Lemma 13.17.2. Let $\mathcal{A}$ be an abelian category. Let $\mathcal{B} \subset \mathcal{A}$ be a Serre subcategory. Then $D(\mathcal{A}) \to D(\mathcal{A}/\mathcal{B})$ is essentially surjective.

**Proof.**
We will use the description of the category $\mathcal{A}/\mathcal{B}$ in the proof of Homology, Lemma 12.10.6. Let $(X^\bullet , d^\bullet )$ be a complex of $\mathcal{A}/\mathcal{B}$. This means that $X^ i$ is an object of $\mathcal{A}$ and $d^ i : X^ i \to X^{i + 1}$ is a morphism in $\mathcal{A}/\mathcal{B}$ such that $d^ i \circ d^{i - 1} = 0$ in $\mathcal{A}/\mathcal{B}$.

For $i \geq 0$ we may write $d^ i = (s^ i, f^ i)$ where $s^ i : Y^ i \to X^ i$ is a morphism of $\mathcal{A}$ whose kernel and cokernel are in $\mathcal{B}$ (equivalently $s^ i$ becomes an isomorphism in the quotient category) and $f^ i : Y^ i \to X^{i + 1}$ is a morphism of $\mathcal{A}$. By induction we will construct a commutative diagram

where the vertical arrows $X^ i \to (X')^ i$ become isomorphisms in the quotient category. Namely, we first let $(X')^1 = \mathop{\mathrm{Coker}}(Y^0 \to X^0 \oplus X^1)$ (or rather the pushout of the diagram with arrows $s^0$ and $f^0$) which gives the first commutative diagram. Next, we take $(X')^2 = \mathop{\mathrm{Coker}}(Y^1 \to (X')^1 \oplus X^2)$. And so on. Setting additionally $(X')^ n = X^ n$ for $n \leq 0$ we see that the map $(X^\bullet , d^\bullet ) \to ((X')^\bullet , (d')^\bullet )$ is an isomorphism of complexes in $\mathcal{A}/\mathcal{B}$. Hence we may assume $d^ n : X^ n \to X^{n + 1}$ is given by a map $X^ n \to X^{n + 1}$ in $\mathcal{A}$ for $n \geq 0$.

Dually, for $i < 0$ we may write $d^ i = (g^ i, t^{i + 1})$ where $t^{i + 1} : X^{i + 1} \to Z^{i + 1}$ is an isomorphism in the quotient category and $g^ i : X^ i \to Z^{i + 1}$ is a morphism. By induction we will construct a commutative diagram

where the vertical arrows $(X')^ i \to X^ i$ become isomorphisms in the quotient category. Namely, we take $(X')^{-1} = X^{-1} \times _{Z^0} X^0$. Then we take $(X')^{-2} = X^{-2} \times _{Z^{-1}} (X')^{-1}$. And so on. Setting additionally $(X')^ n = X^ n$ for $n \geq 0$ we see that the map $((X')^\bullet , (d')^\bullet ) \to (X^\bullet , d^\bullet )$ is an isomorphism of complexes in $\mathcal{A}/\mathcal{B}$. Hence we may assume $d^ n : X^ n \to X^{n + 1}$ is given by a map $d^ n : X^ n \to X^{n + 1}$ in $\mathcal{A}$ for all $n \in \mathbf{Z}$.

In this case we know the compositions $d^ n \circ d^{n - 1}$ are zero in $\mathcal{A}/\mathcal{B}$. If for $n > 0$ we replace $X^ n$ by

then the compositions $d^ n \circ d^{n - 1}$ are zero for $n \geq 0$. (Similarly to the second paragraph above we obtain an isomorphism of complexes $(X^\bullet , d^\bullet ) \to ((X')^\bullet , (d')^\bullet )$.) Finally, for $n < 0$ we replace $X^ n$ by

and we argue in the same manner to get a complex in $\mathcal{A}$ whose image in $\mathcal{A}/\mathcal{B}$ is isomorphic to the given one. $\square$

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