Corner-adapted strong approximations preserve orthant sides

Structure of a zero set. Fix a coordinate and an original simplex, writing $u={\rho }_{i}F$ and $Z=u^{-1}(0)\cap \sigma$. The zero set is a finite union of faces. Its maximal faces are pairwise disjoint: if two met at $b$, the hypothesis would identify their union of tangent cones with the zero face of the convex cone $C_b$ under the supporting form $d{u}_b$. This zero face is convex and corresponds to a face $\tau$ of $\sigma$ containing $b$: the supporting form vanishes on the direction space of the minimal face containing $b$, is nonnegative on every vertex direction $v-b$, and its zero cone is generated by that direction space and precisely the zero-valued vertex directions. A relative interior direction of that face must belong to one of the finitely many zero-face cones, forcing that face to contain $\tau$; hence $\tau \subset Z$, contrary to maximality of two distinct intersecting faces. Thus, near any maximal zero face $\tau$, the only zero face is $\tau$ itself. If $Z=\sigma$, plane preservation already forces ${\rho }_{i}G=0$ throughout $\sigma$; assume otherwise that $Z$ is nonempty and proper. For every $z\in \tau$ and every vertex $v$ outside $\tau$, the tangent hypothesis gives $d{u}_z(v-z)>0$. Their finite minimum over the compact face $\tau$ is a number ${\kappa }_{\tau }>0$. This is a bound on those transverse vertex directions, not on all unit directions outside the zero cone.

Distance lower bound. Near a proper maximal zero face $\tau$, write $x=(1-t)z+ty$ by barycentric projection, with $z\in \tau$, $y$ in the opposite face and $t$ the sum of coordinates outside $\tau$. For $x$ sufficiently near $\tau$, $t<1\mathrm{/}2$ and $\mathrm{\mid }x-z\mathrm{\mid }\le t\mathrm{diam}\sigma$ is arbitrarily small, uniformly along $\tau$. Step 1.1 gives $d{u}_z(y-z)\ge {\kappa }_{\tau }$ by expressing $y$ as a convex combination of outside vertices. Uniform continuity of $du$ on $\sigma$ follows directly by covering it with half-radius continuity balls, taking a finite subcover, and using the triangle inequality on each corresponding full-radius ball. It gives $d{u}_{z+s(x-z)}(y-z)\ge {\kappa }_{\tau }\mathrm{/}2$ for $0\le s\le 1$, after choosing the neighbourhood small enough. Segment integration yields $u(x)\ge t{\kappa }_{\tau }\mathrm{/}2\ge [{\kappa }_{\tau }\mathrm{/}(2\mathrm{diam}\sigma )]\mathrm{dist}(x,Z)$, since $z\in Z$ and $\mathrm{\mid }x-z\mathrm{\mid }\le t\mathrm{diam}\sigma$. Take the finite minimum of these positive constants over the maximal zero faces. If the compact remainder outside their neighbourhoods is nonempty, the continuous positive ratio $u(x)\mathrm{/}\mathrm{dist}(x,Z)$ has a positive minimum there by [F2]; if it is empty, omit this bound. Consequently there is $a>0$ with $u(x)\ge a\mathrm{dist}(x,Z)$ on the whole simplex. No such bound is needed when $Z$ is empty or the whole simplex.

Perturbing the coordinate functions. For $x\in \sigma$ choose a closest point $z\in Z_{i,\sigma }$ when the zero set is nonempty and proper; it exists by compactness of $Z_{i,\sigma }$ and the segment $[z,x]$ lies in the convex simplex. Since ${\rho }_{i}G(z)={\rho }_{i}F(z)=0$ on the zero subcomplex, segment integration of the piecewise smooth difference along $[z,x]$, split into the finitely many refined pieces and added, gives $\mathrm{\mid }{\rho }_{i}G(x)-{\rho }_{i}F(x)\mathrm{\mid }\le \delta \mathrm{\mid }x-z\mathrm{\mid }=\delta \mathrm{dist}(x,Z_{i,\sigma })$, using the derivative bound of the strong approximation. Choose $\delta <a\mathrm{/}2$ with $a$ from step 2.1. Then ${\rho }_{i}G(x)\ge (a\mathrm{/}2)\mathrm{dist}(x,Z_{i,\sigma })$, so ${\rho }_{i}G\ge 0$ on $\sigma$ with equality exactly on $Z_{i,\sigma }$. If $Z_{i,\sigma }=\sigma$, plane preservation gives ${\rho }_{i}G=0$ there. If $Z_{i,\sigma }$ is empty, then ${\rho }_{i}F$ has a positive minimum $m_{\sigma }$ on the compact simplex by [F2], and the value error $\mathrm{\mid }G-F\mathrm{\mid }<\delta$ with $\delta <m_{\sigma }\mathrm{/}2$ keeps ${\rho }_{i}G$ positive on all of $\sigma$.

Conclusion. Choose $\delta >0$ below ${\delta }_0$ of [F1], every $a\mathrm{/}2$ for a proper nonempty zero set, and every $m_{\sigma }\mathrm{/}2$ for an empty zero set, omitting nonexistent cases. Every strong $\delta$-approximation $G$ preserving the zero subcomplexes then has ${\rho }_{i}G\ge 0$ with zero set exactly $Z_{i,\sigma }$ on every simplex, for every $i$: all coordinate functions are nonnegative and the coordinate zero sets are exactly the prescribed ones, so $G$ maps $K$ into the same closed orthant with the same boundary structure. By [F1] $G$ is again a whole-star embedding. This supplies the side control that affine-plane preservation alone does not give, as the half-plane warning in the cited Munkres exercise requires.