Open and connected subset in $\mathbb{R}^2$

Question

If $U$ is an open and connected subset in $\mathbb{R}^2$, then it is path connected. In further, we assume that $U$ is in a unit ball.

If its complement $U^c$ is connected, then show that $U$ is simply connected.

[Add] Definition : $X$ is simply connected if there is a continuous map $h: D^2\rightarrow X$ with $h|\partial D^2=c$ when $c: S^1\rightarrow X$ is continuous map.

Answer 1

I'm not sure if this should be an answer or a comment, since it is an idea, or "sketch of proof ".

Consider the equivalent definition: $X$ is a simply connected space if it is path-connected and for all $x\in X$, any loop with basepoint $x$ (i.e. continuous $\alpha : I \rightarrow X$ such that $\alpha (0)=\alpha (1)=x$) is homotopy equivalent to the constant path $x$.

Suppose $U$ is not simply connected and take a simple closed curve (this is a loop), not homotopy equivalent to a constant path. This curve contained in $U$ divides the plane in two regions $A$ and $B$, one of them bounded, say $A$. The interior of $A$ has points of $U^c$, otherwise you could shrink the loop to a point staying within $U$. Observe that $B\cap U^c \neq \varnothing$ as well, since $U$ is bounded.

Now to show that $U^c$ is not connected, consider the pair of disjoint nonempty open sets of $U^c$ given by taking intersection of $U^c$ and the interiors of $A$ and $B$.

Answer 2

Consider $U$ inside the Riemann sphere. Since $U$ is contained in the unit ball, its complement $U^c$ in the complex plane remains connected when we add the point at infinity (i.e., when we consider the complement in the Riemann sphere instead). From now on, we then shift the problem to $U$ an open subset of a Riemann sphere which has as complement a connected set.

By Poincaré-Alexander duality, we then have that

$$H_1(U) \simeq \widetilde{\check{H}}{}^0(U^c) \simeq 0.$$

Since open subsets of the plane have free fundamental groups (free groups, not free abelian) and $H_1$ is the abelianization of the fundamental group, it follows that $U$ is simply connected (the abelianization of a free group with $\lambda$ generators is the free abelian group with $\lambda$ generators. Thus, if the abelianization is $0$, then so is the original).

OBS: $0$-th Cech-cohomology counts quasicomponents of the space. Since $U^c$ is compact and Hausdorff, this coincides with the components.