Gluing lemma for closed subsets: Difference between revisions
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Let <math>A</math> and <math>B</math> be [[closed subset]]s of a [[topological space]] <math>X</math> whose union is <math>X</math>, and <math>f:A \to Y</math> and <math>g:B \to Y</math> be [[continuous map]]s such that <math>f(x) = g(x) \ \forall \ x \in A \cap B</math>. Then there exists a unique continuous map from <math>X</math> to <math>Y</math> whose restriction to <math>A</math> is <math>f</math> and to <math>B</math> is <math>g</math>. | Let <math>A</math> and <math>B</math> be [[closed subset]]s of a [[topological space]] <math>X</math> whose union is <math>X</math>, and <math>f:A \to Y</math> and <math>g:B \to Y</math> be [[continuous map]]s such that <math>f(x) = g(x) \ \forall \ x \in A \cap B</math>. Then there exists a unique continuous map from <math>X</math> to <math>Y</math> whose restriction to <math>A</math> is <math>f</math> and to <math>B</math> is <math>g</math>. | ||
The result can be modified to handle finitely many closed sets which cover <math>X</math>; however, it does ''not'' cater to arbitrarily many closed sets which cover <math>X</math>. This is in contrast with the gluing lemma for open subsets. | |||
==Related results== | ==Related results== | ||
Revision as of 18:05, 11 December 2007
Statement
Let and be closed subsets of a topological space whose union is , and and be continuous maps such that . Then there exists a unique continuous map from to whose restriction to is and to is .
The result can be modified to handle finitely many closed sets which cover ; however, it does not cater to arbitrarily many closed sets which cover . 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 union of two closed subsets is closed
Fill this in later