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To show that equivalence classes of ker f are concentric circles

  1. May 8, 2005 #1
    I just want to make sure that I am doing this right or if I am on the right track.

    To find a map f :R ---> R so that ker f ={(x,y): x^4=y^4}

    pi_R: (x,y) ---> {(x,y): x^4=y^4}
    pi_R: x ---> {R(x): x^4 is an element in X}
    pi_R: Y ---> {R(y): y^4 is an element in Y}

    Define f: R x R ---> by f(x,y) = x^2+y^2

    To find ker f
    ker f: {(x,y): x^2+y^2

    To show that equivalence classes of ker f are concentric circles.

    f(x): {y:(x,y) E x^2+y^2} (E stands for the element symbol)
    f(y): {x:(x,y) E x^2+y^2}

    if x=y for n >= 1

    f(x): {y:(x,y) E nx^2+ny^2}
    f(y): {x:(x,y) E nx^2+ny^2}

    x=y, there is a (x,y) E ker f
    then there is some y=f(x) for all y E Y f(y)=y

    Find a bijection between the equivalence classes of ker f and {r:r E R and r >= 0}

    ker f=ker r
    f=r, the function is one to one

    if y E Y, such that y=f(x)... onto?

    Attached Files:

    Last edited: May 8, 2005
  2. jcsd
  3. May 8, 2005 #2
    I thougt ker(f) was a subset of the domain of definition of f which is subset of R here...so how can you put pairs (x,y) in R ?
  4. May 8, 2005 #3
    I guess I don't put pairs (x,y) in R, then.

    Do I use the fact that...

    pi : R ---> R/ Ker f is the (surjective) projection map?
    Last edited: May 8, 2005
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