How do I prove $e^z$ is a covering map using this fact? I have proven that $p:\mathbb{R}\rightarrow S^1:t\mapsto (\cos 2\pi t,\sin 2\pi t)$ is a covering map and $S^1$ and $\mathbb{C}\setminus\{0\}$ are homotopically equivalent.
Using these facts, how do I prove that $q:\mathbb{C}\rightarrow \mathbb{C}\setminus\{0\}:z\mapsto e^z$ is a covering map?
(I can prove this directly, but I want to know whether I can ease the proof using given facts.)
 A: You can use the fact that if $\pi : E \to B$ and $\pi' :E' \to B'$ are covering maps, then so is $\pi \times \pi' : E \times E' \to B \times B'$. 
Then note that $\mathbb{C} = \mathbb{R}^2$ topologically, say with coordinates $(x,y)$. Similarly $\mathbb{C} \setminus \{0\} = \mathbb{R}_{> 0} \times S^1$ topologically with coordinates $(r,s)$ where $r$ is the radius and $s \in S^1$ is a coordinate on the unit circle. For convenience we just take $s = (a,b)$ where $a^2 + b^2 = 1$. Then the map $e^z = e^x(\cos y + \sin y)$ is given in these coordinates by
$$
(x,y) \to (e^x,\cos y, \sin y) = (\pi \times \pi')(x,y)
$$
where $\pi : \mathbb{R} \to \mathbb{R}_{>0}$ on the first copy is just $x \mapsto e^x$ and $\pi' : \mathbb{R} \to S^1$ on the second copy is (up to scaling) the covering map you gave. $\pi$ is a covering map (in fact a homeomorphism) and so is $\pi'$ since scaling is also a homeomorphism so it doesn't change the covering map property. Thus the product $z \mapsto e^z$ is a covering. 
A: $\mathbb{C}\setminus{0}$ is homeomorphic to $\mathbb{R}^+\times S_1$ and $\|e^z\|=e^{\Re z}$. Given that $z=a+ib$, $a\to e^a$ is a covering map from $\mathbb{R}$ to $\mathbb{R}^+$ while $b\to e^{ib}$ is a covering map from $\mathbb{R}$ to $S^1$, hence $z\to e^z$ is a covering map from $\mathbb{C}\simeq\mathbb{R}^2$ to $\mathbb{C}\setminus\{0\}\simeq\mathbb{R}^+\times S^1.$
Also notice that $z\to e^z$ is an open map since $\frac{d}{dz}e^z = e^z\neq 0$ ensures local invertibility.
