Recently I started reading Ethan Bloch's "A First Course in Geometric Topology and Differential Geometry" and I came upon this exercise to deduce the Jordan Curve Theorem from the Schönflies Theorem:
Schönflies Theorem: Let $C\subseteq \mathbb{R}^2$ be a 1-sphere (homeomorphic image of $S^1$). Then there is a homeomorphism $H: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ such that $H(S^1)=C$ and $H$ is the identity map outside a disk (homeomorphic image of $D^2$).
Jordan Curve Theorem: Let $C\subseteq \mathbb{R}^2$ be a 1-sphere. Then $(1)$ the set $\mathbb{R}^2 \setminus C$ has pricisely two components, one of which is bounded and one of which is unbounded and $(2)$ the union of $C$ and the bounded component is a disk, of which $C$ is the boundary.
My attempt so far:
(1) $\mathbb{R}^2 \setminus C = H(\mathbb{R}^2 \setminus S^1) = H(\mathbb{R}^2 \setminus D^2)\cup H(intD^2)$ (disjoined)
- $H(intD^2) \subseteq H(D^2)$ which is a continuous image of a compact set and therefore is compact (especially bounded). Hence $H(intD^2)$ is also bounded.
- Since the unbounded $\mathbb{R}^2 = H(\mathbb{R}^2 \setminus D^2)\cup H(intD^2) \cup C$ and $C$ is compact (therefore bounded) we must have that $H(\mathbb{R}^2 \setminus D^2)$ is unbounded.
(2) Since $H^{-1}(C \cup H(intD^2)) = S^1 \cup intD^2 = D^2$ we have that $C \cup H(intD^2)$ is a disk. Also, $\partial[C \cup H(intD^2)] = \partial H(D^2) = H(\partial D^2) = H(S^1) = C$.
Is there something wrong with my solution? I feel like I'm missing something because I didn't use the second part of Schönflies Theorem, but I can't find any gap in my reasoning. Is there something missing?
Thank you in advance!