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)



Braided coverings and braided monodromy
Here is one possible interpretation of your question, I'm not sure it's what you're looking for.


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).


