Lemma 66.16.7. Let $S$ be a scheme. Let $f : X \to Y$ be a separated morphism of algebraic spaces over $S$. Let $V \subset Y$ be an open subspace such that $V \to Y$ is quasi-compact. Let $s : V \to X$ be a morphism such that $f \circ s = \text{id}_ V$. Let $Y'$ be the scheme theoretic image of $s$. Then $Y' \to Y$ is an isomorphism over $V$.

Proof. By Lemma 66.8.9 the morphism $s : V \to X$ is quasi-compact. Hence the construction of the scheme theoretic image $Y'$ of $s$ commutes with restriction to opens by Lemma 66.16.3. In particular, we see that $Y' \cap f^{-1}(V)$ is the scheme theoretic image of a section of the separated morphism $f^{-1}(V) \to V$. Since a section of a separated morphism is a closed immersion (Lemma 66.4.7), we conclude that $Y' \cap f^{-1}(V) \to V$ is an isomorphism as desired. $\square$

Comment #2306 by Luca on

A typo: I think it should be $s\colon V\rightarrow X$ in the first line of the proof.

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