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Let $E, M$ be connected manifolds and $\pi: E \to M$ a surjective local diffeomorphism with finite fibers. Is $\pi$ necessarily a covering map?

I know that this is true if $E$ is compact, or more generally if $\pi$ is a proper map or, equivalently in this case, a closed map. This is also true if every fiber has the same cardinality (Local Homeomorphism with constant fiber is a covering map).

I tried to mess around with coverings of $S^1$ and $S^1 \times \mathbb{R}$ and modify them slightly to get a counterexample, but I couldn't get it to work: either I'd get infinite fibers or no longer a local diffeomorphism.

J. C.
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    Hint: Consider maps of an open interval to the circle. https://math.stackexchange.com/questions/447168/local-homeomorphisms-which-are-not-covering-map – Moishe Kohan Jun 28 '24 at 19:09
  • @MoisheKohan Yes thank you! I dismissed the covering $\mathbb{R} \to S^1$ too quickly for not having finite fibers, but that's fixable by restricting appropriately. – J. C. Jun 28 '24 at 21:25

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