**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.
$\square$

## Comments (2)

Comment #425 by Artem Prihodko on

Comment #426 by Johan on

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