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Let $F:X\rightarrow Y$ be a non constant holomorphic map between compact Riemann surfaces , If we delet the branch points (in $Y$) of $F$ and all their pre-images(in $X$), we obtain a map $F:U\rightarrow V$ between 2- manifolds which is a covering map in the sense of topology: every point of the target $V$ has an open neighborhood $N\subseteq V$ such that the inverse image of $N$ under $F$ breaks into a disjoint open sets $M_i\subseteq U$ with the map $F$ sending each $M_i$ homeomorphically onto $N$, I do not understand why this is a covering map and I can not visualize why inverse image of $N$ is like that, please could any one make me understand with an example of map between known compact riemann surfaces?

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Do you understand why $F : U \to V$ is a local homeomorphism? Do you know that a proper local homeomorphism is a covering map? –  PseudoNeo Jul 26 '12 at 10:08
NO I do not understand and know –  Une Femme Douce Jul 26 '12 at 10:18

1 Answer 1

The important point is that, once you have deleted all the branch points and their preimages, the function $F : U \to V$ is a differentiable function whose derivative is everywhere invertible. So, by the implicit function theorem, $F$ is a local homeomorphism.

Now, local homeomorphism is a notion quite close to “covering map”: a map is a local homeomorphism if for every $x \in U$ you have neighbourhoods $\mathscr V_x$ and $\mathscr V_{y}$ such that $f$ is an homeomorphism $\mathscr V_x \to \mathscr V_y$ ($y$ being $f(x)$). Roughly, it being a covering map means that for every $y \in V$ you can find a neighbourhood $\mathscr V_y$ such that for every preimage $x$ there is a neighbourhood $\mathscr V_x$ such that $f$ induces an homeomorphism $f : \mathscr V_x \to \mathscr V_y$. (I'm conscious that this “definition” isn't equivalent to the correct one. But that's what is important: in a covering map, you have one neighbourhood in the target which works uniformly for every preimage). With this slogan in mind, it is quite easy to cook up examples of local homeomorphism which aren't covering maps when a point has infinitely many preimages. But it's basically the only problem, as shows the classical proposition : a proper map $f$ between locally compact spaces is a covering map if and only if it is a local homeomorphism. I'm sure you can find a proof in any textbook about covering maps, but the key is that properness allows you to have a uniform control on all the preimages of your $\mathscr V_y$.

So, your map $F : U \to V$ is a local homeomorphism, $F : X \to Y$ was clearly proper because $X$ is compact, so now you only have to check that deleting the branch points and all their preimages hasn't destructed properness. (If you think of a proper map as a map such that $f(x)$ goes to infinity as soon as $x$ does, it should be quite clear because of the continuity of $F : X \to Y$. And you have your proof!

(I reckon that what I have said is nothing like a rigourous proof. But you can find such a proof with all the necessary details in any textbook about compact Riemann surfaces, so I thought explaining to you why this result is reasonable was the best thing to do.)

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