Is an open, connected subset of a compact surface with connected boundary determined (up to homeomorphism) by its fundamental group?
If we weaken the hypotheses, I can see how this can fail:
- A cylinder and circle are both subsets of the torus with $F_1$ as their fundamental group, so the requirement that the subset be open is necessary.
- A torus with a point removed and a twice-puntured disk both have $F_2$ as their fundamental group, so the result can fail if we allow subsets with disconnected boundaries.
- Since a disconnected spaces does not generally have a single uniquely defined fundamental group (we have to specify where the base point lies), requiring connectedness seems reasonable.
It seems like open subsets of a compact surface with a connected boundary are sufficiently 'nice' that this result could hold.
This question came to me when I was answering this question: if it were true, then we could enumerate the possible faces of a graph embedding in any surface $S$ by simply enumerating the subgroups of $S \setminus \{p\}$.
I interpret your question as "if $X$ is an open subset of a closed surface, and has only one end, is it determined by its fundamental group?"
Yes. In fact, a surface $\Sigma$ with one end and fundamental group $F_{2g}$ is homeomorphic to the punctured genus $g$ surface, and this is the only possible fundamental group. To prove this use the observation of this answer and exploit the fact that, because $\Sigma$ is an open subspace of a closed surface, it cannot have infinite genus.