# If a manifold suspends to a sphere…

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.

• 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
• @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 ? – user21230 Oct 1 '18 at 11:35

## 1 Answer

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.

• 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