# Is there a domain in $\mathbb{R}^3$ with finite non-trivial $\pi_1$ but $H_1=0$?

The exterior of the Alexander Horned Sphere has $H_1=0$ but $\pi_1\neq 0$, in fact, $\pi_1$ is infinite. (See Hatcher p.171-172). Is there an example of a domain (connected open set) in $\mathbb{R}^3$ where $\pi_1$ is non-trivial but finite, and $H_1=0$?

(This question was stimulated by the question Example of a domain in R^3, with trivial first homology but nontrivial fundamental group.)

• Are there any subsets of $\mathbb R^3$ with nontrivial finite fundamental group? This may be an open problem. See mathoverflow.net/questions/4478/… Feb 18, 2014 at 19:36
• This is indeed difficult for arbitrary subsets, but for open subsets it follows from one of the theorems in Hempel's book that such domains do not exist. Moreover, each open subset in 3d space has torsion free fundamental group. (Each maximal torsion subgroup splits off as a free factor via a connected sum decomposition.) Feb 18, 2014 at 20:31
• I will write a detailed answer when I have access to Hempel's book. Theorem which I quote, I think, is due to Epstein. Feb 20, 2014 at 5:25

Theorem. If $\Omega$ is an open connected subset of $R^3$ then $\pi_1(\Omega)$ is torsion-free.
Proof. Suppose not. It is a theorem of D.B.A. Epstein (see theorem 9.8 in the book "3-manifolds" by J.Hempel) that if $M$ is a connected oriented 3-manifold whose fundamental group has nontrivial elements of finite order, then $M$ splits as a connected sum $M=M_1 \# M_2$, where $M_1$ is a compact manifold with $\pi_1(M_1)$ finite and nontrivial. Now, applying this theorem to the domain $\Omega$ we obtain that $S^3$ contains a compact submanifold $N$ with nonempty boundary, such that $\pi_1(N)$ is finite and nontrivial. It follows that the boundary of $N$ is a disjoint union of 2-spheres (see my answer here). However, attaching balls to the spherical boundary components of a manifold of dimension $\ge 3$ does not change its fundamental group. Therefore, $\pi_1(S^3)\cong \pi_1(N)$ is finite and nontrivial. Contradiction. qed
• Oh, I see: in fact, because $N \subset S^3$ and $S^3$ is irreducible, the complementary $S^3 \backslash N$ is just a disjoint union of balls whose boundary covers $\partial N$. My question was not relevant, sorry. Oct 25, 2014 at 7:24