Lemma 67.52.2 (0ABS)—The Stacks project

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Table of contents
Part 4: Algebraic Spaces
Chapter 67: Morphisms of Algebraic Spaces
Section 67.52: Zariski's Main Theorem (representable case)
Lemma 67.52.2 (cite)

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Lemma 67.52.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ . Assume $f$ is quasi-finite and separated. Let $Y'$ be the normalization of $Y$ in $X$ . Picture:

$\xymatrix{ X \ar[rd]_ f \ar[rr]_{f'} & & Y' \ar[ld]^\nu \\ & Y & }$

Then $f'$ is a quasi-compact open immersion and $\nu$ is integral. In particular $f$ is quasi-affine.

Proof. By Lemma 67.51.1 the morphism $f$ is representable. Hence we may apply Lemma 67.52.1. Thus there exists an open subspace $U' \subset Y'$ such that $(f')^{-1}(U') = X$ (!) and $X \to U'$ is an isomorphism! In other words, $f'$ is an open immersion. Note that $f'$ is quasi-compact as $f$ is quasi-compact and $\nu : Y' \to Y$ is separated (Lemma 67.8.9). Hence for every affine scheme $Z$ and morphism $Z \to Y$ the fibre product $Z \times _ Y X$ is a quasi-compact open subscheme of the affine scheme $Z \times _ Y Y'$ . Hence $f$ is quasi-affine by definition. $\square$

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