Lemma: Exact decomposition

The displayed active cells cover the graph, with disjoint face interiors. The incoming core faces are triangles; the outgoing $V_i$ are simple pentagons and may be nonconvex.

Proof. For $s\le 0$, $s+k_{\mu }\le h_{\mu }$, because $h_{\mu }-k_{\mu }\ge c\rho \ge 0$. Thus the uncapped core is $T=s+h_{\mu }$.

For $s>0$, in the half-sector $\xi \ge \zeta$, equality of the nonconstant branches is

Its endpoints are $(\beta s,0)$ and $(\chi s,\chi s)$. The other half-sector has boundary

The dominance region is the union of the two stated inner triangles. Its complement within $\rho \le r$ is the stated pentagon. The inequalities $a-b>0$ and $\rho (C_i)=s/c<r$ establish the placement and prevent a cap intersection. Outside $\rho =r$, the graph is resting.