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

$(A^\bullet \to B^\bullet \to Q_1^\bullet \to A^\bullet [1]) \longrightarrow (A^\bullet \to C^\bullet \to Q_2^\bullet \to A^\bullet [1])$

and

$(A^\bullet \to C^\bullet \to Q_2^\bullet \to A^\bullet [1]) \longrightarrow (B^\bullet \to C^\bullet \to Q_3^\bullet \to B^\bullet [1]).$

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

$(Q_1^\bullet \to Q_2^\bullet \to Q_3^\bullet \to Q_1^\bullet [1])$

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. $\square$

Comment #8370 by on

Typo: in the equality $C_n=B^n\oplus Q_3^n$, in the LHS the subscript should be a superscript.

Also, regarding the reason of why we have morphisms of triangles

and why the morphism $\delta: Q_3^\bullet\to Q_1^\bullet[1]$ equals the composite $Q_3^\bullet\to B^\bullet[1]\to Q_1^\bullet[1]$. I think one could make the argument more explicit by simply mentioning https://stacks.math.columbia.edu/tag/011J#comment-8369. The case for the morphism of triangles is clear, whereas the claim on $\delta$ follows from application of the comment to the morphism of termwise splitting s.e.s.

Comment #8975 by on

Well, I think you are explaining in more detail exactly what the reader has to do when verifying the proof as given in the text. Although I think your comments are helpful for the reader who chances upon this web location, I am going to leave the text as is for now.

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).