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Math Help - Explicit definition of an equivalence relation

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by Siknature View Post
    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?
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    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.
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