0708-1300/Class notes for Thursday, November 1
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The information below is preliminary and cannot be trusted! (v)
Today's Agenda
- HW4 and TE1.
- Continue with Tuesday's agenda:
- Debt on proper functions.
- Complete the proof of the "tubular neighborhood theorem".
- Prove that "the sphere is not contractible".
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].
