Let $M\_{1},M\_{2}$ be (possibly non-compact) 2-dimensional, connected, smooth, orientable manifolds of finite topological type. Suppose you have smooth, surjective map $F:M\_{1} \rightarrow M\_{2}$, and the pre-image of each point in $M\_{2}$ is finite. Furthermore suppose that there exists $K>0$ such that $|F^{-1}(p)| \leq K$ for all $p \in M\_{2}$. Must $F$ locally be a branched covering?


(I already asked this question here <https://math.stackexchange.com/questions/2660388/finite-pre-images-implies-branched-cover> and had no response.)



Finite pre-images implies (local) branch cover?
No. Under your assumptions the map can have fold singularities (and also other kinds of singularities), namely those of the form $f(x,y) = (x, y^2)$ in local coordinates. 


In order to have a local branched covering, the map needs to be open. In addition, for having a branched covering, you need also a completeness condition in the style of R. Fox. See references:


[Church](https://projecteuclid.org/euclid.mmj/1029002116)


Fox, Covering spaces with singularities, in “Algebraic Geometry and
Topology. A symposium in honour of S. Lefschetz”, Princeton University Press
1957, 243–257.


