is an isomorphism, it follows from the second equation that 𝐺𝐹 (𝜂𝐺 (𝐵) ) = 𝜂𝐺𝐹𝐺 (𝐵) . Thus the composite map above is equal to 𝐺 (𝜀𝐵 ) ◦ 𝐺𝐹 (𝜂𝐺−1(𝐵) ) ◦ 𝐺𝐹𝐺 (𝜀𝐵−1 ) ◦ 𝜂𝐺 (𝐵) = 𝐺 (𝜀𝐵 ) ◦ 𝐺𝐹 (𝜂𝐺 (𝐵) ) −1 ◦ 𝜂𝐺𝐹𝐺 (𝐵) ◦ 𝐺 (𝜀𝐵−1 ) = 𝐺 (𝜀𝐵 ) ◦ 𝐺𝐹 (𝜂𝐺 (𝐵) ) −1 ◦ 𝐺𝐹 (𝜂𝐺 (𝐵) ) ◦ 𝐺 (𝜀𝐵 ) −1 = 1𝐺 (𝐵) , as desired. Note that since 𝜂𝐺 (𝐵) is an isomorphism, it follows that 𝐺 (𝜀𝐵0 ) = 𝜂𝐺−1(𝐵) . Now we are ready to show that 𝜂 : 1𝒜 ⇒ 𝐺𝐹 is such that 𝜂𝐴 : 𝐴 → 𝐺𝐹 (𝐴) is initial in (𝐴 ⇒ 𝐺) for all 𝐴 ∈ 𝒜. Let 𝐴 ∈ 𝒜 be arbitrary. Given (𝐵, ℎ) ∈ (𝐴 ⇒ 𝐺) we want to show that there is a unique arrow 𝑞 : (𝐹 (𝐴), 𝜂𝐴 ) → (𝐵, ℎ) in (𝐴 ⇒ 𝐺). That is, we want a unique 𝑞 : 𝐹 (𝐴) → 𝐵 such that 𝐺 (𝑞) ◦ 𝜂𝐴 = ℎ. To show uniqueness assume that 𝑞 satisfies the given equation. Then 𝐺 (𝑞) = ℎ ◦ 𝜂𝐴−1 = 𝜂𝐺−1(𝐵) ◦ 𝐺𝐹 (ℎ) = 𝐺 (𝜀𝐵0 ) ◦ 𝐺𝐹 (ℎ) = 𝐺 (𝜀𝐵0 ◦ 𝐹 (ℎ)). Since 𝐺 is full and faithful (Exercise 1.3.32) we have 𝑞 = 𝜀𝐵0 ◦ 𝐹 (ℎ), which shows that 𝑞 is unique. It remains to see that defining 𝑞 = 𝜀𝐵0 ◦ 𝐹 (ℎ) gives indeed an arrow (𝐹 (𝐴), 𝜂𝐴 ) → (𝐵, ℎ), i.e. that the diagram 𝜂𝐴

𝐴 ℎ

𝐺𝐹 (𝐴) 𝐺 (𝑞)=𝐺 (𝜀𝐵0 )◦𝐺𝐹 (ℎ)

𝐺 (𝐵) commutes. This holds since 𝐺 (𝜀𝐵0 ) = 𝜂𝐺−1(𝐵) and 𝜂 is natural. We conclude, by Theorem 2.3.6, that 𝐹 is left adjoint to 𝐺 . 29

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Adjunctions via initial objects

Exercise 2.3.11. For the sake of contradiction suppose that there exists a set 𝑆 such that 𝜂𝑆 : 𝑆 → 𝑈 𝐹 (𝑆) is not injective. Then there exist 𝑥, 𝑦 ∈ 𝑆 such that 𝑥 ≠ 𝑦 and 𝜂𝑆 (𝑥) = 𝜂𝑆 (𝑦). We will prove that this implies that for all 𝑇 ∈ Set, the map 𝜂𝑇 : 𝑇 → 𝑈 𝐹 (𝑇 ) is constant. So let 𝑇 ∈ Set. If |𝑇 | ≤ 1 there is nothing to prove, so assume |𝑇 | ≥ 2. Let 𝑡 1 ≠ 𝑡 2 ∈ 𝑇 . Let 𝑓 : 𝑆 → 𝑇 be any function sending 𝑥 to 𝑡 1 and 𝑦 to 𝑡 2 . By naturality of 𝜂, the diagram 𝜂𝑆

𝑈 𝐹 (𝑆)

𝑆

𝑈 𝐹 (𝑓 )

𝑓

𝑇

𝑈 𝐹 (𝑇 )

𝜂𝑇

commutes. Since 𝜂𝑆 (𝑥) = 𝜂𝑆 (𝑦), we have 𝜂𝑇 (𝑡 1 ) = 𝜂𝑇 (𝑓 (𝑥)) = 𝜂𝑇 (𝑓 (𝑦)) = 𝜂𝑇 (𝑡 2 ). It follows that 𝜂𝑇 : 𝑇 → 𝑈 𝐹 (𝑇 ) is constant, as asserted. Now let 𝐴 ∈ 𝒜 be such that 𝑈 (𝐴) has at least two elements, say 𝑏 1 ≠ 𝑏 2 ∈ 𝑈 (𝐴). Let 𝑓 : 𝑈 (𝐴) → 𝑈 (𝐴) be any map such that 𝑓 (𝑏 1 ) = 𝑏 2 and 𝑓 (𝑏 2 ) = 𝑏 1 . By Lemma 2.3.5 there exists a unique 𝑞 : 𝐹𝑈 (𝐴) → 𝐴 such that the diagram 𝑈 (𝐴)

𝜂𝑈 (𝐴)

𝑈 𝐹𝑈 (𝐴) 𝑈 (𝑞)

𝑓

𝑈 (𝐴) commutes. But 𝜂𝑈 (𝐴) is constant, say with value 𝑧 ∈ 𝑈 𝐹𝑈 (𝐴). Then 𝑈 (𝑞) (𝑧) = 𝑓 (𝑏 1 ) = 𝑏 2 and 𝑈 (𝑞)(𝑧) = 𝑓 (𝑏 2 ) = 𝑏 1, a contradiction. It follows that 𝜂𝑆 : 𝑆 → 𝑈 𝐹 (𝑆) is injective for all 𝑆 ∈ Set. In the case of the usual adjunction 𝐹 a 𝑈 between Grp and Set, where 𝑈 : Grp → Set is the forgetful functor and 𝐹 : Set → Grp is the free functor, this means that for each set 𝑆, the underlying set of the free group 𝐹 (𝑆) contains 𝑆. Exercise 2.3.12. Define a functor 𝐺 : Par → Set∗ as follows. For 𝐴 ∈ Par let 𝐺 (𝐴) = 𝐴 q {∗} with basepoint ∗, i.e. we take the (disjoint) union of 𝐴 and the one-point set {∗} and take as basepoint the point the unique element of the latter. If (𝑆, 𝑓 ) : 𝐴 → 𝐵 is a map in Par let 𝐺 (𝑆, 𝑓 ) : 𝐴 q {∗} → 𝐵 q {∗} ( 𝑓 (𝑎), 𝑎 ∈ 𝑆, 𝑎 ↦→ ∗, 𝑎 ∉ 𝑆. If (𝑆, 𝑓 ) : 𝐴 → 𝐵 and (𝑇 , 𝑓 ) : 𝐵 → 𝐶 are maps in Par then their composition is (𝑇 , 𝑔) ◦ (𝑆, 𝑓 ) = (𝑆 ∩ 𝑓 −1 (𝑇 ), 𝑔 ◦ 𝑓 |𝑆∩𝑓 −1 (𝑇 ) ). Thus 𝐺 ((𝑇 , 𝑔) ◦ (𝑆, 𝑓 )) and 𝐺 (𝑇 , 𝑔) ◦ 𝐺 (𝑆, 𝑓 ) both send an element 𝑎 ∈ 𝑆 ∩ 𝑓 −1 (𝑇 ) to 𝑔𝑓 (𝑎) ∈ 𝐶, and an element 𝑎 ∉ 𝑆 ∩ 𝑓 −1 (𝑇 ) to ∗ ∈ 𝐶. Thus 𝐺 : Par → Set∗ is a functor. We will show that 𝐺 is an equivalence. First we show that it is full and faithful. Given 𝐴, 𝐵 ∈ Par we want to prove that the map 𝐺 (−)

Par(𝐴, 𝐵) −−−−→ Set∗ (𝐴 q ∗, 𝐵 q ∗) 30

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3. Interlude on sets is a bijection. Given any pointed map 𝑓 : 𝐴 q ∗ → 𝐵 q ∗, it restricts to a map 𝑓 0 : 𝐴 → 𝐵, and 𝐺 (𝐴, 𝑓 0) = 𝑓 . Thus 𝐺 (−) is surjective. Now assume that 𝐺 (𝑆, 𝑓 ) = 𝐺 (𝑇 , 𝑔) for (𝑆, 𝑓 ), (𝑇 , 𝑔) : 𝐴 → 𝐵. Then both 𝑆 and 𝑌 are the complement of 𝐺 (𝑆, 𝑓 ) −1 (∗) = 𝐺 (𝑇 , 𝑔) −1 (∗) in 𝐴 q {∗}, so 𝑆 = 𝑇 . Moreover, 𝑓 = 𝑔 on 𝑆 = 𝑇 by definition of 𝐺 on morphisms. Thus 𝐺 (−) is injective, hence a bijection. It remains to prove that 𝐺 is essentially surjective on objects: if (𝐴, 𝑎) is any pointed set then 𝐺 (𝐴 \ 𝑎) = ((𝐴 \ 𝑎) q {∗}, ∗)  (𝐴, 𝑎) as pointed sets via the bijection 𝑓 : 𝐺 (𝐴 \ 𝑎) → (𝐴, 𝑎) whose restriction to 𝐴 \ 𝑎 is the identity and such that 𝑓 (∗) = 𝑎. We conclude that 𝐺 is an equivalence. Now we describe Set∗ as a coslice category. Let {∗} ∈ Set be the one-point set and consider the coslice category {∗}/Set = ({∗} ⇒ Set). It has objects pairs (𝐴, ℎ) where ℎ : ∗ → 𝐴, and a morphism (𝐴, ℎ) → (𝐵, 𝑘) is a map 𝑓 : 𝐴 → 𝐵 such that the diagram ℎ

{∗}

𝐴 𝑓 𝑘

𝐵 commutes. A map ℎ : {∗} → 𝐴 is just a choice of basepoint ℎ(∗) ∈ 𝐴, and a morphism (𝐴, ℎ) → (𝐵, 𝑘) is precisely a basepoint-preserving map 𝑓 : (𝐴, ℎ(∗)) → (𝐵, 𝑘 (∗)). Thus {∗}/Set  Set∗ .

3 3.1

Interlude on sets Constructions with sets

Exercise 3.1.1. First we find a left adjoint 𝐹 to Δ. We want 𝐹 : Set × Set → Set such that for all 𝑋, 𝑌, 𝑍 ∈ Set, a map 𝐹 (𝑋, 𝑌 ) → 𝑍 is the same thing as a pair of maps 𝑋 → 𝑍, 𝑌 → 𝑍 . On objects define 𝐹 (𝑋, 𝑌 ) = 𝑋 q𝑌 . On morphisms, send a pair of maps 𝑓 : 𝑋 → 𝑋 0, 𝑔 : 𝑌 → 𝑌 0 to 𝐹 (𝑓 , 𝑔) = 𝑓 q 𝑔; that is, 𝐹 (𝑓 , 𝑔) is the unique map 𝑋 q 𝑌 → 𝑋 0 q 𝑌 0 whose restriction to 𝑋 is 𝑓 (followed by the inclusion 𝑋 0 → 𝑋 0 q 𝑌 0) and whose restriction to 𝑌 is 𝑔 (followed by the inclusion 𝑌 0 → 𝑋 0 q 𝑌 0). For any 𝑋, 𝑌 , 𝑍 ∈ Set we then have a canonical map 𝜑𝑋 ,𝑌 ,𝑍 : Set(𝑋 q 𝑌 , 𝑍 ) → Set × Set((𝑋, 𝑌 ), (𝑍, 𝑍 )) given by 𝑝 ↦→ (𝑝 |𝑋 , 𝑝 |𝑌 ), which is a bijection with inverse given by sending a pair of maps 𝑓 : 𝑋 → 𝑍, 𝑔 : 𝑌 → 𝑍 to the unique map 𝑝 : 𝑋 q 𝑌 → 𝑍 whose restriction to 𝑋 is 𝑝𝑋 = 𝑓 and whose restriction to 𝑌 is 𝑝𝑌 = 𝑔. Naturality amounts to proving that given maps of sets 𝑓 : 𝑋 0 → 𝑋, 𝑔 : 𝑌 0 → 𝑌 and ℎ : 𝑍 → 𝑍 0, the diagram Set(𝑋 q 𝑌 , 𝑍 )

𝜑𝑋 ,𝑌 ,𝑍

Set × Set((𝑋, 𝑌 ), (𝑍, 𝑍 ))

(𝑓 q𝑔) ∗ ◦ℎ ∗

(𝑓 ,𝑔) ∗ ◦(ℎ,ℎ)∗

Set(𝑋 0 q 𝑌 0, 𝑍 0) 𝜑𝑋 0,𝑌 0,𝑍 0 Set × Set((𝑋 0, 𝑌 0), (𝑍 0, 𝑍 0)) 31

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3.1

Constructions with sets

commutes. By definition, if 𝑝 : 𝑋 q 𝑌 → 𝑍 is a map, then both paths in the above diagram send 𝑝 to the pair of maps (ℎ ◦ 𝑝 |𝑋 ◦ 𝑓 , ℎ ◦ 𝑝 |𝑌 ◦ 𝑔). It follows that 𝜑𝑋 ,𝑌 ,𝑍 is a natural isomorphism and therefore 𝐹 a Δ. We now find a right adjoint to Δ. This is a functor 𝐺 : Set × Set → Set such that for all 𝑋, 𝑌, 𝑍 ∈ Set, a map 𝑍 → 𝐺 (𝑋, 𝑌 ) is the same thing as a pair of maps 𝑍 → 𝑋, 𝑍 → 𝑌 . Define 𝐺 (𝑋, 𝑌 ) = 𝑋 × 𝑌 on objects and 𝐺 (𝑓 , 𝑔) = 𝑓 × 𝑔 on morphisms. Given 𝑋, 𝑌 , 𝑍 ∈ Set there is a canonical map 𝜓𝑋 ,𝑌 ,𝑍 : Set × Set((𝑍, 𝑍 ), (𝑋, 𝑌 )) → Set(𝑍, 𝑋 × 𝑌 ) sending 𝑓 : 𝑍 → 𝑋, 𝑔 : 𝑍 → 𝑌 to the map 𝑘 : 𝑍 → 𝑋 ×𝑌 given by 𝑘 (𝑧) = (𝑓 (𝑧), 𝑔(𝑧)), and this is a bijection with inverse sending 𝑞 : 𝑍 → 𝑋 ×𝑌 to the pair of maps (𝜋𝑋 ◦𝑞 : 𝑍 → 𝑋, 𝜋𝑌 ◦𝑞 : 𝑍 → 𝑌 ), where 𝜋𝑋 and 𝜋𝑌 are the projections. If 𝑓 : 𝑋 → 𝑋 0, 𝑔 : 𝑌 → 𝑌 0 and ℎ : 𝑍 0 → 𝑍 are maps of sets then the diagram Set × Set((𝑍, 𝑍 ), (𝑋, 𝑌 ))

𝜓𝑋 ,𝑌 ,𝑍

(ℎ×ℎ) ∗ ◦( 𝑓 ×𝑔)∗

Set(𝑍, 𝑋 × 𝑌 ) ℎ ∗ ◦( 𝑓 ×𝑔)∗

Set × Set((𝑍 0, 𝑍 0), (𝑋 0, 𝑌 0))

𝜓𝑋 0,𝑌 0,𝑍 0

Set(𝑍 0, 𝑋 0 × 𝑌 0)

commutes, for if (𝑝, 𝑞) is a pair of maps 𝑝 : 𝑍 → 𝑋, 𝑞 : 𝑍 → 𝑌 , then both paths in the above diagram send (𝑝, 𝑞) to the map 𝑘 : 𝑍 0 → 𝑋 0 × 𝑌 0 given by 𝑘 (𝑧) = (𝑓 ◦ 𝑝 ◦ ℎ(𝑧), 𝑔 ◦ 𝑞 ◦ ℎ(𝑧)). Thus 𝜓𝑋 ,𝑌 ,𝑍 is a bijection natural in 𝑋, 𝑌 , 𝑍, and we conclude that Δ a 𝐺 . Exercise 3.1.2. (This is also Mac Lane’s Exercise I.5.8.) Let 𝒞 be the category with objects triples (𝑋, 𝑒, 𝑡) where 𝑋 is a non-empty set, 𝑒 ∈ 𝑋 and 𝑡 : 𝑋 → 𝑋 is a map. If (𝑋, 𝑒, 𝑡), (𝑋 0, 𝑒 0, 𝑡 0) are two objects, a map (𝑋, 𝑒, 𝑡) → (𝑋 0, 𝑒 0, 𝑡 0) is defined to be function 𝑓 : 𝑋 → 𝑋 0 such that 𝑓 (𝑒) = 𝑒 0 and 𝑓 ◦𝑡 = 𝑡 0 ◦ 𝑓 . Then the triple (N, 0, 𝑠) is initial in 𝒞. Indeed, given a non-empty set 𝑌 , an element 𝑦0 ∈ 𝑌 and a function 𝑡 : 𝑌 → 𝑌 , we shall prove there exists a unique function 𝑓 : N → 𝑌 such that 𝑓 (0) = 𝑦0 and the diagram 𝑓

N

𝑌

𝑠

N

𝑡

𝑌 𝑓

commutes. This is constructed by induction, starting with 𝑓 (0) = 𝑦0 . If 𝑓 (𝑛) has been defined for some 𝑛 ∈ N, define 𝑓 (𝑛 + 1) = 𝑓 (𝑠 (𝑛)) = 𝑡 (𝑓 (𝑛)). This shows that 𝑓 indeed exists and is unique.

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3.2

3.2

Small and large categories

Small and large categories

Exercise 3.2.12. (a) We will prove 𝜃 (𝑆) = 𝑆. On the one hand, © Ø ª Ø ­ ® 𝜃 (𝑆) = 𝜃 ­ 𝑅® = 𝜃 (𝑅) ⊂ ­ ® 𝑅 ∈𝒫 (𝐴) 𝑅 ∈𝒫 (𝐴) « 𝑅 ⊂𝜃 (𝑅) ¬ 𝑅 ⊂𝜃 (𝑅)

Ø

𝑅 = 𝑆.

𝑅 ∈𝒫 (𝐴) 𝑅 ⊂𝜃 (𝑅)

On the other hand, for each 𝑅 ∈ 𝒫(𝐴) such that 𝑅 ⊂ 𝜃 (𝑅) we have 𝜃 (𝑅) ⊂ 𝜃 2 (𝑅) since 𝜃 is order-preserving, so that 𝑅 0 = 𝜃 (𝑅) in turn satisfies 𝑅 0 ⊂ 𝜃 (𝑅 0). It follows that 𝑆 ⊂ 𝜃 (𝑆) and therefore 𝜃 (𝑆) = 𝑆. (b) Let 𝜃 : 𝒫(𝐴) → 𝒫(𝐴) be given by 𝜃 (𝑆) = 𝐴 \ 𝑔(𝐵 \ 𝑓 (𝑆)) for all 𝑆 ⊂ 𝐴. We claim that 𝜃 is order-preserving. Let 𝑆 ⊂ 𝑆 0 ⊂ 𝐴. Then 𝑓 (𝑆) ⊂ 𝑓 (𝑆 0), so that 𝐵 \ 𝑓 (𝑆 0) ⊂ 𝐵 \ 𝑓 (𝑆). Thus 𝑔(𝐵 \ 𝑓 (𝑆 0)) ⊂ 𝑔(𝐵 \ 𝑓 (𝑆)) and therefore 𝜃 (𝑆) ⊂ 𝜃 (𝑆 0). It follows from part (a) that there exists 𝑆 ∈ 𝒫(𝐴) such that 𝜃 (𝑆) = 𝑆, that is, 𝑔(𝐵 \ 𝑓 (𝑆)) = 𝐴 \ 𝑆. (c) Let 𝐴 and 𝐵 be sets and assume that |𝐴| ≤ |𝐵| ≤ |𝐴|. We want to find a bijection ℎ : 𝐴 → 𝐵. By assumption, there exists injective maps 𝑓 : 𝐴 → 𝐵 and 𝑔 : 𝐵 → 𝐴. Let 𝑆 be a subset of 𝐴 such that 𝑔(𝐵 \ 𝑓 (𝑆)) = 𝐴 \ 𝑆, which exists by part (a). Define ( 𝑓 (𝑎) 𝑎 ∈ 𝑆, ℎ : 𝐴 → 𝐵, 𝑎 ↦→ −1 𝑔 (𝑎) 𝑎 ∈ 𝐴 \ 𝑆. Note that this is well-defined since 𝐴 \𝑆 is contained in the image of 𝑔. It remains to show that ℎ is a bijection. Let 𝑎, 𝑎 0 ∈ 𝐴 and suppose that ℎ(𝑎) = ℎ(𝑎 0). If 𝑎 and 𝑎 0 both belong to 𝑆, then 𝑎 = 𝑎 0 by injectivity of 𝑓 . Similarly 𝑎 = 𝑎 0 if both 𝑎 and 𝑎 0 belong to 𝐴\𝑆. If 𝑎 ∈ 𝑆 and 𝑎 0 ∈ 𝐴\𝑆 = 𝑔(𝐵 \ 𝑓 (𝑆)), then 𝑓 (𝑎) = 𝑔−1 (𝑎 0) = 𝑏 for some 𝑏 ∈ 𝐵 \ 𝑓 (𝑆), a contradiction. It follows that ℎ is injective. Now let 𝑏 ∈ 𝐵 be arbitrary; we want to prove that 𝑏 = ℎ(𝑎) for some 𝑎 ∈ 𝐴. If 𝑏 ∈ 𝑓 (𝐴) this is clear. If 𝑏 ∉ 𝑓 (𝐴), then 𝑔(𝑏) ∈ 𝑔(𝐵 \ 𝑓 (𝑆)) = 𝐴 \𝑆, so that ℎ(𝑔(𝑏)) = 𝑏. It follows that ℎ is bijective. We conclude that 𝐴  𝐵. Exercise 3.2.13. (a) Suppose that 𝑓 is surjective. Let 𝐵 = {𝑎 ∈ 𝐴 | 𝑎 ∉ 𝑓 (𝑎)} ∈ 𝒫(𝐴). By assumption, there exists 𝑎 ∈ 𝐴 such that 𝑓 (𝑎) = 𝐵. Then either 𝑎 ∈ 𝑓 (𝑎) = 𝐵 or 𝑎 ∉ 𝑓 (𝑎) = 𝐵, but either case leads to a contradiction. It follows that 𝑓 is not surjective. (b) Since the function 𝐴 → 𝒫(𝐴) given by 𝑎 ↦→ {𝑎} is injective, we have |𝐴| ≤ |𝒫(𝐴)|. As there is no surjective function 𝐴 → 𝒫(𝐴), it follows that |𝐴| < |𝒫(𝐴)|. Exercise 3.2.14. (a) Let 𝐼 be a set and (𝐴𝑖 )𝑖 ∈𝐼 a family of objects of 𝒜. Then the set ! Ø 𝑈 (𝐴𝑖 ) 𝑆=𝒫 𝑖 ∈𝐼

33

Solutions by positrón0802

3.2

Small and large categories

satisfies |𝑈 (𝐴𝑖 )| < |𝑆 | for all 𝑖 ∈ 𝐼 . By Exercise 2.3.11, the unit map 𝑆 → 𝑈 𝐹 (𝑆) is injective, so that |𝑈 (𝐴𝑖 )| < |𝑆 | ≤ |𝑈 𝐹 (𝑆)|. It follows that 𝐹 (𝑆) is an object of 𝒜 which is not isomorphic to 𝐴𝑖 for any 𝑖 ∈ 𝐼, for an isomorphism 𝐹 (𝑆)  𝐴𝑖 would induce an isomorphism 𝑈 𝐹 (𝑆)  𝑈 (𝐴𝑖 ). (b) It follows from part (a) that the class of isomorphism classes of objects of 𝒜 is large, so 𝒜 is not essentially small. (c) For all of the categories Set, Vect𝑘 , Grp, Ab, Ring, and Top we have a forgetful functor 𝑈 into the category Set (whose left adjoint is a free functor) which satisfies the assumption of (a). It follows from (b) that none of these categories is essentially small. Exercise 3.2.15. (a) The category Mon is not even essentially small by Exercise 3.2.14(b), as the forgetful functor 𝑈 : Mon → Set satisfies the assumption of Exercise 3.2.14(a). It is locally small. (b) The group Z viewed as a one-object category is small as it has only one object with Z as set of morphisms. (c) The ordered set of integers Z has a set of objects and at most one arrow between any two objects, hence it is small. (d) The category Cat of small categories is not even essentially small by Exercise 3.2.14(b), by considering the functor 𝑂 : Cat → Set sending a small category to its set of objects and applying Exercise 3.2.16. Nevertheless, Cat is locally small: if 𝒞 and 𝒟 are small categories with sets of objects ob(𝒞) and ob(𝒟) and sets of morphisms mor(𝒞) and mor(𝒟), respectively, then the class of functors 𝒞 → 𝒟 is a subset of the set Set(ob(𝒞), ob(𝒟)) × Set(mor(𝒞), mor(𝒟)), so it is a set. (d) The multiplicative monoid of cardinals has one object whose class of morphisms is the monoid of cardinals. This class is not a set, for given a set of cardinals {𝜅𝑖 }𝑖 ∈𝐼 there exists a cardinal 𝜅 such that 𝜅 > 𝜅𝑖 for all 𝑖 ∈ 𝐼 . Therefore this category is not locally small. Exercise 3.2.16. First we find a right adjoint 𝐼 to 𝑂. Given a set 𝑆, let 𝐼 (𝑆) be the indiscrete category whose set of objects is 𝑆. That is, for any two 𝑠, 𝑠 0 ∈ 𝑆, there is precisely one arrow 𝑠 → 𝑠 0 . If ℎ : 𝑆 → 𝑆 0 is a function of sets, let 𝐼 (ℎ) : 𝐼 (𝑆) → 𝐼 (𝑆 0) have object function ℎ and send the unique morphism 𝑠 → 𝑠 0 in 𝐼 (𝑆) to the unique morphism ℎ(𝑠) → ℎ(𝑠 0) in 𝐼 (𝑆 0) for all 𝑠, 𝑠 0 ∈ 𝑆. Then 𝐼 : Set → Cat is a functor. We claim 𝑂 a 𝐼 . We need a bijection 𝜑 𝒞,𝑆 : Set(𝑂 (𝒞), 𝑆) → Cat(𝒞, 𝐼 (𝑆)) natural in 𝒞 ∈ Cat and 𝑆 ∈ Set. Let 𝜑 𝒞,𝑆 send a function ℎ : 𝑂 (𝒞) → 𝑆 to the functor 𝒞 → 𝐼 (𝑆) whose object function is ℎ and sending a morphism 𝑐 → 𝑐 0 in 𝒞 to the unique morphism ℎ(𝑐) → ℎ(𝑐 0) in 𝐼 (𝑆). Then 𝜑 𝒞,𝑆 is bijective with inverse given by sending a functor 𝒞 → 𝐼 (𝑆) to its object function. It remains to prove that 𝜑 𝒞,𝑆 is natural in 𝒞 and 𝑆. So let 𝐹 : 𝒞 0 → 𝒞 be a functor between small categories, ℎ : 𝑆 → 𝑆 0 be a function between sets and consider the diagram Set(𝑂 (𝒞), 𝑆)

𝜑 𝒞,𝑆

Set(𝑂 (𝐹 ),ℎ)

Set(𝑂 (𝒞 0), 𝑆 0)

Cat(𝒞, 𝐼 (𝑆)) Cat(𝐹,𝐼 (ℎ))

𝜑 𝒞0,𝑆 0

34

Cat(𝒞 0, 𝐼 (𝑆 0)) . Solutions by positrón0802

3.2

Small and large categories

If 𝑘 : 𝑂 (𝒞) → 𝑆 is a function, then both paths in the above diagram send 𝑘 to the functor 𝒞 0 → 𝐼 (𝑆 0) whose object function is ℎ ◦ 𝑘 ◦ 𝑂 (𝐹 ) and whose morphism function sends an arrow 𝑐 → 𝑐 0 in 𝒞 0 to the unique arrow ℎ ◦ 𝑘 ◦ 𝐹 (𝑐) → ℎ ◦ 𝑘 ◦ 𝐹 (𝑐 0) in 𝐼 (𝑆 0). It follows that the bijection 𝜑 𝒞,𝑆 is natural in 𝒞 ∈ Cat and 𝑆 ∈ Set, and therefore 𝑂 a 𝐼 . We now find a left adjoint 𝐷 to 𝑂. Given a set 𝑆, let 𝐷 (𝑆) be the discrete category whose set of objects is 𝑆 (that is, the only morphisms are the identity morphisms). This gives a functor 𝐷 : Set → Cat. To show that 𝐷 a 𝑂, consider, for 𝑆 ∈ Set and 𝒞 ∈ Cat, the function 𝜑𝑆,𝒞 : Cat(𝐷 (𝑆), 𝒞) → Set(𝑆, 𝑂 (𝒞)) given by sending a functor 𝐷 (𝑆) → 𝒞 to its object function. Then 𝜑𝑆,𝒞 is clearly bijective. Moreover, if 𝐹 : 𝒞 → 𝒞 0 is a functor between small categories and ℎ : 𝑆 0 → 𝑆 a set function, then the diagram Cat(𝐷 (𝑆), 𝒞)

𝜑𝑆,𝒞

Cat(𝐷 (ℎ),𝐹 )

Cat(𝐷 (𝑆 0), 𝒞 0)

Set(𝑆, 𝑂 (𝒞)) Set(ℎ,𝑂 (𝐹 ))

𝜑𝑆 0,𝒞0

Set(𝑆 0, 𝑂 (𝒞 0))

commutes. Indeed, given a functor 𝐺 : 𝐷 (𝑆) → 𝒞, both paths in the above diagrams send 𝐺 to the function 𝑂 (𝐹 ) ◦ 𝑂 (𝐺) ◦ ℎ : 𝑆 0 → 𝑂 (𝒞 0). It follows that 𝜑𝑆,𝒞 is a bijection natural in 𝑆 ∈ Set and 𝒞 ∈ Cat, so that 𝐷 a 𝑂. Finally, we find a left adjoint 𝐶 to 𝐷. Given a small category 𝒞, let 𝐶 (𝒞) be the quotient of the set of objects of 𝒞 by the equivalence relation generated by 𝑐 ∼ 𝑐 0 if there exists an arrow 𝑐 → 𝑐 0 in 𝒞. For 𝑐 ∈ 𝒞, write [𝑐] ∈ 𝐶 (𝒞) for its equivalence class. If 𝐹 : 𝒞 → 𝒞 0 is a functor between small categories and 𝑐 → 𝑐 0 is a morphism in 𝒞, then 𝐹 (𝑐) → 𝐹 (𝑐 0) is a morphism in 𝒞 0, hence the object function of 𝐹 descends to a function 𝐶 (𝐹 ) : 𝐶 (𝒞) → 𝐶 (𝒞 0). Thus 𝐶 : Cat → Set is a functor. For 𝐶 ∈ Cat and 𝑆 ∈ Set, consider the function 𝜑 𝒞,𝑆 : Set(𝐶 (𝒞), 𝑆) → Cat(𝒞, 𝐷 (𝑆)) sending a map ℎ : 𝐶 (𝒞) → 𝑆 to the functor 𝒞 → 𝐷 (𝑆) whose object function sends an element 𝑐 ∈ 𝒞 to ℎ[𝑐] ∈ 𝑆, and whose morphism function sends an arrow 𝑐 → 𝑐 0 in 𝒞 to the identity arrow of ℎ[𝑐] = ℎ[𝑐 0] ∈ 𝑆. If 𝐹 : 𝒞 → 𝐷 (𝑆) is a functor, then for every arrow 𝑐 → 𝑐 0 in 𝒞 the arrow 𝐹 (𝑐 → 𝑐 0) is an identity arrow, so in particular 𝐹 (𝑐) = 𝐹 (𝑐 0) ∈ 𝑆. Thus the object function of 𝐹 descends to a map 𝐶 (𝒞) → 𝑆. This assignment gives an inverse for 𝜑 𝒞,𝑆 , which is therefore bijective. It remains to prove naturality on 𝒞 ∈ Cat and 𝑆 ∈ Set, which amounts to commutativity of the diagram Set(𝐶 (𝒞), 𝑆)

𝜑 𝒞,𝑆

Set(𝐶 (𝐹 ),ℎ)

Set(𝐶 (𝒞 0), 𝑆 0)

Cat(𝒞, 𝐷 (𝑆)) Cat(𝐹,𝐷 (ℎ))

𝜑 𝒞0,𝑆 0

35

Cat(𝒞 0, 𝐷 (𝑆 0)) Solutions by positrón0802

3.3

Historical remarks

for all 𝐹 : 𝒞 0 → 𝒞 and ℎ : 𝑆 → 𝑆 0 . If 𝑘 : 𝐶 (𝒞) → 𝑆 is a function, both compositions above send 𝑘 to the functor 𝒞 0 → 𝐷 (𝑆 0) whose object function is the composite map 𝑂 (𝐹 )

𝑞

𝑘

ℎ

𝑂 (𝒞 0) −−−−→ 𝑂 (𝒞) → − 𝐶 (𝒞) → − 𝑆→ − 𝑆 0, where 𝑞 is the quotient map, and whose morphism function sends an arrow 𝑐 → 𝑐 0 in 𝒞 0 to the identity arrow of ℎ ◦ 𝑘 [𝐹 (𝑐)] = ℎ ◦ 𝑘 [𝐹 (𝑐 0)] ∈ 𝑆 0 . It follows that 𝜑 𝒞,𝑆 is natural in 𝒞 and 𝑆, so that 𝐶 a 𝐷. This completes the chain of adjoint functors 𝐶 a 𝐷 a 𝑂 a 𝐼.

3.3

Historical remarks No Exercises.

4 4.1

Representables Definitions and examples

Exercise 4.1.26. • Let 𝒞 be the category of CW complexes, where for 𝑋, 𝑌 ∈ 𝒞, the set 𝒞(𝑋, 𝑌 ) is that of homotopy classes of maps from 𝑋 to 𝑌 . Then, given a positive integer 𝑛 and a group 𝐺, there exists a space 𝐾 (𝐺, 𝑛) representing the functor 𝑈 ◦𝐻 𝑛 (−; 𝐺) : Tophop → Set sending a CW complex to (the underlying set of) its 𝑛th cohomology group with values in 𝐺 . The space 𝐾 (𝐺, 𝑛) is usually called an Eilenberg-MacLane space. • Let 𝒞 denote the category of connected pointed CW complexes, where for 𝑋, 𝑌 ∈ 𝒞, the set 𝒞(𝑋, 𝑌 ) is that pointed homotopy classes of pointed maps from 𝑋 to 𝑌 . A functor 𝐸e∗ : 𝒞 op → AbZ from 𝒞 into the category AbZ of Z-graded abelian groups is called a cohomology theory. Given a positive integer 𝑛, the composition of 𝐸e∗ with the canonical projection AbZ → Ab, 𝐴 ↦→ 𝐴𝑛 , gives a functor 𝐸f𝑛 : 𝒞 op → Ab. Brown’s representability theorem asserts that 𝐸e∗ and any positive integer 𝑛, there exists 𝐾𝑛 ∈ 𝒞 such that 𝐸f𝑛 is represented by 𝐾𝑛 . • Generalising Example 4.1.8 of the fundamental group 𝜋 1, for all 𝑛 ≥ 2 the composition of 𝑛th homotopy group functor 𝜋 𝑛 : Toph∗ → Ab with the forgetful functor 𝑈 : Grp → Set is represented by (𝑆 𝑛 , ∗) ∈ Toph∗ . • Let 𝐹 : Mat → Set be the functor sending 𝑛 ↦→ R𝑛 , and 𝐴 ∈ Mat(𝑛, 𝑚) to the map R𝑛 → R𝑚 given by left multiplication by the 𝑚 × 𝑛 matrix 𝐴. Then R𝑚 = 𝐹 (𝑚)  Mat(1, 𝑚) naturally in 𝑚, i.e. 𝐹 is represented by 1 ∈ Mat.

36

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4.1

Definitions and examples

Exercise 4.1.27. By assumption, there exists a bijection 𝜑 𝐵 : 𝒜(𝐵, 𝐴) → 𝒜(𝐵, 𝐴 0) natural in 𝐵 ∈ 𝒜. By considering 𝐵 = 𝐴 we have a map 𝑓 = 𝜑𝐴 (1𝐴 ) : 𝐴 → 𝐴 0, and by considering 𝐵 = 𝐴 0 we have a map 𝑔 = 𝜑𝐴−10 (1𝐴0 ) : 𝐴 0 → 𝐴. By naturality, there following diagrams are commutative: 𝒜(𝐴 0, 𝐴)

𝜑𝐴0

𝑓∗

𝒜(𝐴, 𝐴)

𝜑𝐴

𝒜(𝐴 0, 𝐴 0)

𝒜(𝐴, 𝐴)

𝑓∗ 𝜑𝐴

𝑔∗

𝒜(𝐴, 𝐴 0) ,

𝒜(𝐴, 𝐴 0) 𝑔∗

𝒜(𝐴 0, 𝐴)

𝜑𝐴0

𝒜(𝐴 0, 𝐴 0) .

Thus 𝜑𝐴 (𝑔◦ 𝑓 ) = 𝜑𝐴 ◦ 𝑓 ∗ (𝑔) = 𝑓 ∗ ◦𝜑𝐴0 (𝑔) = 𝑓 ∗ (1𝐴0 ) = 𝑓 = 𝜑𝐴 (1𝐴 ). Since 𝜑𝐴 is injective, 𝑔◦ 𝑓 = 1𝐴 . Furthermore, 𝑓 ◦ 𝑔 = 𝑔∗ ◦ 𝜑𝐴 (1𝐴 ) = 𝜑𝐴0 ◦ 𝑔∗ (1𝐴 ) = 𝜑𝐴0 (𝑔) = 1𝐴0 . It follows that 𝑓 : 𝐴 → 𝐴 0 is an isomorphism. Exercise 4.1.28. Recall that 𝑈𝑝 is given on objects by 𝑈𝑝 (𝐺) = {elements of 𝐺 of order 1 or 𝑝}. If 𝜑 : 𝐺 → 𝐺 0 is a group homomorphism then there is an induced homomorphism 𝑈𝑝 (𝜑) : 𝑈𝑝 (𝐺) → 𝑈𝑝 (𝐺 0). Indeed, if 𝑥 ∈ 𝑈𝑝 (𝐺) is not the identity then 𝑥 has order 𝑝. Thus 𝜑 (𝑥) 𝑝 = 𝜑 (𝑥 𝑝 ) = 𝜑 (1) = 1 and since 𝑝 is prime, either 𝜑 (𝑥) is the identity or has order 𝑝. For all 𝐺 ∈ Grp define 𝜂𝐺 : 𝑈𝑝 (𝐺) → Grp(Z/𝑝Z, 𝐺) 𝑥 ↦→ (𝜓𝑥 : 1 ↦→ 𝑥). Note that 𝜂𝐺 is well-defined by definition of 𝑈𝑝 (𝐺). It is of course injective. Moreover, it is surjective: any group homomorphism Z/𝑝Z → 𝐺 is determined by its value at 1, and this value must have order 1 or 𝑝, i.e. must be an element of 𝑈𝑝 (𝐺). To show that 𝜂 : 𝑈𝑝 ⇒ Grp(Z/𝑝Z, −) is natural take a group homomorphism 𝜑 : 𝐺 → 𝐺 0 and consider the diagram 𝑈𝑝 (𝐺)

𝜂𝐺

Grp(Z/𝑝Z, 𝐺)

𝑈𝑝 (𝜑)

𝑈𝑝 (𝐺 0)

𝜑∗ 𝜂𝐺 0

Grp(Z/𝑝Z, 𝐺 0) .

If 𝑥 ∈ 𝑈𝑝 (𝐺), then 𝜑 ∗ ◦𝜂𝐺 (𝑥) and 𝜂𝐺 0 ◦𝑈𝑝 (𝜑) (𝑥) both send 1 to 𝜑 (𝑥). Thus the diagram commutes. It follows that 𝜂 is a natural isomorphism 𝑈𝑝  Grp(Z/𝑝Z, −), and hence 𝑈𝑝 is represented by Z/𝑝Z. Exercise 4.1.29. Let 𝑈 denote the forgetful functor CRing → Set. By Exercise 0.13(a), for each 𝑅 ∈ CRing there is a bijective map 𝜂𝑅 : 𝑈 (𝑅) → CRing(Z[𝑥], 𝑅) given by sending an element 𝑟 ∈ 𝑅 to the unique morphism 𝜓𝑟 : Z[𝑥] → 𝑅 such that 𝜓𝑟 (𝑥) = 𝑟 . If 𝜑 : 𝑅 → 𝑅 0 is a homomorphism of commutative rings and 𝑟 ∈ 𝑅, then 𝜑 ∗ ◦ 𝜂𝑅 (𝑟 ) and 𝜂𝑅0 ◦ 𝑈 (𝜑), which are maps Z[𝑥] → 𝑅 0, both send 𝑥 to 𝜑 (𝑟 ), so by the uniqueness they are equal. Thus 𝜂 : 𝑈 ⇒ CRing(Z[𝑥], −) is a natural isomorphism. 37

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4.1

Definitions and examples

Exercise 4.1.30. Write 𝑆 = {𝑎, 𝑏} for the Sierpiński space, where {𝑎} is open. Let 𝑋 by a topological space. Given a subset 𝑈 ⊂ 𝑋 there is a function 𝑓𝑈 : 𝑋 → 𝑆 given by 𝑓𝑈 (𝑥) = 𝑎 if 𝑥 ∈ 𝑈 , and 𝑓𝑈 (𝑥) = 𝑏 if 𝑥 ∉ 𝑈 . The function 𝑓𝑈 is continuous since 𝑈 is open. Furthermore, it is clear that the map 𝜂𝑋 : 𝒪(𝑋 ) → Topop (𝑆, 𝑋 ) = Top(𝑋, 𝑆) 𝑈 ↦→ 𝑓𝑈 is bijective with inverse given by sending a continuous map 𝑓 : 𝑋 → 𝑆 to the open subset 𝑓 −1 (𝑎) of 𝑋 . Now let 𝑋, 𝑌 ∈ Top and 𝑓 ∈ Topop (𝑋, 𝑌 ) = Top(𝑌 , 𝑋 ) and consider the diagram 𝒪(𝑋 )

𝜂𝑋

𝑓∗

𝒪 (𝑓 )

𝒪(𝑌 )

Top(𝑋, 𝑆)

𝜂𝑌

Top(𝑌 , 𝑆) .

If 𝑈 ∈ 𝒪(𝑋 ), then both 𝑓 ∗ ◦ 𝜂𝑋 (𝑈 ) and 𝜂𝑌 ◦ 𝒪(𝑓 ) (𝑈 ) send all of 𝑓 −1 (𝑈 ) to 𝑎 and all of 𝑌 \ 𝑓 −1 (𝑈 ) to 𝑏. Thus the diagram commutes. It follows that 𝜂 : 𝒪 → Top(−, 𝑆) is a natural isomorphism, so that 𝒪 is represented by 𝑆. Exercise 4.1.31. Let 𝐼 be the category with only two objects 𝑖, 𝑖 0 ∈ 𝒞 and precisely one nonidentity morphism, 𝑓 : 𝑖 → 𝑖 0 . For any 𝒜 ∈ Cat, a functor 𝐹 : 𝐼 → 𝒜 consists of a pair of objects 𝐹 (𝑖) = 𝐴, 𝐹 (𝑖 0) = 𝐴 0 together with a morphism 𝐹 (𝑓 ) : 𝐴 → 𝐴 0 . Thus, for 𝒜 ∈ Cat define 𝜂 𝒜 : 𝑀 (𝒜) → Cat(𝐼, 𝒜) to be the function sending an element ℎ ∈ 𝒜(𝐴, 𝐴 0) ⊂ 𝑀 (𝒜) to the functor 𝐹 : 𝐼 → 𝒜 given by 𝐹 (𝑐) = 𝐴, 𝐹 (𝑐 0) = 𝐴 0 and 𝐹 (𝑓 ) = ℎ. Then 𝜂 𝒜 is bijective. Moreover, consider a functor 𝐺 : 𝒜 → ℬ between small categories and the diagram 𝑀 (𝒜)

𝜂𝒜

𝑀 (𝐺)

𝑀 (ℬ)

Cat(𝐼, 𝒜) 𝐺∗

𝜂ℬ

Cat(𝐼, ℬ) .

Given ℎ ∈ 𝒜(𝐴, 𝐴 0) ⊂ 𝑀 (𝒜), both 𝐺 ∗ ◦ 𝜂 𝒜 (ℎ) and 𝜂 ℬ ◦ 𝑀 (𝐺) (ℎ) are the functor 𝐼 → ℬ sending 𝑐 ↦→ 𝐺 (𝐴), 𝑐 0 ↦→ 𝐺 (𝐴 0) and 𝑓 ↦→ 𝐺 (ℎ). Thus 𝜂 : 𝑀 ⇒ Cat(𝐼, −) is a natural isomorphism. We conclude that 𝑀 is represented by 𝐼 . Exercise 4.1.32. 𝐹 is left adjoint to 𝐺 if and only if there exists bijections 𝜓𝐴,𝐵 : ℬ(𝐹 (𝐴), 𝐵)  𝒜(𝐴, 𝐺 (𝐵)) natural in 𝐴 ∈ 𝒜 and 𝐵 ∈ ℬ, which by Exercise 2.1.14 means that for all 𝑝 : 𝐴 0 → 𝐴, 𝑓 : 𝐴 → 𝐺 (𝐵) and 𝑞 : 𝐵 → 𝐵 0 we have 𝐺 (𝑞) ◦ 𝜑𝐴,𝐵 (𝑓 ) ◦ 𝑝 = 𝜑𝐴0,𝐵0 (𝑞 ◦ 𝑓 ◦ 𝐹 (𝑝)).

38

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4.2

The Yoneda lemma

This is precisely the statement that the diagram ℬ(𝐹 (𝐴), 𝐵)

𝜑𝐴,𝐵

𝒜(𝐴, 𝐺 (𝐵))

ℬ(𝐹 (𝑝),𝑞)

𝒜 (𝑝,𝐺 (𝑞))

ℬ(𝐹 (𝐴 0), 𝐵 0)

𝜑𝐴0,𝐵0

𝒜(𝐴 0, 𝐺 (𝐵 0))

commutes, i.e. that 𝜓 is a natural isomorphism ℬ(𝐹 (−), −)  𝒜(−, 𝐺 (−)).

4.2

The Yoneda lemma

Exercise 4.2.2. (Recall that 𝐻 𝐴 = 𝒜(𝐴, −).) Dual of the Yoneda lemma. Let 𝒜 be a locally small category. Then [𝒜, Set] (𝐻 𝐴, 𝑋 )  𝑋 (𝐴) naturally in 𝐴 ∈ 𝒜 and 𝑋 ∈ [𝒜, Set]. Exercise 4.2.3. (a) Let ∗ denote the unique object of 𝑀. Recall that a functor 𝐹 : 𝑀 op → Set corresponds to the right 𝑀-set 𝐹 (∗) where the map 𝐹 (∗) × 𝑀 → 𝐹 (∗) is given by (𝑥, 𝑚) ↦→ 𝐹 (𝑚)(∗). Since 𝑀 op has only one object ∗, there is only one representable functor 𝑀 op → Set, namely 𝑀 (−, ∗). This functor sends ∗ ↦→ 𝑀 (∗, ∗) = 𝑀 and 𝑀 ∈ 𝑚 ↦→ 𝑀 (𝑚, ∗) = 𝑓𝑚 : 𝑀 → 𝑀 where 𝑓𝑚 (𝑥) = 𝑥𝑚. This is precisely 𝑀. (b) Assume 𝛼 : 𝑀 → 𝑋 exists. Then, given 𝑚 ∈ 𝑀 we have 𝛼 (𝑚) = 𝛼 (1 ·𝑚) = 𝛼 (1) ·𝑚 = 𝑥 ·𝑚. Hence 𝛼 is unique. Moreover, 𝛼 : 𝑀 → 𝑋 given by 𝛼 (𝑚) = 𝑥 · 𝑚 preserves the right 𝑀-action. It follows that 𝛼 → 𝛼 (1) gives a bijection {maps 𝑀 → 𝑋 of right 𝑀-sets} → 𝑋 . (c) Let ℳ denote the full subcategory of Cat consisting of the one-object categories. Recall from Example 3.2.11 that Mon ' ℳ. Furthermore, from Example 1.3.4, a natural transformation between functors 𝑀 op → Set, where 𝑀 ∈ Mon, is the same thing as a map of right 𝑀-sets under the identification of functors 𝑀 op → Set with right 𝑀-sets. In (a) and (b) we have shown that given a one-object category 𝑀 ∈ ℳ and a functor 𝑋 : 𝑀 op → Set, there is a bijection 𝜑𝑋 : [𝑀 op, Set] (𝑀, 𝑋 )  𝑋 (∗), where 𝑀 = 𝐻 ∗, with ∗ ∈ 𝑀 the unique object of 𝑀. The proof of the Yoneda lemma for ℳ is complete once we prove that 𝜑𝑋 is natural in 𝑋, i.e. that if 𝑋, 𝑋 0 ∈ [𝑀 op, Set] and 𝜂 : 𝑋 ⇒ 𝑋 0 then the diagram [𝑀 op, Set] (𝑀, 𝑋 )

𝜑𝑋

𝑋 (∗) 𝜂 (∗)

𝜂∗

[𝑀 op, Set] (𝑀, 𝑋 0)

𝜑𝑋 0

𝑋 0 (∗)

commutes. If 𝛼 : 𝑀 → 𝑋 is a map of 𝑀-sets, then 𝜑𝑋 (𝛼) = 𝛼 (1) and hence 𝜂 (∗) (𝛼 (1)) = 𝜂 ◦ 𝛼 (1). On the other hand, 𝜂 ∗ (𝛼) = 𝜂 ◦ 𝛼, so 𝜑𝑋 0 (𝜂 ◦ 𝛼) = 𝜂 ◦ 𝛼 (1). Thus the diagram indeed commutes and the bijection 𝛼𝑋 is natural in 𝑋 . We conclude that the Yoneda holds on ℳ. 39

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4.3

4.3

Consequences of the Yoneda lemma

Consequences of the Yoneda lemma

Exercise 4.3.15. (a) Let 𝑓 : 𝐴 → 𝐴 0 be a map in 𝒜. If 𝑓 : 𝐴 → 𝐴 0 is an isomorphism in 𝒜, then 𝐽 (𝑓 ) : 𝐽 (𝐴) → 𝐽 (𝐴 0) is an isomorphism in ℬ by functoriality of 𝐽 . Conversely, assume that 𝐽 (𝑓 ) is an isomorphism. Let 𝑔 : 𝐴 0 → 𝐴 be the unique map 𝐴 0 → 𝐴 such that 𝐽 (𝑔) = 𝐽 (𝑓 ) −1, which exists (and is unique) since 𝐽 is full and faithful. Then 𝐽 (𝑔 ◦ 𝑓 ) = 1 𝐽 (𝐴) and 𝐽 (𝑓 ◦ 𝑔) = 1 𝐽 (𝐴0) , and since 𝐽 is full and faithful, it follows 𝑔 ◦ 𝑓 = 1𝐴 and 𝑓 ◦ 𝑔 = 1𝐴0 . Thus 𝑓 is an isomorphism. (a).

(b) There is exactly one map 𝑓 : 𝐴 → 𝐴 0 such that 𝐽 (𝑓 ) = 𝑔, which is an isomorphism by part (c) This follows immediately from parts (a) and (b).

Exercise 4.3.16. (a) Let 𝐴, 𝐴 0 ∈ 𝒜. Assume that 𝑓 , 𝑔 ∈ 𝒜(𝐴 0, 𝐴) are such that 𝐻 𝑓 = 𝐻𝑔 : 𝐻𝐴0 ⇒ 𝐻𝐴 . Then for all 𝐵 ∈ 𝒜, (𝐻 𝑓 )𝐵 = (𝐻𝑔 )𝐵 : 𝒜(𝐵, 𝐴 0) → 𝒜(𝐵, 𝐴). In particular, by considering 𝐵 = 𝐴 0 we obtain 𝑓 = (𝐻 𝑓 )𝐴0 (1𝐴0 ) = (𝐻𝑔 )𝐴0 (1𝐴0 ) = 𝑔. It follows that 𝐻 • is faithful. (b) Let 𝐴, 𝐴 0 ∈ 𝒜 and 𝜂 : 𝐻𝐴0 ⇒ 𝐻𝐴 . We want to find 𝑓 : 𝐴 0 → 𝐴 such that 𝐻 𝑓 = 𝜂. Consider 𝑓 = 𝜂𝐴0 (1𝐴0 ). Given any 𝐵 ∈ 𝒜 and 𝑔 ∈ 𝒜(𝐵, 𝐴 0), by naturality of 𝜂 the diagram 𝒜(𝐴 0, 𝐴 0)

𝜂𝐴0

𝑔∗

𝒜(𝐵, 𝐴 0)

𝒜(𝐴 0, 𝐴) 𝑔∗

𝜂𝐵

𝒜(𝐵, 𝐴)

commutes. Thus 𝜂𝐵 (𝑔) = 𝜂𝐵 ◦ 𝑔∗ (1𝐴0 ) = 𝑔∗ ◦ 𝜂𝐴0 (1𝐴0 ) = 𝑓 ◦ 𝑔 = (𝐻 𝑓 )𝐵 (𝑔) and hence 𝐻 𝑓 = 𝜂. It follows that that 𝐻 • is full. (c) We are given an object 𝐴 ∈ 𝒜, a functor 𝑋 : 𝒜 op → Set and an element 𝑢 ∈ 𝑋 (𝐴) such that for each 𝐵 ∈ 𝒜 and 𝑥 ∈ 𝑋 (𝐵), there is a unique map 𝑥 : 𝐵 → 𝐴 such that 𝑋 (𝑥) (𝑢) = 𝑥 . Thus, for each 𝐵 ∈ 𝒜 the map 𝜑 𝐵 : 𝑋 (𝐵) → 𝐻𝐴 (𝐵) = 𝐻 (𝐵, 𝐴) given by 𝑥 ↦→ 𝑥 is bijective with inverse 𝐻 (𝐵, 𝐴) → 𝑋 (𝐵) given by ℎ ↦→ 𝑋 (ℎ) (𝑢). It remains to prove that the bijection 𝜑 𝐵 is natural in 𝐵. So let 𝑓 : 𝐵 0 → 𝐵 be a morphism and consider the diagram 𝑋 (𝐵)

𝜑𝐵

𝑓∗

𝑋 (𝑓 )

𝑋 (𝐵 0)

𝐻 (𝐵, 𝐴)

𝜑 𝐵0

𝐻 (𝐵 0, 𝐴) .

Let 𝑥 ∈ 𝑋 (𝐵). On the one hand we have 𝑓 ∗ ◦ 𝜑 𝐵 (𝑥) = 𝑥 ◦ 𝑓 , and on the other hand we have 𝜑 𝐵0 ◦ 𝑋 𝑓 (𝑥) = (𝑋 𝑓 ) (𝑥). By definition, (𝑋 𝑓 ) (𝑥) is the unique map 𝐵 0 → 𝐴 such that 𝑋 ((𝑋 𝑓 ) (𝑥)) (𝑢) = (𝑋 𝑓 )(𝑥). Since 𝑋 (𝑥 ◦ 𝑓 ) (𝑢) = (𝑋 𝑓 ) (𝑋𝑥) (𝑢) = 𝑋 𝑓 (𝑥), we have (𝑋 𝑓 ) (𝑥) = 𝑥 ◦ 𝑓 . Thus the diagram commutes. It follows that 𝜑 : 𝑋 ⇒ 𝐻𝐴 is a natural isomorphism. Exercise 4.3.17. Let 𝒜 be a discrete category. Then 𝒜 = 𝒜 op . A presheaf 𝑋 : 𝒜 op → Set on 𝒜 is simply an assignment of a set 𝑋 (𝐴) ∈ Set for each 𝐴 ∈ 𝒜. If 𝐴 ∈ 𝒜, the presheaf represented 40

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4.3

Consequences of the Yoneda lemma

by 𝐴 if given by

( 𝐻 𝐴 (𝐵) = 𝐻𝐴 (𝐵) =

{1𝐴 }, 𝐵 = 𝐴, ∅, 𝐵 ≠ 𝐴.

Now we analyse the Yoneda lemma. Let 𝑋 be a presheaf on 𝒜 and 𝜂 : 𝐻𝐴 → 𝑋 a natural transformation, for some 𝐴 ∈ 𝒜. Then 𝜂𝐵 , for 𝐵 ≠ 𝐴, is the empty function, so we must have 𝑋 (𝐵) = ∅ for all such 𝐵. (In particular, representable presheaves are precisely those 𝑋 such that for some 𝐴 ∈ 𝒜, 𝑋 (𝐵) = for all 𝐵 ≠ 𝐴 and 𝑋 (𝐴) is a singleton.) So a natural transformation 𝜂 : 𝐻𝐴 → 𝑋 is determined by 𝜂𝐴 : {1𝐴 } → 𝑋 (𝐴), an element of 𝑋 (𝐴). So the Yoneda lemma is clear in this case. Finally, consider Corollary 4.3.2. It says that a representation of 𝑋 consists of an object 𝐴 ∈ 𝒜 and an element 𝑢 ∈ 𝑋 (𝐴) such that 𝑋 (𝐵) = ∅ for all 𝐵 ≠ 𝐴, and for all 𝑥 ∈ 𝑋 (𝐴), there is a unique map 𝑥 : 𝐴 → 𝐴 with 𝑋 (𝑥) (𝑢) = 𝑥 . As there is only one map 𝐴 → 𝐴, namely 1𝐴, the last condition says that 𝑥 is the unique element of 𝑋 (𝐴), so 𝑋 (𝐴) is a singleton (as we deduced in the paragraph above). Similarly for Corollary 4.3.3. Exercise 4.3.18. (a) Let 𝐹, 𝐹 0 ∈ [ℬ, 𝒞]. We want to show that the map 𝐽

[ℬ, 𝒞] (𝐹, 𝐹 0) → − [ℬ, 𝒟] (𝐽 𝐹, 𝐽 𝐹 0) is bijective. First suppose that 𝛼, 𝛼 0 : 𝐹 ⇒ 𝐹 0 are natural transformations such that 𝐽 𝛼 = 𝐽 𝛼 0 Then, given 𝐵 ∈ ℬ, 𝐽 (𝛼 𝐵 ) = 𝐽 (𝛼 𝐵0 ) : 𝐽 𝐹 (𝐵) → 𝐽 0𝐹 (𝐵). Since 𝐽 is faithful there is at most one arrow 𝑓 : 𝐹 (𝐵) → 𝐹 0 (𝐵) such that 𝐽 (𝑓 ) = 𝐽 (𝛼 𝐵 ) = 𝐽 (𝛼 𝐵0 ). Since both 𝛼 𝐵 and 𝛼 𝐵0 are such arrows, we have 𝛼 𝐵 = 𝛼 𝐵0 . It follows that 𝛼 = 𝛼 0, so the above map is injective. Now we prove that it is also surjective. Let 𝛽 : 𝐽 𝐹 ⇒ 𝐽 𝐹 0 be natural. Since 𝐽 is full and faithful, for each 𝐵 ∈ ℬ there is precisely one map 𝛼 𝐵 : 𝐹 (𝐵) → 𝐹 0 (𝐵) such that 𝐽 (𝛼 𝐵 ) = 𝛽𝐵 : 𝐽 𝐹 (𝐵) → 𝐽 𝐹 0 (𝐵). It remains to prove that 𝛼 = (𝛼 𝐵 )𝐵 ∈ℬ is natural, for then 𝐽 (𝛼) = 𝛽. Let 𝑓 : 𝐵 → 𝐵 0 be a map in ℬ and consider the diagram 𝐹 (𝐵)

𝛼𝐵

𝐹 0 (𝑓 )

𝐹 (𝑓 )

𝐹 (𝐵 0)

𝐹 0 (𝐵)

𝛼 𝐵0

𝐹 0 (𝐵 0) .

We want to show it is commutative. Now 𝐹 0 (𝑓 ) ◦ 𝛼 𝐵 and 𝛼 𝐵0 ◦ 𝐹 (𝑓 ) are such that 𝐽 (𝐹 0 (𝑓 ) ◦ 𝛼 𝐵 ) = 𝐽 𝐹 0 (𝑓 ) ◦ 𝛽𝐵 = 𝛽𝐵0 ◦ 𝐹 𝐽 (𝑓 ) = 𝐽 (𝛼 𝐵0 ◦ 𝐹 (𝑓 )) by naturality of 𝛽. Since 𝐽 is faithful it follows that 𝐹 0 (𝑓 ) ◦ 𝛼 𝐵 = 𝛼 𝐵0 ◦ 𝐹 (𝑓 ). Thus 𝛼 is natural. We conclude that 𝐽 ◦ − is full and faithful. (b) Since 𝐺 and 𝐺 0 are both maps such that 𝐽 ◦ 𝐺  𝐽 ◦ 𝐺 0 and 𝐽 ◦ − is full and faithful, then 𝐺  𝐺 0 by Lemma 4.3.8(a) (which was proved in Exercise 4.3.15.) (c) We have bijections 𝒜(𝐴, 𝐺 (𝐵))  ℬ(𝐹 (𝐴), 𝐵)  𝒜(𝐴, 𝐺 0 (𝐵)) natural in 𝐴 ∈ 𝒜 and 𝐵 ∈ ℬ. Consider the Yoneda embedding 𝐻 • : 𝒜 → [𝒜 op, Set], which is full and faithful by 41

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5. Limits Corollary 4.3.7. It follows from part (a) that the functor 𝐻 • ◦ − : [ℬ, 𝒜] → [ℬ, [𝒜 op, Set]] is full and faithful. Note that 𝐻 • ◦ 𝐺 : ℬ → [𝒜, Set] is the functor 𝐵 ↦→ 𝐻𝐺 (𝐵) = 𝒜(−, 𝐺 (𝐵)), and similarly for 𝐻 • ◦ 𝐺 0, so that 𝐻 • ◦ 𝐺  𝐻 • ◦ 𝐺 0 . Then 𝐺  𝐺 0 follows from (b).

5 5.1

Limits Limits: definition and examples

Exercise 5.1.33. This was done in Exercise 0.14(a). Exercise 5.1.34. We will show that if 𝐸 is an equaliser then it is not necessarily a pullback. If the above square is a pullback then it has the following universal property: ∀𝑗

∀𝐸 0 ∃!ℎ

𝑖

𝐸 ∀𝑘

𝑋 𝑔

𝑖

𝑋 𝑓

𝑌.

If the maps 𝑘 and 𝑗 are equal, then ℎ indeed exists and is unique since 𝐸 is a coequaliser. However, we do not expect this to be true if 𝑘 ≠ 𝑗 in general. Indeed, consider for instance the category Set, 𝑋 = {1, 2}, 𝑌 = {1} and 𝑓 , 𝑔 the unique possible maps. Then the pullback is (𝑋 × 𝑋, pr1, pr2 ), and the equaliser is (𝑋, 1𝑋 ). Since 𝑋 × 𝑋 is not isomorphic to 𝑋, these are not equal. On the other hand, the converse does hold: if the given square is a pullback then (𝐸, 𝑖) is the equaliser of 𝑓 and 𝑔. To show this, consider the diagram above which illustrates the universal property of the pullback. By taking 𝑘 and 𝑗 to be equal, we see that for any 𝐸 0 and a map 𝑖 0 : 𝐸 0 → 𝑋 such that 𝑓 𝑖 0 = 𝑔𝑖 0, there is a unique map ℎ : 𝐸 0 → 𝐸 such that 𝑖ℎ = 𝑖 0 . Hence (𝐸, 𝑖) is the equaliser of 𝑓 and 𝑔. Exercise 5.1.35. (This is also Mac Lane’s Exercise III.4.8.) Assume we have a commutative diagram 𝑓

𝐴

𝑔

𝐵

𝑗

𝐶

𝑘

𝐷

𝑙

𝐸

𝐹 𝑖

ℎ

in some category 𝒞 such that the right-hand square is a pullback. First assume that the left-hand square is also a pullback. Let 𝐻 be an object together with maps 𝑡 1 : 𝐻 → 𝐻, 𝑡 2 : 𝐻 → 𝐶 such that 𝑖ℎ𝑡 1 = 𝑙𝑡 2 . We want to find a unique 𝑝 : 𝐻 → 𝐴 fitting in a 42

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5.1

Limits: definition and examples

commutative diagram as below: 𝑡2

𝐻 𝑝 𝑓

𝐴

𝑔

𝐵

𝐶

𝑡1 𝑗

𝑘

𝐷

𝑙

𝐸 𝑖

ℎ

𝐹.

Since the right-hand square is a pullback there is a unique 𝑝 0 : 𝐻 → 𝐵 such that 𝑘𝑝 0 = ℎ𝑡 1 and 𝑔𝑝 0 = 𝑡 2 . Then, since the left-hand square is a pullback there is a unique 𝑝 : 𝐻 → 𝐴 such that 𝑗𝑝 = 𝑡 1 and 𝑝 0 = 𝑓 𝑝. Then 𝑝 satisfies 𝑗𝑝 = 𝑡 1 and 𝑔𝑓 𝑝 = 𝑡 2, and is unique as such. We deduce that the outer rectangle is a pullback. Now assume that the outer rectangle is a pullback. Let 𝐻 be an object of 𝒞 and 𝑡 1 : 𝐻 → 𝐷, 𝑡 2 : 𝐻 → 𝐵 be such that 𝑘𝑡 2 = ℎ𝑡 1 . We want to find a unique 𝑝 : 𝐻 → 𝐴 filling the diagram 𝐻

𝑡2 𝑝 𝑓

𝐴

𝑔

𝐵

𝐶

𝑡1 𝑗

𝑘

𝐷

𝑙

𝐸 ℎ

𝑖

𝐹.

Since 𝑙𝑔𝑡 2 = 𝑖𝑘𝑡 2 = 𝑖ℎ𝑡 1 and the right-hand square is a pullback, there exists a unique 𝑝 0 : 𝐻 → 𝐵 such that 𝑔𝑝 0 = 𝑔𝑡 2 and 𝑘𝑝 0 = ℎ𝑡 1 . As 𝑡 2 satisfies these equations, we have that 𝑝 0 = 𝑡 2 . Now, since the outer rectangle is a pullback there exists a unique 𝑝 : 𝐻 → 𝐴 such that 𝑔𝑓 𝑝 = 𝑔𝑡 2 and 𝑗𝑝 = 𝑡 1 . Then 𝑔𝑓 𝑝 = 𝑔𝑡 2 and 𝑘 𝑓 𝑝 = ℎ 𝑗𝑝 = ℎ𝑡 1, i.e. 𝑓 𝑝 satisfies the equations for 𝑝 0, and hence 𝑓 𝑝 = 𝑡 2 by uniqueness. It follows that the left-hand square is a pullback. 𝑝 𝐼 ◦ℎ

𝑢

Exercise 5.1.36. (a) Note that (𝐴 −−−→ 𝐷 (𝐼 ))𝐼 ∈I is a cone on 𝐷. Indeed, if 𝐼 → − 𝐽 in I then 𝐷𝑢 ◦ 𝑝 𝐼 ◦ ℎ = 𝑝 𝐽 ◦ ℎ since 𝐿 is a cone. By the universal property of the limit, there is precisely one map ℎe: 𝐴 → 𝐿 such that 𝑝 𝐼 ◦ ℎe = 𝑝 𝐼 ◦ ℎ for all 𝐼 ∈ I. Since both ℎ and ℎ 0 satisfy this condition, it follows that ℎ = ℎ 0 . (b) A diagram 𝐷 : I → Set is a pair of sets 𝑋, 𝑌 . Its limit is the product 𝑋 × 𝑌 together with the projections pr𝑋 , pr𝑌 . A map 1 → 𝑋 × 𝑌 is an element (𝑥, 𝑦) ∈ 𝑋 × 𝑌 . Thus (a) in this case means that given (𝑥, 𝑦), (𝑥 0, 𝑦 0) ∈ 𝑋 × 𝑌 , if 𝑥 = 𝑥 0 and 𝑦 = 𝑦 0 then (𝑥, 𝑦) = (𝑥 0, 𝑦 0). Exercise 5.1.37. Let 𝐿 denote the given set 𝑢

{(𝑥 𝐼 )𝐼 ∈I | 𝑥 𝐼 ∈ 𝐷 (𝐼 ) for all 𝐼 ∈ I and (𝐷𝑢) (𝑥 𝐼 ) = 𝑥 𝐽 for all 𝐼 → − 𝐽 in I}, 43

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5.1

Limits: definition and examples 𝑝𝐼

and 𝑝 𝐽 : 𝐿 → 𝐷 (𝐽 ) be given by 𝑝 𝐽 ((𝑥 𝐼 )𝐼 ∈I ) = 𝑥 𝐽 for all 𝐽 ∈ I. First note that (𝐿 −→ 𝐷 (𝐼 ))𝐼 ∈I is a 𝑓𝐼

cone on 𝐷 by definition. Now let (𝐴 − → 𝐷 (𝐼 ))𝐼 ∈I be any cone on 𝐷. We need to show that there is a unique map ℎ : 𝐴 → 𝐿 such that 𝑓 𝐽 = 𝑝 𝐽 ◦ ℎ for all 𝐽 ∈ I. Suppose such ℎ exists. Given 𝑎 ∈ 𝐴 write ℎ(𝑎) = (ℎ𝐼 (𝑎))𝐼 ∈I . Then, for all 𝐽 ∈ I, 𝑓 𝐽 (𝑎) = 𝑝 𝐽 ◦ ℎ(𝑎) = ℎ 𝐽 (𝑎). This shows that if ℎ exists then it is unique. Moreover, define ℎ : 𝐴 → 𝐿 by ℎ(𝑎) = (𝑓𝐼 (𝑎))𝐼 ∈I . Note that 𝑓𝐼 (𝑎) ∈ 𝐷 (𝐼 ) for 𝑢 each 𝐼, and if 𝐼 → − 𝐽 is an arrow in I we have (𝐷𝑢) (𝑓𝐼 (𝑎)) = 𝑓 𝐽 (𝑎) since ℎ is a cone on 𝐷. Thus ℎ 𝑝𝐼

well-defined, and moreover satisfies 𝑓 𝐽 = 𝑝 𝐽 ◦ ℎ for all 𝐽 ∈ I. It follows that (𝐿 −→ 𝐷 (𝐼 ))𝐼 ∈I is a limit cone. 𝑝𝐼

𝑢

Exercise 5.1.38. (a) First note that (𝐿 −→ 𝐷 (𝐼 ))𝐼 ∈I is a cone on 𝐷. Indeed, if 𝐼 → − 𝐽 is an arrow in I we have 𝑝 𝐽 = pr 𝐽 ◦𝑝 = 𝐷𝑢 ◦ pr𝐼 ◦𝑝 = 𝐷𝑢 ◦ 𝑝 𝐼 as (𝐿, 𝑝) is the equaliser of 𝑠 and 𝑡 . Now suppose 𝑓𝐼

that (𝐴 − → 𝐷 (𝐼 ))𝐼 ∈I is an arbitrary cone on 𝐷. We shall find a unique map ℎ : 𝐴 → 𝐿 such that Î 𝑓𝐼 = 𝑝 𝐼 ◦ ℎ for all 𝐼 ∈ I. The family (𝑓𝐼 )𝐼 ∈I induces a unique map 𝑓 : 𝐴 → 𝐼 ∈I 𝐷 (𝐼 ) such that 𝑢 pr𝐼 ◦𝑓 = 𝑓𝐼 for all 𝐼 ∈ I. Given 𝐼 → − 𝐽 an arrow in I then pr 𝐽 ◦𝑓 = 𝑓 𝐽 = 𝐷𝑢 ◦ 𝑓𝐼 = 𝐷𝑢 ◦ pr𝐼 ◦𝑓 , so 𝑓 is a map such that 𝑠 ◦ 𝑓 = 𝑡 ◦ 𝑓 . Since (𝐿, 𝑝) is an equaliser there is a unique map ℎ : 𝐴 → 𝐿 such that 𝑝 ◦ ℎ = 𝑓 , and this last equation is equivalent to satisfying 𝑝 𝐼 ◦ ℎ = 𝑓𝐼 for all 𝐼 ∈ I. We 𝑝𝐼

conclude that (𝐿 −→ 𝐷 (𝐼 ))𝐼 ∈I is a limit cone on 𝐷. (b) Existence of a terminal object is equivalent to existence of the empty product. Assuming binary products, we have all finite products by iteration. Hence the same proof above as in (a) applies, assuming the category I to be finite. Exercise 5.1.39. (This is also Mac Lane’s Exercise III.4.10.) Denote by ∗ the terminal object of the category 𝒞. Given two objects 𝑋, 𝑌 of 𝒞, consider the pullback square 𝑝𝑋

𝐸

𝑋

𝑌

∗.

𝑝𝑌

We claim that (𝐸, 𝑝𝑋 , 𝑝𝑌 ) is the product of 𝑋 and 𝑌 . Indeed, 𝐴 ∈ 𝒞 and a pair of maps 𝑓𝑋 : 𝐴 → 𝑋 and 𝑓𝑌 : 𝐴 → 𝑌 , then the respective compositions with 𝑋 → ∗ and 𝑌 → ∗ are equal since ∗ is terminal, so there is a unique map 𝑓 : 𝐴 → 𝐸 such that 𝑝𝑋 ◦ 𝑓 = 𝑓𝑋 and 𝑝𝑌 ◦ 𝑓 = 𝑓𝑌 . That is, (𝐸, 𝑝𝑋 , 𝑝𝑌 ) satisfies the universal property of the product. Now, given parallel arrows 𝑓 , 𝑔 : 𝑋 → 𝑌 , consider the following pullback diagram 𝑒

𝐸

𝑋

𝑒0

𝑋

(1𝑋 ,𝑓 ) (1𝑋 ,𝑔)

𝑋 ×𝑌 .

Then 𝑒 = 𝑒 0 and 𝑒 : 𝐸 → 𝑋 is the equaliser of 𝑓 and 𝑔 (see Exercise 5.1.34.) 44

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5.1

Limits: definition and examples

Thus our category has all binary products, equalisers and a terminal object, so it has all finite limits by Proposition 5.1.26(b) (proved in Exercise 5.1.38(b)). Exercise 5.1.40. (a) Monics in Set are precisely the injective maps (Example 5.1.30). Now 𝑚 : 𝑋 → 𝐴 and 𝑚 0 : 𝑋 0 → 𝐴 are isomorphic in Monic(𝐴) if and only if there exists maps 𝑓 : 𝑋 → 𝑋 0 and 𝑔 : 𝑋 0 → 𝑋 such that 𝑓 𝑔 = 1𝑌 , 𝑔𝑓 = 1𝑋 and 𝑚 0 𝑓 = 𝑚, 𝑚𝑔 = 𝑚 0 . These conditions are equivalent to existence of a bijection 𝑓 : 𝑋 → 𝑋 0 such that 𝑚 0 𝑓 = 𝑚. If such 𝑓 exists then 𝑚 and 𝑚 0 have the same image since 𝑓 is a bijection. Conversely, if 𝑚 and 𝑚 0 have the same image, then they are bijections onto the subset 𝑚(𝑋 ) = 𝑚(𝑋 0) of 𝐴. Then we can define 𝑓 : 𝑋 → 𝑋 0 by 𝑓 (𝑥) = 𝑚 −1 (𝑚(𝑥)), and this is a bijection such that 𝑚 0 𝑓 = 𝑚. It follows that an isomorphism class of objects in Monic(𝐴) corresponds precisely to a subset of 𝐴, namely the image of any representative of the class. (b) Monics in each of these categories are precisely the injective morphisms, and the same proof as before applies, since the map 𝑓 (𝑥) = 𝑚 −1 (𝑚(𝑥)) defined at the end is indeed a morphism in the respective category. Therefore subobjects of Grp, Ring and Vect𝑘 are subgroups, subrings and subspaces respectively. (c) As in the category of sets, a map 𝑓 in Top is monic if and only if it is injective. Now, for 𝐴 ∈ Top and monics 𝑚 : 𝑋 → 𝐴, 𝑚 0 : 𝑋 0 → 𝐴 representing objects of Monic(𝐴), a map from 𝑚 to 𝑚 0 is a homeomorphism 𝑓 : 𝑋 → 𝑋 0 such that 𝑚 0 𝑓 = 𝑚. In this case 𝑓 : 𝑋 → 𝑋 0 given by 𝑓 (𝑥) = 𝑚 −1 (𝑚(𝑥)) may not be continuous. Hence subobjects of 𝐴 are not its subspaces. They are subsets 𝑈 ⊂ 𝐴 equipped with a topology finer than the subspace topology inherited from 𝐴. Exercise 5.1.41. (This is also (the dual of) Mac Lane’s Exercise III.4.4.) First assume that 𝑓 is monic and consider the diagram 𝑋

1𝑋

1𝑋

𝑋 𝑓

𝑋 𝑓

𝑌.

Suppose we are given arrows 𝑔, 𝑔 0 : 𝑍 → 𝑋, such that 𝑓 ◦ 𝑔 = 𝑓 ◦ 𝑔 0 . We must show there is a unique ℎ : 𝑍 → 𝑋 such that 1𝑋 ◦ ℎ = 𝑔 and 1𝑋 ◦ ℎ = 𝑔 0 . But 𝑓 is monic, so ℎ = 𝑔 = 𝑔 0 works. Conversely, assume the diagram above is a pullback and let 𝑔, 𝑔 0 : 𝑍 → 𝑋 be such that 𝑓 ◦ 𝑔 = 𝑓 ◦ 𝑔 0 . Then there is a unique ℎ : 𝑍 → 𝑋 such that 1𝑋 ◦ ℎ = 𝑔 and 1𝑋 ◦ ℎ = 𝑔 0, so 𝑔 = 𝑔 0 . Exercise 5.1.42. (This is also Mac Lane’s Exercise III.4.5.) Let 𝑔, 𝑔 0 : 𝑌 → 𝑋 0 be arrows such that 𝑚 0 ◦𝑔 = 𝑚 0 ◦𝑔 0 . Then 𝑚 ◦ 𝑓 0 ◦𝑔 = 𝑓 ◦𝑚 0 ◦𝑔 = 𝑓 ◦𝑚 0 ◦𝑔 0 = 𝑚 ◦ 𝑓 0 ◦𝑔 0, so 𝑓 0 ◦𝑔 = 𝑓 0 ◦𝑔 0 as 𝑚 is monic. Since the square is a pullback and 𝑚 ◦ 𝑓 0 ◦𝑔 = 𝑓 ◦𝑚 0 ◦𝑔, there is a unique ℎ : 𝑌 → 𝑋 0 such that 𝑚 0 ◦ ℎ = 𝑚 0 ◦ 𝑔 and 𝑓 0 ◦ ℎ = 𝑓 0 ◦ 𝑔. Both 𝑔 and 𝑔 0 satisfy the equations for ℎ, so 𝑔 = 𝑔 0 . It follows that 𝑚 0 is monic.

45

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5.2

5.2

Colimits: definition and example

Colimits: definition and example 1𝑋

Exercise 5.2.21. First assume that 𝑠 = 𝑡 . Consider 𝑋 −−→ 𝑋 . Then 𝑠1𝑋 = 𝑡1𝑋 . Moreover, any map 𝑒 𝑍→ − 𝑋 satisfies the condition 𝑠𝑒 = 𝑡𝑒, and given any such map then 𝑒 is the unique map 𝑍 → 𝑋 such that 1𝑋 𝑒 = 𝑒. Thus (𝑋, 1𝑋 ) is the equaliser of 𝑠 and 𝑡 . Similarly, (𝑌 , 1𝑌 ) is the coequaliser of 𝑠 and 𝑡 . Now suppose that the equaliser (𝑍, 𝑒) of 𝑠 and 𝑡 is an isomorphism. Then 𝑠𝑒 = 𝑡𝑒 and 𝑠 = −1 𝑠𝑒𝑒 = 𝑡𝑒𝑒 −1 = 𝑡 . Similarly, if the coequaliser of 𝑠 and 𝑡 is an isomorphism then 𝑠 = 𝑡 . Exercise 5.2.22. (a) By Example 5.2.9, the coequaliser of 𝑓 and 1𝑋 is the quotient 𝑋 /∼ where ∼ is the equivalence relation generated by 𝑓 (𝑥) ∼ 𝑥 for all 𝑥 ∈ 𝑋, together with the quotient map 𝑋 → 𝑋 /∼ . This is 𝑋 /∼ where 𝑥 ∼ 𝑦 if there exists 𝑛 ≥ 1 such that 𝑓 𝑛 (𝑥) = 𝑦 or 𝑓 𝑛 (𝑦) = 𝑥 . (b) In Top, the coequaliser is again the quotient 𝑋 /∼ as before, equipped with the quotient topology. Let 𝜃 ∈ [0, 2𝜋] be irrational and consider the map 𝑓 : 𝑆 1 → 𝑆 1 given by 𝑒 𝑖𝑡 ↦→ 𝑒 𝑖 (𝑡 +𝜃 ) . Let 𝑘 : 𝑋 → 𝑌 be the coequaliser of 𝑓 and 1𝑋 , where 𝑌 = 𝑋 /∼ is the quotient of 𝑋 by the relation generated by 𝑓 (𝑥) ∼ 𝑥 for all 𝑥 ∈ 𝑋 . Then 𝑌 is uncountable, for each equivalence class [𝑥] is countable. Let 𝑈 ⊂ 𝑌 be a non-empty open set, and [𝑥] ∈ 𝑈 . Since 𝑘 −1 (𝑈 ) is open in 𝑋, there exists 𝜀 > 0 such that 𝐵(𝑥, 𝜀) ∩ 𝑆 1 ⊂ 𝑘 −1 (𝑈 ). Then 𝐵(𝑧, 𝜀) ∩ 𝑆 1 ⊂ 𝑘 −1 (𝑈 ) for all 𝑧 ∈ 𝑋 such that [𝑧] = [𝑥]. Now, let 𝑧 ∈ 𝑋 be arbitrary. As 𝜃 is irrational, the set 𝐴 = {𝑓 𝑛 (𝑥 0) | 𝑛 ∈ Z} is dense in 𝑋, so there exists 𝑤 ∈ 𝐴 ∩ 𝐵(𝑥, 𝜀) ∩ 𝑆 1 . Then [𝑤] = [𝑧] and 𝑤 ∈ 𝑘 −1 (𝑈 ), so 𝑘 (𝑧) = 𝑘 (𝑤) ∈ 𝑈 . Thus 𝑘 −1 (𝑈 ) = 𝑋, so 𝑈 = 𝑌 . It follows that 𝑌 has the indiscrete topology. Exercise 5.2.23. (a) Let 𝑖 denote the inclusion (N, +, 0) ↩− → (Z, +, 0). Let (𝑀, +𝑀 , 0𝑀 ) be a monoid and 𝑓 , 𝑔 : (Z, +, 0) → (𝑀, +𝑀 , 0𝑀 ) morphisms of monoids such that 𝑓 𝑖 = 𝑔𝑖. Then 𝑓 (𝑛) = 𝑔(𝑛) for all 𝑛 ≥ 0. Let 𝑛 < 0. Then 0𝑀 = 𝑓 (0) = 𝑓 (𝑛 − 𝑛) = 𝑓 (𝑛) +𝑀 𝑓 (−𝑛), and similarly 0𝑀 = 𝑓 (−𝑛) +𝑀 𝑓 (𝑛), so 𝑓 (𝑛) is the inverse of 𝑓 (−𝑛) in 𝑀. Similarly 𝑔(𝑛) is the inverse of 𝑔(−𝑛) in 𝑀. Since inverses are unique and 𝑓 (−𝑛) = 𝑔(−𝑛), we have 𝑓 (𝑛) = 𝑔(𝑛). Thus 𝑓 (𝑛) = 𝑔(𝑛) for all 𝑛 < 0 and therefore 𝑓 = 𝑔. (b) (This is also Mac Lane’s Exercise I.5.4.) Let 𝑖 denote the inclusion Z ↩− → Q. Let 𝑅 be a ring and 𝜑,𝜓 : Q → 𝑅 be morphisms such that 𝜑𝑖 = 𝜓𝑖. Then 𝜑 = 𝜓 on Z. If        𝑞 ∈ Z \ {0} then 𝑞 𝑝 1 1 1 −1 −1 1 = 𝜑 (1) = 𝜑 𝑞 = 𝜑 (𝑞)𝜑 𝑞 , so 𝜑 𝑞 = 𝜑 (𝑞) = 𝜓 (𝑞) = 𝜓 𝑞 . Thus 𝜑 𝑞 = 𝜑 (𝑝)𝜑 (𝑞) −1 =   𝑝 𝑝 𝜓 (𝑝)𝜓 (𝑞) −1 = 𝜓 𝑞 for all 𝑞 ∈ Q, so 𝜑 = 𝜓 . It follows that 𝑖 is epic (and not surjective!). Exercise 5.2.24. (a) Epics in Set are precisely the surjective maps (Example 5.2.18). In general, a surjective function 𝑓 : 𝑋 → 𝑌 induces an equivalence relation on 𝑋 given by 𝑥 ∼ 𝑥 0 if 𝑓 (𝑥) = 𝑓 (𝑥 0). Given epics 𝑒 : 𝐴 → 𝑋, 𝑒 0 : 𝐴 → 𝑋 0, an isomorphism from 𝑒 to 𝑒 0 is a map 𝑓 : 𝑋 → 𝑋 0 such that 𝑓 𝑒 = 𝑒 0 and there is a map 𝑔 : 𝑋 0 → 𝑋 with 𝑔𝑒 0 = 𝑒 and 𝑔𝑓 = 1, 𝑓 𝑔 = 1. This amounts to a bijection 𝑓 : 𝑋 → 𝑋 0 such that 𝑓 𝑒 = 𝑒 0 . So assume there is such bijection. Let 𝑎, 𝑎 0 ∈ 𝐴. Then 𝑒 (𝑎) = 𝑒 (𝑎 0) =⇒ 𝑓 𝑒 (𝑎) = 𝑓 𝑒 (𝑎 0) =⇒ 𝑒 0 (𝑎) = 𝑒 0 (𝑎 0), and similarly (using 𝑓 −1 ) 𝑒 0 (𝑎) = 𝑒 0 (𝑎 0) =⇒ 𝑒 (𝑎) = 𝑒 (𝑎 0). Thus the equivalence relations induced on 𝐴 by 𝑒 and 𝑒 0 coincide. Conversely, assume that 𝑒 (𝑎) = 𝑒 (𝑎 0) ⇐⇒ 𝑒 0 (𝑎) = 𝑒 0 (𝑎 0) for all 𝑎, 𝑎 0 ∈ 𝐴. Define 𝑓 : 𝑋 → 𝑋 0 by 46

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Colimits: definition and example

𝑓 (𝑥) = 𝑒 0 (𝑎) where 𝑎 is any preimage of 𝑥 under 𝑒. This is well-defined by assumption, and is a bijection such that 𝑓 𝑒 = 𝑒 0 . Thus 𝑒 and 𝑒 0 are isomorphic in Epic(𝐴). We deduce that the quotient objects of 𝐴 are in canonical one-to-one correspondence with the equivalence relations of 𝐴, namely a quotient object represented by some 𝑒 : 𝐴 → 𝑋 corresponds to the equivalence relation on 𝐴 such that 𝑎 ∼ 𝑎 0 if and only if 𝑒 (𝑎) = 𝑒 (𝑎 0). (b) (The proof that epic =⇒ surjective on Grp can be found in Mac Lane’s Exercise I.5.5.) 𝜑 𝜑 An epimorphism 𝐺 − → 𝐻 induces an isomorphism 𝜑 e: 𝐺/ker 𝜑 → 𝐻 . Two epimorphisms 𝐺 − → 𝐻, 𝜓

𝐺− → 𝐻 0 are isomorphic in Epic(𝐺) if and only if there is an isomorphism 𝜃 : 𝐻 → 𝐻 0 such that 𝜃𝜑 = 𝜓 . If there is such 𝜃 then ker 𝜑 = ker𝜓 . Conversely, if ker 𝜑 = ker𝜓, let 𝜃 = (𝜓e) −1𝜑 e: 𝐻 → 𝐻 0 . Then 𝜃 is an isomorphism, and is given by 𝜑 (𝑔) ↦→ 𝜓 (𝑔) for 𝑔 ∈ 𝐺, so 𝜃𝜑 = 𝜓 . Thus 𝜑,𝜓 ∈ Epic(𝐺) are isomorphic if and only if ker 𝜑 = ker𝜓 . Hence the correspondence 𝜑 ↔ ker 𝜑 is one-to-one between quotient objects of 𝐺 and normal subgroups of 𝐺 . (Any kernel is normal, and any normal subgroup 𝑁 of 𝐺 is the kernel of 𝐺 → 𝐺/𝑁 .) Exercise 5.2.25. (a) Let 𝑚 : 𝐴 → 𝐵 be a morphism. First suppose that 𝑚 is split monic and let 𝑒 : 𝐵 → 𝐴 be such that 𝑒𝑚 = 1𝐴 . Consider the maps 𝑚𝑒, 1𝐵 : 𝐵 → 𝐵. Then 𝑚𝑒𝑚 = 𝑚1𝐴 = 1𝐵𝑚. Moreover, if ℎ : 𝐶 → 𝐵 is a map such that 𝑚𝑒ℎ = 1𝐵 ℎ = ℎ, then 𝑒ℎ : 𝐶 → 𝐴 satisfies 𝑚(𝑒ℎ) = ℎ, and if ℎ 0 : 𝐶 → 𝐴 satisfies 𝑚ℎ 0 = ℎ then ℎ 0 = 𝑒𝑚ℎ 0 = 𝑒ℎ. Thus 𝑚 is the equaliser of the maps 𝑚𝑒 : 𝐵 → 𝐵 and 1𝐵 : 𝐵 → 𝐵, so it is regular monic. Now assume that 𝑚 is regular monic, and let 𝐶 be an object and 𝑓 , 𝑔 : 𝐵 → 𝐶 maps of which 𝑚 is an equaliser. Suppose that ℎ, ℎ 0 : 𝑋 → 𝐴 are maps such that 𝑚ℎ = 𝑚ℎ 0 . Since 𝑚 is an equaliser, given any map 𝑘 : 𝑋 → 𝐵 such that 𝑓 𝑘 = 𝑔𝑘 then there exists a unique ℎe: 𝑋 → 𝐴 such that e so ℎ = ℎ 0 by 𝑚ℎe = 𝑘. Taking 𝑘 = 𝑚ℎ = 𝑚ℎ 0 then both ℎ and ℎ 0 satisfy the condition for ℎ, uniqueness. Thus 𝑚 is monic. (b) Let 𝜑 : 𝐴 → 𝐵 be monic in Ab, that is, 𝜑 is a monomorphism. Consider 𝐶 = 𝐵/im 𝜑, and the maps 𝜋 : 𝐵 → 𝐶, 0 : 𝐵 → 𝐶, where 𝜋 is the projection. Then 𝜋𝜑 = 0𝜑 = 0, and is universal as such. Indeed, if 𝜓 : 𝑋 → 𝐵 is a homomorphism such that 𝜋𝜓 = 0, then 𝜓 (𝑋 ) ⊂ 𝜑 (𝐴), and we can define a map 𝜃 : 𝑋 → 𝐴 by 𝜃 (𝑥) = 𝑎 if 𝜓 (𝑥) = 𝜑 (𝑎). This is well-defined since 𝜑 is injective. Moreover, it is the unique map with the property 𝜑𝜃 = 𝜓 . Thus 𝜑 : 𝐴 → 𝐵 is the equaliser of 𝜋 : 𝐵 → im 𝜑 and 0 : 𝐵 → 𝐶. It follows that all monics in Ab are regular monics. To show that not all monics in Ab are split monics consider 𝜑 : Z → Z given by 𝜑 (1) = 2. Then 𝜑 is a monomorphism, but there is no 𝜓 : Z → Z such that 𝜓𝜑 = 1Z . Indeed, any such 𝜓 would send 2 to 1, which is not possible. (c) We will show that the regular monics in Top are precisely the embeddings, i.e. the injective maps which are homeomorphisms onto their image. Let ℎ : 𝑋 → 𝑌 be regular monic in Top. In particular it is monic, i.e. it is injective. Let ℎ : 𝑋 → ℎ(𝑋 ) be obtained from ℎ by restricting its codomain. Let 𝑓 , 𝑔 : 𝑌 → 𝑍 be maps of which ℎ is an equaliser. Consider the inclusion map 𝑖 : ℎ(𝑋 ) ↩− → 𝑌 . Then 𝑓 𝑖 = 𝑔𝑖, so there is a unique 𝑘 : ℎ(𝑋 ) → 𝑋 such that 𝑖 = ℎ𝑘 = 𝑖ℎ𝑘. As 𝑖 is monic, it follows that ℎ𝑘 = 1ℎ (𝑋 ) . Thus, the inverse function 𝑘 : ℎ(𝑋 ) → 𝑋 of ℎ is continuous, so 47

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5.2

Colimits: definition and example

ℎ is an embedding. Conversely, assume that ℎ : 𝑋 → 𝑌 is an embedding. Consider the pushout ℎ

𝑋

𝑌 𝑔

ℎ

𝑌 𝑓

𝑍.

By the construction of pushouts in Top, 𝑍 = 𝑌 q 𝑌 /∼ where an element ℎ(𝑥) ∈ 𝑌 in the first summand is identified with the same element ℎ(𝑥) ∈ 𝑌 in the second summand, and 𝑓 , 𝑔 are the inclusions. The equaliser of 𝑓 , 𝑔 is then the inclusion 𝑌 0 → 𝑌 of the subspace 𝑌 0 of 𝑌 where both 𝑓 and 𝑔 are equal, i.e. ℎ(𝑋 ) → − 𝑌 . Since ℎ is an embedding this is the same as ℎ : 𝑋 → 𝑌 . Exercise 5.2.26. A regular epic is thus a map which is a coequaliser, and a split epic is a map with a right inverse. As in Exercise 5.2.25(a), split epic =⇒ regular epic =⇒ epic. (a) Let 𝑓 : 𝐴 → 𝐵 be map in a category. If 𝑓 is an isomorphism, then the equations 𝑓 −1 𝑓 = 1 and 𝑓 𝑓 −1 = 1 imply that 𝑓 is split epic and split monic, in particular monic and regular epic. Conversely, assume that 𝑓 is both monic and regular epic. Let 𝑔, ℎ : 𝐶 → 𝐴 be maps of which 𝑓 is a coequaliser. Then 𝑓 𝑔 = 𝑓 ℎ, so 𝑔 = ℎ as 𝑓 is monic. By Exercise 5.2.21, 𝑓 is an isomorphism (b) It suffices to prove that epic =⇒ split epic in Set. Assume 𝑓 : 𝐴 → 𝐵 is epic in Set, i.e. 𝑓 is surjective. By the axiom of choice it follows that 𝑓 has a section, that is, there exists 𝑔 : 𝐵 → 𝐴 such that 𝑓 𝑔 = 1𝐵 . Thus 𝑓 is split epic. (c) In Top the epimorphisms are precisely the surjective (continuous) maps. Let 𝑋 ∈ Top and suppose that 𝑋 0 ∈ Top is another space with the same underlying set as 𝑋, but whose topology is strictly finer. Then the identity 𝑋 0 → 𝑋 is a map in Top, and is epic, but is is not split since the identity 𝑋 → 𝑋 0 is not continuous. It follows from (b) that Top does not satisfy the axiom of choice. In Grp the epics are precisely the surjections (c.f. Exercise 5.2.24 and Mac Lane’s Exercise I.5.5.) Consider the quotient map 𝜋 : Z ∗ Z → Z ⊕ Z of the free group Z ∗ Z on two generators onto its abelianisation Z ⊕ Z. Then 𝜋 is epic, but is not split. Indeed, no map Z ⊕ Z → Z ∗ Z is as since Z ⊕ Z is countable and Z ∗ Z is uncountable. It follows from (b) that Grp does not satisfy the axiom of choice. Exercise 5.2.27. First we analyse stability under pullbacks. By Exercise 5.1.42, monics are stable under pullbacks. Now consider the category generated by the graph •

𝑎

• 𝑐

𝑏

𝑓

•

𝑒

•

• 𝑔

with the relations 𝑓 𝑒 = 𝑔𝑒, 𝑒𝑏 = 𝑐𝑎 and 𝑓 𝑒𝑏 = 𝑔𝑒𝑏. Then 𝑒 is the equaliser of 𝑓 and 𝑔, the square is a pullback, but 𝑎 is not an equaliser, as can be check. It follows that regular monics are not stable 48

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5.2

Colimits: definition and example

under pullbacks. Finally, note that split monics are not stable under pullbacks either: let 𝑋 and 𝑌 be disjoint empty subsets of a set 𝑍 . Then ∅

𝑋

𝑌

𝑍

is a pullback on Set, where 𝑋 ↩− → 𝑍 is split monic but ∅ ↩− → 𝑌 is not. Next, epics are not stable under pullbacks. Consider the category Haus of Hausdorff topological spaces. If 𝐴 ⊂ 𝐵 is an inclusion of a dense subspace in Haus, it is epic, for if two maps into a Hausdorff space agree on a dense subset, they are equal. It follows that in the pullback diagram ∅

Q

R\Q

R

in Haus the map Q ← −↪ R is epic, but ∅ → R \ Q is clearly not. Now, an example similar to the one of regular monics above (where now 𝑓 and 𝑔 are maps into the left-bottom corner) shows that regular epics are not stable under pullbacks either. Finally, note that split epics are stable under pullbacks. Indeed, let 𝑓0

𝑊

𝑌

𝑔0

𝑔

𝑋

𝑍 𝑓

be a pullback square in some category, where 𝑓 is split epic. Then there exists 𝑠 : 𝑍 → 𝑋 a right inverse of 𝑓 . Then the maps 1𝑌 : 𝑌 → 𝑌 and 𝑠𝑔 : 𝑌 → 𝑋 satisfy 𝑔1𝑌 = 𝑓 𝑠𝑔, so by the universal property of the pullback there is a map 𝑠 0 : 𝑌 → 𝑊 such that 𝑓 0𝑠 0 = 1𝑌 . Thus 𝑓 0 is split epic. Now we analyse stability under composition. If 𝑓 and 𝑔 are monic and ℎ 1, ℎ 2 are maps such that 𝑓 𝑔ℎ 1 = 𝑓 𝑔ℎ 2, we have 𝑔ℎ 1 = 𝑔ℎ 2 since 𝑓 is monic, and in turn ℎ 1 = ℎ 2 since 𝑔 is monic. Thus monics are stable under composition. Dually, epics are stable under composition too. Finally, regular monics are not stable under compositions. Consider the full subcategory FHaus of Haus spanned by the functionally Hausdorff spaces (or completely Hausdorff spaces): those spaces 𝑋 such that for any 𝑥, 𝑦 ∈ 𝑋, there exists a continuous map 𝑓 : [0, 1] → 𝑋 such that 𝑓 (0) = 𝑥 and 𝑓 (1) = 𝑦. Let 𝐴 = {1/𝑛 | 𝑛 ∈ Z+ }, 𝐵 = 𝐴 ∪ {0}, and 𝐶 the subspace whose underlying set is R and having basis 𝒯 ∪ {R\𝐴} for its topology, where 𝒯 is a basis for the standard topology on R. Then it can be prove that the inclusions 𝐴 ⊂ 𝐵 and 𝐵 ⊂ 𝐶 are regular monics in FHaus, but their composition is not. Details would take us too far into analysis and hence are omitted. Since regular epics in a category 𝒞 are precisely regular monics in 𝒞 op, it follows that regular epics are not stable under composition either. 49

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5.3

5.3

Interactions between functors and limits

Interactions between functors and limits

Exercise 5.3.8. Let 𝐹 : 𝒜 × 𝒜 be given on objects by 𝐹 (𝑋, 𝑌 ) = 𝑋 × 𝑌 (for our previously made choice). For a morphism (𝑓 , 𝑔) : (𝑋, 𝑌 ) → (𝑋 0, 𝑌 0) in 𝒜 × 𝒜 define 𝐹 (𝑓 , 𝑔) : 𝑋 ×𝑌 → 𝑋 0 ×𝑌 0 to be the unique map 𝑋 ×𝑌 → 𝑋 0 ×𝑌 0 induced by the maps 𝑓 ◦𝑝 1𝑋 ,𝑌 : 𝑋 ×𝑌 → 𝑋 0, 𝑔 ◦𝑝 2𝑋 ,𝑌 : 𝑋 ×𝑌 → 𝑌 0 using the universal property of the product 𝑋 0 × 𝑌 0 . Then clearly 𝐹 (1𝑋 , 1𝑌 ) = 1𝑋 ,𝑌 . Consider two maps (𝑓 , 𝑔) : (𝑋, 𝑌 ) → (𝑋 0, 𝑌 0) and (ℎ, 𝑘) : (𝑋 0, 𝑌 0) → (𝑋 00, 𝑌 00) in 𝒜 × 𝒜. Then 𝐹 (ℎ𝑓 , 𝑔𝑘) is the 00 00 00 00 unique map 𝑋 ×𝑌 → 𝑋 00 ×𝑌 00 such that 𝑝 1𝑋 ,𝑌 ◦ 𝐹 (ℎ𝑓 , 𝑘𝑔) = ℎ ◦ 𝑓 ◦ 𝑝 1𝑋 ,𝑌 and 𝑝 2𝑋 ,𝑌 ◦ 𝐹 (ℎ𝑓 , 𝑘𝑔) = 00 00 0 0 𝑘 ◦ 𝑔 ◦ 𝑝 2𝑋 ,𝑌 . Since 𝑝 1𝑋 ,𝑌 ◦ 𝐹 (ℎ, 𝑘) = ℎ ◦ 𝑝 1𝑋 ,𝑌 we have 𝑝 1𝑋 00

00,𝑌 00

0

0

◦ 𝐹 (ℎ, 𝑘) ◦ 𝐹 (𝑓 , 𝑔) = ℎ ◦ 𝑝 1𝑋 ,𝑌 ◦ 𝐹 (𝑓 , 𝑔) = ℎ ◦ 𝑓 ◦ 𝑝 1𝑋 ,𝑌 .

00

Similarly 𝑝 2𝑋 ,𝑌 ◦ 𝐹 (ℎ, 𝑘) ◦ 𝐹 (𝑓 , 𝑔) = 𝑘 ◦ 𝑔 ◦ 𝑝 1𝑋 ,𝑌 . By uniqueness it follows that 𝐹 (ℎ𝑓 , 𝑔𝑘) = 𝐹 (ℎ, 𝑘) ◦ 𝐹 (𝑓 , 𝑔). Thus 𝐹 is indeed a functor. Exercise 5.3.9. Given 𝐴, 𝑋, 𝑌 ∈ 𝒜 define 𝜑𝐴,𝑋 ,𝑌 : 𝒜(𝐴, 𝑋 × 𝑌 ) → 𝒜(𝐴, 𝑋 ) × 𝒜(𝐴, 𝑌 ) 𝑓 ↦→ (pr𝑋 ◦𝑓 , pr𝑌 ◦𝑓 ). Then 𝜑𝐴,𝑋 ,𝑌 is bijective with inverse given by (𝑔, ℎ) ↦→ 𝑓 , where 𝑓 is the unique map 𝐴 → 𝑋 × 𝑌 induced by the universal property of the product. Given 𝑔 : 𝐴 0 → 𝐴, ℎ : 𝑋 → 𝑋 0 and 𝑘 : 𝑌 → 𝑌 0 consider the diagram 𝒜(𝐴, 𝑋 × 𝑌 )

𝜑𝐴,𝑋 ,𝑌

𝒜(𝐴, 𝑋 ) × 𝒜(𝐴, 𝑌 )

𝑔∗ (ℎ×𝑘)∗

(𝑔∗ℎ ∗ ,𝑔∗𝑘 ∗ )

𝒜(𝐴 0, 𝑋 0 × 𝑌 0) 𝜑𝐴0,𝑋 0,𝑌 0 𝒜(𝐴 0, 𝑋 0) × 𝒜(𝐴 0, 𝑋 0) . For any 𝑓 : 𝐴 → 𝑋 × 𝑌 we have (𝑔∗ℎ ∗, 𝑔∗𝑘 ∗ ) ◦ 𝜑𝐴,𝑋 ,𝑌 (𝑓 ) = (ℎ ◦ pr𝑋 ◦𝑓 ◦ 𝑔, 𝑘 ◦ pr𝑌 ◦𝑓 ◦ 𝑔) and

𝜑𝐴0,𝑋 0,𝑌 0 ◦ 𝑔∗ (ℎ × 𝑘)∗ (𝑓 ) = (pr𝑋 0 ◦(ℎ × 𝑘) ◦ 𝑓 ◦ 𝑔, pr𝑌 0 ◦(ℎ × 𝑘) ◦ 𝑓 ◦ 𝑔).

Thus, proving naturality of 𝜑 reduces to proving commutativity of the diagrams 𝑋 ×𝑌

ℎ×𝑘

𝑋0 ×𝑌0

𝑋 ×𝑌

pr𝑋 0

pr𝑋

𝑋 ℎ

ℎ×𝑘

pr𝑌 0

pr𝑌

𝑋0,

𝑋0 ×𝑌0

𝑌 𝑘

𝑌0.

But ℎ × 𝑘 is by definition the unique map such that the above diagrams commute. It follows that 𝜑𝐴,𝑋 ,𝑌 is an isomorphism 𝒜(𝐴, 𝑋 × 𝑌 )  𝒜(𝐴, 𝑋 ) × 𝒜(𝐴, 𝑌 ) natural in 𝐴, 𝑋, 𝑌 ∈ 𝒜. 50

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5.3

Interactions between functors and limits

Exercise 5.3.10. Let 𝐹 : 𝒜 → ℬ create limits. Let 𝐷 : I → 𝒜 be a diagram. Then, for any limit 𝑞𝐼

𝑝𝐼

cone (𝐵 −→ 𝐹 𝐷 (𝐼 ))𝐼 ∈I on 𝐹 𝐷, there is a unique cone (𝐴 −→ 𝐷 (𝐼 ))𝐼 ∈I on 𝐷 such that 𝐹 (𝐴) = 𝐵 and 𝑝𝐼

𝑝𝐼

𝐹 (𝑝 𝐼 ) = 𝑞𝐼 for all 𝐼 ∈ I, and the cone (𝐴 −→ 𝐷 (𝐼 ))𝐼 ∈I is a limit cone. In particular, if (𝐴 −→ 𝐷 (𝐼 ))𝐼 ∈I 𝐹 (𝑝 𝐼 )

𝑝𝐼

is a cone on 𝐷 such that (𝐹 (𝐴) −−−−→ 𝐹 𝐷 (𝐼 ))𝐼 ∈I is a limit cone on 𝐹 𝐷, then (𝐴 −→ 𝐷 (𝐼 ))𝐼 ∈I is the unique cone given by the property of creating limits, and hence it is a limit cone. Therefore 𝐹 reflects limits. Exercise 5.3.11. (a) If 𝐷 : I → Set is a diagram then 𝑢

− 𝐽 in I} 𝐿 = lim 𝐷  {(𝑥 𝐼 )𝐼 ∈I | 𝑥 𝐼 ∈ 𝐷 (𝐼 ) for all 𝐼 ∈ I and (𝐷𝑢) (𝑥 𝐼 ) = 𝑥 𝐽 for all 𝐼 → →I 𝑝𝐽

is the limit cone on 𝐷 with projections 𝐿 −−→ 𝐷 (𝐽 ) given by 𝑝 𝐽 ((𝑥 𝐼 )𝐼 ∈I ) = 𝑥 𝐽 for all 𝐽 ∈ I. This formula was given in Example 5.1.22 and proven in Exercise 5.1.37. 𝑞𝐼 Let 𝐷 : I → Grp be a diagram in Grp and (𝐵 −→ 𝑈 𝐷 (𝐼 ))𝐼 ∈I be a limit cone on 𝑈 𝐷 in Set. Then 𝑢

− 𝐽 in I} ∈ Set 𝐵 0 = {(𝑥 𝐼 )𝐼 ∈I | 𝑥 𝐼 ∈ 𝑈 𝐷 (𝐼 ) for all 𝐼 ∈ I and (𝑈 𝐷𝑢) (𝑥 𝐼 ) = 𝑥 𝐽 for all 𝐼 → 𝑞0𝐽

is also a limit cone on 𝑈 𝐷 with projections −−→ 𝑈 𝐷 (𝐽 ) given by 𝑞 0𝐽 ((𝑥 𝐼 )𝐼 ∈I ) = 𝑥 𝐽 for all 𝐽 ∈ I, so there is a unique isomorphism ℎ : 𝐵 → 𝐵 0 such that 𝑞 0𝐽 ℎ = 𝑞 𝐽 for all 𝐽 ∈ I. Î Endow the set 𝐵 0 with a canonical group structure as a subgroup of 𝑖 ∈I 𝐷 (𝐼 ) ∈ Grp, call this group 𝐴 0, and let 𝑝 𝐽0 : 𝐴 0 → 𝐷 (𝐽 ), for 𝐽 ∈ I, denote the projection homomorphism whose underlying set function is 𝑞 0𝐽 . Now, the bijection ℎ : 𝐵 → 𝐵 0 endows 𝐵 with a group structure, and call this group 𝐴, so that ℎ : 𝐴 → 𝐴 0 is an isomorphism of groups. For 𝐽 ∈ I, let 𝑝 𝐽 : 𝐴 → 𝐷 (𝐽 ) be the projection, so that its underlying set function is 𝑞 𝐽 . Then 𝑈 (𝐴) = 𝐵 and 𝑈 (𝑝 𝐼0 ) = 𝑝 𝐼 for all 𝐼 ∈ I, and clearly (𝐴, (𝑝 𝐼0 )𝐼 ∈I ) is unique as such. 𝐵0

𝑝 𝐼0

𝑓𝐼

It remains to show that (𝐴 −→ 𝐷 (𝐼 ))𝐼 ∈I is a limit cone on 𝐷. So let (𝐶 − → 𝐷 (𝐼 ))𝐼 ∈I be a cone 𝑈 (𝑓𝐼 )

on 𝐷. Then (𝑈 (𝐶) −−−−→ 𝑈 𝐷 (𝐼 ))𝐼 ∈I is a cone on 𝑈 𝐷, so there is a unique map 𝑔 : 𝑈 (𝐶) → 𝐵 such that 𝑝 𝐼 𝑔 = 𝑈 (𝑓𝐼 ) for all 𝐼 ∈ I. What we must prove is that 𝑔 is a group homomorphism as a map 𝐶 → 𝐴. But composing with ℎ gives a map ℎ𝑔 : 𝑈 (𝐶) → 𝐵 0 which is the unique one given by the universal property. This ℎ𝑔 is a group homomorphism when viewed as a map 𝐶 → 𝐴 0 . Since we defined the group structure on 𝐴 declaring ℎ to be an isomorphism of groups, it follows that 𝑔 : 𝐶 → 𝐴 is indeed a homomorphism, and is of course the unique such that 𝑝 𝐼0𝑔 = 𝑓𝐼 for all 𝐼 ∈ I. 𝑝 𝐼0

Thus (𝐴 −→ 𝐷 (𝐼 ))𝐼 ∈I is a limit cone on 𝐷. We conclude that 𝑈 : Grp → Set creates limits. (b) The above proof works when Grp is replaced by Ab, Ring or Vect𝑘 . 𝑞𝐼

Exercise 5.3.12. Let 𝐷 : I → 𝒜 be a diagram. Then there exists a limit cone (𝐵 −→ 𝐹 𝐷 (𝐼 ))𝐼 ∈I on 𝑝𝐼

𝐹 𝐷. Since 𝐹 creates limits there is a unique cone (𝐴 −→ 𝐷 (𝐼 ))𝐼 ∈I such that 𝐹 (𝐴) = 𝐵, 𝐹 (𝑝 𝐼 ) = 𝑞𝐼 𝑝𝐼

for all 𝐼 ∈ I, and (𝐴 −→ 𝐷 (𝐼 ))𝐼 ∈I is a limit cone on 𝐷. It follows 𝒜 has limits of shape I. 51

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5.3

Interactions between functors and limits 𝑝𝐼

𝑞𝐼

Now suppose (𝐴 −→ 𝐷 (𝐼 ))𝐼 ∈I is a limit cone on 𝐷. There exists a limit cone (𝐵 −→ 𝐹 𝐷 (𝐼 ))𝐼 ∈I 𝑝 𝐼0

𝑝 𝐼0

on 𝐹 𝐷, and a unique cone (𝐴 0 −→ 𝐷 (𝐼 ))𝐼 ∈I such that 𝐹 (𝐴 0) = 𝐵, 𝐹 (𝑝 𝐼0 ) = 𝑞𝐼 for all 𝐼 ∈ I, and (𝐴 0 −→ 𝐷 (𝐼 ))𝐼 ∈I is a limit cone on 𝐷. Thus there is a unique isomorphism ℎ : 𝐴 0 → 𝐴 such that 𝑝 𝐼 ℎ = 𝑝 𝐼0 𝐹 (𝑝 𝐼 )

for all 𝐼 ∈ I. Then 𝐹 (ℎ) : 𝐵 → 𝐹 (𝐴) is an isomorphism. Consider the cone (𝐹 (𝐴) −−−−→ 𝐹 𝐷 (𝐼 ))𝐼 ∈I . 𝑓𝐼

Given any cone (𝐶 − → 𝐹 𝐷 (𝐼 ))𝐼 ∈I, there is a unique 𝑓 : 𝐶 → 𝐵 such that 𝑞𝐼 𝑓 = 𝑓𝐼 for all 𝐼 ∈ I. Then 𝐹 (ℎ)𝑓 : 𝐶 → 𝐹 (𝐴) satisfies 𝐹 (𝑝 𝐼 )𝐹 (ℎ)𝑓 = 𝐹 (𝑝 𝐼0 ) 𝑓 = 𝑓𝐼 for all 𝐼 ∈ I. Moreover, if ℎ 0 : 𝐶 → 𝐹 (𝐴) satisfies 𝐹 (𝑝 𝐼 )ℎ 0 = 𝑓𝐼 for each 𝐼 then 𝐹 (ℎ) −1ℎ 0 satisfies 𝑞𝐼 𝐹 (ℎ) −1ℎ 0 = 𝐹 (𝑝 𝐼0ℎ −1 )ℎ 0 = 𝐹 (𝑝 𝐼 )ℎ 0 = 𝑓𝐼 , so we must have 𝐹 (ℎ) −1ℎ 0 = 𝑓 by uniqueness of 𝑓 , i.e. ℎ 0 = 𝐹 (ℎ)𝑓 is unique. It follows that 𝐹 (𝑝 𝐼 )

(𝐹 (𝐴) −−−−→ 𝐹 𝐷 (𝐼 ))𝐼 ∈I is a limit cone on 𝐹 𝐷. We conclude that 𝐹 preserves limits. Exercise 5.3.13. (a) Let 𝑆 ∈ Set be arbitrary and let 𝑓 : 𝐵 → 𝐵 0 be epic in ℬ. Then 𝐺 (𝑓 ) : 𝐺 (𝐵) → 𝐺 (𝐵 0) is epic in Set. By Exercise 5.2.26(b), 𝐺 (𝑓 ) has a right inverse, say ℎ : 𝐺 (𝐵 0) → 𝐺 (𝐵). Then the induced map 𝐺 (𝑓 )∗ : Set(𝑆, 𝐺 (𝐵)) → Set(𝑆, 𝐺 (𝐵 0)) has a right inverse, namely ℎ ∗ : Set(𝑆, 𝐺 (𝐵 0)) → Set(𝑆, 𝐺 (𝐵)). In particular, 𝐺 (𝑓 )∗ is surjective. Since 𝐺 a 𝐹 there is an isomorphism 𝜑𝑆,𝐵 : ℬ(𝐹 (𝑆), 𝐵) → 𝒜(𝑆, 𝐺 (𝐵)) natural in 𝑆 ∈ Set and 𝐺 (𝐵), so that we have a commutative square ℬ(𝐹 (𝑆), 𝐵)

𝜑𝑆,𝐵

Set(𝑆, 𝐺 (𝐵)) 𝐺 (𝑓 )∗

𝑓∗

ℬ(𝐹 (𝑆), 𝐵 0)

𝜑𝑆,𝐵0

Set(𝑆, 𝐺 (𝐵 0)) .

Since 𝜑𝑆,𝐵 , 𝜑𝑆,𝐵0 are isomorphisms and 𝐺 (𝑓 )∗ is surjective, it follows that ℬ(𝐹 (𝑆), 𝑓 ) is surjective, i.e. it is epic. Therefore 𝐹 (𝑆) is projective. We conclude that 𝐹 (𝑆) is projective for all sets 𝑆. (b) Recall that a map in Ab is epic if and only if it is surjective (Example 5.2.19). Consider Z/2Z ∈ Ab. Then the unique non-trivial map 𝑓 : Z → Z/2Z is epic in Ab, but 𝑓∗ : Ab(Z/2, Z) → Ab(Z/2Z, Z/2Z) is not epic, for Ab(Z/2Z, Z) = 0 and Ab(Z/2Z, Z/2Z)  Z/2Z. Thus Z/2Z is not projective in Ab. op (c) Let 𝑘 be a vector space and 𝑈 ∈ Vect𝑘 arbitrary. Let 𝑓 op : 𝑊 → 𝑉 be epic in Vect𝑘 , that is, 𝑓 : 𝑉 → 𝑊 is monic (i.e. injective) in Vect𝑘 . We shall prove that 𝑓 ∗ : Vect𝑘 (𝑊 , 𝑈 ) → Vect𝑘 (𝑉 , 𝑈 ) is surjective. Given a linear map 𝐿 : 𝑉 → 𝑈 we need e 𝐿 : 𝑊 → 𝑈 such that the diagram 𝑉 𝐿 𝑓

𝑊

𝑈 e 𝐿

commutes. Taking a basis {𝑣𝑖 }𝑖 ∈𝐽 for 𝑉 then {𝑓 (𝑣𝑖 )}𝑖 ∈𝐽 is linear independent in 𝑊 , so extends to a basis for 𝑊 . Then we may define e 𝐿 by sending each 𝑓 (𝑣𝑖 ) to 𝐿(𝑣𝑖 ) and all other basis elements to 0. Thus 𝑈 is injective. Since 𝑈 was arbitrary, it follows that every 𝑘-vector space is injective. Now consider Z ∈ Ab and 𝑓 : Z → Q the inclusion map, which is monic. Then 𝑓 ∗ : Ab(Q, Z) → Ab(Z, Z) is not epic, for Ab(Q, Z) = 0 and Ab(Z, Z) = Z. It follows that Z is not injective in Ab. 52

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6. Adjoints, representables and limits

6 6.1

Adjoints, representables and limits Limits in terms of representables and adjoints

Exercise 6.1.5. Let I be the discrete category with two object 𝑖 0 and 𝑖 1 . Let 𝒜 be a category. A functor 𝐷 : I → 𝒜 is simply a pair of objects 𝐷 (𝑖 0 ) and 𝐷 (𝑖 1 ) of 𝒜. For 𝐴 ∈ 𝒜, the diagonal functor Δ𝒜 → [I, 𝒜] = 𝒜 × 𝒜 is simply given by Δ(𝐴) = (𝐴, 𝐴). For 𝐷 : I → 𝒜 and 𝐴 ∈ 𝒜, a cone on 𝐷 with vertex 𝐴 is a pair of maps 𝐴 → 𝐷 (𝑖 0 ) and 𝐴 → 𝐷 (𝑖 1 ). Proposition 6.1.1 tells us that a limit of 𝐷 is a product (𝐷 (𝑖 0 ) × 𝐷 (𝑖 1 ), (𝑝𝑘 : 𝐷 (𝑖 0 ) × 𝐷 (𝑖 1 ) → 𝐷 (𝑖𝑘 ))𝑘=0,1 ); this is Example 5.1.21(a). Corollary 6.1.2 tells us that products are unique up to isomorphism. Lemma 6.1.3(a) tells us that given two products (𝐷 (𝑖 0 )×𝐷 (𝑖 1 ), (𝑝𝑘 )𝑘=0,1 ) and (𝐷 0 (𝑖 0 )×𝐷 0 (𝑖 1 ), (𝑝𝑘0 )𝑘=0,1 ) in 𝒜 and maps 𝛼𝑘 : 𝐷 (𝑖𝑘 ) → 𝐷 0 (𝑖𝑘 ), 𝑘 = 0, 1, there is a unique map 𝛼 : 𝐷 (𝑖 0 ) × 𝐷 (𝑖 1 ) → 𝐷 0 (𝑖 0 ) × 𝐷 0 (𝑖 1 ) such that 𝛼𝑝𝑘0 = 𝑝𝑘 for 𝑘 = 0, 1. (cf. Exercise 5.3.8). Lemma 6.1.3(b) says that if we further have objects 𝐴, 𝐴 0 ∈ 𝒜 with maps 𝑓𝑘 : 𝐴 → 𝐷 (𝑖𝑘 ) and 𝑓𝑘0 : 𝐴 0 → 𝐷 0 (𝑖𝑘 ) for 𝑘 = 0, 1, and a map 𝑠 : 𝐴 → 𝐴 0 such that 𝛼𝑘 𝑓𝑘 = 𝑓𝑘0𝑠 for 𝑘 = 0, 1, then 𝛼 𝑓 = 𝑓 0 𝑠, where 𝑓 and 𝑓 0 are induced by the 𝑓𝑘 and the 𝑓𝑘0 respectively. Proposition 6.1.4 says that taking products is a functor (after taking choices, see Exercise 5.3.8) is right adjoint to the diagonal functor 𝒜 → 𝒜 × 𝒜. This was shown in Exercise 3.1.1. Exercise 6.1.6. If I = 𝐺 is a group (with unique object ∗), then [𝐺, Set] is the category of left 𝐺-sets and 𝐺-equivariant maps (Example 1.3.4). The diagonal functor Δ : Set → [𝐺, Set] equips a set 𝐴 with the trivial left 𝐺-action, such that 𝑔 · 𝑎 = 𝑎 for all 𝑔 ∈ 𝐺 and 𝑎 ∈ 𝐴. Let 𝐷 : 𝐺 → Set be a left 𝐺-set and write 𝑋 = 𝐷 (∗). A cone on 𝐷 consists of a set 𝑌 together with a function 𝑓 : 𝑌 → 𝑋 such that 𝑔 · 𝑓 (𝑥) = 𝑓 (𝑥) for all 𝑥 ∈ 𝑋 and 𝑔 ∈ 𝐺, that is, the image of 𝑓 is contained in the 𝐺-fixed point subset 𝑋 𝐺 = {𝑥 ∈ 𝑋 | 𝑔 · 𝑥 = 𝑥 for all 𝑔 ∈ 𝐺 } of 𝑋 . It follows that 𝑋 𝐺 together with the inclusion 𝑖 : 𝑋 𝐺 → 𝑋 is a limit cone on 𝐷. Proposition 6.1.4 tells us that the functor (−)𝐺 : [𝐺, Set] → Set sending a 𝐺-set 𝑋 to its 𝐺-fixed point subset 𝑋 𝐺 is right adjoint to the functor Δ : Set → [𝐺, Set] which equips a set with the trivial left 𝐺-action. Now we consider the dual. A cocone on 𝐷 consists of a set 𝑌 together with a function 𝑓 : 𝑋 → 𝑌 such that 𝑓 (𝑔 · 𝑥) = 𝑥 for all 𝑔 ∈ 𝐺 and 𝑥 ∈ 𝑋, that is, 𝑓 factors through the set 𝑋 /𝐺 = {𝐺 · 𝑥 | 𝑥 ∈ 𝑋 } of orbits of 𝑋 under 𝐺 . It follows that 𝑋 /𝐺 together with the quotient 𝑞 : 𝑋 → 𝑋 /𝐺 is a colimit of 𝐷. The dual of Proposition 6.1.4 tells us that the functor (−)/𝐺 : [𝐺, Set] → Set sending a 𝐺-set 𝑋 to its set 𝑋 /𝐺 of orbits under 𝐺 is left adjoint to the functor Δ : Set → [𝐺, Set] which endows a set with the trivial left 𝐺-action.

6.2

Limits and colimits of presheaves

Exercise 6.2.20. (a) Let 𝛼 : A → 𝒮 be a natural transformation. First assume that each 𝛼𝐴, 𝐴 ∈ A, is monic in 𝒮. If 𝛽, 𝛾 : A → 𝒮 are natural transformations such that 𝛼𝛽 and 𝛼𝛾 are defined and equal, so that 𝛼𝐴 𝛽𝐴 = 𝛼𝐴𝛾𝐴 for all 𝐴 ∈ 𝒜, we have 𝛽𝐴 = 𝛾𝐴 for all 𝐴 as each 𝛼𝐴 is monic. Thus 𝛽 = 𝛾 and it follows that 𝛼 is monic in [A, 𝒮]. 53

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6.2

Limits and colimits of presheaves Conversely, assume that 𝛼 is monic in [A, 𝒮]. By Lemma 5.1.32, the square 𝑋

1𝑋

𝑋

1𝑋

𝛼

𝑋

𝛼

𝑌

is a pullback in [A, 𝒮]. Since 𝒮 has pullbacks, by Corollary 6.2.6 it follows that 𝑋 (𝐴)

1𝑋 (𝐴)

1𝑋 (𝐴)

𝑋 (𝐴) 𝛼𝐴

𝑋 (𝐴)

𝛼𝐴

𝑌 (𝐴)

is a pullback in 𝒮 for all 𝐴 ∈ 𝒜. By Lemma 5.1.32 again, we deduce that each 𝛼𝐴, 𝐴 ∈ 𝒜, is monic in 𝒮. (b) By part (a), 𝛼 ∈ [Aop, Set] is monic (respectively epic) if and only if 𝛼𝐴 is monic (respectively epic) in Set for all 𝐴 ∈ A. (c) Let 𝛼 : 𝑋 ⇒ 𝑌 be a natural transformation between functors Aop → Set. If each 𝛼𝐴 is monic (epic), then the proof given in (a) that 𝛼 is monic (epic) in [Aop, Set] did not rely on the fact that (co)limits of presheaves are computed pointwise. It remains to prove that if 𝛼 is monic (epic), then so is 𝛼𝐴 for each 𝐴 ∈ Aop, without relying on this fact. We consider the case of 𝛼 monic, the case of 𝛼 epic is dual. Suppose for the sake of contradiction that there exists 𝐴 ∈ A such that 𝛼𝐴 is not monic. Then there exist 𝑎, 𝑎 0 ∈ 𝑋 (𝐴) such that 𝑎 ≠ 𝑎 0 and 𝛼𝐴 (𝑎) = 𝛼𝐴 (𝑎 0) ∈ 𝑌 (𝐴). By the Yoneda Lemma, we have an isomorphism 𝑋 (𝐴)  [Aop, Set] (𝐻𝐴, 𝑋 ) under which 𝑎 and 𝑎 0 correspond to functors 𝛽 : 𝐻𝐴 ⇒ 𝑋 and 𝛽 0 : 𝐻𝐴 ⇒ 𝑋, respectively, determined uniquely by the conditions 𝛽𝐴 (1𝐴 ) = 𝑎 and 𝛽𝐴0 (1𝐴 ) = 𝑎 0 . Thus 𝛽 ≠ 𝛽 0 and 𝛼𝛽 = 𝛼𝛽 0, contradicting the fact that 𝛼 is monic. We deduce that if 𝛼 is monic in [Aop, Set], then 𝛼𝐴 is monic in Set for all 𝐴 ∈ Aop . Exercise 6.2.21. We will write the sum 𝑋 + 𝑌 in coproduct notation 𝑋 q 𝑌 . By definition, given functor 𝑋, 𝑌 : 𝒜 op → Set then 𝑋 q𝑌 is the functor 𝒜 op → Set given by (𝑋 q𝑌 ) (𝐴) = 𝑋 (𝐴)q𝑌 (𝐴), the disjoint union of 𝑋𝐴 and 𝑌𝐴, for all 𝐴 ∈ 𝒜, and similarly (𝑋 q 𝑌 ) (𝑓 ) = 𝑋 (𝑓 ) q 𝑌 (𝑓 ) on morphisms. Suppose that 𝐴 ∈ 𝒜 is such that there exists an isomorphism 𝛼 : 𝐻𝐴 → 𝑋 q 𝑌 . Then for every 𝐵 ∈ 𝒜, the function 𝛼 𝐵 : 𝒜(𝐵, 𝐴) → 𝑋 (𝐵) q 𝑌 (𝐵) is bijective. Considering 1𝐴 ∈ 𝒜(𝐴, 𝐴), we must have either 1𝐴 ∈ 𝛼 −1 (𝑋 (𝐴)) or 1𝐴 ∈ 𝛼𝐴−1 (𝑌 (𝐴)). Suppose that 1𝐴 ∈ 𝛼𝐴−1 (𝑋 (𝐴)); we will prove that 𝑌 is constant with value ∅. Let 𝐵 ∈ 𝒜. If 𝒜(𝐵, 𝐴) = ∅, then 𝑌 (𝐵) = ∅ as 𝛼 𝐵 is bijective. So assume 𝒜(𝐵, 𝐴) ≠ ∅ and let 𝑓 : 𝐵 → 𝐴 be any map. SBy naturality of 𝛼, the following diagram is commutative: 𝛼𝐴

𝒜(𝐴, 𝐴)

𝑋 (𝐴) q 𝑌 (𝐴)

𝑓∗

𝒜(𝐵, 𝐴)

𝑋 (𝑓 ) q𝑌 (𝑓 ) 𝛼𝐴

𝑋 (𝐵) q 𝑌 (𝐵) . 54

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6.2

Limits and colimits of presheaves

Since 1𝐴 ∈ 𝛼𝐴−1 (𝑋 (𝐴)), it follows that 𝛼𝐴 (𝑓 ) ∈ 𝑋 (𝐵). Since 𝛼𝐴 is a bijection and 𝑓 ∈ 𝒜(𝐵, 𝐴) was arbitrary, we deduce that 𝑌 (𝐵) = ∅. We conclude that 𝑌 is the constant functor ∅. Similarly, if 1𝐴 ∈ 𝛼𝐴−1 (𝑌 (𝐴)) then 𝑋 is constant with value ∅. (b) By part (a), it suffices to prove that no representable functor is constant with value ∅. For this, note that for an object 𝐴 in a locally small category 𝒜 we have 𝐻𝐴 (𝐴) = 𝒜(𝐴, 𝐴) ≠ ∅ as 1𝐴 ∈ 𝒜(𝐴, 𝐴). Exercise 6.2.22. Let A be a category and 𝑋 : Aop → Set a functor. Let {∗} be a one-point set and 𝐹 : 1 → Set the functor sending the unique object of 1 to {∗}. Then the category of element E(𝑋 ) of 𝑋 is precisely the comma category (𝐹 ⇒ 𝑋 ). Indeed, an object of (𝐹 ⇒ 𝑋 ) is a pair consisting of an object 𝐴 ∈ 𝒜 and an arrow 𝑥 : {∗} → 𝑋 (𝐴), the latter being the same as an object 𝑥 ∈ 𝑋 (𝐴). An arrow in (𝐹 ⇒ 𝑋 ) from (𝐴 0, 𝑥 0) to (𝐴, 𝑥) is an arrow 𝑓 : 𝐴 0 → 𝐴 in 𝒜 such that 𝑋 (𝑓 ) ◦𝑥 = 𝑥 0, that is, when 𝑥 and 𝑥 0 are considered as objects of 𝑋 (𝐴) and 𝑋 (𝐴 0) respectively, we have 𝑋 (𝑓 ) (𝑥) = 𝑥 0 . Exercise 6.2.23. Let 𝑋 be a presheaf on a locally small category 𝒜. First assume that 𝑋 is representable; we will prove that E(𝑋 ) h as a terminal object. It suffices to consider the case 𝑋 = 𝐻𝐴 for some 𝐴 ∈ 𝒜. In this case (𝐴, 1𝐴 ) is terminal in E(𝐻𝐴 ), for given (𝐵, 𝑓 : 𝐵 → 𝐴) ∈ E(𝐻𝐴 ) then 𝑓 : 𝐵 → 𝐴 is an arrow such that 𝑓 ∗ (1𝐴 ) = 𝑓 , the unique as such. Conversely, suppose that E(𝑋 ) has a terminal object (𝐴, 𝑥). This is an object 𝐴 ∈ 𝒜 and an element 𝑥 ∈ 𝑋 (𝐴) such that for any 𝐵 ∈ 𝒜 and 𝑦 ∈ 𝑋 (𝐵), there exists a unique map 𝑓 : 𝐵 → 𝐴 such that 𝑋 (𝑓 ) (𝑥) = 𝑦. It follows from Corollary 4.3.2 that 𝑋 is representable. Exercise 6.2.24. Let A be a small category and 𝑋 a presheaf on A. Consider the category B = E(𝑋 ) of elements of 𝑋 . Since A is small, so is E(𝑋 ). We claim there is an equivalence of categories [Aop, Set]/𝑋 ' [E(𝑋 ) op, Set]. By Proposition 1.3.18, it suffices to find a functor 𝐹 : [Aop, Set]/𝑋 → [E(𝑋 ) op, Set] full, faithful and essentially surjective on objects. Let (𝑌 , 𝛼) be an object of [Aop, Set]/𝑋, that is, 𝑌 is a presheaf on A and 𝛼 is a natural transformation 𝑌 ⇒ 𝑋 . Define 𝐹 (𝑌, 𝛼) : E(𝑋 ) op → Set as follows. Given (𝐴, 𝑥) ∈ E(𝑋 ), let 𝐹 (𝑌, 𝛼) send (𝐴, 𝑥) to the subset 𝛼𝐴−1 (𝑥) of 𝑌 (𝐴). If 𝑓 : (𝐴, 𝑥) → (𝐵, 𝑦) is an arrow in E(𝑋 ) and 𝑧 ∈ 𝛼 𝐵−1 (𝑦) ⊂ 𝑌 (𝐵), we claim 𝑌 (𝑓 ) : 𝑌 (𝐵) → 𝑌 (𝐴) sends 𝑧 to an element of 𝛼𝐴−1 (𝑥). Indeed, 𝛼𝐴 ◦ 𝑌 (𝑓 ) (𝑧) = 𝑋 (𝑓 ) ◦ 𝛼 𝐵 (𝑧) = 𝑋 (𝑓 ) (𝑦) = 𝑥 by naturality of 𝛼 and the fact that 𝑓 is an arrow (𝐴, 𝑥) → (𝐵, 𝑦) in E(𝑋 ). It follows that 𝐹 (𝑌 , 𝛼) is a functor E(𝑋 ) op → Set, given on morphisms by sending 𝑓 : (𝐴, 𝑥) → (𝐵, 𝑦) to the function 𝛼 𝐵−1 (𝑦) → 𝛼𝐴−1 (𝑥) obtained by restricting the domain and codomain of 𝑌 (𝑓 ). This defines a map 𝐹 : [Aop, Set]/𝑋 → [E(𝑋 ) op, Set] on objects, now we define 𝐹 on morphisms. Given a map 𝜃 : (𝑌 , 𝛼) → (𝑌 0, 𝛼 0) in [Aop, Set]/𝑋, then for each 𝐴 ∈ 𝒜, 𝜃 𝐴 is a function 𝑌 (𝐴) → 𝑌 0 (𝐴). Let (𝐴, 𝑥) ∈ E(𝑋 ) and 𝑧 ∈ 𝛼𝐴−1 (𝑥). Then 𝛼𝐴0 ◦ 𝜃 𝐴 (𝑧) = 𝛼𝐴 (𝑧) = 𝑥, so that 𝜃 𝐴 (𝑧) ∈ (𝛼𝐴0 ) −1 (𝑥). Thus, we obtain a function 𝐹 (𝜃 )(𝐴,𝑥) : 𝛼𝐴−1 (𝑥) → (𝛼𝐴0 ) −1 (𝑥) by restricting the domain and codomain of 𝜃 𝐴 . Moreover, given a map 𝑓 : (𝐴, 𝑥) → (𝐵, 𝑦) in E(𝑋 ) then the square on the left below commutes, being obtained from the diagram on the right by restricting

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domains and codomains: 𝛼 𝐵−1 (𝑦)

𝐹 (𝜃 ) (𝐵,𝑦)

(𝛼 𝐵0 ) −1 (𝑦) 𝐹 (𝑌 0,𝛼 0 ) (𝑓 )

𝐹 (𝑌 ,𝛼) ( 𝑓 )

𝛼𝐴−1 (𝑥)

𝑌 (𝐵)

𝐹 (𝜃 ) (𝐴,𝑥 )

𝜃𝐴

𝑌 0 (𝑓 )

𝑌 (𝑓 )

(𝛼𝐴0 ) −1 (𝑥) ,

𝑌 0 (𝐵)

𝑌 (𝐴) 𝜃𝐵

𝑌 0 (𝐴) .

Thus 𝐹 (𝜃 ) is a natural transformation 𝐹 (𝑌 , 𝛼) ⇒ 𝐹 (𝑌 0, 𝛼 0). It follows that 𝐹 is a functor from [Aop, Set]/𝑋 to [E(𝑋 ) op, Set]. First we show that 𝐹 is essentially surjective on objects. Let 𝑊 be a presheaf on E(𝑋 ). Define 𝑌 : Aop → Set as follows. Let 𝑌 (𝐴) = q𝑥 ∈𝑋 (𝐴)𝑊 (𝐴, 𝑥) on objects 𝐴 ∈ 𝒜. If 𝑓 : 𝐴 → 𝐵 is a morphism in 𝒜, then for each 𝑦 ∈ 𝑋 (𝐵) it is a morphism 𝑓 : (𝐴, 𝑥) → (𝐵, 𝑦), where 𝑥 = 𝑋 (𝑓 ) (𝑦), so let 𝑌 (𝑓 ) : q𝑦 ∈𝑋 (𝐵) 𝑊 (𝐵, 𝑦) → q𝑥 ∈𝑋 (𝐴)𝑊 (𝐴, 𝑥) be the function whose restriction summand 𝑊 (𝐵, 𝑦), 𝑦 ∈ 𝑋 (𝐵), is the map 𝑊 (𝑓 ) : 𝑊 (𝐵, 𝑦) → 𝑊 (𝐴, 𝑋 (𝑓 ) (𝑦)) follows by the inclusion into q𝑥 ∈𝑋 (𝐴)𝑊 (𝐴, 𝑥). Since 𝑊 is a functor, so is 𝑌 . Moreover, for each 𝐴 ∈ 𝒜 we have a function 𝛼𝐴 : 𝑌 (𝐴) → 𝑋 (𝐴) whose restriction to the summand𝑊 (𝐴, 𝑥), 𝑥 ∈ 𝑋 (𝐴), is constant with value 𝑥 . By definition of 𝑌 on morphisms, for 𝑓 : 𝐴 → 𝐵 a map in 𝒜, the diagram 𝛼𝐵 q𝑦 ∈𝑋 (𝐵)𝑊 (𝐵, 𝑦) 𝑋 (𝐵) 𝑋 (𝑓 )

𝑌 (𝑓 )

q𝑥 ∈𝑋 (𝐴)𝑊 (𝐴, 𝑥)

𝛼𝐴

𝑋 (𝐴)

commutes. Thus 𝛼 is a natural transformation 𝑌 ⇒ 𝑋, and therefore (𝑌 , 𝛼) is an object of [Aop, Set]/𝑋 . Moreover, by definition 𝐹 (𝑌 , 𝛼) sends an object (𝐴, 𝑥) ∈ E(𝑋 ) to 𝛼𝐴−1 (𝑥) = 𝑊 (𝐴, 𝑥) and a morphism 𝑓 : (𝐴, 𝑥) → (𝐵, 𝑦) to 𝑊 (𝑓 ), so that 𝐹 (𝑌 , 𝛼) = 𝑊 . It follows that 𝐹 is surjective on objects. It is left to prove that 𝐹 is full and faithful. For this purpose, let (𝑌 , 𝛼), (𝑌 0, 𝛼 0) ∈ [Aop, Set]/𝑋 . We shall prove that given a natural transformation 𝜀 : 𝐹 (𝑌 , 𝛼) ⇒ 𝐹 (𝑌 0, 𝛼 0), there is a unique map 𝜃 : (𝑌, 𝛼) → (𝑌 0, 𝛼 0) such that 𝐹 (𝜃 ) = 𝜀. To show this, note that for 𝐴 ∈ 𝒜, the set 𝑌 (𝐴) is the disjoint union q𝑥 ∈𝑋 (𝐴) 𝛼𝐴−1 (𝑥) (and similarly for 𝑌 0) so there is precisely one way to define 𝜃 𝐴 : 𝑌 (𝐴) → 𝑌 0 (𝐴) so that 𝐹 (𝜃 )(𝐴,𝑥) = 𝜀 (𝐴,𝑥) for all (𝐴, 𝑥) ∈ E(𝑋 ), namely the restriction of 𝜃 𝐴 to each summand 𝛼𝐴−1 (𝑥), 𝑥 ∈ 𝑋 (𝐴), of 𝑌 (𝐴) must be the function 𝜀 (𝐴,𝑥) : 𝛼𝐴−1 (𝑥) → (𝛼𝐴0 ) −1 (𝑥) followed by the inclusion (𝛼𝐴0 ) −1 (𝑥) ⊂ 𝑌 0 (𝐴). Moreover, 𝜃 : 𝑌 ⇒ 𝑌 0 is natural as 𝜀 is natural, and is a map (𝑌 , 𝛼) → (𝑌 0, 𝛼 0). It follows that 𝐹 is full and faithful. We conclude that 𝐹 is an equivalence of categories [Aop, Set]/𝑋 ' [E(𝑋 ) op, Set]. Exercise 6.2.25. (a) We must define Lan𝐹 𝑋 on morphisms. Let 𝑓 : 𝐵 → 𝐵 0 be a map in B. For (𝐴, ℎ) ∈ (𝐹 ⇒ 𝐵), let 𝑝 (𝐴,ℎ) : 𝑋 (𝐴) → Lan𝐹 𝑋 (𝐵) be the canonical map into the colimit, and similarly for (𝐴, 𝑘) ∈ (𝐹 ⇒ 𝐵 0) let 𝑞 (𝐴,𝑘) : 𝑋 (𝐴) → Lan𝐹 𝑋 (𝐵 0) be the canonical map into 56

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the colimit. Given (𝐴, ℎ) ∈ (𝐹 ⇒ 𝐵), then (𝐴, 𝑓 ℎ : 𝐹 (𝐴) → 𝐵 0) ∈ (𝐹 ⇒ 𝐵 0). Furthermore, if 𝑔 : (𝐴, ℎ) → (𝐴 0, ℎ 0) is a map in (𝐹 ⇒ 𝐵), then 𝑔 is also a map (𝐴, 𝑓 ℎ) → (𝐴 0, 𝑓 ℎ 0), so 𝑞 (𝐴0,𝑓 ℎ0) ◦ 𝑋 (𝑔) = 𝑞 (𝐴,𝑓 ℎ) . Thus Lan𝐹 𝑋 (𝐵 0) together with the maps 𝑞 (𝐴,𝑓 ℎ) is a cocone on 𝑋 ◦ 𝑃𝐵 . By the universal property of the colimit, there is a unique map Lan𝐹 𝑋 (𝑓 ) : Lan𝐹 𝑋 (𝐵) → Lan𝐹 𝑋 (𝐵 0) such that Lan𝐹 𝑋 (𝑓 ) ◦ 𝑝 (𝐴,ℎ) = 𝑞 (𝐴,𝑓 ℎ) for all (𝐴, ℎ) ∈ (𝐹 ⇒ 𝐵). Now let 𝑌 : B → 𝒮 be a functor. We want to prove that there is a canonical bijection [B, 𝒮] (Lan𝐹 𝑋, 𝑌 )  [A, 𝒮] (𝑋, 𝑌 𝐹 ). Let 𝛼 : Lan𝐹 𝑋 ⇒ 𝑌 be a natural transformation. Then, for each 𝐴 ∈ A, we have a map 𝛼 𝐹 (𝐴) : Lan𝐹 𝑋 (𝐹 (𝐴)) → 𝑌 𝐹 (𝐴), or to say the same thing, a cocone on 𝑋 ◦ 𝑃 𝐹 (𝐴) with vertex 𝑌 𝐹 (𝐴), say with structure maps 𝛽 𝐴(𝐶,ℎ) : 𝑋 (𝐶) → 𝑌 𝐹 (𝐴) for (𝐶, ℎ) ∈ (𝐹 ⇒ 𝐹 (𝐴)). In particular we have a map 𝛼e𝐴 = 𝛽 𝐴(𝐴,1 ) : 𝑋 (𝐴) → 𝑌 𝐹 (𝐴). We want to 𝐹 (𝐴) show that the family 𝛼e𝐴, 𝐴 ∈ A, defines a natural transformation. So let 𝑓 : 𝐴 → 𝐴 0 be a map in A. Then 𝑓 is a map (𝐴, 𝐹 (𝑓 )) → (𝐴 0, 1𝐹 (𝐴0) ), so by definition of cocone the following diagram commutes: 𝑋 (𝐴)

𝑋 (𝑓 )

𝑋 (𝐴 0) 0

𝛽𝐴 (𝐴0,1

0 𝛽𝐴 (𝐴,𝐹 (𝑓 ) )

𝐹 (𝐴0 ) )

𝑌 𝐹 (𝐴 0) ,

0

that is, 𝛽 𝐴(𝐴,𝐹 (𝑓 )) = 𝛼e𝐴0 ◦ 𝑋 (𝑓 ). Now consider the diagram 𝛽𝐴 (𝐴,1

𝑋 (𝐴)

𝐹 (𝐴) )

Lan𝐹 𝑋 (𝐹 (𝐴))

𝛼 𝐹 (𝐴)

Lan𝐹 𝑋 (𝐹 (𝑓 ))

𝑋 (𝐴)

𝑌 𝐹 (𝐴) 𝑌 𝐹 (𝑓 )

Lan𝐹 𝑋 (𝐹 (𝐴 0))

𝛼 𝐹 (𝐴0 )

𝑌 𝐹 (𝐴 0)

0

𝛽𝐴 (𝐴,𝐹 ( 𝑓 ) )

where the two horizontal morphisms on the left are the canonical inclusions into the colimit. The left-hand side square commutes by definition of Lan𝐹 𝑋 on morphisms, and the right-hand side 0 square commutes by naturality of 𝛼 . Thus 𝛽 𝐴(𝐴,𝐹 (𝑓 )) = 𝑌 𝐹 (𝑓 ) ◦ 𝛽 𝐴(𝐴,1 ) = 𝑌 𝐹 (𝑓 ) ◦ 𝛼e𝐴 . It follows 𝐹 (𝐴) that 𝛼e is a natural transformation 𝑋 ⇒ 𝑌 𝐹 . We thus have defined a map [B, 𝒮] (Lan𝐹 𝑋, 𝑌 ) → [A, 𝒮] (𝑋, 𝑌 𝐹 ). We now define a map in the other direction. Let 𝜀 : 𝑋 ⇒ 𝑌 𝐹 be a natural transformation. For 𝐵 ∈ ℬ, we shall define a map e 𝜀𝐵 : Lan𝐹 𝑋 (𝐵) → 𝑌 (𝐵). Given (𝐴, ℎ) ∈ (𝐹 ⇒ 𝐵), consider 𝑌 (ℎ) ◦ 𝜀𝐴 : 𝑋 (𝐴) → 𝑌 (𝐵). Given a map 𝑓 : (𝐴, ℎ) → (𝐴 0, ℎ 0) in (𝐹 ⇒ 𝐵), the triangle in the

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diagram 𝑋 (𝐴)

𝜀𝐴

𝑌 𝐹 (𝐴)

𝑋 (𝑓 )

𝑌 (ℎ)

𝑌 𝐹 (𝑓 )

𝑋 (𝐴 0)

𝑌 𝐹 (𝐴 0)

𝜀𝐴0

𝑌 (𝐵)

𝑌 (ℎ0 )

commutes by definition, and the square commutes by naturality of 𝜀. It follows that 𝑌 (𝐵) together with the collection of maps 𝑌 (ℎ) ◦ 𝜀𝐴, for (𝐴, ℎ) ∈ (𝐹 ⇒ 𝐵), is a cocone on 𝑋 ◦ 𝑃𝐵 , and we let e 𝜀𝐵 : Lan𝐹 𝑋 (𝐵) → 𝑌 (𝐵) be the unique map induced by the universal property of the colimit Lan𝐹 𝑋 (𝐵). We claim that e 𝜀 is a natural transformation. Indeed, to prove this it suffices to prove that the diagram 𝑋 (𝐴)

𝑌 (ℎ)◦𝜀𝐴

𝑌 (𝐵) 𝑌 (𝑔)

𝑌 (𝑔ℎ)◦𝜀𝐴

𝑌 (𝐵 0) commutes for all 𝑔 : 𝐵 → 𝐵 0 in B and (𝐴, ℎ) ∈ (𝐹 ⇒ 𝐵), and this is clear. We thus have defined a map [A, 𝒮] (𝑋, 𝑌 𝐹 ) → [B, 𝒮] (Lan𝐹 𝑋, 𝑌 ), 𝜀 ↦→ e 𝜀 , and looking at the definition it is easily checked that this map is a two-sided inverse of the map [B, 𝒮] (Lan𝐹 𝑋, 𝑌 ) → [A, 𝒮] (𝑋, 𝑌 𝐹 ), 𝛼 ↦→ 𝛼e, defined above. (b) First, note that Lan𝐹 is a functor [A, 𝒮] → [B, 𝒮] as follows. Let 𝜂 ⇒ 𝑋 0 → 𝑋 be a natural transformation between functors A → 𝒮. Let 𝐵 ∈ A. For each (𝐴, ℎ) ∈ (𝐹 ⇒ 𝐵), let 𝑞 (𝐴,ℎ) : 𝑋 (𝐴) → Lan𝐹 𝑋 (𝐵) denote the inclusion into the colimit. Given a map 𝑓 : (𝐴, ℎ) → (𝐴 0, ℎ 0) in (𝐹 ⇒ 𝐵), we have a commutative diagram 𝑋 0 (𝐴)

𝜂𝐴

𝑋 0 (𝑓 )

𝑋 (𝐴) 𝑞 (𝐴,ℎ)

𝑋 (𝑓 )

𝑋 0 (𝐴 0)

𝜂𝐴0

𝑋 (𝐴 0)

𝑞 (𝐴0,ℎ0 )

Lan𝐹 𝑋 (𝐵) .

Thus Lan𝐹 𝑋 (𝐵) together with the collection of maps 𝑞 (𝐴,ℎ) ◦𝜂𝐴, for (𝐴, ℎ) ∈ (𝐹 ⇒ 𝐵), is a cocone on 𝑋 0 ◦ 𝑃𝐵 . Let Lan𝐹 (𝜂)𝐵 : Lan𝐹 𝑋 0 (𝐵) → Lan𝐹 𝑋 (𝐵) be the unique map induced by the universal property of the colimit Lan𝐹 𝑋 0 (𝐵). For proving Lan𝐹 (𝜂) : Lan𝐹 𝑋 0 ⇒ Lan𝐹 𝑋 is natural, for a map 𝑔 : 𝐵 → 𝐵 0 in B we shall prove that the diagram Lan𝐹 𝑋 0 (𝐵) Lan𝐹

Lan𝐹 (𝜂)𝐵

Lan𝐹 𝑋 0 (𝐵 0)

Lan𝐹 𝑋 (𝐵) Lan𝐹 𝑋 (𝑔)

𝑋 0 (𝑔) Lan𝐹 (𝜂)𝐵0

Lan𝐹 𝑋 (𝐵 0)

commutes. Let (𝐴, ℎ) ∈ (𝐹 ⇒ 𝐵), and let 𝑝 (𝐴,ℎ) : 𝑋 0 (𝐴) → Lan𝐹 𝑋 0 (𝐵) and 𝑞 (𝐴,ℎ) : 𝑋 (𝐴) → Lan𝐹 𝑋 (𝐵 0) denote the inclusions into the respective colimits. It then follows from the definitions 58

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Limits and colimits of presheaves

Lan𝐹 (𝜂)𝐵0 ◦ Lan𝐹 𝑋 0 (𝑔) ◦ 𝑝 (𝐴,ℎ) = 𝑞 (𝐴,ℎ) ◦ 𝜂𝐴 = Lan𝐹 𝑋 (𝑔) ◦ Lan𝐹 (𝜂) ◦ 𝑝 (𝐴,ℎ)

and thus, as these holds for all (𝐴, ℎ), we deduce that the diagram above commutes. Therefore Lan𝐹 (𝜂) : Lan𝐹 𝑋 0 ⇒ Lan𝐹 𝑋 is natural. Now, in part (a) we have defined for 𝑋 ∈ [A, 𝒮] and 𝑌 ∈ [B, 𝒮] a bijection [A, 𝒮] (𝑋, 𝑌 𝐹 ) → [B, 𝒮] (Lan𝐹 𝑋, 𝑌 ), 𝜀 ↦→ e 𝜀. It remains to prove that this bijection, call it 𝜑𝑋 ,𝑌 , is natural in 𝑋 and 𝑌 . So let 𝜂 : 𝑋 0 ⇒ 𝑋 and 𝜃 : 𝑌 ⇒ 𝑌 0 be natural transformations. We shall prove that the diagram [A, 𝒮] (𝑋, 𝑌 𝐹 )

𝜑𝑋 ,𝑌

(𝜃 𝐹 )∗ ◦𝜂 ∗

[A, 𝒮] (𝑋 0, 𝑌 0𝐹 )

[B, 𝒮] (Lan𝐹 𝑋, 𝑌 ) 𝜃 ∗ ◦Lan𝐹 (𝜂) ∗

𝜑𝑋 0,𝑌 0

[B, 𝒮] (Lan𝐹 𝑋 0, 𝑌 0)

is commutative. Let 𝜀 : 𝑋 ⇒ 𝑌 𝐹 and let 𝐵 ∈ B. Let (𝐴, ℎ) ∈ (𝐹 ⇒ 𝐵) and let 𝑝 (𝐴,ℎ) : 𝑋 0 (𝐴) → Lan𝐹 𝑋 0 (𝐵) denote the inclusion. It follows from the definitions that (𝜃 ◦ 𝜑𝑋 ,𝑌 (𝜀) ◦ Lan𝐹 (𝜂))𝐵 ◦ 𝑝 (𝐴,ℎ) = 𝜃 𝐵 ◦ 𝑌 (ℎ) ◦ 𝜀𝐴 ◦ 𝜂𝐴 and

𝜀𝑋 0,𝑌 0 (𝜃 𝐹 ◦ 𝜀 ◦ 𝜂)𝐵 ◦ 𝑝 (𝐴,ℎ) = 𝑌 0 (ℎ) ◦ 𝜃 𝐹 (𝐴) ◦ 𝜀𝐴 ◦ 𝜂𝐴,

and these two are equal by naturality of 𝜃 . It follows that the diagram above commutes, so that 𝜑𝑋 ,𝑌 is natural in 𝑋 and 𝑌 . We conclude that the functor − ◦ 𝐹 : [B, 𝒮] → [A, 𝒮] is right adjoint to the functor Lan𝐹 : [A, 𝒮] → [B, 𝒮]. (c) Let 𝒮 = Set be the category of sets. First let 𝐹 be the unique functor 1 → 𝐺 . Then − ◦ 𝐹 : [𝐺, Set] → [1, Set] = Set is precisely the forgetful functor 𝑈 from 𝐺-sets to sets. From part (b) we know 𝑈 has both left and right adjoints, and in Exercise 2.1.16 we described these adjoints explicitly. Its left adjoint 𝐺 : Set → [𝐺, Set] is given on objects by sending 𝑋 ∈ Set to 𝐺 × 𝑋 ∈ [𝐺, Set], where the 𝐺-action on 𝐺 × 𝑋 is given by ℎ · (𝑔, 𝑥) = (ℎ𝑔, 𝑥) for all 𝑔, ℎ ∈ 𝐺 and 𝑥 ∈ 𝑋, and a map 𝑓 : 𝑋 → 𝑌 of sets to the map 𝐺 (𝑓 ) : 𝐺 × 𝑋 → 𝐺 × 𝑌 of 𝐺-sets given by (𝑔, 𝑥) ↦→ (𝑔, 𝑓 (𝑥)). Its right adjoint 𝐻 : Set → [𝐺, Set] sends 𝑋 ∈ Set to 𝐻 (𝑋 ) = 𝑋 𝐺 ∈ [𝐺, Set], the set of functions 𝐺 → 𝑋, with 𝐺-action given by (𝑔 · 𝑓 ) (ℎ) = 𝑓 (ℎ𝑔) for all 𝑔, ℎ ∈ 𝐺, 𝑓 ∈ 𝑋 𝐺 , and a function 𝑝 : 𝑋 → 𝑌 to the map 𝐻 (𝑝) : 𝑋 𝐺 → 𝑌 𝐺 given by [𝐻 (𝑝) (𝑓 )] (ℎ) = 𝑝 ◦ 𝑓 (ℎ). Now consider 𝐹 to be the unique functor 𝐺 → 1. Then − ◦ 𝐹 : Set → [𝐺, Set] is precisely the functor Δ that equips a set with the trivial 𝐺-action. From part (b) we know Δ has both left and right adjoints, and in Exercise 2.1.16 we described these adjoints explicitly. Its left adjoint is the functor (−)/𝐺 : [𝐺, Set] → Set sending a 𝐺-set 𝑋 to the set 𝑋 /𝐺 = {𝐺 · 𝑥 | 𝑥 ∈ 𝑋 } of orbits of 𝑋 under 𝐺, and its right adjoint the functor (−)𝐺 : [𝐺, Set] → Set sending a 𝐺-set 𝑋 to its 𝐺-fixed point subset 𝑋 𝐺 = {𝑥 ∈ 𝑋 | 𝑔 · 𝑥 = 𝑥 for all 𝑔 ∈ 𝐺 }. 59

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6.3

6.3

Interactions between adjoint functors and limits

Interactions between adjoint functors and limits

Exercise 6.3.21. (a) The initial object of Grp is the zero group, whose underlying set has precisely one element, and the initial object of Set is the empty set. Thus 𝑈 does not preserve colimits, so it has no right adjoint by Theorem 6.3.1. (b) Recall from Exercise 3.2.16 that 𝐶 : Cat → Set sends a small category 𝒞 be the quotient 𝐶 (𝒞) of the set of objects of 𝒞 by the equivalence relation generated by 𝑐 ∼ 𝑐 0 if there exists an arrow 𝑐 → 𝑐 0 in 𝒞. We shall prove that 𝐶 has no left adjoint. For this purpose, first we claim that right adjoints preserve monics. Indeed, suppose 𝐺 : 𝒟 → ℰ is a functor right adjoint to a functor 𝐹, with the adjunction given by the natural isomorphism 𝜑 𝐸,𝐷 : 𝒟(𝐹 (𝐸), 𝐷) → ℰ(𝐸, 𝐺 (𝐷)). Let 𝑓 : 𝐷 → 𝐷 0 be monic in 𝒟 and ℎ 1, ℎ 2 : 𝐸 → 𝐺 (𝐷) two maps in 𝒞 such that 𝐺 (𝑓 ) ◦ ℎ 1 = 𝐺 (𝑓 ) ◦ ℎ 2 . It follows from equation (2.2) that −1 𝑓 ◦ 𝜑 𝐸,𝐷 (ℎ 1 ) = 𝜑 𝐸,𝐷 0 (𝐺 (𝑓 ) ◦ ℎ 1 ) = 𝜑 𝐸,𝐷 0 (𝐺 (𝑓 ) ◦ ℎ 2 ) = 𝑓 ◦ 𝜑 −1 (ℎ 2 ), −1 (ℎ ) = 𝜑 −1 (ℎ ) as 𝑓 is monic, and therefore ℎ = ℎ . The claim follows. Thus, for so that 𝜑 𝐸,𝐷 1 2 1 2 proving that 𝐶 is not a right adjoint, we will prove that 𝐶 does not preserve monics. Let 𝒞 be a discrete category with two objects 𝑐 1, 𝑐 2 and 𝒟 be the indiscrete category with two objects 𝑑 1, 𝑑 2, that is, we have precisely two non-identity morphisms in 𝒟, one 𝑑 1 → 𝑑 2 and one 𝑑 2 → 𝑑 1 . We have a functor 𝐹 : 𝒞 → 𝒟 sending 𝑐𝑖 to 𝑑𝑖 , 𝑖 = 1, 2. Assume ℰ is a category and 𝐺, 𝐺 0 : ℰ → 𝒞 are functors such that 𝐹𝐺 = 𝐹𝐺 0 . Then 𝐺 and 𝐺 0 must the the same on objects, and must send a morphism 𝑒 → 𝑒 0 in ℰ to the identity map of 𝐺 (𝑒) = 𝐺 (𝑒 0) = 𝐺 0 (𝑒) = 𝐺 0 (𝑒 0). It follows that 𝐹 is monic. But 𝐶 (𝐹 ) is not monic, for 𝐶 (𝒞) has two objects while 𝐶 (𝒟) has only one. We deduce that 𝐶 is not a right adjoint. Now, recall from Exercise 3.2.16 that 𝐼 : Set → Cat sends a set 𝑆 to the indiscrete category 𝐼 (𝑆) whose set of objects is 𝑆. We shall prove that 𝐼 is not a left adjoint. By Theorem 6.3.1, it suffices to note that 𝐼 does not preserve coproducts. If {𝑎} denotes a one-element set, note that {𝑎} q {𝑎} is simply a set with precisely two elements 𝑎 1, 𝑎 2, so that the category 𝐼 ({𝑎} q {𝑎}) has a morphism 𝑎 1 → 𝑎 2 . But, on the other hand, the category 𝐼 ({𝑎}) q 𝐼 ({𝑎}) has no non-identity morphisms. It follows that 𝐼 has no right adjoint.

(c) Let 𝑋 be a topological space. If 𝑋 is empty, then in the chain of adjunctions in Exercise 2.1.17 we have Λ = Δ = ∇ and Π = Γ, so we have two functors which are left and right adjoint to each other and thus the chain extends infinitely at both ends. Now assume that 𝑋 is non-empty. Recall from Exercise 2.1.17 that Λ : Set → [𝒪(𝑋 ) op, Set] sends 𝐴 ∈ Set to the functor Λ(𝐴) : 𝒪(𝑋 ) op → Set given on objects by Λ(𝐴) (∅) = 𝐴 and Λ(𝐴)(𝑉 ) = ∅ if 𝑉 ≠ ∅, and on morphisms by Λ(𝐴) (1 ∅ ) = 1𝐴 and Λ(𝐴) (𝑟𝑉 ,𝑈 ) = the empty function if 𝑉 ≠ ∅. We claim that Λ has no left adjoint. By Theorem 6.3.1, it suffices to prove that Λ does not preserve limits. For this purpose, consider the one-point set {∗}, which is terminal in Set. Since limits in [𝒪(𝑋 ) op, Set] are computed point-wise, if 𝐹 is a terminal object in [𝒪(𝑋 ) op, Set] then 𝐹 (𝑈 ) must be a singleton for all 𝑈 ∈ 𝒪(𝑋 ). As Λ({∗}) ({∗}) = ∅, it follows that Λ({∗}) is not terminal. We deduce that Λ has no left adjoint. 60

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Interactions between adjoint functors and limits

Now, recall from Exercise 2.1.17 that ∇ : Set → [𝒪(𝑋 ) op, Set] is given on objects by sending 𝐴 ∈ Set to the functor ∇𝐴 : 𝒪(𝑋 ) op → Set given by ∇𝐴(𝑋 ) = 𝐴 and ∇𝐴(𝑉 ) = {∗} for all 𝑉 ≠ 𝑋 . Now ∅ is the initial object of Set, but (as 𝑋 is non-empty) the set ∇∅(∅) = {∗} is non-empty. It follows from Theorem 6.3.1 that ∇ has no right adjoint. Exercise 6.3.22. (a) First assume that 𝑈 has a left adjoint 𝐹 : Set → 𝒜 and let 𝜑𝑆,𝐴 : 𝒜(𝐹 (𝑆), 𝐴) → Set(𝑆, 𝑈 (𝐴)) be the isomorphism natural in 𝑆 ∈ Set and 𝐴 ∈ 𝒜 determining the adjunction. In particular, by considering the one-element set 𝑆 = {∗}, we have an isomorphism 𝜑 {∗},𝐴 : 𝒜(𝐹 ({∗}), 𝐴) → Set({∗}, 𝑈 (𝐴)) = 𝑈 (𝐴) natural in 𝐴 ∈ 𝒜. It follows that 𝑈 is representable. Now, if 𝑈 is representable, then it preserves limits by Proposition 6.2.2. (b) Suppose that 𝒜 has sums and is representable. Without loss of generality, we may assume that 𝑈 = 𝐻 𝐴 = 𝒜(𝐴, −) for some 𝐴 ∈ 𝒜. Define a functor 𝐹 : Set → 𝒜 as follows. For 𝑆 ∈ Set, let 𝐹 (𝑆) = q𝑠 ∈𝑆 𝐴 (a chosen coproduct). For a function 𝑓 : 𝑆 0 → 𝑆, we have a unique map 𝐹 (𝑓 ) : 𝐹 (𝑆 0) → 𝐹 (𝑆) induced by the universal property of the coproduct. That is, for 𝑠 0 ∈ 𝑆 0, denote by 𝑗𝑠 0 : 𝐴 → q𝑠 0 ∈𝑆 0 𝐴 inclusion corresponding to the summand 𝑠 0 ∈ 𝑆 0, and for 𝑠 ∈ 𝑆 denote by 𝑙𝑠 : 𝐴 → q𝑠 ∈𝑆 𝐴 the inclusion corresponding to the summand 𝑠 ∈ 𝑆. Then 𝐹 (𝑓 ) is the unique map such that 𝐹 (𝑓 ) ◦ 𝑗𝑠 0 = 𝑙 𝑓 (𝑠 0) . We claim that 𝐹 is left adjoint to 𝑈 . For 𝑆 ∈ 𝒮 and 𝐵 ∈ 𝒜, define 𝜑𝑆,𝐵 : 𝒜(𝐹 (𝑆), 𝐵) → Set(𝑆, 𝑈 (𝐵)) as follows. Let ℎ : 𝐹 (𝑆) → 𝐵. Let 𝜑𝑆,𝐵 (ℎ) : 𝑆 → 𝑈 (𝐵) send 𝑠 to ℎ ◦ 𝑗𝑠 : 𝐴 → 𝐵. Conversely, define 𝜓𝑆,𝐵 : Set(𝑆, 𝑈 𝐵) → 𝒜(𝐹 (𝑆), 𝐵) as follows. For 𝑘 : 𝑆 → 𝑈 (𝐵), let 𝜓𝑆,𝐵 : 𝐹 (𝑆) → 𝐵 be induced from the maps 𝑘 (𝑠) : 𝐴 → 𝐵, 𝑠 ∈ 𝑆, by the universal property of the coproduct. It is clear that 𝜑𝑆,𝐵 and 𝜓𝑆,𝐵 are inverses to each other, so that 𝜑𝑆,𝐵 is a bijection. It remains to prove that 𝜑𝑆,𝐵 is natural in 𝑆 ∈ Set and 𝐵 ∈ 𝒜. Let 𝑓 : 𝑆 0 → 𝑆 be a function of sets and 𝑔 : 𝐵 → 𝐵 0 be a map in 𝒜. We shall prove that the diagram 𝒜(𝐹 (𝑆), 𝐵)

𝜑𝑆,𝐵

𝑔∗ ◦𝐹 (𝑓 ) ∗

𝒜(𝐹 (𝑆 0), 𝐵 0)

Set(𝑆, 𝑈 (𝐵)) 𝑓∗ ◦𝑈 (𝑔) ∗

𝜑𝑆 0,𝐵0

Set(𝑆 0, 𝑈 (𝐵 0))

commutes. Let ℎ : 𝐹 (𝑆) → 𝐵. For 𝑠 ∈ 𝑆 and 𝑠 0 ∈ 𝑆 0, let 𝑙𝑠 and 𝑗𝑠 0 be defined as in the first paragraph above. Then 𝑓 ◦ 𝜑𝑆,𝐵 ◦ 𝑈 (𝑔) sends 𝑠 0 ∈ 𝑆 0 to 𝑔 ◦ ℎ ◦ 𝑙 𝑓 (𝑠 0) , and 𝜑𝑆 0,𝐵0 (𝑔 ◦ ℎ ◦ 𝐹 (𝑓 )) sends 𝑠 0 ∈ 𝑆 0 to 𝑔 ◦ ℎ ◦ 𝐹 (𝑓 ) ◦ 𝑗𝑠 0 , and these two are equal since 𝐹 (𝑓 ) ◦ 𝑗𝑠 0 = 𝑙 𝑓 (𝑠 0) . It follows that the diagram above commutes, so that 𝜑𝑆,𝐵 is natural. We conclude that 𝐹 is left adjoint to 𝑈 . Exercise 6.3.23. (a) Let 𝑃 be a preordered set. Let 𝑄 be the quotient of 𝑃 by the equivalence relation given by 𝑝 ∼ 𝑝 0 if 𝑝 ≤ 𝑝 0 and 𝑝 0 ≤ 𝑝. For 𝑝 ∈ 𝑃, let [𝑝] ∈ 𝑄 denote its equivalence class. 61

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Interactions between adjoint functors and limits

Then 𝑄 is an ordered set via [𝑝] ≤ [𝑝 0] if 𝑝 ≤ 𝑝 0 . Let 𝐹 : 𝑃 → 𝑄 be the quotient function. Then 𝐹 is a functor. Moreover, 𝐹 is clearly full, faithful and essentially surjective on objects, hence an equivalence by Proposition 1.3.18. (b) A map 𝑓 : 𝐴 → 𝐵 𝐼 from 𝐴 into the product of |𝐼 | copies of 𝐵 is the same thing as a collection of maps 𝑓𝑖 : 𝐴 → 𝐵, 𝑖 ∈ 𝐼 . Since there exist distinct maps 𝑓 , 𝑔 : 𝐴 → 𝐵 in 𝒜, there are at least two possibilities for each 𝑓𝑖 . Thus, we have at least 2 |𝐼 | maps 𝐴 → 𝐵 𝐼 in 𝒜. In particular, it follows that 𝒜 is not small. (c) If 𝒜 is small with small products, it follows from part (b) that 𝒜 is a preoder, and in turn from part (a) that 𝒜 is equivalent to an ordered set 𝑄. Note that 𝑄 has all small products, being equivalent to 𝒜. Moreover, trivially any ordered set has equalisers. It follows from Proposition 5.1.26(a) that 𝑄 is complete. (d) Let 𝒜 be a finite category with finite products. From the analogous statement of (b) it follows that 𝒜 is a preorder. Thus, by part (a) we have that 𝒜 is equivalent to a finite ordered set 𝑄 with all finite products. Note that any finite ordered set has a maximal element, i.e. 𝑄 has a terminal object. We deduce from Proposition 5.1.26(b) that 𝑄 is complete. Exercise 6.3.24. (a) An element of the subgroup of 𝐺 generated by {𝑔𝑎 | 𝑎 ∈ 𝐴} is a finite string of elements of the set {𝑔𝑎 , 𝑔𝑎−1 | 𝑎 ∈ 𝐴}. it follows that this subgroup has cardinality at most |N| · |𝐴| = max{|N|, |𝐴|}. (b) Let 𝑆 be a set. Let 𝐺 be a group with cardinality |𝐺 | ≤ |𝑆 |. Then there is an injection 𝑖 : 𝐺 → 𝑆 which endows 𝑖 (𝐺) ⊂ 𝑆 with a group structure such that 𝐺  𝑖 (𝐺). Thus, we may assume that 𝐺 ⊂ 𝑆. Now the group structure of 𝐺 is determined by a function 𝐺 × 𝐺 → 𝐺, which is a subset of 𝐺 × 𝐺 × 𝐺 ⊂ 𝑆 × 𝑆 × 𝑆. It follows that there is at most |𝒫(𝑆)| · |𝒫(𝑆 × 𝑆 × 𝑆)| isomorphism classes of groups of cardinality at most |𝑆 |, so this collection is small. (c) If 𝐴 is infinite, let 𝑆 = 𝐴, otherwise, let 𝑆 = N. Consider the collection S of all elements of ∈ (𝐴 ⇒ 𝑈 ) such that 𝑈 (𝐺 0) ⊂ 𝑆. It follows from part (b) that S is a set. Let (𝐺, ℎ) ∈ (𝐴 ⇒ 𝑈 ) and let 𝐿 denote the subgroup of 𝐺 generated by {ℎ(𝑎) | 𝑎 ∈ 𝐴}. By part (a), 𝐿 has cardinality at most |𝑆 |, so we have a bijection 𝜑 : 𝑆 0 → 𝑈 (𝐿) from 𝑆 0 a subset of 𝑆. This bijection endows the set 𝑆 0 with a unique group structure such that 𝜑 is an isomorphism; call this group 𝐺 0, so that 𝑈 (𝐺 0) = 𝑆 0 . If 𝑖 : 𝐿 → 𝐺 denotes the inclusion, then 𝑓 = 𝑖 ◦ 𝜑 : 𝐺 0 → 𝐺 is a monomorphism of groups. Now let ℎ 0 : 𝐴 → 𝑆 0 be given by ℎ 0 (𝑎) = 𝜑 −1 ◦ ℎ(𝑎). Then 𝑓 : (𝐺 0, ℎ 0) → (𝐺, ℎ) is a map in (𝐴 ⇒ 𝑈 ), where (𝐺 0, ℎ 0) ∈ S. It follows that S is a weakly initial set in (𝐴 ⇒ 𝑈 ). (𝐺 0, ℎ 0)

(d) Note that Set is complete and Grp is locally small. In Exercise 5.3.11 we proved that 𝑈 creates arbitrary limits. Thus, by Lemma 5.3.6 (proved in Exercise 5.3.12), Grp is complete and 𝑈 preserves limits. Finally, in (c) we have prove that (𝐴 ⇒ 𝑈 ) has a weakly initial set for every 𝐴 ∈ Set. We deduce from the General adjoint functor theorem (Theorem 6.3.10) that 𝑈 has a left adjoint. b = [Aop, Set] for the presheaf category. Let 𝐻 • : A → A b denote the Exercise 6.3.25. Write A Yoneda embedding. By assumption, A has finite products and for each 𝐵 ∈ A, the functor − × 62

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Interactions between adjoint functors and limits

𝐵 : A → A has a right adjoint (−) 𝐵 : 𝒜 → 𝒜. Given 𝑋, 𝑌 ∈ 𝒜, there is a natural isomorphism 𝐻𝑋 ×𝑌 = 𝒜(−, 𝑋 × 𝑌 )  𝒜(−, 𝑋 ) × 𝒜(−, 𝑌 ) = 𝐻𝑋 × 𝐻𝑌 given at 𝐴 ∈ A by sending a map 𝑓 : 𝐴 → 𝑋 × 𝑌 into the pair (𝑝𝑋 ◦ 𝑓 , 𝑝𝑌 ◦ 𝑓 ), where 𝑝𝑋 and 𝑝𝑌 denote the projections. Thus 𝐻 • preserves finite products; it remains to prove that it preserves b For 𝐴, 𝐵 ∈ A, exponentials. (Recall from the proof of Theorem 6.3.20 the exponentials in A.) b • × 𝐻𝐵 , 𝐻𝐴 )  A(𝐻 b •×𝐵 , 𝐻𝐴 ). (𝐻𝐴 ) 𝐻𝐵 = A(𝐻 Since 𝐻 • is fully faithful, we have b •×𝐵 , 𝐻𝐴 )  (𝐻𝐴 ) 𝐻𝐵 . 𝐻𝐴𝐵 = A(−, 𝐴𝐵 )  A(− × 𝐵, 𝐴)  A(𝐻 It follows that 𝐻 • preserves the whole cartesian closed structure. Exercise 6.3.26. (a) Let 𝑓 : 𝐴 0 → 𝐴 be a map in 𝒜. Let 𝑚 : 𝑋 → 𝐴 be a monic into 𝐴, representing a class [𝑚] ∈ Sub(𝐴). By assumption, the pullback 𝑋0

𝑓0

𝑋

𝑚0

𝑚

𝐴0

𝐴 𝑓

exists in 𝒜. It follows from Exercise 5.1.42 that 𝑚 0 : 𝑋 0 → 𝐴 0 is a monic into 𝐴 0, and we set Sub(𝑓 ) [𝑚] = [𝑚 0]. It is clear that this gives a well-defined map Sub(𝑓 ) : Sub(𝐴) → Sub(𝐴 0). (b) We follow the hint. It is clear that Sub(1𝐴 ) = 1Sub(𝐴) for any 𝐴 ∈ 𝒜. Now let 𝑓 : 𝐴 00 → 𝐴 0 and 𝑔 : 𝐴 0 → 𝐴 and be composable maps in 𝒜. Let 𝑚 : 𝑋 → 𝐴 be a monic into 𝐴. Consider a diagram 𝑋 00

𝑓0

𝑚00

𝑋0

𝑔0

𝑋

𝑚0

𝐴 00

𝐴0 𝑓

𝑚 𝑔

𝐴

where both squares are pullbacks. Then Sub(𝑔) ◦ Sub(𝑓 ) [𝑚] = Sub(𝑔) [𝑚 0] = [𝑚 00] by definition. On the other hand, the outer rectangle is also a pullback by Exercise 5.1.35, so that Sub(𝑔 ◦ 𝑓 ) [𝑚 00] = [𝑚]. We deduce that Sub is a functor 𝒜 op → Set. (c) Consider the set 2 = {0, 1} with two elements. Given 𝑆 ∈ Set, recall from Exercise 5.1.40(a) that subobjects of 𝑆 are in canonical one-to-one correspondence with subsets of 𝑆, where the class of a monic 𝑚 : 𝑋 → 𝑆 into 𝑆 corresponds to the image 𝑚(𝑋 ) ⊂ 𝑆. Under this correspondence, for a function 𝑓 : 𝑆 0 → 𝑆, the map Sub(𝑓 ) : Sub(𝑆) → Sub(𝑆 0) is simply the inverse image functor 𝒫(𝑆) → 𝒫(𝑆 0), 𝑋 ↦→ 𝑓 −1 (𝑋 ). It follows that Sub  𝐻 {0,1} . 63

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b = [Aop, Set] for the presheaf category. Exercise 6.3.27. Write A b Then Sub(𝑋 )  𝐻 Ω (𝑋 ) = (a) Suppose that Ω : Aop → Set is a subobject classifier of A. b b In particular, this holds for representable presheaves, so for all 𝐴 ∈ A we A(𝑋, Ω) for all 𝑋 ∈ A. b have Sub(𝐻𝐴 )  A(𝐻𝐴, Ω)  Ω(𝐴), the last isomorphism by Yoneda. b op denote the Yoneda emdedding. Inspired from (a), set (b) Let 𝐻 • : Aop → ( A) Ω = Sub ◦𝐻 • : Aop → Set. We shall prove that Ω is a subobject classifier. We shall prove that Ω is a subobject classifier. First b are precisely the natural transformations which recall from Exercise 6.2.20(b) that monics in A are pointwise monic. It follows from the description of subobjects in Set (see Exercise 5.1.40(a)) b Sub(𝑋 ) can be identified with collection of subpresheaves of 𝑋, that is, that for a presheaf 𝑋 ∈ A, b presheaves 𝐹 ∈ A such that 𝐹 (𝐴) ⊂ 𝑋 (𝐴) for all 𝐴 ∈ A and such that for all maps 𝑓 : 𝐴 0 → 𝐴 in A, the function 𝑋 (𝑓 ) : 𝑋 (𝐴) → 𝑋 (𝐴 0) restricts to a function 𝐹 (𝐴) → 𝐹 (𝐴 0). b and define a map 𝜑𝑋 : Sub(𝑋 ) → Now, we want a natural isomorphism Sub  𝐻 Ω . Let 𝑋 ∈ A b A(𝑋, Ω) as follows. Let 𝐹 ∈ Sub(𝑋 ) be a subpresheaf of 𝑋 . Let 𝐴 ∈ A. Given 𝑎 ∈ 𝑋 (𝐴), let 𝜂𝑎 : 𝐻𝐴 ⇒ 𝑋 be the natural transformation corresponding to 𝑎 via the Yoneda Lemma, and define 𝜑𝑋 (𝐹 )𝐴 : 𝑋 (𝐴) → Sub(𝐻𝐴 ) to send 𝑎 ∈ 𝑋 (𝐴) to the subpresheaf 𝐹 0 of 𝐻𝐴 given by the pullback 𝐹0

𝐹 𝜂𝑎

𝐻𝐴

𝑋,

where the right-hand side map is the inclusion 𝐹 ⇒ 𝑋 . Explicitly, 𝐹 0 : Aop → Set is given by 𝐹 0 (𝐵) = {𝑓 ∈ A(𝐵, 𝐴) | (𝜂𝑎 )𝐵 (𝑓 ) = 𝑋 (𝑓 ) (𝑎) ∈ 𝐹 (𝐵)} ⊂ A(𝐵, 𝐴) on objects and it is the restriction of 𝐻𝐴 on morphisms. We claim that 𝜑𝑋 (𝐹 )𝐴 is natural in 𝐴 ∈ A. Let 𝑓 : 𝐴 0 → 𝐴 be a map in A. We shall prove that the diagram 𝑋 (𝐴)

𝜑𝑋 (𝐹 )𝐴

Sub(𝐻 𝑓 )

𝑋 (𝑓 )

𝑋 (𝐴 0)

Sub(𝐻𝐴 )

𝜑𝑋 (𝐹 )𝐴0

Sub(𝐻𝐴0 )

commutes. Let 𝑎 ∈ 𝑋 (𝐴). By definition Sub(𝐻 𝑓 ) ◦ 𝜑𝑋 (𝐹 )𝐴 (𝑎) is the subpresheaf 𝐹 0 ⊂ 𝐻𝐴0 given by 𝐹 0 (𝐵) = {𝑔 ∈ A(𝐵, 𝐴 0) | 𝑓 ◦ 𝑔 ∈ 𝜑𝑋 (𝐹 )𝐴 (𝑎) (𝐵)} = {𝑔 ∈ A(𝐵, 0 𝐴) | 𝑋 (𝑓 ◦ 𝑔) (𝑎) ∈ 𝐹 (𝐵)}. On the other hand, by definition 𝜑𝑋 (𝐹 )𝐴0 ◦ 𝑋 (𝑓 ) (𝑎) is the subpresheaf 𝐹 00 ⊂ 𝐻𝐴0 given by 𝐹 00 (𝐵) = {𝑔 ∈ A(𝐵, 𝐴) | 𝑋 (𝑔) ◦ 𝑋 (𝑓 ) (𝑎) ∈ 𝐹 (𝐵)}. 64

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Thus 𝐹 0 = 𝐹 00 and it follows that 𝜑𝑋 (𝐹 ) : 𝑋 ⇒ Sub ◦𝐻 • is natural. b Conversely, define 𝜓𝑋 : A(𝑋, Ω) → Sub(𝑋 ) as follows. Let 𝜀 : 𝑋 ⇒ Ω be natural. Let 𝜓𝑋 (𝜀) be the subpresheaf of 𝑋 given by 𝜓𝑋 (𝜀) (𝐴) = {𝑎 ∈ 𝑋 (𝐴) | 1𝑎 ∈ 𝜀𝐴 (𝑎) (𝐴) ⊂ A(𝐴, 𝐴)} ⊂ 𝑋 (𝐴) on objects. We shall prove that 𝜓𝑋 is a two-sided inverse of 𝜑𝑋 . Let 𝐹 ∈ Sub(𝑋 ). By definition, 𝜓𝑋 ◦ 𝜑𝑋 (𝐹 ) is the subpresheaf 𝐹 0 of 𝑋 given by 𝐹 0 (𝐴) = {𝑎 ∈ 𝑋 (𝐴) | 1𝑎 ∈ 𝜑𝑋 (𝐹 )𝐴 (𝑎) (𝐴)}. Since 𝜑𝑋 (𝐹 )𝐴 (𝑎) (𝐴) ⊂ A(𝐴, 𝐴) consists of those arrows 𝑓 : 𝐴 → 𝐴 such that 𝑋 (𝑓 ) (𝑎) ∈ 𝐹 (𝐴), it follows that 𝐹 0 (𝐴) = 𝐹 (𝐴) for all 𝐴 and therefore 𝜓𝑋 ◦ 𝜑𝑋 (𝐹 ) = 𝐹 . On the other hand, let 𝜀 : 𝑋 ⇒ Ω. By definition, 𝜑𝑋 (𝜓𝑋 (𝜀))𝐴 (𝑎) (𝐵) = {𝑓 ∈ A(𝐵, 𝐴) | 𝑋 (𝑓 ) (𝑎) ∈ 𝜓𝑋 (𝜀) (𝐵)}. Unravelling the definitions, we thus see that for proving 𝜑𝑋 ◦ 𝜓𝑋 (𝜀) = 𝜀 it suffices to prove the following: for any 𝑓 : 𝐵 → 𝐴 in A and 𝑎 ∈ 𝐴, we have 𝑓 ∈ 𝜀𝐴 (𝑎) (𝐵) if and only if 1𝐵 ∈ 𝜀𝐵 (𝑋 (𝑓 ) (𝑎)) (𝐵). This follows from naturality of 𝜀, by considering the commutative diagram 𝑋 (𝐴)

Sub(𝐻𝐴 )

𝜀𝐴

Sub(𝐻 𝑓 )

𝑋 (𝑓 )

𝑋 (𝐵)

Sub(𝐻𝐵 )

𝜀𝐵

evaluated at 𝑎 ∈ 𝑋 (𝐴). We deduce that 𝜓𝑋 is a two-sided inverse of 𝜑𝑋 . Lastly, we shall prove that 𝜑𝑋 is natural in 𝑋 . So let 𝛼 : 𝑋 0 ⇒ 𝑋 be a natural transformations between presheaves on A and consider the diagram Sub(𝑋 )

𝜑𝑋

Sub(𝛼)

Sub(𝑋 0)

b A(𝑋, Ω) b A(𝛼,Ω)

𝜑𝑋 0

b 0, Ω) . A(𝑋

Let 𝐹 be a subpresheaf of 𝑋 . Let 𝐴, 𝐵 ∈ A and 𝑎 0 ∈ 𝑋 0 (𝐴). On the one hand, we have 𝜑𝑋 (𝐹 )𝐴 (𝛼𝐴 (𝑎 0)) (𝐵) = {𝑓 ∈ A(𝐵, 𝐴) | 𝑋 (𝑓 ) (𝛼𝐴 (𝑎 0)) ∈ 𝐹 (𝐵)}. On the other hand, 𝜑𝑋 0 (Sub(𝛼) (𝐹 ))𝐴 (𝑎 0) (𝐵) = {𝑓 ∈ A(𝐵, 𝐴) | 𝑋 0 (𝑓 ) (𝑎 0) ∈ Sub(𝛼) (𝐹 ) (𝐵)} = {𝑓 ∈ A(𝐵, 𝐴) | 𝛼 𝐵 ◦ 𝑋 0 (𝑓 ) (𝑎 0) ∈ 𝐹 (𝐵)}. It follows from naturality of 𝛼 that the above two sets are equal. We conclude that 𝜑 : Sub ⇒ 𝐻 Ω b is a natural isomorphism, so that Ω is a subobject classifier for A. b is cartesian closed and has all (c) From Theorems 6.2.5 and 6.3.20 we now that the category A b is a topos. limits. By part (b), it has a subobject classifies. Thus A 65

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Appendix: Proof of the general adjoint functor theorem Exercise A.3. (a) For each 𝐵 ∈ ℬ, let 0𝐵 : 0 → 𝐵 denote the unique map from 0 to 𝐵. Given a map 0𝐵

𝑓 : 𝐵 → 𝐵 0, we have 𝑓 ◦ 0𝐵 = 0𝐵0 : 0 → 0𝐵0 by uniqueness, so (0 −−→ 𝐵)𝐵 ∈ℬ is a indeed a cone in ℎ𝐵

1ℬ : ℬ → ℬ. Now let 𝐶 ∈ ℬ and let (𝐶 −−→ 𝐵)𝐵 ∈ℬ be a cone on 1ℬ . In particular, ℎ 0 : 𝐶 → 0 is a map such that 0𝐵 ◦ ℎ 0 = ℎ𝐵 for all 𝐵 ∈ ℬ. Moreover, ℎ 0 is the unique such map, for if ℎ : 𝐶 → 0 satisfies 0𝐵 ◦ ℎ = ℎ𝐵 for all 𝐵 ∈ ℬ, in particular for 𝐵 = 0 we have 00 ◦ ℎ = ℎ 0 ; since 00 = 10, we 0𝐵

have ℎ = ℎ 0 . It follows that (0 −−→ 𝐵)𝐵 ∈ℬ is a limit cone on 1ℬ . 𝑝𝐵

(b) Since (𝐿 −−→ 𝐵)𝐵 ∈ℬ is a cone on 1ℬ, for all 𝐵 ∈ ℬ we have 𝑝 𝐵 ◦ 𝑝 𝐿 = 𝑝 𝐵 . Since we also have 𝑝 𝐵 ◦ 1𝐿 = 𝑝 𝐵 for all 𝐵, it follows by the uniqueness on the universal property of the limit that 𝑝 𝐿 = 1𝐿 . Now let 𝐵 ∈ ℬ. If 𝑓 : 𝐿 → 𝐵 is a map in ℬ, we must have 𝑝 𝐵 = 𝑓 ◦ 𝑝 𝐿 = 𝑝 ◦ 1𝐿 , so 𝑝 𝐵 is the unique map 𝐿 → 𝐵. We deduce that 𝐿 is initial. Exercise A.4. (a) By definition 𝑆 ⊂ 𝐶 is a weakly initial set if for all 𝑐 ∈ 𝐶, there exists 𝑠 ∈ 𝑆 such that 𝑠 ≤ 𝑐. Ó (b) Let 𝑠 0 = 𝑠 ∈𝑆 𝑠. By definition of meet (greatest lower bound), we have 𝑠 0 ≤ 𝑠 for all 𝑠 ∈ 𝑆. Given 𝑐 ∈ 𝐶, there exists 𝑠 ∈ 𝑆 such that 𝑠 ≤ 𝑐, and thus 𝑠 0 ≤ 𝑐. It follows that 𝑠 0 is the least element of 𝐶. Exercise A.5. (a) (We do not need 𝐺 to be limit-preserving in this part.) Let 𝐸 : I → (𝐴 ⇒ 𝐺) be a diagram. By definition, for each 𝑖 ∈ I we are given an object (𝐸𝑖 , ℎ𝑖 ) ∈ (𝐴 ⇒ 𝐺), that is, an object 𝐸𝑖 ∈ ℬ together with a morphism ℎ𝑖 : 𝐴 → 𝐺 (𝐸𝑖 ). Furthermore, for each map 𝑓 : 𝑖 → 𝑗 in I we have a map (𝐸𝑖 , ℎ𝑖 ) → (𝐸 𝑗 , ℎ 𝑗 ), that is, a map 𝐸 (𝑓 ) : 𝐸𝑖 → 𝐸 𝑗 in ℬ such that 𝐸 (𝑓 ) ◦ ℎ𝑖 = ℎ 𝑗 : 𝐴 → 𝐺 (𝐸 𝑗 ). This is precisely the information of a diagram 𝐸 : I → ℬ in ℬ together with a cone on 𝐺 ◦ 𝐺 with vertex 𝐴. 𝑞𝑖

(b) Let 𝐷 : I → (𝐴 ⇒ 𝐺) be a diagram. For each 𝑖 ∈ I, let 𝐷 (𝑖) = (𝐵𝑖 , ℎ𝑖 ). Let (𝐵 −→ 𝑃𝐴 𝐷 (𝑖) = 𝐺 (𝑞𝑖 )

𝐵𝑖 )𝑖 ∈I be a limit cone on 𝑃𝐴 𝐷. Since 𝐺 is limit-preserving, (𝐺 (𝐵) −−−−→ 𝐺 (𝐵𝑖 ))𝑖 ∈I is a limit cone ℎ𝑖

on 𝐺𝑃𝐴 𝐷. Now, from part (a), (𝐴 −→ 𝐺 (𝐵𝑖 ))𝑖 ∈I is a cone on 𝐺𝑃𝐴 𝐷, so there is a unique map 𝑓 : 𝐴 → 𝐺 (𝐵) such that 𝐺 (𝑞𝑖 ) ◦ 𝑓 = ℎ𝑖 for all 𝑖 ∈ I. Now (𝐵, 𝑓 ) ∈ (𝐴 ⇒ 𝐺), and for each 𝑖 ∈ 𝐼, 𝑞𝑖 is 𝑞𝑖 a map (𝐵, 𝑓 ) → (𝐵𝑖 , ℎ𝑖 ) in (𝐴 ⇒ 𝐺). So ((𝐵, 𝑓 ) −→ 𝐷 (𝑖))𝑖 ∈I is a cone on 𝐷 such that 𝑃𝐴 (𝐵, 𝑓 ) = 𝐵 𝑞𝑖 and 𝑃𝐴 (𝑞𝑖 ) = 𝑞𝑖 for all 𝑖 ∈ I. Moreover, ((𝐵, 𝑓 ) −→ 𝐷 (𝑖))𝑖 ∈I is unique as such by uniqueness of 𝑞𝑖

𝑘𝑖

𝑓 . Finally, we shall prove that ((𝐵, 𝑓 ) −→ 𝐷 (𝑖))𝑖 ∈I is a limit cone. So let ((𝐵 0, 𝑓 0) −→ 𝐷 (𝑖))𝑖 ∈I be 𝑘𝑖

a cone on 𝐷. Then (𝐵 0 −→ 𝐵𝑖 )𝑖 ∈I ) is a cone on 𝑃𝐴 𝐷, so there exists a unique map 𝑡 : 𝐵 → 𝐵 0 such 𝐺 (𝑞𝑖 )

that 𝑞𝑖 ◦ 𝑡 = 𝑘𝑖 for all 𝑖 ∈ I. It then follows from the fact that (𝐺 (𝐵) −−−−→ 𝐺 (𝐵𝑖 ))𝑖 ∈I is a limit cone that 𝑡 is a map (𝐵 0, 𝑓 0) → (𝐵, 𝑓 ), the unique such that 𝑞𝑖 ◦ 𝑡 = 𝑘𝑖 (now in (𝐴 ⇒ 𝐺)) for all 𝑖 ∈ I. We conclude that the projection 𝑃𝐴 : (𝐴 ⇒ 𝐺) → ℬ is limit-preserving.

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