
## 13.13 Triangulated subcategories of the derived category

Let $\mathcal{A}$ be an abelian category. In this section we are going to look for strictly full saturated triangulated subcategories $\mathcal{D}' \subset D(\mathcal{A})$ and in the bounded versions.

Here is a simple construction. Let $\mathcal{B} \subset \mathcal{A}$ be a weak Serre subcategory, see Homology, Section 12.9. We let $D_\mathcal {B}(\mathcal{A})$ the full subcategory of $D(\mathcal{A})$ whose objects are

$\mathop{\mathrm{Ob}}\nolimits (D_\mathcal {B}(\mathcal{A})) = \{ X \in \mathop{\mathrm{Ob}}\nolimits (D(\mathcal{A})) \mid H^ n(X) \text{ is an object of }\mathcal{B}\text{ for all }n\}$

We also define $D^{+}_\mathcal {B}(\mathcal{A}) = D^{+}(\mathcal{A}) \cap D_\mathcal {B}(\mathcal{A})$ and similarly for the other bounded versions.

Lemma 13.13.1. Let $\mathcal{A}$ be an abelian category. Let $\mathcal{B} \subset \mathcal{A}$ be a weak Serre subcategory. The category $D_\mathcal {B}(\mathcal{A})$ is a strictly full saturated triangulated subcategory of $D(\mathcal{A})$. Similarly for the bounded versions.

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 13.4.15 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 12.9.1) show that $H^ n(Z)$ is an object of $\mathcal{B}$ for all $n$. Thus $Z$ is an object of $D_\mathcal {B}(\mathcal{A})$. $\square$

An interesting feature of the situation of the lemma is that the functor $D(\mathcal{B}) \to D(\mathcal{A})$ factors through a canonical exact functor

13.13.1.1
$$\label{derived-equation-compare} D(\mathcal{B}) \longrightarrow D_\mathcal {B}(\mathcal{A})$$

After all a complex made from objects of $\mathcal{B}$ certainly gives rise to an object of $D_\mathcal {B}(\mathcal{A})$ and as distinguished triangles in $D_\mathcal {B}(\mathcal{A})$ are exactly the distinguished triangles of $D(\mathcal{A})$ whose vertices are in $D_\mathcal {B}(\mathcal{A})$ we see that the functor is exact since $D(\mathcal{B}) \to D(\mathcal{A})$ is exact. Similarly we obtain functors $D^+(\mathcal{B}) \longrightarrow D^+_\mathcal {B}(\mathcal{A})$ etc for the bounded versions. A key question in many cases is whether the displayed functor is an equivalence.

Now, suppose that $\mathcal{B}$ is a Serre subcategory of $\mathcal{A}$. In this case we have the quotient functor $\mathcal{A} \to \mathcal{A}/\mathcal{B}$, see Homology, Lemma 12.9.6. In this case $D_\mathcal {B}(\mathcal{A})$ is the kernel of the functor $D(\mathcal{A}) \to D(\mathcal{A}/\mathcal{B})$. Thus we obtain a canonical functor

$D(\mathcal{A})/D_\mathcal {B}(\mathcal{A}) \longrightarrow D(\mathcal{A}/\mathcal{B})$

by Lemma 13.6.8. Similarly for the bounded versions.

Lemma 13.13.2. 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 12.9.6. 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

$\xymatrix{ & (X')^1 \ar@{..>}[r] & (X')^2 \ar@{..>}[r] & \ldots \\ X^0 \ar@{..>}[ru] & X^1 \ar@{..>}[u] & X^2 \ar@{..>}[u] & \ldots \\ Y^0 \ar[u]_{s^0} \ar[ru]_{f^0} & Y^1 \ar[u]_{s^1} \ar[ru]_{f^1} & Y^2 \ar[u]_{s^2} \ar[ru]_{f^2} & \ldots }$

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

$\xymatrix{ \ldots & Z^{-2} & Z^{-1} & Z^0 \\ \ldots & X^{-2} \ar[u]_{t_{-2}} \ar[ru]_{g_{-2}} & X^{-1} \ar[u]_{t_{-1}} \ar[ru]_{g_{-1}} & X^0 \ar[u]_{t^0} \\ \ldots & (X')^{-2} \ar@{..>}[u] \ar@{..>}[r] & (X')^{-1} \ar@{..>}[u] \ar@{..>}[ru] }$

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

$(X')^ n = X^ n/\sum \nolimits _{0 < k \leq n} \mathop{\mathrm{Im}}(\mathop{\mathrm{Im}}(X^{k - 2} \to X^ k) \to X^ n)$

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

$(X')^ n = \bigcap \nolimits _{n \leq k < 0} (X^ n \to X^ k)^{-1}\mathop{\mathrm{Ker}}(X^ k \to X^{k + 2})$

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. $\square$

Lemma 13.13.3. 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 13.13.2. 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). $\square$

Comment #511 by Keenan Kidwell on

In 06UQ, "abelian" should be "triangulated."

Comment #898 by Charles Rezk on

In particular, "abelian" should be "triangulated" in the statement of Lemma 13.13.1.

Comment #3558 by YiLinWu on

The differential matrix \ref{Lemma 06XL.} is not square zero in the proof of Lemma 13.13.2,

Comment #3559 by YiLinWu on

The differential matrix \ref{Lemma 06XL.} is not square zero in the proof of Lemma 13.13.2,

Comment #3683 by on

OK, thanks very much. This proof was complete nonsense. I have replaced it by a more honest proof. If you want to be mentioned as a contributor to the Stacks project, can you tell me your name? I don't know how to split your name into 2 or more parts. The changes are here.

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