8 quick postings on the homeomorphism example..........DMD _________________________________________ 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. ________________________________________ 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 __________________________________________ 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 ________________________________________ 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 ____________________________________________ 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 ______________________________________ 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 _________________________________________ 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 ___________________________________________ 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.