Punctured Disks, Mapping Classes, and Point Pushing — Examples

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. ∎

The clockwise loop. For $t\in I$ put $u(t):=\cos 2\pi t-i\sin 2\pi t$ and $\gamma (t):=q_1+2hu(t)$. Then $\gamma$ is continuous, because $t\mapsto 2\pi t$ is continuous, $\sin$ and $\cos$ are continuous by [F9], and the field operations of $\mathbb{C}$ are continuous by [F10]; further $\mathrm{\mid }u(t){\mathrm{\mid }}^2=u(t)\overline{u(t)}={\cos }^{2}2\pi t+{\sin }^{2}2\pi t=1$ by [F9] and [F10], so $\mathrm{\mid }u(t)\mathrm{\mid }=1$, and with $q_1=-h$, $q_2=h$ of [F8] this gives $\mathrm{\mid }\gamma (t)-q_1\mathrm{\mid }=\mathrm{\mid }2hu(t)\mathrm{\mid }=2h$ and $\mathrm{\mid }\gamma (t)\mathrm{\mid }\le \mathrm{\mid }{q}_1\mathrm{\mid }+2h=3h=\frac{1}{4}<1$; similarly $u(0)=1$ and $u(1)=\cos 2\pi -i\sin 2\pi =1$ by [F9], the latter since $\sin 2\pi =\sin (0+2\pi )=-\sin (0+\pi )=0$ and $\cos 2\pi =\cos (0+2\pi )=-\cos (0+\pi )=1$, so $\gamma (0)=q_1+2h=q_2=\gamma (1)$. Hence $\gamma$ is a continuous based loop of $Y_2=\mathrm{int}{D}^2\setminus \{q_1\}$ at $q_2$, because $\gamma (t)\mathrm{\ne }{q}_1$ and $\gamma (t)\in \mathrm{int}{D}^2$ for every $t$.

The rigid rotation loop is the inverse square of the sliced half twist. Put $\alpha :=S({\sigma }_1{)}^{-1}$, the reversed raw slice loop of the standard positive half twist; by [F2] and [F4] its underlying unordered loop is $s\mapsto S({\sigma }_1)(1-s)=[\rho (1-s),-\rho (1-s)]$, a loop at $[Q_2]$ since $\{\rho (1),-\rho (1)\}=\{q_2,q_1\}=\{\rho (0),-\rho (0)\}$, and by [F7] its class is $[\alpha ]=[S({\sigma }_1){]}^{-1}$. Let $P(t):=-\rho (1-2t)$ for $0\le t\le \frac{1}{2}$ and $P(t):=\rho (2-2t)$ for $\frac{1}{2}\le t\le 1$, a continuous path with $P(0)=-\rho (1)=q_1$, $P(\frac{1}{2})=-\rho (0)=q_2$, $P(1)=\rho (0)=q_1$ and $\mathrm{\mid }P(t)\mathrm{\mid }\le h$ by [F2]; a direct substitution of the definitions shows that $(P(t),-P(t))$ is exactly the ordered lift of $\alpha \ast \alpha$ from $Q_2$: on $[0,\frac{1}{2}]$ the pair is $(-\rho (1-2t),\rho (1-2t))$, the lift of the reversed slice from $Q_2$, and on $[\frac{1}{2},1]$ it is $(\rho (2-2t),-\rho (2-2t))$, the continuation of that lift from the swapped tuple $(-\rho (0),\rho (0))=(h,-h)=(\rho (1),-\rho (1))$ of [F2]. Let $Q(t):=-hu(t)=u(t)q_1$, so that $(Q(t),-Q(t))$ is the ordered lift of ${\rho }_{-}$ by [F8], and consider the linear interpolation $W_r(t):=(1-r)P(t)+rQ(t)$, $(r,t)\in I\times I$, which is continuous by [F9] and [F10]. It never vanishes: at $t=0$ and $t=1$ one has $P=Q=q_1$, so $W_r=q_1\mathrm{\ne }0$; at $t=\frac{1}{2}$ one has $u(\frac{1}{2})=\cos \pi -i\sin \pi =-1$ and $Q(\frac{1}{2})=-hu(\frac{1}{2})=h=q_2=P(\frac{1}{2})$ by [F9] and [F2]; and for $0<t<\frac{1}{2}$ both $P(t)$ and $Q(t)$ have second coordinate strictly positive, while for $\frac{1}{2}<t<1$ both have second coordinate strictly negative. Indeed the second coordinate of $\rho (u)$ is $-2uh$ for $u\le \frac{1}{2}$ and $2h(u-1)$ for $u\ge \frac{1}{2}$ by [F2], which is strictly negative for $0<u<1$, so $P$ has second coordinate strictly positive for $0<t<\frac{1}{2}$ and strictly negative for $\frac{1}{2}<t<1$, while $Q(t)=-h\cos 2\pi t+ih\sin 2\pi t$ has second coordinate $h\sin 2\pi t$, positive for $0<t<\frac{1}{2}$ by [F9] and negative for $\frac{1}{2}<t<1$ by [F9] since $\sin 2\pi t=-\sin (2\pi t-\pi )$ with $0<2\pi t-\pi <\pi$. Moreover $\mathrm{\mid }{W}_r(t)\mathrm{\mid }\le (1-r)\mathrm{\mid }P(t)\mathrm{\mid }+r\mathrm{\mid }Q(t)\mathrm{\mid }\le h$ by [F10], so $(W_r(t),-W_r(t))$ is a continuous family in $F_2(\mathrm{int}{D}^2)$ whose initial tuple is $(W_r(0),-W_r(0))=(q_1,q_2)$ and whose terminal tuple is $(W_r(1),-W_r(1))=(q_1,q_2)$, independently of $r$; hence it is a path homotopy relative to $\{0,1\}$ from the ordered lift of $\alpha \ast \alpha$ to the ordered lift of ${\rho }_{-}$, and passing to $C_2(\mathrm{int}{D}^2)$ by [F8] gives $[\alpha \ast \alpha ]=[{\rho }_{-}]$, that is $[S({\sigma }_1){]}^{-2}=[\alpha {]}^2=[\alpha \ast \alpha ]=[{\rho }_{-}]$ by [F7].