BMW has the descending Kauffman tangle basis

By [F2] and [F3], the tangle skein algebra has a basis indexed by the perfect matchings of the $2n$ endpoints. There are $(2n-1)!!$ such matchings: choose the partner of the first endpoint in $2n-1$ ways and apply the same count to the remaining $2n-2$ endpoints, with one empty matching at $n=0$. The unique matching at $n=1$ is the straight strand. Hence the tangle algebra has the stated rank and boundary bases.

Fix a standard matching connector and let $k$ be the number of top-top pairs. The number of bottom-bottom pairs is also $k$, because the remaining endpoints must be paired top-to-bottom. If $k=0$, the connector is a permutation diagram, so its standard descending representative is a braid diagram; reading its finitely many crossing levels expresses it as a word in the crossing images $G_i^{\pm 1}={\mathrm{\Phi }}_n(g_i^{\pm 1})$. If $k>0$, choose a permutation braid $B_1$ that moves the $k$ top pairs to adjacent positions $(1,2),(3,4),\dots ,(2k-1,2k)$, and a permutation braid $B_2$ that routes the corresponding bottom pairs to those positions. The remaining through-strands form a braid $B$ on $n-2k$ strands. With the shifted copy of $B$ placed beside the cup-cap block, the tangle has factorization $B_1[(E_1{E}_3\cdots E_{2k-1})\otimes B]B_2\mathrm{.}$ Each $B_i$ and $B$ is a finite word in the $G_j^{\pm 1}$, each $E_j={\mathrm{\Phi }}_n(e_j)$ is a cup-cap generator, and the shifted factors use indices among $1,\dots ,n-1$. The permutation braid pieces are obtained by reading the existing upper and lower layers of $T_m$; moving the horizontal pairs to adjacent positions is a boundary-fixing braid isotopy, so it preserves the crossing data and total descending order. The factorization therefore has the same connector and is regularly isotopic to the standard representative. Uniqueness for a fixed connector from the spanning proof identifies its class with that representative. Thus every basis tangle is in the image of ${\mathrm{\Phi }}_n$, proving surjectivity by [F2].

Let $N$ be the number of matchings and choose the $N$ source basis elements supplied by [F5]. They define a surjection $u:T^N\to BM{W}_n(T)$, and step 2.1 gives the surjection ${\mathrm{\Phi }}_n:BM{W}_n(T)\to {\mathcal{K}\mathcal{T}}_n(T)$. Identify the target with $T^N$ using step 1.1. The composite $A={\mathrm{\Phi }}_{n}u:T^N\to T^N$ is a surjective square matrix over the commutative ring $T$. Choose a preimage of each of its finitely many standard basis vectors to obtain $B$ with $AB=I$. Then $\det (A)\det (B)=1$, so the adjugate formula makes $A$ invertible. If $x\in \ker {\mathrm{\Phi }}_n$, write $x=u(v)$; then $Av=0$, hence $v=0$ and $x=0$. Thus ${\mathrm{\Phi }}_n$ is injective as well as surjective. Its inverse transports the standard matching basis to the abstract BMW algebra, proving all the stated freeness and rank assertions, including $n=0,1$. No implication from target freeness alone is used: the finite source spanning bound [F5] is essential.