The combed geometric decomposition is unique

The combing identity, by free cancellation. For $1\le j\le n-1$ put ${\delta }_j:={\sigma }_{n-1}{\sigma }_{n-2}\cdots {\sigma }_{j+1}$ and ${\gamma }_j:={\sigma }_{j+1}{\sigma }_{j+2}\cdots {\sigma }_{n-1}$ (empty words for $j=n-1$); then ${\delta }_j={\delta }_{j+1}{\sigma }_{j+1}$ and ${\gamma }_j={\sigma }_{j+1}{\gamma }_{j+1}$ as words, and by [F1] the Artin word ${\tilde{A}}_{jn}={\delta }_j{\sigma }_j^2{\delta }_j^{-1}$ lifts the geometric generator $A_{jn}$, while $x_j={\gamma }_j^{-1}{\sigma }_j^2{\gamma }_j$ is the combing word. First, for every $j$ the Artin word ${\tilde{Q}}_j:={\tilde{A}}_{j,n}{\tilde{A}}_{j+1,n}\cdots {\tilde{A}}_{n-1,n}$ freely reduces to ${\delta }_j{\sigma }_j^2{\gamma }_j$: this is a downward induction on $j$, in which both words are ${\sigma }_{n-1}^2$ for $j=n-1$, and in which the substitution of ${\tilde{A}}_{j,n}={\delta }_j{\sigma }_j^2{\delta }_j^{-1}$ and of the reduced form ${\tilde{Q}}_{j+1}\equiv {\delta }_{j+1}{\sigma }_{j+1}^2{\gamma }_{j+1}$ gives ${\tilde{Q}}_j\equiv {\delta }_j{\sigma }_j^2\left({\delta }_j^{-1}{\delta }_{j+1}\right){\sigma }_{j+1}^2{\gamma }_{j+1}={\delta }_j{\sigma }_j^2{\sigma }_{j+1}^{-1}\left({\delta }_{j+1}^{-1}{\delta }_{j+1}\right){\sigma }_{j+1}^2{\gamma }_{j+1},$ where the block ${\delta }_{j+1}^{-1}{\delta }_{j+1}$ cancels freely to the empty word -- its middle pair ${\sigma }_{n-1}^{-1}{\sigma }_{n-1}$ cancels first, after which the next pair is adjacent, and so on outwards -- leaving ${\delta }_j{\sigma }_j^2{\gamma }_j$. For $1\le i\le n-2$ this gives ${\tilde{P}}_i:={\tilde{Q}}_{i+1}\equiv {\delta }_{i+1}{\sigma }_{i+1}^2{\gamma }_{i+1}$, and the inverse word is ${\tilde{P}}_i^{-1}\equiv {\gamma }_{i+1}^{-1}{\sigma }_{i+1}^{-2}{\delta }_{i+1}^{-1}$; substituting these reduced forms into ${\tilde{P}}_i^{-1}{\tilde{A}}_{in}{\tilde{P}}_i$, and using ${\delta }_{i+1}^{-1}{\delta }_i\equiv {\sigma }_{i+1}$ and ${\delta }_i^{-1}{\delta }_{i+1}\equiv {\sigma }_{i+1}^{-1}$ (the same middle-outward cancellation applied to ${\delta }_i={\delta }_{i+1}{\sigma }_{i+1}$), gives ${\tilde{P}}_i^{-1}{\tilde{A}}_{in}{\tilde{P}}_i\equiv {\gamma }_{i+1}^{-1}{\sigma }_{i+1}^{-2}\left({\delta }_{i+1}^{-1}{\delta }_i\right){\sigma }_i^2\left({\delta }_i^{-1}{\delta }_{i+1}\right){\sigma }_{i+1}^2{\gamma }_{i+1}\equiv {\gamma }_{i+1}^{-1}{\sigma }_{i+1}^{-1}{\sigma }_i^2{\sigma }_{i+1}{\gamma }_{i+1}={\gamma }_i^{-1}{\sigma }_i^2{\gamma }_i=x_i\mathrm{.}$ The remaining index $i=n-1$ has the empty product ${\tilde{P}}_{n-1}$ and $x_{n-1}={\tilde{A}}_{n-1,n}={\sigma }_{n-1}^2$, so $x_{n-1}={\tilde{P}}_{n-1}^{-1}{\tilde{A}}_{n-1,n}{\tilde{P}}_{n-1}$ holds trivially. No Artin relation is used in this step, only free insertions and deletions. Applying $\phi$ gives the stated geometric identity.

The two models of a lower-rank word. Put $a:=n\mathrm{/}(n+1)$ and consider the similarity $A(w)=aw-h_n$ with $h_n=\frac{1}{4(n+1)}$, so that $A(w)=aw-\frac{1}{4(n+1)}$. One computes $A(q_j^{\mathrm{\prime }})=q_j$ for $1\le j\le n-1$, $A(m_i^{\mathrm{\prime }})=m_i$ for $1\le i\le n-2$ where $m_i^{\mathrm{\prime }}=q_i^{\mathrm{\prime }}+(h^{\mathrm{\prime }},0)$ is the (n−1)-strand midpoint, and $a{\rho }^{\mathrm{\prime }}(t)=\rho (t)$ for the two displacement paths, since $a{h}^{\mathrm{\prime }}=h$ and each diamond-path coordinate is linear in its size. The reflected paths for the negative letters satisfy the same scaling identity. Since $A$ is affine, $A(m_i^{\mathrm{\prime }}\pm {\rho }^{\mathrm{\prime }}(t))=A(m_i^{\mathrm{\prime }})\pm a{\rho }^{\mathrm{\prime }}(t)=m_i\pm \rho (t)$. Thus for every $1\le i\le n-2$ and both signs, $A(({\sigma }_i^{(n-1)}{)}_j(t))=({\sigma }_i^{(n)}{)}_j(t)(1\le j\le n-1),A(q_j^{\mathrm{\prime }})=q_j,$ where ${\sigma }_i^{(k)}$ is the elementary half twist of the $k$-strand model of [F4]. Every lower-rank moving point stays in the support disc $U_i^{\mathrm{\prime }}$ of the letter currently being run, or is one of the fixed base points. Relative to $c:=(-\frac{1}{4},0)$, these base points have radii $\mathrm{\mid }{q}_j^{\mathrm{\prime }}-c\mathrm{\mid }=j\mathrm{/}(2n)\le (n-1)\mathrm{/}(2n)$, and the support disc $U_i^{\mathrm{\prime }}$ has center $m_i^{\mathrm{\prime }}$ with $\mathrm{\mid }{m}_i^{\mathrm{\prime }}-c\mathrm{\mid }=(2i+1)\mathrm{/}(4n)$ and radius $3\mathrm{/}(8n)$; hence every such path point has radius at most $(4n-3)\mathrm{/}(8n)$. This local agreement is all that will be used: $A$ maps the lower standard-generator paths to the corresponding $n$-strand paths, but it is not used to map arbitrary lower braids into the complement of $q_n$. Consequently, for every word $V$ in ${\sigma }_1^{\pm 1},\dots ,{\sigma }_{n-2}^{\pm 1}$, if $z=(z_1,\dots ,z_{n-1})$ is its (n−1)-strand coordinate motion and $Z=(Z_1,\dots ,Z_n)$ its n-strand coordinate motion, then $Z_j=A\circ z_j(1\le j\le n-1),Z_n\equiv q_n,$ by induction on its letters: the generator paths agree by the displayed identity, and stacking has the same coupling permutation on $\{1,\dots ,n-1\}$ in both models and fixes $n$. Thus ${\phi }_n(V)$ is represented by $(A\circ z_1,\dots ,A\circ z_{n-1},q_n)$.

The straight-strand extension is a well-defined injective homomorphism. Let $c:=(-\frac{1}{4},0)$, $a:=n\mathrm{/}(n+1)$, $R_0:=\frac{1}{2}-\frac{1}{4n}$ and $r_n:=n\mathrm{/}(2(n+1))$. For $\theta \in \mathbb{R}$ let $u_{\theta }:=(\cos \theta ,\sin \theta )$ and let $R(\theta ):=\frac{1}{4}\cos \theta +\sqrt{1-\frac{1}{16}{\sin }^2\theta },$ the distance from $c$ to the boundary of $D^{\circ }$ along the ray $c+r{u}_{\theta }$; in particular $R(\theta )\ge \frac{3}{4}>R_0$. Define the strictly increasing radial function $\lambda (r,\theta ):=\{\begin{array}{ll}ar,& 0\le r\le R_0,\\ a{R}_0+(r_n-a{R}_0){\displaystyle \frac{r-R_0}{R(\theta )-R_0}},& R_0\le r<R(\theta ),\end{array}$ and define $E(c+r{u}_{\theta }):=c+\lambda (r,\theta )u_{\theta }$, with $E(c):=c$. Since $a{R}_0<r_n$, each ray is mapped increasingly onto the ray segment of length $r_n$; thus $E$ is a homeomorphism from $D^{\circ }$ onto the open disk $L_n:=B(c,r_n)$. The closure of $L_n$ lies in $D^{\circ }$ because $\parallel c{\parallel }_2+r_n<\frac{3}{4}<1$, and $q_n$ lies on $\mathrm{\partial }{L}_n$ because $\mathrm{\mid }{q}_n-c\mathrm{\mid }=n\mathrm{/}(2(n+1))=r_n$. Moreover, $E=A$ whenever $\mathrm{\mid }w-c\mathrm{\mid }\le R_0$. The basepoints satisfy $\mathrm{\mid }{q}_j^{\mathrm{\prime }}-c\mathrm{\mid }=j\mathrm{/}(2n)\le (n-1)\mathrm{/}(2n)<R_0$, and for $n\ge 3$ every lower support disc satisfies ${\max }_{1\le i\le n-2}{\sup }_{w\in U_i^{\mathrm{\prime }}}\mathrm{\mid }w-c\mathrm{\mid }\le \frac{4n-3}{8n}<R_0$ (for $n=2$ there are no such support discs). Therefore $E(Q^{(n-1)})=(q_1,\dots ,q_{n-1})$, and [F4] with step 1.2 shows that $E$ maps every lower-rank standard-generator path to its corresponding n-strand path. If $\beta \in G_{n-1}^{\mathrm{p}\mathrm{u}\mathrm{r}\mathrm{e}}$ with representative coordinate loop $z=(z_1,\dots ,z_{n-1})$, define $\check{s}(\beta )$ to be the class of $(E\circ z_1,\dots ,E\circ z_{n-1},q_n)$. The tuple is a pure n-strand braid: $E$ is injective, its image $L_n$ avoids $q_n$, and the endpoints are $E(Q^{(n-1)})=(q_1,\dots ,q_{n-1})$. Applying $E$ to a braid isotopy preserves pairwise distinctness and avoids $q_n$, so $\check{s}$ is well defined on isotopy classes; applying $E$ to the stacking formula shows it is a homomorphism. To prove injectivity, first note that the induced map $E_{\ast }:{\pi }_1(F_{n-1}(D^{\circ }),Q^{(n-1)})\to {\pi }_1(F_{n-1}(D^{\circ }),E(Q^{(n-1)}))$ is injective. Indeed, for $0\le t\le 1$ let ${\lambda }_t(r,\theta ):=(1-t)r+t\lambda (r,\theta )$ and $E_t(c+r{u}_{\theta }):=c+{\lambda }_t(r,\theta )u_{\theta }$. For each $t$, ${\lambda }_t$ is strictly increasing along every ray and, when $r<R(\theta )$, ${\lambda }_t(r,\theta )<(1-t)R(\theta )+t{r}_n\le R(\theta ),$ so $E_t$ is an embedding $D^{\circ }\to D^{\circ }$, with $E_0=\mathrm{id}$ and $E_1=E$. Hence $F_{n-1}(E_t)$ is a homotopy on the ordered configuration space from the identity to $F_{n-1}(E)$, whose basepoint follows $\eta (t):=F_{n-1}(E_t)(Q^{(n-1)})$; by [F5], ${\mathrm{id}}_{\ast }={\beta }_{\eta }\circ E_{\ast }$, and since ${\beta }_{\eta }$ is an isomorphism, $E_{\ast }$ is injective. If $\check{s}(\beta )=1$, [F3] and the coordinate-projection description of [F2] imply $E_{\ast }[z]=1$; injectivity gives $[z]=1$, and [F3] implies $\beta =1$.

The conjugating identity in the geometric group. Applying $\phi$ to the word identity of step 1.1 gives $\phi (x_i)=P_i^{-1}{A}_{in}{P}_i\text{in }{G}_n,$ where $P_i=A_{i+1,n}\cdots A_{n-1,n}$ is the geometric product defined in the Statement. Each factor $A_{j,n}$ of $P_i$ and $A_{in}$ itself lies in the pure subgroup and maps under ${\mathrm{\Psi }}_n$ into $\ker \rho$ by [F2, F3]. Since $\ker \rho$ is a subgroup, ${\mathrm{\Psi }}_n(\phi (x_i))$ lies in $\ker \rho$; in particular $\phi (x_i)$ is pure.

The straight-strand extensions meet the kernel trivially. By [F2] the forgetting map acts on coordinate loops by dropping the last coordinate. For the straight-strand extension of step 2.1 the coordinate loop is $(E\circ z_1,\dots ,E\circ z_{n-1},q_n)$, whose dropped loop is $E\circ z$. Thus $\rho ({\mathrm{\Psi }}_n(\check{s}(\beta )))={\iota }_{\ast }^F(E_{\ast }[z]{)}^{-1}$ at the basepoint $(q_1,\dots ,q_{n-1})=E(Q^{(n-1)})$. If $\gamma ={\mathrm{\Psi }}_n(\check{s}(\beta ))$ also lies in $\ker \rho$, this class is trivial. The inclusion map ${\iota }_{\ast }^F$ is injective at that basepoint by [F3], so $E_{\ast }[z]=1$. Injectivity of $E_{\ast }$ (step 2.1) gives $[z]=1$, and [F3] gives $\beta =1$ and $\gamma =1$. In particular ${\mathrm{\Psi }}_n(\mathrm{im}\check{s})\cap \ker \rho =\{1\}$, which is the uniqueness statement used below.

The free basis by the left-inverse argument. Let $K:=\ker \rho \le P{B}_n$ be the free kernel, and let $F:=F(a_1,\dots ,a_{n-1})$ be the abstract free group. The map $\mathrm{ȷ}:F\to K$ given by $\mathrm{ȷ}(a_j):={\mathrm{\Psi }}_n(A_{jn})$ is an isomorphism by the free-basis clause of [F2]. Write $w_i:=a_{i+1}{a}_{i+2}\cdots a_{n-1}\in F$ and $t_i:=w_i^{-1}{a}_i{w}_i\in F$. By step 2.2 and the homomorphism property of ${\mathrm{\Psi }}_n$, $\mathrm{ȷ}(t_i)={\mathrm{\Psi }}_n(\phi (x_i))$. First, the elements $t_1,\dots ,t_{n-1}$ generate $F$: downward induction on $i$ shows $⟨t_i,a_{i+1},\dots ,a_{n-1}⟩=⟨a_i,\dots ,a_{n-1}⟩$, because $a_i=w_i{t}_i{w}_i^{-1}$ lies in the left side and $t_i$ lies in the right side; at $i=n-1$ one has $w_{n-1}=1$ and $t_{n-1}=a_{n-1}$. Second, they are independent: define an endomorphism $\theta$ of $F$ by descending recursion on $i$ by $\theta (a_i):=\theta (w_i)a_i\theta (w_i{)}^{-1}$, where $\theta (w_i)$ is the image of the word $w_i$ under the already defined $\theta$ on the generators $a_{i+1},\dots ,a_{n-1}$ (at $i=n-1$ this gives $\theta (a_{n-1})=a_{n-1}$); this prescription specifies an element $\theta (a_i)\in F$ for every free generator, hence defines a unique endomorphism $\theta \text{\hspace{0.17em}⁣}:F\to F$. Then $\theta (t_i)=\theta (w_i{)}^{-1}\theta (a_i)\theta (w_i)=\theta (w_i{)}^{-1}\theta (w_i)a_i\theta (w_i{)}^{-1}\theta (w_i)=a_i$ for every $i$. If ${\rho }^{\mathrm{\prime }}\text{\hspace{0.17em}⁣}:F\to F$ denotes the homomorphism with ${\rho }^{\mathrm{\prime }}(a_i)=t_i$, then $\theta ({\rho }^{\mathrm{\prime }}(a_i))=\theta (t_i)=a_i$, so $\theta \circ {\rho }^{\mathrm{\prime }}$ is the identity on a free basis, hence $\theta \circ {\rho }^{\mathrm{\prime }}={\mathrm{id}}_F$ and ${\rho }^{\mathrm{\prime }}$ is injective. Therefore $t_1,\dots ,t_{n-1}$ freely generate $F$, and applying ${\mathrm{\Psi }}_n^{-1}\circ \mathrm{ȷ}$ carries this free basis to the free basis $\phi (x_1),\dots ,\phi (x_{n-1})$ of ${\mathrm{\Psi }}_n^{-1}(K)$.

The three claims. (i) is step 2.2 together with step 3.2, the free-basis clause of [F2] being what turns ${\mathrm{\Psi }}_n(A_{1n}),\dots ,{\mathrm{\Psi }}_n(A_{n-1,n})$ into a basis of the free kernel. For (ii) and (iii), note first that $\phi (W_1)$ is a product of the pure braids $\phi (x_i)$ of step 2.2 and that $\phi (W_2)=\phi (W_1{)}^{-1}$ because $\phi (W_1{W}_2)=1$; by (i) the elements $\phi (x_1),\dots ,\phi (x_{n-1})$ generate $\ker \rho$, so both $\phi (W_1)$ and $\phi (W_2)$ lie in the free kernel. The lower-rank word $W_2$ has the same reading in the two models, so the endpoint permutation of its n-strand realization is the identity on the labels $1,\dots ,n-1$ (the last strand is fixed) and the permuted labels $1,\dots ,n-1$ agree with those of its (n-1)-strand realization; since the permutation of $\phi (W_2)$ is trivial, the (n-1)-strand realization is a pure braid and step 1.2 identifies $\phi (W_2)=\check{s}({\phi }_{n-1}(W_2))$ in $G_n^{\mathrm{p}\mathrm{u}\mathrm{r}\mathrm{e}}$. Hence ${\mathrm{\Psi }}_n(\phi (W_2))\in {\mathrm{\Psi }}_n(\mathrm{im}\check{s})\cap \ker \rho =\{1\}$ by step 3.1, so $\phi (W_2)=1$; then injectivity of $\check{s}$ (step 2.1) gives ${\phi }_{n-1}(W_2)=1$, which is (iii), and $\phi (W_1)=\phi (W_2{)}^{-1}=1$. Finally, since $x_1,\dots ,x_{n-1}$ are carried to a free basis of $\ker \rho$ by (i), the homomorphism from the free group on the letters $x_1,\dots ,x_{n-1}$ to $G_n^{\mathrm{p}\mathrm{u}\mathrm{r}\mathrm{e}}$ sending $x_i\mapsto \phi (x_i)$ is injective; a word in this alphabet whose image is $\phi (W_1)=1$ is therefore freely trivial, and each cancellation of an adjacent pair $x_i^{\pm 1}{x}_i^{\mp 1}$ expands, after writing out both expanded words $x_i^{\pm 1}$, into a sequence of permitted deletions of adjacent inverse $\sigma$-pairs. This is (ii). ∎