First de Rham cohomology vs fundamental group of an open subset of the Euclidean space

Question

If I understand properly, wikipedia claims the following:

Let $U\subseteq \mathbb R^3$ be a path-connected open subset. Then $H^1_\mathrm{dR}(U) =0$ if and only if $U$ is simply-connected.

I certainly agree that if $U$ is simply connected, then first de Rham cohomology vanishes. I am however unsure why the converse would be true. (For a general $3$-manifold it is false – there are homology $3$-spheres which have non-trivial fundamental group and trivial first and second de Rham cohomology).

As far as I can say, the argument in wikipedia proceeds as follows:

1. A general loop in the fundamental group of $U$ can be deformed into a piecewise-linear one. (Why? I have never seen the proof of this).
2. Therefore $H_1(U)$ vanishes if and only if $\pi_1(U)$ vanishes. (Why?)
3. Moreover, $H_1(U)$ is torsion-free. (Why?)
4. Hence, we can use Universal Coefficient Theorem for cohomology (and de Rham theorem) to get $$ H^1_\mathrm{dR}(U)\simeq \mathrm{Hom}_{\mathbb Z}( H_1(U), \mathbb R ). $$ Assuming 3. we know that either both $H^1_\mathrm{dR}(U)$ and $H_1(U)$ vanish or neither of them. (This is true if $H_1(U)$ is finitely generated. I am unsure if this holds in general).
5. Using 4. and 2. we get the claim.

Therefore my question is:

Is the claim even true? Is there any reference providing a detailed proof of it? (Or at least the proofs of steps 1–3, which look suspicious to me?)

Edit: Roberto Frigerio's comment under this answer suggests that the claim is not true unless one puts additional conditions on $U$...

Answer 1

The claim as stated is false. What is true is that if $U \subset \mathbb{R}^3$ is open, then $H_1(U;\mathbb{Z})=0$ if and only if $H^1_{dR}(U)=0$. Basically, the error lies in statement $2.$

So let's first talk about why the claim is false. It is known that there exists an embedding $K$ of the disk $D^2$ in $S^3$ such that $S^3-K$ is not simply connected. However, every embedding of any disk of any dimension in any sphere of any dimension is such that the complement is acyclic, i.e. has vanishing (reduced) singular homology. (You can see this in the chapter about Jordan's curve theorem in Bredon's book Topology and Geometry.) We can embed $S^3-K$ in $\mathbb{R}^3$ by picking a stereographic projection based on an element of $K$, and this gives us our open set that serves as a counterexample.

However, the claim that $H_1(U;\mathbb{Z})$ is torsion-free is true, and with that the result I mentioned follows, as you seem to observe in the question. If I recall correctly, a proof goes as follows: embed $U$ again in the sphere and let $C$ be the complement. By Alexander duality (also in Bredon's book), $$H_1(U;\mathbb{Z})\simeq \check{H}^1(C;\mathbb{Z}). $$ Now let $X$ be any space. By the universal coefficients theorem, $H^1(X;\mathbb{Z}) \simeq \mathrm{Hom}(H_1(X);\mathbb{Z})$, since the $\mathrm{Ext}$ part vanishes as $H_0$ is free abelian. On the other hand, $\mathrm{Hom}(H_1(X);\mathbb{Z})$ is torsion-free. Thus, $H^1(X;\mathbb{Z})$ is torsion-free. It follows that $\check{H}^1(C;\mathbb{Z})$ is a direct limit of torsion-free abelian groups, and therefore is also torsion-free. (If I recall correctly, one way to see this is by noting that over $\mathbb{Z}$ this is equivalent to being flat, and flatness is preserved by direct limits.) So $H_1(U;\mathbb{Z})$ is torsion-free.

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Some further points:

• In the comments, Moishe Kohan mentions that the wikipedia article first assumes that $U$ is an open subset of $\mathbb{R}^3$ and then switches mid-proof to being an open subset of $\mathbb{R}^2$. In $\mathbb{R}^2$, it is true that $H_1=0 \iff \pi_1=0$, and thus by making minor adaptations it is true that $\pi_1=0 \iff H^1_{dR}=0$. This is due to the fact that an open subset of $\mathbb{R}^2$ must have a free fundamental group, and thus $H_1$ is the free abelian group with the same number of generators of the $\pi_1$ by Hurewicz's theorem. This, of course, does not exempt the sloppiness on the wikipedia article.
• The result that $H_1=0 \iff H^1_{dR} =0$ does not hold for general open subsets of Euclidean spaces. The simplest example is obtained by embedding $\mathbb{R}P^2$ in some $\mathbb{R}^n$ and taking a tubular neighbourhood of it. This will have trivial $H^1_{dR}$, but $H_1=\mathbb{Z}_2$. This also shows that the result is not true as soon as $n=4$, since $\mathbb{R}P^2$ embeds in $\mathbb{R}^4$.