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.


You've used the fact that a morphism of distinguished triangles is an isomorphism as soon as two of its components are. Remember the reason that this is true: one uses the five lemma on the long exact sequences induced by maps from an arbitrary object to the triangles and concludes via Yoneda. But more triangles than just the distinguished ones have this property that the representables send them to long exact sequences. The Stacks project calls such triangles "special", and you can check that the sum of special triangles is special.

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

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