Proof.

Let $\phi : h_A \stackrel{\bullet}{\to} F$ be a natural transformation. Then for each object $X$ of $\mathcal{C}$, $\phi$ gives us an arrow $\phi_X : h_A(X) \to F(X)$ in the category $\mathbf{Set}$, i.e., a function $\phi_X: \text{Hom}(A,X) \to F(X)$, such that the diagram,

commutes for any arrow $f : X \to Y$ in $\mathcal{C}$. In particular, we have,

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commutes for any arrow $f : A \to X$ in $\mathcal{C}$. In particular, we have,

$$(\phi_X\circ h_A (f))(\text{Id}_A) = ((F(f))\circ \phi_A) ( \text{Id}_A) $$

But $h_A(f)$ is the map $\alpha \mapsto f \circ \alpha$ from $\text{Hom}(A,A)$ to $\text{Hom}(A,X)$. We thus have,$$\phi_X (f) = \phi_X (f \circ \text{Id}_A) = (F(f)) (\phi_A( \text{Id}_A))$$

Thus for every object $X$ of $\mathcal{C}$, the morphism $\phi_X$ is uniquely determined by the element $\phi_A (\text{Id}_A)$. Thus the map $\phi \mapsto \phi_A \text{Id}_A$ is an injection from $\text{Nat}(h_A, F)$ to $F(A)$. Here $\text{Nat}(h_A, F)$ denotes the set of natural transformations from $h_A$ to $F$.

It thus remains to be shown that for any $x \in F(A)$, the maps $$\phi_X : f \mapsto (F(f))(x)$$

for every object $X$ of $\mathcal{C}$ define a natural transformation. But this is true, as for any objects $X$ and $Y$ of $\mathcal{C}$, any morphism $f : X \to Y$ of $\mathcal{C}$, and any element $\alpha$ of $h_A(X) = \text{Hom}(A,X)$, we have,

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