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Let $\pi:M_1\rightarrow M_2$ be a surjective $C^{\infty}$ map between two connected manifolds with $d\pi$ an isomorphism.

If $M_1$ is compact, it is seen that $|\pi^{-1}(m_2)|$ is finite, so $\pi$ is a covering. If we only have $M_2$ compact, do we have a counterexample showing that $\pi$ doesn't have to be a cover?

Anonymous Coward
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1 Answers1

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The map $t\mapsto\exp(2\pi it)$ is a standard covering map from the line to the unit circle of the complex plane. If you restrict it to an interval like $(0,2)$, you get a counterexample for your question.

Andreas Blass
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