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I am reading proof of Theorem $5.3$ in Chapter $2$ of "Cohomology of groups by Brown" on page number $42$. I have a problem with understanding the following consequence given in the proof:

Let $G= F/R$ where $F$ is a free group over the set $S$. Let $Y$ be a wedge sum of $|S|$ many circles and $\tilde{Y}$ to be the connected regular covering of $Y$ corresponding to the normal subgroup $R$ of $F$. Consider the exact sequence $C_1{\tilde{Y}} \to C_0{\tilde{Y}} \to \mathbb{Z} \to 0$ of free $\mathbb{Z}[G]$ modules. Then $H_2(G) \cong \ker\{ (H_1(\tilde{Y})_G \to H_1(Y)\}.$

I don't understand why $H_2(G) \cong \ker\{ (H_1(\tilde{Y})_G \to H_1(Y)\}$?

The author has used the following result: Let $F_n \to \cdots \to F_0 \to \mathbb{Z} \to 0$ be an exact sequence of $\mathbb{Z}[G]$-modules where each $F_i$ is projective. Then $$0 \to H_{n+1}(G) \to (H_nF)_G \to H_n(F_G) \to H_n(G) \to 0$$ is exact.

So using the above result I can only see that $$H_2(G) \cong \ker\{ H_1(\tilde{Y})_G \to \ker\big({C_1\tilde{Y}\otimes_{\mathbb{Z}[G]} \mathbb{Z} \to C_0\tilde{Y}\otimes_{\mathbb{Z}[G]} \mathbb{Z}\big)}\}.$$

Can someone tell what I am missing?

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    $\begingroup$ I'm pretty sure that the thing you wrote on the right is $H_1(Y)$, so I don't think you are wrong $\endgroup$ Commented Feb 26, 2020 at 18:25
  • $\begingroup$ @AndresMejia: I don't know how to prove it. $\endgroup$
    – eyp
    Commented Feb 27, 2020 at 6:05

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