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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.

flag	asked May 5 2011 at 17:58 James Cranch 1,176●3●14

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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 5 2011 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 5 2011 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 5 2011 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 6 2011 at 0:37

Igor Belegradek · Accepted Answer · 2011-05-05 19:06:15Z UTC

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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.

answered May 5 2011 at 19:06

Igor Belegradek
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You need $n>4$, though, don't you? – Benoît Kloeckner May 5 2011 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 5 2011 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 5 2011 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 5 2011 at 20:11

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@Willie: but then what do we call it? It's not any one person's theorem... – Pete L. Clark May 5 2011 at 21:25

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