From: "Daniel Giaimo"
Subject: Re: Homeomorphisms of the sphere
Date: Sat, 4 Dec 1999 23:03:54 -0800
Newsgroups: sci.math
Keywords: Invariance of Domain theorems
Romain Brette wrote in message
news:99120507431502.00650@localhost.localdomain...
>Le dim, 05 déc 1999, vous avez écrit :
>>Romain Brette wrote in message
>>news:944261904.589352346@news.fnac.net...
>>> Does anybody know how to prove that the n-dimensional sphere is not
>>> homeomorphic to a (strict) part of itself ? It's quite easy for the circle,
>>> but turns out to be quite hard in general.
>>
>> Suppose S^n is homeomorphic to a strict subset of itself.
>> Let f:S^n->f(S^n) be such a homeomorphism. Then f(S^n) != S^n,
>>therefore there exists a point x_0 in S^n such that x_0 is not in f(S^n).
>>Therefore f actually maps into S^n\{x_0} which is homeomorphic to R^n. Now,
>>by Invariance of Domain, f(S^n) is open in R^n as f is 1-1 and continuous.
>
> I guess you mean the projection of f(S^n) in R^n. But what is "Invariance of
>Domain" ? I can't see why this should be open. The thing is if A is open and f
>is a 1-1 and continuous, then f(A) is open _relatively_ to the range, which is
>f(S^n) (which means, it's the intersection of an open set with f(S^n)). So it
>doesn't tell anything about the image of the domain. For example, take a
>homeomorphism from the disc to a part of it (a contraction)
>Then the image of the domain is not open. Can you state what you mean by
>"Invariance of Domain" ?
Invariance of Domain is a theorem which states that if f:M->N is a map
between n-dimensional manifolds without boundary which is 1-1 and
continuous, then f is an open map. In fact, you really only need the weaker
form which says that if f:S^n->S^n is open and continuous then it is open.
--
--Daniel Giaimo
Remove nospam. from my address to e-mail me. |
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^^^^^^^^^<-(Remove)
|--------BEGIN GEEK CODE BLOCK--------| Ros: I don't believe in it anyway.
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==============================================================================
From: "Daniel Giaimo"
Subject: Re: Homeomorphisms of the sphere
Date: Sun, 5 Dec 1999 12:19:16 -0800
Newsgroups: sci.math
Edward C. Hook wrote in message
news:82egin$nk9$5@sun500.nas.nasa.gov...
[deletia; similar to above --djr]
>|> by Invariance of Domain, f(S^n) is open in R^n as f is 1-1 and continuous.
>
> How do you derive this from Invariance of Domain ?? AFAICR, that
> theorem asserts that either every embedding of a space X into R^n
> has an open image or none of them do.
There are several theorems known as Invariance of Domain. The one I am
using is the theorem that if M and N are topological n-manifolds without
boundary, and f:M->N is 1-1 and continuous, then f is open.
--
--Daniel Giaimo
[sig deleted; see above --djr]
==============================================================================
From: lrudolph@panix.com (Lee Rudolph)
Subject: Re: Topology question in R^n
Date: 14 Jun 1999 11:35:18 -0400
Newsgroups: sci.math
>> Let A and B be two subsets of R^n. Let
>> f: A -> B be a homeomorphism.
>> If A is an open set (for the topology induced by R^n),
>> is B always open?
[deletia --djr]
In fact, the question as understood by me (and, I believe, as meant by
the questioner) is not trivial (at least not for n>1); that the answer
is "yes" (Brouwer's theorem on "invariance of domain") for all n depends
on properties of R^n that are not shared by all spaces.
>A homeomorphism is a bijective continuous map whose inverse is also
>continuous, so it maps open sets to open sets and closed sets to closed
>sets by definition of the continuity of the inverse map.
Consider the space X = the union in R^2 of the x-axis and the y-axis,
with the topology induced by R^2. Let A be the (strictly) positive
x-axis. Then A is an open subset of the space X. Let B be the
entire x-axis. Then B is not an open subset of the space X.
Yet A and B (with their topologies induced by X, which are
of course their topologies induced by R^2) are homeomorphic.
Notice that X does not share various salient properties with R^1.
Notably, the local homology of X at the point (0,0) is different
from its local homology at other points. This is relevant.
Lee Rudolph