The Artin Action on a Free Group

Removing the noncompact puncture neighborhoods. In $B_i\setminus \{q_i\}$ write $z=q_i+r{e}^{\mathrm{i}\theta }$, $0<r\le {\epsilon }_i$, and send it at time $u$ to $q_i+((1-u)r+u{\epsilon }_i)e^{\mathrm{i}\theta }$. Keep the complement of the disk interiors fixed. This formula is continuous on $X=D^2\setminus Q_n$ (no extension to $q_i$ is asserted), fixes $C_i$, and ends in $S$. The finitely many formulas agree on their boundary circles, so they give a strong deformation retraction $X\to S$ fixing the truncated flower $W$.

Cutting the compact holed disk. Open $S$ along the $t_i$, separating the sectors at $d$. The resulting compact surface $K$ is a disk: thicken the straight tethers into thin rectangular strips from the outer boundary to their respective circular holes; the remaining planar region has a single Jordan polygonal boundary with circular detours. First flatten the outer circle near $d$ while fixing every tether. For small $\mathrm{\mid }x\mathrm{\mid }$ write its top as $y_b(x)=\sqrt{1-x^2}$ and put $w(x)=\mathrm{\mid }x\mathrm{\mid }\mathrm{/}2$. At a point of the tether to $q_i\mathrm{\ne }0$ one has $x=t{q}_i$, $y=1-t=1-\mathrm{\mid }x\mathrm{\mid }\mathrm{/}\mathrm{\mid }{q}_i\mathrm{\mid }\le 1-\mathrm{\mid }x\mathrm{\mid }$, since $\mathrm{\mid }{q}_i\mathrm{\mid }<1$; a tether with $q_i=0$ lies on $x=0$. For sufficiently small nonzero $\mathrm{\mid }x\mathrm{\mid }$, $1-y_b(x)<\mathrm{\mid }x\mathrm{\mid }\mathrm{/}2$, so $y_b(x)-w(x)>1-\mathrm{\mid }x\mathrm{\mid }$ and the collar $y_b(x)-w(x)<y<y_b(x)+w(x)$ misses every tether. On each vertical fiber map $y_b(x)$ to $y_b(x)+\chi (x)(1-y_b(x))$, fixing its two collar endpoints and interpolating linearly on the two pieces; take $\chi =1$ near zero and $\chi =0$ outside a slightly larger small interval. Since $0\le 1-y_b(x)<w(x)$, these fiber maps are increasing. At $x=0$ use the identity; the displacement bound $1-y_b(x)\to 0$ proves continuity of both maps and inverses there. The outer boundary becomes flat near $d$ and all tethers stay fixed. Away from that flat segment the outer boundary is at positive distance from the flower, so finitely many ordinary boundary collar charts replace its remaining circular pieces by close polygonal chords, fixing the flower. Next straighten each inner circular boundary portion by an explicit radial collar map: choose a sufficiently fine inscribed polygon with the tether contact as one vertex, let $R(\theta )>0$ be its radial boundary function, and on the outer annular collar interpolate monotonically from radius $R(\theta )$ at the old circle radius to the unchanged outer collar radius. Choose the polygon fine enough that the interpolation stays strictly increasing. On the tether direction $R$ equals the original circle radius, so the tether is fixed. All boundaries are now finite polygons and the tethers remain straight. Open their narrow vertex disks and edge strips using [F5]; the boundary trace is a single simple polygon, and [F5] supplies its disk parametrization by finite diagonal splitting. No general Jordan–Schönflies extension or arbitrary plane-arc theorem is used for this fixed circular/straight geometry. This supplies a disk coordinate compatible with the side collars, so opening the zero-width tethers has the same disk topology. Its boundary is the union of the single outer arc $A$ and its complementary closed arc $P$. The arc $P$ consists, in order, of all the tether shores and all the circles opened at their tether endpoints. For $n\ge 1$, $A$ and $P$ meet just at their two endpoints, and all paired shores lie in $P$. The quotient $\kappa :K\to S$ identifies matching tether shores and the sector copies of $d$, and its image of $P$ is precisely $W$.

The quotient-compatible retraction. In the disk coordinate of step 1.2 take $(K,P)$ to $([0,1{]}^2,[0,1]\times \{0\})$, using the prescribed PL boundary extension of [F5], and returning through the explicit collar coordinates of step 1.2. The homotopy $(x,y)\mapsto (x,(1-u)y)$ strongly retracts the square onto its bottom edge. Transport it to $K$, where it fixes $P$ pointwise. For paired points $z,z^{\mathrm{\prime }}$ one has $\kappa (R_u(z))=\kappa (z)=\kappa (z^{\mathrm{\prime }})=\kappa (R_u(z^{\mathrm{\prime }}))$, since both lie in $P$. Thus $(z,u)\mapsto \kappa (R_u(z))$ descends through $\kappa \times {\mathrm{id}}_I$. This is a quotient map because $K\times I$ is compact and $S\times I$ is Hausdorff. The descended continuous homotopy strongly retracts $S$ onto $W$. Composing with step 1.1 proves the deformation-retract clause. When $n=0$, take $W=\{d\}$ and use the straight-line contraction of $D^2$ to $d$.

The meridian basis. Contract the finite tether tree $T$ to $d$ along its edges, fixing $d$. By [F1], the collapse $c:W\to W\mathrm{/}T$ is a based homotopy equivalence. The quotient graph $W\mathrm{/}T$ is a wedge of the $n$ circles $C_i$, and $c\circ x_i$ traverses its $i$-th circle once positively. Hence [F2] gives a free basis $[x_i]$ of ${\pi }_1(W,d)$ and an isomorphism to ${\pi }_1(X,d)$. The flower itself is a lollipop graph, rather than homeomorphic to the wedge.

Higher homotopy. The universal cover of the wedge graph has vertices the reduced words and an edge from $w$ to $w{x}_i$ for every $i$. Local stars map homeomorphically to the star of its wedge vertex, so this is a covering; uniqueness of reduced words [F3] implies the cover is a tree. Give each edge length one and contract along the unique geodesic to the root, sending distance $r$ to $(1-u)r$. The locally finite graph metric gives the graph topology, and this contraction is continuous and fixes the root. For $j\ge 2$, [F4] lifts any based sphere map to this contractible tree, where it contracts; projecting makes the original map nullhomotopic. Thus the wedge has vanishing higher homotopy, and [F2] with the based equivalence of step 3.1 gives the same for $W$. All coordinate selections and triangulations concern finitely many supplied straight segments and circles; no choice axiom is used.

The universal-property map. Let $u:\{x_1,\dots ,x_n\}\to {\pi }_1(X,d)$, $u(x_i):=[x_i]$. By [F3] there is a unique homomorphism $\widehat{u}:F_n\to {\pi }_1(X,d)$ with $\widehat{u}(x_i)=[x_i]$. The inclusion $i:W\hookrightarrow X$ and the retraction $\rho :X\to W$ of [F1] are based at $d$, and by [F2] the induced maps $i_{\ast }:{\pi }_1(W,d)\to {\pi }_1(X,d)$ and ${\rho }_{\ast }$ are mutually inverse isomorphisms.

The basis map. By [F1], ${\pi }_1(W,d)$ is free on the classes $[x_1],\dots ,[x_n]$, so the assignment $x_i\mapsto [x_i]\in {\pi }_1(W,d)$ extends by [F3] to an isomorphism $\varphi :F_n\to {\pi }_1(W,d)$: it is the unique homomorphism with $\varphi (x_i)=[x_i]$, and the universal property applied to the inverses shows it is bijective (equivalently, $F_n$ and ${\pi }_1(W,d)$ are free on the same set, so [F3]'s uniqueness clause gives the isomorphism).

The case $n=0$. For $n=0$ the configuration is empty, $D^2\setminus Q_0=D^2$ is contractible (the straight-line homotopy to the origin), the empty basis is a basis of the trivial group, and the unique map from the trivial free group is an isomorphism; the argument above also covers this case with empty index sets.

Comparison. The composite $i_{\ast }\circ \varphi :F_n\to {\pi }_1(X,d)$ is a homomorphism with $x_i\mapsto i_{\ast }[x_i]=[x_i]$, since $i$ is the inclusion of the subspace containing the loops $x_i$. By the uniqueness clause of [F3] applied to $u$, $i_{\ast }\circ \varphi =\widehat{u}$. Since $i_{\ast }$ and $\varphi$ are bijections, $\widehat{u}$ is a group isomorphism.

Conclusion. Steps 1.1, 1.2 and 2.1 exhibit the isomorphism $\widehat{u}:F_n\to {\pi }_1(D^2\setminus Q_n,d)$ with $x_i\mapsto [x_i]$, and step 1.3 covers the empty case; hence $[x_1],\dots ,[x_n]$ is a free basis of ${\pi }_1(D^2\setminus Q_n,d)$.

A concrete model of a slit tip. Choose pairwise disjoint small round disks $B_i$ about the punctures, meeting only their own stems. In polar coordinates about $q_i$, put the stem radius at angle $0$. The cut of $B_i\setminus \{q_i\}$ is $(0,R_i]\times [0,2\pi ]$, not a compact annulus or disk. Adjoin its missing tip by forming $E_i=([0,R_i]\times [0,2\pi ])\mathrm{/}(\{0\}\times [0,2\pi ])$: the whole zero-radius edge is collapsed to one point. This is a closed disk (a rectangle with one edge collapsed, equivalently a triangle), whose boundary is the outer circular arc and the two radial sides joined at the tip. The map $(r,\theta )\mapsto q_i+r{e}^{\mathrm{i}\theta }$ extends continuously to the tip and, on identifying the two radial sides, gives the filled disk $B_i$. Deleting the tip before regluing gives $B_i\setminus \{q_i\}$.

The outer piece. Put $\mathrm{\Omega }=D^2\setminus {\bigcup }_i\mathrm{int}{B}_i$. Open it along the truncated stems from $d$ to $C_i=\mathrm{\partial }{B}_i$, separating all sectors at their common endpoint $d$. To see that the result is a disk, first thicken each truncated stem to a narrow strip, with strips disjoint away from a small half-disk at $d$. The region left between these strips and the $B_i$ has one polygonal Jordan boundary: tracing it visits the outer boundary once and makes one detour along both sides of each strip and around its associated hole. It is a closed disk by [F1]. Shrinking the strip widths gives the same cut topology, since each strip-side collar has a rectangular coordinate chart and changing its width is a homeomorphism. Each opened $C_i$ is a closed boundary arc $A_i$; at $d$ there are $n+1$ sector copies when $n>0$, rather than just two copies for the whole star. For $n=0$ the outer piece is $D^2$.

Attaching the completed tips. Glue the circular boundary arc of $E_i$ to $A_i$ with matching radial-side endpoints. Gluing two disks along a proper closed boundary arc gives a disk: map the two disks to the upper and lower half-disks, with the glued arcs as their common diameter, and use these maps on the quotient. Applying this construction finitely many times to the outer disk and the $E_i$ yields a compact disk $H$. Its boundary consists of the outer boundary arc and the two sides of every stem, with each pair joined at $q_i$ and with consecutive sides joined at the appropriate sector copy of $d$.

The quotient and its topology. Identify matching points on each pair of stem sides, including the copies of $d$. The quotient map $\pi :H\to D^2$ is the ordinary coordinate map off the cuts and the polar map of step 1.1 at a tip. It is a continuous surjection whose fibers are exactly these prescribed identifications; compactness of $H$ and the Hausdorff property of $D^2$ show that its quotient is the filled disk $D^2$. Removing the images $q_i$ of the added tips recovers $X=D^2\setminus Q_n$. Thus the compact disk assertion concerns the completed cut, with both filled and punctured quotients accounted for.

Lifting maps. A homeomorphism $f$ as in the statement preserves each local side of every stem: an orientation-reversing map would reverse the outer boundary, which $f$ fixes pointwise, and an orientation-preserving map fixing an oriented stem cannot interchange its sides. Thus it lifts on the open cut surface, fixing both stem-side copies and all sector copies of $d$. Near a tip, continuity of $f$ at $q_i$ implies that points with radius tending to zero have image radius tending to zero, uniformly in their angle; the collapsed-edge model therefore extends the lift continuously by fixing the tip. The same argument applies to $f^{-1}$, so the lift is a boundary-fixed homeomorphism of $H$.

Regluing maps and isotopies. A boundary-fixed homeomorphism $g$ of $H$ respects every fiber of $\pi$ and induces a homeomorphism of the filled disk, with inverse induced by $g^{-1}$. It fixes the outer boundary, stems and punctures pointwise. If $G:H\times I\to H$ is a boundary-fixed isotopy, the map $(z,t)\mapsto \pi (G(z,t))$ is constant on the fibers of $\pi \times {\mathrm{id}}_I$; this map is a quotient map because its domain is compact and its target is Hausdorff. It therefore induces a continuous isotopy $D^2\times I\to D^2$, including at every puncture uniformly in time. This proves the asserted correspondence and the isotopy version needed by consumers.

Constructing the cut disk. Suppose $n\ge 1$ and put $S=D^2\setminus {\bigcup }_i\mathrm{int}{B}_i$. We establish the needed cut geometry directly. Near $d$, the top outer boundary is $y_b(x)=\sqrt{1-x^2}$, whereas every nonvertical tether has $y=1-\mathrm{\mid }x\mathrm{\mid }\mathrm{/}\mathrm{\mid }{q}_i\mathrm{\mid }\le 1-\mathrm{\mid }x\mathrm{\mid }$. Use the collar between $y_b(x)\pm \mathrm{\mid }x\mathrm{\mid }\mathrm{/}2$, which misses the tethers for small nonzero $\mathrm{\mid }x\mathrm{\mid }$ because $1-y_b(x)<\mathrm{\mid }x\mathrm{\mid }\mathrm{/}2$. On each vertical fiber send $y_b(x)$ to $y_b(x)+\chi (x)(1-y_b(x))$, where $\chi =1$ near zero and vanishes outside a small interval; fix the collar endpoints and interpolate linearly. The target lies strictly between those endpoints, so each fiber map is increasing. Extend by the identity outside the collar and on $x=0$; the displacement tends to zero there, proving continuity of the map and its inverse, including on the vertical tether if present. This flattens the outer boundary near $d$ and fixes all tethers. Away from this segment the outer boundary has positive distance from them, so finite circle collar charts replace it by polygonal chords. For each inner circle choose a fine inscribed polygon with $p_i$ as a vertex. If $R_i(\theta )$ is its radial boundary function, map radius ${\epsilon }_i$ to $R_i(\theta )$ and a slightly larger collar radius to itself by increasing linear interpolation. The collars can be disjoint and miss other tethers; on its own tether direction $R_i={\epsilon }_i$, so that tether stays fixed. Thus all boundaries become polygons while the tethers stay straight. Open each tether using the vertex-sector and edge-strip coordinates of [F3], separating the sectors at $d$. The boundary trace follows the outer boundary once and makes one detour down and back along each slit and around its hole. This is a single simple polygon after the shores have been separated: distinct tethers have disjoint interiors, distinct holes are disjoint and meet only their own tether, and the finitely many sectors at $d$ are distinct. By [F3] its enclosed region is a disk. The side and sector coordinates identify this region with the zero-width cut surface $K$, giving $K$ disk topology. Its boundary splits into the outer arc $A$ and the complementary arc $P$; $P$ contains all tether shores and all opened inner circles. Regluing the paired shores and sector copies of $d$ gives a continuous quotient $\kappa :K\to S$, with $\kappa (P)=W$.

The two boundary paths of the cut disk. Orient $A$ by the positive outer boundary traversal, from its initial sector copy of $d$ to its terminal sector copy. Orient the complementary arc $P$ in the same initial-to-terminal direction, opposite to its direction as a piece of the oriented boundary of $K$. In a convex disk coordinate for $K$, linear interpolation between these paths gives a homotopy relative to their endpoints. Composing with $\kappa$ and the inclusion $S\hookrightarrow X$ gives a based homotopy between $\mathrm{\partial }$ and the image of $P$.

Tracing $P$ after regluing. With this direction, $P$ runs out along the first tether, counterclockwise around its circle, back along its other shore, and repeats for each tether in order. The sign follows from boundary orientation: inner circles of the oriented holed disk are clockwise, whereas $P$ traverses them opposite to that boundary direction. Its order is $1,\dots ,n$: from $d=(0,1)$ the positive outer boundary starts toward the left, and the distinct downward tether rays to $q_1<\cdots <q_n$ occur from left to right. After quotienting the paired shores, these successive paths are exactly $t_i{c}_i{t}_i^{-1}=x_i$. Hence the image of $P$ is the concatenation $x_1\cdots x_n$, up to harmless parametrization and constant intervals.

Conclusion. The based homotopy of step 2.1 and the traversal of step 3.1 imply the asserted identity by [F2]. For $n=0$ the straight-line homotopy from $\mathrm{\partial }(t)$ to $d$ contracts the boundary relative to its basepoint, giving the empty product; for $n=1$ the same cut-disk argument is the positive outer/inner circle homotopy in the annulus. All collar charts, polygonal subdivisions and strips are finite, so no choice principle is used.

Passing to an actual wedge. Combine the deformation retraction with the collapse equivalence of [F1]. They give based maps $a:X\to R$, $b:R\to X$ with $ba\simeq {\mathrm{id}}_X$ and $ab\simeq {\mathrm{id}}_R$ relative to their basepoints. For $g=afb:R\to R$, [F2] and $f_{\ast }=\mathrm{id}$ imply $g_{\ast }=a_{\ast }{f}_{\ast }{b}_{\ast }=a_{\ast }{b}_{\ast }=\mathrm{id}$.

Homotoping the wedge map. Restrict $g$ to each actual circle summand of $R$. Its based-loop class equals that summand's standard generator because $g_{\ast }=\mathrm{id}$. By [F3], it has a based homotopy to that summand's inclusion. The finitely many homotopies agree at the wedge vertex at every time and hence glue continuously on the finite quotient $R\times I$. They give $g\simeq {\mathrm{id}}_R$ relative to the vertex.

Returning to the punctured disk. Compose the based homotopies to obtain $f\simeq bafba=bga\simeq ba\simeq {\mathrm{id}}_X$, all relative to $d$. For $n=0$ the wedge is a point and the same argument is the based contraction of the disk. This is a homotopy of maps on $X$; it claims no extension to any puncture. Only finitely many based-loop homotopies and the specified finite graph equivalences occur, so no choice principle is used.

The normalized arc and its double lift. Identify the plane with $\mathbb{C}$ and its compactification with the sphere. The Möbius homeomorphism $M(z)=(z-A)\mathrm{/}(z-B)$ takes the endpoints to $0,\mathrm{\infty }$. Its value at the original plane-infinity is $1$, which is not on the normalized arc $b=M\circ a$. For $0<t<1$, $b(t)$ lies in ${\mathbb{C}}^{\ast }$. Lift its argument continuously on that interval and set $c(t)=\mathrm{\mid }b(t){\mathrm{\mid }}^{1\mathrm{/}2}\exp (\mathrm{i}\arg (b(t))\mathrm{/}2)$. Such an argument is obtained by continuing the elementary local argument on successive compact subintervals. Then $c(t{)}^2=b(t)$, and $c$ is injective since $b$ is. Extend $c$ at the endpoints by $c(0)=0,c(1)=\mathrm{\infty }$: convergence of its modulus proves continuity even if its angle has no endpoint limit. The two arcs $c$ and $-c$ have disjoint interiors; equality $c(t)=-c(u)$ would give $b(t)=b(u)$, hence $t=u$ and $c(t)=0$, impossible in the interior. Their union $J$ is a Jordan curve on the sphere. The involution $\tau (z)=-z$ fixes $0,\mathrm{\infty }$ and exchanges these two arcs.

Why the involution exchanges the complementary disks. By [F1], the two closed complementary regions of $J$ are disks. Parametrize $J$ by $c(t)$ on one semicircle and $-c(t)$ on the other, with equal $t$ at reflected circle parameters. Its restriction $\tau {\mathrm{\mid }}_J$ is therefore circle reflection and acts as $-1$ on $H_1(J;\mathbb{Z})$ (reverse the oriented circle cycle). In contrast $\tau$ is a sphere rotation homotopic to the identity through $z\mapsto e^{\mathrm{i}\pi u}z$, so its action on $H_2(S^2;\mathbb{Z})$ is $+1$. If it preserved one complementary disk $U$, it would preserve the other disk $V$. Give the sphere its two-disk CW structure using [F1]. The pair sequence gives an isomorphism $H_2(S^2)\to H_2(S^2,\bar{V})$ since $\bar{V}$ is contractible. CW excision identifies this relative group with $H_2(\bar{U},J)$, and its boundary map to $H_1(J)$ is an isomorphism since $\bar{U}$ is a disk. Naturality [F2] would then force the action of $\tau$ on $H_1(J)$ to be $+1$, a contradiction. Hence $\tau$ exchanges the two complementary disks.

An equivariant relative extension. Set $r(t)=t\mathrm{/}(1-t)$, with $r(0)=0,r(1)=\mathrm{\infty }$, and prescribe $f(c(t))=r(t)$ and $f(-c(t))=-r(t)$. This is a homeomorphism $J\to \mathbb{R}\cup \{\mathrm{\infty }\}$ commuting with $\tau$. Choose one source disk $U$ and extend $f$ from its boundary to the closed upper hemisphere by [F1]: take the stereographic pole in the other source disk and a target pole in the lower hemisphere, so both relevant regions are bounded Jordan disks in their plane charts. Call this extension $F_{+}$. On the other source disk define $F_{-}=\tau \circ F_{+}\circ \tau$. Step 2.1 ensures this definition has the right domain and maps it to the lower hemisphere. On $J$ it agrees with $F_{+}$ because $f\tau =\tau f$. Pasting the two maps and their inverses gives a sphere homeomorphism $F$ commuting with $\tau$ and fixing $0,\mathrm{\infty }$.

Descending and restoring the plane point. The quotient of the sphere by $\tau$ is the sphere through the map $p(z)=z^2$, with $p(\mathrm{\infty })=\mathrm{\infty }$. Its fibers are exactly $\{z,-z\}$, and compactness makes $p$ a quotient map. Therefore $F$ and $F^{-1}$ descend to inverse sphere homeomorphisms $g$ with $g(b(t))=r(t{)}^2$. The point $v=g(1)$ lies outside the positive real ray $[0,\mathrm{\infty }]$, because $1\mathrm{\notin }b([0,1])$. Move $v$ to $-1$ by a homeomorphism $L$ fixing that ray pointwise. Here is an explicit existence construction: the ray complement is the slit plane with polar angle in $(0,2\pi )$. Rotate the polar angle of $v$ within this interval to $\pi$, then change its radius along the negative real axis to reach $-1$. Approximate this compact path by a finite polygonal path inside the open slit plane. Choose $\delta >0$ less than one third of the distance from this compact polygonal path to the closed positive ray. Subdivide its finitely many segments so each displacement $w$ has length below $\delta \mathrm{/}4$. At the current path vertex $x_0$, use $\eta (x)=\max (0,1-\parallel x-x_0\parallel \mathrm{/}\delta )$ and the map $x\mapsto x+w\eta (x)$. It moves $x_0$ to the next vertex, is supported in the closed $\delta$-ball about $x_0$, and $\parallel w\parallel \mathrm{Lip}(\eta )<1\mathrm{/}4$. The finitely many supports form a compact subset of the ray complement. They are injective by this bound and surjective by [F3] applied to the contraction equation $x=y-w\eta (x)$, Their inverses are Lipschitz with constant at most $1\mathrm{/}(1-\parallel w\parallel \mathrm{Lip}(\eta ))$, by the same lower distance bound. Thus finite small translations move the point along the path and fix its complement. This constructs $L$ supported away from the ray. Define the Möbius homeomorphism $T(z)=z\mathrm{/}(1+z)$, with $T(-1)=\mathrm{\infty }$ and $T(\mathrm{\infty })=1$. Now $TLgM$ fixes the original sphere-infinity, since its successive images are $1,v,-1,\mathrm{\infty }$, hence restricts to a plane homeomorphism, and takes $a(t)$ to $r(t{)}^2\mathrm{/}(1+r(t{)}^2)$.

The prescribed parameter and neighborhoods. The increasing homeomorphism $\lambda (t)=r(t{)}^2\mathrm{/}(1+r(t{)}^2)$ of $[0,1]$ has fixed endpoints. Extend ${\lambda }^{-1}$ to an increasing homeomorphism $\mathrm{\Lambda }$ of $\mathbb{R}$ equal to the identity outside $[0,1]$. Postcompose step 4.1 with $(x,y)\mapsto (\mathrm{\Lambda }(x),y)$ to obtain $G(a(t))=(t,0)$. Transport straight rectangular and endpoint-sector neighborhoods by $G^{-1}$. This proves the conclusion, with AC used precisely in the relative Jordan–Schönflies extensions of steps 2.1 and 3.1. No collar was inferred merely from connectivity of an arc complement.

A relative graph containing each arc. We first treat one supplied arc $a$. If its endpoints are distinct, extend each interior marked endpoint to a distinct point of the outer circle by an auxiliary arc, disjoint from $a$ and all other marked points. The access at an interior endpoint comes from [F2]'s rectangular neighborhood: leave that endpoint in an unused sector inside a small disk missing all other marked points. The required complement inside the disk is connected. Indeed [F6] connects points in the plane minus the compact arc; replace the finitely many excursions of a polygonal path outside the round disk by outer-boundary paths avoiding its possible boundary endpoint, then push those compact paths slightly inward. Their distance from the original arc is positive. Remove other marked points by small finite detours. Joining the endpoint-access germs by a polygonal path and retaining the portion between its last contact with the initial germ and first subsequent contact with the terminal germ gives a simple auxiliary arc; loop erasure handles the polygonal portion. After the first auxiliary arc, the union is again an embedded interval with only one boundary endpoint, so the same argument supplies the second. The resulting crosscut together with the outer circle is a finite 2-connected plane graph. For coincident endpoints, the arc image is a Jordan loop. If the loop lies in the interior, add two disjoint bridges to distinct outer-boundary points, with distinct loop contacts; if it meets the outer boundary at its common endpoint, add one bridge with different contacts. Access at a loop contact comes from [F3]'s circle coordinates. To justify the second bridge, straighten the inner Jordan loop by [F3]; after the first bridge its complement in the disk is connected by the interval argument above. Polygonal paths there can be moved outside the straightened inner circle by detours along that circle with its bridge contact omitted. Those compact circle arcs have positive distance from the bridge and outer boundary. The same detour argument proves connectivity of the region between a loop touching the outer circle and that circle, omitting their one contact. Thus the bridges exist in the required complementary region. Subdivide the cycles and paths to remove loop edges and parallel edges. In each case deleting any vertex leaves the graph connected, so it is 2-connected.

A marked-set-relative smooth representative. Reproduce the rooted ear decomposition of this graph from the outer circle in a target disk, using polygonal ears. At each stage an ear lies in one source Jordan face; its union with either boundary path gives two Jordan curves, so [F3] splits that face into two disks. Choose a polygonal target crosscut in the corresponding target face, subdivide it at the new vertices, and repeat. This constructs the same finite planar graph with the same face incidence and cyclic orders; prescribe the identity on the outer circle and corresponding parametrizations on all edges. Place target copies of the other marked points in their corresponding faces. The finitely many target marked points can be moved to the prescribed $Q_n$ by [F7]: choose fresh intermediate buffer points, move the points one at a time along polygonal paths avoiding the stationary points, then move the buffers to their assigned targets. Round and smoothly reparametrize each finite path to be constant at the joins. Separation and the boundary margin stay positive, so [F7] applies. The resulting target graph is piecewise smooth and has the prescribed marked vertices and face labels. Extend the graph map over each Jordan face by [F3]; the extensions agree on the graph, so they paste to a boundary-fixed disk homeomorphism. In each target face correct its images of the remaining marked points by the same buffer construction, using small supported Lipschitz translations inside that face as in the plane-arc supplier's point-normalization construction. These corrections fix the entire graph. We obtain a homeomorphism $h$ fixing every marked point and the outer boundary, and carrying $a$ to a piecewise smooth arc $e$. By [F7], the class of $h^{-1}$ has a diffeomorphism representative $g$ and an isotopy from $h^{-1}$ to $g$ fixing the outer boundary and preserving $Q_n$ setwise. Each individual point track is constant because it lies in the discrete set $Q_n$ and starts at that same point. Applying this isotopy to $e$ gives a relative-endpoint isotopy from $a$ to the piecewise smooth arc $g(e)$. No topological isotopy-extension theorem was assumed.

Finite transverse position with fixed endpoints. Perform step 2.1 for both arcs. At an interior marked endpoint separate their finitely many tangent germs by a small rotation $z\mapsto q+e^{\mathrm{i}t\theta \chi (\mathrm{\mid }z-q\mathrm{\mid })}(z-q)$, with cutoff supported in a small disk missing the other marks. At a boundary endpoint use a boundary-flattening coordinate $(x,y)$ with $y\ge 0$ and the small shear $(x,y)\mapsto (x+tby\chi (x,y),y)$; the Lipschitz bound of the supported displacement can be made below one, as in the point-translation construction, and $y=0$ is fixed. Pick the rotation angle or shear coefficient outside the finitely many values aligning tangent germs. Endpoint germs enter the disk transversely in the graph construction, so these shears separate them. Each regular smooth germ is a graph over its tangent coordinate; multiplying its graph height by a cutoff interpolation straightens a smaller terminal germ without changing the endpoint or leaving its narrow interior cone. This is an isotopy of embedded arcs and extends by the corresponding small normal-coordinate displacement. Now the germs are distinct straight rays. On the compact remaining parts, finite normal strips and sufficiently fine polygonal interpolation give embedded piecewise polygonal representatives; the near-parameter injectivity follows from the graph coordinate, and far-parameter pieces have positive separation by compactness. Interpolate the graph heights in those strips, retaining the endpoint rays. A generic finite perturbation of their vertices avoids tangencies and overlapping edges. The two arcs now have finitely many transverse interior intersections and disjoint fixed endpoint germs. All changes fix the endpoints and avoid the other marked points.

Ordinary bigons and the projection argument. The homotopy class has a representative disjoint in its interior from the fixed arc: push that arc to one side in its narrow strip, fixing its endpoints. Thus a homotopy reducing intersection to zero exists. First normalize its terminal strips. By compact-square continuity, a common strip $0<u<\epsilon$ maps into a small disk about its endpoint $q$ containing no other mark. For an interior endpoint, lift its angle continuously on the contractible rectangle $(0,\epsilon ]\times I$, and write $H(u,t)-q=R(u,t)e^{\mathrm{i}\mathrm{\Theta }(u,t)}$ with $R>0$. Reparametrize the initial and final straight germs so their radii are linear in $u$. With a cutoff $\eta (u)$ equal to1 for $u\le \epsilon \mathrm{/}2$ and0 for $u\ge 3\epsilon \mathrm{/}4$, interpolate the radius to $uR(\epsilon ,t)\mathrm{/}\epsilon$ and angle to $\mathrm{\Theta }(\epsilon ,t)$. At $t=0,1$ these models already equal the straight germs, so the modification fixes both endpoint arcs; positivity avoids $q$ and both original and model radii tend uniformly to0. In a boundary half-disk use the same interpolation with angle in $(0,\pi )$, so the interior stays inside. Treat each terminal strip separately even at coincident ends. Approximate the positive radius and real lifted angle of the resulting straight-germ family by finitely piecewise linear functions of $t$, keeping their end values and using a generic angle perturbation. The bounded angle range then meets the fixed endpoint ray in only finitely many isolated times. On the remaining compact parameter square all images have positive distance from the forbidden marked points and boundary. Triangulating that square and making a sufficiently small generic perturbation of its finitely many vertex images, with the prescribed terminal-germ boundary data retained, makes the homotopy piecewise transverse. Thus its inverse-image intersections have a finite polygonal description, including the endpoint sectors. A returning component of the inverse image gives a nullhomotopic loop with one side in each arc. Use the cover formed by gluing copies of the compact holed cut disk along paired shores, indexed by reduced meridian words, and append the lifted puncture collars; its adjacency tree gives a simply connected cover. Compact pieces meet only finitely many cut disks and bounded collar rectangles. Enlarge this finite subtree to include every corner star met by the piece. Successive gluing along boundary intervals produces a disk neighborhood, or a half-disk neighborhood at an actual outer-boundary point; thus the compact lifted Jordan curves and their bounded regions lie in a disk exhaustion where the plane Jordan theorem applies. The families of lifted arcs are locally finite: properness bounds their parameter ranges over a compact set away from deleted ends, and finitely many covering charts then admit only finitely many lifts meeting a smaller compact neighborhood. Lift the nullhomotopic loop. Inside a compact disk neighborhood choose an innermost bigon across ALL lifts of both arcs. If another lift enters such a disk, an outermost component of its intersection with the disk, together with the appropriate existing side, cuts off a smaller bigon; repeat this reduction. Transverse position and local finiteness give finitely many intersections in that compact disk, so the reduction terminates with no other lifted arc entering its interior. Its boundary projects injectively: each side is embedded, the two corner signs are opposite, and an identification between different sides would give another transverse lifted crossing on a side. One branch of that transverse curve would enter the disk, contradicting the innermost choice. For a nonidentity deck map $\varphi$, its boundary and the original boundary are disjoint: if $x=\varphi (y)$ were on both, injectivity of the boundary projection would give $x=y$, contradicting the fixed-point-free deck action at an actual covering point. This excludes shared sides and tangencies as well as crossings. If disk interiors overlapped, disjoint Jordan boundaries would imply $\varphi (D)\subset D$ or ${\varphi }^{-1}(D)\subset D$. [F4] would give a deck fixed point, again impossible. Hence the entire disk projects injectively and is a compact puncture-free ordinary bigon in the actual punctured surface.

Endpoint-sector bigons use a different actual surface. A returning inverse-image component ending at a fixed endpoint $q$ bounds a parameter sector adjacent to that endpoint edge, with the opposite endpoint edge omitted. If $q$ is a puncture, fill just $q$ and keep every other puncture deleted; if $q$ is a boundary point, no puncture is filled. The homotopy on this sector maps into that surface because all its nonendpoint tracks avoid every marked point, and its endpoint edge is constantly $q$. Lift this compact sector to the simply connected cover of that surface. The lift of $q$ is an actual surface point, not an ideal end. Choose an innermost disk between the two lifted arc sides; either it is an ordinary bigon, or it has that lift of $q$ as one corner. At this corner the chosen endpoint germs are distinct, and their projections hit $q$ nowhere else. The same finite-subtree disk exhaustion and outermost-subdisk reduction across all lifts apply in this filled surface; properness is retained at every other deleted end and the filled corner has an ordinary covering chart. The boundary-injectivity and disjoint-deck-boundary argument of step 4.1 then applies, and [F4] now applies on this actual filled surface. The projected disk contains no other puncture, and contains $q$ only at its boundary corner: injectivity of the whole disk excludes a second interior preimage of $q$. Thus this is a legitimate endpoint-sector disk in the filled model. No fixed-point-free assertion was made about a completed puncture tip in the original punctured cover. A half-bigon using an outer-boundary interval between two different fixed endpoints is not removed; only a sector at the single fixed endpoint is used here.

Finite ambient reductions. For an ordinary bigon, prescribe the arc push across a slightly enlarged compact disk neighborhood, fixing its outer boundary. The moving subarc is a crosscut there, so its two complementary Jordan pieces extend the prescription by [F3]. The Alexander isotopy [F5] gives the ambient disk move. At a marked endpoint corner choose a slightly enlarged disk neighborhood containing that endpoint in its interior and no other marked point. Extend the moving arc from that endpoint by an auxiliary arc to the neighborhood boundary, on a free side, as in step 1.1; this makes a crosscut. Do the same for its prescribed image after the sector push. The relative graph/face construction of step 2.1, with the endpoint vertex mapped to itself and the outer boundary mapped identically, gives the required neighborhood homeomorphism. Choose its disk coordinate centered at this fixed marked point, using [F3]’s pointed extension, and apply the Alexander formula; that center stays fixed throughout. At a boundary endpoint use a half-disk neighborhood and keep its outer-boundary edge fixed. These moves fix all other marked points and the outer boundary. Each ordinary move removes two transverse corners and each endpoint-sector move removes one interior corner. Consequently the finite number of interior intersections decreases until the arc interiors are disjoint.

The last disk and parametrization. For distinct endpoints, the disjoint arcs form a Jordan curve in the filled disk. Their relative homotopy gives winding number zero about every marked point other than the endpoints, so its bounded region contains none. By [F3] take its endpoints to $(-1,0),(1,0)$ and its two sides to the upper and lower semicircles. The arcs $(x,(1-2t)\sqrt{1-x^2})$ give an isotopy from one side to the other with fixed endpoints. For coincident ends, the two Jordan loops meet only at their common endpoint. Their winding numbers about the other marks agree. If their bounded disks are disjoint, both disks contain no other mark, so shrink the loops toward the common endpoint in disk coordinates by positive homotheties centered at that boundary point. The images remain simple and never collapse to a point. Match the two resulting small images inside a common endpoint neighborhood by the relative graph/face homeomorphism construction of step 2.1, and then [F5]’s Alexander isotopy: center the coordinate at an interior marked endpoint, or keep the boundary edge fixed at an outer endpoint. If they are nested, the region between them contains no mark; split the common contact into two boundary copies in the polygonal representatives. Its completed cut is a polygonal disk, and the same disk-side isotopy descends after identifying those fixed copies. These are isotopies of unoriented images. For distinct ordered endpoints, the final parametrization differs from the target by an increasing endpoint-fixing homeomorphism $f$ of $[0,1]$; interpolation $(1-t)f+t\mathrm{id}$ corrects it through embedded parametrized arcs. Compose the finite moves with steps 2.1 and 3.1. Compact-domain continuity includes each filled puncture endpoint. AC is used precisely through the plane-arc neighborhoods, relative Jordan extensions, finite point-motion extensions and smooth representative clause. This proves the asserted general topological conclusion.