Lemma 13.17.3. Let $\mathcal{A}$ be an abelian category. Let $\mathcal{B} \subset \mathcal{A}$ be a Serre subcategory. Suppose that the functor $v : \mathcal{A} \to \mathcal{A}/\mathcal{B}$ has a left adjoint $u : \mathcal{A}/\mathcal{B} \to \mathcal{A}$ such that $vu \cong \text{id}$. Then

\[ D(\mathcal{A})/D_\mathcal {B}(\mathcal{A}) = D(\mathcal{A}/\mathcal{B}) \]

and similarly for the bounded versions.

**Proof.**
The functor $D(v) : D(\mathcal{A}) \to D(\mathcal{A}/\mathcal{B})$ is essentially surjective by Lemma 13.17.2. For an object $X$ of $D(\mathcal{A})$ the adjunction mapping $c_ X : uvX \to X$ maps to an isomorphism in $D(\mathcal{A}/\mathcal{B})$ because $vuv \cong v$ by the assumption that $vu \cong \text{id}$. Thus in a distinguished triangle $(uvX, X, Z, c_ X, g, h)$ the object $Z$ is an object of $D_\mathcal {B}(\mathcal{A})$ as we see by looking at the long exact cohomology sequence. Hence $c_ X$ is an element of the multiplicative system used to define the quotient category $D(\mathcal{A})/D_\mathcal {B}(\mathcal{A})$. Thus $uvX \cong X$ in $D(\mathcal{A})/D_\mathcal {B}(\mathcal{A})$. For $X, Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A}))$ the map

\[ \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})/D_\mathcal {B}(\mathcal{A})}(X, Y) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A}/\mathcal{B})}(vX, vY) \]

is bijective because $u$ gives an inverse (by the remarks above).
$\square$

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