1. Yoneda Lemma Let F, G : C→D Be Functors Between Two Categories C

Home , Yoneda lemma

1. Yoneda Lemma Let F,G : C → D be functors between two categories C and D. A natural transformation η : F → G is a family of morphisms ηX : F (X) → G(X) parametrized by objects of C such that for every morphism f : X → Y, the following diagram commutes: η F (X) −−−−→X G(X)   (1.1)   F (f)y G(f)y η F (Y ) −−−−→Y G(Y ). Similarly, we can define natural transformations between contravariant functors: in the above definition, we replace f : X → Y by a morphism f : Y → X. Let C be a category. Given an object X of C, we define a contravariant functor

hX : C → Sets from C to the category of sets as follows. For each object Y, define hX (Y ) = HomC(Y,X) and for each f : Z → Y, we define hX (f) : HomC(Y,X) → HomC(Z,X) by hX (f)(u) = u◦f, where u : Y → X is any morphism in C. Definition 1.1. A contravariant functor F : C → Sets is representable if there exists an ∼ object X of C so that F = hX . The object X is called a representative of F. We will show that such X is unique up to an isomorphism.

A functor F : C → C0 is faithful if for any objects X,X0 of C, the map 0 0 HomC(X,X ) → HomC0 (F (X),F (X )) is injective and full if it is surjective. A functor is fully faithful if the above map is a bijection.

Let C∨ be the category of all contravariant functors from C to the category of sets. Then we obtain a functor ∨ h : C → C ,X 7→ hX . Proposition 1.1. Let h : C → C∨ be as above. (1) For any object X of C, we have a bijection

HomC∨ (hX ,F ) ⇐⇒ F (X). (2) The functor h is fully faithful.

Proof. Let φ : hX → F be a natural transformation. Then ηX : hX (X) → F (X) is a map of sets. Since hX (X) = HomC(X,X), 1X belongs to hX (X). Therefore ηX (1X ) gives an element of F (X). Deﬁne a map

α : HomC∨ (hX ,F ) → F (X) by η 7→ ηX (1X ). Let s ∈ F (X). For any object Y of C, and any morphism ψ ∈ hX (Y ) = HomC(Y,X), we have F (ψ) ∈ HomSets(F (X),F (Y )). Since s ∈ F (X), we obtain F (ψ)(s) ∈ F (Y ). This observation allows us to deﬁne a map

β : F (X) → HomC∨ (hX ,F ) as follows. For each s ∈ F (X), β(s) deﬁnes a natural transformation from hX to F by

βY (s) : HomC(Y,X) → F (Y ), ψ 7→ F (ψ)(s) 1 2 for each object Y of C. Check that α, β are inverse to each other. To do this, let us consider the following diagram: given η ∈ HomC∨ (hX ,F ), and φ : Y → X, ηX HomC(X,X) −−−−→ F (X)   (1.2)   hX (φ)y F (φ)y ηY HomC(Y,X) −−−−→ F (Y ).

Using this diagram, we ﬁnd ηY ◦ hX (φ)(1X ) = F (φ) ◦ ηX (1X ). Since hX (φ)(1X ) = φ, we ﬁnd ηY (φ) = F (φ)(ηX (1X )) = βY (α(η))(φ), ∀φ. This shows that ηY = βY (α(η)) for all objects Y of C, and hence η = (β ◦ α)(η) for all η ∈ HomC∨ (hX ,F ). This shows that β ◦ α = 1 . HomC∨ (hX ,F ) Now, let s ∈ F (X), then

(α ◦ β)(s) = α(β(s)) = βX (s)(1X ) = F (1X )(s) = 1F (X)(s) = s. This shows that α ◦ β = 1F (X). We ﬁnd that α, β are inverse to each other. Therefore α is a bijection. 0 0 To prove (2), let us choose F = hX0 for any object X in C . (1) implies the bijection of the set 0 HomC(X,X ) ⇐⇒ HomC∨ (hX , hX0 ). ∨ By deﬁnition, h : C → C sending X → hX is fully faithful. Corollary 1.1. Let C be a category. Suppose that X and X0 are objects of C so that ∼ 0 hX = hX0 . Then X and X are isomorphic.

Proof. Suppose that hX and hX0 are isomorphic. Then there is an isomorphism

η : hX → hX0 −1 whose inverse is denoted by η : hX0 → hX . Denote the map 0 αX,X0 = HomC∨ (hX , hX0 ) → HomC(X,X ). Then by deﬁnition, −1 −1 −1 −1 αX,X (η ◦ η) = αX0,X (η ) ◦ αX,X0 (η), αX0,X0 (η ◦ η ) = αX,X0 (η) ◦ αX0,X (η ).

By αX,X (1 ) = 1X and α 0 0 (1 0 ) = 1 0 , we ﬁnd hX X ,X hX X −1 −1 αX0,X (η ) ◦ αX,X0 (η) = 1X , αX,X0 (η) ◦ αX0,X (η ) = 1X0 . 0 −1 0 This shows that αX,X0 (η): X → X and αX0,X (η ): X → X are inverse to each other. 0 Therefore X and X are isomorphic. This Corollary shows that any two representatives of a representable functor are isomorphic.