Proof that $p:\mathbb{C}\setminus\{0\}\rightarrow\mathbb{C}\setminus\{0\}$ is a covering map, with $p(z)=z^2$ Prove that $p:\mathbb{C}\setminus\{0\}\rightarrow\mathbb{C}\setminus\{0\}$ is a covering map, with $p(z)=z^2$
Let $X=\mathbb{C}\setminus\{0\}$
Let $b\in X$ write $b=re^{j\theta}$ (with $0\le\theta<2\pi$)
For $\theta\ne0$, we can have the [NOT SURE WHAT TO CALL IT - open ball?]
$U=(\frac{1}{2}r,\frac{3}{2}r)\times(\frac{3\theta}{4},\frac{5\theta}{4})$, this is the product of two open balls (one about $r$ of radius $\frac{r}{2}$ the other about theta in the circle (so angles wrap around))
[Not sure how to phrase this part]
We see that $q:\mathbb{R^+}\rightarrow\mathbb{R^+}$ with $q(x)=x^2$ maps $(\sqrt{\frac{r}{2}},\sqrt{\frac{3r}{2}})$ homeomorphiclly to $(\frac{r}{2},\frac{3r}{2})$ (I want to say isomorphically by the strictly monotonic property of squared on $\mathbb{R^+}$)
We also see that $h:\mathbb{S^1}\rightarrow\mathbb{S^1}$ ($h$ deals with angles) given by $h(\theta)=2\theta$ is a covering map itself!
$$h^{-1}\left(\frac{3\theta}{4},\frac{5\theta}{4}\right)=\left(\frac{3\theta}{8},\frac{5\theta}{8}\right)\cup\left(\frac{3\theta}{8}+\pi,\frac{5\theta}{8}+\pi\right)$$
[not sure what comes here]
Using the theorem that the product of two covering maps is a covering map, we see:
$p=q\times h:\mathbb{R^+1}\times\mathbb{S^1}\rightarrow\mathbb{R^+1}\times\mathbb{S^1}$ is a covering map
With $p^{-1}(U)=\left(\sqrt{\frac{r}{2}},\sqrt{\frac{3r}{2}}\right)\times\left(\frac{3\theta}{8},\frac{5\theta}{8}\right)\cup\left(\sqrt{\frac{r}{2}},\sqrt{\frac{3r}{2}}\right)\times\left(\frac{3\theta}{8}+\pi,\frac{5\theta}{8}+\pi\right)$
You then need to do this for $\theta=0$ (just choose a range of angles $(\frac{-\pi}{8},\frac{\pi}{8})$ and use $-\frac{\pi}{2}\le\theta<\frac{3\pi}{2}$ as the range for angles to show this.
How do I fill in my blanks, also:
$z:\mathbb{R^+}\times\mathbb{S^1}\rightarrow\mathbb{C}\setminus\{0\}$ given by $z(r,\theta)=re^{j\theta}$ is a homeomorphism, but it's just like projection, so I couldn't prove this. 
I've got the gist, I need to order this into a proof.
 A: Your map $q$ above is a homeomorphism $\mathbb R^+\to\mathbb R^+$ since it is continuous and has a continuous inverse (the square root function). "Isomorphism" (or "topological isomorphism") is just another way to say "homeomorphism".
Your map $z:\mathbb R^+\times\mathbb S^1\to\mathbb C\setminus\{0\}$ has as it's inverse the map $\varphi$ which sends $z\in\mathbb C\setminus \{0\}$ to the point $\left(\|z\|,\frac{z}{\|z\|}\right)$, where $\mathbb S^1$ can be thought of as the subset $\{z\in\mathbb C:\|z\|=1\}$ of $\mathbb C\setminus\{0\}$. This is continuous (each component is continuous), so $z:\mathbb R^+\times\mathbb S^1\to\mathbb C\setminus\{0\}$ is homeomorphism.
Aside from a few details, your proof looks great. You could also try using a different set as your neighborhood of $z$, say $U=\mathbb C\setminus X$ where $X=\{\lambda z:\lambda\le 0\text{ is real}\}$. It will get rid of the two cases you need to consider: $\theta=0$ and $\theta\not=0$. Just, note that the inverse image of $U$ will be two disjoint "half planes" on each of which $z\mapsto z^2$ is one-to-one, surjective onto $U$, and continuous (just a branch of the square root function).
