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Let $M$ be a , connected, compact, $\mathbb{Z}$-orientable topological $n$-manifold. The latter $\mathbb{Z}$-orientable means that $H_n(M; \mathbb{Z})=\mathbb{Z} $ holds.

Let futhermore $M$ be simply connected and it is homotopy equivalent to a suspesion $\Sigma X$ of path connected space $X$. (btw: I'm not sure if we need here this information)

Why in this case $M$ has the same homology groups like $S^n$?

Therefore $H_i(M, \mathbb{Z})= \mathbb{Z}$ if $i=0,n$ and otherwise $H_i(M, \mathbb{Z})=0$.

My attemps:

Let abbreviate $H_i(M; \mathbb{Z}) = H_i(M)$.

Connectedness implies $H_0(M)=\mathbb{Z}$. Since $M$ simple connected Hurewicz implies $0 =\pi_1(M) \twoheadrightarrow H_1(M)$.

Futhermore - since $M$ orientable - $H_n(M)$ and $H_{n-1}(M) =\mathbb{Z}^{k}$ free. But from here I stuck in calculating other homology groups.

I guess I can argue by induction on $i$. The cases $i =0,1,n$ are ok and wlog I can assume that $n >2$ otherwise I finish.

Can anybody help me to get the induction step? I tried using Universal Coefficient Thm reducing the calculation to $H_i(X; F)$ where $F= \mathbb{Z}/p$ field in order to use Poincaré. But also here without success.

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    $\begingroup$ A suspension has no nontrivial cup products. But the cohomology of a closed orientable manifold satisfies Poincare duality. $\endgroup$ – Qiaochu Yuan Oct 21 '18 at 21:51
  • $\begingroup$ @QiaochuYuan: Ah you mean it in the sense that because by Poincare duality the cup product $\cup$ gives a perfect pairing $\cup: H^p(M) \times H^{n-p}(M) \to H^n(M) =\mathbb{Z}$ but on the other hand $\cup$ is trivial, then all $H^p(M)$ are already trivial? $\endgroup$ – KarlPeter Oct 21 '18 at 22:13
  • $\begingroup$ Basically, but you need to be a little careful about torsion. $\endgroup$ – Qiaochu Yuan Oct 21 '18 at 22:51
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    $\begingroup$ @QiaochuYuan: Yes, right, in case of coefficients in $\mathbb{Z}$ they should be quotiented out. But I can firstly use this for $H^p(M; F)=0$ with $F= \mathbb{Z}/p$ for all primes $p$. Using UCT and classification thm for finitely generated abelian groups I can conclude that $H_p(M)$ should also vannish. The only problem seems to show that all $H_p(M)$ are finitely generated. But I guess that compactness of $M$ makes the job... $\endgroup$ – KarlPeter Oct 21 '18 at 22:59
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    $\begingroup$ Yes, that's a complete proof. $\endgroup$ – user98602 Oct 22 '18 at 1:02

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