# Help in understanding a sheaf of modules over $\mathcal{R}$

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 $$s$$

• 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? – Damo Sep 29 '19 at 3:30
• 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 $i \neq j$... – Alex Wertheim Sep 29 '19 at 3:47
• ...By gluing these sections together, we thus obtain an element of $\mathcal{F}(U)$ which we call $f \cdot s$. – Alex Wertheim Sep 29 '19 at 3:48

## 1 Answer

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.