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suppose $\pi:\widetilde{M} \longrightarrow M$ be a smooth covering map and $M$ and $\widetilde{M}$ are smooth manifolds. I can't undestand why $\pi$ became both an immersion and a submersion.

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 Unless this is just a misunderstanding (a smooth covering map is not just a topological covering map which happens to be smooth), I suggest you look at mathoverflow.net/faq to understand why this question is not appropriate here. – a-fortiori Jun 4 at 13:31
A supplementary example for a-fortiori's comment: the map from the real numbers to itself given by $x \mapsto x^3$ is a covering if we are thinking of the reals as a topological manifold, but it doesn't count as a smooth covering since the definition of smooth covering requires that it be an immersion. – Kevin Walker Jun 4 at 13:43

closed as too localized by Deane Yang, Ryan Budney, Igor Rivin, Kevin Walker, Misha Jun 4 at 13:30

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