Exchange surfaces inside normal disk-copy products are vertical

First straighten the face strips relative to their end and side data. In a face strip the disjoint arcs of $A$ join its opposite horizontal sides. They match their endpoints in order: two matched pairs in reversed order would alternate on the boundary of the strip and would have to intersect, by [F2]. Keep the lower horizontal arc as the base interval. Choose an increasing PL parametrization of the upper horizontal arc matching these finitely many marked endpoints. Keep the two vertical sides parametrized by their prescribed original edge intervals; their endpoints already agree. Compare this marked arc arrangement with the vertical intervals over the marked lower endpoints in a standard rectangle. Choose its interval maps on the marked arcs and extend over each complementary disk by [F2]. This supplies a product parametrization of the entire face strip making all its $A$ arcs vertical, preserving each of its four boundary sides and agreeing with the selected end and edge parametrizations. When strips share an original edge interval, use that one fixed interval parametrization on both, so the maps agree there.

Fix one tetrahedral prism $B$ and let its lower disk be its base. The frontier arcs of its $A$ rectangles are disjoint proper arcs in that disk. The face-strip maps just constructed match their endpoints to the corresponding endpoints in the upper disk. Each rectangle connects precisely its two lower endpoints to its two upper endpoints; the distinct-face hypothesis identifies the endpoints individually by their original faces. Thus the same boundary-order matching carries each lower arc's endpoint pair to its upper counterpart. These arrangements have the same complementary regions: disjoint proper arcs in a disk have pairings determined by their boundary endpoints, and cutting successively along them produces the same finite disk-region incidence tree. Map the lower disk to itself, map the upper disk to that base by extending its prescribed boundary and arc maps across these regions using [F2], and use the maps of step 1.1 on all side strips. We have a prescribed homeomorphism of the entire prism boundary to the boundary of the standard product of its lower disk with the interval. It maps each actual rectangle boundary to the boundary of the vertical rectangle over its lower frontier arc.

Extend the last boundary map over the prism with its rectangle arrangement, rather than presuming the arrangement is vertical already. The actual rectangles are disjoint proper PL disks in the parametrized ball $B$, so [F1] cuts $B$ into parametrized balls. The comparison vertical rectangles cut the standard product into the balls given by its lower disk regions times the interval. Their boundary incidence trees agree by the endpoint and arc correspondence in step 2.1. Choose maps on each actual rectangle to its vertical counterpart extending the already prescribed boundary map, using [F2]. The outer prism map and these rectangle maps now prescribe the complete boundary map on each cut ball. Extend over each ball by [F1] and paste. This gives a PL product parametrization of $B$ agreeing with every prescribed face-strip and horizontal disk map and making its part of $A$ vertical. Its fibers stay inside this tetrahedron, and on its side strips stay in their original faces.

Perform this construction in every tetrahedral prism. On a common face strip their maps agree with the single map already constructed there, including the horizontal boundary intervals. Their lower disks glue to $D_0$, so the resulting maps paste to a parametrization of all of $P$ by $D_0\times [0,1]$. Each actual exchange rectangle is vertical and their gluings agree through the faces. Consequently $A$ is exactly $\beta \times [0,1]$, where $\beta$ is its frontier circle or proper arc on $D_0$. No original simplex is crossed by a fiber: this holds in the prism interiors and was already imposed on their face and edge strata.

If $\beta$ is a circle, it bounds a disk cap $C$ in the disk $D_0$ by [F2]. The supplied subproduct $C\times [0,1]$ is a product between that cap and the cap on $D_1$, with vertical boundary $A$. If $\beta$ is a proper arc, cutting $D_0$ along it gives two disk caps $C,C^{\mathrm{\prime }}$ by [F2]. Their products give both corresponding cap pairs. The part of each vertical boundary other than $\beta \times [0,1]=A$ is over a subarc of $\mathrm{\partial }{D}_0$, hence lies in the prescribed vertical boundary of $P$ in $\mathrm{\partial }M$. Following the fibers supplies compatible normal deformations of all cap pieces with these precise frontier tracks. This is the asserted normal parallelism along $A$, proved from the actual marked prism data.