Fix a stable $\infty$-category $\mathcal{C}$ and two (co)fibre sequences $a \rightarrow b \rightarrow c$ and $x \rightarrow y \rightarrow z$ in $\mathcal{C}$. Now suppose we are given a map $a \rightarrow x$ such that the composite $c[-1] \rightarrow a \rightarrow x \rightarrow y$ is homotopic to zero.

From this data, I would like to extend the map $a \rightarrow x$ to a map of (co)fibre sequences

$\begin{array}{cccccc} & a & \rightarrow & b & \rightarrow & c\\\ \\\ & \downarrow & & \downarrow & & \downarrow\\\ \\\ & x & \rightarrow & y & \rightarrow & z\\\ \end{array}$

and I would like to know exactly what choices are involved in such an extension.

I believe that such an extension exists: the universal property of cofibre sequences should give a map $b \rightarrow y$ so that we have the first two vertical arrows in a map of (co)fibre sequences, and then the argument that the homotopy category of $\mathcal{C}$ is triangulated allows us to construct the third vertical arrow $c \rightarrow z$.

Now the question is the following: *What choices are involved in the construction of the vertical arrows?*

For the second vertical arrow, $b \rightarrow y$, I think that the universal property of cofibre sequences should say that I need to actually specify a homotopy from the composite $c[-1] \rightarrow a \rightarrow x \rightarrow y$ to zero. Is that correct?

To construct the third vertical arrow, $c \rightarrow z$, we would likewise need to specify a homotopy from the composite $a \rightarrow b \rightarrow y \rightarrow z$ to zero, but it seems to me that this is already given by the previous choices. Is that correct?

If so, then we can fill in the third vertical arrow more or less canonically. Is this one of the ways in which stable $\infty$-categories are richer than triangulated categories, in which we have only the existence of *some* third vertical arrow $c \rightarrow z$?

(EDIT: Thanks to David White for providing the above diagram for the map of cofibre sequences.)