Let $Y_n$ be the category whose objects are pairs $(x,y)$, where $x$ belongs to the braid group $B_n$ and $y$ is a parenthesizing of the non-associative product of $n$ elements, for instance, $(x_1x_2)x_3$; and there is exactly one arrow between any two objects.
We have a free right action of $B_n$ given by $(x,y)b=(xb,y)$. In particular, we have an action of the pure braid group $PB_n$. If $|N(Y_n)|$ is the geometric realization, the action on the category induces another action on the geometric realization.
I want to understand the quotient $|N(Y_n)|/PB_n$, finding an homotopy equivalent space.
I believe that freeness of the action at the level of categories implies freeness on the geometric realizacion. But I have another category $X_n$ related to it, and I'd like to show that we have $|N(Y_n)|/PB_n\simeq |N(X_n)|$.
The category $X_n$ is defined to have objects $(\sigma,\pi)$, where $\sigma\in\Sigma_n$ is a permutation and $\pi$ is a parenthesizing of the non-associative product of $n$ elements (just like before). Now, the morphisms are a bit trickier. If one has pairs $(\sigma_1,\pi_1)$, $(\sigma_2,\pi_2)$, consider the tuples $(\sigma_1(1), \dots, \sigma_1(n))$ and $(\sigma_2(1),\dots, \sigma_2(n))$. A morphism $(\sigma_1,\pi_1)\to (\sigma_2,\pi_2)$ is a braid $b\in B_n$ which joins $\sigma_1(i)$ with $\sigma_2(j)$ if $\sigma_1(i)=\sigma_2(j)$. Another way to say this is that $Hom((\sigma_1,\pi_1),(\sigma_2,\pi_2))=p^{-1}(\sigma_2^{-1}\sigma_1)$, where $p:B_n\to\Sigma_n$ is the usual projection with kernel $PB_n$.
Now, the way I see the quotient of the action of $PB_n$ on $Y_n$ relating it to $X_n$ is that one is quotienting out objects of $Y_n$ so that now there is not only one morphism between objects, but there are as much as braids inducing the same permutation, which are the same morphisms we have in $X_n$. But this is the quotient $Y_n/PB_n$, which I'm not sure how it relates to $|N(Y_n)|/PB_n$, so the following question rises
Is $|N(Y_n)|/PB_n\simeq |N(Y_n/PB_n)|\simeq|N(X_n)|?$.
I don't know if this is a general result about geometric realizations and group actions, but if it is true in this case it would help me. If this is not true, how can I show that $|N(Y_n)|/PB_n\simeq |N(X_n)|$?
Reference: My question comes from this article in which a category $PaB_n$ (my $X_n$) is defined in section 3.1 and then another category $PaB_n'$ is defined in section 3.2 (my $Y_n$), and they're supposed to satisfy the relationship I'm asking for.