When is the rank of Jacobian constant? 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 $C^{\infty}$?

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.
