[proofplan] We construct the inverse map explicitly. Starting from an element $u \in F(A)$, we define a natural transformation $\alpha_u: H_A \to F$ by sending a morphism $f: A \to X$ to $F(f)(u) \in F(X)$. Functoriality of $F$ proves naturality, and the two inverse identities follow from evaluating at $\operatorname{id}_A$ and from the naturality square for an arbitrary natural transformation. [/proofplan] [step:Construct a natural transformation from an element of $F(A)$] Let $u \in F(A)$. For each object $X \in \operatorname{Ob}(\mathcal C)$, define a function \begin{align*} (\alpha_u)_X: \operatorname{Hom}_{\mathcal C}(A,X) &\to F(X) \\ f &\mapsto F(f)(u). \end{align*} Here $f: A \to X$ is a morphism in $\mathcal C$, so $F(f): F(A) \to F(X)$ is a function because $F$ is a covariant functor. We verify that the family $\alpha_u = ((\alpha_u)_X)_{X \in \operatorname{Ob}(\mathcal C)}$ is natural. Let $g: X \to Y$ be a morphism in $\mathcal C$, and let $f \in \operatorname{Hom}_{\mathcal C}(A,X)$. The hom-functor $H_A$ sends $g$ to the function $H_A(g): \operatorname{Hom}_{\mathcal C}(A,X) \to \operatorname{Hom}_{\mathcal C}(A,Y)$ given by $H_A(g)(f)=g \circ f$. Therefore \begin{align*} F(g)\bigl((\alpha_u)_X(f)\bigr) &= F(g)\bigl(F(f)(u)\bigr) \\ &= (F(g) \circ F(f))(u) \\ &= F(g \circ f)(u) \\ &= (\alpha_u)_Y(g \circ f) \\ &= (\alpha_u)_Y\bigl(H_A(g)(f)\bigr). \end{align*} The third equality is functoriality of $F$. Hence \begin{align*} F(g) \circ (\alpha_u)_X = (\alpha_u)_Y \circ H_A(g), \end{align*} so $\alpha_u: H_A \to F$ is a natural transformation. Define \begin{align*} \Psi: F(A) &\to \operatorname{Nat}(H_A,F) \\ u &\mapsto \alpha_u. \end{align*} [guided] We want to reverse the evaluation map $\Theta$. If an element $u \in F(A)$ is supposed to determine a natural transformation $H_A \to F$, then its value on a morphism $f: A \to X$ is forced: since $F$ is covariant, $F(f): F(A) \to F(X)$ carries $u$ to an element of $F(X)$. Thus for every object $X$ we define \begin{align*} (\alpha_u)_X: \operatorname{Hom}_{\mathcal C}(A,X) &\to F(X) \\ f &\mapsto F(f)(u). \end{align*} Now we check naturality. Let $g: X \to Y$ be a morphism in $\mathcal C$. Naturality requires the equality of functions \begin{align*} F(g) \circ (\alpha_u)_X = (\alpha_u)_Y \circ H_A(g) \end{align*} from $\operatorname{Hom}_{\mathcal C}(A,X)$ to $F(Y)$. To prove equality of functions, evaluate both sides at an arbitrary morphism $f: A \to X$. The hom-functor acts by postcomposition, so $H_A(g)(f)=g \circ f$. Therefore \begin{align*} F(g)\bigl((\alpha_u)_X(f)\bigr) &= F(g)\bigl(F(f)(u)\bigr) \\ &= (F(g) \circ F(f))(u) \\ &= F(g \circ f)(u) \\ &= (\alpha_u)_Y(g \circ f) \\ &= (\alpha_u)_Y\bigl(H_A(g)(f)\bigr). \end{align*} The equality $F(g) \circ F(f)=F(g \circ f)$ is exactly covariant functoriality. Hence $\alpha_u$ is a natural transformation, and this construction defines a function \begin{align*} \Psi: F(A) &\to \operatorname{Nat}(H_A,F) \\ u &\mapsto \alpha_u. \end{align*} [/guided] [/step] [step:Show evaluation after construction recovers the original element] Let $u \in F(A)$. By definition of $\Theta$ and $\Psi$, \begin{align*} (\Theta \circ \Psi)(u) &= \Theta(\alpha_u) \\ &= (\alpha_u)_A(\operatorname{id}_A) \\ &= F(\operatorname{id}_A)(u) \\ &= \operatorname{id}_{F(A)}(u) \\ &= u. \end{align*} The fourth equality uses functoriality of $F$ on identity morphisms. Thus $\Theta \circ \Psi = \operatorname{id}_{F(A)}$. [/step] [step:Show construction after evaluation recovers the original natural transformation] Let $\alpha \in \operatorname{Nat}(H_A,F)$, and define $u := \Theta(\alpha)=\alpha_A(\operatorname{id}_A) \in F(A)$. We prove that $\Psi(u)=\alpha$. Let $X \in \operatorname{Ob}(\mathcal C)$ and let $f \in \operatorname{Hom}_{\mathcal C}(A,X)$. Naturality of $\alpha$ for the morphism $f: A \to X$ gives \begin{align*} F(f) \circ \alpha_A = \alpha_X \circ H_A(f). \end{align*} Evaluating both sides at $\operatorname{id}_A \in \operatorname{Hom}_{\mathcal C}(A,A)$ yields \begin{align*} 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 $u=\alpha_A(\operatorname{id}_A)$, the left-hand side is \begin{align*} F(f)(u)=(\alpha_u)_X(f). \end{align*} Therefore $(\alpha_u)_X(f)=\alpha_X(f)$ for every object $X$ and every morphism $f: A \to X$. Hence $\alpha_u=\alpha$, so $\Psi \circ \Theta=\operatorname{id}_{\operatorname{Nat}(H_A,F)}$. [guided] Let $\alpha: H_A \to F$ be an arbitrary natural transformation. The evaluation map extracts the element \begin{align*} u := \Theta(\alpha)=\alpha_A(\operatorname{id}_A) \in F(A). \end{align*} We must show that rebuilding a natural transformation from this $u$ gives back the original $\alpha$. Fix an object $X \in \operatorname{Ob}(\mathcal C)$ and a morphism $f: A \to X$. Naturality of $\alpha$ applied to the morphism $f: A \to X$ says that the following two functions from $\operatorname{Hom}_{\mathcal C}(A,A)$ to $F(X)$ are equal: \begin{align*} F(f) \circ \alpha_A = \alpha_X \circ H_A(f). \end{align*} Now evaluate this equality at the identity morphism $\operatorname{id}_A$. Since $H_A(f)$ acts by postcomposition, we have $H_A(f)(\operatorname{id}_A)=f \circ \operatorname{id}_A=f$. Therefore \begin{align*} 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*} Because $u=\alpha_A(\operatorname{id}_A)$, this says \begin{align*} (\alpha_u)_X(f)=F(f)(u)=\alpha_X(f). \end{align*} The object $X$ and morphism $f: A \to X$ were arbitrary, so every component function of $\alpha_u$ agrees with the corresponding component function of $\alpha$. Hence $\alpha_u=\alpha$, which is precisely $\Psi(\Theta(\alpha))=\alpha$. [/guided] [/step] [step:Conclude that evaluation is a bijection] The maps \begin{align*} \Theta: \operatorname{Nat}(H_A,F) &\to F(A), & \Psi: F(A) &\to \operatorname{Nat}(H_A,F) \end{align*} satisfy \begin{align*} \Theta \circ \Psi &= \operatorname{id}_{F(A)}, & \Psi \circ \Theta &= \operatorname{id}_{\operatorname{Nat}(H_A,F)}. \end{align*} Therefore $\Theta$ is bijective, with inverse $\Psi$. This proves that natural transformations $\operatorname{Hom}_{\mathcal C}(A,-) \to F$ are in bijection with elements of $F(A)$. [/step]