[Reference Request] The Definition of Adjoint Functors between dg-categories Let $A$ and $B$ be two dg-categories, $F: A \rightarrow B$ and $G: B \rightarrow A$ are two functors. Then what is the definition that $F$ and $G$ form an adjoint pair?
In my mind $F\dashv G$ requires that there is a quasi-isomorphism between the cochain complext $\text{Hom}_B(Fx, y) $ and $\text{Hom}_A(x, Gy)$ for any $x\in \text{Obj}(A)$ and  $y\in \text{Obj}(B)$. Can we make it more precise?
 A: A good notion of adjointness in the dg-world should perhaps involve derived categories and quasi-functors.
Definition: A quasi-functor $F: \mathcal A \to \mathcal B$ is a dg-bimodule $F \in \mathcal A \text{-dgm-} \mathcal B$ (written $F=F(B,A)$, covariant in $A$ and contravariant in $B$), such that for all $A \in \mathcal A$, $F(-,A)$ is quasi-isomorphic (that is, isomorphic in the derived category $\mathbf{D}(\mathcal B)$) to a representable right $\mathcal B$-dg-module. As a matter of notation, I will sometimes write $F(A)$ as a shorthand for $F(-,A)$.
Note that any bimodule $F \in \mathcal A \text{-dgm-} \mathcal B$ can be viewed as a dg-functor $F: \mathcal B^{\text{op}} \otimes \mathcal A \to \mathbf{C}_{\text{dg}}(k)$, where $\mathbf{C}_{\text{dg}}(k)$ is the dg-category of cochain complexes over our ground commutative ring $k$. So, $F \in \text{dgm-}(\mathcal B \otimes \mathcal A^{\text{op}})$, a right $\mathcal (\mathcal B \otimes \mathcal A^{\text{op}})$-dg-module.
The category $\mathrm{rep}(\mathcal A,\mathcal B)$ of quasi-functors is defined to be the full subcategory of the derived category $\mathbf D(\mathcal B \otimes \mathcal A^{\text{op}})$ whose objects are the quasi-functors $\mathcal A \to \mathcal B$. Composition of quasi-functors $F: \mathcal A \to \mathcal B$ and $G: \mathcal B \to \mathcal C$ is obtained by the derived tensor product:
\begin{equation}
G \circ F = F \otimes_{\mathcal B}^{\mathbb L} G.
\end{equation}
As expected, given a quasi-functor $F: \mathcal A \to \mathcal B$, one has "composition functors" such as
\begin{equation*}
F^* : \mathrm{rep}(\mathcal B, \mathcal C) \to \mathrm{rep}(\mathcal A,\mathcal C).
\end{equation*}
Now, let $\tau : F \to G$ be a morphism of quasi-functors $\mathcal A \to \mathcal B$. We can speak of its "components" $\tau_A : F(A) \to G(A)$ (morphisms in $\mathbf{D}(\mathcal A)$), for any object $A \in \mathcal A$. In fact, an object $A \in \mathcal A$ can be viewed as an element of $\mathrm{rep}(\mathbf 1 , \mathcal A)$, where $\mathbf 1$ is the dg-category with an object and endomorphism ring $k$: we have the derived Yoneda embedding
\begin{equation*}
h_{-} : H^0(\mathcal A) \hookrightarrow \mathbf{D}(\mathcal A),
\end{equation*}
and we may view $h_A$ as a bimodule, $h_A \in \mathbf{1}\text{-dgm-}\mathcal A$. Next, recall that, given $F: \mathcal A \to \mathcal B$ any quasi-functor (or even bimodule), we have
\begin{equation*}
h_A \otimes_{\mathcal A}^{\mathbb L} F = h_A \otimes_{\mathcal A} F = F(A),
\end{equation*}
without the need of taking the derived tensor product, since $h_A$ is cofibrant as a right $\mathcal A$-module. In other words:
\begin{equation*}
h_A^*(-) = h_A \otimes_{\mathcal A} - : \mathrm{rep}(\mathcal A, \mathcal B) \to \mathrm{rep}(\mathbf 1,\mathcal B),
\end{equation*}
and this functor sends a morphism $\tau: F \to G$ to what we call its component $\tau_A : F(A) \to G(A)$, as needed.
Finally, I may write down my definition of adjoint quasi-functors. If $F: \mathcal A \to \mathcal B$ and $G: \mathcal B \to \mathcal A$ are quasi-functors, then we could say that $F \dashv G$ if there is a morphism of quasi functors $\varepsilon : 1 \to GF$ such that the following composition
\begin{equation}
\mathrm{rep}(\mathbf 1, \mathcal B)(F(A), B) \xrightarrow{G_*} \mathrm{rep}(\mathbf 1, \mathcal A)(GF(A), G(B)) \xrightarrow{\mathrm{Hom}(\varepsilon_A, G(B))} \mathrm{rep}(\mathbf 1, \mathcal A)(A, G(B))
\end{equation}
is an isomorphism, for all $A \in \mathcal A, B \in \mathcal B$.
where $A \in \mathcal A$ and $B \in \mathcal B$ are again shorthands for $\mathcal A(-,A) \in \mathrm{rep}(\mathbf 1, \mathcal A)$ and $\mathcal B(-,B) \in \mathrm{rep}(\mathbf 1, \mathcal B)$.
