Lemma: Dominance regions give the cell decomposition

The listed polygons cover the graph and have disjoint face interiors. The cellulation is constant on each open epoch $s<0$ and $s>0$. For $s>0$, the new cells lie strictly inside $\rho =3$ and are not cut by the horizontal cap.

Proof. For $s\le 0$, the bound $k\le h$ gives $s+k\le h$. The core is therefore the three triangles $V_i$.

For $s>0$, in the half-sector $\xi \ge \zeta$,

This segment joins $B_i$ to $C_i$. In the other half-sector,

joining $C_i$ to $B_{i+1}$. The two triangles are precisely the regions where $s+k$ dominates; their complement in the outer triangle is the listed simple pentagon.

At the new vertices,

Thus the new cells remain away from the crest. On them,

so the cap does not remove a new cell. Outside $\rho =3$, A:lem:family gives the horizontal faces. These exact dominance regions have disjoint interiors and exhaust the graph.