Minimum boundary exchange cycles force an impossible zero-edge normal component

There are finitely many original $P$ leaves of minimum positive weight $m$. Map each to the unique minimum leaf contained in its opposite cap, as in [F1]. It cannot map to itself, because the two end caps are disjoint. Iterating this finite function gives a cycle of distinct leaves $E_1,\dots ,E_r$ with $r\ge 2$, with indices read modulo $r$. Write $B_i$ for the opposite cap of $E_i$ and ${\alpha }_i$ for its original exchange arc. Then $B_i$ contains precisely the original minimum leaf $E_{i+1}$ and has weight $m$. The arcs ${\alpha }_i$ are distinct: one original double arc cuts off only one leaf of $P$ when there are at least two double arcs. Otherwise it would cut $P$ into its two leaves with no room for another cut arc.

The caps $B_i$ are pairwise disjoint. By [F5], non-disjoint caps must be nested or cover $F$ in a pair. Nesting $B_i\subset B_j$ puts the minimum leaf $E_{i+1}$ inside $B_j$, so uniqueness in [F1] forces $E_{i+1}=E_{j+1}$, contrary to distinct cycle indices. A pair covering $F$ gives $w(F)\le w(B_i)+w(B_j)=2m$, contrary to [F2]. Thus neither overlap is possible.

Put $R_i=\overline{B_i\setminus E_{i+1}}$, with closure in $F$. This is the zero-edge corridor of [F1]. It is nonempty. Its outer trace is the opposite-cap trace of ${\alpha }_i$, while its inner trace is the original leaf trace of the distinct arc ${\alpha }_{i+1}$. These are disjoint distinct traces by [F3] and [F5], so the two caps cannot be equal. The corridors $R_i$ have pairwise disjoint interiors by step 2.1. Their union carries no original edge point.

Remove the caps $B_i$ from $F$ and retain the closure $F_0$ of their complement. It is a disk with the outer trace of each $B_i$ as a boundary interval: successively removing disjoint end caps from a disk has this effect. All original $E_i$ lie in the removed caps, since $E_i\subset B_{i-1}$. Across the rectangular exchange surface for ${\alpha }_i$, join $E_i$ to $F_0$ at the outer trace of $B_i$. These attachments are disjoint by [F3]. Attaching each disk along its one interval, with its rectangular collar, gives an embedded disk made of $F_0$, the $E_i$ and the attaching rectangles. Its discarded complementary pieces are precisely the corridors $R_i$, with the other local continuations at these exchange arcs. The next step identifies both continuations in the original diagram.

Fix $i$. Label the original rays in cyclic order $0,1,2,3$ so that $0$ is $E_i$'s $P$ ray and $1$ is $B_i$'s $H$ ray. These are adjacent, as proved by the leaf/opposite-cap supplier in [F1]. The other $P$ ray is $2$, and the other $H$ ray is $3$. The regular full resolution pairs $0$ with $3$ and $1$ with $2$: the other adjacent pairing would join the two end caps and is irregular. The trace between $0,3$ is the boundary of $E_i$, and that between $1,2$ is the boundary of $B_i$. Near the first trace, ray $3$ lies in $R_{i-1}$; near the second, ray $1$ lies in $R_i$ and ray $2$ lies in $F_0$. Thus the attachment of step 4.1 pairs retained rays $0,2$, while the discarded corridors pair rays $1,3$. These are exactly the two original unswitched continuations. There is no irregular switch in this construction.

At every arc outside the cycle retain the original regular resolution. Neither the retained/discarded division nor its frontier crosses either trace of such an arc, because its frontier consists solely of the cycle traces used in step 5.1. Thus each local regular output has both of its incident parts retained or both discarded. At every cycle arc, use the two unswitched continuations of step 5.1. Hence this is precisely the original partial regular resolution that leaves the cycle arcs unswitched and resolves all others regularly. Its normalization splits into a retained union containing the disk in step 4.1 and a discarded union consisting of the nonempty $R_i$ with their indicated continuations. The local pairing check ensures neither union has a frontier in that normalization; intersections of their images along an unswitched arc do not identify their parametrizing sheets.

Every discarded piece and its short continuing strip misses $T^{(1)}$, by step 3.1 and [F3]. Their union is nonempty and is a union of normalization components by step 6.1. Choose one such component. It is a nonempty zero-edge component of an actual partial regular resolution, contradicting [F4]. This contradiction also covers the case where the cycle contains every arc: that partial resolution is just the original unswitched pair, whose tetrahedral disks still have normal vertices. Therefore the hypothetical reduced boundary pair does not occur.