In the book I'm reading now the author says that it's easy to prove the equality $\operatorname{Tor}_i(A, B) = \operatorname{Tor}_i(B, A)$ by doing these manipulations:
(i) Let's consider projective resolutions $\mathcal P$ and $\mathcal Q$ of $A$ and $B$, respectively.
(ii) We can show that $\operatorname{Tor}_i(A, B) = H_i(\mathcal P \otimes \mathcal Q)$.
(iii) Now "commutativity" of $\rm Tor$ is an easy corollary of the fact that $\mathcal P \otimes \mathcal Q = \mathcal Q \otimes \mathcal P$.
The proposition (ii) isn't clear for me... Can anyone help me with the understanding of its proof?
Here's an outline. Let $A$ and $B$ be $R$-modules and $\mathcal{P}$ and $\mathcal{Q}$ be projective resolutions of them respectively. We have quasi-isomorphisms $$ \mathcal{P} \otimes_R B \leftarrow \mathrm{Tot}(\mathcal{P} \otimes_R \mathcal{Q}) \rightarrow A \otimes_R \mathcal{Q} $$ which in turn induce isomorphisms $$ H_i(\mathcal{P} \otimes_R B) \cong H_i(\mathrm{Tot}(\mathcal{P} \otimes_R \mathcal{Q})) \cong H_i(A \otimes_R \mathcal{Q}). $$ But then by definition of Tor, we have $\mathrm{Tor}_i^R(A, B) \cong H_i(\mathrm{Tot}(\mathcal{P} \otimes_R \mathcal{Q}))$.