Lemma 6.30.12. Let $X$ be a topological space. Let $\mathcal{B}$ be a basis for the topology on $X$. Let $\mathcal{O}$ be a sheaf of rings on $\mathcal{B}$. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}$-modules on $\mathcal{B}$. Let $\mathcal{O}^{ext}$ be the sheaf of rings on $X$ extending $\mathcal{O}$ and let $\mathcal{F}^{ext}$ be the abelian sheaf on $X$ extending $\mathcal{F}$, see Lemma 6.30.9. There exists a canonical map

$\mathcal{O}^{ext} \times \mathcal{F}^{ext} \longrightarrow \mathcal{F}^{ext}$

which agrees with the given map over elements of $\mathcal{B}$ and which endows $\mathcal{F}^{ext}$ with the structure of an $\mathcal{O}^{ext}$-module.

Proof. It suffices to construct the multiplication map on the level of presheaves of sets. Perhaps the easiest way to see this is to prove directly that if $(f_ x)_{x \in U}$, $f_ x \in \mathcal{O}_ x$ and $(m_ x)_{x \in U}$, $m_ x \in \mathcal{F}_ x$ satisfy $(*)$, then the element $(f_ xm_ x)_{x \in U}$ also satisfies $(*)$. Then we get the desired result, because in the proof of Lemma 6.30.6 we construct the extension in terms of families of elements of stalks satisfying $(*)$. $\square$

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