\xi_0^{-2} \in \pi_1(
\mathcal{M}(r+1)/\mathfrak{S}_r, x)$, we have $\tau'_1 = \psi^{-1}(\xi'_1)
\in \pi_1(\mathcal{M}^{\#}(d,r) / G(d,1,r),y)$. As before we let $\zeta = \exp(2 \ii \pi/d)$ and $g_i \in G(d,1,r)$ being defined by $g_i.(z_1,\dots,z_r) = (z_1,
z_2,\dots,\zeta z_i, \dots, z_r)$. We let $b_i = a_i^{1/d}$. $$\xymatrix{
(b_1,\dots,b_r) \ar@/^1pc/[d]^{\xi_0^{-2}} \ar@/_5pc/@{-->}[dd]_{g_1^{-1} s_1}
\ar@/_10pc/@{-->}[ddd]_{g_1^{-1}s_1g_1}\ar@/^10pc/[ddd]_{\xi'_1}\\
(\zeta^{-1} b_1,b_2,\dots,b_r) \ar@/^1pc/[d]^{g_1^{-1}.\xi_1}\\
(\zeta^{-1} b_2, b_1,\dots,b_r) \ar@/^1pc/[d]^{g_1^{-1} s_1 . \xi_0^2}\\
(\zeta^{-1} b_2, \zeta b_1,\dots,b_r) }$$ In order to generate ${{\mathrm{B}}}(e,e,r) = \pi_1(\mathcal{M}(e,r)/\mathfrak{S}_r, x)$, and letting $e=d$, we only need to take the image of $\tau'_1,\tau_1,\dots,\tau_{r-1}$ under $i^*$ where $i : \mathcal{M}^{\#}(e,r) \to \mathcal{M}(e,r)$ is the natural inclusion. We will use the following definition.

Let $X$ be the complement of an hyperplane arrangement $\mathcal{A}$ in $\C^l$, and $v,v' \in X$. A line segment from $v$ to $v'$ is $t \mapsto (1-t) v + t v'$ for $t \in [0,1]$, If this line segment crosses exactly one hyperplane of $\mathcal{A}$ at one point, a positive *detour* from $v$ to $v'$ is a path of the form $\gamma(t) = (1-t) v + t v' + \ii t (1-t) (v-v') \eps$ for $\eps > 0$ small enough so that it and the similar paths $\gamma'$ for $0 < \eps' < \eps$ do not cross any hyperplane 