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For $n \in \{1,2\}$, is there a subset $E \subset \Bbb R^n$ such that one of its (singular) homology groups $H_k(E)$ has an element of finite order?

According to Is the fundamental group of every subset of $\mathbb{R}^2$ torsion free?, $\pi_1(E)$ is torsion-free, but $H_1(E) = \pi_1(E)/[\pi_1(E),\pi_1(E)]$ could have torsion elements. According to this link on MO (or this one), a theorem of Eda states that this is impossible to find such subsets of $\Bbb R^n$ with $n=2$ (then also for $n=1$). [By the way, I'm not sure to know from this MO thread if the answer is completely known for $n=3$].

I don't have a copy of this article, but the title "Fundamental group of subsets of the plane" suggests that it only proves the fact that $\pi_1(E)$ is torsion-free. I don't see how this can answer my question on homology groups. Related questions: (1), (2).

Thank you for your comments!

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Watson
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    If $E$ is a subset of $\Bbb R^2$ which is compact and locally-connected (aka sufficiently nice), then Alexander duality will tell you $E$ has no torsion on $H_1$ (and $H_2$ vanishes). – Balarka Sen Aug 18 '16 at 11:58
  • Related:https://math.stackexchange.com/questions/1281426/ – Watson Jun 12 '18 at 08:21

1 Answers1

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I don't know about Eda's paper, but Fischer-Zastrow prove that for any subset of a closed oriented surface $X$, $\pi_1(X)$ is (many things, but in particular) locally free: that every finitely-generated subgroup is free. But if there were a nonzero element $x$ with $x^n \in [\pi_1, \pi_1]$ (but $x$ not so), we could write $x^n = [g_1, h_1] \cdots [g_n, h_n]$. This would provide a nontrivial relation on the subgroup generated by $x, [g_i, h_i]$, since $x$ cannot possibly be expressed in terms of the latter elements alone.

  • Thank you for your answer! So this disproves the existence of closed subsets $E \subset \Bbb R^n$ (with $n=1$ or $n=2$) such that $H_1(E)$ has torsion (but gives no information about $H_k(E)$ for $k≥2$), is it right? – Watson Aug 18 '16 at 16:31
  • @Watson Sorry, I guess I missed the part of your question about $H_k(E)$. That is almost certainly always zero, but I don't have a proof right now. –  Aug 18 '16 at 16:31
  • No problem! This already gives an excellent answer for $H_1(E)$, where $E$ is a closed subset of $\Bbb R^n$ ($n≤2$). – Watson Aug 18 '16 at 16:42
  • @Watson: A path-connected subset of $\Bbb R$ is an interval, which you can use to conclude the general result that $H_i = 0$ for $i > 0$ in that case. –  Aug 18 '16 at 16:44
  • @Watson Two things: 1) in my comment above I mean for the case of subsets of $\Bbb R$, and 2) The result of this answer does not actually assume that the subset is closed! –  Aug 18 '16 at 18:20
  • ok 2) Ah, yes! I misinterpreted the word "closed". Thank you for the clarification.
    • – Watson Aug 18 '16 at 18:23