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Let $X$ be a free $G$-CW-complex with $G$-equivariant cell filtration by $n$-skeleta $X_0 \subset \dots \subset X_n \subset \dots \subset X$ (for rigorous definition see Chap. II, p. 98 in linked source below).

Question: Why holds

$$ H_n(X_n, X_{n-1}) \otimes_{\mathbb{Z}G} \mathbb{Z} \cong H_n(X_n/G, X_{n-1}/G) $$

for the relative homology groups (which by construction carry $\mathbb{Z}G$-module structure)? Clearly, the relative complexes $C_*(X_n, X_{n-1})$ are free $\mathbb{Z}G$-modules more less by assumption on freeness of $G$ action & compatibility of cell filtration with $G$. But does this hold for homology groups too? (here I have the universal coefficient theorem in mind)

If the relative homology group is not $\mathbb{Z}G$-free, why should else the isomorphism above be true?

Source: This was stated without a proof in Proposition 9.7. (ii) (page 165) in Transformation Groups by Tammo tom Dieck.
This suggests that at all this should follow as a "triviality" but honestly I not see an immediate reason for this.

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    $\begingroup$ The differential on $C_*(X_n,X_{n-1})$ is zero, since everything is concentrated in dimension $n$. Thus you can freely replace $H_n$ by $C_n$. $\endgroup$
    – HenrikRüping
    Commented Sep 19, 2023 at 10:01
  • $\begingroup$ Henrik's comment looks like an answer to me. $\endgroup$
    – IJL
    Commented Sep 21, 2023 at 11:38
  • $\begingroup$ Ok I post it as an answer, so that the question is closable. $\endgroup$
    – HenrikRüping
    Commented Sep 21, 2023 at 13:35

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The differential on $C_*(X_n,X_{n−1})$ is zero, since everything is concentrated in dimension $n$. Thus you can freely replace $H_n$ by $C_n$.

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