The $A_{in}$ are meridian generators of the forgetful free kernel

The free kernel to be identified. By [F1] the choice hypothesis [A1] supplies DC, so [F2] gives an injection $\kappa$ with image $\ker \phi$. The fibre $Y_n$ has free fundamental group of rank $n-1$ by [F3]. It remains to identify the individual $A_{in}$ with a single coherent meridian basis, with the inverse sign from [F6].

The two-point winding calculation. For any $0<{ϵ}_i\le h\mathrm{/}10$, let $C_i=\{z:\mathrm{\mid }z-q_i\mathrm{\mid }={ϵ}_i\}$, and let $c_i=q_i+({ϵ}_i,0)$ on it. For any simple local stem ${\tau }_i$ from $q_{i+1}$ to $c_i$ in $U_i\setminus \{q_i\}$, let ${\lambda }_i$ be the based loop that follows ${\tau }_i$, goes once counterclockwise around $C_i$, and returns along ${\bar{\tau }}_i$. The straight segment is one possible stem. For a pair in $U_i=B(m_i,R)$, $R=3h\mathrm{/}2$, write $u=(z_2-z_1)\mathrm{/}\mathrm{\mid }{z}_2-z_1\mathrm{\mid }\in S^1$, $r=\mathrm{\mid }{z}_2-z_1\mathrm{\mid }>0$, and $w=\bar{u}((z_1+z_2)\mathrm{/}2-m_i)$. Then $F_2(U_i)\cong S^1\times C,C=\{(w,r)\in \mathbb{C}\times (0,\mathrm{\infty }):\mathrm{\mid }w-r\mathrm{/}2\mathrm{\mid }<R,\mathrm{\mid }w+r\mathrm{/}2\mathrm{\mid }<R\}\mathrm{.}$ The inverse coordinate map is $(u,w,r)\mapsto (m_i+u(w-r\mathrm{/}2),m_i+u(w+r\mathrm{/}2))$, so the displayed coordinates give a homeomorphism. The condition $r>0$ and the two inverse images of the convex disk $B(0,R)$ make $C$ convex and nonempty. Hence the winding of $u$ induces an isomorphism ${\pi }_1(F_2(U_i))\cong \mathbb{Z}$. During the first positive half twist, the relative vector $z_2-z_1$ starts at $2h$, follows the upper half of the diamond path to $-2h$, and its argument increases by $\pi$. In the second positive half twist, stacking matches the exchanged endpoints, so the same ordered pair follows the lower half of the diamond from $-2h$ to $2h$; its argument increases by another $\pi$. Thus the ordered loop of ${\sigma }_i^2$ has relative winding $+1$. The point motion that fixes $q_i$ and moves $q_{i+1}$ out along the stem to $C_i$, once counterclockwise around $C_i$, and back along the stem also has relative winding $+1$: the outgoing and return stem paths cancel in winding and the circle contributes $+1$. These two loops therefore have the same class in ${\pi }_1(F_2(U_i))$. Inserting all other stationary points gives the same equality in the full ordered configuration space, because $U_i$ contains no other base point. By [F7] and the inverse-endpoint definition in [F8], the braid-to-mapping-class isomorphism ${\mathrm{\Psi }}_n^{\mathrm{m}\mathrm{c}}:G_n\to \mathrm{Mod}(D^2,Q;\mathrm{\partial }{D}^2)$ satisfies ${\mathrm{\Psi }}_n^{\mathrm{m}\mathrm{c}}([{\sigma }_i^2])={\mathrm{Push}}_{q_{i+1}}([{\lambda }_i]{)}^{-1}$.

Conjugation transports the local point push. Fix radii and local stems satisfying step 1.2. For a marked point $p\in Q$ and a based loop $\alpha$ in $\mathrm{int}{D}^2\setminus (Q\setminus \{p\})$ at $p$, let ${\mathrm{Push}}_p([\alpha ])$ be the mapping class of $F_1^{-1}$, where $F_t$ is an ambient isotopy starting at the identity, fixing the boundary and every point of $Q\setminus \{p\}$, and satisfying $F_t(p)=\alpha (t)$. This class is well defined: it is the inverse-endpoint boundary-map value of the full configuration loop with the other marked points fixed, as in [F8]. It agrees with [F8] when $p=q_n$. If $h$ represents a mapping class preserving $Q$ setwise and taking $p$ to $p^{\mathrm{\prime }}$, then $h{F}_t{h}^{-1}$ is an ambient lift of $h\circ \alpha$ fixing every other marked point; its endpoint inverse is $h{F}_1^{-1}{h}^{-1}$. Therefore, with the paths typed in the respective puncture complements, $h{\mathrm{Push}}_p([\alpha ])h^{-1}={\mathrm{Push}}_{p^{\mathrm{\prime }}}([h\circ \alpha ])\mathrm{.}$ Now put $g_i={\sigma }_{n-1}\cdots {\sigma }_{i+1}$ and $h_i=H_{n-1}\circ \cdots \circ H_{i+1}$. The rightmost factor acts first: $H_{i+1}$ takes $q_{i+1}$ to $q_{i+2}$, then $H_{i+2}$ takes it to $q_{i+3}$, and so on, while every $H_j$ fixes $q_i$ and the circle $C_i$. Since $\mathrm{\mid }{m}_i-q_i\mathrm{\mid }=h$ and ${ϵ}_i\le h\mathrm{/}10$, $C_i\subset U_i$. For $j\ge i+2$, the supports $U_j$ and $U_i$ are disjoint by [F4]. For $j=i+1$, the center of $U_{i+1}$ is at distance $3h$ from $q_i$, so every point of $C_i$ is at distance at least $3h-h\mathrm{/}10>3h\mathrm{/}2$ from that center; hence this circle too is outside the support. Thus all the stated fixes are pointwise. In particular $h_i(q_{i+1})=q_n$, $h_i(q_i)=q_i$, and $h_i(C_i)=C_i$. If ${\tau }_i$ is the local stem used in step 1.2, put ${\widehat{\tau }}_i:=h_i\circ {\tau }_i$; this is a stem from $q_n$ to $C_i$ in $Y_n$. Let ${\widehat{\gamma }}_i$ be the based loop following ${\widehat{\tau }}_i$, going counterclockwise around $C_i$, and returning along its reverse. The first-under-second word order in [F5] and the mapping-class homomorphism [F9] give the following. Put ${\mathrm{\Theta }}_n:={\mathrm{\Psi }}_n^{\mathrm{m}\mathrm{c}}{\mathrm{\mid }}_{G_n^{\mathrm{p}\mathrm{u}\mathrm{r}\mathrm{e}}}\circ {\mathrm{\Psi }}^{-1}$, the isomorphism $P{B}_n\to \mathrm{PMod}(D^2,Q;\mathrm{\partial }{D}^2)$ of [F8]. The conjugation naturality just proved yields ${\mathrm{\Theta }}_n(\mathrm{\Psi }([A_{in}]))={\mathrm{\Psi }}_n^{\mathrm{m}\mathrm{c}}([A_{in}])=h_i{\mathrm{Push}}_{q_{i+1}}([{\lambda }_i]{)}^{-1}{h}_i^{-1}={\mathrm{Push}}_n([{\widehat{\gamma }}_i]{)}^{-1}\mathrm{.}$ By [F8], this is ${\mathrm{\Theta }}_n(({\kappa }_{\ast }[{\widehat{\gamma }}_i]{)}^{-1})$. Injectivity of ${\mathrm{\Theta }}_n$ proves $\mathrm{\Psi }([A_{in}])=({\kappa }_{\ast }[{\widehat{\gamma }}_i]{)}^{-1}$ for each $i<n$.

A coherent noncrossing system of transported stem classes. For $2\le m\le n$ and $i<m$, put $h_i^{(m)}:=H_{m-1}\circ \cdots \circ H_{i+1}$, with the empty composition for $i=m-1$, and let ${\widehat{\tau }}_i^{(m)}:=h_i^{(m)}\circ {\tau }_i$ be the transported stem of step 2.1 at rank $m$. Here the local stem ${\tau }_1$ is chosen at the base case, and each ${\tau }_m$ is chosen when the rank-$m+1$ fan is added; thus the same coherent family is used in steps 1.2 and 2.1. We construct, by induction on $m$, a fan of embedded arcs ${\eta }_1^{(m)},\dots ,{\eta }_{m-1}^{(m)}$ whose relative endpoint-fixed homotopy classes are exactly those of ${\widehat{\tau }}_1^{(m)},\dots ,{\widehat{\tau }}_{m-1}^{(m)}$. At $m=2$, take $C_1$ of radius $h\mathrm{/}10$ and the straight stem from $q_2$ to $c_1$.

These meridians are a free basis. Let $\mathcal{T}$ be the final fan tree formed by the stems from $q_n$ to $c_i$. It is an embedded tree whose stems have pairwise disjoint interiors, and it meets each $C_i$ only at $c_i$. Take a closed regular neighborhood $N$ of $\mathcal{T}\cup C_1\cup \cdots \cup C_{n-1}$ in $Y_n$. It is a disk with $n-1$ holes: the tree joins the $n-1$ disjoint inner circles and has no cycles, so thickening it adds no further hole. Write $B_i$ for its $i$-th inner boundary and $B_0$ for its outer boundary. The puncture $q_i$ lies inside $B_i$, and $B_i$ is parallel to $C_i$ through an annular collar in $N$. The part of $Y_n$ inside each $B_i$ is a punctured disk and radially deformation retracts onto $B_i$; the outside complement between $B_0$ and $\mathrm{\partial }{D}^2$ is an annular collar that deformation retracts onto $B_0$. These retractions are the identity on the corresponding boundary circles, so they glue to a deformation retraction of $Y_n$ onto $N$. Choose $N$ as vertex disks and edge strips for the finite polygonal graph; collapsing each strip across its width to its core and each vertex disk onto its incident radial arcs gives a deformation retraction onto $\mathcal{T}\cup C_1\cup \cdots \cup C_{n-1}$. Collapsing the tree gives a wedge of $n-1$ circles. Each counterclockwise loop on $C_i$ is homotopic in its collar in $N$ to the counterclockwise loop on $B_i$, and the based loop that follows the $i$-th stem, traverses $C_i$ counterclockwise, and returns therefore follows the corresponding circle summand once. Thus $[{\widehat{\gamma }}_1],\dots ,[{\widehat{\gamma }}_{n-1}]$ form a free basis of $F_{n-1}$; their inverses form a free basis as well. Since $\kappa$ identifies this group with $\ker \phi$ by [F2], steps 2.1 and 3.1 prove both the free-basis assertion and the clockwise sign in the Statement.

The sign and conclusion. For $n=2$, the calculation in step 1.2 says that the raw ordered loop of $A_{12}={\sigma }_1^2$ has relative winding $+1$; the published identification [F6] inverts that loop, so its image is the clockwise meridian of winding $-1$. The same inverse is exactly the inverse-endpoint convention in steps 1.2 and 2.1, and no injectivity of an Artin presentation is used. This completes the proof. ∎