# An exercise on components of $\mathbb{S}^2$ as a closed combinatorial surface.

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. [In fact, $\mathbb{S}^2\backslash C$ must have 2 components, but we will not attempt to prove this.]

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 $\overline{E}$.

Would it be possible for a hint on how to approach this? I thought about using the connectedness of $\overline{E}$, which gives an edge path between any two vertices of $\overline{E}$, but can't seem to make it work.

The full question leads on to a proof of a weaker Jordan Curve Theorem if that helps.

You assume that $S^2 - C$ consists of two components $E_1, E_2$. Since the interior of every 2-simplex $s\subset S^2$ is connected, if $int(s)\cap E_i\ne \emptyset$, then $int(s)\subset E_i$. Since $E_1\cap E_2=\emptyset$, $int(s)$ cannot have nonempty intersection with both $E_1, E_2$. Pick a point $p\in int(s)$. Since $S^2 - C= E_1\cup E_2$, $p$ lies either in $E_1$ or in $E_2$. Suppose it is $E_1$. Then, by the above, $int(s)\subset E_1$. Hence, $s\subset cl(E_1)$. Since $int(s)\cap E_2=\emptyset$, $s$ cannot be contained in $cl(E_2)$.
Let $e$ be an edge of $s$ contained in $C$ and that $s'$ is another simplex of $S^2$ containing this edge. I claim that $s'$ cannot be contained in $E_1$. Suppose it is contained in $E_1$ as well. Let $T^*$ be the triangulation of $S^2$, dual to the original triangulation $T$: Vertices of $T^*$ are at barycenters of $T$, edges of $T^*$. Since $s'$ is in $s'$, there is an injective polygonal path $p$ in 1-skeleton of $T^*$ which is contained in $E_1$ and which connects the barycenters $b, b'$ of $s, s'$. Now, add to $p$ the edge $e^*$ of $T^*$ connecting $b, b'$. The result is a polygonal circle $C^*$ contained in the 1-skeleton of $T^*$ which crosses $C$ exactly once. This is impossible since both loops are null-homologous and the algebraic intersection number of null-homologous loops is zero.
• More precisely, $s$ lies in the closure of precisely one $E_i$. That's what I proved. – Moishe Kohan Apr 26 '17 at 15:25