Can we divide $\mathbb{R}^2$ into two path connected parts such that each part is not simply-connected?

Solution 1:

If there is an example then it is nasty. With some mild hypotheses it's not possible. For notation consider $S^2 = \mathbb{R}^2 \cup \{\infty\}$.

Suppose that $\mathbb{R}^2 = X \cup Y$ and that $X$ is path connected but not simply connected. Suppose in addition that $X$ is a compact CW-complex. Then it deformation retracts onto a graph, so its fundamental group is free (and nontrivial by hypothesis), thus its first cohomology group is nontrivial. Then by Alexander duality, $\tilde{H}_0(S^2 \setminus X) \cong \tilde{H}^1(X) \neq 0$.

Note that $S^2 \setminus X = Y \cup \{\infty\}$. Since $X$ is compact, it follows that $Y$ is a neighborhood of $\infty$. Removing a point from an open subset of $S^2$ does not change its connectivity. It follows that $\tilde{H}_0(Y) = \tilde{H}_0(S^2 \setminus X) \neq 0$ and therefore $Y$ is not path-connected.

There is probably a way around the hypothesis that $X$ is compact (if $X$ is noncompact then add $\infty$ to it and then $S^2 \setminus X = Y$). However I'm not sure how to get rid of the assumption that $X$ is nice; it has to be at least locally contractible to apply Alexander duality. Perhaps one can work something out with Čech cohomology, but then I don't know how to see that $\check{H}^1(X) \neq 0$.

Solution 2:

Both $\{ (x,\sin (1/x)) \mid x \neq 0 \} \cup \{(0,0)\}$
and it's complement are connected.
Since they are not path connected,
they are not simply connected.

It is an example of two connected sets that pass through each other.

Are there two path connected, not simply connected sets whose union is the plane?

Solution 3:

The answer to the modified question:

Can we divide $\mathbb R^2$ into two path connected parts such that each part is not simply connected?

is no, this is not possible. In fact, if $A$ and $B$ are disjoint sets with $A\cup B=\mathbb R^2$ then, if $A,B$ are path connected but $A$ is not simply connected, then $B$ must be bounded. So, if $A$ and $B$ are both path connected but not simply connected, then they must both be bounded, giving a contradiction.

I will make use of the following lemma, from the paper The fundamental groups of subsets of closed surfaces inject into their first shape groups.

Lemma 13: Let any set $A\subseteq\mathbb R^2$, and map $\alpha\colon S^1\to A$ be given. Let $U$ be the unbounded connected component of $\mathbb R^2\setminus{\rm Im}(\alpha)$. If $\alpha\colon S^1\to A$ is null-homotopic, then so is $\alpha\colon S^1\to A\setminus U$.

So, suppose $A$ is connected but not simply connected. Then there is a curve $\alpha\colon S^1\to A$ which is not null-homotopic. The complement $\mathbb R^2\setminus {\rm Im}(\alpha)=\bigcup_iU_i$ decomposes as the union of its connected components $U_i$. As $B\subseteq\bigcup_iU_i$ is connected, it will be contained within one of the $U_i$. We show that $U_i$ is bounded: As $\mathbb R^2\setminus U_i\subseteq A$, $\alpha$ is not null-homotopic in $\mathbb R^2\setminus U_i$. If $U_i$ were unbounded, then the quoted result says that $\alpha$ is not null-homotopic in $\mathbb R^2$, a contradiction.

Solution 4:

So here's my partial answer. Assume additionally that

In each space there is a Jordan curve that is not null homotopic.

Let's assume $\lambda$ is such curve in $X$. With that we apply the Jordan curve theorem. Say $\mathcal{I}$ is the interior of $\lambda$ and $\mathcal{E}$ the exterior. If $\mathcal{I}\cap Y\neq\emptyset$ and $\mathcal{E}\cap Y\neq\emptyset$ then $Y$ cannot be path connected because any path (between exterior and interior) would have to cross $\lambda$.

On the other hand if $\mathcal{I}\cap Y=\emptyset$ then $\mathcal{I}\subseteq X$ and so $\lambda$ is null homotopic in $X$ because by the Jordan–Schoenflies theorem the interior of $\lambda$ is homeomorphic to a ball.

So the only possibility is that $\mathcal{E}\cap Y=\emptyset$, i.e. $\mathcal{E}\subseteq X$.

The same reasoning can be applied to $Y$ in order to get that the exterior of some Jordan curve has to be fully contained in $Y$. This contradicts them being disjoint.

So I couldn't find an example of a path-connected, not simply-connected subset of $\mathbb{R}^2$ which doesn't have a not null homotopic Jordan curve. But I also couldn't prove that this is not possible. Any help appreciated. Note that such subsets do exist in $\mathbb{R}^3$ (see here).

Side note: If we don't require that both have to be path-connected, but at least one, then there's a neat counterexample of $X=\mathbb{Q}^2$ and $Y=\mathbb{R}^2\backslash\mathbb{Q}^2$. Note that $Y$ is path-connected but not simply connected.