8 quick postings on the homeomorphism example..........DMD
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Subject: Re: Subject: Question on homeomorphism
Date: Tue, 29 Jan 2002 09:52:32 -0800
From: "W. Dale Hall"
>
> Subject: Question on homeomorphism
> Date: Mon, 28 Jan 2002 20:56:43 +0530
> From: "Priyavrat C. Deshpande"
>
> Can following situation happen ?
>
> X and Y are 2 topological spaces which are not homeomorphic, but XxI and
> YxI
> are homeomorphic. ( I = [0,1], x = Cartesian product )
> How to dis/prove it ?
Here's a simple example of X and Y satisfying the condition. The two
figures below should be understood as constituting the outlines of the
obvious surfaces (view with fixed-width font):
+-------------+ +--------------+
| | | |
\ \---/ /| | | +--+ +--+ |
\ \ / / | | | | | | | |
\ / / | | | | | | | |
\/ / | | | | | | | |
/ / \ | | | | | | | |
/ /\ \ | | | | | | | |
/ /--\ \| | | +--+ +--+ |
| | | |
+-------------+ +--------------+
Note that the figure on the left has a single boundary component, while
the one on the right has three boundary components, so they are not
homeomorphic. However, upon taking the Cartesian product with the unit
interval, both become the 3-ball with 2 (solid) handles attached.
Dale Hall.
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Subject: Re: 3 postings
Date: Tue, 29 Jan 2002 13:47:55 -0500 (EST)
From: Robert Bruner
This is classic: Let X and Y be annuli
{ x in R^2 | 1 <= |x| <= 2}
with two whiskers attached. On X attach both whiskers to the outer
boundary, on Y attach one to the inner and one to the outer boundary.
Then X is not homeomorphic to Y, but after crossing with I they are.
Bob
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Subject: Re: 3 postings
Date: Tue, 29 Jan 2002 13:56:51 -0500 (EST)
From: Adam Sikora
I enclose an answer to the last question:
>Can following situation happen ?
>X and Y are 2 topological spaces which are not homeomorphic, but XxI and
>YxI are homeomorphic. ( I = [0,1], x = Cartesian product )
>How to dis/prove it ?
This can happen. Take X=punctured torus (torus with a disc removed),
Y=twice punctured disc. They are not homeomorphic since their boundaries
have different numbers of components. On the other hand X x I and Y x I
are homeomorphic.
This follows from the fact that both X and Y can be presented as a disc
with two ``bands'' attached. (In X these bands ``overlap'').
-- Adam
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Subject: Re: 3 postings
Date: Tue, 29 Jan 2002 13:05:10 -0600 (CST)
From: Dan Kahn
For a simple example: first let A be the annulus in the xy-plane defined
by 1 <= x^2+y^2 <= 2. Then define X as the union of A with the line
segments on the x-axis [-3,-2] and [2,3]. Define Y as the union of A
with the line segments on the x-axis [-1,0] and [2,3]. It's an easy
homework exercise to see that X and Y are not homeomorphic (for example,
use local homology groups). To see that XxI and YxI are homeomorphic,
observe that each is homeomophic to a solid torus with two "fins"
attatched. You can then twist the inside fin of YxI to match an outside
fin of XxI.
Daniel S. Kahn
kahn@math.northwestern.edu
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Subject: X and Y are homeomorphic, X x I and Y x I are not.
Date: Tue, 29 Jan 2002 12:21:57 -0700 (MST)
From: Ross Staffeldt
Some easy examples are found among surfaces.
The smallest ones I know: Let X = torus - open disc
and let Y = closed disc - (two open discs in the interior)
Then X and Y are not homeomorphic (look at the number of boundary
components) but X x I and Y x I are.
>From the point of view of handle bodies, both X and Y are discs with
two one handles attached. In the case of X the attaching spheres of
the cores of the handles are linked, whereas in the case of Y the
attaching spheres of the cores of the handles are not linked.
When one takes cartesian product with I, both X x I and Y x I are
still discs (now dim = 3) with two one handles attached, but now
the attaching spheres of the cores in the case of X x I may be
unlinked (by isotopy).
It is easy to illustrate if one has access to Play-Dough.
Ross Staffeldt
New Mexico State University
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Subject: Re: 3 postings
Date: Tue, 29 Jan 2002 14:09:49 -0600 (MST)
From: Jesus Gonzalez
The paper "Cancellation laws in topological products" by Edardo Santillan
gives such an example (X and Y subspaces of the plane). On the other hand,
the same paper shows (among other things) that the above situation is
impossible when X and Y are subspaces of the real line.
The paper is available at: http://chucha.math.cinvestav.mx/morfismos/
(Vol 2 No 2)
Jesus Gonzalez
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Subject: Re: 3 postings
Date: Tue, 29 Jan 2002 15:16:04 -0500 (EST)
From: Walter Neumann
Punctured torus (ie torus minus an open disk) and thrice puncture sphere
both have XxI homeomorphic to a solid handlebody of genus 3.
-walter neumann
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Subject: Re: 3 postings
Date: Tue, 29 Jan 2002 14:28:08 -0600 (CST)
From: "Eduardo Sant. Zeron"
Another reference and more examples can be found in my paper:
E. S. Zeron, Cancellation laws in topological products.
Houston J. Math. 27 (2001), no. 1, 67--74.
Sincerely yours: ESZeron.