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.
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.
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..
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.
$N= \{0\}$
. The complement is contractible, but the boundary of $M$ is not. $\endgroup$ – S. Carnahan♦ Jun 20 '10 at 15:36