Inverse Function Theorem on the plane I have a function $F:ℝ^2 → ℝ^2$  defined by $F(x,y)=(e^x\cos(y),e^x\sin(y))$. $F$ is continuously differentiable and the determinant of the derivative matrix is always non-zero. So we may apply the inverse function theorem. However $F$ is not one-to-one. Doesn't it contradict the inverse function theorem?
 A: This is a common mistake. The inverse function theorem doesn't say the function is globally invertible. It says the function is locally invertible. If $x_0$ is a point where $f$ is continuously differentiable and the determinant of the derivative matrix is non zero then there is an open neighborhood $U$ of $x_0$ such that $f(U)$ is open and the restriction $f|_U:U\to f(U)$ is a diffeomorphism. 
A: Under the identification $(x,y) = x+iy= z$ of $\mathbb R^2$ with $\mathbb C$, your function $f : \mathbb C\to\mathbb C$ is
$$ f(z) = e^z.$$
the Inverse Function Theorem can in fact be very carefully applied to define logarithms, and the failure to define a global logarithm should be well understood. 
In particular, note how important it is to have a particular $z_0$ chosen in the domain that maps to $w_0=e^{z_0}$; the Inverse Function Theorem will guarantee the existence of a smooth inverse $\log:N_{w_0}\to N_{z_0}$ defined on a neighbourhood of $w_0$, mapping to a neighbourhood of $z_0$, with $\log'(w) = \frac{1}{e^{\log(w)}} = \frac1w$.
