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Exactly what the title says. Can you find some topological space $X\subset\mathbb R^2$ such that $\pi_1(X)\neq0$, but $\mathrm H_1(X,\mathbb Z)=0$?

I've been told that this paper shows that the fundamental group of any subspace of $\mathbb R^2$ has a torsion-free fundamental group, so at the very least, for such a space to exist there has to be some torsion-free group with trivial abelianization. I do not know if such a group exists.

Edit: It turns out that simple torsion-free groups exist, so there are torsion-free groups with trivial abelianization.

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By the references in this answer, fundamental groups of subsets of the plane are residually free, and in particular, if $\pi_1(X)$ is nontrivial it surjects onto a nontrivial free group. Because free groups have nontrivial abelianization, we see that $\pi_1(X)$ surjects onto an abelian group, and hence $H_1(X) = \pi_1(X)^{\text{ab}}$ is nontrivial.

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  • $\begingroup$ It looks like this only applies to closed subsets of the plane. Do you know of similar constraints on the fundamental groups of open (or, even better, arbitrary) subsets of the plane? $\endgroup$ – Niven Jan 22 at 7:57
  • $\begingroup$ @Niven Looks like I gave the wrong reference. The result applies to arbitrary subsets. See Fischer-Zastrow, "The fundamental groups of subsets of closed surfaces inject into their first shape groups". $\endgroup$ – user98602 Jan 22 at 13:55
  • $\begingroup$ If you were only interested in open subsets this becomes much easier: it is much more elementary to prove that every open subset of the plane has free fundamental group, and in fact, this is true of any noncompact surface. $\endgroup$ – user98602 Jan 22 at 14:04
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    $\begingroup$ I see now that is the reference I gave. The MO post just asked for closed sets, but the paper itself never makes that restriction. $\endgroup$ – user98602 Jan 22 at 20:29

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