I have a topological manifold whose suspension is homeomorphic to the sphere $S^{k+1}$. Is it necessarily itself homeomorphic to $S^k$?

I know that this is not true if I replace "suspension" with "double suspension", because I found the helpfully named http://en.wikipedia.org/wiki/Double_suspension_theorem.

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    $\begingroup$ This is way outside my area of expertise, so perhaps someone can explain why the answer does not follow from the double suspension theorem: start with the Poincare dodecahedral space $M$ (a homology 3-sphere with nontrivial fundamenatal group) and suspend it once. If you get something homeomorphic to $S^4$, then $M$ is a counterexample. If not, then by DST $SM$ is homeomorphic to $S^5$ so $SM$ is a counterexample. $\endgroup$ Commented May 5, 2011 at 18:29
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    $\begingroup$ Interesting plan! But it is not obvious to me that the suspension of M (or any other space obtained by a similar method) is a topological manifold. $\endgroup$ Commented May 5, 2011 at 18:54
  • $\begingroup$ Okay, so that's what I was missing: that the suspension of a manifold might or might not be a manifold. Like I said: not my area of expertise. (I guess the upvotes on my previous comment mean: "yes, I was wondering that too...") $\endgroup$ Commented May 5, 2011 at 20:15
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    $\begingroup$ Yes, it's pretty easy that the suspension of a space $X$ cannot possibly be an $n+1$-manifold unless $X$ is homotopy equivalent to $S^n$. $\endgroup$ Commented May 6, 2011 at 0:37
  • $\begingroup$ @TomGoodwillie Yes, because there are two singular points in top and bottom which neighborhood is "cone over M" and not disk. Do you (or someone) know whether we can remove neighborhoods of these two points and glue sth which make $SM$ a manifold ? In the same time result should be similar to suspension of $SM$ e.g. it should be 1-connected. Is such construction known ? $\endgroup$
    – user21230
    Commented Oct 1, 2018 at 11:35

1 Answer 1


Suppose $M$ is a closed $n$-manifold whose suspension is homeomorphic to $S^{n+1}$. Removing the two "singular" points from the suspension gives $M\times \mathbb R$, while removing two points from $S^{n+1}$ gives $S^n\times\mathbb R$. Thus $M\times \mathbb R$ and $S^n\times\mathbb R$ are homeomorphic, which easily implies that $M$ and $S^n$ are h-cobordant, and hence $M$ and $S^n$ are homeomorphic.

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    $\begingroup$ You need $n>4$, though, don't you? $\endgroup$ Commented May 5, 2011 at 19:10
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    $\begingroup$ Topological h-cobordism theorem for simply-connected manifolds holds in all dimensions (due to Freedman in dimension 4, to Perelman in dimension 3, and to Newman in dimensions >4). $\endgroup$ Commented May 5, 2011 at 19:15
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    $\begingroup$ To see that $M$ and $S^n$ are h-cobordant consider a homeomorphism $h$ of their products with $\mathbb R$, and use excision in homology to show that the submanifolds $S^n\times 0$ and $h(M\times t)$ bound an h-cobordism, where $t$ need to be sufficiently large to ensure that the submanifolds are disjoint. $\endgroup$ Commented May 5, 2011 at 19:38
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    $\begingroup$ In fact, one need not involve h-cobordisms at all: just note that $M$ and $S^n$ are homotopy equivalent and use Poincare's conjecture. I guess, I just like to advertize that fact that if two closed manifolds become homeomorphic after multiplying by $\mathbb R$, then they are $h$-cobordant. :) $\endgroup$ Commented May 5, 2011 at 20:11
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    $\begingroup$ @Willie: but then what do we call it? It's not any one person's theorem... $\endgroup$ Commented May 5, 2011 at 21:25

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