The Yoneda Lemma
on our choice of A and F , we will from now on call it α A,F . With this notation, it becomes somewhat clearer why we refer to the isomorphism as ‘natural in A and F ’. To complete the proof of the Yoneda Lemma, we must establish how the functors LHS and RHS act on morphisms, and show that α satisfies the naturality condition. This will establish that α is a natural transformation, and since we have shown that each component α A,F is an isomorphism, we will have at once that α is a natural isomorphism as required. Note that we also have the inverse transformation α − 1 given by ( α − 1 ) A,F = ( α A,F ) − 1 , whose components we may unambiguously denote 2.4.1 The application functor We discuss the application functor RHS first since it appears simpler than the other. The application functor takes an object and a functor (more formally, an ordered pair of an object and a functor) and applies the functor to the object. Since it takes an ordered pair of an object X ∈ Ob( C ) and a functor F : C → Set , the domain of RHS is C × Set C , while its codomain is simply Set . To establish how it acts on morphisms, we must first observe that a morphism in this category is an ordered pair ( f, Φ) of a morphism in C and a natural transformation between functors C → Set . Thus we want RHS to take a morphism f : X → Y and a natural transformation Φ : F → G and return a single morphism RHS ( f, Φ) : F ( X ) → G ( Y ). The naturality condition on Φ tells us that F ( f ) ◦ Φ Y = Φ X ◦ G ( f ), so the two natural choices for our morphism RHS ( f, Φ) are in fact equal. We therefore choose this morphism RHS ( f, Φ) = F ( f ) ◦ Φ Y . To see that this is a functor, we observe that if Φ is an identity transformation then every component Φ X will be the identity morphism id X , and that F (id X ) = id F ( X ) . Thus our expression F ( f ) ◦ Φ Y reduces to id X ◦ id Y with X = Y . To see that RHS preserves composition, we make further use of the naturality condition, observing that for morphisms ( f, Φ) : ( X,F ) → ( Y,G ) and ( g, Ψ) : ( Y,G ) → ( Z,H ) we have
RHS (( f, Φ) ◦ ( g, Ψ)) = RHS ( f ◦ g, Φ ◦ Ψ) = F ( f ◦ g ) ◦ (Φ ◦ Ψ) Z
= F ( f ) ◦ F ( g ) ◦ Φ Z ◦ Ψ Z = F ( f ) ◦ Φ Y ◦ G ( g ) ◦ Ψ Z = RHS ( f, Φ) ◦ RHS ( g, Ψ)
This establishes that RHS satisfies all the conditions necessary to be a functor.
2.4.2 The other functor The functor LHS is slightly trickier to work with. We know it must have the same domain and codomain as RHS so that α can be a natural transformation between them, so we have LHS : C × Set C → Set . We also have that LHS ( A,F ) = Nat(Hom A ,F ). Unlike with the application functor, there is no clear way this functor should act on morphisms. However, we know more about it than just that: we know that we want α to specify a natural isomorphism between LHS and RHS . The naturality condition on α would then tell us that for a morphism ( f, Φ) : ( X,F ) → ( Y,G ) we have
LHS ( f, Φ) ◦ α Y,G = α X,F ◦ RHS ( f, Φ)
We can isolate LHS ( f, Φ) in this equation by composing with
to yield
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