Setwise boundary preservation kills a nontrivial braid

The full rotation loop in the unordered configuration space. For $t\in I$ put ${\rho }_{-}(t):=[e^{-2\pi it}{q}_1,e^{-2\pi it}{q}_2]\in C_2(\mathrm{int}{D}^2)$. The two coordinates are distinct and both of modulus $\mathrm{\mid }{q}_j\mathrm{\mid }=h=\frac{1}{12}<1$, so the tuple $(e^{-2\pi it}{q}_1,e^{-2\pi it}{q}_2)$ lies in $F_2(\mathrm{int}{D}^2)$ for every $t$, and the map $t\mapsto (e^{-2\pi it}{q}_1,e^{-2\pi it}{q}_2)$ is continuous because complex multiplication is; hence ${\rho }_{-}$ is a well-defined continuous path in $C_2(\mathrm{int}{D}^2)$ with ${\rho }_{-}(0)=[Q_2]={\rho }_{-}(1)$, since $e^0=e^{-2\pi i}=1$. So ${\rho }_{-}$ is a based loop at $[Q_2]$, and its class lies in ${\pi }_1(C_2(\mathrm{int}{D}^2),[Q_2])$, the group of [F11].

The quotient test map. Define $\tilde{u}:F_2(\mathrm{int}{D}^2)\to S^1$ by $\tilde{u}(x_1,x_2):=((x_1-x_2)\mathrm{/}\mathrm{\mid }{x}_1-x_2\mathrm{\mid }{)}^2$, a map into the unit circle, since $x_1\mathrm{\ne }{x}_2$ on the ordered configuration space and the difference, the modulus, division by a positive modulus and squaring are continuous. Swapping the two coordinates replaces $x_1-x_2$ by $-(x_1-x_2)$, so $\tilde{u}(x_2,x_1)=(-(x_1-x_2)\mathrm{/}\mathrm{\mid }{x}_1-x_2\mathrm{\mid }{)}^2=\tilde{u}(x_1,x_2)$: the map is constant on the $S_2$-orbits. By [F9] applied to the quotient map $p_2:F_2(\mathrm{int}{D}^2)\to C_2(\mathrm{int}{D}^2)$ there is a unique continuous $u:C_2(\mathrm{int}{D}^2)\to S^1$ with $u([x_1,x_2])=\tilde{u}(x_1,x_2)$ for every ordered pair; in particular $u(\{x,y\})=((x-y)\mathrm{/}\mathrm{\mid }x-y\mathrm{\mid }{)}^2$ is well defined as a function of the unordered pair.

The sliced half twist and the explicit half-turn loop. Let $S({\sigma }_1)$ be the unordered slice of the standard positive half twist, so that $S({\sigma }_1)(s)=[{\sigma }_1(s)]=[m_1+\rho (s),m_1-\rho (s)]$ for $s\in I$; with $m_1=(0,0)$ this is the unordered pair $\{\rho (s),-\rho (s)\}$, and the reversed path is $S({\sigma }_1{)}^{-1}(s)=S({\sigma }_1)(1-s)=\{\rho (1-s),-\rho (1-s)\}$, a loop at $[Q_2]$ because $\rho (0)=(-h,0)$ and $\rho (1)=(h,0)$ give $\{\rho (1),-\rho (1)\}=\{q_2,q_1\}=[Q_2]=\{\rho (0),-\rho (0)\}$. Put $\alpha (s):=\{h(\cos \pi s,-\sin \pi s),-h(\cos \pi s,-\sin \pi s)\}=\{\pm h{e}^{-\pi is}\}$ for $s\in I$, again a loop at $[Q_2]$. First, the concatenation $\alpha \ast \alpha$ is the loop $t\mapsto [e^{-2\pi it}{Q}_2]$: for $t\le \frac{1}{2}$ one has $(\alpha \ast \alpha )(t)=\alpha (2t)=\{\pm e^{-2\pi it}h\}=\{e^{-2\pi it}{q}_1,e^{-2\pi it}{q}_2\}$, and for $t\ge \frac{1}{2}$ one has $(\alpha \ast \alpha )(t)=\alpha (2t-1)=\{\pm e^{-\pi i(2t-1)}h\}=\{\pm e^{-2\pi it}h\}$, the middle pair being unchanged because the sign $e^{\pi i}=-1$ is absorbed by $\pm$. Second, $S({\sigma }_1{)}^{-1}$ is path-homotopic to $\alpha$ relative to $\{0,1\}$: the interpolation $w_r(s):=(1-r)\rho (1-s)+rh(\cos \pi s,-\sin \pi s)$ satisfies $w_0(s)=\rho (1-s)$ and $w_1(s)=h(\cos \pi s,-\sin \pi s)$, and it never vanishes, because for $0<s<1$ the second coordinate of $\rho (1-s)$ is $-2(1-s)h$ for $s\ge \frac{1}{2}$ and $-2sh$ for $s\le \frac{1}{2}$, both strictly negative, the second coordinate $-h\sin \pi s$ of $h(\cos \pi s,-\sin \pi s)$ is strictly negative, so the convex combination has strictly negative second coordinate, while at $s=0$ and $s=1$ the two endpoints coincide and equal $(h,0)$ and $(-h,0)$; since $\rho (1-s)$ has modulus at most $h$ and $h(\cos \pi s,-\sin \pi s)$ has modulus exactly $h$, every $w_r(s)$ has modulus at most $h<1$ and the unordered pairs $\{w_r(s),-w_r(s)\}$ lie in $C_2(\mathrm{int}{D}^2)$ and depend continuously on $(r,s)$.

The lift and its endpoint. By [F2], which is available under the Axiom of Choice [F13], the loop ${\rho }_{-}$ of step 1.1 lifts to a continuous path $P:I\to E$ with $P_0=\mathrm{id}$ and $\mathrm{ev}(P_t)={\rho }_{-}(t)$ for all $t$; put $f:=P_1^{-1}$. Since $\mathrm{ev}(P_1)={\rho }_{-}(1)=[Q_2]$, the homeomorphism $P_1$ lies in the fibre $F$ of [F2], so $f\in F$ and its class $[f]$ lies in $\mathrm{Mod}(D^2,Q_2;\mathrm{\partial }{D}^2)={\pi }_0(F)$ by [F1]; as the inverse of an element of $E$, the homeomorphism $f$ fixes $\mathrm{\partial }{D}^2$ pointwise, and it preserves $Q_2$ setwise because $P_1$ does. For every $t$ the tuples $P_t(Q_2)$ and $e^{-2\pi it}{Q}_2$ have the same image under $\mathrm{ev}$, hence lie in the same $S_2$-orbit, so by [F8] their underlying coordinate sets agree: $P_t(\{q_1,q_2\})=\{e^{-2\pi it}{q}_1,e^{-2\pi it}{q}_2\}$ as sets. Applying $P_t^{-1}$ to the rotated set gives $P_t^{-1}(\{e^{-2\pi it}{q}_1,e^{-2\pi it}{q}_2\})=\{q_1,q_2\}$. No commutation with rotation is assumed.

The rotation loop is not nullhomotopic. For every $t$ one has $u({\rho }_{-}(t))=\tilde{u}(e^{-2\pi it}{q}_1,e^{-2\pi it}{q}_2)=(e^{-2\pi it}(q_1-q_2)\mathrm{/}\mathrm{\mid }{q}_1-q_2\mathrm{\mid }{)}^2$; since $q_1-q_2=(-2h,0)$ is a negative real number, $(q_1-q_2)\mathrm{/}\mathrm{\mid }{q}_1-q_2\mathrm{\mid }=-1$, so $u({\rho }_{-}(t))=(-e^{-2\pi it}{)}^2=e^{-4\pi it}=(\cos 2\pi (-2)t,\sin 2\pi (-2)t)$. By [F10] the class of this loop in ${\pi }_1(S^1,(1,0))$ corresponds to $-2$, which is not zero, so $[u\circ {\rho }_{-}]$ is not the identity. Since $u_{\ast }$ is a group homomorphism with $u_{\ast }([{\rho }_{-}])=[u\circ {\rho }_{-}]$ by [F12], a class $[{\rho }_{-}]$ equal to the identity would give the identity here; hence $[{\rho }_{-}]\mathrm{\ne }1$ in ${\pi }_1(C_2(\mathrm{int}{D}^2),[Q_2])$.

The rotation loop is the inverse square of the sliced half twist. By step 1.3 the loop $\alpha$ satisfies $[S({\sigma }_1{)}^{-1}]=[\alpha ]$, and the concatenation $\alpha \ast \alpha$ is the loop $t\mapsto [e^{-2\pi it}{Q}_2]$ of step 1.1, so in the group ${\pi }_1(C_2(\mathrm{int}{D}^2),[Q_2])$ of [F11] one has $[{\rho }_{-}]=[\alpha \ast \alpha ]=[\alpha ][\alpha ]=[S({\sigma }_1{)}^{-1}{]}^2=[S({\sigma }_1){]}^{-2}$.

The setwise isotopy from the identity to the full twist. Put $R_t(x):=e^{-2\pi it}x$ and $G_t:=P_t^{-1}\circ R_t$. The map $(t,x)\mapsto G_t(x)$ is continuous: $t\mapsto P_t^{-1}$ is a continuous path by [F1], its joint evaluation is continuous, and $(t,x)\mapsto (t,R_t(x))$ is continuous. Every $G_t$ is a homeomorphism of $D^2$ with inverse $R_t^{-1}\circ P_t$. On the boundary $P_t^{-1}$ is the identity, so $G_t$ acts there as $R_t$ and preserves $\mathrm{\partial }{D}^2$ setwise. Step 2.1 gives $G_t(\{q_1,q_2\})=P_t^{-1}(R_t(\{q_1,q_2\}))=\{q_1,q_2\}$, so the marked set is preserved at every time. Since $R_0=R_1=\mathrm{id}$ and $P_0=\mathrm{id}$, we have $G_0=\mathrm{id}$ and $G_1=P_1^{-1}=f$. This is the required setwise-boundary isotopy.

The witness is the nontrivial positive full twist. By [F3] and step 2.1 the boundary map evaluates on the rotation loop as $\delta ([{\rho }_{-}])=[P_1^{-1}]=[f]$, and $\delta$ is injective by [F4], so step 2.2 gives $[f]\mathrm{\ne }[\mathrm{id}]$ in $\mathrm{Mod}(D^2,Q_2;\mathrm{\partial }{D}^2)$. Moreover $[f]=\mathrm{\Psi }([{\sigma }_1{]}^2)$, the positive full twist: writing $[{\sigma }_1{]}^2=[{\sigma }_1\star {\sigma }_1]$, [F6] gives $\mathrm{\Phi }([{\sigma }_1{]}^2)=({\iota }_{\ast }^C[S({\sigma }_1\star {\sigma }_1)]{)}^{-1}=({\iota }_{\ast }^C[S({\sigma }_1){]}^2{)}^{-1}$ and $[S({\sigma }_1\star {\sigma }_1)]=[S({\sigma }_1){]}^2$, while [F7] makes ${\iota }_{\ast }^C$ an isomorphism and hence $({\iota }_{\ast }^C{)}^{-1}(({\iota }_{\ast }^C[S({\sigma }_1){]}^2{)}^{-1})=([S({\sigma }_1){]}^2{)}^{-1}=[S({\sigma }_1){]}^{-2}$; since $\mathrm{\Psi }=\delta \circ ({\iota }_{\ast }^C{)}^{-1}\circ \mathrm{\Phi }$ by [F5], this yields $\mathrm{\Psi }([{\sigma }_1{]}^2)=\delta ([S({\sigma }_1){]}^{-2})=\delta ([{\rho }_{-}])=[f]$ by step 2.3, with $\mathrm{\Psi }([{\sigma }_1])$ the class of the positive half twist by [F5].

Conclusion. The homeomorphism $f=P_1^{-1}$ fixes $\mathrm{\partial }{D}^2$ pointwise and preserves $Q_2$ setwise, and by step 3.1 it is isotopic to the identity through homeomorphisms preserving $\mathrm{\partial }{D}^2$ and $Q_2$ setwise, but by step 3.2 it is not isotopic to the identity rel $\mathrm{\partial }{D}^2$, where it represents the positive full twist $\mathrm{\Psi }([{\sigma }_1{]}^2)$. So the pointwise-boundary and setwise-boundary conventions do not define the same isotopy classes: the assignment that views a pointwise-boundary class as a setwise-boundary class sends the nontrivial class $[f]$ to the class of the identity, and the refuted claim fails. ∎