Lemma 67.45.7 (0415)—The Stacks project

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Table of contents
Part 4: Algebraic Spaces
Chapter 67: Morphisms of Algebraic Spaces
Section 67.45: Integral and finite morphisms
Lemma 67.45.7 (cite)

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Lemma 67.45.7. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ . The following are equivalent

$f$ is integral, and
$f$ is affine and universally closed.

Proof. In both cases the morphism is representable, and you can check the condition after a base change by an affine scheme mapping into $Y$ , see Lemmas 67.45.3, 67.20.3, and 67.9.5. Hence the result follows from Morphisms, Lemma 29.44.7. $\square$

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