Compatible normal disk boundary paths have a unique type-preserving switch

For each disk, the two cut paths partition its cyclic list of edge symbols. To form a new word, follow its chosen $f$-to-$g$ path and then the reverse of the chosen $f$-to-$g$ path of the other disk. Once the paths paired in one output have been chosen, the two remaining paths must be paired in the second. Thus there are exactly two pairings. We now check the compatible type cases. The edge sets displayed below also inherit their orders from [F1], so their concatenations determine the indicated cyclic words.

If the two types are equal, their two path edge sets are the same nonempty complementary sets $U,V$. Pairing $U$ with $V$ in each output recovers the original word twice. Pairing $U$ with $U$ and $V$ with $V$ repeats every edge in the corresponding nonempty set, so [F2] excludes that pairing. This covers equal triangles and equal quadrilaterals, for every pair of common faces, including the two-edge paths between opposite faces of a quadrilateral boundary.

For distinct triangles, relabel so they are $T_A,T_B$. Their common faces are exactly $ABC,ABD$. The paths of $T_A$ have edge sets $\{AB\}$ and $\{AC,AD\}$; those of $T_B$ have $\{AB\}$ and $\{BC,BD\}$. Pair the singleton in one disk with the two-edge path of the other. The resulting words are $T_B$ and $T_A$. The other pairing repeats $AB$ in one output, so is excluded by [F2].

For a triangle and quadrilateral, relabel them $T_A,Q_{AB\mid CD}$. If the common faces are $ABC,ABD$, their common edge $AB$ is avoided by the quadrilateral. The triangle paths have edge sets $\{AB\}$ and $\{AC,AD\}$; the quadrilateral paths have $\{AC,AD\}$ and $\{BC,BD\}$. Pair $\{AB\}$ with the quadrilateral's $\{AC,AD\}$, giving $T_A$. The remaining two paths concatenate in the order $(AC,AD,BD,BC)$, giving the same quadrilateral. The other pairing joins the two copies of $\{AC,AD\}$ in one output, repeating both edges, and is excluded.

The other common-face pairs are $ABC,ACD$ and $ABD,ACD$, whose common edges $AC,AD$ are crossed by the quadrilateral. Interchanging $C,D$ identifies these two cases, so consider $ABC,ACD$. The triangle paths have $\{AC\}$ and $\{AB,AD\}$; the quadrilateral paths have $\{AC\}$ and $\{AD,BD,BC\}$. Pair the singleton $\{AC\}$ from the triangle with the three-edge path of the quadrilateral, giving the quadrilateral word up to reversal. Pair the two-edge triangle path with the other singleton, giving $T_A$. The other pairing repeats $AC$, so is excluded. Vertex relabeling covers any other triangle–quadrilateral pair.

The allowed pairs consist of equal types, distinct triangles, and a triangle with a quadrilateral; distinct quadrilateral types were excluded in the hypothesis. Steps 2.1–2.4 cover them all and prove both existence and uniqueness of the type-preserving pairing, with the stated repeated-edge obstruction for its alternative. All lists and relabelings are finite and refer to the original seven types.