Is an entire function, with nowhere vanishing derivative, always a covering map? Assume that $f:\mathbb C\to\mathbb C$ is entire, and also that $f'(z)\ne 0$, for all $z\in\mathbb C$. Does that imply that $f$ is a covering map of $f[\mathbb C]$?
Clearly, $f$ is a local homeomorphism. Hence, the question reduces to examining whether such an $f$ has the curve lifting property.  
 A: To put the questions and answers in a wider context, the set of singular values $S(f)$ of an entire function $f$ is the set of points $w$ near which $f$ is not a covering map.
That is, $w$ is regular if there is a neighbourhood $U$ of $w$ such that $f\colon V\to U$ is a homeomorphism for every connected component $V$ of $f^{-1}(U)$, and singular if $w$ is not regular.
So $S(f)$ is the smallest closed set $S$ such that $\newcommand{\C}{\mathbb{C}}f\colon \C\setminus f^{-1}(S)\to \C\setminus S$ is a covering map.
As discussed in the answers above, an entire function with $S(f)=\emptyset$ must be affine, and an entire function with $\# S(f)=1$ is an exponential map (by basic covering theory). In particular, no other function can be a covering over its range, by Picard's theorem. 
On the other hand, the set of functions having two singular values is extremely rich. 
(The above notions generalise immediately to functions between arbitrary Riemann surfaces. Of course here it is possible to have covering maps with larger numbers of omitted values.)
A: Answer. If $\,f:\mathbb C\to\mathbb C$ is an entire analytic function, which is a covering of $\,f[\mathbb C]$, then either $\,f(z)=az+b$, with $a\ne 0$ or $\,f(z)=\exp(az+b)+c$, with $a\ne 0$.
Proof. We shall use the fact that if $\,f:\mathbb C\to\mathbb C$ is a covering, then it is a universal covering as well, since $\mathbb C$ is simply connected. Due to Picard's Little Theorem, there are only two possibilities:
a. $\,f[\mathbb C]= \mathbb C$. In such case, $g(z)=z$ is another universal covering, and the definition of the universal covering implies that $f$ is biholomorphic, and hence linear.
b. $\,f[\mathbb C]\ne \mathbb C$. Then $f[\mathbb C]= \mathbb C\!\smallsetminus\!\{c\}$, for some $c\in\mathbb C$, due to Picard's Little Theorem. Then $g(z)=\exp(z)+c\,$ is another universal covering of 
$\mathbb C\!\smallsetminus\!\{c\}$. 
Once again the definition of the universal covering implies the existence of a biholomorphic $\,h:\mathbb C\to\mathbb C$, such that $\,f=g\circ h$, and as $h$ has to be linear then
$$\,f(z)=g\big(h(z)\big)=\exp\big(h(z)\big)+c=\exp(az+b)+c.$$  
A: The function $f(z)=\exp(\exp(z))$ gives a counterexample. The image of $f$ is $\mathbb{C}\backslash\{0\}$, but the curve 
$$
\begin{align*}
\gamma:[0,1]&\to\mathbb{C}\backslash\{0\},\\
t&\mapsto \exp(t),
\end{align*}
$$
does not have a lift $\tilde{\gamma}$ satisfying $\tilde{\gamma}(1)=0$ (the lift would have to be $\tilde{\gamma}(t)=\log(t)$, cannot be defined at $t=0$).
A: An entire function cannot be a covering map, unless $f(z)=az+b$ with $a\neq 0$.
Indeed, the complex plane is simply connected, so every covering of the plane is
a bijection, and the ony entire functions that are bijections are polynomials of degree $1$.
According to Picard, the image of an non-constant entire function is either the plane or a plane with one puncture. In the case of one puncture, an entire function can be a covering over the image, and
this happens if and only if $f(z)=e^{az+b}+c,$ where $c$ is the puncture.
In general, let $f:S\to S$ be a holomorhic map between Riemann surfaces. Then either $f$ is a covering map, or it has a critical point or it has an asymptotic curve.
An asymptotic curve $\gamma:[0,1)\to S$ is such a curve that a) for every compact $K\in S$ there exists $t_0\in(0,1)$ such that $\gamma(t)\cap K=\emptyset$ for
$t\in (t_0,1)$ and b) There exists a limit $\lim f(\gamma(t))\in S$. This limit is called an asymptotic value.
EDIT. Example $f(z)=\int_0^z e^{-\zeta^2}d\zeta$ is surjective, so the image is $C$, and $f'(z)\neq 0$ but this
is not a covering: it has two asymptotic values.
