From: Xiao-Song Lin Subject: Summary of replies to the question X^2=Y^2 ==> X=Y? Date: Tue, 09 Feb 1999 08:32:01 -0800 Newsgroups: sci.math.research Keywords: Cancellation in topological categories Thanks to all who replied to my posting: >Is it true that if X and Y are two topological spcaes, then X\times X = >Y\times Y implis X = Y? I was talking with Cho-Ho Chu of University > of London last December in Beijing and he asked me this question. >I know an example that shows X\times Y = Z\times Y but X and Z are not >homeomorphic. We may take W to be the Whitehead manifold, which is a >open 3-manifold obtained as the complement of the intersection of a >certain nested sequence of solid tori in S^3. Then W\times R = R^4 but W >is not equal to R^3 (see Dale Rolfsen "Knots and Links", page 82). Are >there any other simpler examples of this sort? Is it true that W^2 = R^6? The following is a summary of the answer I've received: 1. Simpler examples that WxY=WxZ but Y=/=Z. Daniel Giaimo : Here's a simpler example: [0,1)x[0,1) = [0,1]x[0,1), but [0,1) is not equal to [0,1]. Martin Goldstern : Simpler: 2 x Y = 1 x Y for any discrete infinite Y. John D. McCarthy : There are simple examples of the latter type where Y and Z are compact surfaces with boundary and W is the unit interval I = [0,1]. The idea is that for a compact surface X the product X \times I is in fact a standard handlebody of euler characteristic equal to the euler characteristic of X. So it suffices to find two nonhomeomorphic compact surfaces with boundary, Y and Z, which have the same euler characteristic. This is simple. Lee Rudolph : The simplest example is the pair-of-pants (connected planar surface with three boundary circles) and the torus with an open disk removed: non-homeomorphic, but crossed with I theyboth become solid handlebodies of genus 2. 2. Examples of X^2 = Y^2 but X=/= Y. Martin Goldstern : Let C be a connected space, consisting of more than one point, such that C = C x C. (For example, the indiscrete topology on any infinite set will have that property, but there are probably nicer examples. It is not really necessary that C is connected, as long as C and C+C are sufficiently distinct.) Let Z be a discrete space on infinitely many points. Let 1,2,3,4, etc be the discrete spaces on 1,2,3,4, etc points, respectively. Note that Z+1 = Z+Z = ZxZ. Then C + C + Z is not homoeomorphic to C + Z (since C + C + Z has two nontrivial connected components, C+Z only one. ). However, the squares of these spaces are homeomorphic: (C+C+Z) x (C+C+Z) = CxCx4 + CxZx4 + ZxZ = CxZ + Z. (C+Z) x (C+Z) = CxC + CxZx2 + ZxZ = CxZ + Z. Reinhard Schultz : ... this is a problem that Slawomir Kwasik and I have been looking at for a while and we have a couple of papers ready for publication. It turns out that the answer to the question is yes for polyhedra of dimension 1 or 2 and false in all higher dimensions. The simplest negative examples are 3dim lens spaces, and two such manifolds with isomorpic fundamental groups have diffeomorphic squares. Fred Galvin : A nice counterexample was given by R. H. Fox, "On a problem of S. Ulam concerning Cartesian products", Fund. Math. 34 (1947), 278-287. 3. Is it true that W^2 = R^6, where W is the the Whitehead manifold? Lee Rudolph : Yes. [HTML deleted. Reformatted. -- djr] ============================================================================== From: lrudolph@panix.com (Lee Rudolph) Subject: Re: Summary of replies to the question X^2=Y^2 ==> X=Y? Date: 9 Feb 1999 13:44:05 -0500 Newsgroups: sci.math.research Xiao-Song Lin writes: ... >3. Is it true that W^2 = R^6, where W is the the Whitehead manifold? > >Lee Rudolph : Yes. ... I told X-SL I'd get back with a more detailed answer, and didn't. Here it is. Following McMillan, define a W-space to be a contractible open 3-manifold each of whose compact subsets can be embedded in R^3 (so W is a W-space, and there are uncountably many pairwise non-homeomorphic W-spaces which embed in R^W, as well as uncountably many pairwise non-homeomorphic W-spaces which do not embed in R^3). Then theorems of Glimm and of McMillan show that the cartesian product of any two W-spaces is R^6; in particular, W^2 = R^6. Lee Rudolph