Proving a weaker version of the Jordan Curve Theorem (by induction)

Question

I'm trying to solve the following question:

Suppose that the sphere $ \mathbb{S}^2 $ is given the structure of a closed combinatorial surface. Let $C$ be a subcomplex that is a simplicial circle. Suppose that $ \mathbb{S}^2\backslash C$ has two components. Indeed, suppose that this is true for every simplicial circle in $ \mathbb{S}^2 $ . Let $E$ be one of these components.

The aim is to show that $\bar{E}$ is homeomorphic to a disc. Prove this statement by induction on the number of 2-simplices in $\bar{E}$.

(1): Let $\sigma _1$ be a 1-simplex in $C$ . Since $\mathbb{S}^2$ is a closed combinatorial surface, $\sigma _1$ is adjacent to two 2-simplices. Show that precisely one of these 2-simplices lies in $\bar{E}$.

(2): Start the induction by showing that if $\bar{E}$ contains at most one 2-simplex, then $\bar{E}=\sigma_2$

(3): Let $v$ be the vertex of $\sigma_2$ not lying in $\sigma_1$. Let’s suppose that $v$ does not lie in $C$ . Show how to construct a subcomplex $C'$ of $\mathbb{S}^2$ , that is a simplicial circle, and that has the following properties:

• $\mathbb{S}^2\backslash C'$ has two components
• one of these two components $F$ is a subset of $E$
• $\bar{F}$ has fewer 2-simplices than $\bar{E}$

Show in this case that $\bar{E}$ is homeomorphic to $\mathbb{D}^2$, with the homeomorphism taking $C$ to $\partial\mathbb{D}^2.$

(4): Suppose now that $v$ lies in $C$. How do we complete the proof in this case?

I've managed to complete the first three parts, but am now stuck on (4).

I know you can work your way round the 1-simplices in $C$ using each in place of $\sigma _1$ in turn, and if you come across one where it's associated $v$ is not in $C$ then just apply the result from (3).

Surely though it would be possible that the corresponding 2-simplex for each 1-simplex in $C$ had it's third vertex in $C$ - how else would you show it's homeomorphic, because I can't thing of a way to make the same trick work?

I'd really appreciate any help you could offer!

Answer 1

I suggest to prove the following by induction:

Let $C$ be a simplicial circle in $S^2$. If $E$ is a component of $S^2 \setminus C$ such that the number of $2$-simplices in $\overline{E}$ is at most $n$, then $\overline{E}$ is homeomorphic to a disc.

Obviously $n=0$ is impossible. The case $n=1$ has been considered in your question.

Now let us consider the case that that the number of $2$-simplices in $\overline{E}$ is $n+1$, where $n \ge 1$.

We take a $1$-simplex $\sigma_1 = (v_1,v_2)$ in $C$ and consider the unique $2$-simplex $\sigma_2 = (v_1,v_2,v)$ belonging to $\overline{E}$ which exists by (1). The case that $v$ is not in $C$ has been settled and you ask what happens if $v$ is in $C$.

Since $C$ is a simplicial circle, we can arrange its $1$-simplices in a chain $\sigma^1_1,\dots,\sigma^m_1$ beginning an ending at $v_1$ such that $\sigma^1_1 = \sigma_1$. There is one $\sigma^i_1$ ending at $v$. We get two simplicial circles $C_1$ and $C_2$ given by $(v,v_1), \sigma^1_1,\dots,\sigma^i_1$ and $\sigma^{i+1}_1,\dots,\sigma^m_1,(v_1,v)$. They have the following properties:

• $S^2\setminus C_k$ has two components
• one of these two components $F_k$ is a subset of $E$
• $\bar{F}_k$ has fewer 2-simplices than $\bar{E}$, i.e. at most $n$

Hence $\bar{F}_k$ are (simplicial) discs and we have $\bar{F}_1 \cup \bar{F}_2 = \bar{E}$. $\bar{F}_1$ and $\bar{F}_2$ intersect in a $1$-simplex lying in the boundaries of both. It is then easy to show that their union is a homeomorphic to a disc.