Definition 13.10.1. Let $\mathcal{A}$ be an additive category. A triangle $(X, Y, Z, f, g, h)$ of $K(\mathcal{A})$ is called a *distinguished triangle of $K(\mathcal{A})$* if it is isomorphic to the triangle associated to a termwise split exact sequence of complexes, see Definition 13.9.9. Same definition for $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, and $K^ b(\mathcal{A})$.

## 13.10 Distinguished triangles in the homotopy category

Since we want our boundary maps in long exact sequences of cohomology to be given by the maps in the snake lemma without signs we define distinguished triangles in the homotopy category as follows.

Note that according to Lemma 13.9.14 a triangle of the form $(K^\bullet , L^\bullet , C(f)^\bullet , f, i, -p)$ is a distinguished triangle. This does indeed lead to a triangulated category, see Proposition 13.10.3. Before we can prove the proposition we need one more lemma in order to be able to prove TR4.

Lemma 13.10.2. Let $\mathcal{A}$ be an additive category. Suppose that $\alpha : A^\bullet \to B^\bullet $ and $\beta : B^\bullet \to C^\bullet $ are split injections of complexes. Then there exist distinguished triangles $(A^\bullet , B^\bullet , Q_1^\bullet , \alpha , p_1, d_1)$, $(A^\bullet , C^\bullet , Q_2^\bullet , \beta \circ \alpha , p_2, d_2)$ and $(B^\bullet , C^\bullet , Q_3^\bullet , \beta , p_3, d_3)$ for which TR4 holds.

**Proof.**
Say $\pi _1^ n : B^ n \to A^ n$, and $\pi _3^ n : C^ n \to B^ n$ are the splittings. Then also $A^\bullet \to C^\bullet $ is a split injection with splittings $\pi _2^ n = \pi _1^ n \circ \pi _3^ n$. Let us write $Q_1^\bullet $, $Q_2^\bullet $ and $Q_3^\bullet $ for the “quotient” complexes. In other words, $Q_1^ n = \mathop{\mathrm{Ker}}(\pi _1^ n)$, $Q_3^ n = \mathop{\mathrm{Ker}}(\pi _3^ n)$ and $Q_2^ n = \mathop{\mathrm{Ker}}(\pi _2^ n)$. Note that the kernels exist. Then $B^ n = A^ n \oplus Q_1^ n$ and $C_ n = B^ n \oplus Q_3^ n$, where we think of $A^ n$ as a subobject of $B^ n$ and so on. This implies $C^ n = A^ n \oplus Q_1^ n \oplus Q_3^ n$. Note that $\pi _2^ n = \pi _1^ n \circ \pi _3^ n$ is zero on both $Q_1^ n$ and $Q_3^ n$. Hence $Q_2^ n = Q_1^ n \oplus Q_3^ n$. Consider the commutative diagram

The rows of this diagram are termwise split exact sequences, and hence determine distinguished triangles by definition. Moreover downward arrows in the diagram above are compatible with the chosen splittings and hence define morphisms of triangles

and

Note that the splittings $Q_3^ n \to C^ n$ of the bottom split sequence in the diagram provides a splitting for the split sequence $0 \to Q_1^\bullet \to Q_2^\bullet \to Q_3^\bullet \to 0$ upon composing with $C^ n \to Q_2^ n$. It follows easily from this that the morphism $\delta : Q_3^\bullet \to Q_1^\bullet [1]$ in the corresponding distinguished triangle

is equal to the composition $Q_3^\bullet \to B^\bullet [1] \to Q_1^\bullet [1]$. Hence we get a structure as in the conclusion of axiom TR4. $\square$

Proposition 13.10.3. Let $\mathcal{A}$ be an additive category. The category $K(\mathcal{A})$ of complexes up to homotopy with its natural translation functors and distinguished triangles as defined above is a triangulated category.

**Proof.**
Proof of TR1. By definition every triangle isomorphic to a distinguished one is distinguished. Also, any triangle $(A^\bullet , A^\bullet , 0, 1, 0, 0)$ is distinguished since $0 \to A^\bullet \to A^\bullet \to 0 \to 0$ is a termwise split sequence of complexes. Finally, given any morphism of complexes $f : K^\bullet \to L^\bullet $ the triangle $(K, L, C(f), f, i, -p)$ is distinguished by Lemma 13.9.14.

Proof of TR2. Let $(X, Y, Z, f, g, h)$ be a triangle. Assume $(Y, Z, X[1], g, h, -f[1])$ is distinguished. Then there exists a termwise split sequence of complexes $A^\bullet \to B^\bullet \to C^\bullet $ such that the associated triangle $(A^\bullet , B^\bullet , C^\bullet , \alpha , \beta , \delta )$ is isomorphic to $(Y, Z, X[1], g, h, -f[1])$. Rotating back we see that $(X, Y, Z, f, g, h)$ is isomorphic to $(C^\bullet [-1], A^\bullet , B^\bullet , -\delta [-1], \alpha , \beta )$. It follows from Lemma 13.9.16 that the triangle $(C^\bullet [-1], A^\bullet , B^\bullet , \delta [-1], \alpha , \beta )$ is isomorphic to $(C^\bullet [-1], A^\bullet , C(\delta [-1])^\bullet , \delta [-1], i, p)$. Precomposing the previous isomorphism of triangles with $-1$ on $Y$ it follows that $(X, Y, Z, f, g, h)$ is isomorphic to $(C^\bullet [-1], A^\bullet , C(\delta [-1])^\bullet , \delta [-1], i, -p)$. Hence it is distinguished by Lemma 13.9.14. On the other hand, suppose that $(X, Y, Z, f, g, h)$ is distinguished. By Lemma 13.9.14 this means that it is isomorphic to a triangle of the form $(K^\bullet , L^\bullet , C(f), f, i, -p)$ for some morphism of complexes $f$. Then the rotated triangle $(Y, Z, X[1], g, h, -f[1])$ is isomorphic to $(L^\bullet , C(f), K^\bullet [1], i, -p, -f[1])$ which is isomorphic to the triangle $(L^\bullet , C(f), K^\bullet [1], i, p, f[1])$. By Lemma 13.9.17 this triangle is distinguished. Hence $(Y, Z, X[1], g, h, -f[1])$ is distinguished as desired.

Proof of TR3. Let $(X, Y, Z, f, g, h)$ and $(X', Y', Z', f', g', h')$ be distinguished triangles of $K(\mathcal{A})$ and let $a : X \to X'$ and $b : Y \to Y'$ be morphisms such that $f' \circ a = b \circ f$. By Lemma 13.9.14 we may assume that $(X, Y, Z, f, g, h) = (X, Y, C(f), f, i, -p)$ and $(X', Y', Z', f', g', h') = (X', Y', C(f'), f', i', -p')$. At this point we simply apply Lemma 13.9.2 to the commutative diagram given by $f, f', a, b$.

Proof of TR4. At this point we know that $K(\mathcal{A})$ is a pre-triangulated category. Hence we can use Lemma 13.4.14. Let $A^\bullet \to B^\bullet $ and $B^\bullet \to C^\bullet $ be composable morphisms of $K(\mathcal{A})$. By Lemma 13.9.15 we may assume that $A^\bullet \to B^\bullet $ and $B^\bullet \to C^\bullet $ are split injective morphisms. In this case the result follows from Lemma 13.10.2. $\square$

Remark 13.10.4. Let $\mathcal{A}$ be an additive category. Exactly the same proof as the proof of Proposition 13.10.3 shows that the categories $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, and $K^ b(\mathcal{A})$ are triangulated categories. Namely, the cone of a morphism between bounded (above, below) is bounded (above, below). But we prove below that these are triangulated subcategories of $K(\mathcal{A})$ which gives another proof.

Lemma 13.10.5. Let $\mathcal{A}$ be an additive category. The categories $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, and $K^ b(\mathcal{A})$ are full triangulated subcategories of $K(\mathcal{A})$.

**Proof.**
Each of the categories mentioned is a full additive subcategory. We use the criterion of Lemma 13.4.15 to show that they are triangulated subcategories. It is clear that each of the categories $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, and $K^ b(\mathcal{A})$ is preserved under the shift functors $[1], [-1]$. Finally, suppose that $f : A^\bullet \to B^\bullet $ is a morphism in $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, or $K^ b(\mathcal{A})$. Then $(A^\bullet , B^\bullet , C(f)^\bullet , f, i, -p)$ is a distinguished triangle of $K(\mathcal{A})$ with $C(f)^\bullet \in K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, or $K^ b(\mathcal{A})$ as is clear from the construction of the cone. Thus the lemma is proved. (Alternatively, $K^\bullet \to L^\bullet $ is isomorphic to an termwise split injection of complexes in $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, or $K^ b(\mathcal{A})$, see Lemma 13.9.6 and then one can directly take the associated distinguished triangle.)
$\square$

Lemma 13.10.6. Let $\mathcal{A}$, $\mathcal{B}$ be additive categories. Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor. The induced functors

are exact functors of triangulated categories.

**Proof.**
Suppose $A^\bullet \to B^\bullet \to C^\bullet $ is a termwise split sequence of complexes of $\mathcal{A}$ with splittings $(s^ n, \pi ^ n)$ and associated morphism $\delta : C^\bullet \to A^\bullet [1]$, see Definition 13.9.9. Then $F(A^\bullet ) \to F(B^\bullet ) \to F(C^\bullet )$ is a termwise split sequence of complexes with splittings $(F(s^ n), F(\pi ^ n))$ and associated morphism $F(\delta ) : F(C^\bullet ) \to F(A^\bullet )[1]$. Thus $F$ transforms distinguished triangles into distinguished triangles.
$\square$

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