TITLE: Is every map of rank smaller than r dominated by a constant rank map?
QUESTION [5 upvotes]: 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.
The intuition is that a map of rank $\le r$ is "more constant" than a map of rank exactly $r$.

REPLY [7 votes]: 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.)