[proofplan] Choose a [natural isomorphism](/page/Natural%20Isomorphism) between the two [representable functors](/page/Representable%20Functor) and let its inverse be the inverse [natural transformation](/page/Natural%20Transformation). Evaluating the first transformation at the identity morphism of $A$ gives a morphism $f:A\to B$, and evaluating the inverse transformation at the identity morphism of $B$ gives a morphism $g:B\to A$. Naturality, together with the fact that the two natural transformations are inverse, shows that $g\circ f=\operatorname{id}_A$ and $f\circ g=\operatorname{id}_B$. [/proofplan] [step:Extract the candidate inverse morphisms from the natural isomorphism] Let \begin{align*} \eta:\operatorname{Hom}_{\mathcal C}(-,A)\Longrightarrow \operatorname{Hom}_{\mathcal C}(-,B) \end{align*} be a [natural isomorphism](/page/Natural%20Isomorphism), and let \begin{align*} \theta:\operatorname{Hom}_{\mathcal C}(-,B)\Longrightarrow \operatorname{Hom}_{\mathcal C}(-,A) \end{align*} denote its inverse [natural transformation](/page/Natural%20Transformation). Thus, for every object $X$ of $\mathcal C$, the component \begin{align*} \eta_X:\operatorname{Hom}_{\mathcal C}(X,A)\to \operatorname{Hom}_{\mathcal C}(X,B) \end{align*} is a bijection, the component \begin{align*} \theta_X:\operatorname{Hom}_{\mathcal C}(X,B)\to \operatorname{Hom}_{\mathcal C}(X,A) \end{align*} is its inverse, and \begin{align*} \theta_X\circ \eta_X=\operatorname{id}_{\operatorname{Hom}_{\mathcal C}(X,A)}, \qquad \eta_X\circ \theta_X=\operatorname{id}_{\operatorname{Hom}_{\mathcal C}(X,B)}. \end{align*} Define morphisms \begin{align*} f&:=\eta_A(\operatorname{id}_A)\in \operatorname{Hom}_{\mathcal C}(A,B),\\ g&:=\theta_B(\operatorname{id}_B)\in \operatorname{Hom}_{\mathcal C}(B,A). \end{align*} These are the only possible candidates for the isomorphism and its inverse, because they are the images of the identity morphisms under the given natural transformations. [/step] [step:Use naturality of the inverse transformation to prove $g\circ f=\operatorname{id}_A$] Since $\theta_A$ is inverse to $\eta_A$, we have \begin{align*} \theta_A(f) = \theta_A(\eta_A(\operatorname{id}_A)) = \operatorname{id}_A. \end{align*} Now apply naturality of $\theta$ to the morphism $f:A\to B$. For the contravariant representable functors, naturality says that for every morphism $u:X\to Y$ in $\mathcal C$ and every morphism $\beta:Y\to B$, \begin{align*} \theta_X(\beta\circ u)=\theta_Y(\beta)\circ u. \end{align*} Taking $u=f:A\to B$ and $\beta=\operatorname{id}_B:B\to B$ gives \begin{align*} \theta_A(f) = \theta_A(\operatorname{id}_B\circ f) = \theta_B(\operatorname{id}_B)\circ f = g\circ f. \end{align*} Combining the two equalities yields \begin{align*} g\circ f=\operatorname{id}_A. \end{align*} [guided] We first use that $\theta$ is the inverse of $\eta$. The morphism $f:A\to B$ was defined by \begin{align*} f=\eta_A(\operatorname{id}_A). \end{align*} Because $\theta_A$ is inverse to $\eta_A$, applying $\theta_A$ to $f$ recovers the original element $\operatorname{id}_A\in \operatorname{Hom}_{\mathcal C}(A,A)$: \begin{align*} \theta_A(f) = \theta_A(\eta_A(\operatorname{id}_A)) = \operatorname{id}_A. \end{align*} We now reinterpret the same element $\theta_A(f)$ using naturality. Since $\theta$ is a natural transformation \begin{align*} \theta:\operatorname{Hom}_{\mathcal C}(-,B)\Longrightarrow \operatorname{Hom}_{\mathcal C}(-,A), \end{align*} naturality for a morphism $u:X\to Y$ in $\mathcal C$ says that precomposition by $u$ commutes with the components of $\theta$. Explicitly, for every $\beta:Y\to B$, \begin{align*} \theta_X(\beta\circ u)=\theta_Y(\beta)\circ u. \end{align*} We apply this with $X=A$, $Y=B$, $u=f:A\to B$, and $\beta=\operatorname{id}_B:B\to B$. Then $\beta\circ u=\operatorname{id}_B\circ f=f$, so \begin{align*} \theta_A(f) = \theta_A(\operatorname{id}_B\circ f) = \theta_B(\operatorname{id}_B)\circ f. \end{align*} By the definition of $g$, namely $g=\theta_B(\operatorname{id}_B)$, this becomes \begin{align*} \theta_A(f)=g\circ f. \end{align*} Since we already proved $\theta_A(f)=\operatorname{id}_A$, we conclude \begin{align*} g\circ f=\operatorname{id}_A. \end{align*} [/guided] [/step] [step:Use naturality of the original transformation to prove $f\circ g=\operatorname{id}_B$] Since $\eta_B$ is inverse to $\theta_B$, we have \begin{align*} \eta_B(g) = \eta_B(\theta_B(\operatorname{id}_B)) = \operatorname{id}_B. \end{align*} Apply naturality of $\eta$ to the morphism $g:B\to A$. For every morphism $u:X\to Y$ in $\mathcal C$ and every morphism $\alpha:Y\to A$, \begin{align*} \eta_X(\alpha\circ u)=\eta_Y(\alpha)\circ u. \end{align*} Taking $u=g:B\to A$ and $\alpha=\operatorname{id}_A:A\to A$ gives \begin{align*} \eta_B(g) = \eta_B(\operatorname{id}_A\circ g) = \eta_A(\operatorname{id}_A)\circ g = f\circ g. \end{align*} Combining the two equalities yields \begin{align*} f\circ g=\operatorname{id}_B. \end{align*} [guided] We prove the identity $f\circ g=\operatorname{id}_B$ by computing $\eta_B(g)$ in two ways, using first the inverse relation between $\eta_B$ and $\theta_B$, and then the naturality of $\eta$. Since $g$ was defined by \begin{align*} g=\theta_B(\operatorname{id}_B), \end{align*} and since $\eta_B$ is inverse to $\theta_B$, applying $\eta_B$ to $g$ gives \begin{align*} \eta_B(g) = \eta_B(\theta_B(\operatorname{id}_B)) = \operatorname{id}_B. \end{align*} We now compute $\eta_B(g)$ by naturality. The natural transformation \begin{align*} \eta:\operatorname{Hom}_{\mathcal C}(-,A)\Longrightarrow \operatorname{Hom}_{\mathcal C}(-,B) \end{align*} has the following naturality property: for every morphism $u:X\to Y$ and every morphism $\alpha:Y\to A$, \begin{align*} \eta_X(\alpha\circ u)=\eta_Y(\alpha)\circ u. \end{align*} Take $X=B$, $Y=A$, $u=g:B\to A$, and $\alpha=\operatorname{id}_A:A\to A$. Then $\alpha\circ u=\operatorname{id}_A\circ g=g$, and naturality gives \begin{align*} \eta_B(g) = \eta_B(\operatorname{id}_A\circ g) = \eta_A(\operatorname{id}_A)\circ g. \end{align*} By the definition of $f$, namely $f=\eta_A(\operatorname{id}_A)$, this is \begin{align*} \eta_B(g)=f\circ g. \end{align*} Since $\eta_B(g)=\operatorname{id}_B$, we obtain \begin{align*} f\circ g=\operatorname{id}_B. \end{align*} [/guided] [/step] [step:Conclude that the reconstructed morphisms exhibit an isomorphism] The morphisms $f:A\to B$ and $g:B\to A$ satisfy \begin{align*} g\circ f=\operatorname{id}_A, \qquad f\circ g=\operatorname{id}_B. \end{align*} Therefore $f$ is an [isomorphism](/page/Isomorphism) in $\mathcal C$ with inverse $g$. Hence $A$ and $B$ are isomorphic objects of $\mathcal C$. [/step]