Boundary-fixed mapping class group of a punctured disk

Throughout this page $n\in \mathbb{N}$ is a natural number and

$I:=[0,1],D^2:=\{z\in \mathbb{C}:\mathrm{\mid }z\mathrm{\mid }\le 1\},\mathrm{\partial }{D}^2:=\{z\in \mathbb{C}:\mathrm{\mid }z\mathrm{\mid }=1\},\mathrm{int}{D}^2:=\{z\in \mathbb{C}:\mathrm{\mid }z\mathrm{\mid }<1\}$

is the closed unit disc, its boundary circle and its interior, with the subspace topologies of $\mathbb{C}\cong {\mathbb{R}}^2$. The fixed base configuration is the tuple

$Q_n=(q_1,\dots ,q_n),q_j:=(\frac{2j-n-1}{4(n+1)},0)\in \mathrm{int}{D}^2(1\le j\le n),$

which is exactly the tuple denoted $Q$ in Geometric braids in the disc with setwise endpoints: the points $q_1,\dots ,q_n$ are pairwise distinct, are listed strictly from left to right, and satisfy $q_{j+1}-q_j=(2h_0,0)$ with $h_0:=\frac{1}{4(n+1)}$ and $\parallel q_j{\parallel }_2\le (n-1)h_0<\frac{1}{4}$. A homeomorphism of $D^2$ is a bijection $f:D^2\to D^2$ that is continuous in both directions (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

The boundary-fixed disc homeomorphism group. Write

${\mathrm{Homeo}}^{+}(D^2,\mathrm{\partial }{D}^2):=\{f:D^2\to D^2\text{ a homeomorphism }:f{\mathrm{\mid }}_{\mathrm{\partial }{D}^2}={\mathrm{id}}_{\mathrm{\partial }{D}^2}\}$

for the set of homeomorphisms of $D^2$ that fix the boundary circle pointwise, with composition as its multiplication. This is a group: the identity fixes $\mathrm{\partial }{D}^2$ pointwise, the composite of two boundary-fixing homeomorphisms fixes $\mathrm{\partial }{D}^2$ pointwise, and the inverse of a boundary-fixing homeomorphism fixes $\mathrm{\partial }{D}^2$ pointwise.

Boundary fixing is the primitive condition here; no separate orientation test is imposed. The punctured-disc group discussed in the literature is obtained by restricting the setwise stabilizer of the marked set, defined below, to $D^2\setminus \{q_1,\dots ,q_n\}$. A boundary-fixed homeomorphism that moves a marked point outside that set does not restrict to a self-homeomorphism of this punctured disc.

Isotopies and the compact-open topology. The set of all continuous maps $D^2\to D^2$ carries the compact-open topology (The compact-open topology on $C(X,Y)$ for arbitrary topological spaces), and ${\mathrm{Homeo}}^{+}(D^2,\mathrm{\partial }{D}^2)$ carries the subspace topology. The domain $D^2$ is a nonempty compact metric space, so by On a nonempty compact metric domain, the compact-open topology is the uniform topology this subspace topology is the topology of uniform convergence: basic neighbourhoods of $f$ are the sets $\{g:{\sup }_{x\in D^2}\parallel g(x)-f(x){\parallel }_2<\epsilon \}$. Composition and inversion are continuous for this topology (verified below), so the group is a topological group.

A path $t\mapsto f_t$ in this group transposes to a continuous map $H:D^2\times I\to D^2$, $H(x,t):=f_t(x)$: continuity follows from $\parallel f_t(x)-f_{t_0}(x_0){\parallel }_2\le d(f_t,f_{t_0})+\parallel f_{t_0}(x)-f_{t_0}(x_0){\parallel }_2$. Conversely, a continuous $H$ on the compact metric space $D^2\times I$ is uniformly continuous, so $t\mapsto H(-,t)$ is continuous in the uniform topology. Thus paths correspond exactly to isotopies through homeomorphisms fixing $\mathrm{\partial }{D}^2$ pointwise. We say $f$ and $g$ are isotopic rel $\mathrm{\partial }{D}^2$ when some such path joins them.

Composition and inversion are continuous. For $f,g,f^{\mathrm{\prime }},g^{\mathrm{\prime }}\in {\mathrm{Homeo}}^{+}(D^2,\mathrm{\partial }{D}^2)$, define $d(u,v):={\sup }_{z\in D^2}\parallel u(z)-v(z){\parallel }_2$ and let ${\omega }_f$ be a modulus of uniform continuity for $f$ on $D^2$. For every $x\in D^2$, inserting $f(g^{\mathrm{\prime }}(x))$ gives $\parallel f(g(x))-f^{\mathrm{\prime }}(g^{\mathrm{\prime }}(x)){\parallel }_2\le \parallel f(g(x))-f(g^{\mathrm{\prime }}(x)){\parallel }_2+\parallel f(g^{\mathrm{\prime }}(x))-f^{\mathrm{\prime }}(g^{\mathrm{\prime }}(x)){\parallel }_2\le {\omega }_f(d(g,g^{\mathrm{\prime }}))+d(f,f^{\mathrm{\prime }})\mathrm{.}$ Taking the supremum in $x$ gives $d(f\circ g,f^{\mathrm{\prime }}\circ g^{\mathrm{\prime }})\le {\omega }_f(d(g,g^{\mathrm{\prime }}))+d(f,f^{\mathrm{\prime }})$, which tends to zero as $(f^{\mathrm{\prime }},g^{\mathrm{\prime }})\to (f,g)$.

For inversion, let $f_k\to f$ uniformly in the group and suppose the sequence $f_k^{-1}$ did not converge uniformly to $f^{-1}$. Then for some $\epsilon >0$ there are $x_k\in D^2$ with $\parallel f_k^{-1}(x_k)-f^{-1}(x_k){\parallel }_2\ge \epsilon$ for infinitely many $k$. Passing to a subsequence, $x_k\to x$ for some $x\in D^2$ by compactness. Put $y_k:=f_k^{-1}(x_k)$, so $f_k(y_k)=x_k$; by compactness pass to a further subsequence with $y_k\to y$. Then uniform convergence gives $f(y)={\lim }_k{f}_k(y_k)={\lim }_k{x}_k=x$, so $y=f^{-1}(x)$. Hence $f_k^{-1}(x_k)\to f^{-1}(x)$ along this subsequence, contradicting $\parallel f_k^{-1}(x_k)-f^{-1}(x_k){\parallel }_2\ge \epsilon$ for $f^{-1}(x_k)\to f^{-1}(x)$. A sequence argument of this kind rules out non-uniform convergence, so inversion is continuous. (The same estimates show that the group operations of every subgroup described below are continuous in the subspace topology.)

The setwise stabilizer of $Q_n$. Put

${\mathrm{Homeo}}^{+}(D^2,\mathrm{\partial }{D}^2;Q_n):=\{f\in {\mathrm{Homeo}}^{+}(D^2,\mathrm{\partial }{D}^2):f(\{q_1,\dots ,q_n\})=\{q_1,\dots ,q_n\}\},$

the subgroup of homeomorphisms of the disc that fix the boundary pointwise and preserve the marked set $Q_n$ setwise. The requirement is preservation of the set, not of every marked point: an element may permute $q_1,\dots ,q_n$. If $f$ maps $Q_n$ onto itself, the induced map on the finite set $\{q_1,\dots ,q_n\}$ is a permutation, so there is a unique $\pi (f)\in S_n$ with $f(q_j)=q_{\pi (f)(j)}$ for all $j$; this permutation is computed from the labelling of $Q_n$ fixed above. As in Geometric braids in the disc with setwise endpoints, $S_n$ acts on labels $1,\dots ,n$ through $\kappa (j)=j-1$: the displayed $\pi (f)(j)$ means ${\kappa }^{-1}(\pi (f)(\kappa (j)))$. Equivalently, without this shorthand, $f(q_j)=q_{\pi (f)(j-1)+1}$ for $\pi (f)\in \mathrm{Sym}(\{0,\dots ,n-1\})$.

The mapping class group. The boundary-fixed mapping class group of the punctured disc is the set of path components

$\mathrm{Mod}(D^2,Q_n;\mathrm{\partial }{D}^2):={\pi }_0({\mathrm{Homeo}}^{+}(D^2,\mathrm{\partial }{D}^2;Q_n)),$

the set of isotopy classes rel $\mathrm{\partial }{D}^2$ of homeomorphisms of $D^2$ fixing $\mathrm{\partial }{D}^2$ pointwise and preserving $Q_n$ setwise. Path components are computed in the subspace topology of the compact-open topology, that is, $[f]=[g]$ exactly when $f$ and $g$ are joined by an isotopy whose every time is a boundary-fixing homeomorphism preserving $Q_n$ setwise. The class of $f$ is written $[f]$.

Multiplication is ordinary composition. Composition and inversion are continuous, so ${\pi }_0$ of this topological group is a group: if $f_s$ is a path from $f$ to $f^{\mathrm{\prime }}$ and $g_s$ a path from $g$ to $g^{\mathrm{\prime }}$, then $f_s\circ g_s$ is a path from $f\circ g$ to $f^{\mathrm{\prime }}\circ g^{\mathrm{\prime }}$ and $f_s^{-1}$ is a path from $f^{-1}$ to $f^{\mathrm{\prime }-1}$. Hence the formulas $[f][g]:=[f\circ g],[f{]}^{-1}:=[f^{-1}]$ are well defined, are independent of the chosen representatives, and give $\mathrm{Mod}(D^2,Q_n;\mathrm{\partial }{D}^2)$ the structure of a group with identity $[{\mathrm{id}}_{D^2}]$; associativity, the identity law and the inverse law are those of composition of maps, passed to classes. In particular the multiplication on $\mathrm{Mod}(D^2,Q_n;\mathrm{\partial }{D}^2)$ is ordinary composition of representatives, not stacking of braids.

Elementary cases. For $n=0$ the tuple $Q_0$ is empty, the setwise condition is vacuous, and ${\mathrm{Homeo}}^{+}(D^2,\mathrm{\partial }{D}^2;Q_0)={\mathrm{Homeo}}^{+}(D^2,\mathrm{\partial }{D}^2)$; the statement of this page includes $n=0$ and the later isomorphism theorems state their results for all $n\ge 0$. For $n=1$ the marked set is the single point $q_1$, so setwise and pointwise preservation of the marked set coincide and $Q_1$ is fixed by every element of the stabilizer.

• Why the boundary is fixed pointwise rather than setwise. The definition above fixes $\mathrm{\partial }{D}^2$ pointwise and lets an element move the marked points only within $D^2$. Both requirements are load-bearing later: a boundary rotation can absorb disc twisting, and setwise preservation of $Q_n$ alone does not force the identity permutation of the marked points. The companion page exhibits both phenomena as counterexamples.

• Relation to the punctured disc. Elements of the setwise stabilizer restrict to self-homeomorphisms of $D^2\setminus Q_n$. The punctured-disc isotopies used here are restrictions of ambient isotopies fixing $\mathrm{\partial }{D}^2$ pointwise and preserving $Q_n$ setwise at every time. This specifies the boundary and puncture conventions in the mapping class group just defined.

• Orientation. For the disc, the identity component of the group of all homeomorphisms of $D^2$ is the group of orientation-preserving homeomorphisms; since every element above lies in the identity component after the boundary is fixed pointwise, no additional orientation condition is imposed or needed. This convention is used consistently on this page and its companion.