## 13.17 Triangulated subcategories of the derived category

Let $\mathcal{A}$ be an abelian category. In this section we look at certain strictly full saturated triangulated subcategories $\mathcal{D}' \subset D(\mathcal{A})$.

Let $\mathcal{B} \subset \mathcal{A}$ be a weak Serre subcategory, see Homology, Definition 12.10.1 and Lemma 12.10.3. 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 13.17.1. 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 13.4.16 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 $D_\mathcal {B}(\mathcal{A})$. $\square$

We continue to assume that $\mathcal{B}$ is a weak Serre subcategory of the abelian category $\mathcal{A}$. Then $\mathcal{B}$ is an abelian category and the inclusion functor $\mathcal{B} \to \mathcal{A}$ is exact. Hence we obtain a derived functor $D(\mathcal{B}) \to D(\mathcal{A})$, see Lemma 13.16.9. Clearly the functor $D(\mathcal{B}) \to D(\mathcal{A})$ factors through a canonical exact functor

13.17.1.1
$$\label{derived-equation-compare} D(\mathcal{B}) \longrightarrow D_\mathcal {B}(\mathcal{A})$$

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 $D^+(\mathcal{B}) \to D^+_\mathcal {B}(\mathcal{A})$, $D^-(\mathcal{B}) \to D^-_\mathcal {B}(\mathcal{A})$, and $D^ b(\mathcal{B}) \to D^ b_\mathcal {B}(\mathcal{A})$ 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 12.10.6. 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 13.17.2. 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

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

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 13.17.3. 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). $\square$

For certain Serre subcategories $\mathcal{B} \subset \mathcal{A}$ we can prove that the functor $D(\mathcal{B}) \to D_\mathcal {B}(\mathcal{A})$ is fully faithful.

Lemma 13.17.4. Let $\mathcal{A}$ be an abelian category. Let $\mathcal{B} \subset \mathcal{A}$ be a Serre subcategory. Assume that for every surjection $X \to Y$ with $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ and $Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$ there exists $X' \subset X$, $X' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$ which surjects onto $Y$. Then the functor $D^-(\mathcal{B}) \to D^-_\mathcal {B}(\mathcal{A})$ of (13.17.1.1) is an equivalence.

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. $\square$

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.

Comment #4609 by Aolong on

I fail to follow the proof of the claim in Lemma 0FCL. I don't think the injectivity of the induced map on cohomology is obvious.

The construction of E^i is also confusing to me. For, take i=n, which is the largest number that X^i is nonzero. What is E^n?

So I was wondering it might be a good idea to add more details...

Comment #4610 by on

Oops, this is very confusing, because there is a mistake, namely, we should replace the condition that $d^i(E^i) = (D^{i + 1} + E^{i + 1}) \cap \text{Ker}(d^{i + 1})$ by the condition that $d^i(E^i) = (D^{i + 1} + E^{i + 1}) \cap \text{Im}(d^i)$. Thanks for pointing this out. I will fix this very soon.

Let me explain a bit more: for any choice of $E^i \subset X^i$ with $d^i(E^i) \subset D^{i + 1} + E^{i + 1}$ 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 $d^i(E^i) = (D^{i + 1} + E^{i + 1}) \cap \text{Im}(d^i)$ then we see that the map on cohomology is also injective. So now start with some really large index $n$. Then we take $E^n$ equal to $0$. By descending induction we can choose $E^i$ for all $i$ with the desired property.

For example, for $i$ the largest integer such that $X^i$ is nonzero we can choose $E^i = 0$ because this will surject onto $(D^{i + 1} + E^{i + 1}) \cap \text{Im}(d^i) = 0$.

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