A supported half-twist homeomorphism

The collar function. By [F3] the function $\sigma$ is smooth on $\mathbb{R}$ with values in $[0,1]$, equals $0$ on $(-\mathrm{\infty },0]$ and equals $1$ on $[1,\mathrm{\infty })$; the argument $r\mapsto (\frac{11h}{8}-r)\mathrm{/}\frac{h}{8}$ is smooth and affine on $[0,\mathrm{\infty })$ with $\frac{11h}{8}-r\ge \frac{h}{8}\cdot 1$, that is $r\le \frac{5h}{4}$, exactly when $\sigma$ is evaluated at an argument at least $1$, and $\frac{11h}{8}-r\le 0$, that is $r\ge \frac{11h}{8}$, exactly when it is evaluated at an argument at most $0$. Hence $\theta (r)=\sigma ((\frac{11h}{8}-r)\mathrm{/}\frac{h}{8})$ is smooth on $[0,\mathrm{\infty })$ with values in $[0,1]$, equals $1$ for $r\le \frac{5h}{4}$ and equals $0$ for $r\ge \frac{11h}{8}$.

The half rotation, its support and its point motions. Write $r(x):=\parallel x-m_i{\parallel }_2$ and ${\alpha }_s(x):=\pi s\theta (r(x))$ for $x\in D^2$ and $s\in I$, so that $H_s(x)=m_i+R({\alpha }_s(x))(x-m_i)$; the scalar ${\alpha }_s(x)$ is a continuous function of $(s,x)$ because $\theta$ is smooth, and the entries of $R$ are $\cos$ and $\sin$ of that scalar, so $H_s(x)$ depends continuously on $(s,x)$ by [F6], and also $R$ is a rotation, hence preserves norms and is injective. In polar coordinates $x=m_i+(u\cos \phi ,u\sin \phi )$ with $u=r(x)$ one has $H_s(x)=m_i+(u\cos (\phi +\pi s\theta (u)),u\sin (\phi +\pi s\theta (u)))$ and the map $(u,\phi )\mapsto (u,\phi -\pi s\theta (u))$ is a two-sided inverse, so each $H_s$ is a bijection of $D^2$ continuous in both directions, that is a homeomorphism, and its inverse is as displayed. By step 1.1, $\theta (r(x))=0$ whenever $r(x)\ge \frac{11h}{8}$, so $H_s(x)=x$ for every $x$ outside the closed disc of radius $\frac{11h}{8}$ about $m_i$; that closed disc is contained in the open disc $U_i$ of radius $\frac{3}{2}h$ because $\frac{11}{8}<\frac{3}{2}$, and $U_i\subseteq \mathrm{int}{D}^2$ by [F1], so every $H_s$ fixes $\mathrm{\partial }{D}^2$ pointwise and fixes every point outside $U_i$; moreover $H_0=\mathrm{id}$ because ${\alpha }_0=0$ and $R(0)$ is the identity. For the marked points, [F1] gives $r(q_i)=r(q_{i+1})=h\le \frac{5h}{4}$, so $\theta =1$ there and, using the definition of $R$ as rotation about the origin and the shift formulas of [F5] with $x=0$ and $x=\pi$ respectively, $H_s(q_i)=m_i+R(\pi s)(-h,0)=m_i+h(-\cos \pi s,-\sin \pi s),H_s(q_{i+1})=m_i+R(\pi s)(h,0)=m_i+h(\cos \pi s,\sin \pi s);$ the two moving points are always distinct because their difference is $2h(\cos \pi s,\sin \pi s)\mathrm{\ne }0$, and every other base point $q_k$ has $r(q_k)\ge 3h>\frac{11h}{8}$ by [F1], hence is fixed for all $s$. By [F5] one has $H_1(q_i)=m_i+h(1,0)=q_{i+1}$ and $H_1(q_{i+1})=m_i+(-h,0)=q_i$, while all other base points are fixed, so $H_1$ preserves $Q_n$ setwise and by [F4] lies in ${\mathrm{Homeo}}^{+}(D^2,\mathrm{\partial }{D}^2;Q_n)$; the pair $\{H_s(q_i),H_s(q_{i+1})\}=\{m_i\pm h(\cos \pi s,\sin \pi s)\}$ traverses the anticlockwise semicircle of radius $h$ about $m_i$ from $\{q_i,q_{i+1}\}$ at $s=0$, through $\{m_i\pm (0,h)\}$ at $s=\frac{1}{2}$, to $\{q_{i+1},q_i\}$ at $s=1$, and $s\mapsto H_s(Q_n)$ is a continuous path in $F_n(\mathrm{int}{D}^2)$ with $[H_0(Q_n)]=[Q_n]=[H_1(Q_n)]$, so $s\mapsto [H_s(Q_n)]$ is a based loop of $C_n(\mathrm{int}{D}^2)$ at $[Q_n]$ by [F7].

Interpolation to the published diamond half twist. Let $\rho$ be the diamond path of [F1], so that the raw slice loop of the half twist is $S({\sigma }_i)(s)=[m_i+\rho (s),m_i-\rho (s)]$ with all other coordinates equal to $q_k$, and let $w_r(s):=(1-r)\rho (s)+rh(-\cos \pi s,-\sin \pi s)$ for $(r,s)\in I\times I$, a continuous map. For $0<s<1$ the second coordinate of $\rho (s)$ is $-2sh$ for $s\le \frac{1}{2}$ and $2h(s-1)$ for $s\ge \frac{1}{2}$ by [F1], both strictly negative, while the second coordinate of $h(-\cos \pi s,-\sin \pi s)$ is $-h\sin \pi s$, strictly negative because $\sin \pi s>0$; hence the convex combination $w_r(s)$ has strictly negative second coordinate and does not vanish. At $s=0$ one has $\rho (0)=(-h,0)=h(-\cos 0,-\sin 0)$ and at $s=1$ one has $\rho (1)=(h,0)=h(-\cos \pi ,-\sin \pi )$ by [F1] and [F5], so $w_r(0)=(-h,0)\mathrm{\ne }0$ and $w_r(1)=(h,0)\mathrm{\ne }0$ for every $r$. Also $\parallel \rho (s){\parallel }_2\le h$ and $\parallel h(-\cos \pi s,-\sin \pi s){\parallel }_2=h$, so $\parallel w_r(s){\parallel }_2\le h$ and the unordered pairs $\{m_i\pm w_r(s)\}$ lie in $U_i\subseteq \mathrm{int}{D}^2$. Therefore the formula $H^{\mathrm{i}\mathrm{n}\mathrm{t}}(r,s):=[m_i+w_r(s),m_i-w_r(s),q_k(k\mathrm{\notin }\{i,i+1\})]$ defines a continuous map $I\times I\to C_n(\mathrm{int}{D}^2)$, as the composite of a continuous ordered tuple with the continuous quotient map of [F7], whose every slice is collision-free: the two moving points differ by $2w_r(s)\mathrm{\ne }0$ and have distance at most $h$ from $m_i$, while every other base point has distance at least $3h$ from $m_i$ by [F1]. At $r=0$ the slice is the raw slice loop $S({\sigma }_i)$ of [F1] and at $r=1$ it is the loop $s\mapsto [H_s(Q_n)]$ of step 2.1, because $h(-\cos \pi s,-\sin \pi s)$ is the moving coordinate computed there; both loops start and end at $[Q_n]$, so $H^{\mathrm{i}\mathrm{n}\mathrm{t}}$ is a path homotopy relative to $\{0,1\}$ from $S({\sigma }_i)$ to the based loop of step 2.1.

The class of the supported half rotation. By [F2], available under the present hypothesis of the Axiom of Choice [F8], the isomorphism $\mathrm{\Psi }$ sends the class of the standard positive half twist to the class of the explicit boundary-fixed homeomorphism $H_i$ built in that item from the collar function $\theta$ and the rotation formula displayed in step 2.1, which is literally the homeomorphism $H_1$ of step 2.1 and from [F1] has the same supplied data $m_i$, $q_i$, $q_{i+1}$; hence $\mathrm{\Psi }([{\sigma }_i])=[H_1]$ in $\mathrm{Mod}(D^2,Q_n;\mathrm{\partial }{D}^2)$, and $H_1$ is a homeomorphism of $D^2$ fixing $\mathrm{\partial }{D}^2$ pointwise and exchanging the two adjacent punctures, supported in the disc $U_i$. Independently, step 3.1 exhibits the based loop $s\mapsto [H_s(Q_n)]$ as path-homotopic relative to $\{0,1\}$ to the raw slice of the standard positive half twist, so the explicit time-one map $H_1$ represents the standard positive braid generator. ∎