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
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