You may need to re-state your question. Can't understand your post at all. How is your quotient space defined?
hey, i have trouble proving homeomorphisms, i am new to this concept. the problem is:
The circle {(0, x2, x3): (x2-1)^2 + (x3)^2 = 1} lies in the plane x1=0 in R^3. From the quotient space from the square [0,1]x[0,1] contained in R^2, show that this quotient space is homeomorphic to the torus.
work:
so i know the circle has radius 1 and center (0,1,0) and a torus would be a space that is homeomorphic to the subspace of R^3 and can be obtained by rotating the circle around the x3 axis. and identifying the top edge and bottom edge: (s,0)~(s,1), 0<=s<=1 and the left edge with the right edge, (0,t)~(1,t)
can we use the square [0,2pi]x[0,2pi] for this? i dont know how to tie it in into the proof
the quotient space is defined by taking the top and bottom edge of the square and pasting them together so you form a tube like thing, and then taking the right and left ends of the tube and pasting them together so you have a torus. so i want to show that the square is homeomorphic to the torus
then why did you mention the circle? what is the relationship between the circle and your square/torus?
By the way, the square won't be homeomorphic to the torus since a square has a boundary but a torus doesn't. You can travel on a trous forever without hitting an edge but you'll hit an edge for a square.