Lemma 66.37.8. 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 66.22.1 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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