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 \ref{lemma-get-triangle}. 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 \ref{lemma-get-triangle}: $$ d(p_2\beta s_1) = p_2\beta d(s_1) = p_2\beta\alpha\pi_1 d(s_1) = 0 $$ 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 \ref{homology-lemma-additive-cat-biproduct-kernel} and conclude that $$ \Ker(\pi_1\pi_3) = q_1\oplus q_3 $$ in $\mathcal{C}$. Another application of Homology, Lemma \ref{homology-lemma-additive-cat-biproduct-kernel} to $z = x\oplus q_2$ gives $\Ker(\pi_1\pi_3) = q_2$. 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. \medskip\noindent 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 \ref{lemma-get-triangle}, 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.