Lemma 70.9.8. Let $S$ be a scheme. Let $X$ be a quasi-compact and quasi-separated algebraic space over $S$. Let $U \subset X$ be a quasi-compact open. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Let $\mathcal{G} \subset \mathcal{F}|_ U$ be a quasi-coherent $\mathcal{O}_ U$-submodule which is of finite type. Then there exists a quasi-coherent submodule $\mathcal{G}' \subset \mathcal{F}$ which is of finite type such that $\mathcal{G}'|_ U = \mathcal{G}$.

Proof. Denote $j : U \to X$ the inclusion morphism. As $X$ is quasi-separated and $U$ quasi-compact, the morphism $j$ is quasi-compact. Hence $j_*\mathcal{G} \subset j_*\mathcal{F}|_ U$ are quasi-coherent modules on $X$ (Morphisms of Spaces, Lemma 67.11.2). Let $\mathcal{H} = \mathop{\mathrm{Ker}}(j_*\mathcal{G} \oplus \mathcal{F} \to j_*\mathcal{F}|_ U)$. Then $\mathcal{H}|_ U = \mathcal{G}$. By Lemma 70.9.2 we can find a finite type quasi-coherent submodule $\mathcal{H}' \subset \mathcal{H}$ such that $\mathcal{H}'|_ U = \mathcal{H}|_ U = \mathcal{G}$. Set $\mathcal{G}' = \mathop{\mathrm{Im}}(\mathcal{H}' \to \mathcal{F})$ to conclude. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).