0708-1300/Class notes for Thursday, November 1: Difference between revisions
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{{Equation*|<math>f(B)\cap U=f(B\cap f^{-1}(U))\cap U\subset f(B\cap f^{-1}(\bar U))\cap U\subset f(B)\cap U</math>,}} |
{{Equation*|<math>f(B)\cap U=f(B\cap f^{-1}(U))\cap U\subset f(B\cap f^{-1}(\bar U))\cap U\subset f(B)\cap U</math>,}} |
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so that <math>f(B)\cap U=f(B\cap f^{-1}(\bar U))\cap U</math>. But <math>\bar U</math> is compact by choice, so <math>f^{-1}(\bar U)</math> is compact as <math>f</math> is proper, so <math>B\cap f^{-1}(\bar U)</math> is compact as <math>B</math> is closed, so <math>f(B\cap f^{-1}(\bar U))</math> is compact (and hence closed) as a continuous image of a compact set, so <math>f(B)\cap U</math> is the intersection <math>f(B\cap f^{-1}(\bar U))\cap U</math> of a closed set with <math>U</math>, hence it is closed in <math>U</math>. |
so that <math>f(B)\cap U=f(B\cap f^{-1}(\bar U))\cap U</math>. But <math>\bar U</math> is compact by choice, so <math>f^{-1}(\bar U)</math> is compact as <math>f</math> is proper, so <math>B\cap f^{-1}(\bar U)</math> is compact as <math>B</math> is closed, so <math>f(B\cap f^{-1}(\bar U))</math> is compact (and hence closed) as a continuous image of a compact set, so <math>f(B)\cap U</math> is the intersection <math>f(B\cap f^{-1}(\bar U))\cap U</math> of a closed set with <math>U</math>, hence it is closed in <math>U</math>. |
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===Note=== |
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The example of a non-contractible "comb" seen today is, in fact, "Cantor's comb". See, for example, page 25 of www.karlin.mff.cuni.cz/~pyrih/e/e2000v0/c/ect.ps |
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Revision as of 12:12, 1 November 2007
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Today's Agenda
- HW4 and TE1.
- Continue with Tuesday's agenda:
- Debt on proper functions.
- Prove that "the sphere is not contractible".
- Complete the proof of the "tubular neighborhood theorem".
Proper Implies Closed
Theorem. A proper function [math]\displaystyle{ f:X\to Y }[/math] from a topological space [math]\displaystyle{ X }[/math] to a locally compact (Hausdorff) topological space [math]\displaystyle{ Y }[/math] is closed.
Proof. Let [math]\displaystyle{ B }[/math] be closed in [math]\displaystyle{ X }[/math], we need to show that [math]\displaystyle{ f(B) }[/math] is closed in [math]\displaystyle{ Y }[/math]. Since closedness is a local property, it is enough to show that every point [math]\displaystyle{ y\in Y }[/math] has a neighbourhood [math]\displaystyle{ U }[/math] such that [math]\displaystyle{ f(B)\cap U }[/math] is closed in [math]\displaystyle{ U }[/math]. Fix [math]\displaystyle{ y\in Y }[/math], and by local compactness, choose a neighbourhood [math]\displaystyle{ U }[/math] of [math]\displaystyle{ y }[/math] whose close [math]\displaystyle{ \bar U }[/math] is compact. Then
so that [math]\displaystyle{ f(B)\cap U=f(B\cap f^{-1}(\bar U))\cap U }[/math]. But [math]\displaystyle{ \bar U }[/math] is compact by choice, so [math]\displaystyle{ f^{-1}(\bar U) }[/math] is compact as [math]\displaystyle{ f }[/math] is proper, so [math]\displaystyle{ B\cap f^{-1}(\bar U) }[/math] is compact as [math]\displaystyle{ B }[/math] is closed, so [math]\displaystyle{ f(B\cap f^{-1}(\bar U)) }[/math] is compact (and hence closed) as a continuous image of a compact set, so [math]\displaystyle{ f(B)\cap U }[/math] is the intersection [math]\displaystyle{ f(B\cap f^{-1}(\bar U))\cap U }[/math] of a closed set with [math]\displaystyle{ U }[/math], hence it is closed in [math]\displaystyle{ U }[/math].
Note
The example of a non-contractible "comb" seen today is, in fact, "Cantor's comb". See, for example, page 25 of www.karlin.mff.cuni.cz/~pyrih/e/e2000v0/c/ect.ps
