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 2 F (X): For any object Y of C; and any morphism 2 hX (Y ) = HomC(Y; X); we have F ( ) 2 HomSets(F (X);F (Y )): Since s 2 F (X); we obtain F ( )(s) 2 F (Y ): This observation allows us to define a map β : F (X) ! HomC_ (hX ;F ) as follows. For each s 2 F (X); β(s) defines a natural transformation from hX to F by βY (s) : HomC(Y; X) ! F (Y ); 7! F ( )(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 2 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); 8φ. This shows that ηY = βY (α(η)) for all objects Y of C; and hence η = (β ◦ α)(η) for all η 2 HomC_ (hX ;F ): This shows that β ◦ α = 1 : HomC_ (hX ;F ) Now, let s 2 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 isomor- phic..