Gluing lemma for closed subsets: Difference between revisions

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* A union of two closed subsets is closed
* A union of two closed subsets is closed


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===Proof details===
 
'''Given''': A topological space <math>X</math>, closed subsets <math>A,B</math> of <math>X</math>. Continuous functions <math>f,g:A \cup B \to Y</math>, such that <math>f|_{A \cap B} = g|_{A \cap B}</math>.
 
'''To prove''': There is a unique continuous map <math>h:A \cup B \to Y</math> whose restriction to <math>A</math> equals <math>f</math> and whose restriction to <math>B</math> equals <math>g</math>.
 
'''Proof''': Note that since <math>A,B</math> is closed in <math>X</math>, they are also closed in <math>A \cup B</math>.
 
# There is a unique function <math>h</math> on <math>A \cup B \to Y</math> whose restriction to <math>A</math> is <math>f</math> and to <math>B</math> is <math>g</math>: This is set-theoretically obvious.
# This function is continuous: For this, we prove that the inverse image of any closed subset of <math>Y</math> is closed in <math>A \cup B</math>. Let <math>C</math> be a closed subset of <math>Y</math>. <math>h(x) \in C</math> iff <math>f(x) \in C</math> or <math>g(x) \in C</math>. Thus, <math>h^{-1}(C) = f^{-1}(C) \cup g^{-1}(C)</math>. Since <math>f</math> is continuous, <math>f^{-1}(C)</math> is a closed subset of <math>A</math>, which is closed in <math>A \cup B</math>. So <math>f^{-1}(C)</math> is closed in <math>A \cup B</math>. Similarly, <math>g^{-1}(C)</math> is closed in <math>A \cup B</math>. Since a finite union of closed subsets is closed, <math>f^{-1}(C) \cup g^{-1}(C) = h^{-1}(C)</math> is closed in <math>A \cup B</math>.


==Applications==
==Applications==

Latest revision as of 03:44, 17 July 2009

This article is about the statement of a simple but indispensable lemma in topology

Statement

Let A and B be closed subsets of a topological space X, and f:AY and g:BY be continuous maps such that f(x)=g(x)xAB. Then there exists a unique continuous map from AB to Y whose restriction to A is f and to B is g.

The result can be modified to handle finitely many closed sets which cover X; however, it does not cater to arbitrarily many closed sets. This is in contrast with the gluing lemma for open subsets.

Related results

Proof

The proof uses the following key facts:

  • A map is continuous if and only if the inverse image of any closed subset is closed
  • A closed subset of a closed subset is closed. For full proof, refer: Closedness is transitive
  • A union of two closed subsets is closed

Proof details

Given: A topological space X, closed subsets A,B of X. Continuous functions f,g:ABY, such that f|AB=g|AB.

To prove: There is a unique continuous map h:ABY whose restriction to A equals f and whose restriction to B equals g.

Proof: Note that since A,B is closed in X, they are also closed in AB.

  1. There is a unique function h on ABY whose restriction to A is f and to B is g: This is set-theoretically obvious.
  2. This function is continuous: For this, we prove that the inverse image of any closed subset of Y is closed in AB. Let C be a closed subset of Y. h(x)C iff f(x)C or g(x)C. Thus, h1(C)=f1(C)g1(C). Since f is continuous, f1(C) is a closed subset of A, which is closed in AB. So f1(C) is closed in AB. Similarly, g1(C) is closed in AB. Since a finite union of closed subsets is closed, f1(C)g1(C)=h1(C) is closed in AB.

Applications

The gluing lemma for closed subsets is one of the many results in point-set topology which is applied everywhere, often without even consciously realizing it. Here are some examples:

  • The multiplication defined in the fundamental group and higher homotopy grooups, uses the gluing lemma (to argue that a composite of loops is a loop)
  • The fact that homotopies can be composed also uses the gluing lemma
  • Many of the proofs involving manifolds, for instance, the proof that the inclusion of a point in a manifold is a cofibration, or the proof that connected manifolds are homogeneous, uses the gluing lemma; we glue an explicit map in a neighbourhood of the point with a constant map outside.