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Prove that if $F: \mathbb R^n \to \mathbb R^n$, $n \geq 2$, is a smooth proper map with finitely many critical points, then $F$ is surjective.

Here is my work so far: the set $R$ of all regular points of $F$ is open and path-connected, because $n \geq 2$, and hence connected. For any $u \in R$ we have that $F_{*, p}$ for any $p \in F^{-1}(u)$ is surjective linear map from $\mathbb R^n$ to $\mathbb R^n$ so that it's invertible; thus, by the inverse function theorem, $F$ is locally invertible from a neighborhood of $p$ to a neighborhood $U$ of $u$. By the inverse function theorem again, all points of $U$ are regular values.

I am not sure how to proceed from here. I know that $R$ is path connected. If $v \in R$ is another regular point, then there exists is a path entirely contained within $R$ going from $u$ to $v$. IF there exists a sequence of points that are in the image of $F$ that connects $u$ and $v$ in a dense way (I'm being very loose in language here, I hope you will get the picture), then we'll be done by using the compactness of the path and the inverse function theorem. However, I am not sure if this a priori is the case.

I am not sure how to use the fact that $F$ is a proper map. How can I solve this problem?

Squirrel-Power
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    I'd try to prove that the image of $F$ is also closed (by naively using sequences). Being open and closed, it must equal $\Bbb R^n$ by connectedness. – Ivo Terek Jan 17 '24 at 00:32
  • By critical point do you mean a point $x$ where $dF_x$ has rank $<n$? – Ted Shifrin Jan 17 '24 at 00:58
  • @TedShifrin yes that is correct – Squirrel-Power Jan 17 '24 at 00:58
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    What makes you think this claim is true? Is $R^n$ diffeomorphic to the open ball? – Moishe Kohan Jan 17 '24 at 01:00
  • @IvoTerek thank you for your input, but I still am not certain how to do that. For the image of $F$ being open, if $F(x)$ is a regular point, then by what I have shown $F(x)$ has a neighborhood contained in the image of $F$. However, if $F(x)$ is a critical point, we can have a neighborhood of $U$ of $F(x)$ whose only critical points are $F(x)$ and such that there is a sequence of regular points that are in the image converging to $F(x)$. I also am not sure how to show that $F(x)$ is closed. Indeed, if $F(x_m) \to y \in \mathbb R^n$, how can I show that $y$ is in the image? – Squirrel-Power Jan 17 '24 at 01:04
  • @MoisheKohan thank you for your comment. I forgot to state that $F$ must be a proper map. – Squirrel-Power Jan 17 '24 at 01:06
  • Then start by reading my answer https://math.stackexchange.com/questions/2092954/this-function-must-be-open-if-these-points-are-isolated/2097983#2097983 – Moishe Kohan Jan 17 '24 at 01:10
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    This short note includes a proof of your claim. Note that properness implies the image is closed. – Geoffrey Sangston Jan 17 '24 at 01:25
  • Note that in your comment you are confusing points in the domain and points in the range (values). – Ted Shifrin Jan 17 '24 at 02:19

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