From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: Homeomorphisms of the sphere
Date: 4 Dec 1999 00:10:59 GMT
Newsgroups: sci.math
Keywords: Borsuk-Ulam theorem
In article <944261904.589352346@news.fnac.net>,
Romain Brette wrote:
>Does anybody know how to prove that the n-dimensional sphere is not
>homeomorphic to a (strict) part of itself ?
Yes, for example William Fulton does, in his book "Algebraic Topology".
If you have a map f : X -> S^n whose image misses a point p on S^n,
then you may compose it with stereographic projection h : (S^n - {p}) -> R^n
to get a map (h o f) : X -> R^n . Observe that h is one-to-one; indeed,
it is a homeomorphism.
If X is itself the n-sphere, and f is a putative homeomorphism
with a proper subset of the n-sphere, then (h o f) would in particular
be an injection from S^n to R^n. But no such map exists; indeed
for any map g : S^n -> R^n there is a pair of antipodal points
q1 and q2 with g(q1) = g(q2). I guess this is called the Borsuk-Ulam
theorem although I don't really know whether that's the right attribution
except for small n.
Fulton leaves the meat of the proof to an appendix, and offers pointers
to variant proofs. I can't recall having ever seen any really elementary
proofs.
dave
==============================================================================
From: "Daniel Giaimo"
Subject: Re: Homeomorphisms of the sphere
Date: Sat, 4 Dec 1999 21:31:53 -0800
Newsgroups: sci.math
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.
However, S^n is compact, therefore f(S^n) is closed and bounded. But the
only closed and open subsets of R^n are R^n and the null set. R^n is not
bounded therefore f maps S^n to the null set, which is a contradiction as
S^n is not empty.
--
--Daniel Giaimo
Remove nospam. from my address to e-mail me. |
dgiaimo@(nospam.)ix.netcom.com
^^^^^^^^^<-(Remove)
|--------BEGIN GEEK CODE BLOCK--------| Ros: I don't believe in it anyway.
|Version: 3.1 |
|GM d-() s+:+++ a--- C++ UIA P+>++++ | Guil: What?
|L E--- W+ N++ o? K w>--- !O M-- V-- |
|PS? PE? Y PGP- t+(*) 5 X+ R- tv+(-) | Ros: England.
|b+@ DI++++ D--- G e(*)>++++ h->++ !r |
|!y->+++ | Guil: Just a conspiracy of
|---------END GEEK CODE BLOCK---------| cartographers, you mean?
==============================================================================
From: Mathieu.Zaradzki@mines.u-nancy.fr (Mathieu Zaradzki)
Subject: RE :Homeomorphisms of the sphere
Date: 5 Dec 1999 01:40:59 -0500
Newsgroups: sci.math
Desole mais mon anglais n'est pas tres bon donc j'ecris en francais.
Je suppose qu'en dimension deux, ta demonstration est basee sur le
fait que le cercle est "connexe" alors que, une partie sticte du
cercle n'est pas connexe.
En dimension n (n>2), cette remarque est fausse en generale par
contre,
tu peu remplacer la "connexite" par une propriete plus forte:
la "simple connexite" (les courbes trace sur la spheres sont
"homotopes" à un point).
En effet si on enleve un point a une sphere, elle n'est plus
"simplement connexe" (il suffit de considerer une courbe autour du
trou ainsi cree) or la "simple connexite" comme la "connexite" se
conserve par homeomorphisme.
En esperant que le francais ne vous gene pas trop.