I wrote: << In the torus T^2, find an embedded theta curve C (i.e., 2 points joined by each of 3 otherwise disjoint arcs) such that if C is identified to a point, the quotient space is topologically a sphere S^2.
My solution a few lines below Gareth's solution is correct, of course, and essentially the same as mine, which looks a little more symmetrical then the brick arrangement. Namely, consider the tiling of the plane by regular hexagons. The group G of translations preserving this tiling must be a lattice in the plane. Hence the quotient of the plane by the lattice must be a torus, topologically. Call the torus T. Let one of the hexagons be called H. Since every point of the plane is equivalent to some point of H, we can obtain the same quotient torus T by just identifying the points of H that are equivalent by the group G of translations. This means: Take the (closed) hexagon H and identify opposite edges in the simplest way -- identify each point of each edge with its closest poiwwnt on the opposite edge. This identification turns H into the torus T. It's easy to check that the boundary of H becomes a theta curve C on T. It's also clear that identifying the boundary of H to a point gives a topological sphere S^2 as the quotient space. Which means that taking the theta curve C on the torus T, and identifying it to a point, will also give the same topological sphere S^2 as *its* quotient space. Puzzle solved. * * * ((( Interestingly, it's easy to check (from orientability and Euler characteristic) that taking any regular polygon with an even number 2n of sides, and identifying its opposite sides (in the simplest way), has as quotient space the surface having its genus = floor(n/2). I.e., 2n-gon where 2n = 4 6 8 10 12 14 16 18 20 22 24 26 28 30 . . . -------------------------------------------------------------------------------- genus 1 1 2 2 3 3 4 4 6 6 8 8 10 10 . . . To get a highly symmetric *metric* surface of constant curvature, this already works for the torus using a square or regular hexagon in the plane. But for surfaces of higher genus, we need to start with the regular 2n-gon of constant negative curvature (say = -1) whose vertex angles are each pi/n if n is even, and 2pi/n if n is odd (with geodesic edges). These are unique up to isometry. ))) --Dan ________________________________________________________________________________________ It goes without saying that .