# Inverse Function Thm. and Covering Maps.

1. Jan 4, 2014

### WWGD

Hi, All:

Let $f: X → Y$ be a differentiable map , so that $Df(x)≠0$ for all $x$ in $X$. Then the inverse function
theorem guarantees that every point has a neighborhood where $f$ restricts to a homeomorphism.

Does anyone know the conditions under which conditions a map like above is a covering map? I'm thinking of the case of the complex exponential $e^z$ , with $d/dz(e^z)=e^z ≠0$ which is a covering map $\mathbb C^2 → (\mathbb C-{0} )$ , but I can't tell if the condition $df(x)≠ 0$ is enough to guarantee that $f$ is a covering map, nor what conditions would make $f$ into a covering map.

Thanks for any Ideas.

2. Jan 5, 2014

3. Jan 7, 2014

### WWGD

Thanks, Mathwonk.