I’m trying to understand Lurie’s proof that the homotopy category of a stable $\infty$-category is triangulated. In showing TR2, he constructs a diagram $$\require{AMScd} \begin{CD} X @>f>> Y @>>> 0\\ @VVV @VVV @VVV \\ 0’ @>>> Z @>>> W \\ @. @VVV @VuVV \\ @. 0’’ @>>> V \end{CD}$$ in which every square is a pushout in some stable $\infty$-category $\mathcal C$. He then asserts a map between the suspensions $$\require{AMScd} \begin{CD} X @>>> 0\\ @VVV @VVV \\ 0’ @>>> W \end{CD}$$ $$\require{AMScd} \begin{CD} Y @>>> 0\\ @VVV @VVV \\ 0’’ @>>> V \end{CD}$$ giving rise to commutative square $$\require{AMScd} \begin{CD} W @>>> X[1]\\ @VuVV @Vf[1]VV \\ V @>>> Y[1] \end{CD}$$ in $h\mathcal C$. I think in fact what is needed for this commutative square is that the above map between suspensions specifically have components $f$ and $u$ between the initial vertices and terminal vertices, respectively. By repeated application of HTT.4.3.2.15, the first (large) diagram is determined up to contractible choices by $f$, and, if $\mathcal C$ is stable, also determined up to contractible choices by $u$. Similarly, the map of suspensions is determined by the map $X\to Y$ or by the map $X[1]\simeq W\to V\simeq Y[1]$ (as $\Sigma$ is an equivalence). So the data of the large diagram is equivalent to the data of a map between the suspensions, and each is specified two different ways (via $f$/the map on initial vertices or $u$/the map on cocone points). I want to know why these are in correspondence. Another way of saying this is, the map $f$ determines, via the large diagram, a map $u$, which determines a map $X[1]\to Y[1]$ (always up to contractible choices), but why is this map homotopic to $f[1]$? This is equivalent to the assertion that the construction of the large diagram computes the suspension functor $\Sigma$ on edges of $\mathcal C$. One way to see this would be to construct a map between suspensions (a diagram $\Delta^1\times\Delta^1\times\Delta^1\to\mathcal C$) directly from the data of the large diagram, but I don’t see how to fill in the most obvious candidate. (I could do so if I knew how to fill an outer horn $\Delta^3_0\hookrightarrow\mathcal C$ with the property that both interior vertices are zero objects, so every edge is a zero map.) Any tips would be a big help – thank you!

**Edit 5/20**
I can see the claim if it is true that given a cofiber sequence $X\to Y\to Z$ corresponding to a pushout
$$\require{AMScd}
\begin{CD}
X @>>> Y\\
@VVV @VVV \\
0 @>>> Z,
\end{CD}$$
then if I replace the bottom 2-simplex ($X\to 0\to Z$) of the diagram $\Delta^1\times\Delta^1\to\mathcal C$ with any other one with the same vertices and long edge, the resulting diagram $\Delta^1\times\Delta^1\to\mathcal C$ is still a pushout.