Lemma 22.10.2. Let $(A, \text{d})$ be a differential graded algebra. Suppose that $\alpha : K \to L$ and $\beta : L \to M$ are admissible monomorphisms of differential graded $A$-modules. Then there exist distinguished triangles $(K, L, Q_1, \alpha , p_1, d_1)$, $(K, M, Q_2, \beta \circ \alpha , p_2, d_2)$ and $(L, M, Q_3, \beta , p_3, d_3)$ for which TR4 holds.

Proof. Say $\pi _1 : L \to K$ and $\pi _3 : M \to L$ are homomorphisms of graded $A$-modules which are left inverse to $\alpha$ and $\beta$. Then also $K \to M$ is an admissible monomorphism with left inverse $\pi _2 = \pi _1 \circ \pi _3$. Let us write $Q_1$, $Q_2$ and $Q_3$ for the cokernels of $K \to L$, $K \to M$, and $L \to M$. Then we obtain identifications (as graded $A$-modules) $Q_1 = \mathop{\mathrm{Ker}}(\pi _1)$, $Q_3 = \mathop{\mathrm{Ker}}(\pi _3)$ and $Q_2 = \mathop{\mathrm{Ker}}(\pi _2)$. Then $L = K \oplus Q_1$ and $M = L \oplus Q_3$ as graded $A$-modules. This implies $M = K \oplus Q_1 \oplus Q_3$. Note that $\pi _2 = \pi _1 \circ \pi _3$ is zero on both $Q_1$ and $Q_3$. Hence $Q_2 = Q_1 \oplus Q_3$. Consider the commutative diagram

$\begin{matrix} 0 & \to & K & \to & L & \to & Q_1 & \to & 0 \\ & & \downarrow & & \downarrow & & \downarrow & \\ 0 & \to & K & \to & M & \to & Q_2 & \to & 0 \\ & & \downarrow & & \downarrow & & \downarrow & \\ 0 & \to & L & \to & M & \to & Q_3 & \to & 0 \end{matrix}$

The rows of this diagram are admissible short 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

$(K \to L \to Q_1 \to K) \longrightarrow (K \to M \to Q_2 \to K)$

and

$(K \to M \to Q_2 \to K) \longrightarrow (L \to M \to Q_3 \to L).$

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

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

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

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