Yoneda’s Lemma (part 2)

Theorem (Yoneda):

Let \mathcal C be a locally small category. Let A be an object in \mathcal C, and let F:\mathcal C\to\textsc{Set}. Then,

\mbox{Nat}(h^A,F)\cong F(A),

and this isomorphism is natural in F and in A. (we won’t prove this part here)

Proof:

Given a natural transformation \eta from h^A=\hom(A,-) to F, we need to produce an element in F(A). There aren’t many things we could try. We have a map \eta_A:\hom(A,A)\to F(A). We can use that to produce an element in F(A). Of course, we’d have to plug something in to \eta_A, and there could be many choices there. There is an obvious one though. In fact, it’s the only element of \hom(A,A) we know exists: the identity map. So for a natural transformation \eta, the associated element of F(A) we’ll produce is going to be \eta_A(\mbox{id}_A). I’ll write this as function as \Theta. so \Theta(\eta)=\eta_A(\mbox{id}_A). We need to check that \Theta is an isomorphism.

First, let’s check that \Theta has an inverse. One thing we could do is check that if we knew \Theta(\eta), we could recover \eta. So let’s say we have some x\in F(A), and we want to find a natural transformation \eta such that \Theta(\eta)=x. What do we need to specify \eta?

A natural transformation \eta:h^A\to F is defined by all of the maps \eta_B:\hom(A,B)\to F(B) (we have one for each B in \mathcal C). So if I can tell you what each \eta_B is going to be, without making any choices, then I’ve uniquely defined a natural transformation from x\in F(A). Great, so how do I specify \eta_B? Well that’s just a map \eta_B:\hom(A,B)\to F(B), so I need to tell you what it does to each element in the domain (that is, each \phi\in\hom(A,B)).

So I’m going to define the transformation, and then explain why it works. Let B\in\mathcal{C}, and let \phi\in\hom(A,B)=h^A(B). Then \eta_B(\phi) should be an element of F(B). I define it to be:

\eta_B(\phi)=(F\phi)(x)

How do we check if this defines a natural transformation? We need to check the commuting stuff. So take a look at this diagram (for an arbitrary f:B\to C):

Starting with an element \phi\in\hom(A,B), and going down and then across is

(\eta_C\circ h^Af)(\phi)=\eta_C(h^Af(\phi))=\eta_C(f\circ \phi)=F(f\circ \phi)(x)=(Ff)(F\phi)(x)

Going across and then down is

(Ff\circ\eta_B)(\phi)=(Ff)(\eta_B(\phi))=(Ff)(F\phi)(x)

Great, so the diagram commutes. So \eta really is a natural transformation. We just need to check that it’s the inverse of the \Theta map. That is, we need to check that \Theta(\eta)=x. We also need to check the other direction, that if we start with some \eta, compute \Theta(\eta), and then do this construction, we get back the original \eta. I’m lazy, so I’ll only do the first one, though neither are difficult. You simply need to wade through the definitions.

Suppose we construct \eta as above from a given x\in F(A). We hope that \Theta(\eta)=x. Indeed, \Theta(\eta)=\eta_A(\mbox{id}_A). From the definition of \eta,

\Theta(\eta)=\eta_A(\mbox{id}_A)=F\mbox{id}_A(x)=\mbox{id}_{F(A)}(x)=x.

\square

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