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!


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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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