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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 $r$.

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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.)

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  • $\begingroup$ Thanks; In afterthought, perhaps it's more interesting to assume the existence of at least one map $M \to N$ of rank $r$, so we might have a chance for a positive answer. I guess however, that even under this additional assumption, the answer would be negative in general. $\endgroup$
    – Asaf Shachar
    Commented Aug 14, 2017 at 17:14
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    $\begingroup$ @AsafShachar: That's correct. There is an example $f:S^3\to S^2$ that has rank at most 2 everywhere whose fibers are all connected (and hence circles), such that there is no smooth map $\tilde f:S^3\to M^n$ of constant rank $2$ for any manifold $M$ whose fibers are subsets of the fibers of $f$. Yet, there is a smooth map (the Hopf map) from $S^3$ to $S^2$ of constant rank 2. $\endgroup$
    – Robert Bryant
    Commented Aug 14, 2017 at 18:19

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