If $f: \mathbb{R}^n \to \mathbb{R}^n$ is one-to-one, is the differential necessarily one-to-one?

Question

If $f: \mathbb{R}^n \to \mathbb{R}^n$ is one-to-one, $C^1$, and has everyhere non-singular Jacobian, then for any open set $U \subset \mathbb{R}^n$, the image $f(U)$ is open

I proved the above theorem using the inverse function theorem. However, I stumbled upon a proof online that removed the necessity of the Jacobian being everywhere non-singular, instead only specifying $f$ to be one-to-one and $C^1$

The proof makes sense to me, except for one statement: "an injective map has an injective derivative"

If $f: \mathbb{R}^n \to \mathbb{R}^n$ is one-to-one, is the differential necessarily one-to-one?

Here is a link to the mentioned proof: https://gist.github.com/pervognsen/11251717

Answer 1

To make this question answered, I post my comment as answer.

The map $\Bbb R\to\Bbb R,x\mapsto x^3$, is bijective but the derivative at point $0$ is $0$, which is not injective.

Similarly, for higher dimensions, you can take the map $(x_1,...,x_n)\mapsto(x_1^3,...,x_n^3)$ as a counterexample.