How to show that an open map $f $ implies the surjectivity of $f'$ in a dense set Let $f$ be a $C^1$ map from $U\to \mathbb{R}^m$, where $U$ is an open set in $\mathbb{R}^n$, $n\geq m$. Then we know that if $f'$ is surjective everywhere, then $f$ is open.
My question is whether the converse is valid,  if the answer is no, can we prove the following weaker conclusion:
If $f$ is open, $f'$ is surjective in a dense set of $U$
 A: I expect that the answer is negative when $n>m$; that is, $C^1$ regularity is not enough. An important paper on the subject is A singular map of a cube onto a square by Robert Kaufman, J. Differential Geom. 14 (1979), no. 4, 593–594 (1981). In this two-page note Kaufman constructs a $C^1$ map from $[0,1]^3$ onto $[0,1]^2$ whose derivative has rank at most $1$ at every point. Therefore, the set of points where the derivative is surjective is empty. A more elaborate construction  by S.M. Bates in On the image size of singular maps I takes Hölder continuity of the derivative into account.
Although in these examples the image of $f$ contains an open subset of $\mathbb R^m$, I do not know if they are, or can be modified into, open maps. A starting point would be to go through the papers that cite Kaufman's example, some of which are very recent.
A: Provided $f$ is of class $C^k$ for some $k\ge \max(n-m+1,1)$, your weaker conclusion follows from Sard's theorem.  Suppose the set $R$ of points where $df$ is surjective (i.e., the set of regular points of $f$) is not dense in $U$.  Then $V = U\setminus \overline{R}$ is a nonempty open subset of $U$.  Since every point of $V$ is a critical point of $f$, Sard's theorem implies that $f(V)$ has measure zero in $\mathbb R^m$, so it cannot be open.
A: If f is defined on na open subset of Rm with values on Rn and is of class C1, then f open implies that f'(x) is surjective on na open and dense subset of U.
