Riemann Mapping theorem in triangulations I am reading the paper 'Rotation Distance, Triangulations, and Hyperbolic Geometry' by Thurston et al. The authors are constructing a sequence of triangulation from a regular icosahedron. Each face of the icosahedron is divided into $k^2$ equilateral triangles, giving $20k^2$ triangles in all and $n=10k^2+2$ vertices. The vertices of the triangulation of the sphere defined above are mapped into hyperbolic space in such a way that the resulting polyhedron has a large volume. For this purpose, a map is defined by the Riemann mapping theorem. I need to understand the following:

By the Riemann mapping theorem: corresponding to the icosahedron, there is a subdivision of the sphere into triangles bounded by segments of great circles, obtained by projecting the edges of the icosahedron to the spherical triangles, sending vertices of the icosahedron to the corresponding vertices of the spherical triangles.

Can you either explain the above statement or give me some references?
 A: I have no clue for the reference to the "Riemann mapping theorem" here. The rest of that paragraph is unrelated to it. Maybe this is just set up for some later application of the Riemann mapping theorem?
As for the description, that is simple projection. And it works the same in either Euclidean or Hyperbolic geometry. 
Take any polyhedron. If there is a point $p$ in the interior such that the line segment from $p$ to any point on the surface does not intersect the surface except at that point, the polyhedron is said to be starlike about $p$. Now consider a sphere centered at $p$ and large enough to completely contain the polyhedron. Now project the vertices, edges and faces of the polyhedron onto the sphere: for each point $q$ of the polyhedron, map $q$ to the intersection of the ray from $p$ passing through $q$ with the sphere. Think of a light being placed at $p$. This mapping takes each point $q$ to its shadow on the sphere.
Vertices are each projected to unique points on the sphere. Each edge projects to a great-circle segment connecting the projections of its end points. (To see that the edge projects to a great-circle, consider the plane determined by $p$ and the two endpoints of the edge. Since the edge is a line segment connecting two points in the plane, it lies entirely in this plane, as do all the rays that eminating from $p$ and passing through points on the edge. Thus the projection of the edge on the sphere also lies within this plane. But the intersection of a sphere with any plane through its center is a great-circle.) Because the polyhedron is starlike about $p$, these great-circle segments cannot intersect each other except at their endpoints. For any such point of intersection would require the line segment connecting $p$ to one of the points being projected to pass through the other point. A continuity argument shows that the projection of each face is the region of the sphere bounded by the projection of its edges.
This projection is what the paragraph is talking about.
