Lemma 31.5.6. Let $X$ be a scheme. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. If $U \subset X$ is open and $\text{WeakAss}(\mathcal{F}) \subset U$, then $\Gamma (X, \mathcal{F}) \to \Gamma (U, \mathcal{F})$ is injective.

Proof. Let $s \in \Gamma (X, \mathcal{F})$ be a section which restricts to zero on $U$. Let $\mathcal{F}' \subset \mathcal{F}$ be the image of the map $\mathcal{O}_ X \to \mathcal{F}$ defined by $s$. Then $\text{Supp}(\mathcal{F}') \cap U = \emptyset$. On the other hand, $\text{WeakAss}(\mathcal{F}') \subset \text{WeakAss}(\mathcal{F})$ by Lemma 31.5.4. Since also $\text{WeakAss}(\mathcal{F}') \subset \text{Supp}(\mathcal{F}')$ (Lemma 31.5.3) we conclude $\text{WeakAss}(\mathcal{F}') = \emptyset$. Hence $\mathcal{F}' = 0$ by Lemma 31.5.5. $\square$

Comment #7209 by Matthieu Romagny on

I guess that the phrase "If WeakAss(F)⊂U⊂X is open" doesn't mean that one assumes WeakAss(F) to be open, right? The wording is a little misleading.

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