Localized cone triangles satisfy the octahedral axiom

Clear a composable pair $\alpha :QX\to QY$, $\beta :QY\to QZ$ simultaneously to ordinary $f:X\to Y^{\mathrm{\prime }}$, $g:Y^{\mathrm{\prime }}\to Z^{\mathrm{\prime }}$, using denominator isomorphisms of the objects. This includes zero maps or identity maps. Their composite becomes $gf$.

Apply TR4 in $K(\mathcal{A})$ to $f,g$, using cone triangles with structure maps $i_f,p_f$ and similarly for $g,gf$. It gives $a:C_f\to C_{gf}$ and $b:C_{gf}\to C_g$ such that $a{i}_f=i_{gf}g$ on $Y^{\mathrm{\prime }}$, $p_{gf}a=p_f$, $b{i}_{gf}=i_g$, and $p_{g}b=f[1]p_{gf}$. In particular $C_f\stackrel{a}{\to }{C}_{gf}\stackrel{b}{\to }{C}_g\stackrel{i_f[1]p_g}{\to }{C}_f[1]$ is distinguished. These equations, rather than the cone objects alone, are the octahedral data.

Apply the additive functor $Q$ to the entire octahedron: every face equation survives, including $Q(i_f[1]p_g)=Q(i_f)[1]Q(p_g)$, and its fourth triangle is distinguished by definition. Transport along the object isomorphisms from the clearing step.

If the three triangles specified in TR4 are other completions, TR3 compares each with the constructed completion with identity first two components. Such a comparison has invertible third component: applying either representable Hom and the exact sequences established from TR1–TR3 proves this by the five-term argument, then the Hom criterion for invertibility. Transport the octahedral maps along these triangle isomorphisms. This gives TR4 for the originally prescribed completions with all signs unchanged.