This is Exercise 2.3.D. from Vakil's notes.
If $\mathscr{F}$ is a sheaf of abelian groups on $X$. Show that $\mathit{Hom}_{\mathit{Ab}_X}(\underline{\mathbb{Z}},\mathscr{F})\cong\mathscr{F}$.
Background:
I just showed $\mathit{Hom}(\underline{\{p\}},\mathscr{F})\cong \mathscr{F}$. I think this should be similar.
My Attempt:
$\mathit{Hom}_{\mathit{Ab}_X}(\underline{\mathbb{Z}},\mathscr{F})$ is the collection of data that for any open subset $U\subseteq X$, $\mathit{Hom}_{\mathit{Ab}_X}(\underline{\mathbb{Z}},\mathscr{F})(U)$ is the set of morphisms from $\underline{\mathbb{Z}}|_U$ to $\mathscr{F}|_U$. It suffices to check that $\mathit{Hom}_{\mathit{Ab}_X}(\underline{\mathbb{Z}},\mathscr{F})(U)\cong\mathscr{F}(U)$ for any $U$ as abelian groups. First note that $\mathit{Hom}_{\mathit{Ab}_X}(\underline{\mathbb{Z}},\mathscr{F})(U)$ is an abelian group. It is a set of morphisms $\phi: \underline{\mathbb{Z}}_U\rightarrow \mathscr{F}|_U$ that has is a collection of maps $\phi(V):\underline{\mathbb{Z}}|_U(V)\rightarrow \mathscr{F}|_U(V)$, for each $V\subseteq U$ behaving well with the restriction maps. If $\phi_1,\phi_2\in \mathit{Hom}_{\mathit{Ab}_X}(\underline{\mathbb{Z}},\mathscr{F})(U)$, their addition is defined as $\phi_1+\phi_2$, which is a collection of $(\phi_1+\phi_2)(V)$, such that it sends any $z\in \underline{\mathbb{Z}}|_U(V)$ to $\phi_1(z)+\phi_2(z)$, where $z$ is a locally constant map from $V$ to $\mathbb{Z}$.
My Question:
I don't know how to define a map $\sigma: \mathit{Hom}_{\mathit{Ab}_X}(\underline{\mathbb{Z}},\mathscr{F})(U)\rightarrow \mathscr{F}(U)$ which is one-to-one and onto.
For the case $\mathit{Hom}(\underline{\{p\}},\mathscr{F})\cong \mathscr{F}$, there is only one element in ${\{p\}}|_U(V)$. But here it is the integer set.
Any help will be appreciated!