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laminatedevildoll

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I just want to make sure that I am doing this right or if I am on the right track.

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}

i.

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

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.

ker f=ker r

f=r, the function is one to one

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

**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^2i.

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

*ii.*

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

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