How do you give the sphere $S^2$ the structure of a closed combinatorial surface?
Thank you in advance
Edit1: I suppose I should've made it more clear. I am currently studying topology and one of the questions on the problems sheet aims to prove a weaker form of the Jordan curve theorem. The question starts as follows:
"Suppose that the sphere $S^2$ is given the structure of a closed combinatorial surface. Let $C$ be a subcomplex that is a simplicial circle. Suppose that $S^2$ \ $C$ has two components. Indeed, suppose that this is true for every simplicial circle in $S^2$. Let $E$ be one of these components. Our aim is to show that the closure of $E$ is homeomorphic to a disc.
I can't understand how do you give the sphere $S^2$ the structure of a closed combinatorial surface and how exactly do you define the closure of a set in this situation. Do you want to triangulate the sphere first?
Thanks you once again and apologies for the bad phrasing of the question at first
Edit2: A closed combinatorial surface is connected finite simplicial complex $K$ such that for every vertex $v$ of $K$, the link of $v$ is a simplicial circle. The question is from a problem sheet
Yes you want to triangulate the sphere. A triangulation is a combinatorial structure on the sphere.
The closure should be the usual closure of $E\subset \Bbb S^2$ as subspace where $\Bbb S^2$ has the euclidian topology.
Here is an example how to give the sphere a combinatorial structure:
Just take the standard 3-simplex. Which is a tetrahedron. Then its boundary has four $2$-faces and six $1$-faces and four vertices.
Should be easy to find a homeomorphism from the boundary of the 3-simplex to the sphere. Just center it at the origin and then show that the normalization map gives an homeomorphism to $\Bbb S^2$.
Now have a look at the picture on the wikipedia page. If you take a vertex then the link is just the triangle on the opposed side which is a simplicial circle.