Finite nilpotent peripheral presentation from a link diagram

Cut the traversal of component $i$ just before its initial arc. For every other arc $a$ of that component let $v_a$ be the signed product of over-arc symbols encountered before $a$; for the initial arc set $v_a=1$. The nonclosing relations in [F1] successively say $a=v_a^{-1}{x}_i{v}_a$. The final relation on that component is $x_i=W_i^{-1}{x}_i{W}_i$, with $W_i$ the full traversal product. This rewriting follows by induction along the traversal and does not treat an over-arc as an already eliminated free generator.

Define words in $F$ simultaneously by $P_a^{(1)}=x_i$ and $P_a^{(s+1)}=v_a(P^{(s)}{)}^{-1}{x}_i{v}_a(P^{(s)})$, retaining $P_a^{(s)}=x_i$ at each initial arc. Substitution in a finite word preserves congruence modulo a normal subgroup. If the substituends change modulo ${\mathrm{\Gamma }}_s$, the conjugate of a weight-one generator changes only modulo ${\mathrm{\Gamma }}_{s+1}$, because $[F,{\mathrm{\Gamma }}_s]\subseteq {\mathrm{\Gamma }}_{s+1}$ by [F3]. Consequently $P_a^{(s+1)}\equiv P_a^{(s)}(\mathrm{m}\mathrm{o}\mathrm{d}{\mathrm{\Gamma }}_{s+1}F)$: the case $s=1$ is equality in the abelianization, and this observation proves the induction. In $G$, the true arc meridian equals $P_a^{(s)}$ modulo ${\mathrm{\Gamma }}_{s+1}G$, by the same induction starting from conjugate meridians in the abelianization.

Put $P_a=P_a^{(q-1)}$, $W_i=W_i(P^{(q-1)})$ and $l_{i,q}=x_i^{-{\mathrm{wr}}_i}{W}_i$. Step 2.1 and [F2] show that these are longitude lifts in $G\mathrm{/}{\mathrm{\Gamma }}_{q}G$. The relation $[x_i,l_{i,q}]=1$ is equivalent to $[x_i,W_i]=1$. Let $H$ be the quotient of $F$ by these relations and ${\mathrm{\Gamma }}_{q}F$. Its root-meridian map to $G\mathrm{/}{\mathrm{\Gamma }}_{q}G$ is defined because the true meridian and its true peripheral parallel commute on the boundary torus. Every arc meridian is the image of $P_a$, so this map is onto.

Conversely assign every Wirtinger arc the image of $P_a$ in $H$. If an undercrossing does not close its component, its conjugation equation follows from the consecutive partial traversal products evaluated at $P^{(q-1)}$: those products give $P^{(q)}$, and $P^{(q)}\equiv P^{(q-1)}$ modulo ${\mathrm{\Gamma }}_q$ by step 2.1. At the closing crossing the same argument leaves exactly $[x_i,W_i]=1$, imposed in $H$. Thus all crossing relations hold. By [F1] this gives $G\to H$, and it factors through $G\mathrm{/}{\mathrm{\Gamma }}_{q}G$ because $H$ is a quotient of $F\mathrm{/}{\mathrm{\Gamma }}_{q}F$. The two maps are inverse: they fix every root generator of $H$, and step 2.1 says their composite fixes every arc generator of $G\mathrm{/}{\mathrm{\Gamma }}_{q}G$. This proves the displayed presentation and the kernel assertion.

For a fixed finite maximum degree, the iterates in step 2.1 stabilize in every lower quotient; applying the same observation to the finite word $W_i$ proves coherence of $l_{i,q}$. Finally, forming a further quotient adds the prescribed relations to the kernel. Surjectivity in step 3.1 means every conjugating element of $G\mathrm{/}{\mathrm{\Gamma }}_{q}G$ has a lift in $F$, so the added relations are exactly $[u^{-1}{x}_{i}u,v^{-1}{x}_{i}v]$ for words $u,v\in F$. Their entire normal closure is meant; the argument does not assert this infinite family is finite.