A mapping class fixing all standard adjacent edges is a boundary twist power

For $n=1$, the disk mapping-class theorem gives the trivial group $B_1$ and the assertion holds with $k=0$. Assume $n\ge 2$; purity is a hypothesis, so every edge has its two marked endpoints fixed individually. Apply the compatible graph induction of [F2] to the two chains $\sigma (E_i)$ and $E_i$. Here all distinct edge interiors are disjoint, distinct edges have distinct endpoint pairs, and no triple intersects; the only shared vertices are the prescribed marked points. At the induction stage the preceding chain is already arranged. The relative graph/face construction of [F1] puts the next two arcs in finite transverse position, separating their endpoint germs in sectors consistent with the cyclic order of the chain, and retaining the preceding chain as a graph. The relative homotopy between the two next arcs yields ordinary bigons or marked-endpoint sectors by [F1]'s actual-cover argument. A prior graph edge entering an ordinary bigon must join its two opposite sides: a return to one side would itself bound a removable bigon against that side, contrary to the already minimal position of each of the two original disjoint chains. At a shared marked corner retain the prior germ and use the sector not containing that germ. The same return reduction supplies a smaller ordinary bigon if necessary. Thus the portions of the preceding graph in a reduction disk are disjoint through-arcs.

Subdivide a reduction disk along those through-arcs. Prescribe the push of the new arc across it, carrying each through-arc to itself as a set and fixing all graph vertices. Extend the prescription over the resulting Jordan disk faces using [F1]; the supported Alexander moves realize it while preserving the preceding graph. Each push removes an intersection; the final disk region between the two disjoint isotopic arcs is treated by exactly the same subdivision. It contains no other marked point, since the relative homotopy has zero winding about each other mark. The finite induction therefore gives a representative $\tau$ preserving every edge as a set and fixing every marked point. Each restriction to the full straight spine $X=[p_1,p_n]$ is an increasing interval homeomorphism $f$ fixing its marked vertices. In a narrow rectangle about $X$, extend $f^{-1}$ increasingly by the identity past the two endpoints and set $k_s(x,y)=((1-s\chi (y))x+s\chi (y)f^{-1}(x),y),$ where $\chi =1$ on $y=0$ and vanishes near the top and bottom of the rectangle. This is an isotopy of rectangle homeomorphisms, fixed on its boundary and at each marked vertex, extended by identity outside. Its final composition with $\tau$ fixes $X$ pointwise. Only the preservation of the prior graph as a set was required in the induction; the final interval correction fixes all of it pointwise.

For completeness, the compact annulus classification in [F3] is relative to BOTH boundary circles. Lift a boundary-fixed annulus homeomorphism to the strip $\mathbb{R}\times [0,1]$ with its lower boundary fixed. Its upper boundary is $x\mapsto x+k$ for a unique integer $k$ (using angular period one). Compose with the inverse $k$-twist. The image of a supplied radial arc then has the same lifted endpoints as that radial arc, so is relatively homotopic to it. The disk-cover bigon moves of [F1] isotope it to that arc, fixing both boundary circles; the increasing parametrization correction fixes it pointwise. Cutting along it gives a compact rectangle, whose boundary is now fixed. The Alexander contraction on that disk shows that the residual annulus map is isotopic to identity relative to both circles. Reglue the fixed shores; compact quotient continuity gives the annulus isotopy. This proves the stated integer classification.

Complete the slit complement accurately. By an orientation-preserving affine change write the straight spine as $[-a,a]$ about its center (the outer boundary becomes a circle with possibly different center). The Joukowski map $J(w)=\frac{a}{2}(w+w^{-1})$ maps $\mathrm{\mid }w\mathrm{\mid }>1$ homeomorphically to the plane minus this segment: its quadratic inverse has exactly one root of modulus greater than one, and $J^{\mathrm{\prime }}(w)\mathrm{\ne }0$ there. Write the outer circle as $\mathrm{\mid }z-c\mathrm{\mid }=b$, with real $c$. Along $w=r{e}^{\mathrm{i}\theta }$ set $s=r+r^{-1}$; then $\mathrm{\mid }J(w)-c{\mathrm{\mid }}^2=\frac{a^2}{4}(s^2-4{\sin }^2\theta )-acs\cos \theta +c^2$. This upward quadratic is below $b^2$ at $s=2$, because the entire spine is strictly inside the disk, and tends to infinity. It therefore has exactly one root $s>2$, depending continuously on $\theta$, and gives a continuous radial graph $r=R(\theta )>1$. The region $1\le r\le R(\theta )$ is explicitly a compact annulus $\widehat{A}$, parametrized by $r=1+u(R(\theta )-1)$. On its inner boundary $J(e^{\mathrm{i}\theta })=a\cos \theta$, identifying the two shores and giving one preimage at each terminal tip. Its quotient is the FILLED disk $D$; removing the marked images recovers the punctured disk. The homeomorphism $\tau$ fixing $X$ pointwise extends to a homeomorphism of $\widehat{A}$ fixing both boundary components pointwise. At an interior spine point it preserves the two local sides, since it is orientation preserving and fixes the oriented interval; a side swap would reverse the cyclic orientation of a small transverse disk. Uniform continuity on the filled disk then sends approaching points on either shore to that same shore point. At either terminal tip there is only one inner-boundary preimage, so the same compactness argument gives continuity there. Apply this also to ${\tau }^{-1}$ to obtain a homeomorphism, not merely a continuous extension.

By step 1.3 isotope that extension, on the COMPACT annulus and relative to both boundary circles, to the $k$-twist. Every map in this isotopy fixes every inner-boundary point, hence respects the shore identifications. The quotient map $\widehat{A}\times I\to D\times I$ is a closed quotient map (compact source and Hausdorff target), so the descended family is jointly continuous, including the slit and both tips. Its inverse family descends likewise. Thus it is a filled-disk isotopy fixing $\mathrm{\partial }D\cup X$, and hence all marked points. Its twist curve is parallel to the outer boundary. By [F4] that twist is ${\mathrm{\Delta }}^2$, proving $[\sigma ]=({\mathrm{\Delta }}^2{)}^k$. No arbitrary punctured-space homotopy was extended to a tip; the isotopy was compact and boundary fixed before descent.