Compact manifold, regular value

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

• The fiber $F^{-1}(y)$ is compact and discrete, and is thus finite. – Georges Elencwajg Nov 7 '15 at 18:28

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 $x_n \rightarrow x$.
• Why is it possible to find such a nbh around $x$ such that it doesn't contain any other points of the fiber? – Algebear Oct 8 '18 at 13:03
• @GuusPalmer: Since $M$ and $N$ are of the same dimension, a point $y \in N$ is a regular value if for all $p \in F^{-1}(y)$ the differential $dF|_p$ is one-to-one and onto. By the inverse function theorem, any such $p$ has a neighborhood on which $F$ is one-to-one and so in particular it doesn't contain any other point of $F^{-1}(y)$. – levap Oct 9 '18 at 15:26
Here is another approach. Since $F^{-1}(y)$ is a closed subset of the compact space $M$, then $F^{-1}(y)$ is compact. For any $p \in F^{-1}(y)$, we have a nbd $U_p$ of $p$, and nbd $V$ of $y$ such that $F|_{U_p} : U_p \to V$ is a diffeomorphism. Since $F(p)=y$ and $F|_{U_p}$ is one to one, then there is no other points of $F^{-1}(y)$ in $U_p$. So we have $$U_p \cap F^{-1}(y) = \{p\}.$$ This means that $\{p\}$ is open in $F^{-1}(y)$. Since $p \in F^{-1}(y)$ arbitrary, then all points in $F^{-1}(y)$ is open in $F^{-1}(y)$, and we have an open cover $\{\{p\} : \forall p \in F^{-1}(y) \}$ for $F^{-1}(y)$. By compactness, there is finite subcover $\{\{p_i\} : i=1,\dots,k \}$. This clearly can't happen if $F^{-1}(y)$ is infinite. Hence, $$F^{-1}(y) = \{p_1,\dots,p_k\}.$$