# When is a regular map a covering map?

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?

• 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. – Aaron Dec 13 '13 at 2:38
• Could you explain how to prove the statement? – noot Dec 13 '13 at 2:45
• What are $M$ and $N?$ – Igor Rivin Dec 13 '13 at 2:49
• both are n dimensional manifolds – noot Dec 13 '13 at 2:51
• I am guessing they need to be connected... – Igor Rivin Dec 13 '13 at 3:17

• 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? – noot Dec 13 '13 at 3:39
Use the condition on the cardinality of the fiber of $F$ to verify that $F$ is proper. Afterwards, apply Ehresmann's theorem.