Let $p:\mathbb{C}-\{0\}\to \mathbb{C}-\{0\}$ be $z\to z^2$, wich is $a+ib\mapsto (a^2-b^2)+i(2ab)$. By fundamental theorem of algebra we know that $p$ is surjective, and if $w$ is a square root of $z$, also $-w$ is. Thus for each $z\ne 0$ we have two distinct square roots of $z$.
I want to prove that the map $p$ is a covering map.
Now, I'm not acquainted of Complex Analysis, but I have read something here on MSE and on Wikipedia.
If we let $z\in \mathbb{C}-\{0\}$, then we can represent $z$ in polar coordinates in unique way as $z=[r,\theta]$ with $r>0$ and $\theta\in [0,2\pi)$ and by de Moivre formula I have that $w_1=[\sqrt{r},\theta/2]$ and $w_2=[\sqrt{r},\theta/2+\pi]$ are the two square roots of $z$.
Now, in order to define a function, I have to make a choice on who I want as square root of $z$.
So, let's say that my square root map is $\mathbb{C}-\{0\}\to \mathbb{C}-\{0\},\quad[r,\theta] \mapsto [\sqrt{r},\theta/2]$, so I'm taking the root that lies in the upper half space.
The problem now is that I want a continuous map. So this map is not good since points of the form $[r,\theta], [r,2\pi-\theta]$ are close to each other when $\theta$ goes to $0^+$, but their images are not. But if I restrict my map to $\mathbb{C}-([0,+\infty)\times \{0\})$, then it shoud be continuous.
But now, how can I formally prove the that map is truly continuous? By definition the topology on $\mathbb{C}$ is the one that came with the set-indentification with $\mathbb{R}^2$, so by definition a map $\mathbb{C}\to \mathbb{C}$ is continuous if and only if it is continuous as a map $\mathbb{R}^2\to \mathbb{R}^2$.
For example the map $p:\mathbb{C}-\{0\}\to \mathbb{C}-\{0\}$ , $z\to z^2$ is continuous beacuse as a map $p:\mathbb{R}^2-\{0\}\to \mathbb{R}^2-\{0\}$ , $(a,b)\to (a^2-b^2,2ab)$ it is continuous.
So to show that $q:\mathbb{C}-([0,+\infty)\times \{0\})\to \mathbb{C}, \quad [r,\theta]\mapsto [\sqrt{r},\theta/2]$ should I write it as a map from a certain subset of $\mathbb{R}^2$ to a certain subset of $\mathbb{R}^2$ and then control that it is continuous? (As done for $p$, so this include adap the defyning formula of $q$ to "cartesian coordinates")
EDIT
Then, how can i prove that $p$ is a covering map with a strategy like this: pick $z\in \mathbb{C}-\{0\}$, let $U$ be a certain set. Observe that $U$ is open neighborhood of $z$ and is connected. Now observe that $p^{-1}(U)$ has the following components, each of which is mapped homeomorphically to $U$ by $p$
Best Answer
For $M \subset \mathbb{C}$ define $-M = \{ -z \mid z \in M \}$. It is easy to verify that $p^{-1}(p(M)) = M \cup -M$.
Let us prove that $p$ is an open map. Assume that there exists an open $U \subset \mathbb{C}^* = \mathbb{C} \setminus \{ 0 \}$ such that $p(U)$ is not open. This means that there exists a sequence $(z_n )$ in $\mathbb{C}^* \setminus p(U)$ converging to some $z \in p(U)$. Let $(w_n)$ be a sequence in $\mathbb{C}^*$ such that $w_n^2 = z_n$. Since $(z_n)$ is bounded, also $(w_n)$ must be bounded and thus has subsequence converging to some $w \in \mathbb{C}$. W.l.o.g. we may assume that $w_n \to w$. This shows $p(w) = \lim p(w_n) = \lim z_n = z \in p(U)$. Hence $w \in p^{-1}(p(U)) = U \cup -U$, i.e. $w \in U$ or $-w \in U$. Since $-w_n \to -w$, one of the sequences $(w_n)$ and $(-w_n)$ converges to a point in $U$. Hence $w_n \in U$ or $-w_n \in U$ for $n \ge n_0$, therefore $z_n = p(\pm w_n) \in p(U)$ for $n \ge n_0$. This contradicts the fact that all $z_n \notin p(U)$.
We now show that each $z \in \mathbb{C}^* = \mathbb{C} \setminus \{ 0 \}$ has an open neighborhood $U$ which is evenly covered by $p$.
Let us first consider $z = -1$. $U = \mathbb{C} \setminus [0,\infty)$ is an open neighborhood of $-1$ in $\mathbb{C}^*$. We have $p^{-1}(U) = \mathbb{C} \setminus p^{-1}([0,\infty)) = \mathbb{C} \setminus \mathbb{R}$. Let $V_+$ and $V_-$ denote the open halfplanes $\text{Im}(z) > 0$ and $\text{Im}(z) < 0$, respectively. They are disjoint and their union is $\mathbb{C} \setminus \mathbb{R}$. Obviously $p$ maps both $V_\pm$ bijectively onto $U$. Since $p$ is an open map, we see that $U$ is evenly covered with sheets $V_\pm$.
For $z \in \mathbb{C}^*$ define $h_z : \mathbb{C}^* \to \mathbb{C}^*, h_z (\zeta) = -z \zeta$. This is a homeomorphism with inverse $h_{1/z}$. We have $h_z(-1) = z$. If $w$ is a square root of $-z$, we get $p(h_w(\zeta)) = w^2 \zeta^2 = -z \zeta^2 = h_z(p(\zeta))$, i.e. $p \circ h_w = h_z \circ p$.
Now consider an arbitrary $z \in \mathbb{C}^*$. Let $w$ be a square root of $-z$. $U' = h_z(U)$ is an open neighborhood of $z$ and the sets $V'_\pm = h_w(V_\pm)$ are open and disjoint. We have $p(V'_\pm) = p(h_w(V_\pm)) = h_z(p(V_\pm)) = h_z(U) = U'$. Let $p(z_1) = p(z_2)$ with $z_1, z_2$ either in $V'_+$ or in $V'_-$. Let $\zeta_i = h_w^{-1}(z_i) \in V_\pm$. Since $1/w$ is a square root of $-(1/z)$, we get $p(\zeta_i) = p(h_w^{-1}(z_i)) = p(h_{1/w}(z_i)) = h_{1/z}(p(z_i))$, hence $p(\zeta_1) = p(\zeta_2)$ and we conclude $\zeta_1 = \zeta_2$ and thus $z_1 = z_2$. This shows that $p$ is injective on $V'_\pm$. Therefore $p$ maps $V'_\pm$ homeomorphically onto $U'$. It remains to show that $p^{-1}(U') = V'_+ \cup V'_-$. Let $\zeta \in p^{-1}(U')$, i.e. $p(\zeta) \in U' = h_z(U)$. Thus $\zeta^2 = -z \zeta' = w^2 \zeta'$ with some $\zeta' \in U$. We conclude $p(-\zeta/w) = (-\zeta/w)^2 \in U$, hence $-\zeta/w \in V_\pm$. This means $\zeta = h_w(-\zeta/w) \in h_w(V_\pm) = V'_\pm$.
Remark:
This proof is completely elementary. It does not use any theorems of complex analysis. The only ingredients are the continuity of complex multplication and the fact that each complex number has a root (which can be explictly calculated, see How do I get the square root of a complex number? ).