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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

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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 '11 at 18:29
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 '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 5 '11 at 20:15
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 '11 at 0:37
up vote 32 down vote accepted

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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You need $n>4$, though, don't you? – Benoît Kloeckner May 5 '11 at 19:10
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 '11 at 19:15
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 '11 at 19:38
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 '11 at 20:11
@Willie: but then what do we call it? It's not any one person's theorem... – Pete L. Clark May 5 '11 at 21:25

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