# Gluing lemma for open subsets

## Statement

Let $\{U_i\}_{i \in I}$ be a collection of open subsets of a topological space $X$, and $f_i:U_i \to Y$ be continuous maps, such that for $x \in U_i \cap U_j$ we have $f_i(x) = f_j(x)$.

Let $U$ be the union of the $U_i$s. Then there exists a unique map $f:U \to Y$ such that $f|_{U_i} = f_i$.

This is the proof that the presheaf of continuous functions to $Y$, is actually a sheaf.

## Proof

The key facts used in the proof are:

• A map of topological spaces is continuous iff the inverse image of any open set is open
• An open subset of an open subset is open in the whole space
• An arbitrary union of open subsets is open

### Proof details

Given: An open cover $\{ U_i \}_{i \in I}$ of a topological space $X$. Continuous maps $f_i:U_i \to Y$, such that for $x \in U_i \cap U_j$, we have $f_i(x) = f_j(x)$. $U$ is the union of the $U_i$s.

To prove: There exists a unique map $f:U \to Y$ such that $f|_{U_i} = f_i$.

Proof: Note first that the $U_i$s are all open in $X$, hence also in $U$.

1. There exists a unique function $f$ on $U$ such that $f|_{U_i} = f_i$ for all $i$: For any $x \in X$, pick any $i$ such that $x \in U_i$, and define $f(x) = f_i(x)$. Such an $i$ exists because $U$ is the union of the $U_i$s. Further, the definition of $f(x)$ is independent of the choice of $i$ because if $x \in U_i \cap U_j$, $f_i(x) = f_j(x)$. Moreover, this is the only possible way to define $f$.
2. $f$ is continuous, i.e., if $V$ is an open subset of $Y$, $f^{-1}(V)$ is an open subset of $U$: If $f(x) \in V$, then $f_i(x) \in V$ for some $i$. Thus, we have $f^{-1}(V) = \bigcup_i f_i^{-1}(V)$. Since $f_i:U_i \to Y$ is continuous, $f_i^{-1}(V)$ is open in $U_i$. Since open subsets of open subsets are open, and $U_i$ is open in $U$, $f_i^{-1}(V)$ is open in $U$. Thus, the union $f^{-1}(V)$ of all the $f_i^{-1}(V)$ is also an open subset of $U$.