Gluing lemma for open subsets

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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_is. 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.

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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_is.

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

Proof: Note first that the U_is 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_is. 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.