Lemma 76.17.4 (061A)—The Stacks project

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Table of contents
Part 4: Algebraic Spaces
Chapter 76: More on Morphisms of Spaces
Section 76.17: Infinitesimal deformations of maps
Lemma 76.17.4 (cite)

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Lemma 76.17.4. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$ . Let $X \subset X'$ and $Y \subset Y'$ be first order thickenings over $B$ . Assume given a morphism $a : X \to Y$ and a map $A : a^*\mathcal{C}_{Y/Y'} \to \mathcal{C}_{X/X'}$ of $\mathcal{O}_ X$ -modules. For an object $U'$ of $(X')_{spaces, {\acute{e}tale}}$ with $U = X \times _{X'} U'$ consider morphisms $a' : U' \to Y'$ such that

$a'$ is a morphism over $B$ ,
$a'|_ U = a|_ U$ , and
the induced map $a^*\mathcal{C}_{Y/Y'}|_ U \to \mathcal{C}_{X/X'}|_ U$ is the restriction of $A$ to $U$ .

Then the rule

76.17.4.1

$\begin{equation} \label{spaces-more-morphisms-equation-sheaf} U' \mapsto \{ a' : U' \to Y'\text{ such that (1), (2), (3) hold.}\} \end{equation}$

defines a sheaf of sets on $(X')_{spaces, {\acute{e}tale}}$ .

Proof. Denote $\mathcal{F}$ the rule of the lemma. The restriction mapping $\mathcal{F}(U') \to \mathcal{F}(V')$ for $V' \subset U' \subset X'$ of $\mathcal{F}$ is really the restriction map $a' \mapsto a'|_{V'}$ . With this definition in place it is clear that $\mathcal{F}$ is a sheaf since morphisms of algebraic spaces satisfy étale descent, see Descent on Spaces, Lemma 74.7.2. $\square$

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