\\
}$$ Since $\pi_1(\mathcal{M}^{\#}(d,r)/G(d,1,r)) = {{\mathrm{B}}}(d,1,r)$, $\psi$ identifies the latter group with $\pi_1(\mathcal{M}(r+1)
/\mathfrak{S}_r)$. The generators of ${{\mathrm{B}}}(d,1,r)$ are then obtained in [@BMR] by taking the preimages under $\psi$ and the covering of $\mathcal{M}(r+1)/\mathfrak{S}_r \to \mathcal{M}(r+1)/\mathfrak{S}_{r+1}$. Note that this covering provides an injection between fundamental groups, hence an embedding $\tilde{\psi} : {{\mathrm{B}}}(d,1,r) \into {{\mathrm{Br}}}(r+1)$, where ${{\mathrm{Br}}}(r+1)$ denotes the usual braid group on $r+1$ strands. We choose for base point in $\mathcal{M}(r+1)$ the point $x = (0,x_1,\dots,x_r)$ with the $x_i \in \R$ and $x_{i+1} \ll x_i$, and for generators of the usual braid group $\mathcal{M}(r+1)/\mathfrak{S}_{r+1}$ the elements $\xi_0, \xi_1,\dots,\xi_{r-1}$ as described below :

![image](Artingen.pdf)

Then (see [@BMR]), the group $\pi_1(\mathcal{M}(r+1)/\mathfrak{S}_r)$ is generated by $\xi_0^2,\xi_1,\dots,\xi_{r-1}$. The element $\xi_0^2$ is the class in $\mathcal{M}(r+1)$ of the loop $(\frac{x_1}{2}(1-e^{2 \ii \pi t}), \frac{x_1}{2}( e^{2 \ii \pi t} +1),x_2,
\dots,x_r )$. Taking its image by $p$ provides a loop based at $(-x_1,-x_2,\dots,-x_r)$ described by $(-x_1 e^{2 \ii \pi t},
\frac{x_1}{2} (1 - e^{2 \ii \pi t}) -x_2,\dots,\frac{x_1}{2} (1 - e^{2 \ii \pi t}) - x_r)$. Since $|x_i | \ll |x_{i+1}|$, this path is homotopic to $(-x_1 e^{2 \ii \pi t}, -x_2, \dots, -x_r)$, both in $\mathcal{M}^{\#}(r)$ and in $\mathcal{M}^{\#}(r)/\mathfrak{S}_r$. Letting $a_i = -x_i$, we have 