On Thu, May 20, 2010 at 1:49 AM, Dan Asimov <dasimov@earthlink.net> wrote:
I believe the condition that "No two adjacent polygons have the same image under F" takes care of Andy's concern, and in fact that is the reason this condition was included.
Yes, you're right. This problem occurred to me when I read the edge condition, before I got to the regularity condition, which implies that each polygon is a regular polygon, so that the images of two polygons on the same side would be identical. Put another way, if you wanted to define "spherical polygon" rather than "regular spherical polygon", it wouldn't suffice to remove the regularity condition; you'd have to replace ""No two adjacent polygons have the same image under F" with something more like the condition I specified.
* Project the regular tetrahedron to the sphere and choose any two disjoint edges e, e' of it. Now take two identical copies T_1, T_2 of this spherical tetrahedron and "slit" each of them them along each edge corresponding to e and e', creating 8 loose edges. Now identify these in 4 pairs (as when creating a Riemann surface) so that one side of e on T_1 is identified to the other side of e on T_2, etc., so that no loose edges remain.
(In this case the topology of X is that of a torus.)
Puzzle (not that hard, but confused me for a few minutes): What's wrong with the following argument? That can't be, because the resulting map from the torus to the sphere is a double cover. But the sphere has trivial fundamental group, and therefore has no covering spaces. Answer below a f t e r s o m e s p o i l e r s p a c e The cover is ramified at the corners of the tetrahedron. Andy