Lemma 29.6.8. Let $f : X \to Y$ be a separated morphism of schemes. Let $V \subset Y$ be a retrocompact open. 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. The assumption that $V$ is retrocompact in $Y$ (Topology, Definition 5.12.1) means that $V \to Y$ is a quasi-compact morphism. By Schemes, Lemma 26.21.14 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 29.6.3. 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 (Schemes, Lemma 26.21.11), we conclude that $Y' \cap f^{-1}(V) \to V$ is an isomorphism as desired. $\square$

Comment #4726 by Zhaodong Cai on

In the proof $s:V\to Y$ should be $s:V\to X$.

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