Yoneda lemma

There is a natural bijection

natural in both $A$ and $F$, where $\mathrm{Nat}(-,-)$ denotes the set of natural transformations.

Explicit correspondence

• Given $\eta\in \mathrm{Nat}(h^A,F)$, the corresponding element of $F(A)$ is
  $$\eta_A(\mathrm{id}_A)\in F(A),$$
  where $\mathrm{id}_A$ is the identity morphism of $A$.

• Given $x\in F(A)$, define a natural transformation $\eta^x:h^A\Rightarrow F$ by, for each object $X$,
  $$\eta^x_X:\mathrm{Hom}_{\mathcal C}(A,X)\to F(X),\quad f \mapsto F(f)(x).$$
  Naturality follows from functoriality of $F$ and composition in $\mathcal C$.

This bijection is in fact a natural isomorphism between functors in $A$ and $F$.

For a functor $G:\mathcal C^{\mathrm{op}}\to \mathbf{Set}$, there is a natural bijection

(Here $\mathrm{Hom}_{\mathcal C}(-,A)$ is the usual contravariant representable.)

1. Subsets via the power set functor. Take $\mathcal C=\mathbf{Set}$, let $A$ be a set, and let $F=\mathcal P$ be the power set functor $X\mapsto \mathcal P(X)$. The Yoneda lemma gives a bijection
   $$\mathrm{Nat}(\mathrm{Hom}(A,-),\mathcal P)\cong \mathcal P(A),$$
   so natural transformations correspond exactly to subsets $S\subseteq A$. Concretely, $S\subseteq A$ yields $\eta^S_X(f)=f(S)\subseteq X$.

1. Picking an element of a group naturally. Take $\mathcal C=\mathbf{Grp}$, let $A=G$ be a group, and let $F=U:\mathbf{Grp}\to\mathbf{Set}$ be the underlying-set functor. Then
   $$\mathrm{Nat}(\mathrm{Hom}_{\mathbf{Grp}}(G,-),U)\cong U(G),$$
   i.e. natural transformations correspond to elements $g\in G$. The corresponding $\eta^g_H:\mathrm{Hom}(G,H)\to U(H)$ sends a homomorphism $\varphi:G\to H$ to $\varphi(g)\in H$.

1. Recovering maps into a fixed module. Let $\mathcal C=R\text{-}\mathbf{Mod}$ and take $F(X)=\mathrm{Hom}_R(M,X)$ viewed as a set-valued functor (forgetting the abelian group structure). Yoneda yields
   $$\mathrm{Nat}(\mathrm{Hom}_R(A,-),\,\mathrm{Hom}_R(M,-))\cong \mathrm{Hom}_R(M,A),$$
   so a natural way to turn maps $A\to X$ into maps $M\to X$ is the same as choosing a map $M\to A$.