# Braided coverings and braided monodromy

We can map from set of coverings over $X$ to symmetric group $\mathfrak{S}_n$ via monodromy (if we fix a loop at the basepoint). Also we can consider braid group $Br_n(Y)$, allow strands pass through themselves and map $Br_n \to \mathfrak{S}_n$ such a way. Can we construct a category of "braided coverings" over $X$ with morphisms to category of coverings over $X$ and to category of subgroups of fixed subrgoup of $Br_n(Y)$ ("$Y$-braided monodromy" group of $X$) such that the square will be commutative?

(edited at 16:48 UTC, June 18)

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I don't understand the question, but it looks like if I should... –  Fernando Muro Jun 18 '13 at 15:39

An $n$-sheeted covering of $X$ is classified by a map $X\to B\mathfrak{S}_n$. (On taking the induced map on fundamental groups, and picking an element of $\pi_1(X)$, I believe you get the monodromy you describe).
Now the homomorphism $Br_n\to \mathfrak{S}_n$ induces a map $BBr_n\to B\mathfrak{S}_n$ of classifying spaces. Forming the pullback in the category of topological spaces, $$\begin{array}{ccc} Y & \to & BBr_n \newline \downarrow & & \downarrow \newline X & \to & B\mathfrak{S}_n \end{array}$$ we get a space $Y$ with a cover with $Br_n$ monodromy, which fibers over $X$ with fiber $BPr_n$(the classifying space of the pure braid group).
This link, giving explicit models for these classifying spaces, might be helpful: mathoverflow.net/questions/56363/… In particular, it states that a braided cover is just a cover with an embedding up to isotopy in $X \times \mathbb R^2$. –  Will Sawin Jun 18 '13 at 18:22