Solutions to
Basic Category Theory
Tom Leinster
Solutions by positrón0802
https://positron0802.wordpress.com
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
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10
2 Adjoints
2.1 Definition and examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Adjunctions via units and counits . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Adjunctions via initial objects . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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27
3 Interlude on sets
3.1 Constructions with sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Small and large categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Historical remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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36
4 Representables
4.1 Definitions and examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 The Yoneda lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3 Consequences of the Yoneda lemma . . . . . . . . . . . . . . . . . . . . . . . . . .
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5 Limits
5.1 Limits: definition and examples . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Colimits: definition and example . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3 Interactions between functors and limits . . . . . . . . . . . . . . . . . . . . . . .
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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 . . . . . . . . . . . . . . . . . . .
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Appendix: Proof of the general adjoint functor theorem
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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 .
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Solutions by positrón0802
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.
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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.
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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:
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• 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
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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
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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
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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)
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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
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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.
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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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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),
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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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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.
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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
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2.1
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
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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 ,
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Solutions by positrón0802
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) .
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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.
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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 .
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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).
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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 ).
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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 ∗ .
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Solutions by positrón0802
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 .
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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.
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2.3
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 .
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2.3
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 ∗)
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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))
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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
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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)) .
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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
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Cat(𝒞 0, 𝐷 (𝑆 0))
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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.
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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.
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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)).
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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.
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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
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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
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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
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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},
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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.)
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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.
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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
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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
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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
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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.
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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.
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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.
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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.
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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].
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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) .
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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
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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
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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 }.
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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.
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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.
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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 − ×
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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} .
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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)}.
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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
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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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