Prasolov and Sossinsky define a branched cover (of a 3-manifold) as
A continuous map $p: M^3 \to N^3$ such that there exists a one-dimensional subcomplex $L$ in $N^3$ whose inverse image $p^{-1}(L)$ is also a one-dimensional subcomplex, and the restriction of $p$ to $M^3 \setminus p^{-1}(L)$ is a covering
Let $N^3$ be compact, $L \subseteq N^3$ be a one-dimensional subcomplex. I would like to show that there is a 1-1 correspondence $$ \{\text{branched covers of }(N^3,L)\} \leftrightarrow \{\text{(ordinary) covers of } N^3 \setminus L\}$$
Given a branched cover $p: M^3 \to N^3$, by definition, the restriction $p: M^3 \setminus p^{-1}(L) \to N^3 \setminus L$ is an ordinary covering.
I am having some difficulty with the other direction. Given an ordinary cover of $N^3 \setminus L$, I would like to construct a branched cover so that the restriction of the branched cover is the given cover. (that is, construct an inverse of the function defined in the previous paragraph)
In the two dimensional case (where the branching set is a set of points) I tried to do the following: around each branch point find a disc (or a set homeomorphic to a disc) that includes only the chosen branch point. Removing a point from the disc gives us a subset of $N^2 \setminus L$. The given cover restricts to a cover of this punctured disc, which I would like to extend to a map on the entire disc.
I then tried to use this idea in the 3-dimensional case before realizing this extension doesn't necessarily exist.
Am I on the right track? It seems as though we could have something fishy going on around the branch points, and I don't know how to deal with it in the 2-dimensional case, much less the 3-dimensional one.
Also, it's possible I have forgotten some important hypothesis (this came up in a conversation), so feel free to add any if you feel they are necessary.