Components of $\mathbb{S}^2$ as a closed combinatorial surface.

Question

I'm trying to solve the following problem, and I'm not having much luck:

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.

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 $\overline{E}$.

I've managed to show that each of the two 2-simplices adjacent to $\sigma _1$ must be in one of the two components, but I can't work out how to show that they must be in different components (i.e. that if one is in $E$ then the other must be in the other component).

I saw the answer to this question An exercise on components of $\mathbb{S}^2$ as a closed combinatorial surface., but the relevant part of the answer seems much more advanced than any of the material I've been learning (I haven't encountered barycenters or null-homologous loops, for instance).

I'd really appreciate it if someone could give me a hand.

Answer 1

The first thing to observe is the following:

Lemma 1: For every 2-simplex $\tau$ in $\mathbb S^2$ there exists a 2-simplex $\tau'$ which shares an edge with $C$ and such that $\tau$ and $\tau'$ are in the same component of $\mathbb S^2 - C$.

Let me assume that you believe that. As an immediate consequence,

Corollary 2: For each of the two components of $\mathbb S^2 - C$ there exists a 2-simplex in that component that shares an edge with $C$.

Okay, now for the real work. starting from $\sigma_1$, let's list the 1-simplices along $C$ in order as $C = \sigma_1 \cup \sigma_2 \cup\cdots\cup \sigma_K$. Let $v_k = \sigma_{k} \cap \sigma_{k+1}$ (for all $k=1,...,K$, using addition modulo $K$).

The two 2-simplices on opposite sides of $\sigma_1$ can be written $\tau_{1L}$, $\tau_{1R}$ (if you imagine yourself standing on $\sigma_1$ looking towards $\sigma_2$, $\tau_L$ is on your left and $\tau_R$ is on your right).

Now move forward from $\sigma_1$ past $v_1$ to $\sigma_2$, and you'll see that the 2-simplices on opposite sides of $\sigma_2$ can be written as $\tau_{2L}$, $\tau_{2R}$ so that $\tau_{1L}$ and $\tau_{2L}$ are in the same component of $\mathbb S^2 - C$, and so that $\tau_{1R}$ and $\tau_{2R}$ are in the same component of $\mathbb S^2 - C$.

Continue in this fashion by induction. For each $k=1,...,K$ you will then have listed the 2-simplices that share an edge with $C$ as $\tau_{1L},...,\tau_{KL}$ and $\tau_{1R},...,\tau_{KR}$ so that $\tau_{1L},\tau_{2L},...,\tau_{KL}$ are all in the same component of $\mathbb S^2 - C$, and so that $\tau_{1R},\tau_{2R},...,\tau_{KR}$ are all in the same component of $\mathbb S^2 - C$.

If $\tau_{1L}$ and $\tau_{1R}$ were in the same component of $\mathbb S^2 - C$ then it would follow that all 2-simplices sharing an edge with $C$ are in a single component of $\mathbb S^2 - C$, contradicting Corollary 2.

Therefore, $\tau_{1L}$ and $\tau_{1R}$ are in opposite components of $\mathbb S^2 - C$.