I guess that one can make the compact surface by edge-pairing. I'm trying to rearrange the triangles so that they could be considered as parts of a disk.
Let $S=\bigcup_{i=1}^n T_i$. Then we are able to reorder these triangles so that $\bigcup_{i=1}^j T_i$ and $T_{j+1}$ shares at least one edge, $e_j$. By using mathematical induction, I want to show the quotient space of $\bigcup_{i=1}^j T_i$ obtained after gluing $e_1,\cdots,e_{j-1}$ is a topological disk.
However there are two ways of gluing the edges. One is just gluing one edge of a polygon and that of a triangle not in the polygon. The other one is gluing two distinct edges of the polygon. How can I distinguish
How can I use the induction by integrating the above 2 ways? Or is there easier proof without using the mathematical induction?
Any help will be appreciated.
Construct a graph which has one vertex at the center of each triangle and in which two vertices are connected iff the corresponding triangles share a side (this is called the dual graph for the triangulation)
In this graph construct an arbitrary spanning tree. Now cut the surface along the sides of the triangles which do not correspond to sides in the spanning tree. You get a disk, and clearly the surface can be reobtained by identifying things in its boundary.