## 13.3 The definition of a triangulated category

In this section we collect most of the definitions concerning triangulated and pre-triangulated categories.

Definition 13.3.1. Let $\mathcal{D}$ be an additive category. Let $[1] : \mathcal{D} \to \mathcal{D}$, $E \mapsto E[1]$ be an additive functor which is an auto-equivalence of $\mathcal{D}$.

1. A triangle is a sextuple $(X, Y, Z, f, g, h)$ where $X, Y, Z \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ and $f : X \to Y$, $g : Y \to Z$ and $h : Z \to X[1]$ are morphisms of $\mathcal{D}$.

2. A morphism of triangles $(X, Y, Z, f, g, h) \to (X', Y', Z', f', g', h')$ is given by morphisms $a : X \to X'$, $b : Y \to Y'$ and $c : Z \to Z'$ of $\mathcal{D}$ such that $b \circ f = f' \circ a$, $c \circ g = g' \circ b$ and $a[1] \circ h = h' \circ c$.

A morphism of triangles is visualized by the following commutative diagram

$\xymatrix{ X \ar[r] \ar[d]^ a & Y \ar[r] \ar[d]^ b & Z \ar[r] \ar[d]^ c & X[1] \ar[d]^{a[1]} \\ X' \ar[r] & Y' \ar[r] & Z' \ar[r] & X'[1] }$

In the setting of Definition 13.3.1, we write $[0] = \text{id}$, for $n > 0$ we denote $[n]$ the $n$-fold composition of $[1]$, we choose a quasi-inverse $[-1]$ of $[1]$, and we set $[-n]$ equal to the $n$-fold composition of $[-1]$. Then $\{ [n]\} _{n \in \mathbf{Z}}$ is a collection of additive auto-equivalences of $\mathcal{D}$ indexed by $n \in \mathbf{Z}$ such that we are given isomorphisms of functors $[n] \circ [m] \cong [n + m]$.

Here is the definition of a triangulated category as given in Verdier's thesis.

Definition 13.3.2. A triangulated category consists of a triple $(\mathcal{D}, \{ [n]\} _{n\in \mathbf{Z}}, \mathcal{T})$ where

1. $\mathcal{D}$ is an additive category,

2. $[1] : \mathcal{D} \to \mathcal{D}$, $E \mapsto E[1]$ is an additive auto-equivalence and $[n]$ for $n \in \mathbf{Z}$ is as discussed above, and

3. $\mathcal{T}$ is a set of triangles (Definition 13.3.1) called the distinguished triangles

subject to the following conditions

1. Any triangle isomorphic to a distinguished triangle is a distinguished triangle. Any triangle of the form $(X, X, 0, \text{id}, 0, 0)$ is distinguished. For any morphism $f : X \to Y$ of $\mathcal{D}$ there exists a distinguished triangle of the form $(X, Y, Z, f, g, h)$.

2. The triangle $(X, Y, Z, f, g, h)$ is distinguished if and only if the triangle $(Y, Z, X[1], g, h, -f[1])$ is.

3. Given a solid diagram

$\xymatrix{ X \ar[r]^ f \ar[d]^ a & Y \ar[r]^ g \ar[d]^ b & Z \ar[r]^ h \ar@{-->}[d] & X[1] \ar[d]^{a[1]} \\ X' \ar[r]^{f'} & Y' \ar[r]^{g'} & Z' \ar[r]^{h'} & X'[1] }$

whose rows are distinguished triangles and which satisfies $b \circ f = f' \circ a$, there exists a morphism $c : Z \to Z'$ such that $(a, b, c)$ is a morphism of triangles.

4. Given objects $X$, $Y$, $Z$ of $\mathcal{D}$, and morphisms $f : X \to Y$, $g : Y \to Z$, and distinguished triangles $(X, Y, Q_1, f, p_1, d_1)$, $(X, Z, Q_2, g \circ f, p_2, d_2)$, and $(Y, Z, Q_3, g, p_3, d_3)$, there exist morphisms $a : Q_1 \to Q_2$ and $b : Q_2 \to Q_3$ such that

1. $(Q_1, Q_2, Q_3, a, b, p_1[1] \circ d_3)$ is a distinguished triangle,

2. the triple $(\text{id}_ X, g, a)$ is a morphism of triangles $(X, Y, Q_1, f, p_1, d_1) \to (X, Z, Q_2, g \circ f, p_2, d_2)$, and

3. the triple $(f, \text{id}_ Z, b)$ is a morphism of triangles $(X, Z, Q_2, g \circ f, p_2, d_2) \to (Y, Z, Q_3, g, p_3, d_3)$.

We will call $(\mathcal{D}, [\ ], \mathcal{T})$ a pre-triangulated category if TR1, TR2 and TR3 hold.1

The explanation of TR4 is that if you think of $Q_1$ as $Y/X$, $Q_2$ as $Z/X$ and $Q_3$ as $Z/Y$, then TR4(a) expresses the isomorphism $(Z/X)/(Y/X) \cong Z/Y$ and TR4(b) and TR4(c) express that we can compare the triangles $X \to Y \to Q_1 \to X[1]$ etc with morphisms of triangles. For a more precise reformulation of this idea see the proof of Lemma 13.10.2.

The sign in TR2 means that if $(X, Y, Z, f, g, h)$ is a distinguished triangle then in the long sequence

13.3.2.1
$$\label{derived-equation-rotate} \ldots \to Z[-1] \xrightarrow {-h[-1]} X \xrightarrow {f} Y \xrightarrow {g} Z \xrightarrow {h} X[1] \xrightarrow {-f[1]} Y[1] \xrightarrow {-g[1]} Z[1] \to \ldots$$

each four term sequence gives a distinguished triangle.

As usual we abuse notation and we simply speak of a (pre-)triangulated category $\mathcal{D}$ without explicitly introducing notation for the additional data. The notion of a pre-triangulated category is useful in finding statements equivalent to TR4.

We have the following definition of a triangulated functor.

Definition 13.3.3. Let $\mathcal{D}$, $\mathcal{D}'$ be pre-triangulated categories. An exact functor, or a triangulated functor from $\mathcal{D}$ to $\mathcal{D}'$ is a functor $F : \mathcal{D} \to \mathcal{D}'$ together with given functorial isomorphisms $\xi _ X : F(X[1]) \to F(X)[1]$ such that for every distinguished triangle $(X, Y, Z, f, g, h)$ of $\mathcal{D}$ the triangle $(F(X), F(Y), F(Z), F(f), F(g), \xi _ X \circ F(h))$ is a distinguished triangle of $\mathcal{D}'$.

An exact functor is additive, see Lemma 13.4.17. When we say two triangulated categories are equivalent we mean that they are equivalent in the $2$-category of triangulated categories. A $2$-morphism $a : (F, \xi ) \to (F', \xi ')$ in this $2$-category is simply a transformation of functors $a : F \to F'$ which is compatible with $\xi$ and $\xi '$, i.e.,

$\xymatrix{ F \circ [1] \ar[r]_\xi \ar[d]_{a \star 1} & [1] \circ F \ar[d]^{1 \star a} \\ F' \circ [1] \ar[r]^{\xi '} & [1] \circ F' }$

commutes.

Definition 13.3.4. Let $(\mathcal{D}, [\ ], \mathcal{T})$ be a pre-triangulated category. A pre-triangulated subcategory2 is a pair $(\mathcal{D}', \mathcal{T}')$ such that

1. $\mathcal{D}'$ is an additive subcategory of $\mathcal{D}$ which is preserved under $[1]$ and such that $[1] : \mathcal{D}' \to \mathcal{D}'$ is an auto-equivalence,

2. $\mathcal{T}' \subset \mathcal{T}$ is a subset such that for every $(X, Y, Z, f, g, h) \in \mathcal{T}'$ we have $X, Y, Z \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}')$ and $f, g, h \in \text{Arrows}(\mathcal{D}')$, and

3. $(\mathcal{D}', [\ ], \mathcal{T}')$ is a pre-triangulated category.

If $\mathcal{D}$ is a triangulated category, then we say $(\mathcal{D}', \mathcal{T}')$ is a triangulated subcategory if it is a pre-triangulated subcategory and $(\mathcal{D}', [\ ], \mathcal{T}')$ is a triangulated category.

In this situation the inclusion functor $\mathcal{D}' \to \mathcal{D}$ is an exact functor with $\xi _ X : X[1] \to X[1]$ given by the identity on $X[1]$.

We will see in Lemma 13.4.1 that for a distinguished triangle $(X, Y, Z, f, g, h)$ in a pre-triangulated category the composition $g \circ f : X \to Z$ is zero. Thus the sequence (13.3.2.1) is a complex. A homological functor is one that turns this complex into a long exact sequence.

Definition 13.3.5. Let $\mathcal{D}$ be a pre-triangulated category. Let $\mathcal{A}$ be an abelian category. An additive functor $H : \mathcal{D} \to \mathcal{A}$ is called homological if for every distinguished triangle $(X, Y, Z, f, g, h)$ the sequence

$H(X) \to H(Y) \to H(Z)$

is exact in the abelian category $\mathcal{A}$. An additive functor $H : \mathcal{D}^{opp} \to \mathcal{A}$ is called cohomological if the corresponding functor $\mathcal{D} \to \mathcal{A}^{opp}$ is homological.

If $H : \mathcal{D} \to \mathcal{A}$ is a homological functor we often write $H^ n(X) = H(X[n])$ so that $H(X) = H^0(X)$. Our discussion of TR2 above implies that a distinguished triangle $(X, Y, Z, f, g, h)$ determines a long exact sequence

13.3.5.1
$$\label{derived-equation-long-exact-cohomology-sequence} \xymatrix@C=3pc{ H^{-1}(Z) \ar[r]^{H(h[-1])} & H^0(X) \ar[r]^{H(f)} & H^0(Y) \ar[r]^{H(g)} & H^0(Z) \ar[r]^{H(h)} & H^1(X) }$$

This will be called the long exact sequence associated to the distinguished triangle and the homological functor. As indicated we will not use any signs for the morphisms in the long exact sequence. This has the side effect that maps in the long exact sequence associated to the rotation (TR2) of a distinguished triangle differ from the maps in the sequence above by some signs.

Definition 13.3.6. Let $\mathcal{A}$ be an abelian category. Let $\mathcal{D}$ be a triangulated category. A $\delta$-functor from $\mathcal{A}$ to $\mathcal{D}$ is given by a functor $G : \mathcal{A} \to \mathcal{D}$ and a rule which assigns to every short exact sequence

$0 \to A \xrightarrow {a} B \xrightarrow {b} C \to 0$

a morphism $\delta = \delta _{A \to B \to C} : G(C) \to G(A)[1]$ such that

1. the triangle $(G(A), G(B), G(C), G(a), G(b), \delta _{A \to B \to C})$ is a distinguished triangle of $\mathcal{D}$ for any short exact sequence as above, and

2. for every morphism $(A \to B \to C) \to (A' \to B' \to C')$ of short exact sequences the diagram

$\xymatrix{ G(C) \ar[d] \ar[rr]_{\delta _{A \to B \to C}} & & G(A)[1] \ar[d] \\ G(C') \ar[rr]^{\delta _{A' \to B' \to C'}} & & G(A')[1] }$

is commutative.

In this situation we call $(G(A), G(B), G(C), G(a), G(b), \delta _{A \to B \to C})$ the image of the short exact sequence under the given $\delta$-functor.

Note how a $\delta$-functor comes equipped with additional structure. Strictly speaking it does not make sense to say that a given functor $\mathcal{A} \to \mathcal{D}$ is a $\delta$-functor, but we will often do so anyway.

[1] We use $[\ ]$ as an abbreviation for the family $\{ [n]\} _{n\in \mathbf{Z}}$.
[2] This definition may be nonstandard. If $\mathcal{D}'$ is a full subcategory then $\mathcal{T}'$ is the intersection of the set of triangles in $\mathcal{D}'$ with $\mathcal{T}$, see Lemma 13.4.16. In this case we drop $\mathcal{T}'$ from the notation.

## Comments (6)

Comment #359 by Fan on

"Our discussion of TR2 above implies that says that ..." is there an extra verb here?

Comment #446 by on

Write $Y/X$ instead of $Y/Z$ in the sentence : "TR4(a) expresses the isomorphism $(Z/X)/(Y/Z)\simeq Z/Y$".

Comment #8853 by on

Since now $[1]$ is only assumed to be an autoequivalence, in sequence 13.3.2.1 not all four consecutive terms are honest triangles, right? (For instance $T^{-1}Z\to X\to Y\to Z$ is not an actual triangle, since $TT^{-1}Z$ might not be strictly equal to $Z$.)

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