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?
Why this is a covering map between compact Riemann surfaces?
1
$\begingroup$
complex-analysis
algebraic-topology
riemann-surfaces
-
2Do you understand why $F : U \to V$ is a local homeomorphism? Do you know that a proper local homeomorphism is a covering map? – 2012-07-26
-
0NO I do not understand and know – 2012-07-26