## Triangulated categories

We will need to utilize some structure of $K(\mathcal A)$ to facilitate arguments. The abstraction of this structure goes under the name of a triangulated category.

Definition. Let $\mathcal T$ be an additive category equipped with an autoequivalence $[1] : \mathcal T \to \mathcal T$ and with a class of diagrams $$X \overset{\alpha}{\to} Y \overset{\beta}{\to} Z \overset{\gamma}{\to} X[1],$$ called triangles. If the structure satisfies TR1)-TR4) below, then it is called a triangulated category.

TR1)

• For any $X$ $$X \overset{1_X}{\to} X \to 0 \to X[1]$$ is a triangle.
• For any $f: X \to Y$ there exists a triangle $$X \overset{f}{\to} Y \to Z \to X[1]$$
• Given a triangle $$X \to Y \to Z \to X[1]$$ and a commutative diagram

where the vertical maps are isomorphisms, the diagram $$X^\prime \to Y^\prime \to Z^\prime \to X^\prime[1]$$ is also a triangle.

TR2) If $$X \overset{\alpha}{\to} Y \overset{\beta}{\to} Z \overset{\gamma}{\to} X[1]$$ is a triangle, then so are $$Y \overset{\beta}{\to} Z \overset{\gamma}{\to} X[1] \overset{\alpha[1]}{\to} Y[1]$$ and $$Z[-1] \overset{\gamma[-1]}{\to} X \overset{\alpha}{\to} Y \overset{\beta}{\to} Z$$

TR3) Given a commutative diagram

with $$X \to Y \to Z \to X[1]$$ and $$X^\prime \to Y^\prime \to Z^\prime \to X^\prime[1]$$ triangles, there exists a map $h: Z \to Z^\prime$ making the diagram

commute.

TR4) (Octahedral Axiom) Given $\alpha: X \to Y$ and $\beta : Y \to Z$ we have a commutative diagram

where all rows and columns are triangles.

## Basic consequences of the definition

Triangles are automatically complexes.

Lemma. Given a triangle $$X \overset{\alpha}{\to} Y \overset{\beta}{\to} Z \overset{\gamma}{\to} X[1]$$ we have $$\beta\alpha, \gamma\beta, \alpha[1]\gamma = 0.$$

Proof. (Expand to view)

From TR3), we can complete the diagram

giving $$\beta \alpha = 0.$$ The other cases are handled similarly.

Definition. A functor $H : \mathcal T \to \mathcal A$ from a triangulated category to an abelian category is homological if for any triangle $$X \to Y \to Z \to X[1]$$ the sequence $$HX \to HY \to HZ$$ is exact. If a functor $H: \mathcal T^{op} \to \mathcal A$ is homological, we call it cohomological.

The following is an immediate consequence.

Lemma. If $H$ is homological, then there is a long exact sequence $$\cdots \to H(Z[-1]) \to HX \to HY \to HZ \to H(X[1]) \to \cdots$$

Proof. (Expand to view)

This is an immediate consequence of TR2).

Representable functors are (co)homological.

Lemma. For any object $T$ of $\mathcal T$, the functor $\operatorname{Hom}_{\mathcal T}(T,-)$ is homological and $\operatorname{Hom}_{\mathcal T}(-,T)$ is cohomological.

Proof. (Expand to view)

We treat the first case. Since any triangle is a complex, we already know that $$\cdots \to \operatorname{Hom}(T,X) \to \operatorname{Hom}(T,Y) \to \operatorname{Hom}(T,Z) \to \operatorname{Hom}(T,X[1]) \to \cdots$$ is also a complex. Assume that $\beta f = 0$ for some $f: T \to Y$. Then, we can complete the diagram

giving a $g$ with $\alpha g = f$.

Definition. A candidate triangle is a diagram $$X \to Y \to Z \to X[1]$$ such that either $$\cdots \to \operatorname{Hom}(T,X) \to \operatorname{Hom}(T,Y) \to \operatorname{Hom}(T,Z) \to \operatorname{Hom}(T,X[1]) \to \cdots$$ is an exact sequence for all $T$ or $$\cdots \to \operatorname{Hom}(X[1],T) \to \operatorname{Hom}(Z,T) \to \operatorname{Hom}(Y,T) \to \operatorname{Hom}(X,T) \to \cdots$$ is an exact sequence for all $T$.

A useful lemma in detecting triangles is the following.

Lemma. Assume we have a map of candidate triangles

with all maps labeled by $\sim$ isomorphisms. Then $Z \to Z^\prime$ is an isomorphism.

In particular, if at least one of the candidate triangles is an actual triangle, so is the other.

Proof. (Expand to view)

We assume that applying $\operatorname{Hom}(T,-)$ to our candidate triangle yields an exact sequence. We apply $\operatorname{Hom}(T,-)$ to the whole diagram giving

From the five lemma, we have an isomorphism $$\operatorname{Hom}(-,Z^\prime) \cong \operatorname{Hom}(-,Z)$$ and hence by Yoneda $Z^\prime \cong Z$. Thus, we have an isomorphism between our triangle and candidate triangle. By TR1), the candidate triangle is triangle.

Note that the previous lemma applies in the case both rows are triangles. In particular, it implies that the object $Z$ completing the triangle for $f: X \to Y$ is unique up to isomorphism.

Definition. In a triangle $$X \overset{f}{\to} Y \to Z \to X[1]$$ we call $Z$ the cone of the morphism $f$.

Lemma. A retract of a triangle is also a triangle.

Proof. (Expand to view)

Assume we have a diagram

Applying $\operatorname{Hom}(T,-)$ gives a retract of an exact sequence which is also exact. Now complete $f: X^\prime \to Y^\prime$ to a triangle: $$X^\prime \to Y^\prime \to \bar{Z} \to X^\prime[1]$$ We can complete the square
and compose to get a map
By the previous lemma, we have a triangle.

We can play a similar strategy with other functorial constructions that preserve exact sequences.

Lemma. Let $$X_i \to Y_i \to Z_i \to X_i[1]$$ be triangles for $i \in I$.

• If $$\bigoplus_I X_i \to \bigoplus_I Y_i \to \bigoplus_I Z_i \to \bigoplus_I X_i[1]$$
exists, it is a triangle.
• If $$\prod_I X_i \to \prod_I Y_i \to \prod_I Z_i \to \prod_I X_i[1]$$
exists, it is a triangle.
Proof. (Expand to view)

We treat $\bigoplus$ since $\prod$ follows analogously. Take a triangle $$\bigoplus X_i \to \bigoplus Y_i \to Z \to \bigoplus X_i[1]$$ Note that the possibile ambiguity in writing $\bigoplus X_i[1]$ is fine since $[1]$, as an autoequivalence commutes with coproducts. For each $i$ we have a map of triangles

which gives a map
The top row is a candidate triangle as products of exact sequences of abelian groups remain exact. Applying the previous lemma shows it is a triangle.

Finally, we can detect isomorphisms using objects via triangles.

Lemma. The map $f : X \to Y$ is an isomorphism in $\mathcal T$ if and only if $$X \overset{f}{\to} Y \to 0 \to X[1]$$
is a triangle.

Proof. (Expand to view)

For the forward direction assume $f$ is an isomorphism. Then we can construct a morphism of triangles

showing $Z \cong 0$. In the other direction, we can apply $\operatorname{Hom}(T,-)$ and conclude that $$\operatorname{Hom}(-,f) : \operatorname{Hom}(-,X) \to \operatorname{Hom}(-,Y)$$ is an isomorphism. Hence $f: X \to Y$ is also.

## The homotopy category as a triangulated category

For $K(\mathcal A)$ we already have the shift $[1]$. We just need to the collection of triangles.

Definition. Given a chain map $\phi : A \to B$, the cone of $\phi$ is the complex $$C(\phi)^i = A^{i+1} \oplus B^i, \ \delta^i_{C(\phi)} = \begin{pmatrix} -\delta_A^{i+1} & 0 \\ \phi_{i+1} & \delta_B^i \end{pmatrix}$$ There are chain maps given by inclusion $$B \to C(\phi)$$ and by projection $$C(\phi) \to A$$

The triangles will be diagrams isomorphic to a triangle for the form $$A \overset{\phi}{\to} B \to C(\phi) \to A[1]$$ for some $\phi : A \to B$.

Proposition. The homotopy category $K(\mathcal A)$ with $[1]$ and these triangles is a triangulated category.

One important fact to notice: this cone construction is a functorial. Given a square

we have a map of cones $$\begin{pmatrix} \alpha & 0 \\ 0 & \beta \end{pmatrix} : C(f) \to C(g)$$

For a general triangulated category, this is not the case.

Next, let’s check (some of) the axioms in order of convenience.

TR1) Clearly $C(1_A) \neq 0$ but we only need it to be null-homotopic. The map $$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} : A_{i+1} \oplus A_i \to A_i \oplus A_{i-1}$$ provides the null-homotopy.

The other two conditions here are automatic from the definition of the set of triangle:

• Any map $\phi : A \to B$ can be completed to the triangle $$A \to B \to C(\phi) \to A[1]$$
• Anything isomorphic to a triangle is itself a triangle as we declared that the set of triangles to be closed under isomorphism as part of the definition.

TR3) We already mentioned the functoriality of the cone construction. Thus given a commutative square

we can complete it to a map of triangles

TR2) This reduces to the claim that $C(C(\phi) \to A[1])$ is homotopic to $B[1]$ and that $C(B \to C(\phi))$ is homotopic to $A[1]$. We expand on the later claim.

The projection defines a chain map $$\rho : C(B \to C(\phi)) \to A[1].$$ The map in the other direction is $$\ell : \begin{pmatrix} \phi & 1 & 0 \end{pmatrix} A[1] \to C(B \to C(\phi)).$$

The composition $\rho \ell$ is $1_{A[1]}$. The other composition is homotopic to the identity via $$\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} : C(B \to C(\phi)) \to C(B \to C(\phi))$$

Checking the other claim boils down to producing such homotopies similar to some of the following observations that relate the abelian category structure of $Ch(\mathcal A)$ to cones.

Proposition.

• The sequence $$0 \to B \to C(\phi) \to A \to 0$$ is an exact sequence of complexes.
• Let $K = \ker \phi$ and $C = \operatorname{cok} \phi$. Then we can factor the map $K \to A$ as $$K \to C(\phi)[-1] \to A$$
and the map $B \to C$ as $$B \to C(\phi) \to C.$$
• If $\phi: A \hookrightarrow B$, then $C(\phi) \to C$ is a quasi-isomorphism.
• If $\phi: A \twoheadrightarrow B$, the $K \to C(\phi)[-1]$ is a quasi-isomorphism.

We leave the details to the reader.

Finally, we ignore TR4) until we actually need it (as is tradition).