Let $J \subset \Bbb{R}^2$ be homeomorphic to a circle. By the JCT, J separates $\Bbb{R}^2$ into two components. Is the bounded one simply connected?

Question

Question in the title. If you separate the plane with a Jordan curve, is the bounded component of its complement simply connected? Intuitively, you would think that the curve might extend to a homeomorphism of the 2-disc, with the image of the interior contained within the bounded component.

Answer 1

This is the Schoenflies problem for $\mathbb{R}^2$. There is a homeomorphism $f:\mathbb{R}^2\to\mathbb{R}^2$ such that the image of $J$ is the standard unit circle, so the bounded one is not only simply connected, but more strongly homeomorphic to an open disk.