A dualizable link can slide off a meridian disk

Let $q$ be the unique point of $B\cap \mathrm{int}{D}_R$, and put the current attaching circle $G$ transverse to $D_R$. Its intersection set is finite: it is a compact zero-dimensional transverse submanifold, disjoint from $\mathrm{\partial }{D}_R=R$. If it is empty there is nothing to do. Otherwise choose $p\in G\cap \mathrm{int}{D}_R$. A sufficiently close framed parallel $B^{\mathrm{\prime }}$ of $B$ is disjoint from $B\cup G\cup R$ and meets $D_R$ once, at a point $q^{\mathrm{\prime }}$ close to $q$. Indeed near $q$ the transverse arc $B$ has product coordinates in which $D_R$ is $z=0$ and $B$ is a vertical interval; choose the parallel's displacement there tangent to the disk. Away from this coordinate ball the compact sets $B$ and $D_R$ are disjoint, so a sufficiently small parallel creates no other intersection. The displacement can be interpolated in the framed normal disk bundle without changing the prescribed parallel's framing.

Orient the signed parallel $B^{\mathrm{\prime }}$ so that its intersection sign with $D_R$ at $q^{\mathrm{\prime }}$ is opposite to the sign of the intersection of $G$ with $D_R$ at $p$. Reversing the orientation of the parallel changes this sign without changing its underlying framed curve, and [F2] permits either signed parallel. Join $p$ to $q^{\mathrm{\prime }}$ by a smooth embedded arc $a$ in $\mathrm{int}{D}_R$ whose interior misses $G\cap D_R$, $q$, and $q^{\mathrm{\prime }}$. Such an arc is obtained by a polygonal path in the disk minus the finitely many forbidden points, removing loops in its finite subdivision and smoothing its corners in small disjoint disks. A thin strip about $a$ misses the remaining intersections and $B$. Choose the strip's normal product neighborhood so that the short branches of $G$ at $p$ and the chosen oriented $B^{\mathrm{\prime }}$ at $q^{\mathrm{\prime }}$ are vertical, while its interior misses all three retained components. In its positive side, join short intervals of these vertical branches by an embedded band running just above $a$. Give the band its product framing and choose the band-end coordinates by the full seam convention in [F2], so its orientation extends the orientation of $G$ over the selected signed parallel. Band summing $G$ with this $B^{\mathrm{\prime }}$ is therefore an allowed slide over $B$.

One of the new band-sum arcs runs along one edge of the band between $p$ and $q^{\mathrm{\prime }}$. Together with a parallel of $a$ in $D_R$, it bounds an embedded product disk in the positive half-neighborhood. Of the two endpoints, one has positive intersection sign and the other negative sign. Orienting this arc from the positive-sign crossing to the negative-sign crossing, it enters the positive side of $D_R$ at the first endpoint and exits at the second; thus the entire arc lies on the positive side. Its interior misses the slid circle and the retained components: the other edge of the band is chosen on the opposite transverse side of this disk, and the rest of $B^{\mathrm{\prime }}$ and $G$ misses the neighborhood. Push this arc across the disk and slightly into the negative side of $D_R$, rounding its ends. In strip coordinates this replaces the arch above $z=0$ joining the two opposite-sign intersections by an arc below $z=0$; extend the motion to its thin tubular neighborhood. This is a link isotopy supported away from $B\cup R$. It removes precisely $p,q^{\mathrm{\prime }}$ and creates none, while all old intersections away from $p$ persist. Thus a slide followed by this isotopy changes $\mathrm{\mid }G\cap D_R\mathrm{\mid }$ from $N$ to $N+1-2=N-1$. Transport the attaching framing by this isotopy.

Repeat steps 1.1–3.1 for the current attaching circle whenever an intersection remains. Each new parallel and strip can be chosen sufficiently thin for that finite stage, and the fixed $B,R,D_R$ are retained. The nonnegative integer intersection count drops by one at every stage, so after exactly the original number of intersections the current attaching circle is disjoint from $D_R$. The resulting finite sequence consists solely of the permitted framed slides and link isotopies and proves the full statement.