Where is the topology hiding in this theorem on entire functions?
Solution 1:
Where is the topology hiding in the original proof?
It is hiding in the phrase "is well-defined," or, equivalently, in Cauchy's theorem. In a domain that is not simply connected, this "definition" of $g$ in general depends on the path taken from your base point to $z$. (For example, if there are isolated singularities, then different paths could wind around the singularities differently and pick up residues.) The well-definedness depends on Cauchy's theorem for simply connected domains.