Yoneda Lemmma and evaluation map.

The aim of this blog is to provide a trivial but nice example of Yoneda Lemma, and then we prove the Yoneda Lemma.

Let $\mathcal{C}$$\mathcal C$ be a locally small category, that is, $\mathrm{\forall }X,Y\in \mathrm{Ob}(\mathcal{C}),{\mathrm{Hom}}_{\mathcal{C}}(X,Y)$$\forall X,Y\in \mathrm{Ob}(\mathcal C),\mathrm{Hom}_{\mathcal C}(X,Y)$ is a set.

The Yoneda lemma claim that

$$\mathrm{Nat}({\mathrm{Hom}}_{\mathcal{C}}(X,-),F)\cong F(X)$$

Let us look at a simple example. Let $\mathcal{C}$$\mathcal C$ be the category of set, and $F$$F$ be the identity functor.

Then let us consider the natural transformation betwenen ${\mathrm{Hom}}_{\mathcal{C}}(X,-)$$\mathrm{Hom}_{\mathcal C}(X,-)$ and ${\mathrm{Id}}_{\mathcal{C}}$$\mathrm{Id}_{\mathcal C}$.

$$\begin{array}{ccc}{\mathrm{Hom}}_{\mathcal{C}}(X,A)& \stackrel{{\eta }_A}{\to }& A\\ f_{\ast }\downarrow & & \downarrow f& \\ {\mathrm{Hom}}_{\mathcal{C}}(X,B)& \underset{{\eta }_B}{\overset{}{\to }}& B\end{array}$$

What is natural transormation here? How could you map $h:X\to A$$h:X\to A$ to $A$$A$ and make sure the diagram commute?

The evaluation map! Let us pick a $\alpha \in X$$\alpha\in X$ and let ${\eta }_A={\eta }_{\alpha }={\eta }_B$$\eta_A=\eta_\alpha=\eta_B$. Here ${\eta }_{\alpha }$$\eta_{\alpha}$ is the evaluation map at $\alpha$$\alpha$.

$$f\circ {\eta }_{\alpha }(h)=f(h(\alpha ))={\eta }_{\alpha }(f_{\ast }(h))={\eta }_{\alpha }(f\circ h)=f(h(\alpha ))$$

It is not hard to see the map from $X$$X$ to $\mathrm{Nat}({\mathrm{Hom}}_{\mathcal{C}}(X,-),F)$$\mathrm{Nat}(\mathrm{Hom}_{\mathcal C}(X,-),F)$ is injective.

To see it is surjective, let $A=X$$A=X$ and consider ${\eta }_X({\mathrm{Id}}_X)=\alpha$$\eta_X(\mathrm{Id}_X)=\alpha$ with the following diagram.

$$\begin{array}{ccc}{\mathrm{Hom}}_{\mathcal{C}}(X,X)& \stackrel{{\eta }_X}{\to }& X\\ f_{\ast }\downarrow & & \downarrow f& \\ {\mathrm{Hom}}_{\mathcal{C}}(X,A)& \underset{{\eta }_A}{\overset{}{\to }}& A\end{array}$$

The diagram tells us that:

$${\eta }_A(f_{\ast }({\mathrm{Id}}_X))=f({\eta }_X({\mathrm{Id}}_{\mathrm{X}}))$$

i.e. $\mathrm{\forall }A$$\forall A$ and $f:X\to A$$f:X\to A$, we have

$${\eta }_A(f)=f(\alpha )$$

Hence the map from $X$$X$ to $\mathrm{Nat}({\mathrm{Hom}}_{\mathcal{C}}(X,-),F)$$\mathrm{Nat}(\mathrm{Hom}_{\mathcal C}(X,-),F)$ is surjective.

That is a cute example of Yoneda Lemma.

In general, the proof of Yoneda Lemma is literally same idea.

Let $F:\mathcal{C}\to \mathrm{Set}$$F:\mathcal C\to\mathrm{Set}$ be a functor, and considering the following diagram.

$$\begin{array}{ccc}{\mathrm{Hom}}_{\mathcal{C}}(X,X)& \stackrel{{\eta }_X}{\to }& F(X)\\ f_{\ast }\downarrow & & \downarrow F(f)& \\ {\mathrm{Hom}}_{\mathcal{C}}(X,A)& \underset{{\eta }_A}{\overset{}{\to }}& F(A)\end{array}$$

If ${\eta }_X({\mathrm{Id}}_{\mathrm{X}})=u$$\eta_X(\mathrm{Id_X})=u$. The diagram shows that for every ${\eta }_A$$\eta_A$ we have ${\eta }_A(f_{\ast }({\mathrm{Id}}_A))={\eta }_A(f)=F(f)(u)$$\eta_A(f_*(\mathrm{Id}_A))=\eta_A(f)=F(f)(u)$.

Hence each ${\eta }_A$$\eta_A$ (arrow in $\mathrm{Set}$$\mathrm{Set}$, i.e. function) is completely determined by $u$$u$​.

i.e.

$${\eta }_A(f)=F(f)(u)$$

Conversely, pick an element $u\in F(X)$$u\in F(X)$, consider an function ${\eta }_X({\mathrm{Id}}_X)=u$$\eta_X(\mathrm{Id}_X)=u$​

The diagram force us to define ${\eta }_A(f)=F(f)(u)$$\eta_A(f)=F(f)(u)$ for all $A$$A$ since

$${\eta }_A(f_{\ast }({\mathrm{Id}}_{\mathrm{X}}))={\eta }_A(f)=F(f)({\eta }_X({\mathrm{Id}}_{\mathrm{X}}))=F(f)(u)$$

Then we can check that the following diagram commute.

$$\begin{array}{ccc}{\mathrm{Hom}}_{\mathcal{C}}(X,A)& \stackrel{{\eta }_A}{\to }& F(A)\\ f_{\ast }\downarrow & & \downarrow F(f)& \\ {\mathrm{Hom}}_{\mathcal{C}}(X,B)& \underset{{\eta }_B}{\overset{}{\to }}& F(B)\end{array}$$

Pick a $h\in {\mathrm{Hom}}_{\mathcal{C}}(X,A)$$h\in\mathrm{Hom}_{\mathcal C}(X,A)$,

$$F(f)({\eta }_A(h))=F(f)(F(g)(u))=F(f)\circ F(g)(u)=F(f\circ g)(u)$$

Here $F(g)$$F(g)$ is a function from $F(X)$$F(X)$ to $F(A)$$F(A)$, $u\in F(X)$$u\in F(X)$.

$${\eta }_B(f_{\ast }(g))={\eta }_B(f\circ g)=F(f\circ g)(u)$$

Hence the diagram commute.

Hence we have

$$\mathrm{Nat}({\mathrm{Hom}}_{\mathcal{C}}(X,-),F)\cong F(X)◻$$

The ${\mathrm{Set}}^{{\mathcal{C}}^{\mathrm{op}}}$$\mathrm{Set}^{\mathcal C^{\mathrm{op}}}$ case is claimed by duality.