Lemma 17.16.5. Let $(X, \mathcal{O}_ X)$ be a ringed space. For any $\mathcal{O}_ X$-module $\mathcal{F}$ the functor

$\textit{Mod}(\mathcal{O}_ X) \longrightarrow \textit{Mod}(\mathcal{O}_ X) , \quad \mathcal{G} \longmapsto \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{G}$

commutes with arbitrary colimits.

Proof. Let $I$ be a preordered set and let $\{ \mathcal{G}_ i\}$ be a system over $I$. Set $\mathcal{G} = \mathop{\mathrm{colim}}\nolimits _ i \mathcal{G}_ i$. Recall that $\mathcal{G}$ is the sheaf associated to the presheaf $\mathcal{G}' : U \mapsto \mathop{\mathrm{colim}}\nolimits _ i \mathcal{G}_ i(U)$, see Sheaves, Section 6.29. By Lemma 17.16.2 the tensor product $\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{G}$ is the sheafification of the presheaf

$U \longmapsto \mathcal{F}(U) \otimes _{\mathcal{O}_ X(U)} \mathop{\mathrm{colim}}\nolimits _ i \mathcal{G}_ i(U) = \mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}(U) \otimes _{\mathcal{O}_ X(U)} \mathcal{G}_ i(U)$

where the equality sign is Algebra, Lemma 10.12.9. Hence the lemma follows from the description of colimits in $\textit{Mod}(\mathcal{O}_ X)$, see Lemma 17.3.2. $\square$

Comment #6216 by Yuto Masamura on

Typo: in the equation block of the statement, "$\mathcal F\otimes_{\mathcal O}\mathcal G$" should be "$\mathcal F\otimes_{\mathcal O_X}\mathcal G$".

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