Simple connectedness criterion in the plane In Rotman’s An Introduction to Algebraic Topology, chap.3, p.49, he cites the result:
Given an open connected subset $U$ of $S^2$ $(= \mathbb{C}\cup\{\infty\})$ and a base point $u_0$, $\pi_1(U, u_0) =\{1\}$ iff $S^2 \setminus U$ is connected.
He says that this result takes some work to prove & doesn’t seem to prove it in the book but I could be wrong. Can a proof or a reference for this result be suggested?
 A: Since $U$ is open and connected, it must be path-connected, and any two connected components of $S^2 \setminus U$ will be separated. So if $S^2 \setminus U$ has at least 2 connected components, any loop in $U$ that circles one of them will not be null-homotopic.
Picture for reference: Here is an open set $U$ (blue) where the complement $S^2 \setminus U$ (yellow) has multiple connected components:

Take a loop $\gamma$ in $U$ that circles one of the connected components of $S^2 \setminus U$ and pick a point $p \in S^2 \setminus U$:

There cannot be a null-homotopy of $\gamma$ that stays inside $U$ the entire time. If there was, that would imply $\Bbb{R}^2 \setminus \{ p \}$ was simply connected: we could homotope any loop in $\Bbb{R}^2 \setminus \{ p \}$ that circled $p$ so that its image matched the image $\gamma$, and then apply our null homotopy (which deforms this loop entirely within the blue region) to shrink this loop to a point. But we know $\pi_1(\Bbb{R}^2 \setminus \{ p \}) \cong \Bbb{Z}$ is nontrivial.
We started by assuming $S^2 \setminus U$ had multiple connected components and ended by using this fact to show that $U$ contains loops which cannot be shrunk to a point inside $U$. This shows that, when $S^2 \setminus U$ has multiple connected components, $\pi_1(U)$ is nontrivial. This is the contrapositive of Rotman's statement, which is that $\pi_1(U)$ being the trivial group implies $S^2 \setminus U$ is connected.
