Gluing lemma for open subsets
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.
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
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 .