Extremal fork-noodle terms have one sign and cannot cancel

Let $m_{i,j}$ be maximal in the lexicographic order. Then $a_{i,j}$ is maximal among the integers $a_{k,l}$, hence $a_{i,i}\le a_{i,j}$ and $a_{j,j}\le a_{i,j}$. By [F1], $a_{i,j}=\frac{1}{2}(a_{i,i}+a_{j,j})\le \frac{1}{2}(a_{i,j}+a_{i,j})=a_{i,j},$ and the two inequalities must be equalities; this is possible only if $a_{i,i}=a_{j,j}=a_{i,j}$.

To prove $b_{i,i}=b_{i,j}$, suppose $b_{i,i}<b_{i,j}$, the only possible strict inequality by maximality and step 1.1. Let $\alpha$ run from $z_i^{\mathrm{\prime }}$ to $z_j^{\mathrm{\prime }}$ along $T(F^{\mathrm{\prime }})$, and let $\beta$ return along $N$. If $\beta$ misses $z_i$, lifting the loop with one point fixed at $z_i$ gives $b_{i,j}-b_{i,i}=2w$, where $w=\mathrm{wind}(\alpha \beta ,z_i)$. If $\beta$ passes through $z_i$, push it locally so $z_i$ lies to its left; this contributes one positive half twist, giving $1+b_{i,j}-b_{i,i}=2w$. In either case $w>0$.

In the infinite cyclic cover $\pi :{\tilde{D}}_1\to D\setminus \{z_i\}$ lift $\alpha$, then $\beta$ from its endpoint. Choose a clockwise return loop $\gamma$ at $z_i^{\mathrm{\prime }}$, winding $w$ times about $z_i$, nullhomotopic in $D\setminus P$, and meeting $\alpha \cup \beta$ only at its endpoints. It can be drawn in a thin puncture-free neighborhood of an access path to $z_i$; its lifted spiral $\tilde{\gamma }$ is embedded and joins the endpoint of $\tilde{\beta }$ back to the start of $\tilde{\alpha }$. Let ${\tilde{z}}_k^{\mathrm{\prime }}$ be the first intersection of $\tilde{\alpha }$ with $\tilde{\beta }$, and take the initial ${\tilde{\alpha }}^{\mathrm{\prime }}$ and final ${\tilde{\beta }}^{\mathrm{\prime }}$ at this point. Then ${\tilde{\alpha }}^{\mathrm{\prime }}{\tilde{\beta }}^{\mathrm{\prime }}\tilde{\gamma }$ is a Jordan curve. Its bounded disk $\tilde{B}$ lies to its left: the clockwise spiral leaves a noncompact region on the right. The labelled-loop winding computation gives $a_{i,k}-a_{i,i}=\mathrm{\#}(\tilde{B}\cap {\pi }^{-1}(P))$. Maximality forces this nonnegative integer to be zero. Hence the disk misses the lifted punctures and its projected boundary $\delta ={\alpha }^{\mathrm{\prime }}{\beta }^{\mathrm{\prime }}\gamma$ is nullhomotopic in $D\setminus P$. This is the disk calculation in Bigelow 2001's extremal-claim proof, printed p. 481 (the precise claim number is recorded in the source locator).

The projection of ${\alpha }^{\mathrm{\prime }}{\beta }^{\mathrm{\prime }}$ need not be embedded. If it is a Jordan curve, cancellation of the nullhomotopic $\gamma$ shows it bounds a puncture-free digon. Otherwise the portions of ${\pi }^{-1}({\alpha }^{\mathrm{\prime }})$ entering $\tilde{B}$ are disjoint arcs with both endpoints on ${\tilde{\beta }}^{\mathrm{\prime }}$. Choose an innermost such arc ${\tilde{\alpha }}^{\mathrm{\prime }\mathrm{\prime }}$ and its corresponding side ${\tilde{\beta }}^{\mathrm{\prime }\mathrm{\prime }}$. The latter has no further intersections with ${\pi }^{-1}({\alpha }^{\mathrm{\prime }})$, so its projection meets ${\alpha }^{\mathrm{\prime }}$ only at the two endpoints. Thus ${\alpha }^{\mathrm{\prime }\mathrm{\prime }}{\beta }^{\mathrm{\prime }\mathrm{\prime }}$ is an embedded loop. Its lifted disk is contained in the puncture-free $\tilde{B}$, so the projected loop is nullhomotopic and bounds a puncture-free digon by Jordan separation. In both cases the parallel tine and $N$ cobound a digon, which transfers across the narrow parallel strip to $T(F)$ and $N$ and contradicts minimality. Therefore $b_{i,i}=b_{i,j}$.

For the other diagonal keep the second point fixed at $z_j^{\mathrm{\prime }}$ and move the first point: let $\rho$ run from $z_j$ to $z_i$ along $T(F)$ and let $\eta$ return from $z_i$ to $z_j$ along $N$. Maximality and step 1.1 give $b_{j,j}\le b_{i,j}$. Suppose the inequality is strict. If $\eta$ misses $z_j^{\mathrm{\prime }}$, the lift of $\{\rho \eta ,z_j^{\mathrm{\prime }}\}$ compares the lifts at $\{z_j,z_j^{\mathrm{\prime }}\}$ and $\{z_i,z_j^{\mathrm{\prime }}\}$, giving $b_{i,j}-b_{j,j}=2w$ with $w=\mathrm{wind}(\rho \eta ,z_j^{\mathrm{\prime }})$. If $\eta$ passes through $z_j^{\mathrm{\prime }}$, detour with $z_j^{\mathrm{\prime }}$ on its left. This inserts a positive half turn of the moving point about the fixed point relative to the noodle return, so the same lift comparison gives $1+b_{i,j}-b_{j,j}=2w$. Hence $w>0$ in either case. Repeat steps 3.1–4.1 in the cyclic cover of $D\setminus \{z_j^{\mathrm{\prime }}\}$, based at $z_j$, using the clockwise nullhomotopic return near $z_j^{\mathrm{\prime }}$. At the first lifted intersection of $\rho$ and $\eta$, now a lift of some unprimed $z_k$, the oriented disk has puncture count $a_{k,j}-a_{j,j}$. This follows from the same labelled-loop computation, with the first coordinate moving and the second fixed; no coordinate interchange changes the puncture-winding sum. The count is nonnegative and at most zero, since $a_{j,j}=a_{i,j}$ is globally maximal. The disk is therefore puncture-free, and the innermost-arc argument of step 4.1 gives a digon between $T(F)$ and $N$, contradicting minimality. Thus $b_{j,j}=b_{i,j}$, and step 4.1 gives both diagonal monomial equalities. This comparison uses maximality of $m_{i,j}$ and the diagonal $q$-exponents, without assuming maximality of $m_{j,i}$.

By [F2] and step 5.1, if $m_{i,j}$ is maximal then ${ϵ}_{i,j}=-(-1{)}^{b_{i,i}+b_{j,j}+b_{i,j}}=-(-1{)}^{b_{i,j}}$; the same holds for every pair $(i^{\mathrm{\prime }},j^{\mathrm{\prime }})$ whose monomial equals the maximal monomial $m_{i,j}$. Hence all terms of $⟨N,F⟩$ carrying the maximal monomial have the same sign, and since monomials are compared in the lexicographic order, the coefficient of the maximal monomial in $⟨N,F⟩$ is, up to sign, the number of such terms, which is at least one. In particular $⟨N,F⟩\mathrm{\ne }0$.