## Triangulation of compact surfaces

A triangulation of a compact surface $S$ is a finite family of closed subsets $\{ T_1, T_2, \dotsc, T_m\}$ that cover $S,$ and a family of homeomorphisms $\varphi_k: \mathcal{T} \to T_k$, where $\mathcal{T} \subset \mathbb{R}^2$ is a proper triangle in the plane. We say that the sets $T_k$ are triangles as well, and the images by $\varphi_k$ of a vertex (resp. edge) of the triangle $\mathcal{T}$ is also called a vertex (resp. edge). We impose one condition: Given two different triangles $T_k$ and $T_j,$ their intersection must be either void, or a common vertex, or a common edge.

Since the (surface of a) tetrahedron is homeomorphic to the sphere $\mathbb{S}_2,$ we may consider this as a valid triangulation for the latter. This triangulation is formed basically by four triangles, four vertices and six edges.

The following example shows a triangulation of a torus, performed on the representation of $\mathbb{T}$ given by the quotient space of the square $\square_2$ by the proper identification in the border. Note that in this representation, all of the vertices of the square are actually the same vertex, which we denote $P_1.$ We construct a triangulation by placing two more vertices on the horizontal borders, $P_2, P_3,$ two more vertices in the vertical borders, $P_4, P_5,$, four more vertices inside the square, $P_6, P_7, P_8, P_9,$, and joining all of them with edges, as shown in the image below:

This gives the following triangles:

$\begin{array}{ccc}P_1P_2P_5&P_2P_5P_6&P_2P_3P_6\\P_3P_6P_7&P_1P_3P_6&P_1P_5P_7\\P_4P_5P_6&P_4P_6P_8&P_6P_7P_8\\P_7P_8P_9&P_4P_7P_9&P_4P_5P_7\\P_1P_2P_4&P_2P_4P_8&P_2P_3P_8\\P_3P_8P_9&P_3P_4P_9&P_1P_3P_4\end{array}$

Two important observations about triangulations of any compact surface:

• Each edge belongs to exactly two triangles. This is due to the fact that the surfaces have at every point, a neighborhood that is homeomorphic to an open ball.
• Given a vertex $v$ in a proper triangulation, it is possible to sort all the triangles that share than vertex, say clock or counterclockwise, in such a way that two consecutive triangles share a common edge. This is a direct consequence of the previous observation, and the fact that the union of all the triangles sharing a common vertex must be a connected set.