 Solution

The Yoneda embedding is the functor $H_{\bullet }:\mathcal{C}\to [{\mathcal{C}}^{\mathrm{op}},\mathbf{Set}]$ sending $A$ to $H_A=\mathcal{C}(-,A)$ and $f:A\to B$ to postcomposition, $(H_f{)}_U(h)=fh$. Composition and identities are preserved by associativity and the identity laws in $\mathcal{C}$.

Apply the Yoneda lemma with $X=H_B$. It gives a bijection $\mathrm{Nat}(H_A,H_B)\cong \mathcal{C}(A,B)$. The inverse sends $f$ precisely to $H_f$, since $H_B(h)(f)=fh$. Hence every transformation between the representables comes from exactly one morphism of $\mathcal{C}$. This proves that $H_{\bullet }$ is a full and faithful functor.