Lemma 76.15.12. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $h : Z \to X$ be a formally unramified morphism of algebraic spaces over $B$. Let $Z \subset Z'$ be the universal first order thickening of $Z$ over $X$ with structure morphism $h' : Z' \to X$. The canonical map

$\text{d}h' : (h')^*\Omega _{X/B} \to \Omega _{Z'/B}$

induces an isomorphism $h^*\Omega _{X/B} \to \Omega _{Z'/B} \otimes \mathcal{O}_ Z$.

Proof. The map $c_{h'}$ is the map defined in Lemma 76.7.6. If $i : Z \to Z'$ is the given closed immersion, then $i^*c_{h'}$ is a map $h^*\Omega _{X/S} \to \Omega _{Z'/S} \otimes \mathcal{O}_ Z$. Checking that it is an isomorphism reduces to the case of schemes by étale localization, see Lemma 76.15.11 and Lemma 76.7.3. In this case the result is More on Morphisms, Lemma 37.7.9. $\square$

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