[proofplan] We prove the equivalence by constructing the correspondence in both directions. A representing natural isomorphism determines an element $u\in F(A)$ by evaluating the component at $A$ on $\operatorname{id}_A$, and naturality then shows that every element of $F(X)$ is uniquely of the form $F(f)(u)$. Conversely, a universal element defines a componentwise bijection $\operatorname{Hom}_{\mathcal C}(A,X)\to F(X)$, and functoriality of $F$ proves that these bijections assemble into a natural isomorphism. [/proofplan] [step:Extract a universal element from a representing natural isomorphism] Fix an object $A\in\mathcal C$, and define the covariant hom-functor \begin{align*} H_A:\mathcal C&\to \mathbf{Set}\\ X&\mapsto \operatorname{Hom}_{\mathcal C}(A,X). \end{align*} For a morphism $g:X\to Y$, the map $H_A(g):\operatorname{Hom}_{\mathcal C}(A,X)\to \operatorname{Hom}_{\mathcal C}(A,Y)$ is given by \begin{align*} H_A(g)(f)=g\circ f. \end{align*} This is set-valued because $\mathcal C$ is locally small. Suppose $\alpha:H_A\Rightarrow F$ is a natural isomorphism. Define \begin{align*} u:=\alpha_A(\operatorname{id}_A)\in F(A). \end{align*} Let $X\in\mathcal C$ be an object and let $f:A\to X$ be a morphism. Naturality of $\alpha$ with respect to $f:A\to X$ gives \begin{align*} F(f)\circ \alpha_A=\alpha_X\circ H_A(f). \end{align*} Evaluating this equality at $\operatorname{id}_A\in \operatorname{Hom}_{\mathcal C}(A,A)$ gives \begin{align*} F(f)(u) &=F(f)\bigl(\alpha_A(\operatorname{id}_A)\bigr)\\ &=\alpha_X\bigl(H_A(f)(\operatorname{id}_A)\bigr)\\ &=\alpha_X(f\circ \operatorname{id}_A)\\ &=\alpha_X(f). \end{align*} Since $\alpha_X:\operatorname{Hom}_{\mathcal C}(A,X)\to F(X)$ is a bijection, every $x\in F(X)$ has a unique morphism $f:A\to X$ such that $\alpha_X(f)=x$. The displayed identity is exactly $F(f)(u)=x$, so $u$ is a universal element with vertex $A$. [guided] We start from a representation and recover the element that should be universal. Fix an object $A\in\mathcal C$, and let \begin{align*} H_A:\mathcal C&\to \mathbf{Set}\\ X&\mapsto \operatorname{Hom}_{\mathcal C}(A,X) \end{align*} be the covariant hom-functor. For a morphism $g:X\to Y$, this functor acts by postcomposition: \begin{align*} H_A(g):\operatorname{Hom}_{\mathcal C}(A,X)&\to \operatorname{Hom}_{\mathcal C}(A,Y)\\ f&\mapsto g\circ f. \end{align*} The local smallness of $\mathcal C$ ensures that each $\operatorname{Hom}_{\mathcal C}(A,X)$ is a set, so $H_A$ really is a functor into $\mathbf{Set}$. Assume $\alpha:H_A\Rightarrow F$ is a natural isomorphism. The only element of $\operatorname{Hom}_{\mathcal C}(A,A)$ that is distinguished without making any choices is $\operatorname{id}_A$, so define \begin{align*} u:=\alpha_A(\operatorname{id}_A)\in F(A). \end{align*} We now prove that $u$ has the required universal property. Let $X\in\mathcal C$ and let $f:A\to X$ be a morphism. Naturality of $\alpha$ for the morphism $f:A\to X$ says that the square comparing $H_A(f)$ and $F(f)$ commutes, which is the equality \begin{align*} F(f)\circ \alpha_A=\alpha_X\circ H_A(f). \end{align*} Evaluating both sides at $\operatorname{id}_A$ gives \begin{align*} F(f)(u) &=F(f)\bigl(\alpha_A(\operatorname{id}_A)\bigr)\\ &=\alpha_X\bigl(H_A(f)(\operatorname{id}_A)\bigr)\\ &=\alpha_X(f\circ \operatorname{id}_A)\\ &=\alpha_X(f). \end{align*} Thus the element obtained from a morphism $f:A\to X$ by applying the universal-element formula $F(f)(u)$ is exactly the element obtained from the representing bijection $\alpha_X(f)$. Since $\alpha$ is a natural isomorphism, its component \begin{align*} \alpha_X:\operatorname{Hom}_{\mathcal C}(A,X)\to F(X) \end{align*} is a bijection. Therefore, for every $x\in F(X)$, there exists a unique morphism $f:A\to X$ such that $\alpha_X(f)=x$. By the identity just proved, this condition is equivalent to $F(f)(u)=x$. Hence $u$ is a universal element with vertex $A$. [/guided] [/step] [step:Build a representing natural isomorphism from a universal element] Conversely, suppose $A\in\mathcal C$ and $u\in F(A)$ have the stated universal property. For each object $X\in\mathcal C$, define a function \begin{align*} \alpha_X:\operatorname{Hom}_{\mathcal C}(A,X)&\to F(X)\\ f&\mapsto F(f)(u). \end{align*} By the universal property of $u$, for every $x\in F(X)$ there exists a unique morphism $f:A\to X$ with $\alpha_X(f)=x$. Hence each $\alpha_X$ is a bijection. It remains to prove naturality. Let $g:X\to Y$ be a morphism in $\mathcal C$, and let $f:A\to X$ be an element of $\operatorname{Hom}_{\mathcal C}(A,X)$. By functoriality of $F$, \begin{align*} F(g)\bigl(\alpha_X(f)\bigr) &=F(g)\bigl(F(f)(u)\bigr)\\ &=(F(g)\circ F(f))(u)\\ &=F(g\circ f)(u)\\ &=\alpha_Y(g\circ f)\\ &=\alpha_Y\bigl(H_A(g)(f)\bigr). \end{align*} Thus $F(g)\circ \alpha_X=\alpha_Y\circ H_A(g)$ for every morphism $g:X\to Y$, so $\alpha:H_A\Rightarrow F$ is natural. Since every component $\alpha_X$ is a bijection, $\alpha$ is a natural isomorphism. Therefore $F$ is represented by $A$. [guided] Now we reverse the construction. Suppose $A\in\mathcal C$ and $u\in F(A)$ satisfy the universal property: for each object $X\in\mathcal C$ and each element $x\in F(X)$, there is a unique morphism $f:A\to X$ such that $F(f)(u)=x$. For each object $X\in\mathcal C$, define \begin{align*} \alpha_X:\operatorname{Hom}_{\mathcal C}(A,X)&\to F(X)\\ f&\mapsto F(f)(u). \end{align*} This is a well-defined function because $f:A\to X$ is a morphism, so functoriality gives a function $F(f):F(A)\to F(X)$, and therefore $F(f)(u)\in F(X)$. The universal property of $u$ says precisely that this function is bijective. Surjectivity follows because every $x\in F(X)$ is equal to $F(f)(u)$ for at least one morphism $f:A\to X$. Injectivity follows because the morphism producing a given element $x$ is unique. Hence each component \begin{align*} \alpha_X:\operatorname{Hom}_{\mathcal C}(A,X)\to F(X) \end{align*} is a bijection. We must still show that these bijections are compatible with morphisms in $\mathcal C$. Let $g:X\to Y$ be a morphism, and let $f:A\to X$ be a morphism. The naturality condition requires \begin{align*} F(g)\bigl(\alpha_X(f)\bigr)=\alpha_Y\bigl(H_A(g)(f)\bigr). \end{align*} Using the definition of $\alpha_X$, then functoriality of $F$, we compute \begin{align*} F(g)\bigl(\alpha_X(f)\bigr) &=F(g)\bigl(F(f)(u)\bigr)\\ &=(F(g)\circ F(f))(u)\\ &=F(g\circ f)(u). \end{align*} On the other hand, by definition of the hom-functor $H_A$, we have $H_A(g)(f)=g\circ f$, and therefore \begin{align*} \alpha_Y\bigl(H_A(g)(f)\bigr) &=\alpha_Y(g\circ f)\\ &=F(g\circ f)(u). \end{align*} The two expressions are equal, so the naturality square commutes for $g$. Since $g$ was arbitrary, $\alpha:H_A\Rightarrow F$ is a natural transformation. Since each $\alpha_X$ is a bijection, $\alpha$ is a natural isomorphism. Thus $F$ is representable. [/guided] [/step] [step:Verify that the two constructions are inverse to each other] Starting with a natural isomorphism $\alpha:H_A\Rightarrow F$, the first construction gives $u=\alpha_A(\operatorname{id}_A)$. The second construction applied to this $u$ gives maps \begin{align*} \beta_X:\operatorname{Hom}_{\mathcal C}(A,X)&\to F(X)\\ f&\mapsto F(f)(u). \end{align*} From the computation in the first step, $\beta_X(f)=\alpha_X(f)$ for every object $X$ and every morphism $f:A\to X$. Hence $\beta=\alpha$. Conversely, starting with a universal element $u\in F(A)$, the second construction gives $\alpha_X(f)=F(f)(u)$. The first construction then recovers \begin{align*} \alpha_A(\operatorname{id}_A) &=F(\operatorname{id}_A)(u)\\ &=\operatorname{id}_{F(A)}(u)\\ &=u, \end{align*} where $F(\operatorname{id}_A)=\operatorname{id}_{F(A)}$ by functoriality. Thus the two constructions are inverse bijections between representations by $A$ and universal elements of $F$ with vertex $A$. Consequently, $F$ is representable if and only if it has a universal element. [/step]