07081300/Class notes for Thursday, November 1

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 from a topological space to a locally compact (Hausdorff) topological space is closed.
Proof. Let be closed in , we need to show that is closed in . Since closedness is a local property, it is enough to show that every point has a neighbourhood such that is closed in . Fix , and by local compactness, choose a neighbourhood of whose close is compact. Then
so that . But is compact by choice, so is compact as is proper, so is compact as is closed, so is compact (and hence closed) as a continuous image of a compact set, so is the intersection of a closed set with , hence it is closed in .
Note
The example of a noncontractible "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
Typed Notes
Definition 1
X is contractible to if there exists a continuous function where
1)
2)
3)
Example 1
The singleton is contractible
Example 2
, and are all contractible. For instance, is contractible via the function given by H(x,t) = (1t)x.
Example 3
the unit sphere in a Hilbert space is contractible
Example 4
is NOT contractible for all n (as we shall prove)
Example 5
Consider the Cantor Comb consisting of the subset of consisting of the unit interval along the x axis with a spike of height 1 going up perpendicularly to the x axis at the location 1/n for n = 0,1,2,...
This set is contractible to (0,0) via the mapping that shrinks all the spikes on the comb to the real axis in time t=1/2 and then shrinks the interval to the point (0,0) by time t=1.
However, this set is NOT contrabible to (0,1). This is because should we try to flatten to the real axis points on the teeth of the comb within an epsilon neighborhood of (1,0) they would pull (1,0) down to the real axis as well otherwise continuity would be broken. Hence the third requirement of a contraction is broken.
Proposition
is not contractible
Assume not, thus it is contractible with contraction H(x,t).
Consider and consider the spherical shell of diameter less than or equal to 1 inside where each shell comes to rest tangentially to a point on . This is like the idea of making a new spherical shell for each such diameter inside of and letting all the shells "fall" to the bottom.
We now define a retract r on by associating H(,t) with the spherical shell of diameter t. I.e. the outside shell of is associated with H(x,0).
Hence . R is clearly continuous and is a retract. But we proved last class that such retracts are impossible and this establishes the contradiction.
Q.E.D