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Let $X \subset \mathbf{C}$ be any nonempty connected subspace, where $\mathbf{C}$ has the Euclidean topology, and let $x \in X$. Suppose that $\pi_1(X, x)$ is a finitely generated abelian group.

Then my question is whether it is true that $\pi_1(X, x)$ is either trivial or isomorphic to $\mathbf{Z}$.

I think a first step in showing that the answer to my question is positive would be to establish that $\pi_1(X, x)$ has no torsion. This is indeed true for surfaces as can be verified in the given references in this question.

A second step would be to show that the rank of $\pi_1(X, x)$ cannot be greater than 1.

I cannot seem to find a reference for either of these two claims, and I also cannot seem to prove these two claims.

Anton Odina
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1 Answers1

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Here is an answer for an open subset: This is true, but it's not easy.

You can see these issues discussed in this mathoverflow question. And from that question you can see that:

  • $\pi_1(X,x)$ is a free group; the only abelian free groups are the trivial group and the infinite cyclic group.
  • Even more, every connected, noncompact surface (perhaps assume paracompactness) is homotopy equivalent to a graph; the fundamental group of a connected graph is always a free group.

The proof of the latter statement uses the triangulation theorem for surfaces, due to Rado (which needs paracompactness). One can find a complete writeup here, which refers to the original 1931 paper of Johansson.

Lee Mosher
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    It's NOT true that fundamental group of a subspace of a plane is free. Fundamental group of Hawaiian earring is not free, for example; it's analogous to Baer-Specker group $\prod \Bbb Z$, in sense that it is an uncountable group with countably many homomorphisms to $\Bbb Z$. – xsnl Aug 19 '24 at 10:52
  • While it IS true that every subspace of a plane having homotopy type of a CW complex is homotopy equivalent to a graph, it's not true that every subspace has a homotopy type of a CW complex. – xsnl Aug 19 '24 at 11:19
  • What IS true is that fundamental groups of compact subspaces of a plane are locally free+fully residually free; the proof is still nontrivial, but doesn't require much past basic shape theory. So the answer to the OP is "yes for compact subspaces". /// I think that in some papers by Katsuya Eda there was a result along the lines "if a countably based top. space has finitely generated pi_1, then it is locally simply connected"; I'll try to locate it. – xsnl Aug 19 '24 at 11:33
  • I amended my answer. I originally mis-read the question as asking about a subsurface, i.e. an open subset (where the question is already nontrivial). You are of course right about the Hawaiian earring etc. – Lee Mosher Aug 19 '24 at 12:44
  • I think I am missing something in the first point of your list. Why do you claim that the only free abelian group is trivial or infinite cyclic? How about $\mathbf{Z}^2$? – Anton Odina Aug 19 '24 at 20:11
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    @AntonOdina $\Bbb Z^2$ is a free Abelian group. However, it is not a free group. You could say it is a nonfree Abelian group which is a free Abelian group. : ) – FShrike Aug 19 '24 at 20:30
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    @AntonOdina Words in sentences usually do not commute. (Also they often do not associate, and the lack of parentheses may be confusing). Free (abelian group) is not the same as abelian (free group). – xsnl Aug 19 '24 at 20:45
  • @AntonOdina: Presumably, knowing something about algebraic-topology and fundamental-groups, you have studied Van Kampen's Theorem, which for many people (perhaps) is the first place they encounter free groups. – Lee Mosher Aug 20 '24 at 00:07