(Related question: [What part of the fundamental group is captured by the second homology group?](https://mathoverflow.net/questions/40399/what-part-of-the-fundamental-group-is-captured-by-the-second-homology-group))


Suppose I have a path-connected space $X$ for which $\pi\_1(X)=\mathbb{Z}$. Suppose I know $\pi\_2(X)$ as a $\pi\_1(X)$-module, and I want to compute $H\_2(X)=H\_2(X;\mathbb{Z})$.


**Claim:** $H\_2(X)$ is isomorphic to the largest quotient of $\pi\_2(X)$ on which $\pi\_1(X)$ acts trivially (ie the group of coinvariants).


One can prove this by passing to universal covers (assuming $X$ is locally path-connected, blah, blah, blah). The Cartan-Leray spectral sequence of the regular cover $\tilde{X}\to X$ degenerates at the $E\_2$ page (since $K(\mathbb{Z},1)$ is a circle) and it follows that $$H\_2(X)\cong H\_0(\mathbb{Z};H\_2(\tilde{X}))\cong H\_0(\mathbb{Z};\pi\_2(\tilde{X}))\cong H\_0(\mathbb{Z};\pi\_2(X)).$$
This tells me the answer, but teaches me nothing.


**Question:** Is there a more elementary proof of this fact? (I am prepared to accept that "elementary" might still involve Postnikov towers, obstruction theory, etc)


**Edit:** Many thanks for all the helpful answers, I wasn't expecting four alternatives! A remark and a question:


1. This is true with an arbitrary free group replacing $\mathbb{Z}$, and
2. it seems to be well known. Has anyone seen it written down anywhere?



Computing H_2 from pi_1=Z and pi_2
Here's one way. Suppose first that one has a Serre fibration $F\hookrightarrow E \to S^1$ with $F$ simply connected. Then $\pi\_2(E)=\pi\_2(F)$ by the exact sequence of homotopy groups, and $\pi\_2(F)=H\_2(F)$ by Hurewicz.


The map $H\_2(F)\to H\_2(E)$ is surjective, and its kernel is the image of $\mathrm{id}-\phi$, where $\phi$ is the action of the generator of $\pi\_1(S^1)$ on $H\_\ast(F)$. This is by the Wang exact sequence and the vanishing of $H\_1(F)$. So $H\_2(E)$ is isomorphic to $H\_2(F)\_{\pi\_1(S^1)}$ (coinvariants) and hence to $\pi\_2(E)\_{\pi\_1(E)}$.


If $X$ is (say) a path-connected CW complex with $\pi\_1(X)=\mathbb{Z}$ then its universal cover is classified by a map $X\to B\mathbb{Z}=S^1$, and we can convert this by the usual procedure into a Serre fibration $F\hookrightarrow E \to S^1$. The mapping fibre, $F$, is simply connected, and $E$ is homotopy equivalent to $X$. Now apply the previous argument. 


