The manifold boundary of the explicit disk is the unit sphere

Let $m$ be any natural number, $D=\{x\in\mathbb R^{m+1}:\|x\|\leq1\}$, $H=\{y\in\mathbb R^{m+1}:y_0\geq0\}$, and $s=2e_0$. For each unit vector $p$, let $L_p:(-p)^\perp\to\mathbb R^m$ be the coordinate isometry from the orthonormal basis selected by the construction, put $S_p(u)=L_p\bigl(\frac{2}{1+\langle p,u\rangle}(u-\langle p,u\rangle p)\bigr)$ and $U_p(t)=\frac{4L_p^{-1}(t)+(4-\|t\|^2)p}{\|t\|^2+4}$, and define $N_p(0)=p$, $N_p(x)=x/\|x\|$ for $x\ne0$. In these total formulas a reciprocal of zero is zero. The interior chart has forward map $x\mapsto x+s$, inverse $y\mapsto\min\{1,\|y-s\|^{-1}\}(y-s)$, source $\{x\in D:\|x\|<1\}$, and target $\{y\in H:\|y-s\|<1\}$. The chart indexed by $p$ has forward map $x\mapsto(1-\|x\|,S_p(N_p(x)))$, inverse $(a,t)\mapsto\max\{0,\min\{1,1-a\}\}U_p(t)$, source $\{x\in D:x\ne0,\ x\ne\|x\|(-p)\}$, and target $\{(a,t)\in H:a<1\}$. Give $D$ the charted-space structure with that atlas and preferred chart equal to the interior chart when $\|x\|<1$ and otherwise equal to the chart indexed by $N_{e_0}(x)$, with $e_0$ the first standard unit vector. For the half-space model $I:H\to\mathbb R^{m+1}$ given by inclusion, with total inverse replacing the first coordinate by its maximum with zero, the set of points whose preferred extended-chart image lies in the frontier of $I(H)$ is exactly $\{x\in D:\|x\|=1\}$. Equivalently, the boundary defined by this model and these preferred charts consists precisely of the disk points with norm one. Equality is of subsets of the closed disk. No positive lower bound on $m$ is required, so the case $m=0$ is included and the right-hand set consists of the two endpoints of the interval $[-1,1]$.