I am recently trying to learn the language of DG categories and I have a question concerning the notion of quasi-equivalence.
According to the definition, which you can find for instance on Keller's paper "On differential graded categories", for a given DG functor $F \colon \mathcal{A} \rightarrow \mathcal{B}$ to be a quasi-equivalence means that for all $X,Y \in \mathrm{Ob}(\mathcal{A})$ the induced map
$$
F_{X,Y} \colon \mathrm{Hom}_{\mathcal{A}}^{\bullet}(X,Y) \rightarrow \mathrm{Hom}_{\mathcal{B}}^{\bullet}(F(X),F(Y))
$$
of chain complexes is a quasi-isomorphism, and moreover that the induced functor
$$
H^0(F) \colon H^0(\mathcal{A}) \rightarrow H^0(\mathcal{B})
$$
on the level of categories is essentially surjective.
Now, in ordinary category theory, a given functor $F \colon \mathcal{A} \rightarrow \mathcal{B}$ is an equivalence of categories if and only if one can find another functor $G \colon \mathcal{B} \rightarrow \mathcal{A}$ such that $F \circ G \cong \mathrm{id}_{\mathcal{B}}$ and $G \circ F \cong \mathrm{id}_{\mathcal{A}}$. This so-called “quasi-inverse'' turns out to be unique up to natural equivalence.
My question is the following: Given a quasi-equivalence $F \colon \mathcal{A} \rightarrow \mathcal{B}$ is it possible to find a quasi-equivalence $G \colon \mathcal{B} \rightarrow \mathcal{A}$ together with DG natural isomorphisms $F \circ G \cong \mathrm{id}_{\mathcal{B}}$ and $G \circ F \cong \mathrm{id}_{\mathcal{A}}$?
Here, by a DG natural isomorphism between two DG functors $F,G \colon \mathcal{A} \rightarrow \mathcal{B}$, I mean a DG natural transformation $\varphi \colon F \Rightarrow G$ of degree $0$, as defined for instance in Genovese's paper "The uniqueness problem of dg-lifts and Fourier-Mukai kernels", such that $\varphi(X)$ is an isomorphism for all $X \in \mathrm{Ob}(X)$.
Best Answer
LPK's example in the comments can be upgraded to a counterexample to exactly this situation. Consider two DG-categories $A,B$, both with two objects $0,1$ and morphisms $A(0,0)=A(1,1)=B(0,0)=B(1,1)=\mathbb{Z}$, $A(1,0)=B(1,0)=0$, and finally $A(0,1)=\mathbb{Z}\to \mathbb{Z}$ while $B(0,1)=\mathbb{Z}/2$. Then there is a quasi-equivalence $A\to B$ which is the identity on objects and has the usual quasi-isomorphism $A(0,1)\to B(0,1)$ as its only nontrivial morphism action. This is not invertible since $A(0,1)\to B(0,1)$ is not. The problem is that $A$ is not bifibrant as a DG category. This is the model category theoretic condition that, generally, realizes weak equivalences (such quasi-equivalences) as homotopy equivalences-legitimately weakly invertible morphisms. The bifibrant DG-categories are difficult to describe fully explicitly, but at the very least the hom-complexes must be levelwise projective (or perhaps injective, in a different model structure.)