Before applying the inverse function theorem, we must show that it is locally 1:1.
I am confused by two things:
1) what exactly does it mean by "local"? How is it different from "global"?
2) What is the best way to show this?
For instance, consider:
$$F: \mathbb{R}^2 \to \mathbb{R}^2 \; F(x,y) = (e^y\cos(x), e^y\sin(x))$$
How would I show that this is locally 1:1, but not globally 1:1?
When I calculate the determinant of the jacobian, I get:
$|DF(x)| = -e^{2y}[\sin(x)\cos(x) + \cos^2(x)]$
But, I can see this is clearly singular for many points throughout the domain, namely whenever $x = \pi n - \frac{\pi}{4}$ - however, and correct me if I'm wrong, I have a feeling this is a CONSEQUENCE of the failure of "not globally one-to-one".
One thing I thought of was setting:
$F(x=a,y=b)$ and solving for:
$$e^b \cos(a) = e^y \cos(x) \; \; \cdots (1)$$ $$e^b \sin(a) = e^y \sin(x) \; \; \cdots (2)$$
For $(1):\; x = 2 \pi n - \cos^{-1}(\cos(a)), n \in \mathbb{Z}$ is clearly not globally 1:1. But what about locally?
Thanks in advance for any help.