Compact manifold, regular value

Question

I am still trying to convince myself about a fact which I've seen in Milnor's book ''Topology from the differentiable viewpoint'': Let M and N be manifolds of the same dimension. If M is a compact manifold and $y \in N$ is a regular value and $F: M \rightarrow N$ is a smooth map, then $F^{-1} (y)$ is a finite set (possibly empty). I tried thinking on this direction but still I don't see how to conclude that $F^{-1} (y)$ has to be a finite set: M is a manifold so it is an Hausdorff space $\Rightarrow$ every point is a close set, furthermore because $F$ is smooth then is also continuos (since differentiability implies continuity) $ \Rightarrow F^{-1} (y)$ is closed in $M$. But $M$ is compact so every close subset of a compact subset is also compact then also $F^{-1}(y)$ is compact. For Hypothesis $y$ is a regular value which implies that $F^{-1}(y)$ contains only regular points, by the implicit function theorem if $x \in F^{-1}(y)$ then there exist an open neighborhood $U$ of $x$ where F restricted to U is a local diffeomorphism. How can I conclude that $F^{-1}(y)$ is a finite set? I told by using compactness but I think I miss something

Answer 1

Assume that $F^{-1}(y)$ is infinite and pick a sequence $x_n \in F^{-1}(y)$. Since $M$ is compact, $x_n$ has a convergent subsequence (which we will rename to $x_n$) so $x_n \rightarrow x$. Since $F^{-1}(y)$ is closed, $F(x) = y$. However, as you wrote, using the inverse function theorem you see $F$ is a local diffeomorphism in a neighborhood of $x$ and in particular one-to-one, so there is a neighborhood of $x$ that doesn't contain any other point of $F^{-1}(y)$ contradicting the fact that $x_n \rightarrow x$.