Point pushing one puncture around another

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].

The ordered lift and the push. The tuple $L_{\gamma }(t)=(q_1,\gamma (t))$ has pairwise distinct coordinates, since $\gamma (t)\mathrm{\ne }{q}_1$ for all $t$, and both coordinates in $\mathrm{int}{D}^2$, so $L_{\gamma }$ is a continuous loop in $F_2(\mathrm{int}{D}^2)$ with $L_{\gamma }(0)=(q_1,q_2)=L_{\gamma }(1)$; hence $\bar{\gamma }:=p_2\circ L_{\gamma }$ is a based loop of $C_2(\mathrm{int}{D}^2)$ at $[Q_2]$ by [F8], and [F1], available under the present hypothesis of the Axiom of Choice [F12], gives ${\mathrm{Push}}_2([\gamma ])=\delta ([\bar{\gamma }])\in \mathrm{PMod}(D^2,Q_2;\mathrm{\partial }{D}^2)$.

Homotopy to the rigid rotation loop. Define $x_1^s(t):=-h((1-s)+su(t))$ and $x_2^s(t):=x_1^s(t)+2hu(t)$ for $(s,t)\in I\times I$, and let ${\rho }_{-}(t):=[u(t)q_1,u(t)q_2]$ be the clockwise rigid rotation loop of the two marked points. The pair $(x_1^s(t),x_2^s(t))$ is continuous in $(s,t)$ by [F9] and [F10], lies in $F_2(\mathrm{int}{D}^2)$ because $x_2^s(t)-x_1^s(t)=2hu(t)\mathrm{\ne }0$ and because $\mathrm{\mid }{x}_1^s(t)\mathrm{\mid }\le h((1-s)+s\mathrm{\mid }u(t)\mathrm{\mid })=h$ and $\mathrm{\mid }{x}_2^s(t)\mathrm{\mid }\le 3h=\frac{1}{4}<1$ by [F10], and it satisfies $x_1^s(0)=x_1^s(1)=-h$ and $x_2^s(0)=x_2^s(1)=h$, so the initial and terminal tuples are $(q_1,q_2)$ for every $s$; at $s=0$ the pair is $(-h,-h+2hu(t))=(q_1,\gamma (t))=L_{\gamma }(t)$, and at $s=1$ it is $(-hu(t),hu(t))=(u(t)q_1,u(t)q_2)$, the ordered lift of ${\rho }_{-}$ from $Q_2$. Hence $(s,t)\mapsto (x_1^s(t),x_2^s(t))$ is a path homotopy relative to $\{0,1\}$ in $F_2(\mathrm{int}{D}^2)$ from $L_{\gamma }$ to the ordered lift of ${\rho }_{-}$, and composing with the quotient map $p_2$ of [F8] gives a path homotopy relative to $\{0,1\}$ from $\bar{\gamma }$ to ${\rho }_{-}$ in $C_2(\mathrm{int}{D}^2)$; therefore $[\bar{\gamma }]=[{\rho }_{-}]$ in ${\pi }_1(C_2(\mathrm{int}{D}^2),[Q_2])$.

The push is the positive full twist. By [F4] and [F5] and [F11], $\mathrm{\Psi }([{\sigma }_1])=\delta (({\iota }_{\ast }^C{)}^{-1}\mathrm{\Phi }([{\sigma }_1]))=\delta (({\iota }_{\ast }^C{)}^{-1}(({\iota }_{\ast }^C[S({\sigma }_1)]{)}^{-1}))=\delta ([S({\sigma }_1){]}^{-1})$, because the inverse of a group isomorphism preserves inverses; by [F3] this value is $[H_1]$, so $\delta ([S({\sigma }_1){]}^{-2})=\delta ([S({\sigma }_1){]}^{-1}{)}^2=[H_1{]}^2$ by the homomorphism property of [F6]. Combining with steps 2.1, 3.1 and 1.2, ${\mathrm{Push}}_2([\gamma ])=\delta ([\bar{\gamma }])=\delta ([{\rho }_{-}])=\delta ([S({\sigma }_1){]}^{-2})=[H_1{]}^2,$ and $[H_1{]}^2=\mathrm{\Psi }([{\sigma }_1]{)}^2=\mathrm{\Psi }([{\sigma }_1{]}^2)=\mathrm{\Psi }([{\sigma }_1\star {\sigma }_1])$ by [F3], [F4], [F11], the class of the square of the standard positive half twist; this class is pure, as it is a point push by [F1].

The counterclockwise push is the inverse. The loop ${\gamma }^{-}(t):=\gamma (1-t)$ is the reversed loop $\bar{\gamma }$ of [F7] at $q_2$, so $[{\gamma }^{-}]=[\gamma {]}^{-1}$ in ${\pi }_1(Y_2,q_2)$; since ${\mathrm{Push}}_2$ is a group homomorphism by [F1], ${\mathrm{Push}}_2([\gamma {]}^{-1})={\mathrm{Push}}_2([\gamma ]{)}^{-1}=[H_1{]}^{-2}$ by step 3.1, and the traces $t\mapsto q_1+2hu(1-t)$ of $q_2$ under the reversed loop are the counterclockwise parametrisation of the same circle: pushing $q_2$ counterclockwise around $q_1$ gives the inverse of the positive two-strand full twist.