Can we divide $\mathbb{R}^2$ into two connected parts such that each part is not simply-connected?
My attempt
Put $A= \{ (0,0) \} $ and $B$ is the punctured plane.
Since that $S^1$ is a deformation retract of the punctured plane, $B$ is not simply-connected. Thus we can find a division of $\mathbb{R}^2$ such that one part is simply-connected but the other is not.
But how to deal with the problem above which requires that each part is not simply-connected?
It seems to be related to contractible and holes. But I don't know how to convert these ideas into precise mathematical language.
Any hints? Thanks in advance!
Added:
As pointed out in the comment, the counterexample exists.
Now I want to ask another question
Can we divide $\mathbb{R}^2$ into two path connected parts such that each part is not simply connected?
Best Answer
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 $\check{H}^1(X) \neq 0$.