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Let $U$ be a simply connected open set in $\mathbb{R}^2$. If $C$ is a simple closed curve (a space homeomorphic to unit circle $S^1$) lying in $U$, then each bounded component of $\mathbb{R}^2\setminus C$ also lies in $U$.

This is an exercise from Munkres' topology, I don't know how to use the hypothesis that $U$ is simply connected, thanks for any help.

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Arguing by contradiction, suppose there exists $x$ contained in the bounded component of $\mathbb{R}^2 \setminus C$ such that $x \not\in U$.

Choose a base point $p \in C$ and parameterize $C$ by a closed path $\gamma$ based at $p$ and going exactly once around $C$. This closed path represents an element of $\pi_1(U,p)$ which I'll denote $[\gamma]_U$.

The key fact to use is that $\pi_1(\mathbb{R}^2 - \{x\},p)$ is an infinite cyclic group and $\gamma$ represents a generator of that group. I'll denote that generator $[\gamma]_{\mathbb{R}^2- \{x\}}$.

We also know that the inclusion $i : U \to \mathbb{R}^2$ induces a homomorphism $i_* : \pi_1(U,p) \to \pi_1(\mathbb{R}^2,p)$.

Since $i \circ \gamma = \gamma$, it follows that $$i_*\bigl([\gamma]_U\bigr) = [\gamma]_{\mathbb{R}^2- \{x\}} $$ Since the right hand side is a nontrivial element of the group $\pi_1(\mathbb{R}^2,p)$, it follows that $[\gamma]_U$ is a nontrivial element of the trivial group $\pi_1(U,p)$, a contradiction.

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  • $\begingroup$ How to show that $[\gamma]$ is a generator for $\pi_1(\mathbb{R}^2\setminus\{x\},p)$? It is very intuitive, but seems difficult for me to prove. $\endgroup$
    – Xiang Yu
    Mar 25 '16 at 14:48
  • $\begingroup$ I believe this is done in Munkres book for the special case that $C$ is a circle centered on $x$. The general case may be derived from that special case by applying the Jordan-Schonflies theorem. $\endgroup$
    – Lee Mosher
    Mar 25 '16 at 14:53
  • $\begingroup$ By the way, $\mathbb{R}^2-C$ has only one bounded component; that's part of the Jordan Curve Theorem. $\endgroup$
    – Lee Mosher
    Mar 25 '16 at 14:55
  • $\begingroup$ Yes, this exercise is before the section of the Joran curve theorem, so it assumes that you only know the Joran separation theorem, not the full version of the Jordan curve theorem. $\endgroup$
    – Xiang Yu
    Mar 25 '16 at 14:59
  • $\begingroup$ May I ask a question? If we have additional hypothesis that $U$ is also bounded, how to show the complement $U^c$ is connected? $\endgroup$
    – Xiang Yu
    Mar 25 '16 at 16:11
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Taking complement, it suffices to show that $U^c$ lies in the unbounded component of $C$. Since $C$ is a simple closed curve, there is a homeomorphism $g:S^1\to C$ from $S^1$ to $C$. Now we use the assumption that $U$ is simply connected to conclude that the map $g:S^1\to U$ is path-homotopic (with range in $U$) to a constant map, hence nulhomotopic. Since $g$ is a homeomorphsim, we see that the inclusion map $i:C\to U$ is also nulhomotopic. Let $x\in U^c$ be a point not in $U$. Note that the map $i:C\to \mathbb{R}^2\setminus\{x\}$ is continuous injective, and nulhomotopic (note that $\mathbb{R}^2\setminus\{x\}$ contains $U$), it follows from Borsuk lemma (see Munkres' topology Lemma 62.2) that $x$ lies in the bounded component $\mathbb{R}^2\setminus C$.

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