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Is there a subset of $\mathbb{R}^3$ with an element of finite order (not the identity!) in its fundamental group?

I think the real projective plane is such a subset as its fundamental group is isomorphic to $\Bbb{Z}_2$. If $f:\pi_1(\Bbb{R}P^2)\to\Bbb{Z}_2$ is an isomorphism and $[a]\in\pi_1(\Bbb{R}P^2)$ ($[a]$ is not the identity), then $f([a]\cdot[a])=f([a])\cdot f([a])=0$, which means that $[a]\cdot[a]$ is the identity.

The only problem might be the assumption $\Bbb{R}P^2\subset\Bbb{R}^3$. But as we can think of $\Bbb{R}P^2$ as the unit sphere with its antipodal points identified, this seems plausible (visually).

Is my visual intuition wrong?

Xena
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    Unfortunately, yes - $\mathbb{R}P^2$ does not embed in $\mathbb{R}^3$ (but it does embed into $\mathbb{R}^4$). – mdp Sep 03 '13 at 08:51
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    A related topic: http://math.stackexchange.com/questions/36279/the-fundamental-group-of-every-subset-of-mathbbr2-is-torsion-free – Seirios Sep 03 '13 at 08:53
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    What kind of subsets are you interested in? Arbitrary? Open? Simplicial subcomplexes? For arbitrary subsets (even compact ones) this appears to be an open problem. – Moishe Kohan Sep 03 '13 at 12:44
  • I was interested in arbitrary sets. @Matt I googled it and it seems to have a very sophisticated (for me) proof – Xena Sep 03 '13 at 15:09
  • @Seirios Thanks for the reference, I will read it carefully soon. – Xena Sep 03 '13 at 15:12
  • Related: https://mathoverflow.net/questions/4478/torsion-in-homology-or-fundamental-group-of-subsets-of-euclidean-3-space – R. H. Vellstra Apr 05 '25 at 23:47

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This is an open problem, even for compact subsets of $\mathbb R^3$. See the discussion in this MO post.

R. H. Vellstra
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