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It seems an easy problem but I couldn't prove it.

Let $M$ be a manifold with boundary and $N\subset \mathrm{int}(M)$ is a strong deformation retract of $M$.

Then I wonder whether $M-N$ is homotopic equivalent to the boundary of $M$ or not.

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$M=[0,1]$ has a strong deformation retract to $N= \{0\}$. The complement is contractible, but the boundary of $M$ is not. –  S. Carnahan Jun 20 '10 at 15:36
    
Thank you. I corrected the question. –  user6569 Jun 20 '10 at 15:51
    
If $M=\mathbb{R}$, with empty boundary, and $N=[0,+\infty)$, then the answer is no. Do you want $M$ to be compact? –  S. Carnahan Jun 20 '10 at 16:44
    
Yes, and it should have boundary. –  user6569 Jun 21 '10 at 2:22
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2 Answers 2

up vote 6 down vote accepted

Take a compact contactible manifold $C$ whose boundary is not a homotopy sphere sphere (those are no so easy to construct, but it has been done long ago). Then removing a point (or equivalently a small ball) from the interior of $C$ gives a manifold that is homotopy equivalent (by excison in homology) to the boundary of the small ball, which is a sphere. This is a desired counterexample.

EDIT: References to how to built contactible manifolds can by found e.g. in my paper pp.8-9..

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Thank you! Can you let me know or give me a reference how can I construct such a contractible manifold? –  user6569 Jun 21 '10 at 2:21
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The Alexander horned sphere furnishes an example of a subspace of the interior of a compact $3$-ball that is contractible (being a $3$-ball itself) and whose complement is not simply connected.

There are also less pathological examples in high dimensions: If $n$ is at least $5$ or so then it is not hard to make a smooth compact contractible $n$-manifold whose boundary is not simply connected, whereas the complement of a point in a simply connected manifold of dimension $3$ or more must be simply connected.

EDIT: I retract (heh) my first example. I overlooked the requirement that the subset should be a deformation retract; I was using the weaker requirement that the inclusion should have an inverse up to homotopy.

EDIT: Doh! My first example was correct. If i:A-->B is an inclusion of compact metric spaces and A is homeomorphic to a ball, then by the Tietze extension theorem there is a retraction r:B-->A. If B is also a ball, then the resulting "straight-line" homotopy from ir to the identity gives a deformation retraction.

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The 3-ball does not retract onto the 2-sphere, because such a retraction would be onto on 2-dimensional homology. Maybe you meant "horned 2-ball". –  Igor Belegradek Jun 20 '10 at 18:13
    
I did not say that the Alexander horned sphere is an example; I said that it <i>furnishes</i> an example. –  Tom Goodwillie Jun 20 '10 at 19:28
    
How does one construct a deformation retraction of the ambient disk onto a wildly embedded disk? The pair does not seem to satisfy the Homotopy Extension Property. Maybe in the case of Alexander's horned disk the retraction can be constructed by hand. Am I missing an obvious way to do it? –  Igor Belegradek Jun 20 '10 at 20:41
    
My mistake. Answer now edited. –  Tom Goodwillie Jun 20 '10 at 21:27
    
The questioner can find other examples of wild disks in disks of higher dimensions in MathSciNet references to the original example of Fox-Artin arc [Fix, Artin, "Some wild cells and spheres in three-dimensional space", Ann. of Math. (2) 49, (1948). 979--990]. The argument of Tom Goodwillie's second edit applies to all of them. –  Igor Belegradek Jun 21 '10 at 13:45
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