Welcome to MathOverflow

Question

1

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.

edited Jun 20 2010 at 18:10

Igor Belegradek
9,881●1●13●42

asked Jun 20 2010 at 15:08

HJ
385●6

	$M=[0,1]$ has a strong deformation retract to $N= \{0\}$ . The complement is contractible, but the boundary of $M$ is not. – S. Carnahan♦ Jun 20 2010 at 15:36
	Thank you. I corrected the question. – HJ Jun 20 2010 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 2010 at 16:44
	Yes, and it should have boundary. – HJ Jun 21 2010 at 2:22

Igor Belegradek · Accepted Answer · 2010-06-21 10:56:24Z UTC

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..

Tom Goodwillie · Answer 2 · 2010-06-21 10:34:45Z UTC

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.

Is the complement of a strong deformation retract of a manifold M homotopic equivalent with the boundary of M?

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

2 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.