# Easier proof about suspension of a manifold

For what manifolds $M$ is the suspension $\Sigma M$ also a manifold?

By the suspension of a topological space $X$ (not necessarily a manifold), I mean the space $$\Sigma X = (X \times [0,1])/{\sim}$$ where $\sim$ is the equivalence relation which glues $X \times \{0\}$ to a point $p$ and $X \times \{1\}$ to a point $q$. The topology is of course given by the quotient topology.

If $M = S^n$ is a sphere, then it is easy to see that $\Sigma M$ is homeomorphic to $S^{n+1}$, and hence a manifold.

Claim: The only possibility is that $M$ is a sphere.

I can prove this statement, but it relies on some pretty hefty results. So my question is the following:

Is there an easier proof than the following? In particular, can we remove the use of the double suspension theorem or the $h$-cobordism theorem / Poincaré Conjecture?

Proof: Suppose $M$ and $\Sigma M$ are manifolds of dimensions $n$ and $n+1$ respectively. This is obviously true locally everywhere except at $p$ and $q$, so we need only consider that some neighbourhood of $p$ is homeomorphic to a disk (by symmetry, this implies the same result for $q$).

Pick a neighbourhood $U = (M\times [0,\epsilon))/{\sim}$ of $p$. By excision and since $\Sigma M$ is a manifold, we have $$\widetilde{H_*}(S^{n+1}) \cong H_*(\Sigma M, \Sigma M \setminus \{p\}) \cong H_*(U,U \setminus \{p\}).$$ Now, $U \setminus \{p\} = M \times (0,\epsilon)$ deformation retracts to $M$, and so we can compute this by taking the LES for the pair $(U,M)$. Since $U$ is contractible, every third term of this sequence (except along the $0^{\mathrm{th}}$ cohomology) is $0$, and we easily can conclude that $$H_{k+1}(\Sigma M, \Sigma M \setminus \{p\}) \cong H_k(M)$$ for $k > 0$. Therefore, $H_k(M) = H_{k+1}(S^{n+1})$ for $k > 0$. And we know $H_0$ is just $\mathbb{Z}$ (since $H_n \cong \mathbb{Z}$, Poincaré duality implies $H^0 \cong \mathbb{Z}$ and so there is only one component). In particular, this proves that $M$ is a homology sphere.

That's all relatively standard, and all well and good. Here's the part that gets a little bit crazy. Since $M$ is a homology sphere, the double suspension theorem yields that $\Sigma \Sigma M$ is a sphere. So we need that $\Sigma M$ suspends to a sphere. By this MO question, the answer of which appears to rely either on $h$-cobordism or the Poincaré Conjecture (choose your poison), it follows that $\Sigma M$ is therefore a sphere, and applying this same reasoning to $\Sigma M$, it follows that $M$ is a sphere.

• No, there's something a bit more subtle in that last paragraph. In fact, if a homology sphere is not a sphere, then $\Sigma M$ is not a manifold, and so we cannot apply the reasoning of the last paragraph. (See the first comments on the MO question.) The point is that we needed $\Sigma M$ to be a manifold to conclude that it was a sphere, and then for $M$ after that. Commented May 7, 2014 at 13:35
• Yes, you're right, I realized afterwards. Anyway since MO is supposed to be "higher level" than MSE, you might get better answers there, considering you quote results from an MO question... Commented May 7, 2014 at 13:36
• I was more curious if there was some more elementary solution. I figured MSE would be a better place to start for that, but I suppose it could be moved later. Commented May 7, 2014 at 13:37
• I changed (X\times[0,1])/\sim to (X\times[0,1])/{\sim}. Sincee "\sim" is used as a binary relation symbol like "=" or "<", it has preceding and following spaces that should not be there when it's not used as a binary relation symbol. Thus: $(X\times[0,1])/\sim$ versus $(X\times[0,1])/{\sim}$. ${}\qquad{}$ Commented May 7, 2014 at 15:50
• Cheers! TeXing was never my strong point. Commented May 7, 2014 at 16:44

You don't need the double suspension theorem, because $M$ is a homotopy sphere (hence, by the Poincaré conjecture, it is homeomorphic to the sphere). To prove that, it is enough to show that $M$ is simply-connected (assume that $\dim M \geq 2$ as otherwise there is nothing to prove). Then it follows from your computation of homology and the Whitehead theorem that a degree one map $M \rightarrow S^n$ is a homotopy equivalence.
The suspension $\Sigma M$ is simply-connected by the van Kampen theorem, because it is a union of two contractible open cones over $M$, whose intersection $M \times (-\epsilon,\epsilon)$ is connected. Since $\Sigma M$ is a manifold of dimension at least $3$, removing a finite number of points from $\Sigma M$ doesn't change its fundamental group (again, this is a consequence of the van Kampen theorem and the fact that $S^k$ is simply-connected for $k \geq 2$). In particular, $\Sigma M \setminus \{p,q\} = M \times (0,1)$ is simply-connected, and so is $M$.