Ribbon immersions and disk movies without maxima are equivalent

Start with a no-maximum disk movie. On its stipulated boundary collar, $\rho (c_D(y,s))=(1-s{)}^2$, so there are no radial critical points for $0<s<\epsilon$. The critical set is therefore a compact discrete subset of the disk interior, hence finite by [F2]. On each regular radial interval, radial projection to the unit three-sphere gives an isotopy of the level link, extended by [F3]. A local minimum has the usual disk model and creates a split round unknot in a small ball; a local saddle has the rectangle model and changes the link by its framed band move. These assertions follow in coordinates from the Morse equations $x^2+y^2$ and $x^2-y^2$: the first level is a small circle born in a ball, and in the second the two attaching sides of the rectangle are replaced by the other two sides. Critical points sharing a value have disjoint small ambient neighborhoods, so the corresponding local events may be processed successively. Thus there is a finite sequence of births, band attachments, and boundary-link isotopies, whose abstract disk and band handles give exactly the original disk. We will realize this same sequence by immersed surfaces in $S^3$.

For the reverse implication, start with a ribbon immersion $f:D^2\to S^3$. Remove narrow disjoint strip neighborhoods of all its proper preimage arcs. Choose them disjoint from every interior preimage arc; this is possible because all preimage arcs are disjoint compact sets. Cutting along a proper arc of a disk separates it into two disks, so cutting along all these strips leaves finitely many disk components. Their images are embedded and pairwise disjoint, since every original multiple point had one preimage in a removed strip. The removed strips themselves are embedded bands with pairwise disjoint images, and are disjoint from the disk boundaries except along their attaching edges: their remaining intersections with the disk components are exactly the original ribbon arcs, away from these boundaries. Rounding the cut corners gives a finite presentation by disjoint embedded disks and disjoint embedded bands. The boundaries of the disks form an unlink. Indeed each disk has a disjoint small product neighborhood, and shrinking concentric circles in its disk coordinates gives an isotopy to a small round unknot in that neighborhood; [F3] extends these disjoint isotopies. The band surgeries on this unlink recover the original boundary $K$.

The inductive invariant is a compact oriented ribbon-immersed surface with the current oriented level link as embedded boundary. Begin with the empty surface. An isotopy event acts on the whole immersion by [F3]. For a birth, its ball is disjoint from the old boundary link. The old immersed surface has empty interior in $S^3$: in each immersion chart its image is a smooth two-dimensional sheet, and finitely many such charts cover its compact domain. Its image is also closed. Hence there is a small ball inside the birth ball disjoint from the entire old image. Add an embedded disk there. Its boundary is a split unknot isotopic within the birth ball to the required newborn component, and apply [F3] to match the new boundary link. This introduces no multiple point.

For a band event, first shrink its attaching intervals in their boundary collars and slide them along the boundary, if needed, to avoid the finitely many endpoints of old ribbon arcs. This changes the banded boundary by an isotopy supported in its attaching collars. The immersion is embedded on these collars. Keep the prescribed normal framing of the band core. Perturb the core relative to these collars, as in [F4], to avoid the old double arcs and meet each regular old sheet transversely. Away from its endpoints it is compact and disjoint from the old boundary, so all resulting intersections with the old surface lie in its regular interior sheets; transversality makes their inverse image a finite discrete set. The perturbation can be arbitrarily small, so the core remains embedded, the band framing persists, and the boundary band move is unchanged up to isotopy. Shrink the framed band about that core, including its attaching intervals. This shrinking is an isotopy of the banded boundary: in the rectangle coordinates contract the width, extend the same contraction along the two attaching link intervals, and round corners in their disjoint collars. At an intersection with an old sheet the transverse core parameter is a local defining coordinate for that sheet, so the implicit function theorem expresses its intersection with a sufficiently narrow band as a single arc running across the full band width. Compactness away from these finitely many intersections allows one width small enough to exclude every other intersection. Each new double arc consequently has one proper preimage in the new band and one interior preimage in the old surface. The new band is embedded, it avoids all old double arcs, and the neighborhoods of its new intersections are disjoint, so no triple point is introduced. Glue and smooth the band at its attaching collars. The old proper preimage arcs remain proper because those collars avoided all their endpoints. The new proper arcs have endpoints on the two free band sides, hence on the new surface boundary. This proves the invariant at the saddle.

Finite induction over the movie now gives a ribbon-immersed oriented surface with boundary $K$. It is the original abstract surface: births added its zero-handles, saddles added its one-handles in the same order and with the same oriented attaching data, and boundary isotopies changed neither handle topology nor orientation. Thus its domain is a disk, not merely a surface of Euler characteristic one. Step 3.1 also shows that its only multiple points are finitely many disjoint transverse ribbon double arcs and its boundary is embedded. This proves the implication from the disk movie to a ribbon immersion.

Realize that presentation in a radial shell $S^3\times [a,1]\subset B^4$, $a>0$. At early distinct radii create a minimum disk for each unlink component, using its embedded disk to give the trace down to its cap. Between these caps and the bands keep product annuli on the unlink. At later radii attach the disjoint bands by the standard saddle traces of [F2]. Each saddle trace meets the existing surface only at its attaching collars: below that radius the surface contains only the product boundary annuli at those levels, and the band interior is disjoint from the current boundary link. The interiors of the zero-handle disks lie at earlier radii, so band intersections with their projections in $S^3$ cause no intersection in the four-ball. Distinct bands have disjoint images and disjoint attaching intervals, so they can also be placed at successive radii. Finish with a product collar up to radius one. Smooth the finitely many seams in their product charts. This is a smooth embedded disk because its abstract disks and bands reconstruct the original domain disk of step 1.2. Its radius has exactly the chosen minimum and saddle models; squared radius is their composition with an increasing smooth function having positive derivative on $[a,1]$, so it has the same nondegenerate critical points and indices. Its boundary is $K$, and the final product collar makes it proper and neat. This proves the converse and the equivalence.