Question about covering map of punctured unit disk

[MathLearner123]

Let $\alpha \in \mathbb{R}^*,\, p:\mathbb{H} \to \mathbb{D} \setminus \{0\}, \, p(z) = e^\frac{2 \pi i z}{|a|}$. I want to show that $p$ is a covering map but I dont't know how to make this. I think I need to start with an $y \in \mathbb{D} \setminus \{0\}$ and take an open disk $D \subset \mathbb{D} \setminus \{0\}$ to be a neighbourhood of $y$. Now, because disk is open and connected, there exists a holomorphic branch of logarithm in $D$. It's ok this start? How I can continue? Thanks!

 

[FactChecker]

Sorry for my ignorance -- What is $\mathbb{R}^*$? Is $\mathbb{H}$ a half-plane?

 

[MathLearner123]

@FactChecker $\mathbb{R}^* = \mathbb{R} \setminus \{0\}$ and $\mathbb{H}$ is the upper half-plane

 

[FactChecker]

I think you may need to include more at the branch cut. I think that $\mathbb{H}$ does not include any of the real line. Don't you need to include the real line in the domain of $p$ in addition to $\mathbb{H}$ in order to cover $\mathbb{D} \setminus \{0\}$? And doesn't the domain need to include an open set that includes the reals?

CORRECTION: My mistake. The positive reals in $\mathbb{D} \setminus \{0\}$ will be covered by the image of $z=x+|\alpha|i$.