Lemma 6.33.1 (04TN)—The Stacks project

Lemma 6.33.1. Let $X$ be a topological space. Let $X = \bigcup U_ i$ be an open covering. Let $\mathcal{F}$, $\mathcal{G}$ be sheaves of sets on $X$. Given a collection

\[ \varphi _ i : \mathcal{F}|_{U_ i} \longrightarrow \mathcal{G}|_{U_ i} \]

of maps of sheaves such that for all $i, j \in I$ the maps $\varphi _ i, \varphi _ j$ restrict to the same map $\mathcal{F}|_{U_ i \cap U_ j} \to \mathcal{G}|_{U_ i \cap U_ j}$ then there exists a unique map of sheaves

\[ \varphi : \mathcal{F} \longrightarrow \mathcal{G} \]

whose restriction to each $U_ i$ agrees with $\varphi _ i$.

Proof. For each open subset $U \subset X$ define

\[ \varphi _ U : \mathcal{F}(U) \to \mathcal{G}(U), \quad s \mapsto \varphi _ U(s) \]

where $\varphi _ U(s)$ is the unique section verifying

\[ (\varphi _ U(s))|_{U \cap U_ i} = (\varphi _ i)_{U \cap U_ i}(s|_{U \cap U_ i}). \]

Existence and uniqueness of such a section follows from the sheaf axioms due to the fact that

\begin{align*} ((\varphi _ i)_{U \cap U_ i}(s|_{U \cap U_ i}))|_{U \cap U_ i \cap U_ j} & = (\varphi _ i)_{U \cap U_ i \cap U_ j}(s|_{U \cap U_ i \cap U_ j})\\ & = (\varphi _ j)_{U \cap U_ i \cap U_ j}(s|_{U \cap U_ i \cap U_ j})\\ & = ((\varphi _ j)_{U \cap U_ j}(s|_{U \cap U_ j}))|_{U \cap U_ i \cap U_ j}. \end{align*}

This family of maps gives us indeed a map of sheaves: Let $V \subset U \subset X$ be open subsets then

\[ (\varphi _ U(s))|_ V = \varphi _ V(s|_ V) \]

since for each $i \in I$ the following holds

\begin{align*} (\varphi _ U(s))|_{V \cap U_ i} & = ((\varphi _ U(s))|_{U \cap U_ i})|_{V \cap U_ i}\\ & = ((\varphi _ i)_{U \cap U_ i}(s|_{U \cap U_ i}))|_{V \cap U_ i}\\ & = (\varphi _ i)_{V \cap U_ i}(s|_{V \cap U_ i})\\ & = \varphi _ V(s_{V})|_{V \cap U_ i}. \end{align*}

Furthermore, its restriction to each $U_ i$ agrees with $\varphi _ i$ since given $U \subset X$ open subset and $s \in \mathcal{F}(U \cap U_ i)$ then

\begin{align*} \varphi _{U \cap U_ i}(s) & = \varphi _{U \cap U_ i}(s)|_{U \cap U_ i}\\ & = (\varphi _ i)_{U \cap U_ i}(s|_{U \cap U_ i})\\ & = (\varphi _ i)_{U \cap U_ i}(s). \end{align*}

$\square$

Comments (3)

Comment #7470 by Elías Guisado on July 09, 2022 at 10:58

I think it is interesting to remark that the proof only uses the sheaf condition for $\mathcal{G}$ , and never the fact that $\mathcal{F}$ is a sheaf.

Maybe it will be nice to write in the statement “let $\mathcal{F}$ be a presheaf of sets and let $\mathcal{G}$ be a sheaf of sets” to generalize it.

In other words, the hom presheaf $\mathcal{H}om(\mathcal{F},\mathcal{G})$ is a sheaf provided that $\mathcal{G}$ is a sheaf.

Comment #7471 by Laurent Moret-Bailly on July 10, 2022 at 03:04

@#7470: I agree with this comment, but let me also observe that the more general statement (with $\mathcal{F}$ only a presheaf) immediately follows from the sheaf case since if $\mathcal{F}$ is a presheaf and $\mathcal{G}$ is a sheaf we have $\mathcal{Hom}( \mathcal{F},\mathcal{G})=\mathcal{Hom}(\mathcal{F}',\mathcal{G})$ where $\mathcal{F}'$ is the sheafification of $\mathcal{F}$ (the # sign wouldn't print in math mode).

Comment #7620 by Stacks Project on August 14, 2022 at 10:20

OK, I am going to leave this as is.

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7 comment(s) on Section 6.33: Glueing sheaves

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