Lemma: Exact face decomposition

These faces cover the graph and have disjoint face interiors. For $s>0$, the two inner triangles are contained strictly inside the cap boundary $\rho =r$.

Proof. For $s\le 0$, $s+k\le h$, by $h-k\ge d\rho$. For $s>0$, in $\xi \ge \eta$, the interface of the two nonconstant branches is

Its endpoints are $(\beta s,0)$ and $(\chi s,\chi s)$. The other half-sector has the reflected equation. The dominance region is exactly the union of the two inner triangles; its complement in the core triangle is the stated simple pentagon. Lemma lem:bounds excludes contact with the cap. Outside the cap the graph is horizontal.