A question of isomorphism of sheaves

Question

How do I verify the following: If $F$ is a sheaf of abelian groups on $X$, then show that $\operatorname{Hom}_{AbX}(Z,F)$ is isomorphic to $F$ (an isomorphism of sheaves of abelian groups).

Answer 1

Surely, this follows from the existence of a natural isomorphism $\text{Hom}_{\textbf{Ab}}(\mathbb{Z},A)\cong A$ for abelian groups $A$ and the definition of "sheaf morphism" ...

Answer 2

1. Step: Verify that $\mathsf{Ab} \to \mathsf{PSh}(X)$, $A \mapsto (U \mapsto A)$ (constant presheaf) is left adjoint to the global section functor.

2. Step: Conclude that $\mathsf{Ab} \to \mathsf{Sh}(X)$, $A \mapsto \underline{A}$ (constant sheaf) is left adjoint to the global section functor.

3. Step: Recall that $\mathbb{Z} \in \mathsf{Ab}$ represents the forgetful functor $\mathsf{Ab} \to \mathsf{Set}$.

All these steps are quite trivial. But now the claim follows easily:

The sheaf $\underline{\hom}(\underline{\mathbb{Z}},F)$ is given by $U \mapsto \hom(\underline{\mathbb{Z}}|_U,F|_U)$. But we have $\hom(\underline{\mathbb{Z}}|_U,F|_U) \stackrel{2.}{\cong} \hom(\mathbb{Z},\Gamma(U,F)) \stackrel{3.}{\cong} \Gamma(U,F)$.