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