Fork detection transports to arbitrary boundary crosscuts

First choose the closing neighborhoods small enough to miss $M$; this is possible because its compact image misses $P$. Choose $r_0<1$ enclosing $P$, the filled tine, the chosen closed closing neighborhoods, and the projection of the compact support of $c_F$; all are compact in the interior of $D$. Write the two source boundary angles in positive cyclic order and likewise the two target angles of $M$; choose the ordering of the target endpoints accordingly. There is an increasing piecewise linear lift $f:\mathbb{R}\to \mathbb{R}$ with $f(\theta +2\pi )=f(\theta )+2\pi$ taking the two source angles to those targets. Interpolate $f_s=(1-s)\mathrm{id}+sf$. For a radial cutoff $\chi$ equal to zero on $r\le r_0$ and one at $r=1$, set $h_s(r{e}^{\mathrm{i}\theta })=r\exp (\mathrm{i}[(1-\chi (r))\theta +\chi (r)f_s(\theta )])$. The angular maps are strictly increasing degree-one homeomorphisms; their inverses vary continuously by compactness. Thus $h_s$ is a filled-disk isotopy, identity on $r\le r_0$, carrying the source endpoint pair to the target pair. Its configuration-space isotopy lifts from the identity, commutes with every deck transformation by uniqueness of lifts, and fixes $c_F$ pointwise since every track on its support is constant.

The crosscut $h_1^{-1}M$ is an original noodle with endpoints $d_1,d_2$. The closing neighborhoods are fixed by $h_s$, hence also miss $h_1^{-1}M$. Its triangle class exists by the same parameter-gap truncation as any noodle: truncate $0\le u<v\le 1$ by $v-u\ge \delta$, with the new edge in a collision neighborhood by uniform continuity. Transport this class by the lifted isotopy to obtain $y_M$. An arbitrary choice of its lift differs by a deck monomial unit. The finite intersection naturality in [F2] and step 1.1 give $⟨c_F,y_M⟩=u⟨c_F,y_{h_1^{-1}M}⟩=u{\mathrm{\Delta }}_F⟨h_1^{-1}M,F⟩,$ for a Laurent monomial unit $u$. Hence vanishing is equivalent to the last diagram polynomial vanishing, since $u{\mathrm{\Delta }}_F\mathrm{\ne }0$. By [F3] this means that the tine can be isotoped off $h_1^{-1}M$. Conjugate that isotopy by $h_1$: it still fixes $P$ and fixes the outer boundary pointwise at every time, while $h_1$ fixes the tine. This is exactly disjoinability from $M$. The inverse conjugation proves the reverse implication.