[proofplan] We prove the bijection directly. Given an element $a\in F(A)$, we construct a natural transformation by sending each morphism $v:X\to A$ to $F(v)(a)\in F(X)$. Naturality follows from functoriality of the contravariant functor $F$, and the two inverse identities follow by evaluating at $\operatorname{id}_A$ and by applying naturality to the morphism $v:X\to A$. [/proofplan] [step:Construct a natural transformation from an element of $F(A)$] Let $a\in F(A)$. For each object $X$ of $\mathcal C$, define a function \begin{align*} \eta^a_X:\operatorname{Hom}_{\mathcal C}(X,A) &\to F(X) \\ v &\mapsto F(v)(a). \end{align*} This is well-defined because $v:X\to A$ is a morphism in $\mathcal C$, so the contravariant functor $F:\mathcal C^{\mathrm{op}}\to \mathbf{Set}$ sends $v$ to a function \begin{align*} F(v):F(A)\to F(X). \end{align*} We verify that the family $\eta^a=(\eta^a_X)_X$ is natural. Let $f:X\to Y$ be a morphism in $\mathcal C$, and let $v\in \operatorname{Hom}_{\mathcal C}(Y,A)$. The functor $\operatorname{Hom}_{\mathcal C}(-,A)$ sends $f$ to precomposition by $f$, so it sends $v$ to $v\circ f$. Therefore \begin{align*} \eta^a_X(v\circ f) &= F(v\circ f)(a) \\ &= (F(f)\circ F(v))(a) \\ &= F(f)(F(v)(a)) \\ &= F(f)(\eta^a_Y(v)). \end{align*} The second equality is functoriality of $F:\mathcal C^{\mathrm{op}}\to \mathbf{Set}$, since contravariance reverses the composite $v\circ f$. Thus the naturality square commutes for every morphism $f:X\to Y$, and $\eta^a:\operatorname{Hom}_{\mathcal C}(-,A)\Rightarrow F$ is a natural transformation. [guided] Start with an element $a\in F(A)$. We need to turn $a$ into an entire natural transformation from the representable functor $\operatorname{Hom}_{\mathcal C}(-,A)$ to $F$. For an object $X$ of $\mathcal C$, an element of $\operatorname{Hom}_{\mathcal C}(X,A)$ is a morphism $v:X\to A$. Since $F$ is contravariant, this morphism induces a function in the reverse direction: \begin{align*} F(v):F(A)\to F(X). \end{align*} Therefore it is forced that the component at $X$ should send $v$ to $F(v)(a)$. Define \begin{align*} \eta^a_X:\operatorname{Hom}_{\mathcal C}(X,A) &\to F(X) \\ v &\mapsto F(v)(a). \end{align*} Now verify naturality. Let $f:X\to Y$ be a morphism in $\mathcal C$, and let $v:Y\to A$ be an element of $\operatorname{Hom}_{\mathcal C}(Y,A)$. The representable functor $\operatorname{Hom}_{\mathcal C}(-,A)$ acts on $f$ by precomposition, so it sends $v$ to $v\circ f:X\to A$. Hence the left-hand route around the naturality square gives \begin{align*} \eta^a_X(v\circ f)=F(v\circ f)(a). \end{align*} Because $F$ is a functor on $\mathcal C^{\mathrm{op}}$, it reverses composition. Thus \begin{align*} F(v\circ f)=F(f)\circ F(v), \end{align*} as functions from $F(A)$ to $F(X)$. Therefore \begin{align*} \eta^a_X(v\circ f) &= F(v\circ f)(a) \\ &= (F(f)\circ F(v))(a) \\ &= F(f)(F(v)(a)) \\ &= F(f)(\eta^a_Y(v)). \end{align*} This is exactly the naturality condition for the morphism $f:X\to Y$. Since $f$ and $v$ were arbitrary, $\eta^a$ is a natural transformation. [/guided] [/step] [step:Define the two comparison functions] Define \begin{align*} \Phi:\operatorname{Nat}(\operatorname{Hom}_{\mathcal C}(-,A),F) &\to F(A) \\ \eta &\mapsto \eta_A(\operatorname{id}_A) \end{align*} and define \begin{align*} \Psi:F(A) &\to \operatorname{Nat}(\operatorname{Hom}_{\mathcal C}(-,A),F) \\ a &\mapsto \eta^a, \end{align*} where $\eta^a$ is the natural transformation constructed in the previous step. [/step] [step:Show that evaluation after construction recovers the original element] Let $a\in F(A)$. Then \begin{align*} (\Phi\circ \Psi)(a) &= \Phi(\eta^a) \\ &= \eta^a_A(\operatorname{id}_A) \\ &= F(\operatorname{id}_A)(a) \\ &= \operatorname{id}_{F(A)}(a) \\ &= a. \end{align*} The fourth equality uses functoriality of $F$, which sends the identity morphism $\operatorname{id}_A$ in $\mathcal C$ to the identity function on $F(A)$. Hence $\Phi\circ\Psi=\operatorname{id}_{F(A)}$. [/step] [step:Show that construction after evaluation recovers the original natural transformation] Let $\eta:\operatorname{Hom}_{\mathcal C}(-,A)\Rightarrow F$ be a natural transformation. We prove that $\Psi(\Phi(\eta))=\eta$ by checking components. Put \begin{align*} a:=\Phi(\eta)=\eta_A(\operatorname{id}_A)\in F(A). \end{align*} Let $X$ be an object of $\mathcal C$, and let $v\in \operatorname{Hom}_{\mathcal C}(X,A)$. Apply naturality of $\eta$ to the morphism $v:X\to A$. Since $\operatorname{Hom}_{\mathcal C}(-,A)$ sends $v$ to the function \begin{align*} \operatorname{Hom}_{\mathcal C}(A,A) &\to \operatorname{Hom}_{\mathcal C}(X,A) \\ w &\mapsto w\circ v, \end{align*} it sends $\operatorname{id}_A$ to $v$. Therefore the naturality square gives \begin{align*} \eta_X(v) &= \eta_X(\operatorname{id}_A\circ v) \\ &= F(v)(\eta_A(\operatorname{id}_A)) \\ &= F(v)(a) \\ &= \eta^a_X(v). \end{align*} Thus $\eta_X=\eta^a_X$ for every object $X$ of $\mathcal C$, so $\eta=\eta^a=\Psi(\Phi(\eta))$. [guided] Let $\eta:\operatorname{Hom}_{\mathcal C}(-,A)\Rightarrow F$ be any natural transformation. We want to prove that once we know the single element $\eta_A(\operatorname{id}_A)\in F(A)$, every component of $\eta$ is forced. Define \begin{align*} a:=\Phi(\eta)=\eta_A(\operatorname{id}_A)\in F(A). \end{align*} The natural transformation constructed from $a$ has components \begin{align*} \eta^a_X:\operatorname{Hom}_{\mathcal C}(X,A) &\to F(X) \\ v &\mapsto F(v)(a). \end{align*} So we must prove $\eta_X(v)=F(v)(a)$ for every object $X$ and every morphism $v:X\to A$. Apply naturality of $\eta$ to the morphism $v:X\to A$ in $\mathcal C$. The representable functor $\operatorname{Hom}_{\mathcal C}(-,A)$ sends this morphism to the precomposition function \begin{align*} \operatorname{Hom}_{\mathcal C}(A,A) &\to \operatorname{Hom}_{\mathcal C}(X,A) \\ w &\mapsto w\circ v. \end{align*} In particular, this function sends $\operatorname{id}_A$ to $\operatorname{id}_A\circ v=v$. Naturality says that applying $\eta$ after this precomposition agrees with applying $F(v)$ after $\eta_A$. Evaluating that equality at $\operatorname{id}_A$ gives \begin{align*} \eta_X(v) &= \eta_X(\operatorname{id}_A\circ v) \\ &= F(v)(\eta_A(\operatorname{id}_A)) \\ &= F(v)(a) \\ &= \eta^a_X(v). \end{align*} Since $X$ and $v$ were arbitrary, every component of $\eta$ agrees with the corresponding component of $\eta^a$. Therefore $\eta=\eta^a=\Psi(\Phi(\eta))$. [/guided] [/step] [step:Conclude that the evaluation map is a bijection] The previous two steps prove \begin{align*} \Phi\circ\Psi=\operatorname{id}_{F(A)} \qquad\text{and}\qquad \Psi\circ\Phi=\operatorname{id}_{\operatorname{Nat}(\operatorname{Hom}_{\mathcal C}(-,A),F)}. \end{align*} Hence $\Phi$ is a bijection with inverse $\Psi$. By the definition of $\Phi$, the element corresponding to a natural transformation $\eta$ is precisely $\eta_A(\operatorname{id}_A)\in F(A)$. [/step]