# A more general definition of branched covering.

If $$f:X\longrightarrow Y$$ is a holomorphic map between two compact Riemann surfaces, then $$f$$ is called also a branched covering map. This is because the branched points of $$f$$ form a finite set $$S\subseteq Y$$ and the map $$f$$ restricted on $$X\setminus f^{-1}(S)$$ is a topological covering map of $$Y\setminus S$$.

Now I cannot find a standard notion of branched covering for topological spaces, but I think that it should be like the following:

A continuous and surjective map $$f:X\longrightarrow Y$$ between topological spaces is a branched covering map if there exists a dense subset $$S\subseteq Y$$ such that $$f$$ restricted on $$X\setminus f^{-1}(S)$$ is a topological covering map of $$Y\setminus S$$.

Some authors say that $$S$$ should be nowhere dense and others require that $$S$$ is a finite set. So, what is the more standard definition of branched covering in the topological framework?

• I don't know about standard definitions, but requiring $S$ to be finite precludes holomorphic maps branching over varieties of positive dimension (e.g., projection away from a point representing a Fermat hypersurface in $\mathbf{P}^3$ as a branched cover of a hyperplane). – Andrew D. Hwang Mar 22 '14 at 11:06

## 2 Answers

First, in your definition you should say that $S$ is closed with empty interior (instead of $S$ being dense which is surely not what you meant).

Next, I also do not think there is a standard terminology in general. The very least people require is that $f: X\to Y$ is a local homeomorphism away from some nowhere dense closed subset $Z$ of $X$. Nobody (as far as I know) requires $Z$ to be finite (unless dealing with compact Riemann surfaces). The definition you gave (with my correction) probably will not be accepted by most algebraic geometers since it would allow a blow-down map to be regarded as a branched covering.

To remedy this, one can make the following definition (motivated by the notion of the orbifold covering):

Definition. An open continuous map $f: X\to Y$ of topological spaces is called a branched covering if the following conditions hold:

1. There exists an open and dense subset $A\subset Y$ such that $f| f^{-1}(A)\to A$ is a covering map.

2. For every point $x\in X$ there exists an (open) neighborhood $V_x\subset X$ and a finite group of homeomorphisms $G_x$ acting on $V_x$ and a homeomorphism $U_x=f(V_x) \to V_x/G_x$ which makes the following diagram commutative $$\begin{array}{ccc} V_x & \stackrel{id}{\to} &V_x\\ f\downarrow & ~ & \downarrow\\ U_x & \to & V_x/G_x \end{array}$$

In other words, locally, $f$ is modelled on the quotient by a finite group action.

Edit. One last thing: depending on the category you are working in, all maps and group actions should be from this category. For instance, if you are working in the category of complex manifolds, all maps should be holomorphic (or biholomorphic when needed) and group actions should be holomorphic too.

• I would guess that e.g. the projection $(x,e^x) \mapsto e^x$ to the second coordinate of the graph of the exponential would be considered a "topological branched covering" by most people, no? – Stephen Jan 25 at 14:08
• @Stephen: This projection is a homeomorphism to its image but is not surjective. A topologist would require a branched covering to be surjective. Hence, this map would not qualify as a branched covering, but it would be a branched covering to its image. A complex analyst might disagree. As for "most people", they would have no idea what we are talking about. – Moishe Kohan Jan 25 at 15:55
• Well, its image is $\mathbf{C}^\times$, and it's not a homeomorphism onto that, but I (leaving aside the proper ambient space for defining "most people") would definitely regard it as a "branched covering" of $\mathbf{C}^\times$ which is nowhere ramified. So, I am not sure how to interpret your comment. – Stephen Jan 25 at 18:30
• @Stephen: Oh, I see, you are taking $x$ and $y$ to be complex numbers (I was assuming that your $x$ and $y$ are real) and your map then is $exp: {\mathbf C}\to \{\mathbf C}$. I am yet to meet a topologist who would call this a covering map ${\mathbf C}\to \{\mathbf C}$, but, of course, it is a covering map {\mathbf C}\to \{\mathbf C}^\times$. As for complex analysts, it depends on the person. – Moishe Kohan Jan 25 at 18:40 • @Stephen: Here is a sample of definitions of covering maps used in the theory of Riemann surfaces (among books that I happen to have): Springer, Ahlfors-Sario, Farkash-Kra use (as you do!) a local homeomorphism as their definition and have the separate name "unlimited covering" for the topological (standard) definition of a covering map. Donaldson, Jost, Miranda, Forster and Narasimhan use the topological (standard) definition. The bottom line is that newer books tend to use the standard definition. – Moishe Kohan Jan 25 at 22:48 It depends on the setting. When you study surfaces, you usually require that the branch locus (where you are not a covering) is locally finite. In general the brach locus can be a submanifold or even a singuar set. For instance, every 3-manifold is a branched covering over the$S^3$with branch locus the figure-eight knot. So 3d-topologist will tells you that the branch-locus is a proper sub complex. People doing more algebraic stuff will say "analytic subset of codimention$\geq 1\$"

In conclusion, I would say that the more general definition would require the branch locus to be a sub-object in the appropriate category, with codimension at least one.

(note that codimension 1 objects in a holomorphic surfcace setting are points)