Let $M$, $N$ be two manifolds of the same dimension.

A map from $M$ to $N$ is regular provided its tangent map is one to one.

A map from $M$ to $N$ is a covering map provided each point in $N$ has a neighborhood that is evenly covered (its preimage split into disjoint open sets in $M$, each mapped diffeomorphically onto this evenly covered neighborhood).

I need to prove the following:

Let $F:M \to N$ be a regular map. If $F^{-1}(q)$ contains the same finite number of points for all $q \in N$, then $F$ is a covering map.

My questions are, what is the use of the condition '$F^{-1}(q)$ has the same finite number of points for all $q \in N$' other than to show $F$ is onto?

How to prove the statement? In particular, how to prove that the preimages are disjoint?

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    $\begingroup$ If the condition were only being used because it implied that $F$ were onto, then you could "wrap" the open unit interval around a circle (of circumference slightly less than $1$) to get a regular onto map which isn't a covering map. $\endgroup$ – Aaron Dec 13 '13 at 2:38
  • $\begingroup$ Could you explain how to prove the statement? $\endgroup$ – noot Dec 13 '13 at 2:45
  • $\begingroup$ What are $M$ and $N?$ $\endgroup$ – Igor Rivin Dec 13 '13 at 2:49
  • $\begingroup$ both are n dimensional manifolds $\endgroup$ – noot Dec 13 '13 at 2:51
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    $\begingroup$ I am guessing they need to be connected... $\endgroup$ – Igor Rivin Dec 13 '13 at 3:17

Well, use the fact that the spaces are Hausdorff together with the inverse function theorem (invariance of domain) to show that the set of points evenly covered is open and closed.

  • $\begingroup$ what is the use of the condition '$F^{-1}(q)$ has the same finite number of points for all q in N' other than to show F is onto? $\endgroup$ – noot Dec 13 '13 at 3:39
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    $\begingroup$ You won't know until you try to carry out the argument as I had suggested. $\endgroup$ – Igor Rivin Dec 13 '13 at 3:41

Use the condition on the cardinality of the fiber of $F$ to verify that $F$ is proper. Afterwards, apply Ehresmann's theorem.


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