# Question on the fill-in morphism in a triangulated category

Let $$\begin{array}{rcl} A&\to& B\\ \downarrow & &\downarrow\\ A'&\to& B' \end{array}$$ be a commutative diagram in a triangulated category. By the axioms of a triangulated category, this may be 'completed' to $$\begin{array}{rccccl} A&\to& B&\to& C&\to\\ \downarrow & &\downarrow&&\downarrow\\ A'&\to& B'&\to& C'&\to \end{array}$$ where both rows are exact triangles. Let $$\begin{array}{rccccl} A&\to& B&\to& \tilde C&\to\\ \downarrow & &\downarrow&&\downarrow\\ A'&\to& B'&\to& \tilde C'&\to \end{array}$$ be another 'completion' of the initial diagram such that both rows are exact triangles. One gets that $C\cong \tilde C$ and $C'\cong \tilde C'$. My question is: May these two isomorphism be chosen in a way that the diagram $$\begin{array}{rcl} C&\overset{\cong}{\to}& \tilde C\\ \downarrow & &\downarrow\\ C'&\overset{\cong}{\to}& \tilde C' \end{array}$$ commutes?

For example, let the original commutative square be $$\begin{array}{ccc} X[-1]&\to&0\\ \downarrow & &\downarrow\\ 0&\to&X \end{array}$$ for any non-zero object $X$.
This can be completed to a diagram $$\begin{array}{cccccccc} X[-1]&\to& 0&\to& X&\to &X\\ \downarrow & &\downarrow&&\downarrow&&\downarrow\\ 0&\to& X&\to& X&\to& 0 \end{array}$$ with a completely free choice of the vertical map $X\to X$.
• Dear Jeremy Rickard, I am sorry but I don't understand your answer anymore: Isn't only the zero map $X\to X$ a correct completion of the diagram? The identity wouldn't make the diagram commute. Did I understand you correctly? – user8463524 Feb 4 '15 at 12:49