Lemma 22.27.15. In Situation 22.27.2 given admissible monomorphisms $x \xrightarrow {\alpha } y$, $y \xrightarrow {\beta } z$ in $\mathcal{A}$, there exist distinguished triangles $(x,y,q_1,\alpha ,p_1,\delta _1)$, $(x,z,q_2,\beta \alpha ,p_2,\delta _2)$ and $(y,z,q_3,\beta ,p_3,\delta _3)$ for which TR4 holds.

Proof. Given admissible monomorphisms $x\xrightarrow {\alpha } y$ and $y\xrightarrow {\beta }z$, we can find distinguished triangles, via their extensions to admissible short exact sequences,

$\xymatrix{ x\ar@<0.5ex>[r]^{\alpha } & y\ar@<0.5ex>[l]^{\pi _1} \ar@<0.5ex>[r]^{p_1} & q_1 \ar[r]^{\delta _1} \ar@<0.5ex>[l]^{s_1} & x[1] }$

$\xymatrix{ x\ar@<0.5ex>[r]^{\beta \alpha } & z\ar@<0.5ex>[l]^{\pi _1\pi _3} \ar@<0.5ex>[r]^{p_2} & q_2 \ar[r]^{\delta _2} \ar@<0.5ex>[l]^{s_2} & x[1] }$

$\xymatrix{ y\ar@<0.5ex>[r]^{\beta } & z\ar@<0.5ex>[l]^{\pi _3} \ar@<0.5ex>[r]^{p_3} & q_3 \ar[r]^{\delta _3} \ar@<0.5ex>[l]^{s_3} & x[1] }$

In these diagrams, the maps $\delta _ i$ are defined as $\delta _ i = \pi _ i d(s_ i)$ analagous to the maps defined in Lemma 22.27.1. They fit in the following solid commutative diagram

$\xymatrix@C=5pc@R=3pc{ x\ar@<0.5ex>[r]^{\alpha } \ar@<0.5ex>[dr]^{\beta \alpha } & y\ar@<0.5ex>[d]^{\beta } \ar@<0.5ex>[l]^{\pi _1} \ar@<0.5ex>[r]^{p_1} & q_1 \ar[r]^{\delta _1} \ar@<0.5ex>[l]^{s_1} \ar@{.>}[dd]^{p_2\beta s_1} & x[1] \\ & z \ar@<0.5ex>[u]^{\pi _3}\ar@<0.5ex>[d]^{p_3} \ar@<0.5ex>[dr]^{p_2} \ar@<0.5ex>[ul]^{\pi _1\pi _3} & & \\ & q_3\ar@<0.5ex>[u]^{s_3} \ar[d]^{\delta _3} & q_2 \ar@{.>}[l]^{p_3s_2} \ar@<0.5ex>[ul]^{s_2} \ar[dr]^{\delta _2} \\ & y[1] & & x[1]}$

where we have defined the dashed arrows as indicated. Clearly, their composition $p_3s_2p_2\beta s_1 = 0$ since $s_2p_2 = 0$. We claim that they both are morphisms of $\text{Comp}(\mathcal{A})$. We can check this using equations in Lemma 22.27.1:

$d(p_2\beta s_1) = p_2\beta d(s_1) = p_2\beta \alpha \pi _1 d(s_1) = 0$

since $p_2\beta \alpha = 0$, and

$d(p_3s_2) = p_3d(s_2) = p_3\beta \alpha \pi _1\pi _3 d(s_2) = 0$

since $p_3\beta = 0$. To check that $q_1\to q_2\to q_3$ is an admissible short exact sequence, it remains to show that in the underlying graded category, $q_2 = q_1\oplus q_3$ with the above two morphisms as coprojection and projection. To do this, observe that in the underlying graded category $\mathcal{C}$, there hold

$y = x\oplus q_1,\quad z = y\oplus q_3 = x\oplus q_1\oplus q_3$

where $\pi _1\pi _3$ gives the projection morphism onto the first factor: $x\oplus q_1\oplus q_3\to z$. By axiom (A) on $\mathcal{A}$, $\mathcal{C}$ is an additive category, hence we may apply Homology, Lemma 12.3.10 and conclude that

$\mathop{\mathrm{Ker}}(\pi _1\pi _3) = q_1\oplus q_3$

in $\mathcal{C}$. Another application of Homology, Lemma 12.3.10 to $z = x\oplus q_2$ gives $\mathop{\mathrm{Ker}}(\pi _1\pi _3) = q_2$. Hence $q_2\cong q_1\oplus q_3$ in $\mathcal{C}$. It is clear that the dashed morphisms defined above give coprojection and projection.

Finally, we have to check that the morphism $\delta : q_3 \to q_1[1]$ induced by the admissible short exact sequence $q_1\to q_2\to q_3$ agrees with $p_1\delta _3$. By the construction in Lemma 22.27.1, the morphism $\delta$ is given by

\begin{align*} p_1\pi _3s_2d(p_2s_3) = & p_1\pi _3s_2p_2d(s_3) \\ = & p_1\pi _3(1-\beta \alpha \pi _1\pi _3)d(s_3) \\ = & p_1\pi _3d(s_3)\quad (\text{since }\pi _3\beta = 0) \\ = & p_1\delta _3 \end{align*}

as desired. The proof is complete. $\square$

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