Thread: Explicit definition of an equivalence relation

1. Explicit definition of an equivalence relation

A = { (x1 , x2) : 0<= x1<=1 and 0<= x2<=1}

B = { (y1, ,y2, y3) : (y1)^2 + (y2)^2 = 1 and 0<= y3<=1}

each with the usual toplogy.

Let f: A--->B be the funtion given by f(x1,x2) = (cos(2pi x1),sin (2pi x1), x2)

How do i give the explicit definition of the equivalence relation ~ on A ?

Any help is greatly apprecitated.

2. Originally Posted by Siknature
A = { (x1 , x2) : 0<= x1<=1 and 0<= x2<=1}

B = { (y1, ,y2, y3) : (y1)^2 + (y2)^2 = 1 and 0<= y3<=1}

each with the usual toplogy.

Let f: A--->B be the funtion given by f(x1,x2) = (cos(2pi x1),sin (2pi x1), x2)

How do i give the explicit definition of the equivalence relation ~ on A ?

Any help is greatly apprecitated.
I have no idea what you mean here. Just a function from A to B does NOT define an equivalence relation on A. Is there an equivalence relation given on B? Are you saying that p and q in A are equivalent if f(p) and f(q) are equivalent in B?

3. Originally Posted by HallsofIvy
I have no idea what you mean here. Just a function from A to B does NOT define an equivalence relation on A. Is there an equivalence relation given on B? Are you saying that p and q in A are equivalent if f(p) and f(q) are equivalent in B?
I am trying to prove that there is an equivalence relation on A such that f induces a homeomorphism F: A/~ ---> B.

Once i have worked out what the explicit definition for the equivalence relation ~ on A is i can then go on to:

. prove that f induces a bijection F: A/~ ---> B

. then prove that F is continuous

. then finally prove that F is a homeomorphism.