When is the rank of Jacobian constant?

Question

Suppose I've got a function $f : \mathbb{R}^{n} \to \mathbb{R}^{m}$ which I know is bijective.

Considering $\mathcal{J}$, the Jacobian of $\ f$, I want to understand what can be said about the rank of $\mathcal{J}(\mathbf{x})$.

Let's say I evaluate $\mathcal{J}(\mathbf{0})$, and find that the rank of $\mathcal{J}(\mathbf{0})$ is $k$. Does this mean that the rank of $\mathcal{J}(\mathbf{x})$ is $k$ for all $\mathbf{x} \in \mathbb{R}^{n}$?

Is there a theorem regarding this?

EDIT: If this is not enough that $f$ is a bijection, what if $f$ is a homeomorphism? Or of class $C^{\infty}$?

Answer 1

No, this is not true. For example, consider $f : \mathbb{R} \to \mathbb{R}$ given by $f(x) = x^3$. Then $\mathcal{J}(x) = 3x^2$, so the rank of the Jacobian at $x = 0$ is zero, but $f$ has full rank at every other point.

Note, it follows from the Inverse Function Theorem that if $f$ has full rank at $x$, then it has full rank in a neighbourhood of $x$.

Answer 2

If $f$ is a bijection, the Jacobian need not even exist. Let $m=n$ and $f$ be a generic permutation of $\Bbb{R}^m$. A random permutation is unlikely to be continuous anywhere, much less have derivatives.

(Bi-)Continuity (being homeomorphic) isn't sufficient to guarantee a derivative exists. Consider continuous nowhere differentiable functions, for example, the Weierstrass function.

A standard counterexample to $\mathscr{C}^{\infty}$ being sufficient is $f(x) = x^3$, which is rank deficient at $x=0$ and full rank everywhere else.