# Is an open subset of a compact surface with connected boundary completely determined by its fundamental group?

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 $S \setminus \{p\}$.

• Isn't a cylinder homeomorphic to a punctured disc? Commented Aug 2, 2015 at 22:46
• @posilon Indeed it is. I believe my new example should work (since the $1$ point compactification of $T_2 \setminus \{p\}$ is a torus, and the $1$ point compactification of $D_2 \setminus\{p_1, p_1\}$ is not a surface).
– user88319
Commented Aug 2, 2015 at 22:55

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.

• Thank you for your answer (I'll have to think more about it when I get the chance). However, I was looking at the boundary of the open subset as a subset of the larger surface (so, for example, a cylinder could have as its boundary in the torus a single circle).
– user88319
Commented Aug 3, 2015 at 19:02
• @Strants: Do you demand the "boundary" be a manifold? If so, the above answer applies but with one or possible two ends, and it's again determined by the fundamental group. If not, there's no hope of what you want being true.
– user98602
Commented Aug 3, 2015 at 19:57
• (I am interpreting "boundary" here to mean "$\overline{X} \setminus X$".)
– user98602
Commented Aug 3, 2015 at 20:09
• That was the notion of boundary I was intending, yes. I would be fine with requiring that the boundary be a manifold. Thank you for your help!
– user88319
Commented Aug 3, 2015 at 21:28
• @Strants: For a counterexample in the non-manifold case, you can embed the twice-punctured disc into the genus 2 surface such that its boundary is homeomorphic to a circle with an arc through the center. You can embed the punctured torus into this same manifold.
– user98602
Commented Aug 3, 2015 at 21:37