Isomorphism between distinguished triangles in a triangulated category

Question

$\DeclareMathOperator{\id}{id}$ Throughout this post "d.t." stands for "distinguished triangles". I'm studying some category theory and the book I'm following has the following exercise :

Let $X\xrightarrow{f}Y\xrightarrow{g}Z\xrightarrow{h}T(X)$ be a distinguished triangle in a triangulated category $(\mathcal{C},T)$.

Show that if $h=0$ then the d.t. above is isomorphic to the d.t. $X\xrightarrow{}X\oplus Z\xrightarrow{}Z\xrightarrow{0}T(X)$

There is a hint, suggesting to first constructing an arrow $Y\to X\oplus Z$ using the 4th axiom for a triangulated category and then using a proposition that was proved earlier on : Given a morphism of d.t. as below, with $\alpha, \beta$ being isomorphisms, then $\gamma$ is also an isomorphism. $\require{AMScd}$ \begin{CD} X@>f>> Y@>g>> Z @>h>> TX\\ @V \alpha VV @V\beta VV @V\gamma VV @V T\alpha VV\\ X'@>f'>> Y' @> g'>> Z'@>h'>> TX' \end{CD}

The axiom being referred to in the hint is the following :

Given two d.t. and two morphisms as in the diagram below

$\require{AMScd}$ \begin{CD} X@>f>> Y@>g>> Z @>h>> TX\\ @V \alpha VV @V\beta VV @. @V T\alpha VV\\ X'@>f'>> Y' @> g'>> Z'@>h'>> TX' \end{CD}

then there exists a morphism $\gamma : Z \to Z'$ giving rise to a morphism of d.t. .

So supposedly, from these two facts, I should be able to prove the exercise. But I'm not sure how to construct $Y\to X\oplus Z$ because in the axiom, the "missing" arrow is in the third column whereas in the exercise we would have something like $\require{AMScd}$ \begin{CD} X@>f>> Y@>g>> Z @>0>> TX\\ @V \id VV @. @V\id VV @V \id VV\\ X@>>> X\oplus Z @>>> Z@>0>> TX \end{CD}

My guess is that in this case ($h=0$) we have a "2 out of 3" property or something ? I'm very new to triangulated categories so I'm not sure where to go here.

Answer 1

Answering my own question so it can be closed.

What I was missing to complete the question was the following observation :

One of the axioms of triangulated categories is the following statement :

A triangle $X\xrightarrow{f} Y\xrightarrow{g} Z \xrightarrow{h} TX$ is a d.t if and only if $Y\to Z \to TX \xrightarrow{-Tf} TY$ is a d.t..

This statement can be used to fill in the missing arrow, even when that arrow isn’t ‘in the third column’

Suppose we have the following diagram where the rows are d.t. :

$\require{AMScd}$ \begin{CD} X@>f>> Y@>g>> Z @>h>> TX\\ @V \alpha VV @. @V \gamma VV @V T\alpha VV\\ X'@>f'>> Y' @> g'>> Z'@>h'>> TX' \end{CD}

By the axiom above we have the diagram :

$\require{AMScd}$ \begin{CD} X@>f>> Y@>g>> Z @>h>> TX @> -Tf>> TY@> -Tg>> TZ\\ @V \alpha VV @. @V \gamma VV @V T\alpha VV@.@VVV\\ X'@>f'>> Y' @> g'>> Z'@>h'>> TX'@>-Tf’>> TY’@>-Tg’>> TZ’ \end{CD}

Where $Z^{\bullet}\to TX^{\bullet} \to TY^{\bullet}\to TZ^{\bullet}$ are d.t.. Hence we can fill in the arrow $TY\to TY’$, applying $T^{-1}$ (by definition $T$ is an automorphism) we do get a map $Y\to Y’$. This solves the problem I was having.