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.)

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    $\begingroup$ What do you call a branched covering? Take an étale (say, double) covering $F':M'_1\rightarrow M_2$ of compact surfaces, and take for $F$ the restriction of $F'$ to $M'_1$ minus a point. Would you call $F$ a branched covering? $\endgroup$ – abx Feb 22 '18 at 16:51
  • $\begingroup$ Thanks for the example, I guess I would not call this a branched covering. Although I suppose there is a local notion of branched covering ($z \mapsto z^{n}$ in some local chart). I will clarify the question. $\endgroup$ – Nick L Feb 22 '18 at 16:55
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    $\begingroup$ Consider the blow-up of the complex plane at a point $p$, and then remove the whole exceptional divisor minus 2 points. You have an isomorphism outside $p$, whereas the preimage of $p$ consists of exactly two points, so this is clearly not a branched cover. $\endgroup$ – Francesco Polizzi Feb 22 '18 at 16:56
  • $\begingroup$ Nice example. I guess in this example the domain is not orientable. I am mainly interested in this case (apologies for not including this condition). I edit. $\endgroup$ – Nick L Feb 22 '18 at 17:05
  • $\begingroup$ Francesco, your example is not suitable for two reasons: first of all what you obtain is not a manifold, and it is of (topological) dimension $\neq 2$. $\endgroup$ – Daniele Zuddas Feb 22 '18 at 17:37

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:


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

  • $\begingroup$ The image of $f(x,y) = (x,y^2)$ is $\mathbb{R} \times \mathbb{R}_{\geq 0}$ so it cannot surject onto a 2-manifold, or at least it requires some additional reasoning. $\endgroup$ – Nick L Feb 22 '18 at 17:59
  • $\begingroup$ Actually, I see how you can use it to get a counter example. Thanks! $\endgroup$ – Nick L Feb 22 '18 at 18:02
  • $\begingroup$ that fold map $f$ is intended to be a local model near a critical point of some map $F$, which can be surjective. $\endgroup$ – Daniele Zuddas Feb 22 '18 at 18:03

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