Boundary exchange middle disks have positive original edge weight

Suppose $w(K)=0$. The other component of the irregular resolution [F1] is the surface $V$ obtained by joining the two cut-arc sides of the disk $K$. It is connected, embedded and proper; the sole remaining image double arc was resolved. Identifying those two boundary intervals by a joining strip reduces Euler characteristic from one to zero, and its remaining original-boundary intervals are nonempty. Thus $V$ is an annulus or Mobius band by [F2]. It has zero edge weight, since $K$ does and the joining strip misses all edges.

Each component of $\mathrm{\partial }V$ is contained in one open original boundary face. Indeed it is a connected polygonal curve in $\mathrm{\partial }M\setminus T^{(1)}$, whose components are the open boundary triangles. If $V$ were a Mobius band, [F3] would cap its sole boundary circle by a disk $B$ inside that face. The union $V\cup B$ is an embedded projective plane, still missing $T^{(1)}$. It may be pushed into $\mathrm{Int}M$ while preserving that property. Here a finite local construction suffices: take a compact polygonal subdisk of that same open face containing $B$ in its interior, use the adjacent tetrahedron's affine face prism, and move its bottom subdisk a sufficiently small distance inward, tapering the displacement to zero near the outer prism boundary. Triangulate its finitely many graph-bottom cells and cone matching faces to give an injective PL prism map; on the top and outer sides use the identity. The map moves $B$ and the adjoining collar of $V$ inward. The compact surface misses the old edges, so choose the displacement smaller than its positive distance from them. This gives the embedded interior projective plane excluded by [F4]. Thus $V$ is an annulus.

Label the endpoints of $\alpha$ by $p,q$. Before the join, the two cut sides on $K$ each run from a copy of $p$ to a copy of $q$. In the annular case its two complementary original-boundary arcs connect the two $p$ copies and the two $q$ copies respectively: joining the cut sides without interchanging their ends gives an annulus, whereas interchanging the ends gives the Mobius rectangle model. Consequently one boundary circle of $V$ consists of the $p$-connecting old arc and the short joining arc at $p$; the other is the corresponding $q$ circle. By step 2.1 each such circle lies in the open original face of its endpoint. Its two constituent paths are therefore homotopic with endpoints fixed in that convex face.

The regular disk boundary follows the original boundary arc of $E$, one of the two old arcs of $K$, the boundary arc of $E^{\mathrm{\prime }}$, and the other old arc of $K$, with the directions determined by the annular endpoint matching in step 3.1. The irregular disk $W$ follows the same $E,E^{\mathrm{\prime }}$ arcs with the short joining paths at $p,q$ instead. The endpoint-fixed face homotopies in step 3.1 therefore make $\mathrm{\partial }F$ freely homotopic to $\mathrm{\partial }W$, up to reversal. But the former is essential and the latter is null-homotopic by [F1], a contradiction. This excludes $w(K)=0$. Finally [F1] gives $w(E)=w(E^{\mathrm{\prime }})$ and $w(F)=w(E)+w(K)+w(E^{\mathrm{\prime }})$, so $w(F)>2w(E)$ as stated.