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.