Proof Consider the following hyperplane arrangement complement.

$$ \begin{align*} W=\{w\in{\Bbb C}^n\ |\ w_i\neq \pm w_j,\ \text{for all}\ i\neq j; w_k\neq \pm 1,\ \text{for all}\ k\}. \end{align*} $$

In [Reference Callegaro, Moroni and Salvetti2, Section 3], the following homeomorphism was observed. Let ${\Bbb C}^*={\Bbb C}-\{0\}$ .

$$ \begin{align*} {\Bbb C}^*\times W&\simeq X:=\{x\in{\Bbb C}^{n+1}\ |\ x_i\neq \pm x_j,\ \text{for all}\ i\neq j; x_1\neq 0\}.\\&(\lambda, w_1,w_2,\ldots, w_n)\mapsto (\lambda, \lambda w_1,\ldots, \lambda w_n).\end{align*} $$

In [Reference Callegaro, Moroni and Salvetti2, Lemma 3.1], it was then proved that the hyperplane arrangement complement X is simplicial, in the sense of [Reference Deligne3].

From [Reference Huang and Osajda5], it follows that $FICwF$ is true for $\pi _1(X)$ , since X is a finite real simplicial arrangement complement. Hence, $FICwF$ is true for $\pi _1(W)$ , as $\pi _1(W)$ is a subgroup of $\pi _1(X)$ and $FICwF$ has hereditary property (see [Reference Roushon6]).

Next, note that there are the following two finite sheeted orbifold covering maps:

$$ \begin{align*}W\to &PB_n(Z):=\{z\in Z^n\ |\ z_i\neq z_j,\ \text{for all}\ i\neq j\}\\&(w_1,w_2,\ldots, w_n)\mapsto (w_1^2,w_2^2,\ldots, w_n^2) \end{align*} $$

and $PB_n(Z)\to B_n(Z):=PB_n(Z)/S_n$ . Here, $Z={\Bbb C}(1,1;2)$ (see [Reference Roushon6]) is the orbifold whose underlying space is ${\Bbb C}-\{1\}$ , and $0$ is an order $2$ cone point. And, the symmetric group $S_n$ is acting on $PB_n(Z)$ by permuting coordinates.

Therefore, $\pi _1(W)$ embeds in $\pi _1^{orb}(B_n(Z))$ as a finite index subgroup. Hence, $FICwF$ is true for $\pi _1^{orb}(B_n(Z))$ , since $FICwF$ passes to finite index overgroups (see [Reference Roushon6]). Next, recall that in [Reference Allcock1] Allcock showed that ${\cal A}_{\widetilde B_n}$ is isomorphic to a subgroup of $\pi _1^{orb}(B_n(Z))$ , and hence $FICwF$ is true for ${\cal A}_{\widetilde B_n}$ by the hereditary property of $FICwF$ .