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. 



Is an entire function, with nowhere vanishing derivative, always a covering map?
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 $t=0$).


