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{equation} \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} \end{equation}

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.