# Example of a domain in R^3, with trivial first homology but nontrivial fundamental group

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 $\mathbb{R}^3$.

• If you search a bit you will see that something awfully related to this has been asked here and on MO ages ago (with no conclusive answer, btw) Feb 17, 2014 at 19:41
• Here is one from MO that seems to have an answer. Feb 17, 2014 at 19:42
• @JohnHabert: No, that MO post does not answer this question since here the question is about domains in 3d space. Feb 17, 2014 at 20:24
• @studiosus It has been a while since I studied topology but I believe the second answer there should embed into 3-space. Feb 17, 2014 at 20:36

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.

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.
• Your welcome. I've posted a follow-up question, whether there's an example like this but with $\pi_1$ finite. Feb 18, 2014 at 16:34
• This is an immediate consequence of Prop. 2B.1 of Hatcher (p.169). Alternately, Hatcher points out on p.172 that $\pi_1$ has trivial abelianization, and the abelianization of $\pi_1$ is $H_1$ (Theorem 2A.1, p.166). Apr 28, 2017 at 17:10
First, suppose that you have a compact connected submanifold $C$ with nonempty boundary in 3d sphere. If some boundary components are spheres, you add the 3-ball which they bound $C$ without changing 1st homology or fundamental group. Suppose the result, which I will still call $C$, still has nonempty boundary. Recall that $$\chi(C)=\chi(\partial C)/2,$$ Which immediately implies that the 1st Betti number of $C$ is positive. Hence, we are done in this case. The remaining possibility is that the original $C$ was simply connected to begin with.