Lemma 67.48.8 (0824)—The Stacks project

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Part 4: Algebraic Spaces
Chapter 67: Morphisms of Algebraic Spaces
Section 67.48: Relative normalization of algebraic spaces
Lemma 67.48.8 (cite)

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Lemma 67.48.8. Let $S$ be a scheme. Let $f : Y \to X$ be a quasi-compact and quasi-separated morphism of algebraic spaces over $S$ . Suppose that $Y = Y_1 \amalg Y_2$ is a disjoint union of two algebraic spaces. Write $f_ i = f|_{Y_ i}$ . Let $X_ i'$ be the normalization of $X$ in $Y_ i$ . Then $X_1' \amalg X_2'$ is the normalization of $X$ in $Y$ .

Proof. Omitted. $\square$

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