TITLE: dg-categories and fully faithful functor
QUESTION [6 upvotes]: dg: is for differential graded 
Suppose that $F: C\rightarrow D$ is a dg-functor between small dg-categories such that:

F: Objects of $C$ $\rightarrow$ Objects of $D$ is injective. 
$Hom_{C}(a,b)\rightarrow Hom_{D}(F(a),F(b))$ induces an isomorphism in homology for any $a, b \in C$

Let $\hat{C}$ be the category of dg-$C$-modules. Is it true that the induced dg-functor 
$$\hat{F}:\hat{C}\rightarrow \hat{D}$$  is faithfull in the sense that 
for any $x,y\in \hat{C} $, $Hom_{\hat{C}}(a,b)\rightarrow Hom_{\hat{D}}(\hat{F}(a),\hat{F}(b))$ induces an isomorphism in homology. 
Edit: I'm also interested in the particular case when $F:C\rightarrow D$ is an embedding of dg-categories e.g. when $C$ is a full dg-subcategory of $D$.

REPLY [2 votes]: Let me start with an answer for your last claim, namely, that a fully faithful dg-functor $F \colon \mathcal C \to \mathcal D$ induces a fully faithful dg-functor $\hat{F} \colon \hat{\mathcal C} \to \hat{\mathcal D}$ between the dg-categories of right dg-modules. This is true, and the reason lies in general facts about Kan extensions. In any case, a sketch of proof would be as follows.
First, $\hat{F}$ is usually defined as the "induction" dg-functor $\operatorname{Ind}_F$, and in particular
$$
\operatorname{Ind}_F(M)(D) = M \otimes_{\mathcal C} \mathcal D(D,F(-)) = \int^C M(C) \otimes\mathcal D(D,F(C)),
$$
if you also like coends. To show that $\operatorname{Ind}_F$ is fully faithful, we first recall that it is by definition the left adjoint of the restriction dg-functor $\operatorname{Res}_F \colon \hat{\mathcal D} \to \hat{\mathcal C}$, which maps $N$ to $N \circ F$. There is a unit morphism
$$
M \to \operatorname{Res}_F \operatorname{Ind}_F(M),
$$
and if we check that this is an isomorphism, then $\operatorname{Ind}_F$ will be fully faithful. But now we can compute:
\begin{align*}
\operatorname{Res}_F \operatorname{Ind}_F(M)(C) &= M \otimes_{\mathcal C} \mathcal D(F(C),F(-)) \\
&\cong M \otimes_{\mathcal C} \mathcal C(C,-)\\
& \cong M(C),
\end{align*}
where I used fully faithfulness and for the last step the "co-Yoneda lemma" or "density theorem".
Now, your main claim that a quasi-fully faithful dg-functor induces something which is again quasi-fully faithful between the dg-category of dg-modules is a little more nuanced. Working with this "cohomological" assumptions, you'll want derived stuff all around. So, instead of working with dg-modules, it's best to work with h-projective/cofibrant/semi-free dg-modules, namely your favourite version of the "derived dg-category". Let us work with the dg-category of h-projective dg-modules, just to fix ideas. What is true is that if you have a quasi-equivalence $G \colon \mathcal A \to \mathcal B$ between dg-categories, then $\operatorname{Ind}_G$ will induce a quasi-equivalence
$$
\operatorname{Ind}_G \colon \operatorname{h-proj}(\mathcal A) \to \operatorname{h-proj}(\mathcal B).
$$
This is quite a well-known fact: if you want to check it yourself, you can work instead with semi-free dg-modules and prove that the unit and counit maps are quasi-isomorphism "by hand", or you can use the fact that you have compact generators: a reference is Drinfeld's famous article (Remark 4.3). 
Now, that was the hard part. If you have just a quasi-fully faithful dg-functor $F \colon \mathcal C \to \mathcal D$ as in your post (you don't even need injectivity on objects, just that you have quasi-isomorphisms on hom-complexes), then clearly you can factor it as a quasi-equivalence and a (strictly) fully faithful dg-functor. So, in the end, I think you'll end up with a quasi-fully faithful dg-functor $\operatorname{h-proj}(\mathcal C) \to \operatorname{h-proj}(\mathcal D)$.