The rank-n diagram basis is counted by the nth Catalan number

Read the $2n$ boundary labels in their fixed cyclic order. Given a noncrossing pairing, mark the earlier endpoint of each pair by an opening parenthesis and the later endpoint by a closing parenthesis. The resulting word has equal numbers of each kind. In every prefix, each closing endpoint is paired with an earlier opening endpoint in that prefix, so the number of closes never exceeds the number of opens; hence the word is balanced. This construction also applies at $n=0$, where it gives the empty word.

Conversely, scan a balanced word from left to right, placing each opening position on a stack and pairing each closing position with the most recently placed unmatched opening. At the end the stack is empty because the total numbers agree. The pairs are nested or disjoint, so they do not interlace. For a noncrossing pairing, the same rule recovers its original partners: if a close at position $j$ were paired with an earlier open $i$ while a later open $k$ between them remained unmatched until after $j$, then its eventual partner $l>j$ would give interlacing pairs $\{i,j\}$ and $\{k,l\}$, contrary to [F1]. Thus the two constructions are inverse bijections.

Replace each opening bracket by an up-step and each closing bracket by a down-step. Equality of the bracket counts gives total height zero, and the prefix inequality proved in step 1.1 gives nonnegative prefix heights; hence this is a Dyck path of semilength $n$. Reading its step word recovers the bracket word, so [F3] gives a bijection. By [F2] the bracket-word set is finite of size $C_n$. Composing this bijection with the two inverse constructions in steps 1.1 and 2.1 proves that $\mathcal{D}(n,n)$ is in bijection with Dyck paths and has size $C_n$. The free-module definition in [F1] makes these diagrams a basis of ${\mathrm{T}\mathrm{L}}_n(R,\delta )$; no coefficient is changed in this count.