# 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

## 2 Answers

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.

• 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
• You won't know until you try to carry out the argument as I had suggested. – 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.