thus need only consider the case $i = 0$. Let $\alpha = -(b_1+\dots+b_r) \ll 0$, $\underline{b}_0
= (\alpha,b_2,\dots,b_r)$, $X = \mathcal{M}(e,r)$ and $X_0 = \{  (z_1,\dots,z_r) \ | \ z_i \neq z_j, z_1+\dots+z_r = 0 \}$. By [@BMR] we get an embedding $\pi_1(X_0/\mathfrak{S}_r, 
\underline{b}_0) \into \pi_1(X/G(e,e,r),\underline{b})$, natural only up to the choice of a path from $\underline{b}$ to $\underline{b}_0$ in $Y$. The line segment $\gamma$ from $\underline{b}$ to $\underline{b}_0$ provides such a natural choice.

We now need to prove that composing the positive detours from $\underline{b}$ with this path provides the positive detours from $\underline{b}_0$, up to homotopy in $Y = \{  (z_1,\dots,z_r) \ | \ z_i \neq z_j \}$. For $s_3,\dots,s_r$ this is true because the first component of the first path can be homotoped to the second one in $(\R_{\leq b_1}, b_1)$. For $t_0$ we let $\sigma_0$ and $\sigma$ denote the positive detours in $P^0$ from $(b_1,b_2)$ to $(b_2,b_1)$ and from $(\alpha,b_2)$ to $(b_2,\alpha)$, respectively. Let $\gamma, \gamma'$ denote the line segments $(b_1,b_2) \to (\alpha,b_2)$ and $(b_2,\alpha) \to (b_2,b_1)$. We need to prove that $\sigma_0$ is homotopic to $\gamma' \sigma \gamma$ in $\{ (z_1,z_2) \in \C^2 \ | \ 
z_1 \neq z_2 \}$, the other coordinates $z_3,\dots,z_r$ being 