The Stacks project

Lemma 69.12.6. Let $S$ be a scheme. Let $X$ be a locally Noetherian algebraic space over $S$. Let $\mathcal{F}$, $\mathcal{G}$ be coherent $\mathcal{O}_ X$-modules. Let $\varphi : \mathcal{G} \to \mathcal{F}$ be a homomorphism of $\mathcal{O}_ X$-modules. Let $\overline{x}$ be a geometric point of $X$ lying over $x \in |X|$.

  1. If $\mathcal{F}_{\overline{x}} = 0$ then there exists an open neighbourhood $X' \subset X$ of $x$ such that $\mathcal{F}|_{X'} = 0$.

  2. If $\varphi _{\overline{x}} : \mathcal{G}_{\overline{x}} \to \mathcal{F}_{\overline{x}}$ is injective, then there exists an open neighbourhood $X' \subset X$ of $x$ such that $\varphi |_{X'}$ is injective.

  3. If $\varphi _{\overline{x}} : \mathcal{G}_{\overline{x}} \to \mathcal{F}_{\overline{x}}$ is surjective, then there exists an open neighbourhood $X' \subset X$ of $x$ such that $\varphi |_{X'}$ is surjective.

  4. If $\varphi _{\overline{x}} : \mathcal{G}_{\overline{x}} \to \mathcal{F}_{\overline{x}}$ is bijective, then there exists an open neighbourhood $X' \subset X$ of $x$ such that $\varphi |_{X'}$ is an isomorphism.

Proof. Let $\varphi : U \to X$ be an étale morphism where $U$ is a scheme and let $u \in U$ be a point mapping to $x$. By Properties of Spaces, Lemmas 66.29.4 and 66.22.1 as well as More on Algebra, Lemma 15.45.1 we see that $\varphi _{\overline{x}}$ is injective, surjective, or bijective if and only if $\varphi _ u : \varphi ^*\mathcal{F}_ u \to \varphi ^*\mathcal{G}_ u$ has the corresponding property. Thus we can apply the schemes version of this lemma to see that (after possibly shrinking $U$) the map $\varphi ^*\mathcal{F} \to \varphi ^*\mathcal{G}$ is injective, surjective, or an isomorphism. Let $X' \subset X$ be the open subspace corresponding to $|\varphi |(|U|) \subset |X|$, see Properties of Spaces, Lemma 66.4.8. Since $\{ U \to X'\} $ is a covering for the étale topology, we conclude that $\varphi |_{X'}$ is injective, surjective, or an isomorphism as desired. Finally, observe that (1) follows from (2) by looking at the map $\mathcal{F} \to 0$. $\square$

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