Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 25 / 1 / a

For the covariant version of the Yoneda embedding, put $Y(A)=\mathcal{C}(A,-)$. A morphism $u:A\to B$ induces the natural transformation $Y(u):Y(B)\to Y(A)$ whose component sends $v:B\to X$ to $vu$. Thus $Y$ is a functor from the opposite category ${\mathcal{C}}^{\mathrm{op}}$ to the functor category $[\mathcal{C},\mathbf{Set}]$. The locally small category hypothesis makes each value a set.

The Yoneda lemma says that, for a functor $F:\mathcal{C}\to \mathbf{Set}$, evaluation at the identity morphism gives a natural bijectionGiven $a\in F(A)$, define ${\theta }_X^a(v)=F(v)(a)$. For $w:X\to X^{\prime }$, the functor law gives $F(w){\theta }_X^a(v)=F(wv)(a)={\theta }_{X^{\prime }}^a(wv)$, proving naturality. Conversely, naturality of $\theta$ at $v:A\to X$ gives ${\theta }_X(v)=F(v){\theta }_A(1_A)$. These two constructions are inverse, because $F(1_A)(a)=a$.

The bijection is natural in both variables. A natural transformation $\beta :F\to H$ sends the element $a$ to ${\beta }_A(a)$, matching postcomposition by $\beta$. For $u:A\to B$, precomposing with $Y(u)$ sends $\theta$ to a transformation from $Y(B)$ and its distinguished element is ${\theta }_B(u)=F(u)(a)$. This is exactly the required dependence on $A$.

Taking $F=Y(B)$ gives $\mathrm{Nat}(Y(A),Y(B))\cong \mathcal{C}(B,A)$. Consequently the covariant Yoneda embedding is full and faithful, with the reversal of arrows accounted for by its opposite category domain.