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. | 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: "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