A box is not a C1-boundary domain

Present Q by its 2n closed coordinate faces with normals $\pm e_i$ and the set E where at least two coordinates are endpoints. As in the explicit box verification F1, the parameter boundaries are finite unions of bounded coordinate hyperplanes and have measure zero, and every boundary point outside E has a single planar one-sided neighborhood. This verifies the finite piecewise hypotheses.

Let p have at least two endpoint coordinates i and j. Approach p through relative interiors of the i-face, moving every other endpoint coordinate slightly into (0,1); also approach through relative interiors of the j-face. The outward normals along these two sequences are the distinct constants ${\epsilon }_i{e}_i$ and ${\epsilon }_j{e}_j$, with ${\epsilon }_k=-1$ at endpoint 0 and $+1$ at endpoint 1. If a C1 graph chart as in F2 existed at p, its normal would be the continuous function $(-Dh,1)\mathrm{/}\sqrt{1+\mathrm{\mid }Dh{\mathrm{\mid }}^2}$, transformed by the chart rotation and given the unique outward sign. At neighboring planar points this normal must be the corresponding coordinate normal, since orthogonality to the plane and the outward side uniquely determine it. Continuity at p would force those two distinct constant vectors to have the same limit, a contradiction. This proves failure at every edge and vertex, including the origin.