While reading Weibel's Homological Algebra, on pg. 391 he considers an additive functor $F:\mathcal{A}\to\mathcal{B}$ between abelian categories, and writes "If $F$ is not exact, then the induced functor $\mathbb{K}(\mathcal{A})\to\mathbb{K}(\mathcal{B})$ will not preserve quasi-isomorphisms."
I did some digging, and based on this MO post, I assume it is true that
if $F:\mathcal{A}\to\mathcal{B}$ is additive between abelian categories, the induced functor $\mathbb{K}(\mathcal{A})\to\mathbb{K}(\mathcal{B})$ on the homotopy categories will preserve quasi-isomorphisms if and only if $F$ is exact.
It seems mentioned in passing in a lot of sources, but not proven anywhere. Does anyone have a proof of this proposition or reference to a place where it is? (If it's indeed true, that is.)
$\Leftarrow$: if $F$ is exact, then it commutes with taking homology; in particular, it preserves acyclic complexes. Now use the fact that a map is a quasi-isomorphism iff its mapping cone is acyclic and that any functor on chain complexes induced from a functor between abelian categories preserves mapping cones.
$\Rightarrow$: we'll prove the contrapositive. Suppose $0 \to a \to b \to c \to 0$ is a short exact sequence not preserved by $F$, so $F(a) \to F(b) \to F(c)$ is not exact. Now, a three-term complex is exact iff it's acyclic iff it's quasi-isomorphic to $0$, so $F$ does not preserve quasi-isomorphisms.