# derived category of quotient category

If $\mathcal B$ is a Serre subcategory of an abelian category $\mathcal A$,then we have a new abelian category $\mathcal A/B$ and an exact functor $q:\mathcal A \rightarrow \mathcal A/\mathcal B$.

Since $q$ is exact,$q$ maps quasi-isomorphism to quasi-isomorphism. So $q$ induces an exact functor $Dq:D(\mathcal A)\rightarrow D(\mathcal A/\mathcal B)$.

Consider $D(\mathcal B)$ as triangulated subcategory of $D(\mathcal A)$.(That is:the object of $D(\mathcal B)$ is isomorphic to a object in $D(\mathcal A)$

I have the following two question:

1.Is $D(\mathcal B)$ thick subcategory of $D(\mathcal A ）$？

2.there is a natural induced exact functor $q^-：D(\mathcal A)/D(\mathcal B)\rightarrow D(\mathcal A/\mathcal B)$.Is $q^-$ an equivalence?

We can also consider the induced exact functor $Kq:K(\mathcal A)\rightarrow K(\mathcal A/B)$.Is the same two questions true for this case?

Theorem [Miyachi, Thm. 3.2]. Let $$0 \longrightarrow \mathcal{B} \longrightarrow \mathcal{A} \longrightarrow \mathcal{A}/\mathcal{B} \longrightarrow 0\tag{1}\label{eq:abcat}$$ be an exact sequence of abelian categories. Then, $$0 \longrightarrow \mathsf{D}_{\mathcal{B}}^\bullet(\mathcal{A}) \longrightarrow \mathsf{D}^\bullet(\mathcal{A}) \longrightarrow \mathsf{D}^\bullet(\mathcal{A}/\mathcal{B}) \longrightarrow 0\tag{2}\label{eq:dercat}$$ is an exact sequence of triangulated categories for $\bullet \in \{+,-,b\}$.
Here, $\eqref{eq:abcat}$ should be interpreted as saying that $\mathcal{B}$ is a dense subcategory of $\mathcal{A}$ and that $\mathcal{A}/\mathcal{B}$ is the quotient category. Similarly $\eqref{eq:dercat}$ should be interpreted as saying that the subcategory $\mathsf{D}^\bullet_{\mathcal{B}}(\mathcal{A})$ of $\mathsf{D}^\bullet(\mathcal{A})$ consisting of objects with cohomology in $\mathcal{B}$ is an épaisse subcategory of $\mathsf{D}^\bullet(\mathcal{A})$, and that $\mathsf{D}^\bullet(\mathcal{A}/\mathcal{B})$ is equivalent to the quotient category $\mathsf{D}^\bullet(\mathcal{A})/\mathsf{D}^\bullet_{\mathcal{B}}(\mathcal{A})$. See [Miyachi] for precise definitions.
We therefore see that the answer to your question is yes when $\mathsf{D}^\bullet_{\mathcal{B}}(\mathcal{A})$ is equivalent to $\mathsf{D}^\bullet(\mathcal{B})$. Maybe this is true in your situation?