Given a manifold $M$ with boundary $\partial M \neq \varnothing$, when can we form a manifold $\tilde M$ from $M$ by collapsing the boundary? In the examples I've considered it seems like collapsing each component of the boundary to a separate point will result in a manifold.

Obviously we can't collapse a disconnected boundary to a single point, e.g. $M = S^1\times [0,1]$ with $\partial M = S^1 \times \{0,1\}$, since the quotient here will be homeomorphic to a "pinched" torus with one of the noncontractible $S^1$ collapsed. However if we collapse $S^1 \times \{0\}$ and $S^1 \times \{1\}$ to separate points, then the result will be the suspension $SS^1 \approx S^2$.

I believe that $(M,\partial M)$ will be a good pair due to the existence of a collar neighborhood, so we should have $H_i(M,\partial M) \cong \tilde{H}_i(M/ \partial M)$, and if $M$ compact and orientable, then by Lefschetz Duality we also have $H_i(M,\partial M) \cong H^{n-i}(M,\varnothing) \cong H^{n-i}(M)$.

Anyhow, I'm quite sure where that gets us, and I have yet failed to construct a counterexample from my repertoire of spaces.

  • $\begingroup$ What happens when you collapse $\partial(D^2\times S^1)$ to a point? In particular, what is the link of the point to which you collapse the boundary? $\endgroup$
    – Neal
    Commented Jan 5, 2013 at 20:18
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    $\begingroup$ Your repertoire of spaces surely contains some manifold whose boundary is not a disjoint union of spheres! $\endgroup$ Commented Jan 5, 2013 at 20:23

1 Answer 1


From what I understand, you are asking: Giving a manifold with boundary $M$, with $\partial M = N_0 \sqcup \ldots \sqcup N_k$, when does the topological space $\tilde M$ obtained by collapsing each $N_i$ to a point $x_i$ admit a manifold structure?

(1) Unless each $N_i$ is a sphere, $\tilde M$ will not be a manifold. This is because in a manifold, any point needs to have a punctured neighborhood equivalent to $S^{n-1} \times \mathbb{R}$, where $n$ is the dimension of the manifold. But because the collapsed point $x_i \in \tilde M$ corresponding to $N_i$ is (by collaring the boundary, for instance) the cone point of a cone over $N_i$, its deleted neighborhood will be equal to $N_i \times \mathbb{R}$.*

(2) I know you haven't talked about smoothness, but just in case: Even if each $N_i$ is a sphere, there may not be a unique smooth structure on this collapse. For instance, if $M = S^3 \times [0,\infty)$, collapsing the boundary gives a copy of $\mathbb{R}^4$, which admits many smooth structures compatible with the $M$ with which you began.

*Actually, there's a question of whether a manifold $N'$ diffeomorphic to $S^3 \times \mathbb{R}$ can be diffeomorphic to $N_i \times \mathbb{R}$ for some $N_i\not \cong S^3$. I'm not sure whether such $N'$ exist.

  • $\begingroup$ Ahh, so the collar neighborhood is more useful than I realized! That makes sense. So it is true that whenever the boundary is a disjoint union of spheres, then collapsing each to a point will result in a manifold? (At this point I'm not concerned with smoothness). $\endgroup$ Commented Jan 5, 2013 at 20:52
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    $\begingroup$ Can we also say that if the boundary is a homology sphere (or disjoint union of them), that the quotient will be a homology manifold? $\endgroup$ Commented Jan 5, 2013 at 20:53
  • $\begingroup$ To your first comment -- yes, a topological manifold. The open cone over a sphere is locally homeomorphic to Euclidean space. As for your second question, I don't know what a homology manifold is. $\endgroup$
    – user54535
    Commented Jan 5, 2013 at 21:21
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    $\begingroup$ For it to be a homology manifold, I just mean it would satisfy whatever duality a manifold would satisfy. Perhaps the better definition is that the local homology of every point would be $\mathbb{Z}$ in dimension $n$ and $0$ elsewhere. $\endgroup$ Commented Jan 5, 2013 at 21:38
  • $\begingroup$ Then what you say is right! $\endgroup$
    – user54535
    Commented Jan 6, 2013 at 1:45

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