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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
\begin{equation} \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 \end{equation}

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
\begin{equation} \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) } \end{equation}

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 (9)

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 instead of in the sentence : "TR4(a) expresses the isomorphism ".

Comment #8853 by on

Since now is only assumed to be an autoequivalence, in sequence 13.3.2.1 not all four consecutive terms are honest triangles, right? (For instance is not an actual triangle, since might not be strictly equal to .)

Comment #9797 by on

I think it is interesting to give reasons to care for the definition of a -morphism in the category of triangulated categories (given before Definition 13.3.4), instead of just staying with plain natural transformations. I will call a -morphism as the one before Definition 13.3.4 a trinatural transformation (as other authors already do). Given a pre-triangulated category , I will denote to the category of triangles of , and to the full subcategory of distinguished triangles. A triangulated functor of pre-triangulated categories induces functors and . If is another triangulated functor, then a trinatural transformation induces natural transformations and . That is, if is a (distinguished) triangle in then we have a morphism of (distinguished) triangles in . We remark this is possible only if is trinatural, just naturality is not enough. (Note the assignments and become -functors from pre-triangulated categories to categories.) In particular, if is a trinatural isomorphism, then the triangle on top of the last diagram is distinguished if and only if the bottom one is.

Comment #9877 by on

@9797 I just realized the converse is also true. That is, if the diagram commutes for all distinguished triangles of , then is trinatural. Indeed, just apply the hypothesis to the distinguished triangle .

Comment #9889 by on

(Original code here.) For reference, this is the actual result:

Lemma. Let be triangulated functors between pre-triangulated categories. For each object , suppose given a morphism and denote . The following are equivalent:

  1. is a natural transformation and we have a commutative triangle of natural transformations

  1. For every distinguished triangle in , the diagram commutes, where the vertical maps are given by , and the last top and bottom arrows are and , respectively.

Succintly, is a transnatural transformation (a 2-morphism in the category of categories with translation) if and only if it is a trinatural transformation (a 2-morphism in the category of pre-triangulated categories). I wrote the proof in this question (see the Lemma at the end).

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