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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.

James Cranch
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    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.     – Pete L. Clark May 05 '11 at 18:29
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    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. – James Cranch May 05 '11 at 18:54
  • 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...") – Pete L. Clark May 05 '11 at 20:15
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    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 $S^n$. – Tom Goodwillie May 06 '11 at 00:37
  • @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 ? –  Oct 01 '18 at 11:35

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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.

Igor Belegradek
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    You need $n>4$, though, don't you? – Benoît Kloeckner May 05 '11 at 19:10
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    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). – Igor Belegradek May 05 '11 at 19:15
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    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. – Igor Belegradek May 05 '11 at 19:38
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    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. :) – Igor Belegradek May 05 '11 at 20:11
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    I think we can stop calling it a "conjecture" now. :-) – Willie Wong May 05 '11 at 20:58
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    @Willie: but then what do we call it? It's not any one person's theorem... – Pete L. Clark May 05 '11 at 21:25
  • I'm no expert but perhaps it's not necessary to invoke all of these deep theorems to prove this. I'm fairly sure the following is true. Given an $n$-manifold $M$, consider the cone $CM$ on $M$. Then the vertex of $CM$ has a neighbourhood homeomorphic to an open set in $\mathbb{R}^{n+1}$ if and only if $M$ is the $n$-sphere. If this really is true then the only manifold whose suspension is again a manifold is $S^n$. Am I right here? – Joel Fine May 05 '11 at 22:43
  • @Joel Fine: I am not sure how to prove your "only if" assertion. – Igor Belegradek May 05 '11 at 23:54
  • @Igor, no me neither! I guess it's the case that $M$ must be a homotopy sphere and so it follows from Poincaré. But since we are assuming a homeomorphism it feels like we're giving ourselves strictly more information than in the Poincaré conjecture. Although I could easily be wrong on that. – Joel Fine May 06 '11 at 00:30
  • @Igor again, I'm sorry, having reread all of the comments more carefully I see that this line of reasoning was clearly already apparent to you! – Joel Fine May 06 '11 at 00:31
  • @IgorBelegradek: If I'm not mistaken, your argument also works smoothly: if $M$ and $N$ are closed smooth manifolds such that $M\times\mathbb{R}$ and $N\times\mathbb{R}$ are diffeomorphic, then $M$ and $N$ are smoothly $h$-cobordant. In particular, if $\Sigma$ is an exotic $n$-sphere, then $\Sigma\times\mathbb{R}$ and $S^n\times\mathbb{R}$ are not diffeomorphic. Is that correct or have I oversimplified? – Michael Albanese Mar 29 '22 at 09:51
  • @MichaelAlbanese: what you say is correct, except that the "in particular" part needs $n\ge 5$ so that the h-cobordism theorem applies (the argument does not apply to exotic 4-spheres if they exist). In fact, a similar argument applied twice shows that $\Sigma\times\mathbb R^2$ and $S^n\times\mathbb R^2$ are not diffeomorphic: if there were then $\Sigma\times S^1$ and $S^n\times S^1$ would be h-cobordant, hence diffeomorphic, and passing to the universal cover gives a diffeomorphism of $\Sigma\times\mathbb R$ and $S^n\times\mathbb R$. – Igor Belegradek Mar 29 '22 at 13:12
  • @IgorBelegradek: Right, I forgot about the potential existence of exotic 4-spheres. If I'm not mistaken, it follows from the fact that the tangent bundles of $\Sigma$ and $S^n$ are isomorphic that $\Sigma\times\mathbb{R}^{n+1}$ and $S^n\times\mathbb{R}^{n+1}$ are diffeomorphic. I wonder what the smallest value of $k$ is such that $\Sigma\times\mathbb{R}^k$ and $S^n\times\mathbb{R}^k$ are diffeomorphic. By your comment, $k > 2$. – Michael Albanese Mar 29 '22 at 23:08
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    @MichaelAlbanese: if $\Sigma$ bounds a parallelizable manifold, then $k=3$, see e.g. remark 5.8 in https://arxiv.org/abs/0912.4869. In general, the answers are more complicated, see Proposition 8.2 in https://arxiv.org/abs/1104.4136. Certainly, stable range is an overkill; metastable range suffices: if $2k\ge n+3$, then $\Sigma\times\mathbb R^k$ and $S^n\times\mathbb R^k$ are diffeomorphic. – Igor Belegradek Mar 30 '22 at 02:36
  • @MichaelAlbanese Actually my "metastable range" statement contradicts Proposition 8.2 mentioned above, so it is wrong. If $2k\ge n+3$ the obvious homotopy equivalence $\Sigma\to S^n\times\mathbb R^k$ is homotopic to a smooth embedding, and its tubular neighborhood is diffeomorphic to $S^n\times\mathbb R^k$ when $k\ge 3$. The tubular neighborhood is the total space of a vector bundle over $\Sigma$, but the bundle may not be trivial. It only is stably trivial. – Igor Belegradek Mar 30 '22 at 21:30
  • A proof of the fact that Igor Belegradek mentioned in the comments from 2011 (if two closed manifolds become homeomorphic after multiplying by $\mathbb{R}$, then they are h-cobordant), can be found in this answer, written following Igor's suggestions. – Michael Albanese Aug 07 '23 at 03:30