When do regular values form an open set?

Solution 1:

I think I can prove the converse of your result: if every smooth function $f: M \to N$ has regular values that form an open subset of $N$, then $M$ is compact.

I will first prove this statement for $N=\mathbb{R}^n$. Suppose, for contradiction, that $M$ is not compact. Then there is a countable locally-finite open cover $U=\{U_i\}$ of $M$ and a countable set of points $\{x_i\}$ with $x_i\in U_i$ and $x_i \not\in U_j$ for $i\neq j$.

Let $\{\phi_i\}$ be a partition of unity subordinate to $U$. Let $g(x) = \sum_i \phi_i(x)^2$. The map $g$ is a well-defined smooth function on $M$ with $g(x) \leq 1$ and $g(x_i) = 1$ for all $i$. Hence $x_i$ is a critical point of $g$, and therefore a critical point of $\phi_i$, for every $x_i$.

Let $S$ be the set of rational numbers in $[0,1]^n$. The set $S$ is countably infinite, so there exists a bijection $q: \mathbb{N} \to S.$

Finally, let $$f(x) = \sum_i q(i) \phi_i^2(x).$$ $f$ is a smooth function from $M$ to $[0,1]^n$. Since $x_i$ is a critical point of $\phi_i$, it is a critical point of $f$, with $f(x_i) = q(i)$. Hence no element of $S$ is a regular value of $f$. But by Sard's theorem, $f$ has a regular value $v\in[0,1]^n$. Since every neighborhood of $v$ contains a point of $S$, the set of regular values of $f$ can't be open, a contradiction.

Replacing $[0,1]^n$ with a suitable subset of a chart of $N$ gives the result for arbitrary manifold $N$.

Solution 2:

The set of critical points is closed. You want that the image of this set under $f$ be closed. What about demanding that $f$ is closed? A condition that implies that $f$ is closed is to demand that $f$ is proper (i.e. preimages of compact sets are compact).