Suppose $\pi:M_1 \to M_2$ is a $C^\infty$ map of one connected differentiable manifold to another.And suppose for each $p\in M_1$,the differential $\pi_*:T_p M_1 \to T_{\pi(p)}M_2$ is a vector space isomorphism.

(a)Show that if $M_1$ is compact,then $\pi$ is a covering space projection. (b)Given an example where $M_2$ is compact but $\pi:M_1 \to M_2$ is not a covering space (but has the $\pi_*$ isomorphism property).

I want to ask is my solution right? My solution:$\pi(M_1)$ is compact in $M_2$,so $\pi(M_1)$ is closed in $M_2$.For each $x\in \pi(M_1)$,there is an open set $V\subset M_2$ contains $x$ and is diffeomorphic to an open set $U\subset M_1$.So $V$ is an open set in $\pi(M_1)$. so $\pi(M_1)$ is open and closed in $M_2$.It equals to $M_2$.

Use $\pi_*:T_p M_1 \to T_{\pi(p)}M_2$ directly and we get each $x\in M_2$ has an open neighbourhood $U$ which is covered by $\pi$ evenly.

And could the counterexample in this page Question on covering spaces be the example of (b)?

Thank you :))

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Your part (a) looks fine. For part (b), I suggest that you think carefully about what "compactness" buys you, because there is a very simple counterexample. – Aaron Mazel-Gee May 24 '12 at 21:50
@AaronMazel-Gee Thank you for your hint,but I'm so sorry that I can't think out the counterexample.I think the compactness buys me a closed manifold $M_1$ or a manifold without boundry,so if $M_1$ is not compact,we could take it as an open set or an infinite big manifold.But I can't find such an example :( – Jiangnan Yu May 26 '12 at 15:18
No, you're exactly right. If $M_1$ is an open submanifold of $M_2$, then the differential map of the inclusion is a vector space isomorphism at every point, but nevertheless we don't get a covering space. The point here is that compactness disallows such behavior. – Aaron Mazel-Gee May 27 '12 at 0:42
@AaronMazel-Gee Ah,yes!$M_1$ could be an open submanifold of $M_2$.Thank you! :)) – Jiangnan Yu May 27 '12 at 4:58