Minimum boundary exchange leaves have a single-leaf zero-edge corridor

Put $m=w(E)>0$, using [F1]. Cut $E^{\mathrm{\prime }}$ along all exchange traces in its interior. Every other trace lies wholly inside or wholly outside $E^{\mathrm{\prime }}$: its interior cannot cross the sole trace side in the boundary of $E^{\mathrm{\prime }}$, and all trace endpoints are on the original boundary. Thus this cutting is a proper arc cutting of the disk $E^{\mathrm{\prime }}$. Its pieces are precisely the original cut patches, with the harmless half-switch-band collars of [F3]. They are all disks, and their adjacency graph is a finite tree by [F4]. Exactly one piece can meet the outer trace side of $E^{\mathrm{\prime }}$; mark that vertex of the tree as its root.

Every non-root leaf of this tree is an original leaf disk patch of $P$. Its tree degree equals its full number of trace sides, since the added collars in [F3] preserve that number. If that degree is one, it is a boundary disk patch of one of the original summands. It cannot come from $H$: such a patch would be cut off from $H$ by its sole trace arc and make that arc separating, contrary to [F1]. It is therefore one of the original $P$ leaves and has weight at least $m$. If the tree has only its root vertex, the same argument with its sole exterior trace side shows that $E^{\mathrm{\prime }}$ itself is an original $P$ leaf; this is the asserted case $E^{\mathrm{\prime }}=L$.

Suppose now that there is more than one vertex. The tree has a non-root leaf. By step 2.1 its weight is at least $m$, whereas the weight of the entire $E^{\mathrm{\prime }}$ is exactly $m$ by [F2]. All edge counts of pieces are nonnegative and add: no cut meets an original edge point. Hence there is exactly one non-root leaf $L$, it has weight $m$, and every other piece has weight zero. There cannot be two such leaves, since they would contribute at least $2m>m$. Every finite tree with at least two vertices has at least two leaves. Thus its only leaves are $L$ and the root. It is a path: a vertex of degree at least three would give at least three leaves by extending paths along its three branches. Removing the end piece $L$ leaves a chain of zero-edge disk pieces with a single exterior trace side at either end. Its union is a disk corridor $R$. The remaining sides are two original-boundary arcs, since each interior piece of the chain has two trace sides alternating with its two original-boundary sides.

The patch $R$ is a subsurface of the original normal disk $F$, cut off by its two end traces, and misses every original edge by step 3.1. Those traces satisfy the distinct-side condition [F3]. Apply [F4] to give $R$ its original-cell PL product rectangle. This in particular proves the corridor has no unrecorded twist or fold: its two end traces meet exactly the same chain of original normal cells, with corresponding sides in corresponding faces.

To transport the entire cap, add a small product collar of the inner trace inside $L$. It can be chosen in the same normal cells and disjoint from $T^{(1)}$: in each normal disk the trace is a finite proper polygonal arc away from its vertices, so a sufficiently thin polygonal strip on its $L$ side exists; choose compatible endpoint intervals on the finitely many shared normal sides. The resulting extended rectangle has transverse coordinate in $[-\epsilon ,1]$, with the original inner trace at zero and the outer trace at one. For $0\le s\le 1$, compress this transverse interval increasingly onto $[-\epsilon ,s]$, fixing $-\epsilon$, and fix the rest of $L$. Conjugating by the product maps gives embeddings of the cap, with image $L$ plus the corridor up to level $s$. These maps are continuous in $s$ and PL for each $s$; the interval maps may be chosen affine and the finitely many polygonal product pieces triangulated. The family from $s=1$ to $s=0$ is the asserted cap isotopy. Each moving piece stays within its original normal cell, its boundary endpoints stay on the same old sides, and its support misses every original edge point. At $s=0$ its image is exactly $L$.