 Solution

Take $G={\mathbb{C}}^{\ast }$ and the complete analytic function generated by a logarithm germ. Its continuation surface identifies with $\mathbb{C}$ via $w\mapsto [\log {]}_z$ with $z=e^w$ and logarithm value $w$. Thus $\pi (w)=e^w$. It is a universal cover; the deck transformations $w\mapsto w+2\pi in$ act transitively on each fibre, so it is also a regular covering map in the deck-transitive sense. It is not bijective, since every nonzero $z$ has infinitely many logarithm values. It also meets the alternative convention in which “regular” merely requires an evenly covered neighbourhood at each base point.