# The Stacks Project

## Tag 006V

Example 6.7.3. Let $X$, $Y$ be topological spaces. Consider the rule $\mathcal{F}$ which associates to the open $U \subset X$ the set $$\mathcal{F}(U) = \{ f : U \to Y \mid f \text{ is continuous}\}$$ with the obvious restriction mappings. We claim that $\mathcal{F}$ is a sheaf. To see this suppose that $U = \bigcup_{i\in I} U_i$ is an open covering, and $f_i \in \mathcal{F}(U_i)$, $i\in I$ with $f_i |_{U_i \cap U_j} = f_j|_{U_i \cap U_j}$ for all $i, j \in I$. In this case define $f : U \to Y$ by setting $f(u)$ equal to the value of $f_i(u)$ for any $i \in I$ such that $u \in U_i$. This is well defined by assumption. Moreover, $f : U \to Y$ is a map such that its restriction to $U_i$ agrees with the continuous map $U_i$. Hence clearly $f$ is continuous!

The code snippet corresponding to this tag is a part of the file sheaves.tex and is located in lines 553–572 (see updates for more information).

\begin{example}
\label{example-basic-continuous-maps}
Let $X$, $Y$ be topological spaces.
Consider the rule $\mathcal{F}$ which associates to
the open $U \subset X$ the set
$$\mathcal{F}(U) = \{ f : U \to Y \mid f \text{ is continuous}\}$$
with the obvious restriction mappings. We claim that
$\mathcal{F}$ is a sheaf. To see this suppose that
$U = \bigcup_{i\in I} U_i$ is an open covering, and
$f_i \in \mathcal{F}(U_i)$, $i\in I$ with
$f_i |_{U_i \cap U_j} = f_j|_{U_i \cap U_j}$ for all $i, j \in I$.
In this case define $f : U \to Y$ by setting $f(u)$
equal to the value of $f_i(u)$ for any $i \in I$ such that
$u \in U_i$. This is well defined by assumption. Moreover,
$f : U \to Y$ is a map such that its restriction to $U_i$
agrees with the continuous map $U_i$. Hence clearly $f$ is
continuous!
\end{example}

Comment #2806 by Luke on September 14, 2017 a 7:45 am UTC

Simple typo of the word "which" in the first line.

Comment #2907 by Johan (site) on October 7, 2017 a 3:48 pm UTC

THanks, fixed here.

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