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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.

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.

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.

corrected yesterday's edit "f. g." should say "homology"
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Jeremy Brazas
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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 fundamentalfirst 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.

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 fundamental 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.

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.

After seeing this pop up again, I added content to give a more state of the art answer. I also added a newer reference, keeping David's improved formatting for the references.
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Jeremy Brazas
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Fundamental Group: The fundamental group of a closed planar set naturally injects into the first Cech homotopy group, which is an inverse limit of finitely generated 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:

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

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

Easy case. If $X\subset \mathbb{R}^2$$X$ is compact and 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 $H_1(X)\to \check{H}_1(X)$$\phi:H_1(X)\to \check{H}_1(X)$ is surjective but the kernel. 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 be difficultidentify $\ker(\phi)$, then we can "compute" $H_1(X)$.

Now, if $X$ happens to understand even foralso 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 fundamental group tells you that abelianizing $\pi_1$ for these spaces actually kills all of the geometry remembered by $\pi_1$.

For details, see:

In general:

Katsuya Eda, Kazuhiro Kawamura, The surjectivity of the canonical homomorphism from singular homology to CechSingular homology groups of one-dimensional Peano continua, Proc. Amer. Math. Soc.Fundamenta Mathematicae 128232 No. 5Issue 2 (19992016) pp 1487-149599–115, doi:10.1090/S0002-9939-99-05670-110.4064/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.

The fundamental group of a closed planar set naturally injects into the first Cech homotopy group, which is an inverse limit of finitely generated free groups. In particular, the algebraic restrictions gained are: the fundamental group must be locally free, fully residually free, and residually finite.

The first homology pro-group then consists of finitely generated free groups and the first Cech homology group is the inverse limit of these.

When $X\subset \mathbb{R}^2$ is compact and locally path connected, the canonical map $H_1(X)\to \check{H}_1(X)$ is surjective but the kernel can be difficult to understand even for the Hawaiian earring:

In general:

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

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:

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 fundamental 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, 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.

Improved references, fixed link
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David Roberts
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Jeremy Brazas
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