# Systematic approach to triangulation closed combinatorial surfaces

I was wondering whether there is a systematic approach to the triangulation of closed combinatorial surfaces, which we know can be shown to be homeomorphic to polygons with complete set of side identifications on the way to the classification theorem, or just more generally to the manifolds obtained from the sphere by adding $$g$$ handles or $$h$$ crosscaps.

Well, intuitively, we should be able to triangulate a handle/crosscap alone, and then we also know of a simple triangulation of the sphere (two tetrahedrons glued together on a side with the interior of that face removed), but what concerns is me whether such cut-paste of those separate triangulations would generally result in a triangulation of the desired space?

TL;DR: knowing that triangulations of closed combinatorial surfaces exist (by definition), do we know of a simple way to write those triangulations down, knowing the class of the surface given (i. e. $$M_g$$ or $$N_h$$)?

...starting from a polygon with a complete set of side identifications, say?

There is a simple answer to your question which ignores all of your remarks about handles and crosscaps. Starting from the polygon $$P$$, first form the 2nd barycentric subdivision of $$P$$, and then glue the sides of $$P$$. After gluing, the 2nd barycentric subdivision of $$P$$ becomes a triangulation of the surface.
You may wonder: one knows what the 2nd barycentric subdivision of a triangulation is, but what is the 2nd barycentric subdivision of the polygon $$P$$? Let me assume for simplicity that $$P$$ is a regular polygon in the plane. The 1st barycentric subdivision of $$P$$ is the following triangulation of $$P$$. Its $$0$$-skeleton contains the original vertex set of $$P$$, plus some additional vertices, namely the midpoint $$M_E$$ of each edge of $$P$$ and the center $$C$$ of $$P$$. Its $$1$$-skeleton has two edges obtained from each edge $$E$$ by subdividing it at the point $$M_E$$, plus some additional edges, namely $$CV$$ for each vertex $$V$$ of $$P$$, and $$CM_E$$ for each edge of $$E$$. And now, as expected, the 2nd barycentric subdivision of $$P$$ is the 1st barycentric subdivision of its 1st barycentric subdivision.