Gluing lemma for open subsets

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Statement

Let {Ui}iI be a collection of open subsets of a topological space X, and fi:UiY be continuous maps, such that for xUiUj we have fi(x)=fj(x).

Let U be the union of the Uis. Then there exists a unique map f:UY such that f|Ui=fi.

This is the proof that the presheaf of continuous functions to Y, 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 {Ui}iI of a topological space X. Continuous maps fi:UiY, such that for xUiUj, we have fi(x)=fj(x). U is the union of the Uis.

To prove: There exists a unique map f:UY such that f|Ui=fi.

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

  1. There exists a unique function f on U such that f|Ui=fi for all i: For any xX, pick any i such that xUi, and define f(x)=fi(x). Such an i exists because U is the union of the Uis. Further, the definition of f(x) is independent of the choice of i because if xUiUj, fi(x)=fj(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, f1(V) is an open subset of U: If f(x)V, then fi(x)V for some i. Thus, we have f1(V)=ifi1(V). Since fi:UiY is continuous, fi1(V) is open in Ui. Since open subsets of open subsets are open, and Ui is open in U, fi1(V) is open in U. Thus, the union f1(V) of all the fi1(V) is also an open subset of U.