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. $\square$
Comments (0)
There are also: