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I have been studying $3$-manifolds recently and I got stuck in the following situation. For lens spaces the below fact is true.

Let $G$ be a finite group acting freely and orthogonally on $S^3$ so that $S^3/G$ is a spherical $ 3$ manifold. Now construct $K(G,1)$ whose CW structure has a $3$rd skeleton $X^3$ as $S^3/G$. Can I say that the attaching map in the cellular chain $H_4(X^4, X^3) \to H_3(X^3, X^2)$ is zero? I want to show that $H_3(G; \mathbb{Z})\neq 0$.

Any reference will be helpful. 

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    $\begingroup$ Wouldn't this imply that the third homology is $\mathbb Z$ in the $S^3/G$ orientable case? $\endgroup$
    – Will Sawin
    Commented Nov 22, 2022 at 20:10
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    $\begingroup$ @WillSawin: there are no non-orientable spherical 3-manifolds. $\endgroup$
    – Ryan Budney
    Commented Nov 22, 2022 at 21:14
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    $\begingroup$ I don't think it's true for lens spaces. $\pi_3 \mathbb{R}P^3$ is the integers, but $\pi_3 \mathbb{R}P^\infty$ is trivial, similarly, $H_3 \mathbb{R}P^\infty$ is $\mathbb Z_2$. $\endgroup$
    – Ryan Budney
    Commented Nov 22, 2022 at 21:30

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The attaching map has to kill $\pi_3(S^3/G)$, and the map $\mathbb Z =\pi_3(S^3) \to \pi_3 (S^3/G) $ induced by the covering is an isomorphism, so $\pi_3$ is generated by the class of the covering 3-sphere. The attaching map will attach a single 4-cell with boundary the covering 3-sphere.

The induced homology class is the fundamental class of the covering 3-sphere, which is $|G|$ times the generator of $H_3(S_3/G)$.

So I think the map you want is not $0$ but rather is multiplication by $|G|$.

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