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Consider the set of homeomorphisms of the topological n-ball to itself with the compact open topology. Sitting inside this space of homeomorphisms are particular subspaces. The first subspace is those homeomorphisms such that the restriction to the boundary is the identity. A path in this space is the same as an isotopy. Alexander's trick tells us that this space is path connected. In other words, if we are given any two homeomorphisms that restrict to the identity on the boundary, then these two homeomorphisms are isotopic relative to the boundary. If we push Alexander's trick a bit further, we can show that if we have two homeomorphisms of the n-ball such that the homeomorphisms restricted to the boundary are isotopic, then we may extend to an isotopy of the two homeomorphisms on the entire n-ball. My questions are as follows:

  1. What is the homotopy type of the space of homeomorphisms of the n-ball such that the homeomorphism restricted to the boundary is the identity? In particular, is this space contractible?

  2. Same question for the space of homeomorphisms such that the homeomorphisms restricted to the boundary is isotopic to the identity (on the boundary). In other words, what is the homotopy type of the space of homeomorphisms such that the homeomorphisms restricted to the boundary is isotopic to the identity? In particular is this space contractible?

  3. What is a good reference for all of this (including, but not limited to Alexander's trick).

As for Alexander's trick, the Wikipedia page is fairly good, but not a proper source.

see: http://en.wikipedia.org/wiki/Alexander%27s_trick

EDIT

I have found part of the answer in a set of class notes, math.northwestern.edu/~jnkf/classes/mflds/14poincare.pdf .

The formulation of Alexander's trick used in these notes is that

4 The space $homeo(D^n,S^{n-1})$ the space of homeomorphisms such that the restriction to the boundary is the identity is contractible as Ryan has noted in his comment.

5 We have a short exact sequence of topological group, $$1\rightarrow homeo(D^n,S^{n-1})\rightarrow homeo(D^n)\rightarrow homeo(S^{n-1})\rightarrow 1$$. The later map, $$\rightarrow homeo(D^n)\rightarrow homeo(S^{n-1})$$ is a fibration with fiber $homeo(D^n,S^{n-1})$. Therefore, $$\rightarrow homeo(D^n)\rightarrow homeo(S^{n-1})$$ is a homotopy equivalence.

6 Now let us look at those homeomorphisms such that the restriction to the boundary is isotopic to the identity. When you hit these with the restriction map, $ homeo(D^n)\rightarrow homeo(S^{n-1})$ the image of these homeomorphisms are in the path component of $homeo(S^{n-1}$ that contains the identity. But since our map , $ homeo(D^n)\rightarrow homeo(S^{n-1})$ is a homotopy equivalence, we have that the homeomorphisms from (2) is homotopy equivalent to the connected component of $homeo(S^{n-1})$

We therefore have the following: The spaces in (1) are contractible. The spaces in (2) are homotopy equivalent to $homeo(S^{n-1})_e$. In particular if $n=2$, I believe that $homeo(S^{n-1})_e$ has the homotopy type of the circle. The question still stands. What is the homotopy type of $homeo(S^{n-1})_e$? And what are some references?

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(1) Is contractible, the proof is just another variant of Alexander's trick. –  Ryan Budney Nov 7 '12 at 5:48
    
I imagine in 6 you meant to say "the restriction to the boundary is isotopic to the identity". Interesting question! –  tweetie-bird Nov 10 '12 at 0:59
    
Yes that is exactly right. I edited to make the correction. –  Spice the Bird Nov 11 '12 at 0:28
    
I'm not sure about this, but perhaps the original reference for Alexander's trick might be: J.W. Alexander, On the deformation of an $n$-cell, Proc. Nat. Acad. Sci. USA 9 (1923), 406-407. At least this is the reference given by Kirby and Siebenmann for "Alexander's isotopy" on p. 17 in "Foundational essays on topological manifolds, smoothings, and triangulations" (which is a compilation of key papers written by one or both of them). There is no MathSciNet review of this paper from 1923 to check, obviously. –  Patricia Hersh Nov 14 '12 at 19:46

1 Answer 1

I recommand Rob Kirby's video of his first lecture in Edinburgh: http://www.maths.ed.ac.uk/~aar/kirby.htm his lecture begins with Alexander's trick.

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Presumably they will be posted after they are given. –  j.c. Nov 7 '12 at 11:49

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