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.

1$\begingroup$ See this question: mathoverflow.net/q/189323/1345 $\endgroup$ – Ian Agol Dec 10 '14 at 18:32
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) 16551676, doi:10.2140/agt.2005.5.1655, arXiv:math/0512343.
Cech Homology: It follows that the first homology progroup 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 pathconnected and semilocally simply connected, then $H_1(X)\cong \check{H}_1(X)$ is free abelian.
Harder case. Suppose your planar set $X$ is pathconnected but not semilocally simply connected. To simplify things, let's at least suppose $X$ is a Peano continuum (locally pathconnected 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 14871495, doi:10.1090/S0002993999056701
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 onedimensional (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) onedimensional 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 onedimensional Peano continua, Fundamenta Mathematicae 232 Issue 2 (2016) pp 99–115, doi:10.4064/fm23221 (pdf)
Possibly unknown case. As of 2019, I don't think $H_1$ is known for a general 2dimensional planar set (or Peano continuum) because path reduction is not as straightforward as in the 1dimensional case. However, in the end, the answer is likely to be similar to the onedimensional case.
Eda, K. Fundamental group of subsets of the plane. Topology and its Applications Volume 84, Issues 13, 24 April 1998, Pages 283306
This is more or less a duplicate of the math.se thread: https://math.stackexchange.com/questions/36279/thefundamentalgroupofeverysubsetofmathbbr2istorsionfree

4$\begingroup$ A duplicate of that question?!?!?!? Really? Other than the fact that they both concern the topology of subsets of the plane, they seem like rather different questions to me (and the questioner over there seemed really clueless). Are you suggesting that I should have asked it over there? $\endgroup$ – Tina Oct 28 '12 at 3:20

$\begingroup$ I have seen people referring to Eda's paper also in other similar threads, but could you indicate the precise part of the paper where this result is proved? It seems to deal only with the complement of subsets of a line.. Moreover, the real title is Free $σ$products and fundamental groups of subspaces of the plane. $\endgroup$ – Mizar May 17 '15 at 17:22