# When do regular values form an open set?

Let $f:M\to N$ be a $C^\infty$ map between manifolds.

When is the set of regular values of $N$ an open set in $N$?

There is a case which I sort of figured out:

• If $\operatorname{dim} M = \operatorname{dim} N$ and $M$ is compact, it is open by the following argument (fixed thanks to user7530 in the comments):

Let $y\in N$. Suppose $f^{-1}(y)\neq \emptyset$. The stack of records theorem applies: $f^{-1}(y)=\{x_1,\dots,x_n\}$ and there is an open neighborhood $U$ of $y$ such that $f^{-1}(U)=\bigcup_{i=1}^n U_i$, with $U_i$ an open neighborhood of $x_i$ such that the $U_i$ are pairwise disjoint and $f$ maps $U_i$ diffeomorphically onto $U$.

Now every point in $f^{-1}(U)$ is regular, since if $x_i'\in U_i$, then $f|_{U_i}:U_i\to U$ diffeomorphism $\Rightarrow$ $df_{x_i'}$ isomorphism (thanks to user 7530 for simplifying the argument).

Now suppose $f^{-1}(y)=\emptyset$. Then there is an open neighborhood $V$ of $y$ such that every value in $V$ has no preimages. Indeed, the set $N\setminus f(M)$ is open, since $M$ compact $\Rightarrow$ $f(M)$ compact, hence closed. Therefore $V$ is an open neighborhood of $y$ where all values are regular, and we are done.

Can we remove the compactness/dimension assumptions in some way?

• Why do you need $f$ surjective? If $y$ is a regular value, then there exists a neighborhood $U$ of $y$ with every point of $f^{-1}(U)$ a regular point; hence every point in $U$ is a regular value. (Note that you don't need $V$) – user7530 Jan 17 '13 at 20:59
• @user7530: Why does such a $U$ exist? – Bruno Stonek Jan 17 '13 at 21:08
• The Stack of Records Theorem; it's the same $U$ you use. – user7530 Jan 17 '13 at 21:15
• If $M$ is compact and $f$ is continuous, $f(M)$ compact and hence closed, so $N\setminus f(M)$ is open. So the points without a preimage are no problem (but you're right taht they must be considered separately.) – user7530 Jan 17 '13 at 21:29
• You still don't need $V$: since $f$ restricted to $U_i$ is a diffeomorphism, $Df$ has full rank for every $x\in U_i$ and hence $x$ is a regular value. – user7530 Jan 17 '13 at 21:40

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 $$N$$.

• Now generalized to arbitrary $N$. – user7530 Jan 20 '13 at 23:02

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).