The Stacks project

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. $\square$