[proofplan] For the covariant statement, compare the two representations by forming the natural isomorphism $\gamma=\alpha^{-1}\circ\beta$ between the two Hom-functors. Naturality forces every component of $\gamma$ to be postcomposition with the single morphism $u=\gamma_B(\operatorname{id}_B):A\to B$. Applying the same construction to $\gamma^{-1}$ gives a morphism $v:B\to A$, and the component formulas imply that $u$ and $v$ are inverse isomorphisms. The uniqueness follows by evaluating the required identity at $X=B$ and $f=\operatorname{id}_B$. The contravariant case is the same argument with precomposition in place of postcomposition. [/proofplan] [step:Form the comparison natural isomorphism between the two Hom-functors] For each object $X \in \operatorname{Ob}(\mathcal C)$, define a function \begin{align*} \gamma_X:\operatorname{Hom}_{\mathcal C}(B,X) &\to \operatorname{Hom}_{\mathcal C}(A,X) \\ f &\mapsto \alpha_X^{-1}(\beta_X(f)). \end{align*} Since $\alpha$ and $\beta$ are natural isomorphisms, each $\gamma_X$ is a bijection, and the family \begin{align*} \gamma:\operatorname{Hom}_{\mathcal C}(B,-) \Rightarrow \operatorname{Hom}_{\mathcal C}(A,-) \end{align*} is a natural isomorphism. Define the morphism $u:A\to B$ by \begin{align*} u:=\gamma_B(\operatorname{id}_B), \end{align*} where $\operatorname{id}_B:B\to B$ is the identity morphism of $B$. [/step] [step:Use naturality to identify every component as postcomposition with $u$] Let $X \in \operatorname{Ob}(\mathcal C)$, and let $f:B\to X$ be a morphism. Naturality of $\gamma$ with respect to $f:B\to X$ says that the diagram \begin{align*} \operatorname{Hom}_{\mathcal C}(B,B) &\xrightarrow{\gamma_B} \operatorname{Hom}_{\mathcal C}(A,B) \\ \operatorname{Hom}_{\mathcal C}(B,f)\downarrow &\hspace{2.5em} \downarrow \operatorname{Hom}_{\mathcal C}(A,f) \\ \operatorname{Hom}_{\mathcal C}(B,X) &\xrightarrow{\gamma_X} \operatorname{Hom}_{\mathcal C}(A,X) \end{align*} commutes. Here \begin{align*} \operatorname{Hom}_{\mathcal C}(B,f):\operatorname{Hom}_{\mathcal C}(B,B)&\to \operatorname{Hom}_{\mathcal C}(B,X), & h&\mapsto f\circ h, \\ \operatorname{Hom}_{\mathcal C}(A,f):\operatorname{Hom}_{\mathcal C}(A,B)&\to \operatorname{Hom}_{\mathcal C}(A,X), & k&\mapsto f\circ k. \end{align*} Evaluating the commuting square at $\operatorname{id}_B$ gives \begin{align*} \gamma_X(f) &=\gamma_X(f\circ \operatorname{id}_B) \\ &=\gamma_X\bigl(\operatorname{Hom}_{\mathcal C}(B,f)(\operatorname{id}_B)\bigr) \\ &=\operatorname{Hom}_{\mathcal C}(A,f)\bigl(\gamma_B(\operatorname{id}_B)\bigr) \\ &=f\circ u. \end{align*} Therefore, by the definition of $\gamma_X$, \begin{align*} \beta_X(f)=\alpha_X(\gamma_X(f))=\alpha_X(f\circ u). \end{align*} [/step] [step:Construct the inverse morphism from the inverse natural isomorphism] For each object $X \in \operatorname{Ob}(\mathcal C)$, define \begin{align*} \delta_X:\operatorname{Hom}_{\mathcal C}(A,X) &\to \operatorname{Hom}_{\mathcal C}(B,X) \\ g &\mapsto \beta_X^{-1}(\alpha_X(g)). \end{align*} Then \begin{align*} \delta:\operatorname{Hom}_{\mathcal C}(A,-) \Rightarrow \operatorname{Hom}_{\mathcal C}(B,-) \end{align*} is the inverse natural isomorphism to $\gamma$. Define \begin{align*} v:=\delta_A(\operatorname{id}_A), \end{align*} so $v:B\to A$. Applying the preceding naturality calculation to $\delta$ gives, for every object $X \in \operatorname{Ob}(\mathcal C)$ and every morphism $g:A\to X$, \begin{align*} \delta_X(g)=g\circ v. \end{align*} [/step] [step:Show that the two morphisms are inverse isomorphisms] Since $\delta$ is inverse to $\gamma$, we have $\delta_B\circ \gamma_B=\operatorname{id}_{\operatorname{Hom}_{\mathcal C}(B,B)}$. Evaluating at $\operatorname{id}_B$ gives \begin{align*} \operatorname{id}_B &=\delta_B(\gamma_B(\operatorname{id}_B)) \\ &=\delta_B(u) \\ &=u\circ v. \end{align*} Similarly, $\gamma_A\circ \delta_A=\operatorname{id}_{\operatorname{Hom}_{\mathcal C}(A,A)}$. Evaluating at $\operatorname{id}_A$ gives \begin{align*} \operatorname{id}_A &=\gamma_A(\delta_A(\operatorname{id}_A)) \\ &=\gamma_A(v) \\ &=v\circ u. \end{align*} Thus $u:A\to B$ is an isomorphism with inverse $v:B\to A$. [/step] [step:Prove uniqueness by evaluating at the identity of the representing object] Let $u':A\to B$ be a morphism satisfying \begin{align*} \beta_X(f)=\alpha_X(f\circ u') \end{align*} for every object $X \in \operatorname{Ob}(\mathcal C)$ and every morphism $f:B\to X$. Take $X=B$ and $f=\operatorname{id}_B$. Then \begin{align*} \beta_B(\operatorname{id}_B) =\alpha_B(\operatorname{id}_B\circ u') =\alpha_B(u'). \end{align*} By definition of $u$, we also have \begin{align*} \alpha_B(u) =\alpha_B(\gamma_B(\operatorname{id}_B)) =\beta_B(\operatorname{id}_B). \end{align*} Hence $\alpha_B(u')=\alpha_B(u)$. Since $\alpha_B$ is a bijection, it follows that $u'=u$. Therefore the isomorphism $u$ is unique. [/step] [step:Repeat the argument contravariantly with precomposition] Now suppose $F:\mathcal C^{\mathrm{op}}\to \mathbf{Set}$ is represented by both $(A,\alpha)$ and $(B,\beta)$, with \begin{align*} \alpha:\operatorname{Hom}_{\mathcal C}(-,A)\Rightarrow F, \qquad \beta:\operatorname{Hom}_{\mathcal C}(-,B)\Rightarrow F. \end{align*} For each object $X \in \operatorname{Ob}(\mathcal C)$, define \begin{align*} \gamma_X:\operatorname{Hom}_{\mathcal C}(X,B)&\to \operatorname{Hom}_{\mathcal C}(X,A)\\ f&\mapsto \alpha_X^{-1}(\beta_X(f)). \end{align*} Then $\gamma:\operatorname{Hom}_{\mathcal C}(-,B)\Rightarrow \operatorname{Hom}_{\mathcal C}(-,A)$ is a natural isomorphism. Define \begin{align*} u:=\gamma_B(\operatorname{id}_B), \end{align*} so $u:B\to A$. Let $X \in \operatorname{Ob}(\mathcal C)$, and let $f:X\to B$ be a morphism. Naturality of $\gamma$ with respect to $f:X\to B$ gives commutativity of the square whose horizontal maps are $\gamma_B$ and $\gamma_X$, and whose vertical maps are precomposition by $f$. Evaluating at $\operatorname{id}_B$ gives \begin{align*} \gamma_X(f) &=\gamma_X(\operatorname{id}_B\circ f) \\ &=u\circ f. \end{align*} Therefore \begin{align*} \beta_X(f)=\alpha_X(u\circ f). \end{align*} Applying the same construction to the inverse natural isomorphism \begin{align*} \delta:\operatorname{Hom}_{\mathcal C}(-,A)\Rightarrow \operatorname{Hom}_{\mathcal C}(-,B) \end{align*} gives a morphism $v:A\to B$ such that $\delta_X(g)=v\circ g$ for every $g:X\to A$. Since $\delta$ and $\gamma$ are inverse natural isomorphisms, evaluating at identity morphisms gives \begin{align*} u\circ v=\operatorname{id}_A, \qquad v\circ u=\operatorname{id}_B. \end{align*} Thus $u:B\to A$ is an isomorphism. Finally, if $u':B\to A$ also satisfies $\beta_X(f)=\alpha_X(u'\circ f)$ for all $X$ and all $f:X\to B$, then taking $X=B$ and $f=\operatorname{id}_B$ gives \begin{align*} \alpha_B(u')=\beta_B(\operatorname{id}_B)=\alpha_B(u). \end{align*} Since $\alpha_B$ is a bijection, $u'=u$. This proves the contravariant uniqueness statement. [/step]