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If a short question: I have a closed orientable 3-Manifold, which has a perfect fundamental group.

therfore $H_1(M)=0,H_0(M)=H_3(M)=\mathbb{Z}$ somehow it seems like i can conclude that $H_2(M)=0$, but i don't know how to conclude this. by Poincaire duality I could say that $b_2=0$ but why cant there be Torsion.

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  • $\begingroup$ You can also use the universal coefficient theorem. $\endgroup$ – Angina Seng May 7 '17 at 16:08
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The torsion parts of $H^3(M) = \mathbb{Z}$ and $H_2(M)$ coincide by the universal coefficient theorem.

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  • $\begingroup$ More generally, if $M$ is a closed $n$-dimensional orientable manifold, $H_{n-1}(M; \mathbb{Z})$ is free. $\endgroup$ – Michael Albanese May 7 '17 at 17:16
  • $\begingroup$ Sure: The proof is the same. (I think it's also an exercise or minor result in Hatcher.) $\endgroup$ – anomaly May 7 '17 at 22:18

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