Braid group as boundary-fixed punctured-disk mapping classes

The composite is an isomorphism. By [L1] the map $\mathrm{\Phi }$ is a group isomorphism onto ${\pi }_1(C_n(D^2),[Q_n])$, whose basepoint is the orbit of the same tuple $Q_n$ used in the definition of $G_n$. By [L2] the induced map ${\iota }_{\ast }^C$ is a group isomorphism at that configuration, so its inverse is a group isomorphism; by [L3] the boundary map $\delta$ is a group isomorphism onto $\mathrm{Mod}(D^2,Q_n;\mathrm{\partial }{D}^2)$. A composite of group isomorphisms is a group isomorphism, so $\mathrm{\Psi }=\delta \circ ({\iota }_{\ast }^C{)}^{-1}\circ \mathrm{\Phi }$ is one, and this holds for every $n\ge 0$ because [L1], [L2] and [L3] all include the cases $n=0$ and $n=1$.

Reading the inverse endpoint off a lift. Let $\alpha :I\to C_n(\mathrm{int}{D}^2)$ be a based loop at $[Q_n]$ and let $g:I\to {\mathrm{Homeo}}^{+}(D^2,\mathrm{\partial }{D}^2)$ be a lift of $\alpha$ with $g(0)=\mathrm{id}$; write $h:=g(1)\in F$. Define $g^{-}:I\to {\mathrm{Homeo}}^{+}(D^2,\mathrm{\partial }{D}^2)$ by $g^{-}(t):=g(1-t)\circ h^{-1}$; it is continuous by [L7], satisfies $g^{-}(0)=h\circ h^{-1}=\mathrm{id}$ and $g^{-}(1)=\mathrm{id}\circ h^{-1}=h^{-1}$, and it lifts the reversed loop because $\mathrm{ev}(g^{-}(t))=[g(1-t)(h^{-1}(Q_n)\left)\right]=[g(1-t)(Q_n)]=\alpha (1-t)$ for all $t$, the middle equality holding because $h\in F$ preserves $Q_n$ setwise. Since the reversed loop represents the inverse class by [L11], [L4] gives $\delta ([\alpha {]}^{-1})=\delta ([{\alpha }^{-1}])=[(h^{-1}{)}^{-1}]=[h]$. Applying this to $\alpha =S(\beta )$ and using $\mathrm{\Phi }([\beta ])=({\iota }_{\ast }^C[S(\beta )]{)}^{-1}$ from [L1] together with the fact that the group isomorphism $({\iota }_{\ast }^C{)}^{-1}$ carries inverses to inverses, we obtain $\mathrm{\Psi }([\beta ])=\delta ([S(\beta ){]}^{-1})=[h_{\beta }],$ where $h_{\beta }\in F$ is the endpoint of any lift of the raw slice loop $S(\beta )$ with initial value $\mathrm{id}$.

The supported half rotation and its point motion. Put $\theta (r):=\sigma ((11h\mathrm{/}8-r)\mathrm{/}(h\mathrm{/}8))$ for $r\ge 0$, so that $\theta$ is smooth with values in $[0,1]$ by [L6], equals $1$ for $r\le 5h\mathrm{/}4$ and equals $0$ for $r\ge 11h\mathrm{/}8$. For $x\in D^2$ and $s\in I$ let $R(\alpha )$ denote rotation about the origin by the angle $\alpha$ and set $H_s(x):=m_i+R(\pi s\theta (\parallel x-m_i{\parallel }_2))(x-m_i)\mathrm{.}$ Since $\theta =0$ beyond radius $11h\mathrm{/}8$, the map $H_s$ is the identity outside the disc of radius $11h\mathrm{/}8$ about $m_i$, which lies in $U_i\subseteq \mathrm{int}{D}^2$ by [L5]; in polar coordinates about $m_i$ it is $(r,\phi )\mapsto (r,\phi +\pi s\theta (r))$, with inverse $(r,\phi )\mapsto (r,\phi -\pi s\theta (r))$, so each $H_s$ is a homeomorphism of $D^2$ that fixes $U_i$-exterior points and in particular fixes $\mathrm{\partial }{D}^2$ pointwise. The map $(s,x)\mapsto H_s(x)$ is continuous, and $H_0=\mathrm{id}$. Write $H_i:=H_1$ for the time-one map of this family at the fixed adjacent index $i$. For the two adjacent marked points, [L5] gives $\parallel q_i-m_i{\parallel }_2=\parallel q_{i+1}-m_i{\parallel }_2=h\le 5h\mathrm{/}4$, so $\theta =1$ there and $H_s(q_i)=m_i+h(-\cos \pi s,-\sin \pi s),H_s(q_{i+1})=m_i+h(\cos \pi s,\sin \pi s):$ the pair $\{H_s(q_i),H_s(q_{i+1})\}$ is $\{m_i\pm h(\cos \pi s,\sin \pi s)\}$ and describes the lower semicircle of radius $h$ about $m_i$ from $\{q_i,q_{i+1}\}$ at $s=0$ to $\{q_{i+1},q_i\}$ at $s=1$, passing through $\{m_i\pm (0,h)\}$ at $s=\frac{1}{2}$; by [L5] every other base point has distance at least $3h\ge 11h\mathrm{/}8$ from $m_i$ and is fixed throughout. Consequently $H_0(Q_n)=H_1(Q_n)=Q_n$ as unordered marked sets, so $s\mapsto \mathrm{ev}(H_s)=[H_s(Q_n)]$ is a based loop of $C_n(\mathrm{int}{D}^2)$ at $[Q_n]$, and the family $w_r(s):=(1-r)\rho (s)+rh(-\cos \pi s,-\sin \pi s),r,s\in I,$ defines a homotopy of the moving pairs: by [L5], $\rho (s)$ has second coordinate $-2sh$ for $s\le \frac{1}{2}$ and $2h(s-1)$ for $s\ge \frac{1}{2}$, both strictly negative for $0<s<1$, while $-\sin \pi s<0$ for $0<s<1$; hence the linear interpolation $w_r(s)$ has strictly negative second coordinate and is nonzero for $0<s<1$, and $w_r(0)=(-h,0)$, $w_r(1)=(h,0)$ are nonzero, so the interpolated pairs $\{m_i\pm w_r(s)\}$ are collision-free for all $r,s$, lie within distance $h$ of $m_i$, and are separated from all fixed base points by at least $2h$; composing with the quotient map gives a path homotopy relative to $\{0,1\}$ from the raw slice loop $S({\sigma }_i)$ of [L5] to $\mathrm{ev}\circ H$.

The positive half twist maps to the supported half rotation. The element $H_1\in {\mathrm{Homeo}}^{+}(D^2,\mathrm{\partial }{D}^2)$ fixes $\mathrm{\partial }{D}^2$ pointwise by step 1.3 and satisfies $H_1(Q_n)=Q_n$ as a set, because it exchanges $q_i$ and $q_{i+1}$ and fixes every other base point; hence $H_1\in F$ and $[H_1]\in \mathrm{Mod}(D^2,Q_n;\mathrm{\partial }{D}^2)$ is defined. The family $s\mapsto H_s$ is a lift with initial value $\mathrm{id}$ of the based loop $\mathrm{ev}\circ H$, so by [L4] its class satisfies $\delta ([\mathrm{ev}\circ H])=[H_1^{-1}]$; since $s\mapsto \mathrm{ev}(H_s)$ is path homotopic relative to $\{0,1\}$ to $S({\sigma }_i)$ by step 1.3, the well-definedness of $\delta$ from [L4] gives $\delta ([S({\sigma }_i)])=[H_1^{-1}]$, and applying the isomorphism [L3] to inverses gives $\delta ([S({\sigma }_i){]}^{-1})=[H_1]$. Step 1.2 turns the left-hand side into the class $[h_{{\sigma }_i}]$ of the lift endpoint of $S({\sigma }_i)$, so $\mathrm{\Psi }([{\sigma }_i])=[H_1]$: the standard positive half twist maps to the class of the supported half rotation, which is supported in $U_i$ and exchanges the adjacent pair.

Smooth boundary-fixed representatives. Let $[f]\in \mathrm{Mod}(D^2,Q_n;\mathrm{\partial }{D}^2)$ and put $[\beta ]:={\mathrm{\Psi }}^{-1}([f])\in G_n$, so that $[f]=[h_{\beta }]$ with $h_{\beta }$ the endpoint of a lift of $S(\beta )$ from $\mathrm{id}$ by step 1.2. By [L8] the based loop $S(\beta )$ is path homotopic relative to $\{0,1\}$ to a based loop ${\beta }^{\mathrm{\prime }}$ whose unique ordered lift $z$ from $Q_n$ consists of smooth, pairwise collision-free paths, constant on some initial and terminal interval; extending each $z_j$ by its constant values beyond $[0,1]$ gives smooth collision-free paths $z_j:\mathbb{R}\to \mathrm{int}{D}^2$ that are constant on $(-\mathrm{\infty },0]$ and on $[1,\mathrm{\infty })$, so the extension lemma [L9], available under the present assumption by [L10], supplies a smooth $\mathrm{\Phi }:D^2\times [0,1]\to D^2$ with ${\mathrm{\Phi }}_0=\mathrm{id}$, every ${\mathrm{\Phi }}_s$ a diffeomorphism of $D^2$ fixing $\mathrm{\partial }{D}^2$ pointwise, and ${\mathrm{\Phi }}_s(q_j)=z_j(s)$ for all $j$ and $s$, the last identity using $z_j(0)=q_j$. Then $s\mapsto {\mathrm{\Phi }}_s$ is a path in ${\mathrm{Homeo}}^{+}(D^2,\mathrm{\partial }{D}^2)$ from $\mathrm{id}$ that lifts ${\beta }^{\mathrm{\prime }}$, because $\mathrm{ev}({\mathrm{\Phi }}_s)=[{\mathrm{\Phi }}_s(Q_n)]=[z(s)]={\beta }^{\mathrm{\prime }}(s)$; by the computation of step 1.2 its endpoint satisfies $[{\mathrm{\Phi }}_1]=\delta ([{\beta }^{\mathrm{\prime }}{]}^{-1})=\delta ([S(\beta ){]}^{-1})=[f]$, the middle equality because ${\beta }^{\mathrm{\prime }}$ and $S(\beta )$ are path homotopic relative to endpoints and $\delta$ is well defined. Moreover ${\mathrm{\Phi }}_1(Q_n)=z(1)$ is a permutation of $Q_n$, since $[z(1)]={\beta }^{\mathrm{\prime }}(1)=[Q_n]$; hence ${\mathrm{\Phi }}_1\in F$, and ${\mathrm{\Phi }}_1$ is a diffeomorphism fixing $\mathrm{\partial }{D}^2$ pointwise that is the time-one map of the smooth isotopy $\mathrm{\Phi }$ from the identity.

Conclusion and elementary cases. Step 1.1 exhibits the isomorphism $\mathrm{\Psi }$ of the first assertion, step 2.1 identifies $\mathrm{\Psi }([{\sigma }_i])$ with the class of the explicit supported half rotation for every $1\le i\le n-1$, and step 2.2 produces the smooth boundary-fixed representative of every mapping class; this proves all three assertions. For $n=0$ the braid group and the mapping class group are trivial and the arguments above return the isomorphism of trivial groups and the identity as smooth representative; for $n=1$ there is no adjacent index, no half twist is asserted by [L5], and the same isomorphism and smooth-representative arguments apply verbatim.