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I read the following in "Sheaves on Manifolds by Kashiwara" on page 87

The relevant definition is

Let $\mathcal{R}$ be a sheaf of rings on $X$. An $\mathcal{R}$-module $M$ (or a sheaf of modules over $\mathcal{R}$ is a sheaf $M$ such that for each open set $U\subset X$, $M(U)$ is a left $\mathcal{R}(U)$ module, and for any $V\subset U$, the restriction morphism is compatible with the structure of module, that is $\rho_{V,U}(sm)=\rho_{V,U}(s)\cdot \rho_{V,U}(m)$ for any $s\in \mathcal{R}(U), m\in M(U)$. One denotes by $Mod(\mathcal{R})$ the category of left $\mathcal{R}$ modules.

The example they gave was the following. Let $Sh(X)$ be the category of sheaves with values in the category of abelian groups. Let $\mathbb{Z}_{X}$ be the sheaf associated to the presheaf $U\mapsto \mathbb{Z}$. Then we apparently have

$$Sh(X)=Mod(\mathbb{Z}_{X})$$

I don't understand how they got the last equality. $\mathbb{Z}_{X}$, as a sheaf, can be described as follows. for $U\subset X$ we have that $\mathbb{Z}_{X}(U)$ is the ring of locally constant functions from $U$ to $\mathbb{Z}$. Consider $\mathcal{F}\in Sh(X)$, then $\mathcal{F}(U)$ should be a $\mathbb{Z}_{X}(U)$ module. Let $f\in \mathbb{Z}_{X}(U)$ and let $s\in \mathcal{F}(U)$, then how does $f$ act on $s$? I can't see any obvious way to have $f$ act on $s$

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  • $\begingroup$ You're absolutely right about $U$ instead of $X$, I have changed it. I'm not sure If I understand your explicit construction. $f$ is a function on $U$, not $\mathcal{F}(U)$, or are you using some sort of abuse of notation? $\endgroup$ – Damo Sep 29 '19 at 3:30
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    $\begingroup$ Yes, you're absolutely right - sorry about that, my comment was idiotic. If I have not made another dumb mistake, I think the construction is this, which is unfortunately less explicit: let $f \in \mathbb{Z}_{X}(U)$. Since $f$ is locally constant, the collection of sets $\{U_{i} := f^{-1}(i)\}_{i \in \mathbb{Z}}$ is a disjoint open cover of $U$. In particular, given $s \in \mathcal{F}(U)$, we can define the collection $\{s_{i} := i \cdot s\vert_{U_{i}}\}_{i \in \mathbb{Z}}$, which is trivially coherent because the intersections $U_{i} \cap U_{j}$ are empty for $i \neq j$... $\endgroup$ – Alex Wertheim Sep 29 '19 at 3:47
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    $\begingroup$ ...By gluing these sections together, we thus obtain an element of $\mathcal{F}(U)$ which we call $f \cdot s$. $\endgroup$ – Alex Wertheim Sep 29 '19 at 3:48
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The key facts here are:

  1. Every abelian group is a $\mathbb Z$-module.
  2. An $\mathcal R$-module $\mathcal F$ is a sheaf of abelian groups equipped with a morphism $\mathcal R \to \operatorname{End}_{\text{Sh}(X)}(\mathcal F)$.

Let $\mathbb Z_X^-$ be the presheaf $U \mapsto \mathbb Z$ and let $\mathbb Z_X$ be its sheafification. Let $\mathcal F$ be a sheaf of abelian groups over $X$. By (1), $\mathcal F$ is a $\mathbb Z_X^-$-module. The corresponding morphism $\mathbb Z_X^- \to \operatorname{End}_{\text{Sh}(X)}(\mathcal F)$ gives rise to a compatible morphism $\mathbb Z_X \to \operatorname{End}_{\text{Sh}(X)}(\mathcal F)$ by the universal property of sheafification. By (2), this morphism gives $\mathcal F$ its $\mathbb Z_X$-module structure.

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