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.

Related results

• Gluing lemma for closed subsets

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 $U$.