proving homeomorphisms, circles, torus, square

• Mar 14th 2011, 08:29 PM
trle907
proving homeomorphisms, circles, torus, square
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
• Mar 14th 2011, 09:15 PM
xxp9
You may need to re-state your question. Can't understand your post at all. How is your quotient space defined?
• Mar 15th 2011, 05:55 AM
trle907
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
• Mar 15th 2011, 04:49 PM
xxp9
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.