TITLE: Example of a domain in R^3, with trivial first homology but nontrivial fundamental group
QUESTION [6 upvotes]: Let $\Omega \subset \mathbb{R}^3$ be a domain. Is it true that if $H_1(\Omega)$ = 0, then $\pi_1(\Omega) = 0$? For a counterexample, the group $\pi_1(\Omega)$ needs to be a perfect group and so I was trying with the smallest one i.e. $A_5$. But I don't think the standard construction of the space from CW complexes, embeds into $\mathbb{R}^3$.

REPLY [7 votes]: The exterior of the Alexander Horned sphere (Hatcher p.171-172) has $H_1=0$ but $\pi_1\neq 0$. (This is what Hatcher refers to as $\mathbb{R}^3\smallsetminus B$.)
Below is a diagram I made to help with Hatcher's computation of $\pi_1$ of the horned sphere. You'll find a fuller explanation in my Hatcher notes, at diagonalargument.com.
Hatcher remarks, about the computation:

In particular we see explicitly why $\pi_1(\mathbb{R}^3−B)$ has trivial abelianization, because each of its generators is exactly equal to the commutator of two other generators. This inductive construction in which each generator of a free group is decreed to be the commutator of two new generators is perhaps the simplest way of building a nontrivial group with trivial abelianization, and for the construction to have such a nice geometric interpretation is something to marvel at.