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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 $\mathbb{R}^3$.

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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) – Mariano Suárez-Alvarez Feb 17 '14 at 19:41
Here is one from MO that seems to have an answer. – John Habert Feb 17 '14 at 19:42
@JohnHabert: No, that MO post does not answer this question since here the question is about domains in 3d space. – studiosus Feb 17 '14 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. – John Habert Feb 17 '14 at 20:36
up vote 2 down vote accepted

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 $\mathbb{R}^3\smallsetminus B$.)

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Thank you so much! – Siddhant Feb 18 '14 at 13:26
Your welcome. I've posted a follow-up question, whether there's an example like this but with $\pi_1$ finite. – Michael Weiss Feb 18 '14 at 16:34

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.

This answers your question positively in the case of domains which are interiors of compact manifolds with boundary.

However, for general domains the answer is negative. Take a doubly wild arc in 3d space, meaning that it is wild at both ends. Then the complement is in general not simply connected but is acyclic.

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