I am trying to do exercise 9.19 in Rotman's Intro to Homolgical Algebra. Admittedly, I have skipped much of the content leading up to it (which is probably causing my issues). Here's where I'm at:
Let $G$ and $Q$ be groups such that $\mathbb{Z}G \cong \mathbb{Z}Q$, and let $\varphi: \mathbb{Z}G \rightarrow \mathbb{Z}Q$ be a ring $\textbf{homomorphism}$, (even though these rings are isomorphic). For any $Q$-module $K$, we can view $K$ as a $G$-module where the action is induced by $\varphi$, and we denote this $G$-module by ${}^{\varphi}K$. I am tasked with proving that $H_n(G, {}^{\varphi}K) \cong H_n(Q, K)$.
I began with a projective resolution of right $\mathbb{Z}Q$-modules and the corresponding sequence of tensor products: $$... \overset{}{\longrightarrow} R_2 \overset{d_2}{\longrightarrow} R_1 \overset{d_1}{\longrightarrow} R_0 \overset{\epsilon}{\longrightarrow} \mathbb{Z} \longrightarrow 0 $$ $$ (1) ... \overset{}{\longrightarrow} R_2 \otimes_{\mathbb{Z}Q} K\overset{d_2\otimes 1_{K}}{\longrightarrow} R_1\otimes_{\mathbb{Z}Q} K \overset{d_1\otimes 1_k}{\longrightarrow} R_0 \otimes_{\mathbb{Z}Q} K \longrightarrow 0 $$
Now, we can define right projective modules $P_i$ to be the $R_i$ viewed as $\mathbb{Z}G$-modules, which is induced by the isomorphism $\psi: \mathbb{Z}G \overset{\cong}{\rightarrow} \mathbb{Z}Q$, so that $p_i \cdot g := p_i \cdot \psi(g) = r_i \cdot \psi(g)$. Clearly, the $P_i$ are projective and we have a projective resolution and a corresponding sequence of tensor products:
$$... \overset{}{\longrightarrow} P_2 \overset{d'_2}{\longrightarrow} P_1 \overset{d'_1}{\longrightarrow} P_0 \overset{\epsilon'}{\longrightarrow} \mathbb{Z} \longrightarrow 0 $$ $$ (2) ... \overset{}{\longrightarrow} P_2 \otimes_{\mathbb{Z}G} {}^{\varphi}K \overset{d'_2\otimes 1_{{}^{\varphi}K}}{\longrightarrow} P_1\otimes_{\mathbb{Z}G} {}^{\varphi}K \overset{d'_1\otimes 1_{{}^{\varphi}K}}{\longrightarrow} P_0 \otimes_{\mathbb{Z}G} {}^{\varphi}K \longrightarrow 0 $$ Since homology is independent of the choice of resolution, my plan was to show that these two tensor chain complexes (denoted (1) and (2) above) are isomorphic, (or in this case, is it more correct to show quasi-isomorphism?), and then since isomorphic chains produce the same homology, I would be done.
Does this strategy make sense? To prove (quasi-)isomorphism, is it sufficient to find an isomorphism $f_n : R_n \otimes_{\mathbb{Z}Q}K \rightarrow P_n \otimes_{\mathbb{Z}G}{}^{\varphi}K$ for each $n$? What would this even mean, since aren't these modules over two different (albeit isomorphic) rings? My best guess would have to be: $$f_n(r_n \otimes k) = p_n \otimes \tilde{k},$$ where $p_n$ is simply $r_n$ with $\mathbb{Z}G$ action induced by $\psi$ and $\tilde{k}$ is simply $k$ with $\mathbb{Z}G$ action induced by $\varphi$. Wouldn't the fact that $\varphi$ is merely a homomorphism and not an isomorphism prevent this $f_n$ from being an isomorphism?
I can't seem to find much information about (quasi-)isomorphism of chain complexes. Any advice/info/reference would be greatly appreciated.