Lemma 76.14.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:
$f$ is formally unramified, and
$\Omega _{X/Y} = 0$.
Lemma 76.14.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:
$f$ is formally unramified, and
$\Omega _{X/Y} = 0$.
Proof. This is a combination of Lemma 76.14.2, More on Morphisms, Lemma 37.6.7, and Lemma 76.7.3. $\square$
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