Rank of an injective function

Question

We define $f: \mathbb R^m \to \mathbb R^n$ such that $f\in C^1$ and is injective.

Is the following statement true ? If so, how to prove it ?

If $f$ is injective then $D_f$ is injective and has a rank equal to $m$

Answer 1

Consider the function

$f:\Bbb R \to \Bbb R, \; f(x) = x^n, \tag 1$

for odd

$n \in \Bbb N; \tag 2$

then $f(x)$ is injective but

$Df(x) = f'(x)\; dx = nx^{n - 1} \; dx, \tag 3$

whence

$Df(0) = f'(0)\; dx = 0; \tag 4$

thus $Df(0)$ is neither injective nor of rank $1$ at $x = 0$.

Answer 2

Consider $f(x)=x^3$ it is injective, but its differential at $0$ is zero, so it is not injective.