Fundamental groups and homology groups of closed subsets of the plane

Question

Let $X$ be a closed subset of $\mathbb{R}^2$. What restrictions are there on $\pi_1(X)$ and on the homology groups of $X$ (both singular and Cech)? This is elementary if $X$ has reasonable local properties, but the example of the Hawaiian Earring shows that things can be very complicated indeed.

Answer 1

Fundamental Group: The fundamental group of a planar set naturally injects into the first Cech homotopy group, which is an inverse limit of free groups. In particular, the algebraic restrictions gained from this fact are: the fundamental group must be locally free, fully residually free (and thus torsion free), and residually finite. See:

Fischer, H., Zastrow, A., The fundamental groups of subsets of closed surfaces inject into their first shape groups, Algebraic and Geometric Topology 5 (2005) 1655-1676, doi:10.2140/agt.2005.5.1655, arXiv:math/0512343.

Cech Homology: It follows that the first homology pro-group is an inverse system of of finitely generated free groups. The first Cech homology group $\check{H}_1(X)$ is an inverse limit of free abelian groups.

Singular Homology: This is more complicated. Certain cases are understood.

Easy case. If $X$ is locally path-connected and semilocally simply connected, then $H_1(X)\cong \check{H}_1(X)$ is free abelian.

Harder case. Suppose your planar set $X$ is path-connected but not semilocally simply connected. To simplify things, let's at least suppose $X$ is a Peano continuum (locally path-connected compact metric space). It is a general result of Katsuya Eda that for any Peano continuum $X$ (planar or otherwise) the canonical map $\phi:H_1(X)\to \check{H}_1(X)$ is surjective. See:

Katsuya Eda, Kazuhiro Kawamura, The surjectivity of the canonical homomorphism from singular homology to Cech homology, Proc. Amer. Math. Soc. 128 No. 5 (1999) pp 1487-1495, doi:10.1090/S0002-9939-99-05670-1

This result helps because it means that if know $\phi$ splits and we can identify $\ker(\phi)$, then we can "compute" $H_1(X)$.

Now, if $X$ happens to also be one-dimensional (such as the Hawaiian earring or Sierpinski carpet) then we can do exactly that. In particular, it turns out that $\check{H}_1(X)\cong \prod_{n\in\mathbb{N}}\mathbb{Z}$ and $\ker(\phi)\cong \prod_{n\in\mathbb{N}}\mathbb{Z}\Big/ \bigoplus_{n\in\mathbb{N}}\mathbb{Z}$ where the second isomorphism is purely abstract. Moreover, since $\prod_{n\in\mathbb{N}}\mathbb{Z}\Big/ \bigoplus_{n\in\mathbb{N}}\mathbb{Z}$ is algebraically compact, the homomorphism $\phi$ splits. We can conclude that for any (planar) one-dimensional Peano continuum $X$, that $$H_1(X)\cong \prod_{n\in\mathbb{N}}\mathbb{Z}\oplus \left(\prod_{n\in\mathbb{N}}\mathbb{Z}\Big/ \bigoplus_{n\in\mathbb{N}}\mathbb{Z}\right).$$

The fact that all of these spaces have exactly the same first singular homology group tells you that abelianizing $\pi_1$ for these spaces actually kills all of the geometry remembered by $\pi_1$.

For details, see:

Katsuya Eda, Kazuhiro Kawamura, The Singular Homology of the Hawaiian Earring, Journal of the London Mathematical Society 62 Issue 1 (2000) pp 305–310, doi:10.1112/S0024610700001071 (pdf)

Katsuya Eda, Singular homology groups of one-dimensional Peano continua, Fundamenta Mathematicae 232 Issue 2 (2016) pp 99–115, doi:10.4064/fm232-2-1 (pdf)

Possibly unknown case. As of 2019, I don't think $H_1$ is known for a general 2-dimensional planar set (or Peano continuum) because path reduction is not as straightforward as in the 1-dimensional case. However, in the end, the answer is likely to be similar to the one-dimensional case.

Answer 2

Eda, K. Fundamental group of subsets of the plane. Topology and its Applications Volume 84, Issues 1-3, 24 April 1998, Pages 283-306

This is more or less a duplicate of the math.se thread: https://math.stackexchange.com/questions/36279/the-fundamental-group-of-every-subset-of-mathbbr2-is-torsion-free