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

$\begin{matrix} 0 & \to & A^\bullet & \to & B^\bullet & \to & Q_1^\bullet & \to & 0 \\ & & \downarrow & & \downarrow & & \downarrow & \\ 0 & \to & A^\bullet & \to & C^\bullet & \to & Q_2^\bullet & \to & 0 \\ & & \downarrow & & \downarrow & & \downarrow & \\ 0 & \to & B^\bullet & \to & C^\bullet & \to & Q_3^\bullet & \to & 0 \end{matrix}$

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

$(A^\bullet \to B^\bullet \to Q_1^\bullet \to A^\bullet ) \longrightarrow (A^\bullet \to C^\bullet \to Q_2^\bullet \to A^\bullet )$

and

$(A^\bullet \to C^\bullet \to Q_2^\bullet \to A^\bullet ) \longrightarrow (B^\bullet \to C^\bullet \to Q_3^\bullet \to B^\bullet ).$

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 $ in the corresponding distinguished triangle

$(Q_1^\bullet \to Q_2^\bullet \to Q_3^\bullet \to Q_1^\bullet )$

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

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).