Simple positive braids are indexed by permutations

The permutation calculus (a). By [F2], $E(\tau s_i)=E(\tau )\mathrm{△}\{\{\tau (i),\tau (i+1)\}\}$ and $\mathrm{inv}(\tau s_i)=\mathrm{inv}(\tau )\pm 1$, the sign being $+1$ exactly when $\tau (i)<\tau (i+1)$; moreover $(i,j)\mapsto (\tau (j),\tau (i))$ is a bijection $\mathrm{Inv}(\tau )\to \mathrm{Inv}({\tau }^{-1})$ between position inversions, so $\mathrm{inv}(\tau )=\mathrm{inv}({\tau }^{-1})$. Applying the right-multiplication formula to ${\tau }^{-1}$ and using $(s_i\tau {)}^{-1}={\tau }^{-1}{s}_i$ gives $\mathrm{inv}(s_i\tau )=\mathrm{inv}({\tau }^{-1}{s}_i)=\mathrm{inv}({\tau }^{-1})\pm 1=\mathrm{inv}(\tau )\pm 1$, with sign $+1$ exactly when ${\tau }^{-1}(i)<{\tau }^{-1}(i+1)$, that is ${\mathrm{pos}}_{\tau }(i)<{\mathrm{pos}}_{\tau }(i+1)$. Finally, concatenating a reduced word for $\tau$ with one for $\upsilon$ gives a word of length $\mathrm{inv}(\tau )+\mathrm{inv}(\upsilon )$ representing $\tau \upsilon$, so the minimal length satisfies $\mathrm{inv}(\tau \upsilon )\le \mathrm{inv}(\tau )+\mathrm{inv}(\upsilon )$ by [F2].

Reducedness criterion. (iii) $\iff$ (iv): if $\mathrm{\ell }(a)=\mathrm{inv}(\pi (a))$ and $w$ is a word with $[w]=a$, then $\mathrm{\mid }w\mathrm{\mid }=\mathrm{\ell }(a)=\mathrm{inv}(\pi (a))=\mathrm{inv}(\sigma (w))$, so $w$ is reduced and $a=[w]=\widehat{\pi (a)}$; conversely $\mathrm{\ell }(\widehat{\sigma })=\mathrm{inv}(\sigma )$ by [F2].

Every permutation gives a left divisor of $\mathrm{\Delta }$. Let $\sigma \in S_n$ and $\tau :={\sigma }^{-1}{w}_0$. Since ${\tau }^{-1}=w_0^{-1}\sigma =w_0\sigma$, the value-pair inversion set of $\tau$ is computed by $\{u,v\}\in E(\tau )⟺w_0(\sigma (u))>w_0(\sigma (v))⟺\sigma (u)<\sigma (v)$ for $u<v$; hence $E(\tau )$ consists of the $2$-subsets $\{u,v\}$, $u<v$, on which $\sigma$ is increasing, and $\mathrm{\mid }E(\tau )\mathrm{\mid }=\left(\genfrac{}{}{0px}{}{n}{2}\right)-\mathrm{inv}(\sigma )=N-\mathrm{inv}(\sigma )$, because $\mathrm{inv}(\sigma )=\mathrm{\#}\{u<v:\sigma (u)>\sigma (v)\}$ and $\left(\genfrac{}{}{0px}{}{n}{2}\right)=N$. By [F2], $\mathrm{inv}(\tau )=\mathrm{\mid }E(\tau )\mathrm{\mid }$, so $\mathrm{inv}(\sigma )+\mathrm{inv}({\sigma }^{-1}{w}_0)=N$. Take a reduced word $u$ for $\sigma$ and a reduced word $v$ for ${\sigma }^{-1}{w}_0$; the concatenation represents $\sigma {\sigma }^{-1}{w}_0=w_0$ and has length $\mathrm{inv}(\sigma )+\mathrm{inv}({\sigma }^{-1}{w}_0)=N=\mathrm{inv}(w_0)$, so it is a reduced word for $w_0$ by [F2]. By [F2] all reduced words for $w_0$ represent $\widehat{w_0}=\mathrm{\Delta }$, so $\mathrm{\Delta }=[uv]=\widehat{\sigma }\cdot [v]$; in particular $\widehat{\sigma }{\preccurlyeq }_L\mathrm{\Delta }$ for every $\sigma \in S_n$.

Left divisors of $\mathrm{\Delta }$ are lifts (b), forward implication. Let $a,c\in B_n^{+}$ with $ac=\mathrm{\Delta }$. Additivity of $\mathrm{\ell }$ gives $\mathrm{\ell }(a)+\mathrm{\ell }(c)=\mathrm{\ell }(\mathrm{\Delta })=N$, and applying $\pi$ gives $\pi (a)\pi (c)=\pi (\mathrm{\Delta })=w_0$. Hence $N=\mathrm{inv}(w_0)=\mathrm{inv}(\pi (a)\pi (c))\le \mathrm{inv}(\pi (a))+\mathrm{inv}(\pi (c))\le \mathrm{\ell }(a)+\mathrm{\ell }(c)=N$ by step 1.1 and [F2]. All inequalities are equalities, so in particular $\mathrm{\ell }(a)=\mathrm{inv}(\pi (a))$, and $a=\widehat{\pi (a)}$ by step 1.2; the same equality chain also gives $\mathrm{\ell }(c)=\mathrm{inv}(\pi (c))$, so step 1.2 yields $c=\widehat{\pi (c)}$.

Descents (d). Let $b\in B_n^{+}$ satisfy $\mathrm{\ell }(b)=\mathrm{inv}(\pi (b))$ and let ${\sigma }_i{\preccurlyeq }_{L}b$, say $b={\sigma }_{i}c$; by additivity $\mathrm{\ell }(b)=1+\mathrm{\ell }(c)$, and applying $\pi$ gives $\pi (b)=s_i\pi (c)$. If $\mathrm{inv}(s_i\pi (c))=\mathrm{inv}(\pi (c))-1$, then $\mathrm{inv}(\pi (b))\le \mathrm{inv}(\pi (c))-1\le \mathrm{\ell }(c)-1<\mathrm{\ell }(c)+1=\mathrm{\ell }(b)=\mathrm{inv}(\pi (b))$, a contradiction; hence the sign is $+1$ by step 1.1, i.e. $\mathrm{inv}(s_i\pi (c))=\mathrm{inv}(\pi (c))+1$, and then $\mathrm{\ell }(b)=\mathrm{inv}(\pi (b))=1+\mathrm{inv}(\pi (c))$ forces $\mathrm{inv}(\pi (c))=\mathrm{\ell }(c)$. By step 1.1 the sign $+1$ means ${\mathrm{pos}}_{\pi (c)}(i)<{\mathrm{pos}}_{\pi (c)}(i+1)$. Left multiplication by $s_i$ swaps the values $i$ and $i+1$ in the one-line notation, because $(s_i\tau )(x)=s_i(\tau (x))$ by [F2]; therefore ${\mathrm{pos}}_{\pi (b)}(i)={\mathrm{pos}}_{\pi (c)}(i+1)>{\mathrm{pos}}_{\pi (c)}(i)={\mathrm{pos}}_{\pi (b)}(i+1)$, as claimed. If this holds for all $i\in \{1,\dots ,n-1\}$, then ${\mathrm{pos}}_{\pi (b)}(1)>{\mathrm{pos}}_{\pi (b)}(2)>\cdots >{\mathrm{pos}}_{\pi (b)}(n)$, so the one-line notation of $\pi (b)$ is $(n,n-1,\dots ,1)$ and $\pi (b)=w_0$; since $b$ is reduced, step 1.2 gives $b=\widehat{w_0}=\mathrm{\Delta }$ by [F2].

The bijection (b), converse, and (c). If $a=\widehat{\pi (a)}$ then $a{\preccurlyeq }_L\mathrm{\Delta }$ by step 2.1, and if $a{\preccurlyeq }_L\mathrm{\Delta }$ then $a=\widehat{\pi (a)}$ by step 2.2; combined with step 1.2 this proves the equivalence of (i), (iii), (iv) of (b), and shows that the image of $\sigma \mapsto \widehat{\sigma }$ is exactly the set of left divisors of $\mathrm{\Delta }$. That map is injective because $\pi (\widehat{\sigma })=\sigma$ for all $\sigma$ [F2], so it is a bijection onto the left divisors of $\mathrm{\Delta }$, a set of $n!$ elements.

Right divisors coincide with left divisors (b), (c). Let $\rho$ be the reversal anti-automorphism of [F3]. First, $\pi (\rho (b))=\pi (b{)}^{-1}$ for every $b\in B_n^{+}$: the map $\psi :=\pi \circ \rho$ is an anti-homomorphism with $\psi ({\sigma }_i)=s_i$, so $b\mapsto \psi (b{)}^{-1}$ is a homomorphism $B_n^{+}\to S_n$ carrying every ${\sigma }_i$ to $s_i$, hence equals $\pi$ by uniqueness of the homomorphism induced by the atoms [F1]. Second, $\rho (\widehat{\tau })=\widehat{{\tau }^{-1}}$ for every $\tau \in S_n$: applying the first identity, $\pi (\rho (\widehat{\tau }))=\pi (\widehat{\tau }{)}^{-1}={\tau }^{-1}$, while $\rho$ preserves lengths, so $\mathrm{\ell }(\rho (\widehat{\tau }))=\mathrm{inv}(\tau )=\mathrm{inv}({\tau }^{-1})$ and step 1.2 gives $\rho (\widehat{\tau })=\widehat{{\tau }^{-1}}$ (note $\tau \mapsto {\tau }^{-1}$ is a bijection of $S_n$, so the right divisors listed below are again indexed by all of $S_n$). Now $a{\preccurlyeq }_R\mathrm{\Delta }$ means $\mathrm{\Delta }=ca$; applying the involutive anti-automorphism $\rho$ and using $\rho (\mathrm{\Delta })=\mathrm{\Delta }$ and $\rho (ca)=\rho (a)\rho (c)$ this is equivalent to $\mathrm{\Delta }=\rho (a)\rho (c)$, i.e. to $\rho (a){\preccurlyeq }_L\mathrm{\Delta }$, hence by step 3.1 to $\rho (a)=\widehat{\pi (a{)}^{-1}}$, i.e. to $a=\rho (\widehat{\pi (a{)}^{-1}})=\widehat{\pi (a)}$. Therefore $a$ is a right divisor of $\mathrm{\Delta }$ iff $a=\widehat{\pi (a)}$ iff $a$ is a left divisor of $\mathrm{\Delta }$; the two divisor sets coincide and both have the $n!$ elements of step 3.1.

Divisibility in the braid group (e). Let $a\in B_n^{+}$. Since $\mathrm{\Delta }$ is positive, [F4] says that $a{\preccurlyeq }_L\mathrm{\Delta }$ in $B_n$ holds if and only if $a{\preccurlyeq }_L\mathrm{\Delta }$ in $B_n^{+}$, which by (b) is the definition of $a$ being a simple braid; the right-handed statement is identical with ${\preccurlyeq }_R$.

Assembly. Part (a) is step 1.1, part (b) is steps 1.2, 2.2, 3.1 and 4.1, part (c) is steps 3.1 and 4.1, part (d) is step 2.3, and part (e) is step 4.2. The only imported statements about $S_n$ are the inversion calculus, the type-A Matsumoto theorem and the identification $\pi (\mathrm{\Delta })=w_0$ collected in [F2]; no geometric model of braids, no crossing number and no injectivity of a geometric representation is used, so the count $n!$ of simple braids is established purely algebraically. For $n\le 1$ the alphabet is empty, $S_n$ and $B_n^{+}$ are trivial and all assertions are vacuous, as noted in [F1] and statement; every construction above is finite and no choice principle is used. ∎