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I am currently working on the following problem :

Let $X$ be a topological space such that its suspension $\Sigma X$ is a n-manifold. Show that $H_k(X) \cong H_k(\mathcal{S}^{n-1})$ for all $k$.

I already proved that $H_k(X) \cong H_{k+1}(\Sigma X)$ for all $k$, using Mayer-Vietoris. I thought it could help me but then I couldn't find any way to get to my goal. My intuition is telling me that if $\Sigma X$ is a n-manifold, it means that $X$ has to be a (n-1)-manifold too but it has to be such that things are not getting too bad near the points $X \times \{0\}$ and $X \times \{1\}$ in the suspension. I have no idea how to prove it : do someone has a suggestion to do so ? Thank you.

KnitKnot
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Unfortunately $X$ does not have to be a manifold, see this: https://mathoverflow.net/questions/394391/suspension-of-a-topological-space

But we don't need it to be a manifold. Pick $y_0\in \Sigma X$ one of the ends. Then we have the long exact sequence of homologies:

$$\cdots\to H_k(\Sigma X-y_0)\to H_k(\Sigma X)\to H_k(\Sigma X,\Sigma X-y_0)\to H_{k-1}(\Sigma X-y_0)\to\cdots$$

Note that $H_*(\Sigma X-y_0)$ vanishes because $\Sigma X-y_0$ is a cone and thus contractible. This means $H_k(\Sigma X)\simeq H_k(\Sigma X,\Sigma X-y_0)$.

Finally for manifolds the local homologies are well known and easy to calculate via excision theorem:

$$H_k(\Sigma X,\Sigma X-y_0)\simeq H_{k-1}(S^{n-1})$$

which together with the fact $H_{k-1}(X)\simeq H_k(\Sigma X)$ gives us what you want.

freakish
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