The Stacks project

Proof. It is clear that $D_\mathcal {B}(\mathcal{A})$ is an additive subcategory preserved under the translation functors. If $X \oplus Y$ is in $D_\mathcal {B}(\mathcal{A})$, then both $H^ n(X)$ and $H^ n(Y)$ are kernels of maps between maps of objects of $\mathcal{B}$ as $H^ n(X \oplus Y) = H^ n(X) \oplus H^ n(Y)$. Hence both $X$ and $Y$ are in $D_\mathcal {B}(\mathcal{A})$. By Lemma 13.4.16 it therefore suffices to show that given a distinguished triangle $(X, Y, Z, f, g, h)$ such that $X$ and $Y$ are in $D_\mathcal {B}(\mathcal{A})$ then $Z$ is an object of $D_\mathcal {B}(\mathcal{A})$. The long exact cohomology sequence (13.11.1.1) and the definition of a weak Serre subcategory (see Homology, Definition 12.10.1) show that $H^ n(Z)$ is an object of $\mathcal{B}$ for all $n$. Thus $Z$ is an object of $D_\mathcal {B}(\mathcal{A})$. $\square$

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. $\square$

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

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

Proof. Let $X^\bullet $ be a bounded above complex of $\mathcal{A}$ such that $H^ i(X^\bullet ) \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$ for all $i \in \mathbf{Z}$. Moreover, suppose we are given $B^ i \subset X^ i$, $B^ i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$ for all $i \in \mathbf{Z}$. Claim: there exists a subcomplex $Y^\bullet \subset X^\bullet $ such that

1. $Y^\bullet \to X^\bullet $ is a quasi-isomorphism,

2. $Y^ i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$ for all $i \in \mathbf{Z}$, and

3. $B^ i \subset Y^ i$ for all $i \in \mathbf{Z}$.

To prove the claim, using the assumption of the lemma we first choose $C^ i \subset \mathop{\mathrm{Ker}}(d^ i : X^ i \to X^{i + 1})$, $C^ i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$ surjecting onto $H^ i(X^\bullet )$. Setting $D^ i = C^ i + d^{i - 1}(B^{i - 1}) + B^ i$ we find a subcomplex $D^\bullet $ satisfying (2) and (3) such that $H^ i(D^\bullet ) \to H^ i(X^\bullet )$ is surjective for all $i \in \mathbf{Z}$. For any choice of $E^ i \subset X^ i$ with $E^ i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$ and $d^ i(E^ i) \subset D^{i + 1} + E^{i + 1}$ we see that setting $Y^ i = D^ i + E^ i$ gives a subcomplex whose terms are in $\mathcal{B}$ and whose cohomology surjects onto the cohomology of $X^\bullet $. Clearly, if $d^ i(E^ i) = (D^{i + 1} + E^{i + 1}) \cap \mathop{\mathrm{Im}}(d^ i)$ then we see that the map on cohomology is also injective. For $n \gg 0$ we can take $E^ n$ equal to $0$. By descending induction we can choose $E^ i$ for all $i$ with the desired property. Namely, given $E^{i + 1}, E^{i + 2}, \ldots $ we choose $E^ i \subset X^ i$ such that $d^ i(E^ i) = (D^{i + 1} + E^{i + 1}) \cap \mathop{\mathrm{Im}}(d^ i)$. This is possible by our assumption in the lemma combined with the fact that $(D^{i + 1} + E^{i + 1}) \cap \mathop{\mathrm{Im}}(d^ i)$ is in $\mathcal{B}$ as $\mathcal{B}$ is a Serre subcategory of $\mathcal{A}$.

The claim above implies the lemma. Essential surjectivity is immediate from the claim. Let us prove faithfulness. Namely, suppose we have a morphism $f : U^\bullet \to V^\bullet $ of bounded above complexes of $\mathcal{B}$ whose image in $D(\mathcal{A})$ is zero. Then there exists a quasi-isomorphism $s : V^\bullet \to X^\bullet $ into a bounded above complex of $\mathcal{A}$ such that $s \circ f$ is homotopic to zero. Choose a homotopy $h^ i : U^ i \to X^{i - 1}$ between $0$ and $s \circ f$. Apply the claim with $B^ i = h^{i + 1}(U^{i + 1}) + s^ i(V^ i)$. The resulting map $s' : V^\bullet \to Y^\bullet $ is a quasi-isomorphism as well and $s' \circ f$ is homotopic to zero as is clear from the fact that $h^ i$ factors through $Y^{i - 1}$. This proves faithfulness. Fully faithfulness is proved in the exact same manner. $\square$