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.

## 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

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

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

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

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. $\square$

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

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

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

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

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

$B^ i \subset Y^ i$ 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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (9)

Comment #511 by Keenan Kidwell on

Comment #898 by Charles Rezk on

Comment #909 by Johan on

Comment #3558 by YiLinWu on

Comment #3559 by YiLinWu on

Comment #3683 by Johan on

Comment #4609 by Aolong on

Comment #4610 by Johan on

Comment #4779 by Johan on