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I am learning the basics of general triangulated categories, and I have the following question: Let $\mathcal T$ be a triangulated category and let $A,B,C,D$ be objects such that we have two exact triangles $A\xrightarrow{f}B\to C\to$ and $A\xrightarrow{f}B\to D\to$. Then, is it true that $C\cong D$ ?

If this is not true in general, is it at least true when $\mathcal T$ is the bounded derived category of an abelian category?

Thanks in advance.

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Yes, it's true. First, given $$\require{AMScd} \begin{CD} A @>{f}>> B @>>> C\\ @V{=}VV @V{=}VV \\ A @>{f}>> B @>>> D \end{CD} $$ where the rows are exact triangles, there is a map $C \to D$ making the diagram commute (this is one of the axioms of a triangulated category). Second, given a map between exact triangles like this in which two of the three vertical maps are isomorphisms, so is the third: see 13.4.3 in https://stacks.math.columbia.edu/tag/05QN, for example.

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