TITLE: Is an entire function, with nowhere vanishing derivative, always a covering map?
QUESTION [10 upvotes]: 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.

REPLY [6 votes]: 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.)