Lemma 76.17.2. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $(a, a') : (X \subset X') \to (Y \subset Y')$ be a morphism of first order thickenings over $B$. Let

\[ \theta : a^*\Omega _{Y/B} \to \mathcal{C}_{X/X'} \]

be an $\mathcal{O}_ X$-linear map. Then there exists a unique morphism of pairs $(b, b') : (X \subset X') \to (Y \subset Y')$ such that (1) and (2) of Lemma 76.17.1 hold and the derivation $D$ and $\theta $ are related by Equation (76.17.1.1).

**Proof.**
Consider the map

\[ \alpha = (a')^\sharp + D : a^{-1}\mathcal{O}_{Y'} \to \mathcal{O}_{X'} \]

where $D$ is as in Equation (76.17.1.1). As $D$ is an $\mathcal{O}_ B$-derivation it follows that $\alpha $ is a map of sheaves of $\mathcal{O}_ B$-algebras. By construction we have $i_ X^\sharp \circ \alpha = a^\sharp \circ i_ Y^\sharp $ where $i_ X : X \to X'$ and $i_ Y : Y \to Y'$ are the given closed immersions. By Lemma 76.9.2 we obtain a unique morphism $(a, b') : (X \subset X') \to (Y \subset Y')$ of thickenings over $B$ such that $\alpha = (b')^\sharp $. Setting $b = a$ we win.
$\square$

## Comments (0)