Alternative way to prove that the powers $\mathbb{C}^* \to \mathbb{C}^*, z \mapsto z^k$ and $\exp: \mathbb{C} \to \mathbb{C}^*$ are covering maps Prove that the power map, $\mathbb{C}^* \to \mathbb{C}^*, \: z \mapsto z^k$ and the exponential map $\mathbb{C} \to \mathbb{C}^*, \: z \mapsto \exp(z)$ are covering maps using the fact that they are Lie-group homomorphisms.
I have seen proofs that the power map is a covering map before. But only as map from $S^{1}\to S^{1}$. And I have no idea how to use the fact that the maps are Lie-group homomorphisms.
Thanks in advance for help.
 A: $\newcommand{\C}{\mathbb{C}}$
Here is an outline of the proof that $f_k\colon z\in \C^* \mapsto z^k \in \Bbb C^*$ is a $k$-folded covering map:

*

*Let $\alpha >0$ be a real number and $U = \{ re^{i\theta} \mid r >0, \theta \in (-\alpha,\alpha)\}$. Show that
$$
f_k^{-1}(U) = \bigcup_{j=0}^{k-1} U_j
$$
where $U_j = \{ re^{i\theta} \mid r >0, \theta - \frac{2j\pi}{k} \in (-\frac{\alpha}{k},\frac{\alpha}{k}) \}$.

*Find $\alpha$ small enough such that all $U_j$'s are disjoint. You can find some inspiration here, where the green region on the RHS is replaced by the whole sector $U$, and similarly for the blue regions (replaced by the $U_j$'s).

*Show that $f_k$ maps $U_j$ onto $U$ homeomorphically.

*For $\xi \in \C^*$, let $z$ be any $k$th root of $\xi$. Show that $f^{-1}(\xi U) = \bigcup_{j=1}^{k-1} zU_j$ and conclude.

The Lie group homomorphism's properties are used in 4: namely, translations in $\C^*$ (i.e multiplication by $\xi$ and $z$) are homeomorphisms.
The same reasoning applies for $\exp \colon \C \to \C^*$. The difference is that there will be infinitely many connected components in the preimage so that this is an countably-infinite covering. The outline is typically the same:

*

*Find an open neighbourhood $U$ of $1$ in $\Bbb C^*$ such that $\exp^{-1}(U) = \bigcup_{j\in \Bbb Z} U_j$ where $\exp \colon U_j \to U$ is a homeomorphism and $U_j$'s are disjoint.

*For $\xi \in \C^*$, find $z \in \Bbb C$ a preimage of $\xi$ by $\exp$ and show that $\exp^{-1}(\xi U) = \bigcup_{j\in \Bbb Z} zU_j$, and conclude.

These ideas do not come from nowhere, they come from simple drawings. In geometry / topology, sketches are good friends.
