Can you help me with this problem? Thanks

Let $f:M->N$ be a proper nonsingular smooth map between connected manifolds. Dim(M) = dim(N). Show f is a covering map.


So here is what I have so far, sorry for not posting before hand, I wanted hints, had no idea how to start, then thought more and came up with this much, not sure if it's correct though:

Let $p \in N$

$\exists K$ compact subset of N (and such subset exists, because $\exists \phi $ homeomorphism from N to $R^n$, and $R^n$ contains a compact subset containing $\phi (p)$, and since $\phi ^{-1}$ is continuous, then $\phi ^{-1}$ of that compact subset will be compact in N and will contain p).

So K is compact containing p, then there is an open set U that contains p. Then since f is proper, $f^{-1}(K)$ will be compact in M, and $f^{-1}(U) \subset f^{-1}(K)$

$f^{-1}(U)$ has to be finite, since there is finite subcover of $f^{-1}(K)$ and M is Hausdorff. So I can get open sets around each point in $f^{-1}(U)$ and shrink them so that they are disjoint.

So now I need to show that those open sets are mapped into U diffeomorphically. I know that f is injective since it's nonsingular, and dimM=dimN, so f is bijection, so $f^{-1}$ exists, but I am not sure why it's smooth.

In wikipedia it says if f is constant rank, then f bijective means f is diffeo, but I am not sure how to show that f is constant rank.

I'll appreciate any feedback or hints to help me figure this problem out :)

And I am not sure what else I need to show after showing f is diffeo, cause I have not used that N is connected. thanks :)

  • $\begingroup$ I edited my question! $\endgroup$ – lll Oct 24 '14 at 7:16
  • $\begingroup$ What's your definition of "nonsingular"? $\endgroup$ – Jack Lee Oct 24 '14 at 14:14
  • $\begingroup$ @JackLee It was not defined in my class, but online it says if ker{f} = 0, so I am using that $\endgroup$ – lll Oct 24 '14 at 14:44
  • $\begingroup$ That definition only applies to a linear map. I think you need to find out what definition the person who posed that problem had in mind. $\endgroup$ – Jack Lee Oct 24 '14 at 14:46
  • $\begingroup$ Oh, ok, I'll check that. Thanks @JackLee $\endgroup$ – lll Oct 24 '14 at 14:50

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