0708-1300/Class notes for Thursday, November 1

0708-1300/Navigation Panel   Add your name / see who's in! # Week of... Links Fall Semester 1 Sep 10 About, Tue, Thu 2 Sep 17 Tue, HW1, Thu 3 Sep 24 Tue, Photo, Thu 4 Oct 1 Questionnaire, Tue, HW2, Thu 5 Oct 8 Thanksgiving, Tue, Thu 6 Oct 15 Tue, HW3, Thu 7 Oct 22 Tue, Thu 8 Oct 29 Tue, HW4, Thu, Hilbert sphere 9 Nov 5 Tue,Thu, TE1 10 Nov 12 Tue, Thu 11 Nov 19 Tue, Thu, HW5 12 Nov 26 Tue, Thu 13 Dec 3 Tue, Thu, HW6 Spring Semester 14 Jan 7 Tue, Thu, HW7 15 Jan 14 Tue, Thu 16 Jan 21 Tue, Thu, HW8 17 Jan 28 Tue, Thu 18 Feb 4 Tue 19 Feb 11 TE2, Tue, HW9, Thu, Feb 17: last chance to drop class R Feb 18 Reading week 20 Feb 25 Tue, Thu, HW10 21 Mar 3 Tue, Thu 22 Mar 10 Tue, Thu, HW11 23 Mar 17 Tue, Thu 24 Mar 24 Tue, HW12, Thu 25 Mar 31 Referendum,Tue, Thu 26 Apr 7 Tue, Thu R Apr 14 Office hours R Apr 21 Office hours F Apr 28 Office hours, Final (Fri, May 2) Register of Good Deeds Errata to Bredon's Book

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In Preparation

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.
  • Prove that "the sphere is not contractible".
  • Complete the proof of the "tubular neighborhood theorem".

Proper Implies Closed

Theorem. A proper function $f:X\to Y$ from a topological space $X$ to a locally compact (Hausdorff) topological space $Y$ is closed.

Proof. Let $B$ be closed in $X$, we need to show that $f(B)$ is closed in $Y$. Since closedness is a local property, it is enough to show that every point $y\in Y$ has a neighbourhood $U$ such that $f(B)\cap U$ is closed in $U$. Fix $y\in Y$, and by local compactness, choose a neighbourhood $U$ of $y$ whose close ${\bar {U}}$ is compact. Then

$f(B)\cap U=f(B\cap f^{-1}(U))\cap U\subset f(B\cap f^{-1}({\bar {U}}))\cap U\subset f(B)\cap U$,

so that $f(B)\cap U=f(B\cap f^{-1}({\bar {U}}))\cap U$. But ${\bar {U}}$ is compact by choice, so $f^{-1}({\bar {U}})$ is compact as $f$ is proper, so $B\cap f^{-1}({\bar {U}})$ is compact as $B$ is closed, so $f(B\cap f^{-1}({\bar {U}}))$ is compact (and hence closed) as a continuous image of a compact set, so $f(B)\cap U$ is the intersection $f(B\cap f^{-1}({\bar {U}}))\cap U$ of a closed set with $U$, hence it is closed in $U$.