Let $\, f:M \to N$ be a smooth map, with rank $df \le r$ everywhere.


Does there exist a smooth map $\tilde f:M \to N$ of constant rank $r$, such that each level set of $\tilde f$ is contained in some level set of $f$?


Is it true at least locally?


(It is easy to see rank $df \le r$ is a necessary condition for the existence of such $\tilde f$).


**Motivation:** I am trying to visualize maps of small rank. In the case of constant rank maps , this is easy since the level sets form a [foliation of the domain](https://math.stackexchange.com/questions/1728372/do-the-level-sets-of-a-constant-rank-map-give-a-foliation-of-the-domain).


The intuition is that a map of rank $\le r$ is "more constant" than a map of rank exactly $r$.



Is every map of rank smaller than r dominated by a constant rank map?
No. The simplest case is $M = S^1$ and $N = \mathbb{R}$. Then any nonconstant map $f:M\to N$ has rank at most 1, but there is no smooth map from $M$ to $N$ that has constant rank $1$. 


The question would be more interesting if you were considering $\tilde f: M\to \tilde N$ of rank at most $r$ instead of $\tilde f: M \to N$. (The answer would still be 'no', but the 'counterexamples' would be more interesting. For example, any nonconstant smooth map $f:S^2\to \mathbb{R}$ has rank at most $1$, but there is no smooth map from $S^2$ to any manifold that has constant rank $1$, since the tangent bundle of $S^2$ is irreducible.) 


