Homeomorphisms Send Bdries to Bdries

[Bacle]

Hi, All:

Say D^ is a closed disk standardly-embedded in the plane, and D is its interior.

I have an argument to the effect that any homeomorphism h:D^-->D^ sends boundary pts
to boundary points, but it seems kind of clunky, and I wonder if someone has a nicer,
cleaner one:

First, we show interior points are sent to interior points:

We take x in D. Since h is a homeo., h(D) is open, and h(x) lies in h(D). By openness
of h(D), there is an open set Ux containing h(x), so interior points are sent to interior points.

It follows, by considering the exterior of D^ as the complement of the disk, that
exterior points (i.e., those in R^2-D^ )

Now, the kind-of-clunky part:

We show that if x is a boundary point of D , so that every 'hood Vx of x intersects
points both in the interior of D and in the exterior of D, the same is the case for
h(Vx).

So the points in Ext(D^)/\Vx are sent to points in Ext(D^)/\h(Vx) , as a consequence

of the fact that exterior points are sent to exterior points.

Anyone know a nicer way?

Same for the interior points.

 

[henry_m]

A subtlety that you need to be careful about: you initially talk about homeomorphisms \bar{D}\to\bar{D} but then you implicitly assume instead that you have a homeomorphism \mathbb{R}^2\to\mathbb{R}^2 which preserves the disc. Now I'm certain that if you have the former you can extend to to the latter, but you would have to prove it first (I can see a straightforward way to do this but it requires the result you're trying to prove).

If you are talking only about the topological space \bar{D} then you need to firstly define what you mean by 'boundary points', using only this space (not the embedding in the plane). One possibility is 'points that have no open n'hood homeomorphic to \mathbb{R}^2'. Then this will be invariant under homeomorphisms straightforwardly.

 

[Bacle]

Thanks, henry_m:

I was actually trying to rigorize a statement I saw made about homeomorphisms

from D^ to D^ , in which one was asked to show that interior points are sent to

interior points and boundary ones to boundary ones. The statement did not mention

manifold boundaries like the ones you refer to, so I assumed an embedding was

implicit. But your point is a good one, and it brings up that question: I think that

the answer is that every homeo of the disk can be extended; this has to see with

the fact that there are no knots in R^2, and Schoenflies theorem. Others may

improve on this.

 

[henry_m]

I don't think you need to be that sophisticated, and you can get all the way without mentioning an embedding.

It is clear that in the closed disc, topologically there are exactly two distinct types of points. If you use my definition of an interior point (n'hood homeomorphic to the plane), it is quick to see that homeomorphisms map interior points to interior points. Let x be interior. Let U be a n'hood of x homeomorphic to the plane. Then the restriction of h to U is also a homeomorphism between U and h(U), so h(U) is homeomorphic to the plane also. Thus h(x) is interior.

Since the boundary points are then just 'all the other points', the same argument shows that boundary points map to boundary points.

 

[mathwonk]

i think this is not so obvious. i would use something like this: a boundary point has a neighborhood basis of open sets, such that removing the boundary point renders them contractible, but no interior point has such a neighborhood.