To show that equivalence classes of ker f are concentric circles

[laminatedevildoll]

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

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

ii.
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}

f(x)=f(y)
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

iii.
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?

 

[kleinwolf]

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

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 ?

 

[laminatedevildoll]

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 ?

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?