# Restriction of Isotopy is an Isotopy

1. Jul 17, 2014

### WWGD

Hi All,
This is a follow-up to another post. Question is:

Is the restriction of an isotopy that is the identity on the boundary (working with MCG) an isotopy?

First, let me try to answer the case I am most interested in: Isotopies of the (closed) n-disk D^n, and the restriction to its interior, the open disk, say D_^n:

By Alexander's trick http://en.wikipedia.org/wiki/Alexander's_trick, there is only one isotopy class for D^n . I think this restricts to D_^n:
Now, working in this class , specially since , in an isotopy, the boundary is sent onto the boundary in each embedding in the path, (so that the interior is sent to the interior )I think this map restricts to an isotopy in the interior . Is this right ?

Related question :any two contractible subspaces of the same space are homotopic (homotopy is an equiv. rel. ). Are they also isotopic?

2. Jul 18, 2014

### homeomorphic

Yes, it restricts to an isotopy.

The problem in the other thread was that you can't assume all self-homeomorphisms or the open disk are necessarily restrictions of things coming from the closed disk. Unless you can prove they always extend, but I think I have a counter-example to that.

Technically, you should use the word homotopy equivalence, rather than saying that subspaces are homotopic. Only maps are homotopic or isotopic. Spaces are homotopy equivalent or isotopy equivalent. And actually, it's better than that. Any two contractible spaces are homotopy equivalent because the definition of contractible implies that they are homotopy equivalent to a point.

3. Jul 18, 2014

### WWGD

I see, I just thought the other post was getting messy (messi?) witth too many side questions, thanks.

4. Jul 18, 2014

### WWGD

But , homeomorphic, there is a confusing issue that results from this: since the isotopy of D^n restricts to one in the interior, it seems to follow that the isotopy group of the interior is trivial. And the interior is homeomorphic to R^n . But, AFAIK, MCG (R^n) = +/- Id $\neq Id$. So, what is wrong here?

5. Jul 18, 2014

### homeomorphic

As I said, suppose you have a homeo of the open disk that does not come from the closed disk. You haven't proved that that one is isotopic to the identity. You have only proved that the ones that are restrictions of ones from the whole closed disk are isotopic to the identity. If it's a restriction, you extend it and then isotop. If it doesn't extend, you are stuck.

6. Jul 18, 2014

### homeomorphic

Another weird thing is your use of -Id. To me that looks like a 180 degree rotation, in the case of a 2-disk. But what you really want is just an orientation-reversing guy, I think. But usually, when we define the mapping class group, we don't include any orientation-reversing guys.

7. Jul 19, 2014

### WWGD

Yes, I undersand the issue of the restriction; I realized what I was using/assuming does not work: a uniformly continuous map on a dense subset (interior ) has a continuous extension -- way too weak--to the whole space.