I have a question when learning the Jacobian, if given a map:$f:\mathbf{R}^n\rightarrow \mathbf{R}^n$ has a nonzero Jacobian, could it be proved that $f$ is one to one?
My thought:$$\mathbf{d(f(x))}=\frac{\partial \ \mathbf{f}}{\partial \ \mathbf{x}} \mathbf{dx} $$ so when $\mathbf{d(f(x))}=0$, with nonzero determinant, we can conclude the $d(f(x))$ is only the approximation of the $\Delta{f(x)}$, so can you help me to improve the proof or give a counterexample? Thank you in advance!
Here's a counterexample: $f(x, y) = (e^{x} \cos y, e^{x} \sin y)$. This has nonzero Jacobian (check it!) but not one-to-one: $f(0, 0) = f(0, 2 \pi)$. (This function actually the same as the function $\exp: \mathbb{C} \to \mathbb{C}$.)
However, once the Jacobian is nonzero and constant, your statement would be true, but it is a famous conjecture which is still open.