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I'm trying to solve the following problem. Some feedback would be really helpful.

Suppose that $f:M \to N$ is a $C^\infty$ map, $M$ and $N$ are compact connected $n$-manifolds, and $rank(df) = n$. Show that $f$ is a covering map.

My attempt:

By the inverse function theorem, we know that $\forall p \in N: f^{-1}(p)$ is locally well defined. However, I want to show that for each $y \in N$ there is a neighborhood $U$ containing $y$ such that $\pi^{-1}(U)$ consists of $\bigsqcup_{d \in D} V_d$ where each $v_d$ is homeomorphic to $M$. Not sure how to go about constructing this $U$.

Arctic Char
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Math_Day
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  • Don't you have the surjective hypotesis?
  • I don't think you would conclude that $V_d$ is homeomorphic to $M$, once $V_d$ is open and $M$ is a compact connected manifold.
    • – Ygor Arthur Mar 27 '23 at 23:20
  • I suggest translating the statement $f^{-1}(p)$ is locally well-defined by applying the definition. That should give you a big clue on how to find $U$ for each $y$. – Lee Mosher Mar 27 '23 at 23:23
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    I have some ideas in mind.
    1. Prove that for any $p \in N$, $f^{-1}(p)$ is a finite set.
    2. For each $y \in f^{-1}(p)$ you will get $f|_{U_i}: U_i \rightarrow V_i$ diffeo between open sets.
    3. Consider the open set $V=\cap V_i$ and $\tilde{U}_i=U_i \cap f^{-1}(V)$.
     – Ygor Arthur Mar 27 '23 at 23:24
  • @YgorArthur (1) It's possible that the question missed that detail but I'm not sure. (2) Isn't $V_d$ being homeomorphic to $M$ what i means for $f$ to be a covering map? https://en.wikipedia.org/wiki/Covering_space – Math_Day Mar 27 '23 at 23:28
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    It is almost a duplicate of this one. If by covering you meant a topological covering, then it exactly answers your question. Indeed by applying the proof there, one can show that your $f$ is a smooth covering map. – Arctic Char Mar 27 '23 at 23:47
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    Found: topological-covering + local-diffeomorphism-gives-smooth-covering – Arctic Char Mar 27 '23 at 23:51
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    @YgorArthur The image is closed (for $M$ is compact and $f$ continuous) and open (by the inverse function theorem). As $N$ is connected, $f$ is surjective. – Didier Mar 28 '23 at 09:47
  • @ArcticChar I'm pretty sure it's a topological covering, but what other kinds of coverings are there? – Math_Day Mar 31 '23 at 21:09