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As a follow-up to this question, I was wondering what dimension provides the minimal counterexample to the claims:

  1. If $U\subseteq\mathbb{R}^n$ is an open connected set with trivial $H^1(U)$, then $\pi_1(U)$ is trivial;
  2. Same as above, but with a more general $U$.

Let $m$ be the minimal counterexample's ambient space's dimension.

Qiaochu's answer to the above linked question shows $m\leq 4$, since for claim 2, we have $\mathbb{RP}^2$ smoothly imbedded into $\mathbb{R}^4$ providing the example directly, whereas for claim 1 we take a tubular neighborhood of the embedded image of $\mathbb{RP}^2$, which is then open and connected, and deformation retracts (I guess, correct me if I'm wrong) onto $\mathbb{RP}^2$, thus having the same fundamental group and cohomology.

Qiaochu also stated in a comment that the fundamental group of an open subset of $\mathbb{R}^2$ is free, which I found proof-sketched here. He also stated that the Universal Coefficient Theorem (which I will sooner or later be studying for an exam) can be used to prove $H^1(U)=1\iff\mathrm{Hom}(\pi_1(U),\mathbb{R})=1$. These two facts put together imply $m_1\in\{1,3,4\}$, where $m_1$ is the $m$ for claim 1.

In dimension 1, connected subsets are intervals, hence have trivial $\pi_1$. So $m_1\in\{2,3,4\}$ and $m_2\in\{3,4\}$.

The 3-dimensional case seems to have claim 1 as an open problem, so I was wondering if I could get:

  1. An answer about 2 dimensions and claim 2; I'm guessing we can once more take a non open connected set in the plane and say a tubular neighborhoods of it would retract onto it, having the same fundamental group, thus concluding by Qiaochu's comment's "nonobvious fact" that we have no counterexamples in the plane because all fundamental groups are free; is that right?
  2. Something about the 3-dimensional case, if it is possible for claim 2; although the prhasing of that comment about the open problem seems to suggest that "open" does make a differnece, which would contrast with the tubular neighborhood retraction argument.

Edit

In case comments to that answer get trimmed (they are a lot), here are screenshots 1 and 2.

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MickG
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  • What does "more general $U$" mean? –  Mar 10 '16 at 20:29
  • Not necessarily open @Mike. Arbitrary connected subset of $\mathbb R^n$. – MickG Mar 10 '16 at 20:30
  • Note that $H_1(U)$ is torsion free for an open subset of $\Bbb R^3$ (no conditions on $H^1$); this follows from Poincare duality for noncompact manifolds and an excision argument. So torsion must live in the commutator subgroup. I don't have any further ideas. –  Mar 10 '16 at 20:47
  • See this question http://mathoverflow.net/questions/4478/torsion-in-homology-or-fundamental-group-of-subsets-of-euclidean-3-space . I doubt this is any easier. – PVAL-inactive Mar 10 '16 at 20:48

1 Answers1

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In the case of open connected subsets $U$ of $R^3$, indeed, non-simply connected implies $b_1(U)>0$. Hint: First prove it for compact 3-dimensional submanifolds $M$ of $R^3$ (with boundary): Use the formula $\chi(\partial M)= 2\chi(M)$. Then use exhaustion and the fact that direct limit commutes with the $\pi_1$-functor and with the Hurewicz homomorphism.

Moishe Kohan
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