$G$-equivariant quasi-ismorphisms over a field

Question

Let $(A, d_A)$ and $(B, d_B)$ be cochain complexes over a field $\mathbb{k}$, and let $f: A \to B$ be a quasi-isomorphism. It is well-known that over a field, every quasi-isomorphism is chain homotopically invertible, hence there exists a chain map $g: B \to A$ and chain homotopies $h_A: A \to A$ and $h_B: B \to B$ such that $$ fg - \text{id}_B = d_B h_B + h_B d_B$$ $$ gf - \text{id}_A = d_A h_A + h_A d_A$$ My question is about whether this extends to the equivariant setting. Namely let $G$ be a finite group acting on $A$ and $B$ by chain maps. Assume that $f: A \to B$ is a $G$-equivariant quasi-isomorphism. Is $f$ invertible up to a $G$-equivariant homotopy? That is, can $g$, $h_A$, $h_B$ be chosen to be $G$-equivariant maps?

Answer 1

Thanks to Pedro's very helpful comment, I can answer my own question.

Let $\mathbb{k}$ be a field whose characteristic does not divide the order of $G$ (e.g. $\text{char}(\mathbb{k}) = 0$). Then any $\mathbb{k}$-linear map $\phi: C \to D$ between two $G$-vector spaces $C \to D$ can be averaged in the following way: $$ \phi \mapsto \overline{\phi}: C \to D $$ $$ \overline{\phi}(v) = \frac{1}{|G|} \sum_{a \in G}a \phi a^{-1} (v) $$ It is easy to verify that $\overline{\phi}$ is $G$-equivariant. Note that $\frac{1}{|G|}$ is well-defined because $\text{char}(\mathbb{k})$ does not divide $|G|$.

Then given $f, g, h_A, h_B$ as in the question, $\overline{g}, \overline{h_A}, \overline{h_B}$ are equivariant maps that give a chain homotopy inverse to $f$.

In fact, the premise that underpins the fact that quasi-isos over a field are chain homotopy equivalences is the lifting property of projective objects, and it is also easy to show that this lifting property extends to an equivariant lifting property by averaging (when $\text{char}(\mathbb{k})$ does not divide $|G|$). So the entire "story" of quasi-isomorphisms over a field being chain homotopy equivalences extends $G$-equivariantly.