Lemma 67.37.8 (06CP)—The Stacks project

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Table of contents
Part 4: Algebraic Spaces
Chapter 67: Morphisms of Algebraic Spaces
Section 67.37: Smooth morphisms
Lemma 67.37.8 (cite)

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Lemma 67.37.8. A smooth morphism of algebraic spaces is syntomic.

Proof. Let $X \to Y$ be a smooth morphism of algebraic spaces. By definition this means there exists a diagram as in Lemma 67.22.1 with $h$ smooth and surjective vertical arrow $a$ . By Morphisms, Lemma 29.34.7 $h$ is syntomic. Hence $X \to Y$ is syntomic by definition. $\square$

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