# Homology Groups of Oriented Simply Connected Manifold

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.

• A suspension has no nontrivial cup products. But the cohomology of a closed orientable manifold satisfies Poincare duality. – Qiaochu Yuan Oct 21 '18 at 21:51
• @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? – KarlPeter Oct 21 '18 at 22:13
• Basically, but you need to be a little careful about torsion. – Qiaochu Yuan Oct 21 '18 at 22:51
• @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... – KarlPeter Oct 21 '18 at 22:59
• Yes, that's a complete proof. – user98602 Oct 22 '18 at 1:02