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 $f_ i$. Hence clearly $f$ is continuous!

Comment #2806 by Luke on

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

Comment #3428 by on

"...agrees with the continuous map $U_i$." should be "...agrees with the continuous map $f_i$."

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