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I am trying to prove the following, which is a problem set question from a course on differential geometry and topology that I am currently taking:

Suppose $C \subseteq S^2$ is homeomorphic to $S^1$. Prove that $S^2 - C$ has two connected comoponents.

It makes sense, if I draw a closed curve that does not cross itself on a spherical object, the curve will separate two connected regions on the spherical surface.

I feel that I have made some progress, but I am stuck. The symbol $\simeq$ stands for "is homeomorphic to" in this context.


My attempt:

Since $S^2$ is not homeomorphic to $S^1$, we have that $S^2$ is not homeomorphic to $C$. Thus, $S^2 \neq C$, so $S^2 - C$ is non-empty. Choose $p \in S^2 - C$.

Then $S^2 - \{p\}$ is homeomorphic to $\mathbb{R}^2$ via the stereographic projection $\sigma_p : S^2 - \{p\} \to \mathbb{R}^2$. The restriction of $\sigma_p$ to $C$ is a homeomorphism between $C$ and $\sigma_p(C)$. Hence, $\sigma_p(C) \simeq S^1$. I.e. $\sigma_p(C)$ is a Jordan curve.

Applying the Jordan Schoenflies theorem to $\sigma_p(C)$, we have that $\mathbb{R}^2 - \sigma_p(C)$ has two connected components, with each component having boundary $\sigma_p(C)$.

But $\mathbb{R}^2 - \sigma_p(C) = \sigma_p((S^2 - \{p\}) - C)$, and $$(S^2 - \{p\}) - C \simeq \sigma_p((S^2 - \{p\}) - C),$$ by the restriction of $\sigma_p$ to $(S^2 - \{p\}) - C.$

Since the number of connected components is preserved under homeomorphism, we deduce that $(S^2 - \{p\}) - C$ has two connected components, call them $X$ and $Y$. We have

$$(S^2 - \{p\}) - C = X \sqcup Y,$$ where $\sqcup$ denotes a disjoint union.


This is about as far as I can get.

We have $\{p\} \sqcup ((S^2 - \{p\}) - C) = S^2 - C$, and I believe that either $X \sqcup \{p\}$ and $Y$ are the connected components of $S^2 - C$, or $Y \sqcup \{p\}$ and $X$ are the connected components of $S^2 - C$. I'm not sure how to prove this though.

Any ideas on how to proceed with this line of reasoning? It would be great to see other approaches too!

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The key is to identify which of the two connected components should be completed to contain $p$. To do this, let $U_p\subset S^2$ be a neighborhood of $p$ disjoint from $C$; such a neighborhood exists because $C$ is compact. Then $\sigma_p$ will map $U_p\setminus\{p\}$ into one of the two connected components of $\mathbb{R}^2\setminus\sigma_p(C)$, and therefore $U_p\setminus\{p\}$ lies in one of the two connected components $X$ and $Y$ forming $(S^2\setminus\{p\})\setminus C$.

Without loss of generality let us say $U_p\setminus\{p\} \subset X$. Clearly $S^2\setminus C$ is the union of $X\cup\{p\}$ and $Y$. The union is disjoint, because $X$ and $Y$ were disjoint and $p$ is not an element of either $X$ or $Y$. All that remains is to check that $X\cup\{p\}$ is connected. For this, let $X\cup\{p\} = A\cup B$, where $A$ and $B$ are disjoint and open. Suppose without loss of generality that $p\in A$. Then $X = (A\setminus\{p\})\cup B$. Thus $X$ is a union of disjoint open sets. But $X$ is nonempty and connected, so one of these sets must be $X$ and the other empty. In fact it must be that $B$ is empty; for if $B=X$, then $A = \{p\}$, which is not open. We conclude that whenever $X\cup\{p\} = A\cup B$ with $A,B$ disjoint and open, one of the two must be empty and the other must be $X\cup\{p\}$. Therefore $X\cup\{p\}$ is connected and proof is complete.

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  • $\begingroup$ If $U_p$ is disjoint from $C$, then we have $\sigma_p(U_p - \{p\}) \cap \sigma_p(C) = \emptyset$, but this alone doesn't necessarily mean that $\sigma_p(U_p - \{p\})$ has an empty intersection with one of the connected components of $\mathbb{R}^2 - \sigma_p(C)$. Could this be fixed by letting $U_p$ be connected? There would be some additional proof needed then. $\endgroup$ – user657854 Nov 13 '19 at 3:17
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    $\begingroup$ Fair point. Sure, it's enough for $U_p$ to be connected, as then it can only lie in a single connected component. Arranging this isn't hard: embed $S^2$ into $\mathbb{R}^3$, give it the topology induced by the Euclidean metric, and take $U_p$ to be a small punctured disc around $p$. $\endgroup$ – Gyu Eun Lee Nov 13 '19 at 11:52

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