# Gluing lemma for open subsets

From Topospaces

## Statement

Let be a collection of open subsets of a topological space , and be continuous maps, such that for we have .

Let be the union of the s. Then there exists a unique map such that .

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

## Related results

## 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 of a topological space . Continuous maps , such that for , we have . is the union of the s.

**To prove**: There exists a unique map such that .

**Proof**: Note first that the s are all open in , hence also in .

- There exists a unique function on such that for all : For any , pick any such that , and define . Such an exists because is the union of the s. Further, the definition of is independent of the choice of because if , . Moreover, this is the only possible way to define .
- is continuous, i.e., if is an open subset of , is an open subset of : If , then for some . Thus, we have . Since is continuous, is open in . Since open subsets of open subsets are open, and is open in , is open in . Thus, the union of all the is also an open subset of .