Covering spaces of $\mathbb{C}^{\times}$ Let $\mathbb{C}^{\times}=\mathbb{C} \setminus \{0\}$. I'm trying to find out all covering spaces of this space.
Let's start. (I'm using Massey's book: Algebraic Topology, an introduction)
First of all I know that $\mathbb{C}^{\times}$ admits universal covering space, for example $(\mathbb{C}, p=exp(z))$, so I know that Aut$_{\mathbb{C}^{\times}}(\mathbb{C})$ (=group of automorphisms of the covering sapce $\mathbb{C}$) is isomorphic to $\pi_1(\mathbb{C}^{\times})$. However $\mathbb{C}^{\times} \cong  \mathbb{R}^2 \setminus \{0\}$ then $\pi_1(\mathbb{C}^{\times}) \cong \mathbb{Z} \cong $ Aut$_{\mathbb{C}^{\times}}(\mathbb{C})$. Now the problems begin.
The main idea is retracing the proof of Lemma 10.1,chapter 5 on Massey's book. For example, let $G=3\mathbb{Z}$ subgroup of $\mathbb{Z}$. Now there exists a subgroup of Aut$_{\mathbb{C}^{\times}}(\mathbb{C})$ called A$_G$ which is isomorphic to $G$. Let $\tilde X = \mathbb{C}/$A$_G$; I know that $(\mathbb{C},r)$ (with $r \colon \mathbb{C} \to \tilde X$ the natural projection) is a covering sapce of $\tilde X$ and so $(\tilde X, q)$, with $q$ such that $p=q \circ r$, is also a covering space of $\mathbb{C}^{\times}$. Finally, $G$ is normal in $\pi_1(\mathbb{C}^{\times})$ and so, by using Corollary 7.4, I see that Aut$_{\mathbb{C}^{\times}}(\tilde X) \cong \mathbb{Z}/ \mathbb{3Z}$ and $\pi_1(\tilde X) \cong \mathbb{3Z}$.
Ok, that's great, but 
I would like to know how the cover map $r$ actually works; how Aut$_{\mathbb{C}^{\times}}(\mathbb{C})$ actually is (its elements); how I can imagine $\tilde X$ (maybe $\mathbb{S}^1 \times \mathbb{R}$ ? Is this isomorphic to $\mathbb{C}^{\times}$? Right?). Can I suppose $q \colon z \mapsto z^3$? But why?
Aut$_{\mathbb{C}^{\times}}(\tilde X) \cong \mathbb{Z}/ \mathbb{3Z}$ and $\pi_1(\tilde X) \cong \mathbb{3Z}$ are usefull in order to answer the previous questions or are they something extra?
Extra question: Let $(\tilde X,p)$ a covering space of $X$. I have studied the action of the group $\pi_1(X,x)$ on the fiber $p^{-1}(x)$. Is this topic 
concretely used on computing covering spaces or I can see it like theory employed in order to demonstrate the Existence Theorem of covering spaces (Lemma 10.1 and Theorem 10.2)?
 A: The universal cover of $\mathbb{C} - \{0\}$ is $\mathbb{C} $. The covering map is the exponential so it's fundamental group is $\mathbb{Z} $. The coverings correspond to quotient of $\mathbb{C}$ by subgroups of $\mathbb{Z}$. 
A: This is a bit extenuating to write here, I hope it will be at least intelligible.
The first thing you should notice (if you have not already done) is that
$\mathbb{C}^{\times}$ is homotopy equivalent to $S^1$. Hence I think it is more convenient to do the following:
if for $S^1$ you can picture the universal cover as a helix, for $\mathbb{C}^{\times}$ you can picture it as some sort of helicoid.
Say you see everything embedded in $\mathbb{R}^3$ as
$$
\begin{align}
\mathbb{C}^{\times} &= \{(x, y, 0) \mid (x, y) \neq (0, 0)\} \\
U &= \{(r \, cos (2 \pi \, \theta), r \, \sin (2 \pi \, \theta), \theta) 
        \mid r \in \mathbb{R}_{> 0}, \theta \in \mathbb{R}\}.
\end{align}
$$
You can somehow plot $U$ in wolframalpha. Set as universal covering map
$$
\begin{align}
q \colon U &\to \mathbb{C}^{\times} \\
(x, y, z) &\mapsto (x, y).
\end{align}
$$
Then your fibers are of the form $(x, y, z_0 + k)$ where $k \in \mathbb{Z}$, and the automorphisms are of the form
$$(x, y, z) \mapsto (x, y, z + k), \quad k \in \mathbb{Z}.$$
As you can imagine, the homotopy equivalence between $\mathbb{C}^{\times}$ and $S^1$ gives you a dictionary between the covering theories of the two spaces. If you understand the story for $S^1$ then you do for $\mathbb{C}^{\times}$. I will be very imprecise from now.
If you quotient $X$ by $3 \mathbb{Z}$ you get a space $\tilde{X}$ that you can see as "three turns of helix/helicoid" with the two edges identified (just as when you quotient $\mathbb{R}$ by $\mathbb{Z}$ you get $S^1$).
The fibers of your covering $r \colon U \to \tilde{X}$ are of the form $(x, y, z_0 + 3 k)$ where $k \in \mathbb{Z}$, and as you can probably imagine, the automorphisms of the covering $r$ are of the form
$$(x, y, z) \mapsto (x, y, z + 3k).$$
To rigorously prove that, just employ the things you know about quotient spaces / morphisms.
For your extra question, the propositions behind the action of the deck on the fibers are mostly used as you say to prove some theorems later on, and to caracterize galois (or regular) actions. I don't think it is common practice to use them to "compute" coverings (in fact, I don't think it is common practice to compute coverings at all), but it at least a "visual aid" to understand automorphisms of covering maps (you know how an automorphism act on the fibers, and the automorphism has to be continuous, then..).
A: Topologically $\mathbb{C}^\times = \mathbb{C} \backslash \{ 0\}$ is the cylinder and the universal cover of the cylinder is the plane.  The covering spaces are indexed by $\pi_1(\mathbb{C}\backslash \{ 0\}) \simeq \mathbb{Z}$.  What could be the representatives of $\mathbb{Z}$.  These are the exponent maps $z \mapsto z^n$.  
In reverse the square-root map map $z \mapsto \sqrt{z}$ carries an ambiguity since we don't know which one it is, $\pm \sqrt{z}$. The $2:1$ cover of a cylinder is...another cylinder.  However, by continuity the sign $\sqrt{z + \epsilon}$ in very small neighborhoods is consistent:
$$ \sqrt{z+\epsilon} = \sqrt{z}\,\big( 1 + \frac{\epsilon}{z}\big) \approx \sqrt{z} \,\big( 1 + \frac{\epsilon}{2z}\big) $$
Yet as we make a loop around a path $[1] \in \pi_1(\mathbb{C}^\times)$ the sign of the square root finally changes $\sqrt{z} \mapsto -\sqrt{z}$.
The "canonical" recipe for solving this type of process would be to keep a mental note $(\sqrt{z}, \pm 1)$ where $+1$ and $-1$ is the label of the two-sheeted cover.  Once we cross the boundary we switch a $+1$ to a $-1$.
As long as we're not too rigorous we can find relate the fundamental group to other branches such as analysis and algebra.
A: Let us consider any continuous $f : \mathbb{C} \to \mathbb{C}$ such that $p \circ f = p$. This means $exp(f(z)) = exp(z)$ for all $z$, so that $f(z) = 2k_z \pi i$ for some $k_z \in \mathbb{Z}$. In other words, $g(z) = f(z) - z \in 2\pi i \mathbb{Z}$. Since $g$ is continuos, $g(\mathbb{C})$ is connected and this implies that it is constant. Therefore $f = t_k$ = translation by $2k\pi i$ for a unique $k \in \mathbb{Z}$.
This shows that $Aut_{\mathbb{C}^\times}(\mathbb{C}) = \lbrace t_k \mid k \in \mathbb{Z} \rbrace$. This also gives you an explicit isomorphism $Aut_{\mathbb{C}^x}(\mathbb{C}) \to \mathbb{Z}$.
$\mathbb{C}^\times$ is in fact homeomorphic to $\mathbb{R} \times S^1$. This is done via the map $H : \mathbb{R} \times S^1 \to \mathbb{C}^\times, H(t,z) = exp(t)\cdot z$. It is a homeomorphism with inverse $H^{-1}(w) = (ln(\lvert w \rvert),w/\lvert w \rvert)$.  Note that $exp : \mathbb{R} \to (0,\infty)$ is a homeomorphism.
Define $\omega : \mathbb{R} \to S^1, \omega(x) = exp(ix)$. This is the well-known universal covering of $S^1$. We have $p = H \circ ( 1_{\mathbb{R}} \times \omega)$ where we used $\mathbb{C} = \mathbb{R} \times \mathbb{R}$.
Therefore $p$ agrees "essentially" with $1_{\mathbb{R}} \times \omega : \mathbb{R} \times \mathbb{R} \to \mathbb{R} \times S^1$ . The translations $t_k$ can be written as $1_{\mathbb{R}} \times \tau_k$ with translations $\tau_k(y) = y + 2k \pi$ on $\mathbb{R}$. Your quotient space $\tilde{X}$ corresponds to $\mathbb{R} \times (\mathbb{R}/G_3)$, where $G_3$ is the group of translations generated by $\tau_3$. We have $\mathbb{R}/G_3 \approx S^1$, the homeomorphism explicitly given by $\rho_3([x])= exp(x/3)$.
Using this we see that $r$ corresponds to $1_{\mathbb{R}} \times \rho_3 : \mathbb{R} \times \mathbb{R} \to \mathbb{R} \times S^1$.
$q$ corresponds to $1_{\mathbb{R}}  \times \mu_3 : \mathbb{R} \times S^1 \to \mathbb{R} \times S^1$ where $\mu_3(z) = z^3$.
Using the identification of $\mathbb{R} \times S^1$ with $\mathbb{C}^\times$,  you can derive explicit formualae for $r$ and $q$.
