Lemma 29.7.10 (01RH)—The Stacks project

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Table of contents
Part 2: Schemes
Chapter 29: Morphisms of Schemes
Section 29.7: Scheme theoretic closure and density
Lemma 29.7.10 (cite)

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Lemma 29.7.10. Let $S$ be a scheme. Let $X$ , $Y$ be schemes over $S$ . Let $f, g : X \to Y$ be morphisms of schemes over $S$ . Let $U \subset X$ be an open subscheme such that $f|_ U = g|_ U$ . If the scheme theoretic closure of $U$ in $X$ is $X$ and $Y \to S$ is separated, then $f = g$ .

Proof. Follows from the definitions and Schemes, Lemma 26.21.5. $\square$

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