# 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 $\eta$,

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

$\square$