Q:

Why is a triangulation of a surface not a polygonal decomposition

In Wilson's Curved Spaces, he says 'every triangulation of the sphere/torus can be extended to a polygonal decomposition' but I cannot see why, given his definitions, it is not already a polygonal decomposition?
For clarity his definitions are:
Triangulation of a compact metric space X
A finite collection of topological triangles whose union is X, where
1) Two triangles are either disjoint or their intersection is either a common vertex or a common edge
2) Each edge is an edge of exactly two triangles
(He defines a topological triangle as the image of a closed triangle in the plane under a local homeomorphism from an open neighbourhood containing it- the triangle is the image of the bounded set in the plane)
Polygonal decomposition
A finite collection of polygons which cover X and whose interiors are disjoint. The edges in the decomposition correspond to sides of the polygons, and the vertices to vertices of the polygons. The interior of every edge contains no vertices and is a side of just two polygons in the decomposition.
A polygon is defined as the image under local homeomorphism of the bounded connected component of a closed curve consisting of a finite number of straight line segments.
I don't understand why triangulation isn't a specific example of this?

A:

This was a misunderstanding of the definition of a polygonal curve in X, since he doesn't define this properly in the book. A polygon in X is the bounded connected component of the complement of a closed curve consisting of a finite number of 'line segments in X'.
He defines a line segment in X as a curve whose length is the distance between the endpoints, so while the edges of the triangulation can be whacky shapes- depending on the local homeomorphism; the edges of the polygonal decomposition are always geodesics.

