[proofplan] We prove the contravariant statement directly from naturality, without invoking Yoneda as a black box. The two given representations determine a natural isomorphism between the two representable functors, and evaluating it at the identity morphisms produces inverse arrows $A\to B$ and $B\to A$. Naturality then shows that the natural isomorphism is exactly postcomposition with the arrow $A\to B$, which gives both the required compatibility and uniqueness. For the covariant statement, we construct the natural isomorphism between $\operatorname{Hom}_{\mathcal C}(A,-)$ and $\operatorname{Hom}_{\mathcal C}(B,-)$ directly, evaluate it at $\operatorname{id}_A$ to obtain the arrow $B\to A$, and then write out the inverse and uniqueness computations in the covariant composition order. [/proofplan] [step:Construct the natural isomorphism between the representable functors] For each object $X\in\mathcal C$, define a map \begin{align*} \Theta_X:\operatorname{Hom}_{\mathcal C}(X,A)&\to \operatorname{Hom}_{\mathcal C}(X,B) \\ g&\mapsto \Psi_X^{-1}(\Phi_X(g)). \end{align*} Since $\Phi_X$ and $\Psi_X$ are bijections for every $X$, each $\Theta_X$ is a bijection. The family $\Theta=(\Theta_X)_{X\in\mathcal C}$ is natural because it is the composite natural isomorphism \begin{align*} \operatorname{Hom}_{\mathcal C}(-,A)\xrightarrow{\Phi}F\xrightarrow{\Psi^{-1}}\operatorname{Hom}_{\mathcal C}(-,B). \end{align*} Thus for every morphism $u:Y\to X$ in $\mathcal C$ and every morphism $g:X\to A$, \begin{align*} \Theta_Y(g\circ u)=\Theta_X(g)\circ u. \end{align*} [/step] [step:Recover the comparison morphism by evaluating at the identity] Define the morphism $f:A\to B$ by \begin{align*} f:=\Theta_A(\operatorname{id}_A). \end{align*} We claim that $\Theta_X$ is postcomposition with $f$ for every object $X$. Let $X\in\mathcal C$ and let $g:X\to A$ be a morphism. Applying naturality of $\Theta$ to the morphism $g:X\to A$ and the element $\operatorname{id}_A\in\operatorname{Hom}_{\mathcal C}(A,A)$ gives \begin{align*} \Theta_X(\operatorname{id}_A\circ g)=\Theta_A(\operatorname{id}_A)\circ g. \end{align*} Since $\operatorname{id}_A\circ g=g$ and $\Theta_A(\operatorname{id}_A)=f$, this becomes \begin{align*} \Theta_X(g)=f\circ g. \end{align*} By the definition of $\Theta_X$, this is equivalent to \begin{align*} \Psi_X^{-1}(\Phi_X(g))=f\circ g. \end{align*} Applying the bijection $\Psi_X$ to both sides gives \begin{align*} \Phi_X(g)=\Psi_X(f\circ g). \end{align*} This is the required compatibility. [/step] [step:Construct the inverse morphism in the opposite direction] Define, symmetrically, for each object $X\in\mathcal C$ a map \begin{align*} \Lambda_X:\operatorname{Hom}_{\mathcal C}(X,B)&\to \operatorname{Hom}_{\mathcal C}(X,A) \\ k&\mapsto \Phi_X^{-1}(\Psi_X(k)). \end{align*} The family $\Lambda=(\Lambda_X)_{X\in\mathcal C}$ is the inverse natural isomorphism to $\Theta$, because for every object $X$, \begin{align*} \Lambda_X=\Theta_X^{-1}. \end{align*} Define the morphism $e:B\to A$ by \begin{align*} e:=\Lambda_B(\operatorname{id}_B). \end{align*} Let $X\in\mathcal C$ be an object and let $k:X\to B$ be a morphism. Applying naturality of $\Lambda$ to the morphism $k:X\to B$ and the element $\operatorname{id}_B\in\operatorname{Hom}_{\mathcal C}(B,B)$ gives \begin{align*} \Lambda_X(\operatorname{id}_B\circ k)=\Lambda_B(\operatorname{id}_B)\circ k. \end{align*} Since $\operatorname{id}_B\circ k=k$ and $\Lambda_B(\operatorname{id}_B)=e$, this becomes \begin{align*} \Lambda_X(k)=e\circ k. \end{align*} Now evaluate the identities $\Lambda_A\circ\Theta_A=\operatorname{id}_{\operatorname{Hom}_{\mathcal C}(A,A)}$ and $\Theta_B\circ\Lambda_B=\operatorname{id}_{\operatorname{Hom}_{\mathcal C}(B,B)}$ at the identity morphisms. First, \begin{align*} \operatorname{id}_A &=\Lambda_A(\Theta_A(\operatorname{id}_A)) \\ &=\Lambda_A(f) \\ &=e\circ f. \end{align*} Second, \begin{align*} \operatorname{id}_B &=\Theta_B(\Lambda_B(\operatorname{id}_B)) \\ &=\Theta_B(e) \\ &=f\circ e. \end{align*} Therefore $f:A\to B$ is an isomorphism with inverse $e:B\to A$. [/step] [step:Prove uniqueness of the compatible isomorphism] Let $f':A\to B$ be any morphism satisfying \begin{align*} \Phi_X(g)=\Psi_X(f'\circ g) \end{align*} for every object $X\in\mathcal C$ and every morphism $g:X\to A$. Taking $X=A$ and $g=\operatorname{id}_A$, we obtain \begin{align*} \Phi_A(\operatorname{id}_A)=\Psi_A(f'\circ \operatorname{id}_A)=\Psi_A(f'). \end{align*} By the definition of $f$, we also have \begin{align*} \Phi_A(\operatorname{id}_A)=\Psi_A(f). \end{align*} Since $\Psi_A:\operatorname{Hom}_{\mathcal C}(A,B)\to F(A)$ is injective, it follows that \begin{align*} f'=f. \end{align*} Thus the compatible isomorphism $f:A\to B$ is unique. [/step] [step:Construct the covariant comparison morphism directly] Now suppose $F:\mathcal C\to\mathbf{Set}$ is represented by both $(A,\Phi)$ and $(B,\Psi)$, with natural isomorphisms \begin{align*} \Phi:\operatorname{Hom}_{\mathcal C}(A,-)\xrightarrow{\cong}F, \qquad \Psi:\operatorname{Hom}_{\mathcal C}(B,-)\xrightarrow{\cong}F. \end{align*} For each object $X\in\mathcal C$, define a map \begin{align*} \Theta_X:\operatorname{Hom}_{\mathcal C}(A,X)&\to \operatorname{Hom}_{\mathcal C}(B,X) \\ h&\mapsto \Psi_X^{-1}(\Phi_X(h)). \end{align*} Since $\Phi_X$ and $\Psi_X$ are bijections for every object $X$, each $\Theta_X$ is a bijection. The family $\Theta=(\Theta_X)_{X\in\mathcal C}$ is the composite natural isomorphism \begin{align*} \operatorname{Hom}_{\mathcal C}(A,-)\xrightarrow{\Phi}F\xrightarrow{\Psi^{-1}}\operatorname{Hom}_{\mathcal C}(B,-). \end{align*} Thus for every morphism $u:X\to Y$ in $\mathcal C$ and every morphism $h:A\to X$, \begin{align*} \Theta_Y(u\circ h)=u\circ \Theta_X(h). \end{align*} Define the morphism $f:B\to A$ by \begin{align*} f:=\Theta_A(\operatorname{id}_A). \end{align*} Let $X\in\mathcal C$ be an object and let $h:A\to X$ be a morphism. Applying naturality of $\Theta$ to the morphism $h:A\to X$ and the element $\operatorname{id}_A\in\operatorname{Hom}_{\mathcal C}(A,A)$ gives \begin{align*} \Theta_X(h\circ\operatorname{id}_A)=h\circ\Theta_A(\operatorname{id}_A). \end{align*} Since $h\circ\operatorname{id}_A=h$ and $\Theta_A(\operatorname{id}_A)=f$, this becomes \begin{align*} \Theta_X(h)=h\circ f. \end{align*} By the definition of $\Theta_X$, this is equivalent to \begin{align*} \Psi_X^{-1}(\Phi_X(h))=h\circ f. \end{align*} Applying the bijection $\Psi_X$ to both sides gives \begin{align*} \Phi_X(h)=\Psi_X(h\circ f). \end{align*} This is the required covariant compatibility. [/step] [step:Verify that the covariant comparison morphism is an isomorphism] Define, symmetrically, for each object $X\in\mathcal C$ a map \begin{align*} \Lambda_X:\operatorname{Hom}_{\mathcal C}(B,X)&\to \operatorname{Hom}_{\mathcal C}(A,X) \\ k&\mapsto \Phi_X^{-1}(\Psi_X(k)). \end{align*} The family $\Lambda=(\Lambda_X)_{X\in\mathcal C}$ is the inverse natural isomorphism to $\Theta$, because for every object $X$, \begin{align*} \Lambda_X=\Theta_X^{-1}. \end{align*} Define the morphism $e:A\to B$ by \begin{align*} e:=\Lambda_B(\operatorname{id}_B). \end{align*} Let $X\in\mathcal C$ be an object and let $k:B\to X$ be a morphism. Applying naturality of $\Lambda$ to the morphism $k:B\to X$ and the element $\operatorname{id}_B\in\operatorname{Hom}_{\mathcal C}(B,B)$ gives \begin{align*} \Lambda_X(k\circ\operatorname{id}_B)=k\circ\Lambda_B(\operatorname{id}_B). \end{align*} Since $k\circ\operatorname{id}_B=k$ and $\Lambda_B(\operatorname{id}_B)=e$, this becomes \begin{align*} \Lambda_X(k)=k\circ e. \end{align*} Now evaluate the identities $\Lambda_A\circ\Theta_A=\operatorname{id}_{\operatorname{Hom}_{\mathcal C}(A,A)}$ and $\Theta_B\circ\Lambda_B=\operatorname{id}_{\operatorname{Hom}_{\mathcal C}(B,B)}$ at the identity morphisms. First, \begin{align*} \operatorname{id}_A &=\Lambda_A(\Theta_A(\operatorname{id}_A)) \\ &=\Lambda_A(f) \\ &=f\circ e. \end{align*} Second, \begin{align*} \operatorname{id}_B &=\Theta_B(\Lambda_B(\operatorname{id}_B)) \\ &=\Theta_B(e) \\ &=e\circ f. \end{align*} Therefore $f:B\to A$ is an isomorphism with inverse $e:A\to B$. [/step] [step:Prove uniqueness in the covariant case] Let $f':B\to A$ be any morphism satisfying \begin{align*} \Phi_X(h)=\Psi_X(h\circ f') \end{align*} for every object $X\in\mathcal C$ and every morphism $h:A\to X$. Taking $X=A$ and $h=\operatorname{id}_A$, we obtain \begin{align*} \Phi_A(\operatorname{id}_A)=\Psi_A(\operatorname{id}_A\circ f')=\Psi_A(f'). \end{align*} By the definition of $f$, we also have \begin{align*} \Phi_A(\operatorname{id}_A)=\Psi_A(f). \end{align*} Since $\Psi_A:\operatorname{Hom}_{\mathcal C}(B,A)\to F(A)$ is injective, it follows that \begin{align*} f'=f. \end{align*} Thus the compatible isomorphism $f:B\to A$ is unique, and the covariant case completes the proof of the theorem. [/step]