Lemma 37.9.4 (04FH)—The Stacks project

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Table of contents
Part 2: Schemes
Chapter 37: More on Morphisms
Section 37.9: Infinitesimal deformations of maps
Lemma 37.9.4 (cite)

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Lemma 37.9.4. Let $S$ be a scheme. Let $X \subset X'$ and $Y \subset Y'$ be first order thickenings over $S$. 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 open subscheme $U' \subset X'$ consider morphisms $a' : U' \to Y'$ such that

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

Here $U = X \cap U'$. Then the rule

37.9.4.1

\begin{equation} \label{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 $X'$.

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 are defined locally. $\square$

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