Proving the direct sum of two distinguished triangles is a distinguished triangle

Suppose that we are working in a triangulated category and there are two distinguished triangles $X_i\longrightarrow Y_i\longrightarrow Z_i\longrightarrow X_i[1]$ ($i=1,2$). I am stuck in proving that their direct sum $$X_1\oplus X_2\longrightarrow Y_1\oplus Y_2\longrightarrow Z_1\oplus Z_2\longrightarrow X_1\oplus X_2[1]$$ is a distinguished triangle.

Here is my incomplete proof:

Adopting the axiom $(TR1)$ we obtain a distinguished triangle $$X_1\oplus X_2\longrightarrow Y_1\oplus Y_2\longrightarrow Z\longrightarrow X_1\oplus X_2[1]$$ I wish to show that there is an isomorphism between these two triangles: $$\require{AMScd} \begin{CD} X_1\oplus X_2@>>> Y_1\oplus Y_2@>>>Z_1\oplus Z_2@>>>X_1\oplus X_2[1]\\ @VV{\mathrm{id}}V@VV{\mathrm{id}}V @VV{\varphi}V @VV{\mathrm{id}}V\\ X_1\oplus X_2@>>> Y_1\oplus Y_2@>>>Z@>>>X_1\oplus X_2[1]\end{CD}$$ and then $(TR1)$ verifies the statement. Next we contruct the $\varphi$. Applying $(TR3)$ we have a morphism $\varphi_i\colon Z_i\to Z$ ($i=1,2$) such that the diagram below is commutative: $$\require{AMScd} \begin{CD} X_i @>>> Y_i@>>>Z_i@>>>X_i[1]\\ @VVV@VVV @VV{\varphi_i}V @VVV\\ X_1\oplus X_2@>>> Y_1\oplus Y_2@>>>Z@>>>X_1\oplus X_2[1] \end{CD}$$ where the three verticle arrows are given by $\left(\begin{smallmatrix}\mathrm{id}\\ 0\end{smallmatrix}\right)$ for $i=1$ and $\left(\begin{smallmatrix}0\\ \mathrm{id}\end{smallmatrix}\right)$ for $i=2$. Analogously we have a morphism $\psi_i\colon Z\to Z_i$ ($i=1,2$). Define $\varphi=(\varphi_1,\varphi_2)\colon Z_1\oplus Z_2\to Z$ and $\psi=\left(\begin{smallmatrix}\psi_1\\ \psi_2\end{smallmatrix}\right)\colon Z\to Z_1\oplus Z_2$. Then considering the following commutative diagram $$\require{AMScd} \begin{CD} X_1\oplus X_2@>>> Y_1\oplus Y_2@>>>Z@>>>X_1\oplus X_2[1]\\ @VV{\mathrm{id}}V@VV{\mathrm{id}}V @VV{\psi}V @VV{\mathrm{id}}V\\ X_1\oplus X_2@>>> Y_1\oplus Y_2@>>>Z_1\oplus Z_2@>>>X_1\oplus X_2[1]\\ @VV{\mathrm{id}}V@VV{\mathrm{id}}V @VV{\varphi}V @VV{\mathrm{id}}V\\ X_1\oplus X_2@>>> Y_1\oplus Y_2@>>>Z@>>>X_1\oplus X_2[1] \end{CD}$$ since the first and third rows are distinguished triangles, we may see that $\varphi\psi$ is an isomorphism. If moreover $\psi\varphi$ is an isomorphism, then we may conclude that $\varphi$ is an isomorphism and the proof is complete. However I am stuck here, as $$\psi\varphi=\left(\begin{matrix}\psi_1\\\psi_2\end{matrix}\right)(\varphi_1,\varphi_2)=\left(\begin{matrix}\psi_1\varphi_1&\psi_1\varphi_2\\ \psi_2\varphi_1&\psi_2\varphi_2\end{matrix}\right).$$ Similarly, it can be shown that $\psi_1\varphi_1$ and $\psi_2\varphi_2$ are isomorphisms, but it seems that $\psi_2\varphi_1$ and $\psi_1\varphi_2$ have to be $0$ if I want $\psi\varphi$ to admit an inverse.

Any help is appreciated.

• Now I have understood why $\psi\varphi$ is an isomorphism. Thanks for your hint! – josephz Apr 15 '18 at 5:13