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 $\mathcal{D}'$.

## 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 12.9. Before we do the actual construction we briefly discuss kernels of exact functors.

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

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 [Rickard-derived]) 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

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 $[1]$ 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 13.4.15 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

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 $[1]$ 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 13.4.15 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

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

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 $[1]$ 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 13.4.15 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

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

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

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.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

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

$S$ is a saturated multiplicative system,

$\mathcal{D}'$ 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

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

with $s, s' \in S$. By Proposition 13.4.22 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 13.4.10. 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

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.

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 $s : Q_2 \to Q_1[1]$ the composition $Q_2 \to X[1] \to Q_1[1]$. Denote $t : Q_3 \to Q_2[1]$ the composition $Q_3 \to Y[1] \to Q_2[1]$. 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 $s[1]t$ is an element of $S$ as $S$ is closed under compositions and shifts. Note that $s[1]t = 0$ as $Y[1] \to Q_2[1] \to X[2]$ is zero, see Lemma 13.4.1. Hence $Q_3[1] \oplus Q_1[2]$ is isomorphic to an object of $\mathcal{D}'$, see Lemma 13.4.10. 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.

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.

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

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

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

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.,

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

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

$\mathcal{B}(S(\mathcal{B}))$ 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 13.4.8 in the localized category)

in the notation of Categories, Lemma 4.26.21. 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

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

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. $\square$

It is clear that in the lemma above the factorization of $H$ through $\mathcal{D}/\mathop{\mathrm{Ker}}(H)$ is the universal factorization. Namely, if $F : \mathcal{D} \to \mathcal{D}'$ is an exact functor of triangulated categories and if there exists a homological functor $H' : \mathcal{D}' \to \mathcal{A}$ such that $H \cong H' \circ F$, then $F$ factors through the quotient functor $Q : \mathcal{D} \to \mathcal{D}/\mathop{\mathrm{Ker}}(H)$.

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