13.6 Quotients of triangulated categories

Given a triangulated category and a triangulated subcategory we can construct another triangulated category by taking the “quotient”. The construction uses a localization. This is similar to the quotient of an abelian category by a Serre subcategory, see Homology, Section 12.10. Before we do the actual construction we briefly discuss kernels of exact functors.

Definition 13.6.1. Let $\mathcal{D}$ be a pre-triangulated category. We say a full pre-triangulated subcategory $\mathcal{D}'$ of $\mathcal{D}$ is saturated if whenever $X \oplus Y$ is isomorphic to an object of $\mathcal{D}'$ then both $X$ and $Y$ are isomorphic to objects of $\mathcal{D}'$.

A saturated triangulated subcategory is sometimes called a thick triangulated subcategory. In some references, this is only used for strictly full triangulated subcategories (and sometimes the definition is written such that it implies strictness). There is another notion, that of an épaisse triangulated subcategory. The definition is that given a commutative diagram

$\xymatrix{ & S \ar[rd] \\ X \ar[ru] \ar[rr] & & Y \ar[r] & T \ar[r] & X }$

where the second line is a distinguished triangle and $S$ and $T$ isomorphic to objects of $\mathcal{D}'$, then also $X$ and $Y$ are isomorphic to objects of $\mathcal{D}'$. It turns out that this is equivalent to being saturated (this is elementary and can be found in ) and the notion of a saturated category is easier to work with.

Lemma 13.6.2. Let $F : \mathcal{D} \to \mathcal{D}'$ be an exact functor of pre-triangulated categories. Let $\mathcal{D}''$ be the full subcategory of $\mathcal{D}$ with objects

$\mathop{\mathrm{Ob}}\nolimits (\mathcal{D}'') = \{ X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}) \mid F(X) = 0\}$

Then $\mathcal{D}''$ is a strictly full saturated pre-triangulated subcategory of $\mathcal{D}$. If $\mathcal{D}$ is a triangulated category, then $\mathcal{D}''$ is a triangulated subcategory.

Proof. It is clear that $\mathcal{D}''$ is preserved under $$ and $[-1]$. If $(X, Y, Z, f, g, h)$ is a distinguished triangle of $\mathcal{D}$ and $F(X) = F(Y) = 0$, then also $F(Z) = 0$ as $(F(X), F(Y), F(Z), F(f), F(g), F(h))$ is distinguished. Hence we may apply Lemma 13.4.16 to see that $\mathcal{D}''$ is a pre-triangulated subcategory (respectively a triangulated subcategory if $\mathcal{D}$ is a triangulated category). The final assertion of being saturated follows from $F(X) \oplus F(Y) = 0 \Rightarrow F(X) = F(Y) = 0$. $\square$

Lemma 13.6.3. Let $H : \mathcal{D} \to \mathcal{A}$ be a homological functor of a pre-triangulated category into an abelian category. Let $\mathcal{D}'$ be the full subcategory of $\mathcal{D}$ with objects

$\mathop{\mathrm{Ob}}\nolimits (\mathcal{D}') = \{ X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}) \mid H(X[n]) = 0\text{ for all }n \in \mathbf{Z}\}$

Then $\mathcal{D}'$ is a strictly full saturated pre-triangulated subcategory of $\mathcal{D}$. If $\mathcal{D}$ is a triangulated category, then $\mathcal{D}'$ is a triangulated subcategory.

Proof. It is clear that $\mathcal{D}'$ is preserved under $$ and $[-1]$. If $(X, Y, Z, f, g, h)$ is a distinguished triangle of $\mathcal{D}$ and $H(X[n]) = H(Y[n]) = 0$ for all $n$, then also $H(Z[n]) = 0$ for all $n$ by the long exact sequence (13.3.5.1). Hence we may apply Lemma 13.4.16 to see that $\mathcal{D}'$ is a pre-triangulated subcategory (respectively a triangulated subcategory if $\mathcal{D}$ is a triangulated category). The assertion of being saturated follows from

\begin{align*} H((X \oplus Y)[n]) = 0 & \Rightarrow H(X[n] \oplus Y[n]) = 0 \\ & \Rightarrow H(X[n]) \oplus H(Y[n]) = 0 \\ & \Rightarrow H(X[n]) = H(Y[n]) = 0 \end{align*}

for all $n \in \mathbf{Z}$. $\square$

Lemma 13.6.4. Let $H : \mathcal{D} \to \mathcal{A}$ be a homological functor of a pre-triangulated category into an abelian category. Let $\mathcal{D}_ H^{+}, \mathcal{D}_ H^{-}, \mathcal{D}_ H^ b$ be the full subcategory of $\mathcal{D}$ with objects

$\begin{matrix} \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}_ H^{+}) = \{ X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}) \mid H(X[n]) = 0\text{ for all }n \ll 0\} \\ \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}_ H^{-}) = \{ X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}) \mid H(X[n]) = 0\text{ for all }n \gg 0\} \\ \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}_ H^ b) = \{ X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}) \mid H(X[n]) = 0\text{ for all }|n| \gg 0\} \end{matrix}$

Each of these is a strictly full saturated pre-triangulated subcategory of $\mathcal{D}$. If $\mathcal{D}$ is a triangulated category, then each is a triangulated subcategory.

Proof. Let us prove this for $\mathcal{D}_ H^{+}$. It is clear that it is preserved under $$ and $[-1]$. If $(X, Y, Z, f, g, h)$ is a distinguished triangle of $\mathcal{D}$ and $H(X[n]) = H(Y[n]) = 0$ for all $n \ll 0$, then also $H(Z[n]) = 0$ for all $n \ll 0$ by the long exact sequence (13.3.5.1). Hence we may apply Lemma 13.4.16 to see that $\mathcal{D}_ H^{+}$ is a pre-triangulated subcategory (respectively a triangulated subcategory if $\mathcal{D}$ is a triangulated category). The assertion of being saturated follows from

\begin{align*} H((X \oplus Y)[n]) = 0 & \Rightarrow H(X[n] \oplus Y[n]) = 0 \\ & \Rightarrow H(X[n]) \oplus H(Y[n]) = 0 \\ & \Rightarrow H(X[n]) = H(Y[n]) = 0 \end{align*}

for all $n \in \mathbf{Z}$. $\square$

Definition 13.6.5. Let $\mathcal{D}$ be a (pre-)triangulated category.

1. Let $F : \mathcal{D} \to \mathcal{D}'$ be an exact functor. The kernel of $F$ is the strictly full saturated (pre-)triangulated subcategory described in Lemma 13.6.2.

2. Let $H : \mathcal{D} \to \mathcal{A}$ be a homological functor. The kernel of $H$ is the strictly full saturated (pre-)triangulated subcategory described in Lemma 13.6.3.

These are sometimes denoted $\mathop{\mathrm{Ker}}(F)$ or $\mathop{\mathrm{Ker}}(H)$.

The proof of the following lemma uses TR4.

Lemma 13.6.6. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{D}' \subset \mathcal{D}$ be a full triangulated subcategory. Set

13.6.6.1
\begin{equation} \label{derived-equation-multiplicative-system} S = \left\{ \begin{matrix} f \in \text{Arrows}(\mathcal{D}) \text{ such that there exists a distinguished triangle } \\ (X, Y, Z, f, g, h) \text{ of }\mathcal{D}\text{ with } Z\text{ isomorphic to an object of }\mathcal{D}' \end{matrix} \right\} \end{equation}

Then $S$ is a multiplicative system compatible with the triangulated structure on $\mathcal{D}$. In this situation the following are equivalent

1. $S$ is a saturated multiplicative system,

2. $\mathcal{D}'$ is a saturated triangulated subcategory.

Proof. To prove the first assertion we have to prove that MS1, MS2, MS3 and MS5, MS6 hold.

Proof of MS1. It is clear that identities are in $S$ because $(X, X, 0, 1, 0, 0)$ is distinguished for every object $X$ of $\mathcal{D}$ and because $0$ is an object of $\mathcal{D}'$. Let $f : X \to Y$ and $g : Y \to Z$ be composable morphisms contained in $S$. Choose distinguished triangles $(X, Y, Q_1, f, p_1, d_1)$, $(X, Z, Q_2, g \circ f, p_2, d_2)$, and $(Y, Z, Q_3, g, p_3, d_3)$. By assumption we know that $Q_1$ and $Q_3$ are isomorphic to objects of $\mathcal{D}'$. By TR4 we know there exists a distinguished triangle $(Q_1, Q_2, Q_3, a, b, c)$. Since $\mathcal{D}'$ is a triangulated subcategory we conclude that $Q_2$ is isomorphic to an object of $\mathcal{D}'$. Hence $g \circ f \in S$.

Proof of MS3. Let $a : X \to Y$ be a morphism and let $t : Z \to X$ be an element of $S$ such that $a \circ t = 0$. To prove LMS3 it suffices to find an $s \in S$ such that $s \circ a = 0$, compare with the proof of Lemma 13.5.3. Choose a distinguished triangle $(Z, X, Q, t, g, h)$ using TR1 and TR2. Since $a \circ t = 0$ we see by Lemma 13.4.2 there exists a morphism $i : Q \to Y$ such that $i \circ g = a$. Finally, using TR1 again we can choose a triangle $(Q, Y, W, i, s, k)$. Here is a picture

$\xymatrix{ Z \ar[r]_ t & X \ar[r]_ g \ar[d]^1 & Q \ar[r] \ar[d]^ i & Z \\ & X \ar[r]_ a & Y \ar[d]^ s \\ & & W }$

Since $t \in S$ we see that $Q$ is isomorphic to an object of $\mathcal{D}'$. Hence $s \in S$. Finally, $s \circ a = s \circ i \circ g = 0$ as $s \circ i = 0$ by Lemma 13.4.1. We conclude that LMS3 holds. The proof of RMS3 is dual.

Proof of MS5. Follows as distinguished triangles and $\mathcal{D}'$ are stable under translations

Proof of MS6. Suppose given a commutative diagram

$\xymatrix{ X \ar[r] \ar[d]^ s & Y \ar[d]^{s'} \\ X' \ar[r] & Y' }$

with $s, s' \in S$. By Proposition 13.4.23 we can extend this to a nine square diagram. As $s, s'$ are elements of $S$ we see that $X'', Y''$ are isomorphic to objects of $\mathcal{D}'$. Since $\mathcal{D}'$ is a full triangulated subcategory we see that $Z''$ is also isomorphic to an object of $\mathcal{D}'$. Whence the morphism $Z \to Z'$ is an element of $S$. This proves MS6.

MS2 is a formal consequence of MS1, MS5, and MS6, see Lemma 13.5.2. This finishes the proof of the first assertion of the lemma.

Let's assume that $S$ is saturated. (In the following we will use rotation of distinguished triangles without further mention.) Let $X \oplus Y$ be an object isomorphic to an object of $\mathcal{D}'$. Consider the morphism $f : 0 \to X$. The composition $0 \to X \to X \oplus Y$ is an element of $S$ as $(0, X \oplus Y, X \oplus Y, 0, 1, 0)$ is a distinguished triangle. The composition $Y[-1] \to 0 \to X$ is an element of $S$ as $(X, X \oplus Y, Y, (1, 0), (0, 1), 0)$ is a distinguished triangle, see Lemma 13.4.11. Hence $0 \to X$ is an element of $S$ (as $S$ is saturated). Thus $X$ is isomorphic to an object of $\mathcal{D}'$ as desired.

Finally, assume $\mathcal{D}'$ is a saturated triangulated subcategory. Let

$W \xrightarrow {h} X \xrightarrow {g} Y \xrightarrow {f} Z$

be composable morphisms of $\mathcal{D}$ such that $fg, gh \in S$. We will build up a picture of objects as in the diagram below.

$\xymatrix{ & & Q_{12} \ar[rd] & & Q_{23} \ar[rd] \\ & Q_1 \ar[ld]_{\! + \! 1} \ar[ru] & & Q_2 \ar[ld]_{\! + \! 1} \ar[ll]_{\! + \! 1} \ar[ru] & & Q_3 \ar[ld]_{\! + \! 1} \ar[ll]_{\! + \! 1} \\ W \ar[rr] & & X \ar[lu] \ar[rr] & & Y \ar[lu] \ar[rr] & & Z \ar[lu] }$

First choose distinguished triangles $(W, X, Q_1)$, $(X, Y, Q_2)$, $(Y, Z, Q_3)$ $(W, Y, Q_{12})$, and $(X, Z, Q_{23})$. Denote $s : Q_2 \to Q_1$ the composition $Q_2 \to X \to Q_1$. Denote $t : Q_3 \to Q_2$ the composition $Q_3 \to Y \to Q_2$. By TR4 applied to the composition $W \to X \to Y$ and the composition $X \to Y \to Z$ there exist distinguished triangles $(Q_1, Q_{12}, Q_2)$ and $(Q_2, Q_{23}, Q_3)$ which use the morphisms $s$ and $t$. The objects $Q_{12}$ and $Q_{23}$ are isomorphic to objects of $\mathcal{D}'$ as $W \to Y$ and $X \to Z$ are assumed in $S$. Hence also $st$ is an element of $S$ as $S$ is closed under compositions and shifts. Note that $st = 0$ as $Y \to Q_2 \to X$ is zero, see Lemma 13.4.1. Hence $Q_3 \oplus Q_1$ is isomorphic to an object of $\mathcal{D}'$, see Lemma 13.4.11. By assumption on $\mathcal{D}'$ we conclude that $Q_3$ and $Q_1$ are isomorphic to objects of $\mathcal{D}'$. Looking at the distinguished triangle $(Q_1, Q_{12}, Q_2)$ we conclude that $Q_2$ is also isomorphic to an object of $\mathcal{D}'$. Looking at the distinguished triangle $(X, Y, Q_2)$ we finally conclude that $g \in S$. (It is also follows that $h, f \in S$, but we don't need this.) $\square$

Definition 13.6.7. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{B}$ be a full triangulated subcategory. We define the quotient category $\mathcal{D}/\mathcal{B}$ by the formula $\mathcal{D}/\mathcal{B} = S^{-1}\mathcal{D}$, where $S$ is the multiplicative system of $\mathcal{D}$ associated to $\mathcal{B}$ via Lemma 13.6.6. The localization functor $Q : \mathcal{D} \to \mathcal{D}/\mathcal{B}$ is called the quotient functor in this case.

Note that the quotient functor $Q : \mathcal{D} \to \mathcal{D}/\mathcal{B}$ is an exact functor of triangulated categories, see Proposition 13.5.5. The universal property of this construction is the following.

Lemma 13.6.8. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{B}$ be a full triangulated subcategory of $\mathcal{D}$. Let $Q : \mathcal{D} \to \mathcal{D}/\mathcal{B}$ be the quotient functor.

1. If $H : \mathcal{D} \to \mathcal{A}$ is a homological functor into an abelian category $\mathcal{A}$ such that $\mathcal{B} \subset \mathop{\mathrm{Ker}}(H)$ then there exists a unique factorization $H' : \mathcal{D}/\mathcal{B} \to \mathcal{A}$ such that $H = H' \circ Q$ and $H'$ is a homological functor too.

2. If $F : \mathcal{D} \to \mathcal{D}'$ is an exact functor into a pre-triangulated category $\mathcal{D}'$ such that $\mathcal{B} \subset \mathop{\mathrm{Ker}}(F)$ then there exists a unique factorization $F' : \mathcal{D}/\mathcal{B} \to \mathcal{D}'$ such that $F = F' \circ Q$ and $F'$ is an exact functor too.

Proof. This lemma follows from Lemma 13.5.6. Namely, if $f : X \to Y$ is a morphism of $\mathcal{D}$ such that for some distinguished triangle $(X, Y, Z, f, g, h)$ the object $Z$ is isomorphic to an object of $\mathcal{B}$, then $H(f)$, resp. $F(f)$ is an isomorphism under the assumptions of (1), resp. (2). Details omitted. $\square$

The kernel of the quotient functor can be described as follows.

Lemma 13.6.9. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{B}$ be a full triangulated subcategory. The kernel of the quotient functor $Q : \mathcal{D} \to \mathcal{D}/\mathcal{B}$ is the strictly full subcategory of $\mathcal{D}$ whose objects are

$\mathop{\mathrm{Ob}}\nolimits (\mathop{\mathrm{Ker}}(Q)) = \left\{ \begin{matrix} Z \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}) \text{ such that there exists a }Z' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}) \\ \text{ such that }Z \oplus Z'\text{ is isomorphic to an object of }\mathcal{B} \end{matrix} \right\}$

In other words it is the smallest strictly full saturated triangulated subcategory of $\mathcal{D}$ containing $\mathcal{B}$.

Proof. First note that the kernel is automatically a strictly full triangulated subcategory containing summands of any of its objects, see Lemma 13.6.2. The description of its objects follows from the definitions and Lemma 13.5.7 part (4). $\square$

Let $\mathcal{D}$ be a triangulated category. At this point we have constructions which induce order preserving maps between

1. the partially ordered set of multiplicative systems $S$ in $\mathcal{D}$ compatible with the triangulated structure, and

2. the partially ordered set of full triangulated subcategories $\mathcal{B} \subset \mathcal{D}$.

Namely, the constructions are given by $S \mapsto \mathcal{B}(S) = \mathop{\mathrm{Ker}}(Q : \mathcal{D} \to S^{-1}\mathcal{D})$ and $\mathcal{B} \mapsto S(\mathcal{B})$ where $S(\mathcal{B})$ is the multiplicative set of (13.6.6.1), i.e.,

$S(\mathcal{B}) = \left\{ \begin{matrix} f \in \text{Arrows}(\mathcal{D}) \text{ such that there exists a distinguished triangle } \\ (X, Y, Z, f, g, h) \text{ of }\mathcal{D}\text{ with } Z\text{ isomorphic to an object of }\mathcal{B} \end{matrix} \right\}$

Note that it is not the case that these operations are mutually inverse.

Lemma 13.6.10. Let $\mathcal{D}$ be a triangulated category. The operations described above have the following properties

1. $S(\mathcal{B}(S))$ is the “saturation” of $S$, i.e., it is the smallest saturated multiplicative system in $\mathcal{D}$ containing $S$, and

2. $\mathcal{B}(S(\mathcal{B}))$ is the “saturation” of $\mathcal{B}$, i.e., it is the smallest strictly full saturated triangulated subcategory of $\mathcal{D}$ containing $\mathcal{B}$.

In particular, the constructions define mutually inverse maps between the (partially ordered) set of saturated multiplicative systems in $\mathcal{D}$ compatible with the triangulated structure on $\mathcal{D}$ and the (partially ordered) set of strictly full saturated triangulated subcategories of $\mathcal{D}$.

Proof. First, let's start with a full triangulated subcategory $\mathcal{B}$. Then $\mathcal{B}(S(\mathcal{B})) = \mathop{\mathrm{Ker}}(Q : \mathcal{D} \to \mathcal{D}/\mathcal{B})$ and hence (2) is the content of Lemma 13.6.9.

Next, suppose that $S$ is multiplicative system in $\mathcal{D}$ compatible with the triangulation on $\mathcal{D}$. Then $\mathcal{B}(S) = \mathop{\mathrm{Ker}}(Q : \mathcal{D} \to S^{-1}\mathcal{D})$. Hence (using Lemma 13.4.9 in the localized category)

\begin{align*} S(\mathcal{B}(S)) & = \left\{ \begin{matrix} f \in \text{Arrows}(\mathcal{D}) \text{ such that there exists a distinguished} \\ \text{triangle }(X, Y, Z, f, g, h) \text{ of }\mathcal{D}\text{ with }Q(Z) = 0 \end{matrix} \right\} \\ & = \{ f \in \text{Arrows}(\mathcal{D}) \mid Q(f)\text{ is an isomorphism}\} \\ & = \hat S = S' \end{align*}

in the notation of Categories, Lemma 4.27.21. The final statement of that lemma finishes the proof. $\square$

Lemma 13.6.11. Let $H : \mathcal{D} \to \mathcal{A}$ be a homological functor from a triangulated category $\mathcal{D}$ to an abelian category $\mathcal{A}$, see Definition 13.3.5. The subcategory $\mathop{\mathrm{Ker}}(H)$ of $\mathcal{D}$ is a strictly full saturated triangulated subcategory of $\mathcal{D}$ whose corresponding saturated multiplicative system (see Lemma 13.6.10) is the set

$S = \{ f \in \text{Arrows}(\mathcal{D}) \mid H^ i(f)\text{ is an isomorphism for all }i \in \mathbf{Z}\} .$

The functor $H$ factors through the quotient functor $Q : \mathcal{D} \to \mathcal{D}/\mathop{\mathrm{Ker}}(H)$.

Proof. The category $\mathop{\mathrm{Ker}}(H)$ is a strictly full saturated triangulated subcategory of $\mathcal{D}$ by Lemma 13.6.3. The set $S$ is a saturated multiplicative system compatible with the triangulated structure by Lemma 13.5.4. Recall that the multiplicative system corresponding to $\mathop{\mathrm{Ker}}(H)$ is the set

$\left\{ \begin{matrix} f \in \text{Arrows}(\mathcal{D}) \text{ such that there exists a distinguished triangle } \\ (X, Y, Z, f, g, h)\text{ with } H^ i(Z) = 0 \text{ for all }i \end{matrix} \right\}$

By the long exact cohomology sequence, see (13.3.5.1), it is clear that $f$ is an element of this set if and only if $f$ is an element of $S$. Finally, the factorization of $H$ through $Q$ is a consequence of Lemma 13.6.8. $\square$

Comment #4312 by Linyuan Liu on

I don't think the last paragraph is correct. For example, take $\mathcal{D}'=\mathcal{D}$ and let $F$ be the identity functor and $H'=H$. Then $F$ cannot factor through $Q$ as long as $\operatorname{Ker}(H)$ is non trivial.

Comment #4471 by on

Thanks very much. I couldn't find a suitable replacement statement, so I just removed the paragraph. Changes are here.

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).