how to show any edges of a closed surface M is on exactly two triangles of M.

Question

I just started to learn a book about surfaces on graph, here is my definition of closed surface:

a closed surface is a collection $M$ of triangles (in some Euclidean space) such that

(a) $M$ satisfies the intersection condition (i.e. two triangles either are disjoint or have one vertex in common or have two vertices and consequently the entire edge joining them, in common)

(b) $M$ is connected

(c) for every vertex $v$ of a triangle of $M$, the link of $v$ is a simple closed polygon.

I know little about topology, if the proof requires some background about it, please point it out so that I can learn it, thanks.

Answer 1

I asked the question a few days ago, and now I (possibly) figure out the solution.

For any edge $uv$ in the surface, consider $v$'s link, by definition the link is a simple closed polygon. Any triangle containing $uv$ as a edge will contribute another edge as part of the link. In the polygon we know there are two edges have $u$ as a endpoints, thus we have two triangles containing the edge $uv$ in the surface.