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

1. $a'$ is a morphism over $S$,

2. $a'|_ U = a|_ U$, and

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

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. $\square$

## Post a comment

Your email address will not be published.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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 04FH. Beware of the difference between the letter 'O' and the digit '0'.