Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ be a topological manifold with boundary.What is the idea behind the fact that emoving the boundary doesn't change the homotopy type of the manifold; i.e.,that is the manifold $X$ has the same homotopy type of $X-\partial X$?

share|improve this question

1 Answer 1

up vote 3 down vote accepted

In a smooth manifold, there is a neighborhood of the boundary $V$ which is diffeomorphic to $\partial M\times [0,1)$, called a collar neighborhood. (Google "Collar neighborhood theorem.") Removing the boundary would give you $\partial M\times (0,1)$. These are homotopy-equivalent in a way that fixes $\partial M\times [.5,1)$, so the homotopy equivalence extends to the interior of the manifold $M$. This homotopy equivalence can be visualized by collapsing both $\partial M\times (0,1)$ and $\partial M\times [0,1)$ down to $\partial M\times [.5,1)$.

For a topological (non-smoothable) manifold, I believe the collar neighborhood theorem also holds, but I'm not as sure about that.

share|improve this answer
(In response to a no-longer existent comment:) I want to define two homotopy-equivalences. $f\colon \partial M\times [0,1)\to \partial M\times[.5,1)$ and $g\colon \partial M\times (0,1)\to\partial M\times [.5,1)$. Both $f$ and $g$ are defined by $(x,t)\mapsto (x,.5)$ if $t\geq .5$ and $(x,t)\mapsto(x,t)$ otherwise. The reverse maps are just inclusions. One has to check that the compositions in both directions are homotopic to the identity. –  Grumpy Parsnip Oct 12 '12 at 22:26
yes i figured that out that's why i deleted the previous comment and added a new comment. I'msorry for that. –  palio Oct 12 '12 at 22:31
what do you mean by "the homotopy equivalence extends to the interior of $M$? do you mean that the homotopy equivalence $h:\partial M\times [0,1)\to \partial M \times (0,1)$ extends to a homotopy equivalence $\tilde h:M\times [0,1)\to M \times (0,1)$? –  palio Oct 12 '12 at 22:48
what we have showed so far is that the collar neighborhood of the boundary $N(\partial M)$ is homotopy equivalent to $N(\partial M)-\partial M$; the collar neighborhood minus the boundary. How can we generalize this to a statement $M\simeq M-\partial M$? –  palio Oct 13 '12 at 8:37
You define all of your maps to be the identity away from the boundary of the manifold. That way, using the gluing lemma, you can glue together the maps relating $M$ and $M\setminus\partial M$ and the homotopies between their compositions. –  Grumpy Parsnip Oct 13 '12 at 13:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.