Tensoring with flat complexes preserves quasi-isomorphisms

Proof. Given a quasi-isomorphism \(f\colon C \to D\), we need to show that the map \(F_{\bullet } \otimes f\) is again a quasi-isomorphism. To this end, filter \(F_{\bullet }\) by the subcomplexes \(F^{\leq n}\), where \((F^{\leq n})_k = F_k\) for \(k \leq n\) and \(0\) otherwise. We have degreewise split short exact sequences \(0 \to F^{\leq n-1} \to F^{\leq n} \to F_n[n] \to 0\). The map \(F_n[n] \otimes f\) is a quasi-isomorphism since \(F_n\) is flat. By induction on \(n\) and the five lemma, each \(F^{\leq n} \otimes f\) is a quasi-isomorphism. Since \(F_{\bullet }, C_{\bullet }\) and \(D_{\bullet }\) are all bounded below, the sequence \((F^{\leq n} \otimes C)_k\) is eventually constant with value \((F \otimes C)_k\) in every degree \(k\), and similarly for \(D\). The map \(F_{\bullet } \otimes f\) is therefore a degreewise colimit of the maps \(F^{\leq n} \otimes f\) along which homology is eventually constant, and so it is a quasi-isomorphism. □