Processing math: 0%

The Stacks project

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)


Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.