Covering map from sphere with six points removed to doubly-punctured complex plane 
$X$ is $S^2\subset \mathbf{R}^3$ with its intersection points with the coordinate axes removed.
Show that the following map is a covering map. $$\begin{align*}p:X&\longrightarrow \mathbf{C}-\{0,1\} \\ (x,y,z)&\longmapsto \left(\frac{x+iy}{1-z} \right)^4 \end{align*}$$

To check if it is a covering map, I should show that it is (1) continuous, (2) surjective and (3) that for all $w\in\mathbf{C}-\{0,1\}$ there is an open neighbourhood $U$ for which $p^{-1}(U)$ is the disjoint union of subsets $V_\alpha\subset X$ which are all homeomorphic to $U$.
We have never encountered such a difficult covering map before, the only one we have seen in class is the standard covering map $\mathbf{R}\to S^1:t\mapsto e^{it}$ and $S^1\to S^1:z\mapsto z^n$ for the circle $S^1$.
I know that the formula for stereographic projection $S^2-\{(0,0,1) \}\to \mathbf{C}$ from the north pole $(0,0,1)$ is $(x,y,z)\mapsto \frac{x+iy}{1-z}$. This is clearly a bijection $\phi:S^2-\setminus \{0,0,1\}\leftrightarrow\mathbf{C}-\{0\}$. Under $\phi$, the missing points of $X$ are mapped to $0,1,-1,i$ and $-i$. The map $z\mapsto z^4$ then 'rotates' the plane such that the points $1,-1,i,-i$ are sent to $1$.
The function is continuous since it's a formula, and $z\neq 1$ since we removed the point $(0,0,1)$ from the sphere.
I tried proving that it is surjective by using spherical coordinates, but that did not work out. Since $X$ does not contain $(0,0,0)$ we never reach $0$ in the image, but then why don't we reach $1$?
Could someone provide any help?

$\left(\dfrac{x+iy}{1-z} \right)^4 = 1$ means that $\dfrac{x+iy}{1-z} \in \{ 1, -1, i, -i \}$. Let us assume $\dfrac{x+iy}{1-z} = \pm1$. Then $x + iy = \pm(1-z)$ which implies $y = 0$ and $x = \pm(1-z)$. Since $x^2 + z^2 = 1$, we get $(1-z)^2 + z^2 = 1$ which implies $z =0$ or $z=1$. For $z=0$ we get $x = \pm 1$, hence $(x,y,z) = (\pm 1, 0,0) \notin X$,  for $z=1$ we get $x = 0$, hence $(x,y,z) = (0, 0,1) \notin X$. Similarly $\dfrac{x+iy}{1-z} = \pm i$ is impossible.
Now look what $p$ does. As we know, $z = \pm 1$ is impossible since then $(x,y,z) = (0,0,\pm 1)$. For $\zeta \ne \pm 1$ let $X_\zeta = \{(x,y,z)\in X \mid z = \zeta \}$. This is the intersection of $X$ with the plane $P_\zeta = \{(x,y,z) \in \mathbb R^3 \mid z = \zeta \}$.
For $\zeta \ne 0$ it is a circle in $P_\zeta$ with center $(0,0,\zeta)$ and radius $\sqrt{1-\zeta^2}$. It is wrapped by $p$ four times around the circle in $\mathbb C$ with center $0$ and radius $r_\zeta = \dfrac{(1-\zeta^2)^2}{(1-\zeta)^4} = \left( \dfrac{1+\zeta}{1 - \zeta} \right)^2$. For $\zeta \in (-1,0)$ the function $\zeta \mapsto r_\zeta$ is a decreasing homeomorphism onto $(0,1)$, and for $\zeta \in (0,1)$ an increasing homeomorphism onto $(1,\infty)$. This shows that each $w\in \mathbb C$ with $\lvert w \rvert \ne 0,1 $ is in the image of $p$.
For $\zeta = 0$ the set $X_0$ is the disjoint union of four open quarters of the unit circle in $P_0$. Each of them is mapped to the unit circle in $\mathbb C$ minus the point $1$. This shows that $p$ is surjective.
It should now be clear how to verify that $p$ is a covering projection with four sheets. 
