A braid moves points around a disk. A mapping class records how the disk itself has been rearranged, allowing deformations that respect its boundary and marked points. This chapter defines that group. The proof that it is the braid group comes in Chapter 6.
Prerequisites: homeomorphisms, orientation, and quotients
A homeomorphism is a continuous bijection with continuous inverse. An orientation-preserving homeomorphism respects the usual counterclockwise orientation of the plane. No differentiability is required. Composition \(f\circ g\) applies \(g\) first.
A quotient group \(G/N\) identifies elements differing by an element of a normal subgroup \(N\). Normality means \(gNg^{-1}=N\). See Alphabeta Math's normal subgroups and quotient groups. We prove the needed normality below.
Why an isotopy preserves orientation
Orientation can be tested by winding number. For an interior point \(x\), choose a small positively oriented circle \(c\) about \(x\) contained in the disk. A homeomorphism \(f\) preserves orientation when, for every such interior point and small circle, the loop \((f(c(t))-f(x))/|f(c(t))-f(x)|\) has winding number \(+1\). Winding number is the endpoint difference of a lift to \(\mathbb R\) under \(v\mapsto e^{2\pi\mathrm{i}v}\); see the covering argument in Chapter 3. For a continuous family of homeomorphisms, the normalized difference stays defined because no circle point maps to the image of its centre. It gives a homotopy of circle loops, so its integer winding number stays constant by homotopy lifting. Consequently a homeomorphism joined to the identity is orientation preserving. This is the only orientation-invariance fact used for our explicit constructions.
4.1. Boundary-fixed homeomorphisms
Fix \(n\geq3\). In the complex plane, put
\[ D=\{z\in\mathbb C:|z|\leq1\},\qquad q_j=-1+\frac{2j}{n+1}\quad(1\leq j\leq n),\qquad Q=\{q_1,\ldots,q_n\}. \]The points of \(Q\) lie in the interior. The punctured disk is \(D\setminus Q\). We define its mapping classes using the marked disk: keep the points in the disk and specify how maps treat them. Every allowed map restricts to a homeomorphism of \(D\setminus Q\), because it preserves \(Q\) setwise; the allowed punctured-disk isotopies are these ambient isotopies restricted to \(D\setminus Q\).
Definition 4.1. The marked-disk homeomorphism group
Let
\[ \mathcal H_Q=\{f:D\to D\mid f\text{ is an orientation-preserving homeomorphism},\quad f|_{\partial D}=\mathrm{id},\quad f(Q)=Q\}. \]The boundary condition is pointwise: every boundary point stays exactly where it was. The puncture condition is setwise: \(f\) may permute the \(q_j\). These are different requirements. A rigid rotation of the whole disk generally violates the first, even though it carries the boundary circle onto itself.
Proposition 4.2. Composition and inversion
The set \(\mathcal H_Q\) is a group under ordinary composition.
Proof.
Composition makes \(\mathcal H_Q\) a group. Composites and inverses are orientation preserving, fix every boundary point, and carry \(Q\) onto itself. The identity map satisfies all three conditions.
□4.2. Isotopies and mapping classes
Definition 4.3. Boundary-fixed marked isotopy
An isotopy in \(\mathcal H_Q\) from \(f\) to \(g\) is a continuous map \(H:D\times[0,1]\to D\) such that \(H_t(z)=H(z,t)\) belongs to \(\mathcal H_Q\) for every \(t\), with \(H_0=f\) and \(H_1=g\). Thus each intermediate map is a homeomorphism, fixes the boundary pointwise, and preserves the marked set.
A homotopy requires continuous intermediate maps, which need not be homeomorphisms. It can collapse or fold the disk. Our equivalence relation uses isotopies with the stated boundary and puncture conditions. Reversing and concatenating isotopies, along with constant isotopies, proves that this is an equivalence relation.
Worked example for Definition 4.3
Counterexample: A marked homotopy need not be a marked isotopy
Lemma 4.4. Stationary marked-point images
An isotopy in \(\mathcal H_Q\) has a time-independent permutation of \(Q\). An identity isotopy fixes each marked point throughout.
Proof.
For each \(j\), the map \(t\mapsto H_t(q_j)\) is continuous and takes values in the finite discrete set \(Q\). Since \([0,1]\) is connected, that map is constant:
\[ H_t(q_j)=f(q_j)\qquad\text{for every }t. \]Consequently isotopic maps induce the same puncture permutation. If \(H_0=\mathrm{id}\), each puncture stays at its own location throughout the isotopy. Requiring setwise preservation at every time already forces this; it does not let punctures travel between marked locations.
□Definition 4.5. Mapping class group
Write \([f]\) for the isotopy class of \(f\). The mapping class group is
\[ \operatorname{Mod}(D,Q;\partial D)=\mathcal H_Q/{\sim}. \]This is the boundary-fixed convention in Alphabeta Math's definition and in González-Meneses, §1.4. The boundary notation is part of the definition.
4.3. Group operations on mapping classes
Theorem 4.6. The mapping class group law
The operations \([f][g]=[f\circ g]\) and \([f]^{-1}=[f^{-1}]\) are well defined on isotopy classes and make \(\operatorname{Mod}(D,Q;\partial D)\) a group.
Proof.
Suppose \(f_t\) joins \(f\) to \(f'\), and \(g_t\) joins \(g\) to \(g'\), within \(\mathcal H_Q\). Then \(f_t\circ g_t\) joins \(f\circ g\) to \(f'\circ g'\). The inverse family \(f_t^{-1}\) joins \(f^{-1}\) to \((f')^{-1}\). Therefore
\[ [f][g]=[f\circ g],\qquad [f]^{-1}=[f^{-1}] \]are independent of representatives. Associativity and the identity law follow from those for maps. In this multiplication the right factor acts first.
□Prerequisite: topology on the homeomorphism group
Proposition 4.A. Uniform topology and inverse families
On the compact disk, the compact-open and uniform topologies agree. Composition and inversion are continuous. Paths of homeomorphisms are equivalent to jointly continuous isotopies.
Proof.
The map \(\Phi:D\times[0,1]\to D\times[0,1]\), \(\Phi(x,t)=(f_t(x),t)\), is a continuous bijection. Its domain is compact and its target is Hausdorff, so it is a homeomorphism. Its inverse is \((y,t)\mapsto(f_t^{-1}(y),t)\). This proves the required joint continuity without assuming differentiability.
For the topology on the group, use the uniform distance \(d(f,g)=\sup_{x\in D}|f(x)-g(x)|\). On the compact disk this gives the compact-open topology. The compact-open topology has subbasic sets \(\{g:g(K)\subset V\}\), with \(K\subset D\) compact and \(V\subset D\) open. Such a set is uniformly open: the compact set \(f(K)\subset V\) has positive distance from the closed complement of \(V\), unless that complement is empty. Conversely, given \(\varepsilon>0\), choose finitely many compact neighbourhoods \(K_j\) covering \(D\) with \(|f(x)-f(x_j)|<\varepsilon/3\) on \(K_j\). The finitely many conditions \(g(K_j)\subset B(f(x_j),2\varepsilon/3)\cap D\) give a compact-open neighbourhood of \(f\) contained in its uniform \(\varepsilon\)-ball. A jointly continuous isotopy is uniformly continuous on \(D\times[0,1]\), so \(t\mapsto f_t\) is a path continuous in the uniform topology. Conversely, continuity of this path and continuity of each \(f_t\) give joint continuity by
\[ |f_t(x)-f_{t_0}(x_0)|\leq d(f_t,f_{t_0})+|f_{t_0}(x)-f_{t_0}(x_0)|. \]For a fixed continuous map \(u:D\to D\), define its modulus of continuity by
\[\omega_u(\delta)=\sup\{|u(x)-u(y)|:x,y\in D,\ |x-y|\leq\delta\}.\]Compactness implies \(\omega_u(\delta)\to0\) as \(\delta\to0\). Inserting \(f\circ g'\) between two composites gives
\[ d(f'\circ g',f\circ g)\leq d(f',f)+\omega_f(d(g',g)). \]For inversion, put \(x=g^{-1}(y)\). Then
\[|f(x)-y|=|f(x)-g(x)|\leq d(f,g)\]Therefore
\[ |g^{-1}(y)-f^{-1}(y)|=|f^{-1}(f(x))-f^{-1}(y)|\leq\omega_{f^{-1}}(d(f,g)). \]Taking the supremum over \(y\) proves \(d(g^{-1},f^{-1})\leq\omega_{f^{-1}}(d(f,g))\). Thus composition and inversion are continuous: both the unmarked boundary-fixed homeomorphism group and its marked-set stabilizer \(\mathcal H_Q\) are topological groups. This is the topology used in Chapter 6.
□Proposition 4.7. Quotient by the identity path component
The identity path component \(\mathcal H_{Q,0}\) is normal in \(\mathcal H_Q\), and \(\operatorname{Mod}(D,Q;\partial D)=\mathcal H_Q/\mathcal H_{Q,0}\).
Proof.
There is also a quotient-by-subgroup description. Let \(\mathcal H_{Q,0}\) be the maps isotopic to the identity in \(\mathcal H_Q\). With the compact-open topology, an isotopy is exactly a path of homeomorphisms. Thus \(\mathcal H_{Q,0}\) is the identity isotopy component; “component” here means path component.
It is a subgroup because products and inverses of identity isotopies are identity isotopies. It is normal: if \(h_t\) joins the identity to \(h\), then \(f\circ h_t\circ f^{-1}\) joins the identity to \(f\circ h\circ f^{-1}\), still inside \(\mathcal H_Q\). Finally, \(f\sim g\) exactly when \(g^{-1}\circ f\in\mathcal H_{Q,0}\), by composing the relevant isotopy with \(g\) or \(g^{-1}\). Hence
\[ \operatorname{Mod}(D,Q;\partial D)=\mathcal H_Q/\mathcal H_{Q,0}. \] □4.4. Pure mapping classes
Definition 4.8. The puncture permutation and pure subgroup
Define \(\rho([f])\in S_n\) by \(f(q_j)=q_{\rho([f])(j)}\), where \(S_n\) is the group of permutations of \(\{1,\ldots,n\}\). The stationary-puncture argument makes \(\rho\) well defined. Under ordinary permutation composition,
\[ \rho([f][g])=\rho([f])\circ\rho([g]). \]The pure mapping class group is the kernel \(\operatorname{PMod}(D,Q;\partial D)=\ker\rho\). Its representatives fix each \(q_j\) individually. The full group permits permutations. Fixing each puncture at the endpoint does not imply that a map is isotopic to the identity; it only makes its class pure.
Worked example for Definition 4.8
Counterexample: A marked homotopy need not be a marked isotopy
4.5. Supported half twists
A local rotation interchanges a pair of punctures. A radial cutoff makes this rotation compatible with pointwise fixation of the outer boundary.
Definition 4.9. A clockwise half-twist motion
Choose consecutive punctures \(q_i,q_{i+1}\). Set \(c=(q_i+q_{i+1})/2\) and \(a=1/(n+1)\). Choose radii \(a<r_0<R<2a\). The disk \(|z-c|\leq R\) lies inside \(D\) and contains exactly this pair: the next puncture is \(3a\) from \(c\), and the closest outer boundary is at least \(3a\) away.
Let \(b(r)\) be a continuous radial cutoff equal to \(1\) for \(r\leq r_0\), decreasing linearly to \(0\) on \([r_0,R]\), and equal to \(0\) for \(r\geq R\). Define
\[ h_t(c+re^{i\theta})=c+re^{i(\theta-\pi t b(r))},\qquad 0\leq t\leq1. \]Proposition 4.10. The half-twist endpoint
Every \(h_t\) in the preceding definition is a boundary-fixed orientation-preserving homeomorphism. The endpoint \(h_1\) exchanges the chosen pair and fixes all other punctures. For \(0<t<1\), the map \(h_t\) does not preserve \(Q\).
Proof.
At \(r=0\), set \(h_t(c)=c\). Radius is unchanged, and the inverse adds the same angle instead of subtracting it. Each \(h_t\) is therefore a homeomorphism, joined to the identity through homeomorphisms and hence orientation preserving. It is the identity outside the support disk, so it fixes the outer boundary. At \(t=1\) it exchanges the two punctures clockwise and fixes every other puncture.
For every \(0<t<1\), the moving pair is outside \(Q\): both points have nonzero imaginary coordinate, whereas all marked points are real. Thus \(h_t\) is an isotopy of the unmarked disk, but not a permitted isotopy from the identity to \(h_1\) in \(\mathcal H_Q\). Indeed, \(h_1\) cannot be isotopic to the identity there because its puncture permutation is nontrivial. See also the supported half-twist example.
□Solid tracks show the motion so far; faint tracks show its continuation. Drag to rotate, or use the camera sliders. All sliders support arrow keys. Playback is manual; reduced-motion mode advances one time step per press.
Theorem 4.11. The puncture-permutation exact sequence
The sequence
\[ 1\longrightarrow\operatorname{PMod}(D,Q;\partial D) \longrightarrow\operatorname{Mod}(D,Q;\partial D) \xrightarrow{\rho}S_n\longrightarrow1. \]is exact.
Proof.
The classes \([h_1]\) realize every adjacent transposition \((i\ i+1)\). Adjacent transpositions generate \(S_n\): swapping adjacent out-of-order entries repeatedly sorts any permutation into increasing order. Thus \(\rho\) is surjective. We have proved the exact sequence
Its kernel is the pure subgroup by Definition 4.8.
□4.6. Dehn twists and the boundary condition
Definition 4.12. Dehn twist on an annulus
In annular coordinates, a clockwise Dehn twist is the homeomorphism \((r,\theta)\mapsto(r,\theta-2\pi b(r))\), where the continuous cutoff \(b\) equals \(1\) at one boundary and \(0\) at the other. Both boundary circles are fixed; extend the map by the identity beyond the annulus.
Proposition 4.13. The square of a half twist
The map \(h_1^2\) is a clockwise Dehn twist about a curve enclosing the chosen puncture pair. Under the equivalences of Chapters 5 and 6, its class corresponds to \(s_i^2\) and has infinite order.
Alphabeta Math: supported half twist · exponent sum
Worked example for Proposition 4.13
Worked example for Proposition 4.13
Counterexample: A marked homotopy need not be a marked isotopy
Proof.
The square \(h_1^2\) rotates the inner disk by \(-2\pi\), so its endpoint map is the identity there. On the surrounding annulus it makes one full angular turn across the annulus, and outside it is the identity. This is a Dehn twist about a curve enclosing the pair, with clockwise sign in these coordinates. Its permutation is trivial. Here a Dehn twist means an annulus map \((r,\theta)\mapsto(r,\theta-2\pi b(r))\), extended by the identity beyond its boundary circles; the two boundary values of \(b\) are \(1\) and \(0\), so both circles are fixed. This directly identifies the representative. After the isomorphisms in Chapters 5 and 6, \([h_1^2]\) corresponds to the abstract element \(s_i^2\). The exponent-sum homomorphism of Chapter 2 sends \(s_i^{2m}\) to \(2m\), so this class has infinite order. Its trivial permutation alone does not establish that fact.
□Proposition 4.14. A boundary twist and a rotating boundary
A full clockwise twist about all punctures fixes the outer boundary pointwise. It becomes isotopic to the identity when the boundary is only required to be preserved setwise.
Proof.
Likewise, choose \(\max_j|q_j|<r_0<R<1\), and now use a radial cutoff centred at \(0\), equal to \(1\) up to \(r_0\) and \(0\) from \(R\) onward. The endpoint \(T(r e^{\mathrm{i}\theta})=r e^{\mathrm{i}(\theta-2\pi b(r))}\) is a full twist supported in the annulus \(r_0\leq r\leq R\) near the boundary. It fixes the actual boundary pointwise. A rigid rotation is not such a representative. If the boundary were only preserved setwise, a compensating boundary rotation would undo it. For a cutoff \(b\) equal to \(1\) on the inner disk and \(0\) near the boundary, the maps \((r,\theta)\mapsto(r,\theta-2\pi t(b(r)-1))\) fix the punctures throughout, rotate the boundary for intermediate \(t\), and at \(t=1\) give exactly the same full-twist endpoint. This is an allowed isotopy only under the setwise boundary convention. See the boundary-setwise counterexample and González-Meneses, §1.4, p. 7. Under Chapter 6's inverse-endpoint correspondence, the clockwise boundary-twist class is the image of the loop rotating all marked points counterclockwise through \(2\pi\). Indeed the lifted motion \((r,\theta)\mapsto(r,\theta+2\pi t b(r))\) moves those points in exactly this way, and its inverse endpoint is our twist.
□Corollary: nontriviality of the boundary twist
Corollary 4.15. Infinite order of the boundary twist
Under the pointwise boundary convention, the full boundary-twist class has infinite order.
Alphabeta Math: full-twist loop · winding obstruction
Proof.
After Chapter 6 proves that the inverse-endpoint map is injective, the boundary twist has infinite order. Its preimage is the ordered loop \(z_j(t)=e^{2\pi\mathrm{i}t}q_j\), projected to unordered configurations. For any distinct pair \(j,k\), the map \((z_1,\ldots,z_n)\mapsto(z_k-z_j)/|z_k-z_j|\) sends its \(m\)-th power to a circle loop of winding number \(m\). Such a loop contracts only if \(m=0\), by the circle-covering argument of Chapter 3. The covering injection from ordered to unordered configuration groups and the Chapter 6 isomorphism transfer this conclusion to \([T]^m\). Thus the explicit isotopy with rotating boundary really kills a nontrivial class when the pointwise boundary condition is relaxed.
□Remark 4.16. Compatibility of sign and multiplication
Our configuration-space generator in Chapter 3 exchanges the pair counterclockwise: the initially left point travels below, and the initially right point above. For configuration loops, \(\alpha*\beta\) means \(\alpha\) first, then \(\beta\). Mapping classes multiply by usual composition, with the right factor first. To reconcile those orders, Chapter 6 sends a lifted motion \(F_t\) to \([F_1^{-1}]\). Thus that positive braid generator \(\sigma_i\) corresponds to the clockwise half-twist class constructed here. Endpoint assignment \([F_1]\) alone reverses product order; it must not silently be called the same homomorphism.
Sources and further reading
- Juan González-Meneses, Basic results on braid groups, §1.4, printed pp. 6–7: the boundary-fixed mapping-class model and the role of the boundary twist. This chapter specifies its own multiplication and sign conventions.
- Alphabeta Math: Punctured disks, mapping classes, and point pushing, with the directly linked definition and examples above.
Worked examples
These calculations and proofs use this series’ chronological loop products, inverse-endpoint mapping classes, and counterclockwise positive point exchanges. Related statements link to their examples.
Worked example 1. A supported homeomorphism isotopic to the identity with marks fixed
Applies to: Definition 4.1, Proposition 4.2, Definition 4.3, Lemma 4.4, Definition 4.5, Theorem 4.6, Proposition 4.A, Proposition 4.7.
Proof and calculation.
For \(n=3\), choose the empty supporting disk centered at \(c=3i/4\) of radius \(1/16\). It lies in \(D^\circ\) and misses all three real marked points. Put \(\chi(r)=\max(0,1-16r)\) and \(k_t(c+re^{i\theta})=c+re^{i(\theta+t\chi(r))}\), with identity outside. Radius preservation and the inverse angle formula prove every \(k_t\) is a homeomorphism; the formulas match continuously at its boundary and center. The path begins at the identity, so the orientation is positive. Every marked point and every outer boundary point is fixed throughout, giving a genuine isotopy in \(\mathcal H_Q\) and the identity class \([k_1]\).
For all \(s,t\), \(\|k_t-k_s\|_\infty\le(1/16)|t-s|\), since \(|e^{ia}-e^{ib}|\le|a-b|\); the same bound holds for the inverse family. Thus it is also a uniform-topology path. The inverse is \(k_{-t}\) using the same formula, and products add the angle parameters, so \([k_1][k_1^{-1}]=1\) explicitly. For any \(f\in\mathcal H_Q\), \(fk_tf^{-1}\) is an identity isotopy fixing \(Q\) setwise and the boundary pointwise. Joint continuity follows from continuity of \(f,f^{-1}\) on the compact disk. Hence conjugating this element stays in the identity component, exhibiting the normal-component quotient. Finally a path’s puncture images are continuous maps into finite \(Q\): disjoint small neighborhoods separate its points, so each preimage is open and closed in the connected interval. Each image is constant, which verifies the stationary-puncture assertion.
Worked example 2. A clockwise half twist and its pure Dehn-twist square
Applies to: Definition 4.8, Definition 4.9, Proposition 4.10, Theorem 4.11, Definition 4.12, Proposition 4.13.
Proof and calculation.
For the first pair at \(n=3\), take \(c=-1/4,a=1/4,r_0=5/16,R=3/8\). The supporting disk stays inside \(D\), since \(|c|+R=5/8<1\), and misses \(q_3\), whose distance to \(c\) is \(3/4\). The cutoff is 1 up to \(r_0\) and is \((R-r)/(R-r_0)\) on the annulus. Thus \(h_t(c+re^{i\theta})=c+re^{i(\theta-\pi t b(r))}\) has the inverse angle formula, fixes the outer boundary, and is orientation preserving by its path from the identity. At \(t=1\) the first two marks swap; for \(0<t<1\), their nonzero imaginary coordinates show the marked set is not \(Q\). Thus the path is not an isotopy in \(\mathcal H_Q\), even though its endpoint belongs there.
Squaring the endpoint gives angle \(\theta-2\pi b(r)\): it is the identity on the inner disk and outside the support, and on the intervening annulus it is exactly the declared clockwise Dehn twist. Its puncture permutation is the identity, while \(h_1\) realizes \((12)\). Repeating the same construction for the second pair realizes \((23)\); these transpositions generate \(S_3\). The kernel of the endpoint permutation consists precisely of maps fixing each mark, which verifies exactness and the pure-subgroup description. Infinite order of the square follows from the direct argument in the next example.
Worked example 3. Detecting pair twists and boundary twists without an isotopy shortcut
Applies to: Proposition 4.13, Proposition 4.14, Corollary 4.15.
Proof and calculation.
Here is a useful obstruction with proof. Suppose a boundary-fixed disk path \(F_t\), beginning at the identity, traces a configuration loop whose discriminant has nonzero winding. If \(F_1\) were isotopic to the identity through marked-set-preserving boundary-fixed maps, append that isotopy to \(F_t\), obtaining a loop \(L_t\) in the unmarked boundary-fixed homeomorphism group. For \(u>0\), use \(A_u(L_t)(z)=uL_t(z/u)\) on \(|z|\le u\), identity outside, and identity at \(u=0\). The pieces agree on the seam, their inverses are given by replacing \(L_t\) with \(L_t^{-1}\), and their displacement is at most \(2u\). These facts give a continuous based contraction of \(L_t\). Evaluating at \(Q\) would contract its configuration loop, contradicting winding invariance. Thus its endpoint cannot be marked-isotopic to the identity.
For \(h_1^{2m}\), the path of clockwise rotations through angle \(-2m\pi t b(r)\) traces the pair loop with discriminant winding \(-2m\), so these powers are nontrivial for \(m\ne0\). For a boundary twist choose \(b(r)=1\) for \(r\le3/4\), decreasing continuously to 0 at 1, and put \(f_t(z)=e^{-2\pi it b(|z|)}z\). At the three marks, this is total clockwise rotation; each of the three squared pair differences contributes winding \(-2\), total \(-6\). For \(f_1^m\) the total is \(-6m\), so the boundary class also has infinite order under the pointwise convention.
Under setwise boundary preservation, \(g_t(z)=e^{2\pi it(1-b(|z|))}z\) instead runs from the identity to \(f_1\) and fixes all marked points, since \(b=1\) at their radii. On the outer circle it rotates through \(2\pi t\); it is a radius-preserving homeomorphism at every time. This explicitly gives the isotopy that becomes available only after weakening the boundary condition.
Worked example 4. Chronological motion versus ordinary composition
Applies to: Remark 4.16.
Proof and calculation.
Let \(F_t,G_t\) be the positive counterclockwise lifts of \(\sigma_1,\sigma_2\), with endpoints \(F_1=h_1^{-1},G_1=h_2^{-1}\). A lift of “first \(\sigma_1\), then \(\sigma_2\)” first uses \(F_t\), then \(G_t\circ F_1\), because \(F_1(Q)=Q\). Its inverse endpoint is \((G_1F_1)^{-1}=F_1^{-1}G_1^{-1}=h_1h_2\). This homeomorphism has puncture permutation \((12)(23)=(123)\), the inverse of the forward endpoint permutation \((132)\) of the chronological motion. Both inversions and the clockwise sign are forced by these explicit formulas; neither is a matter of suppressing notation.
Counterexamples and distinctions
Each example refutes the stated stronger claim while retaining the hypotheses explicitly checked in its proof.
Counterexample 1. A marked homotopy need not be a marked isotopy
Related statements: Definition 4.3, Definition 4.8, Proposition 4.13.
Claim refuted. Any boundary-fixed homotopy between pure marked homeomorphisms proves their mapping classes equal.
Proof.
Let \(f=h_1^2\) be the supported full pair twist above. It fixes every marked point and the outer boundary. Since the disk is convex, \(H_s(z)=(1-s)z+s f(z)\) is a continuous disk homotopy from the identity to \(f\), fixing every mark and boundary point throughout. But choose a radius in its annulus where \(b(r)=1/2\). On that entire circle \(f(c+re^{i\theta})=c-re^{i\theta}\), so \(H_{1/2}\) sends every point of the circle to \(c\); it is not injective. The winding obstruction just proved shows that no alternative marked isotopy exists either, since \(f\) has nonzero pair-twist winding. Homotopy through arbitrary maps is strictly weaker than isotopy through homeomorphisms here.
Counterexample 2. Setwise boundary fixation changes the group
Related statements: Definition 4.1, Proposition 4.14, Corollary 4.15.
Claim refuted. Replacing pointwise boundary fixation by setwise preservation leaves the same mapping classes.
Proof.
The boundary twist \(f_1\) of the preceding example is nontrivial with pointwise fixation, and every nonzero power is nontrivial by winding \(-6m\). But \(g_t\) is a marked isotopy from the identity to \(f_1\) when the outer circle may rotate. Its boundary points move for \(0<t<1\), so it is excluded by the original convention. Thus the same concrete endpoint map is trivial under one convention and nontrivial under the other. This is a change to the equivalence relation, not a contradiction to the boundary-twist theorem.
Comments