Every geometric braid is braid-isotopic to a smooth braid

Empty, singleton and uniform margins. For $n=0$ choose the empty smooth braid and constant empty isotopy; every assertion is vacuous. Henceforth $n\ge 1$. The finite disk-boundary function $t\mapsto {\min }_j(1-\mathrm{\mid }{z}_j(t)\mathrm{\mid })$ is positive and continuous on $I$, so has positive minimum. If $n\ge 2$, the finite collision function $t\mapsto {\min }_{i<j}\mathrm{\mid }{z}_i(t)-z_j(t)\mathrm{\mid }$ also has positive minimum. Choose $m>0$ at most both minima when $n\ge 2$, and at most the boundary minimum alone when $n=1$. Put $\eta =\min \{\rho \mathrm{/}4,m\mathrm{/}8\}$, so $3\eta <\rho$ and $3\eta <m\mathrm{/}2$. No minimum of an empty pair list is used.

Approximate on a boundaryless domain. Extend the continuous tuple $z:I\to {\mathbb{R}}^{2n}$ to $\mathbb{R}$ by the constant tuple $z(0)$ for $t<0$ and $z(1)$ for $t>1$. This extension is continuous because its values agree at $0,1$. Apply [F2] on the boundaryless smooth manifold $\mathbb{R}$, with constant error $\eta$, and restrict the resulting smooth map to $I$. It gives $g=(g_1,\dots ,g_n)$ with $\mathrm{\mid }{g}_j(t)-z_j(t)\mathrm{\mid }<\eta$ for every $j,t$. Thus no boundaryless Whitney assertion is applied directly to $I$.

Exact endpoint collars, including preserved smooth germs. Choose $\delta \in (0,1\mathrm{/}2)$ so $\mathrm{\mid }{z}_j(t)-q_j\mathrm{\mid }<\eta$ on $[0,\delta ]$ and $\mathrm{\mid }{z}_j(t)-z_j(1)\mathrm{\mid }<\eta$ on $[1-\delta ,1]$ for all $j$. Take smooth cutoffs ${\chi }_0,{\chi }_1$ supported in these disjoint collars and equal to one on the half-sized endpoint collars. Ordinarily use the constant collar data $a_{j0}(t)=q_j$, $a_{j1}(t)=z_j(1)$. If the original strands are smooth in an endpoint neighbourhood, shrink the corresponding collar into that neighbourhood and instead use $a_{j0}(t)=z_j(t)$ or $a_{j1}(t)=z_j(t)$ there. The products with their cutoffs extend smoothly by zero off the collars. Define $h_j=(1-{\chi }_0-{\chi }_1)g_j+{\chi }_0{a}_{j0}+{\chi }_1{a}_{j1}$. This is smooth with $h_j(0)=q_j$, $h_j(1)=z_j(1)$. In the already-smooth case it agrees with $z_j$ throughout the smaller original endpoint neighbourhood, rather than replacing that neighbourhood by a constant.

All collar choices satisfy the same error bound. On either collar, the replacement error $\mathrm{\mid }{a}_{j\mathrm{\ell }}-z_j\mathrm{\mid }$ is less than $\eta$ for constant data and zero for preserved original data. Since the cutoffs have disjoint supports and weights sum to one, $\mathrm{\mid }{h}_j-z_j\mathrm{\mid }\le (1-{\chi }_0-{\chi }_1)\mathrm{\mid }{g}_j-z_j\mathrm{\mid }+{\chi }_0\mathrm{\mid }{a}_{j0}-z_j\mathrm{\mid }+{\chi }_1\mathrm{\mid }{a}_{j1}-z_j\mathrm{\mid }<3\eta \le 3m\mathrm{/}8<m\mathrm{/}2$. This covers both arbitrary continuous endpoints and already-smooth endpoint neighbourhoods.

The repaired motion is a braid, with the same endpoints. For all $i\mathrm{\ne }j$ and all $t$, $\mid h_i(t)-h_j(t)\mid \ge \mid z_i(t)-z_j(t)\mid -\mid h_i(t)-z_i(t)\mid -\mid h_j(t)-z_j(t)\mid >m-\frac{3m}{8}-\frac{3m}{8}=\frac{m}{4}>0,$ so the values $h_1(t),\dots ,h_n(t)$ are pairwise distinct; and $\mid h_j(t)\mid \le \mid z_j(t)\mid +\frac{3m}{8}<1$, so they lie in $D^{\circ }$. Together with $h_j(0)=q_j$ and $\{h_j(1)\}=\{z_j(1)\}=\{q_j\}$ from [F3] and step 3.1, this shows that $h$ is a braid based at $Q$, smooth by step 3.1, whose endpoint permutation equals that of $\beta$ because $h_j(1)=z_j(1)$ for every $j$.

The straight-line isotopy. Define $Z_j(s,t):=(1-s)z_j(t)+s{h}_j(t)$ for $(s,t)\in I\times I$. It is jointly continuous, $Z(0,\cdot )=\beta$ and $Z(1,\cdot )=h$, and every slice is a braid based at $Q$: for all $s,t,i\mathrm{\ne }j$, $\mid Z_j(s,t)-z_j(t)\mid =s\mid h_j(t)-z_j(t)\mid <\frac{3m}{8},\mid Z_j(s,t)\mid \le \mid z_j(t)\mid +\frac{3m}{8}<1,$ so the $n$ values are pairwise distinct (their pairwise distances are at least $m-\frac{3m}{8}-\frac{3m}{8}=\frac{m}{4}>0$) and lie in $D^{\circ }$; moreover $Z_j(s,0)=(1-s)q_j+s{q}_j=q_j$ and $Z_j(s,1)=(1-s)z_j(1)+s{z}_j(1)=z_j(1)$ for every $s$. Hence $Z$ is a braid isotopy from $\beta$ to the smooth braid $h$.

Conclusion. By step 6.1 the smooth braid $h$ is braid-isotopic to $\beta$ through the isotopy $Z$, whose deviation from $\beta$ satisfies $\mid Z_j(s,t)-z_j(t)\mid =s\mid h_j(t)-z_j(t)\mid <3\eta <\rho \text{for all }j,s,t,$ by the choice of $\eta$ in step 1.1. The isotopy keeps every bottom endpoint fixed pointwise and every individual top endpoint fixed, so the endpoint permutation is preserved by step 5.1. Where a smooth original collar was retained in step 3.1, $h=z$ there, so the entire straight-line isotopy is also fixed there. Taking ${\beta }^{\mathrm{\prime }}:=h$ proves the statement for the prescribed $\rho$, and $\rho >0$ was arbitrary.