- Configuration loops
- Local point-motion sections
- Lifted disk motions
- Inverse endpoint class
- Alexander contraction
- Isomorphism
A collision-free motion of marked points can be extended to a motion of the whole disk. Its final homeomorphism preserves the marked set. We prove that its inverse determines a mapping class independent of every choice, and that this construction gives an isomorphism.
Theorem 6.1. The mapping class isomorphism
For \(n\geq3\), the map constructed below is an isomorphism
\[\Psi:\pi_1(C_n(D^\circ),Q)\longrightarrow\operatorname{Mod}(D,Q;\partial D).\]If \(F_t\) is a boundary-fixed disk motion with \(F_0=\mathrm{id}\) and \(F_t(Q)=\alpha(t)\), then \(\Psi([\alpha])=[F_1^{-1}]\).
Definition 6.2. Homeomorphism groups and their topology
Recall \(D=\{z\in\mathbb C:|z|\leq1\}\), \(q_j=-1+2j/(n+1)\), and \(Q=\{q_1,\ldots,q_n\}\). Let
\[\mathcal H=\operatorname{Homeo}^{+}(D;\partial D),\qquad \mathcal K=\{f\in\mathcal H:f(Q)=Q\}.\]Both groups carry the uniform topology on maps, equivalent here to the compact-open topology. Their multiplication is ordinary composition. As in chapter 4, \(\operatorname{Mod}(D,Q;\partial D)=\pi_0(\mathcal K)\), where \(\pi_0\) denotes path components. A path in \(\mathcal H\) may move the marked points; a path in \(\mathcal K\) preserves their set at every time.
Worked example for Definition 6.2
Counterexample: The Alexander contraction does not contract the marked subgroup
6.1. Local extension of point motions
Lemma 6.3. Continuous local point-motion sections
For every \(S_0\in C_n(D^\circ)\), there is an open neighbourhood \(U\) and a continuous map \(S\mapsto g_S\in\mathcal H\) such that \(g_S(S_0)=S\) and \(g_{S_0}=\mathrm{id}\). The maps have support in disjoint interior disks.
Proof.
Fix a configuration \(S_0=\{p_1,\ldots,p_n\}\) in the open disk. Choose \(r>0\) so that the closed radius-\(r\) disks about the \(p_i\) are disjoint and contained in \(D^\circ\). A sufficiently nearby unordered configuration \(S\) has exactly one point \(s_i\) in each radius-\(r/4\) disk. Write \(v_i=s_i-p_i\). Let \(U\) be the open neighbourhood consisting of these configurations. The disjoint-disk covering chart from chapter 3 identifies \(U\) homeomorphically with the product of the radius-\(r/4\) disks, so each coordinate \(S\mapsto s_i\) is continuous.
On the plane define the bump and point-moving map
\[\rho_i(z)=\max\{0,1-|z-p_i|/r\},\qquad h_{i,S}(z)=z+\rho_i(z)v_i.\]Then \(h_{i,S}(p_i)=s_i\), and \(h_{i,S}\) is the identity outside its radius-\(r\) disk. It is a homeomorphism: its perturbation of the identity has Lipschitz constant \(|v_i|/r<1/4\), so
\[|h_{i,S}(z)-h_{i,S}(w)|\geq(1-|v_i|/r)|z-w|.\]For every \(y\), the equation \(z=y-\rho_i(z)v_i\) has a unique solution by the contraction principle. Thus the map is onto as well as one-to-one, and its inverse is Lipschitz. Replacing \(v_i\) by \(uv_i\), \(0\leq u\leq1\), gives a path of homeomorphisms from the identity, so it preserves orientation. Since the complement of the supporting disk is fixed and the map is injective and onto, that disk maps to itself: an inside point cannot map to a fixed outside point, and surjectivity gives the converse inclusion.
Restricting the product to \(D\) gives
\[g_S=h_{1,S}\circ\cdots\circ h_{n,S}\in\mathcal H, \qquad g_S(S_0)=S,\qquad g_{S_0}=\mathrm{id}.\]The supports are disjoint, so the product sends every \(p_i\) to \(s_i\). Moreover, \(\|h_{i,S}-h_{i,T}\|_\infty\leq|s_i-t_i|\), where \(t_i\) is the point of \(T\) in the same small disk. Continuity of composition (chapter 4) therefore proves that \(S\mapsto g_S\) is continuous in the uniform topology. This is a continuous local point-motion section. See the corresponding Alphabeta Math lemma.
□Prerequisite: the contraction principle
Theorem 6.A. Contraction principle
A contraction with constant \(c<1\) on a nonempty complete metric space has a unique fixed point.
Proof.
If \(T\) maps a complete metric space to itself with \(d(Tx,Ty)\leq c\,d(x,y)\), \(c<1\), the iterates \(x_{m+1}=T(x_m)\) satisfy \(d(x_{m+1},x_m)\leq c^m d(x_1,x_0)\). The geometric-series estimate makes them Cauchy. Their limit is a fixed point by continuity; any two fixed points have distance at most \(c\) times their distance, so coincide. Apply this to the complete plane and \(T(z)=y-\rho_i(z)v_i\).
The inverse estimate above gives continuous dependence on the target. Continuous dependence on \(S\) also follows from the explicit maps and continuity of inversion in \(\mathcal H\), proved in chapter 4.
□6.2. Global extension and path connectedness
Lemma 6.4. Extension of a continuous point motion
Every path \(\alpha:I\to C_n(D^\circ)\) with \(\alpha(0)=Q\) admits a path \(F_t\in\mathcal H\) with \(F_0=\mathrm{id}\) and \(F_t(Q)=\alpha(t)\).
Proof.
A continuous configuration path \(\alpha\), starting at \(Q\), lifts to labeled paths \(z_i(t)\) by the covering-space construction in chapter 3. Compactness gives a positive lower bound for the distances between points and from points to \(\partial D\). Uniform continuity allows a finite subdivision of time on which each point moves only a small fraction of this bound.
On each subinterval use the local maps from step 1, centered at its initial configuration. If \(F_{t_j}\) has already been constructed, set \(F_t=g_{\alpha(t)}\circ F_{t_j}\) there. Since the local section equals the identity at \(t_j\), the pieces join continuously. Hence
\[F_0=\mathrm{id},\qquad F_t\in\mathcal H,\qquad F_t(Q)=\alpha(t).\] □Lemma 6.5. Path connectedness
The space \(C_n(D^\circ)\) is path connected. Every configuration is \(f(Q)\) for some \(f\in\mathcal H\).
Proof.
The space \(C_n(D^\circ)\) is path connected. One explicit construction first moves the labeled points one at a time to a temporary set disjoint from both the initial and target configurations, then to their targets. At each step a polygonal path in the disk avoiding the finitely many stationary points exists. For endpoints \(a,b\) outside a finite forbidden set, choose an intermediate point \(w\) in the open disk outside the finitely many lines through \(a\) or \(b\) and a forbidden point. Then the segments \(a\to w\to b\) avoid the forbidden set and remain in the open disk by convexity. Thus every configuration is \(f(Q)\) for some \(f\in\mathcal H\).
□6.3. The evaluation bundle
Definition 6.6. Evaluation on the marked set
Define
\[e:\mathcal H\longrightarrow C_n(D^\circ),\qquad e(f)=f(Q).\]Proposition 6.7. Local triviality of evaluation
The map \(e:\mathcal H\to C_n(D^\circ)\) is a continuous surjective fiber bundle with fiber \(\mathcal K\).
Proof.
It is continuous because \(f\mapsto(f(q_1),\ldots,f(q_n))\) is continuous in the uniform topology and the quotient map to unordered configurations is continuous. Step 2 proves that it is onto; its fiber at \(Q\) is \(e^{-1}(Q)=\mathcal K\). To describe it near \(S_0\), choose \(f_0(Q)=S_0\) and put \(s(S)=g_S\circ f_0\). Then
\[U\times\mathcal K\longrightarrow e^{-1}(U),\qquad (S,k)\longmapsto s(S)\circ k\]is a homeomorphism, with inverse \(f\mapsto(e(f),s(e(f))^{-1}\circ f)\). These formulas are continuous because composition and inversion in \(\mathcal H\) are continuous. Consequently \(e\) is a locally trivial fiber bundle with fiber \(\mathcal K\).
□The local product structure also supplies homotopy lifting. The exact version needed below is proved in the following section. This agrees with the evaluation-bundle lemma; no long exact sequence is required for our proof.
Prerequisite: fiber bundles and lifting a square
Definition 6.B. Locally trivial fiber bundle
A locally trivial fiber bundle \(p:E\to B\) has a homeomorphism \(p^{-1}(U)\cong U\times F\) over each suitable open \(U\subseteq B\), taking \(p\) to projection. In one chart a lift of a path is obtained by keeping its fiber coordinate constant.
Theorem 6.C. Lifting a square
We also need this statement: a map \(H:I\times I\to B\), with a chosen lift along the bottom edge, has a lift on the whole square. Chosen lifts on the vertical edges can be preserved when they agree at the bottom corners. Here \(I=[0,1]\).
Proof.
Pull back the trivializing cover to the compact square. A Lebesgue-number subdivision yields finitely many rectangles, each mapping into one chart. Work row by row and within each row from left to right. On each new rectangle the lift is already specified on its bottom edge, usually also its left edge, and sometimes its right edge. In a product chart the base coordinate is given by \(H\). Extend the fiber coordinate by composing its existing edge values with a retraction of the rectangle onto those edges. A rectangle retracts onto its bottom edge, onto its bottom and left edges, and onto its bottom and both vertical edges. For the L-shaped union in a unit square, use
\[(x,y)\mapsto(\max(x-y,0),\max(y-x,0)).\]For the U-shaped union, continue the ray from \( (1/2,2) \) through each point until it hits a side or the bottom. Explicitly, put
\[\lambda(x,y)=\frac{1}{\max\{1-y/2,\,2|x-1/2|\}},\qquad R(x,y)=\bigl(1/2+\lambda(x,y)(x-1/2),\,2+\lambda(x,y)(y-2)\bigr).\]The denominator is at least \(1/2\). The image lies on the bottom or a vertical side, and \(\lambda=1\) on those edges. Thus this formula is a continuous retraction. Rescaling gives these retractions on every grid rectangle. The resulting lifts agree on shared edges, so the finite pasting lemma gives a continuous lift. This argument includes the relative edge conditions just stated.
This is the square case of Hatcher, Algebraic Topology, Proposition 4.48, pp. 379–380. We use this theorem for bundles, whose lifts need not be unique; covering maps have discrete fibers and unique based path lifts.
□6.4. The inverse-endpoint homomorphism
Proposition 6.8. Well-definedness and multiplication
If \(F_0=\mathrm{id}\) and \(F_t(Q)=\alpha(t)\), the rule
\[\Psi([\alpha])=[F_1^{-1}].\]defines a homomorphism \(\Psi:\pi_1(C_n(D^\circ),Q)\to\operatorname{Mod}(D,Q;\partial D)\), independent of the lift and the loop representative.
Proof.
Let \(\alpha\) be a loop at \(Q\), and choose a lift \(F_t\) in \(\mathcal H\) starting at the identity. Since \(F_1(Q)=Q\), its inverse lies in \(\mathcal K\). Define
Choice of lift. For another such lift \(G_t\), the path \(k_t=F_t^{-1}\circ G_t\) lies in \(\mathcal K\) and starts at the identity. Thus \([k_1]=1\), and \(G_1=F_1\circ k_1\) gives \([G_1^{-1}]=[F_1^{-1}]\).
Choice of loop representative. An endpoint-fixed homotopy \(H(s,t)\) between loops lifts to \(\widetilde H(s,t)\) with \(\widetilde H(s,0)=\mathrm{id}\), by the square-lifting result. At \(t=1\), the path \(s\mapsto\widetilde H(s,1)\) lies in \(\mathcal K\). Its inverse path shows that the two endpoint inverse classes agree. The choice-of-lift argument compares these lifts with any previously chosen ones.
Multiplication. Our product \(\alpha*\beta\) traverses \(\alpha\) first. If \(F_t,G_t\) lift the two loops, a lift of their product first follows \(F_t\), then \(G_t\circ F_1\). Its endpoint is \(G_1\circ F_1\), so
\[\Psi([\alpha][\beta])=[(G_1\circ F_1)^{-1}] =[F_1^{-1}\circ G_1^{-1}]=\Psi([\alpha])\Psi([\beta]).\] □The inverse is required by our convention. Sending a motion to \([F_1]\) would reverse the order of multiplication. Compare the point-motion homomorphism lemma.
6.5. The Alexander contraction
Theorem 6.9. Contractibility of the unmarked homeomorphism group
For \(f\in\mathcal H\) and \(0<u\leq1\), define the family below and set \(A_0(f)=\mathrm{id}\). This family is jointly continuous in \(u\) and \(f\), consists of boundary-fixed orientation-preserving homeomorphisms, and satisfies \(A_0(f)=\mathrm{id}\), \(A_1(f)=f\), and \(A_u(\mathrm{id})=\mathrm{id}\). Hence \(\mathcal H\) is contractible.
\[A_u(f)(z)= \begin{cases}u\,f(z/u), & |z|\leq u,\\ z, & |z|\geq u,\end{cases} \qquad A_0(f)=\mathrm{id}.\]Worked example for Theorem 6.9
Counterexample: The Alexander contraction does not contract the marked subgroup
Proof.
The formulas agree at \(|z|=u\), since \(f\) fixes the unit circle. The inverse is \(A_u(f^{-1})\), so every \(A_u(f)\) is a boundary-fixed homeomorphism. It preserves orientation: on the inner disk it is a conjugate of \(f\), and outside it is the identity. We have \(A_1(f)=f\) and \(A_u(\mathrm{id})=\mathrm{id}\).
The estimate \(\|A_u(f)-\mathrm{id}\|_\infty\leq2u\) proves continuity at \(u=0\), uniformly in \(f\). At positive \(u\), the glued formulas are jointly continuous in \((u,z)\) for fixed \(f\); compactness gives continuity in the uniform norm. Finally,
\[\|A_u(f)-A_u(g)\|_\infty\leq\|f-g\|_\infty\]gives joint continuity in \((u,f)\). Thus \(A_u\) contracts \(\mathcal H\) to the identity. This is the Alexander contraction.
□This contraction need not preserve \(Q\). It contracts the unmarked group \(\mathcal H\), which is precisely what the argument needs.
6.6. Proof of the mapping class isomorphism
Proof of Theorem 6.1.
Surjectivity. Given a mapping class \([k]\), \(k\in\mathcal K\), take the path \(F_t=A_t(k^{-1})\) from the identity to \(k^{-1}\). Its projected configuration path \(\alpha(t)=F_t(Q)\) is a loop. By construction \(\Psi([\alpha])=[k]\).
Injectivity. Suppose \(\Psi([\alpha])=1\). For a chosen lift \(F_t\), this means \(F_1\) is joined to the identity by a path \(k_t\) in \(\mathcal K\). Concatenate \(F_t\) with \(k_t\) to obtain a based loop \(L\) in \(\mathcal H\). Its evaluation is \(\alpha\) followed by a constant loop, so represents \([\alpha]\). The based contraction
\[(u,t)\longmapsto A_u(L(t))\]deforms \(L\) to the constant loop while keeping its endpoints the identity. Evaluating at \(Q\) gives an endpoint-fixed null homotopy of \(\alpha\). Hence \([\alpha]=1\), proving injectivity.
Together with the homomorphism calculation, this proves the theorem. See also the evaluation-bundle isomorphism theorem.
□6.7. The three models and pure subgroups
Corollary 6.10. Comparison of braid groups and pure subgroups
The isomorphism sends \(s_i\) to the clockwise half-twist class and restricts to \(P_n\cong\operatorname{PMod}(D,Q;\partial D)\).
Proof.
A counterclockwise elementary exchange lifts to a counterclockwise half rotation supported near two adjacent marked points. The inverse endpoint is the clockwise half twist used in chapter 4. Combining \(\Psi\) with chapter 5 therefore identifies the Artin generators with those clockwise mapping classes:
The construction restricts to an isomorphism of pure subgroups. Indeed, \(t\mapsto(F_t(q_1),\ldots,F_t(q_n))\) is the unique ordered lift of the loop starting at the base tuple. If its forward endpoint permutation is \(e_\alpha\), then \(F_1(q_j)=q_{e_\alpha(j)}\); the puncture permutation of \(F_1^{-1}\) is \(e_\alpha^{-1}\). Thus \(\Psi\) intertwines exactly the permutation homomorphisms defined in chapters 3 and 4. Their kernels correspond in both directions under the isomorphism, giving \(P_n\cong\operatorname{PMod}(D,Q;\partial D)\).
□Sources
The linked Alphabeta Math items supply the local-section, lifting, contraction, and isomorphism statements. The standard braid interpretation is discussed in González-Meneses, Basic results on braid groups, §1.4. For bundle lifting and its general homotopy-theoretic form, see Hatcher, Algebraic Topology, Proposition 4.48 and Theorem 4.41. Our argument proves bijectivity directly instead of assuming the long exact sequence.
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 quantitatively controlled local section and its inverse
Applies to: Lemma 6.3, Theorem 6.A.
Proof and calculation.
At \(n=3\), choose radius \(r=1/8\) disks about \(q_j\) and perturb only the first mark by \(v=i/64\). Their closures are disjoint and lie in the open disk, and \(|v|/r=1/8<1/4\). Set \(\rho(z)=\max(0,1-|z-q_1|/r)\), \(g(z)=z+\rho(z)v\). The bump has Lipschitz constant \(1/r\) by the reverse triangle inequality and the fact that taking the positive part is 1-Lipschitz. Therefore
\[|g(z)-g(w)|\ge\frac78|z-w|.\]It is injective. Given \(y\), iteration \(z_{m+1}=y-\rho(z_m)v\) has step differences at most \((1/8)^m|z_1-z_0|\), so its tails are bounded by the corresponding geometric series and converge in the complete plane. Passing to the limit gives the unique inverse point; uniqueness follows by multiplying the distance of two solutions by at most \(1/8\). The inverse is Lipschitz with constant \(8/7\). Outside the support, \(g\) is the identity; injectivity and surjectivity therefore imply that the supporting disk maps to itself. Replacing \(v\) by \(tv\) gives orientation-preserving boundary-fixed homeomorphisms, with \(g(q_1)=q_1+v\). For different small displacements, \(\|g_v-g_w\|_\infty\le|v-w|\), proving continuous dependence for this actual local section.
Worked example 2. Extend a short point path and connect two configurations
Applies to: Lemma 6.4, Lemma 6.5.
Proof and calculation.
The path \(\alpha(t)=\{q_1+it/64,q_2,q_3\}\) stays in the local neighborhood above. The formula \(F_t(z)=z+\rho(z)(it/64)\) extends it to disk homeomorphisms, beginning at the identity, and fixes the other marks. This directly exhibits extension without an existence-only appeal.
For a larger path-connectivity example, connect \(Q\) to \(\{-i/4,0,1/2\}\) by moving only the first point along \(z_1(t)=-(1-t)/2-it/4\). The disk is convex, and this segment never hits 0: its real and imaginary parts cannot both vanish at the same \(t\). It never hits \(1/2\), since its real part is nonpositive. Explicitly, \(|z_1(t)|^2=(5t^2-8t+4)/16\ge1/20\), so its distance from the stationary mark zero is at least \(1/\sqrt{20}\); its distance from \(1/2\) is at least \(1/2\), and every point has boundary distance at least \(1/2\). Supporting radius \(r=1/16\) therefore gives disjoint interior disks throughout. Dividing into 64 equal time intervals moves the first point by \(\sqrt5/256<1/64=r/4\) per step. On each interval apply the explicitly proved local map, centered at that interval’s initial configuration, and compose with the already constructed endpoint map. At the join the new map is the identity, so the pieces paste continuously. The final map carries \(Q\) to this concrete target configuration and proves it lies in the same path component.
Worked example 3. A trivializing chart and a square lift, with its fiber coordinate
Applies to: Definition 6.6, Proposition 6.7, Definition 6.B, Theorem 6.C.
Proof and calculation.
Near \(Q\), let \(s(S)=g_S\) be the product of the disjoint small-displacement maps. Evaluation is continuous here because \(|f(q_j)-h(q_j)|\le\|f-h\|_\infty\) for every \(j\), followed by the continuous unordered quotient. Every \(f\) evaluating to \(S\) has a unique decomposition \(f=s(S)k\) with \(k=s(S)^{-1}f\) fixing \(Q\) setwise. Indeed \(k(Q)=s(S)^{-1}S=Q\), and the two formulas compose to the identity. Continuity of these explicit products and inverses gives the local product chart \(U\times\mathcal K\), with evaluation equal to projection. This verifies the bundle definition in the stated application. To check the continuity used here, let \(\omega_f(\delta)=\sup_{|x-y|\le\delta}|f(x)-f(y)|\), tending to zero by uniform continuity on the compact disk. The product and inverse estimates are
\[\|f_jg_j-fg\|_\infty\le\|f_j-f\|_\infty+\omega_f(\|g_j-g\|_\infty),\]\[\|f_j^{-1}-f^{-1}\|_\infty\le\omega_{f^{-1}}(\|f_j-f\|_\infty).\]For the second estimate, substitute \(x=f_j^{-1}(y)\), so \(|f(x)-y|\le\|f_j-f\|_\infty\), and apply \(f^{-1}\). These inequalities prove the needed product and inverse continuity directly.
For a local square take \(H(s,t)=\{q_1+ist/64,q_2,q_3\}\). In this chart \(\widetilde H(s,t)=g_{H(s,t)}\) is a lift with bottom edge the identity. More generally, if the chosen bottom lift has fiber coordinate \(k(s)\), the formula \(g_{H(s,t)}k(s)\) preserves it. For prescribed fiber data \(k\) on the bottom and both vertical edges, use the chapter’s explicit U-edge retraction \(R(s,t)\) and set \(g_{H(s,t)}k(R(s,t))\); it agrees exactly with those edges because \(R\) fixes them. The retraction’s denominator is at least \(1/2\), so the composition is continuous. Thus this example proves the requested relative lifting behavior locally without assuming unique bundle lifts.
Worked example 4. A constant evaluation loop and an explicit Alexander null homotopy
Applies to: Definition 6.2, Theorem 6.9.
Proof and calculation.
Let \(k_t\) be the marked-point-fixing rotation supported in the empty disk from chapter 4, and put \(L_t=k_{\sin(\pi t)}\). Then \(L_0=L_1=\mathrm{id}\), and \(L_t(Q)=Q\) throughout. It is a nonconstant loop of actual maps in \(\mathcal K\). On \(|z|\le u\), \(A_u(L_t)(z)=uL_t(z/u)\); outside it is \(z\). The two values agree on \(|z|=u\) because \(L_t\) fixes the unit circle. The inverse on each slice is \(A_u(L_t^{-1})\), and the displacement is at most \(2u\), proving joint continuity at zero. At positive \(u\) the formulas and their matching seams are continuous; uniform continuity on compact parameter sets supplies uniform-topology continuity. The endpoints in \(t\) remain the identity. Evaluating this square gives a based null homotopy of its configuration loop, explicitly exhibiting the unmarked contraction mechanism. The evaluated starting loop is already constant, but its lifted disk loop is not, so the distinction between these objects is visible.
Worked example 5. The comparison maps on a product and on a pure square
Applies to: Theorem 6.1, Proposition 6.8, Corollary 6.10.
Proof and calculation.
The positive exchanges \(\alpha=\sigma_1,\beta=\sigma_2\) lift to counterclockwise supported maps \(F_t,G_t\). The lift of their chronological product has endpoint \(G_1F_1\); taking its inverse gives \(F_1^{-1}G_1^{-1}=h_1h_2\), so multiplication is preserved. Its puncture permutation is \((12)(23)=(123)\), while the forward strand-endpoint permutation is \((132)\); the inverse exactly matches the chapter-3 quotient map. For \(\alpha^2\), the inverse endpoint is \(h_1^2\). It fixes every mark, is the explicit annular Dehn twist, and has nonzero winding obstruction as proved in chapter 4, so the pure element and the pure mapping class are nontrivial.
Proof of independence for these lifts
If \(G_t'\) is any other lift of the same point path starting at the identity, then \(k_t=F_t^{-1}G_t'\) maps \(Q\) to itself and starts at the identity. It is a continuous marked isotopy by the uniform-topology inverse and composition formulas. Its endpoint is therefore trivial in \(\pi_0(\mathcal K)\), and \((G_1')^{-1}=k_1^{-1}F_1^{-1}\) has the same class as \(F_1^{-1}\). For a homotopy of the point paths with fixed endpoints, lift its square as above; its endpoint edge is a path in \(\mathcal K\), proving the inverse endpoint classes equal. Thus the calculation is of classes independent of choices, not only a coincidence for one selected formula.
Counterexamples and distinctions
Each example refutes the stated stronger claim while retaining the hypotheses explicitly checked in its proof.
Counterexample 1. A bundle lift need not be unique
Related statements: Lemma 6.4, Proposition 6.7, Theorem 6.C.
Claim refuted. As for the configuration covering, an initial homeomorphism uniquely determines a lift through evaluation.
Proof.
For the constant path \(\alpha(t)=Q\), both \(F_t=\mathrm{id}\) and the nonconstant empty-disk supported \(k_t\) from chapter 4 lift it and start at the identity. The latter is not the identity for \(t>0\): at any radius strictly between 0 and its supporting radius the angular change \(t\chi(r)\) is nonzero and less than \(2\pi\). Both lifts fix all marks, so evaluation cannot distinguish them. The fiber is a homeomorphism group with nonconstant paths, unlike the discrete fiber of the configuration covering. The choice-independence proof passes to endpoint classes precisely to remove this nonuniqueness.
Counterexample 2. The inverse endpoint cannot be omitted
Related statements: Theorem 6.1, Proposition 6.8.
Claim refuted. Sending a chronological loop to the class of its forward endpoint is the same homomorphism.
Proof.
For \(\sigma_1\) followed by \(\sigma_2\), the forward endpoint is \(G_1F_1\), whose puncture permutation is \((23)(12)=(132)\). Multiplying the two forward endpoint classes in the written order instead gives \(F_1G_1\), with permutation \((12)(23)=(123)\). These classes differ because a marked isotopy cannot change the puncture permutation. Thus the forward endpoint assignment reverses products and is not this homomorphism. Inverting the endpoint reverses the order again and changes counterclockwise supported rotation to the clockwise generator, as the explicit product calculation proves.
Counterexample 3. The Alexander contraction does not contract the marked subgroup
Related statements: Definition 6.2, Theorem 6.9.
Claim refuted. Because the unmarked group contracts, the same Alexander formula stays inside the marked group.
Proof.
Take \(n=3\) and the half-twist endpoint \(f=h_1\), so \(f(0)=-1/2\). At \(u=1/4\), the Alexander formula gives \(A_u(f)(0)=u f(0)=-1/8\), which is not in \(Q=\{-1/2,0,1/2\}\). The other two marked points are outside \(|z|\le1/4\) and remain fixed. Therefore this perfectly valid boundary-fixed homeomorphism does not preserve \(Q\), so the contraction leaves \(\mathcal K\). Contractibility of \(\mathcal H\) says nothing of that form about \(\mathcal K\); the marked components are exactly where the mapping classes remain.
Counterexample 4. Both fixed-point hypotheses have content
Related statements: Theorem 6.A.
Claim refuted. A strict contraction needs no completeness assumption, or a 1-Lipschitz map always has a fixed point.
Proof.
On the nonempty incomplete metric space \((0,1)\), \(T(x)=x/2\) is a self-map with Lipschitz constant \(1/2\). A fixed point would satisfy \(x=x/2\), hence \(x=0\), outside the space. Its iterates \(x/2^m\) are Cauchy but have no limit there. On the complete real line, \(S(x)=x+1\) has Lipschitz constant 1 and no fixed point. These examples respectively retain strict contraction and completeness, while removing the other required condition. In the local disk extension proof, the complete plane and a constant strictly less than 1 are both checked before invoking iteration.
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