Lemma 13.4.10. Let $\mathcal{D}$ be a pre-triangulated category. Let $(X, Y, Z, f, g, h)$ and $(X', Y', Z', f', g', h')$ be triangles. The following are equivalent

1. $(X \oplus X', Y \oplus Y', Z \oplus Z', f \oplus f', g \oplus g', h \oplus h')$ is a distinguished triangle,

2. both $(X, Y, Z, f, g, h)$ and $(X', Y', Z', f', g', h')$ are distinguished triangles.

Proof. Assume (2). By TR1 we may choose a distinguished triangle $(X \oplus X', Y \oplus Y', Q, f \oplus f', g'', h'')$. By TR3 we can find morphisms of distinguished triangles $(X, Y, Z, f, g, h) \to (X \oplus X', Y \oplus Y', Q, f \oplus f', g'', h'')$ and $(X', Y', Z', f', g', h') \to (X \oplus X', Y \oplus Y', Q, f \oplus f', g'', h'')$. Taking the direct sum of these morphisms we obtain a morphism of triangles

$\xymatrix{ (X \oplus X', Y \oplus Y', Z \oplus Z', f \oplus f', g \oplus g', h \oplus h') \ar[d]^{(1, 1, c)} \\ (X \oplus X', Y \oplus Y', Q, f \oplus f', g'', h''). }$

In the terminology of Remark 13.4.4 this is a map of special triangles (because a direct sum of special triangles is special) and we conclude that $c$ is an isomorphism. Thus (1) holds.

Assume (1). We will show that $(X, Y, Z, f, g, h)$ is a distinguished triangle. First observe that $(X, Y, Z, f, g, h)$ is a special triangle (terminology from Remark 13.4.4) as a direct summand of the distinguished hence special triangle $(X \oplus X', Y \oplus Y', Z \oplus Z', f \oplus f', g \oplus g', h \oplus h')$. Using TR1 let $(X, Y, Q, f, g'', h'')$ be a distinguished triangle. By TR3 there exists a morphism of distinguished triangles $(X \oplus X', Y \oplus Y', Z \oplus Z', f \oplus f', g \oplus g', h \oplus h') \to (X, Y, Q, f, g'', h'')$. Composing this with the inclusion map we get a morphism of triangles

$(1, 1, c) : (X, Y, Z, f, g, h) \longrightarrow (X, Y, Q, f, g'', h'')$

By Remark 13.4.4 we find that $c$ is an isomorphism and we conclude that (2) holds. $\square$

Comment #425 by Artem Prihodko on

Hello!

I'm a bit confused by this proofs. In both $(1) \Rightarrow (2)$ and vice versa after some work we get morphism of triangles of the form (1,1,c). Then by lemma 13.4.3 we conclude that because two of three morphisms are isomorphism then so is third.

But this lemma hold only for morphism of triangles that we already know to be distinguished. So we using fact that we want to prove.

May be I do not understand something? Thank you!

Comment #426 by on

Ha, good catch! Fixed here. If you are ever in NY come by my office and I will give you a stacks project mug (If I haven't run out of them by then).

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