Elementary cup-cap diagrams generate every finite planar pairing endomorphism diagram

For $n=0$ the only diagram is the empty identity; for $n=1$ the only diagram is the single through strand. Both are the empty product of generators. Now let $n\ge 2$. A through strand joins a top point to a bottom point; let $t$ be their number. The other top points are paired into caps and the other bottom points into cups, so there are $p=(n-t)\mathrm{/}2$ caps and $p$ cups. The through strands occur in the same left-to-right order at top and bottom, since a reversed pair of through strands would interlace and violate noncrossing in [F1].

Regard a cap on consecutive positions as one symbol $C$ occupying two positions and a through endpoint as a symbol $T$ occupying one. In each maximal interval with no through endpoints, replace the noncrossing cap matching by consecutive adjacent caps; do the corresponding operation on the bottom cups. This gives a flat diagram with the same through endpoints and the same order-preserving through strands. The baseline product $d_1{d}_3\cdots d_{2p-1}$ has $p$ adjacent caps and cups followed by $t$ through strands, with $p,t$ as in step 1.1. Whenever a top cap at $(i-1,i)$ is followed by a through strand at $i+1$, stacking $d_i$ above that diagram changes the top symbols $CT$ to $TC$: tracing the glued endpoints sends the former through strand to $i-1$ and makes the cap occupy $(i,i+1)$. The cap of the generator and the old through component join into a path meeting the exterior boundary, so no cycle is created. Thus left multiplication by these generators performs each adjacent $CT\mapsto TC$ top swap and leaves the bottom cap/cup pairing unchanged. Sorting from $C^p{T}^t$ to the prescribed top pattern takes finitely many swaps. For the bottom, if a cup at $(i-1,i)$ is followed by a through endpoint at $i+1$, stack $d_i$ below; its top cap joins the old cup path to the through path, moving the through endpoint to $i-1$ and placing the bottom cup at $(i,i+1)$. These are distinct paths, so no cycle forms. Right multiplication therefore performs the corresponding bottom $CT\mapsto TC$ swaps while leaving the top pairing unchanged. Sorting the bottom pattern is finite as well. The resulting flat diagram has both prescribed patterns and the same order-preserving through strands.

It remains to restore each nested cap or cup in the flat diagram of step 2.1. A cap island is the matching on a consecutive interval $i,i+1,\dots ,i+2q-1$ whose endpoints are all paired internally and whose outermost cap joins $i$ to $i+2q-1$; the cup version is defined analogously. In the flat diagram those points have simple caps $(i,i+1),(i+2,i+3),\dots ,(i+2q-2,i+2q-1)$. For $q=1$ no generator is needed. For $q>1$, stacking $d_{i+1}{d}_{i+3}\cdots d_{i+2q-3}$ above them creates the outer cap $(i,i+2q-1)$ and leaves $q-1$ simple caps on the interior endpoints. To verify this, start with the rightmost factor: its glued bottom cup meets endpoints belonging to two distinct cap paths, so their union is a path with exterior endpoints rather than a cycle; it replaces those two adjacent simple caps by one outer cap and one interior simple cap. Each preceding factor performs the same merge with the next simple cap to the left, extending the outer cap by two endpoints while preserving the remaining interior simple caps. Thus no closed component is created at any stage. The reflected product below restores the corresponding bottom cup island by the same path check.

The cap intervals of a noncrossing matching are nested or disjoint, because interlacing pairs are excluded by [F1]. Thus they form a finite nesting forest. Process maximal islands first and then recursively process the smaller islands inside each one; every child interval is strictly shorter than its parent, so induction on interval length terminates. Apply the same construction to top caps and bottom cups. The order-preserving through strands stay fixed, and the final top and bottom pairings are exactly those of $D$. Every step is a stack of one of the $d_i$ and the path checks in steps 2.1 and 3.1 show that it creates no closed cycle.