It doesn't seem obvious at all. But consider eight houses that have collided. (Where each house is a pentagon in the plane.) Suppose they've collided at their apices, in two groups of 4 each. At the center of each quartet there is now one vertex of valence = 4, with the adjacent 8 edges (originally the rooves) identified in pairs. Now identify the 4 houses' bases each with the corresponding base of the other group of 4. At this point we have what is essentially a sphere with 4 square holes that are arranged in a square formation. Now identify opposite square holes with each other. The result is surface of genus 2 made of 8 pentagons, 4 per vertex. (This is *not* a regular polyhedron, because not all mappings of one pentagon to any other in any dihedral manner can extend to a mapping of the entire polyhedron that takes pentagons to pentagons.) I'm not certain but I suspect that any conjectured tiling with the correct Euler characteristic and orientability can be realized as the compact surface with those properties. —Dan ----- Very true. So, let's return to the case in my original question (well, the question wasn't _originally_ mine, but you know what I mean) and consider the possibility of a genus-2 surface with 8 pentagonal faces, 4 of which meet at each of its 10 vertices. Should it be obvious to me that this thing exists as an abstract polyhedron? -----