To show that equivalence classes of ker f are concentric circles

In summary, the author is asking how one can put pairs of (x,y) in the domain of definition of f. The author then states that ker f is the (surjective) projection map, so one can put pairs of (x,y) in the domain of definition of f.
  • #1
laminatedevildoll
211
0
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?
 

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  • #2
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 ?
 
  • #3
kleinwolf said:
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?
 
Last edited:

Related to To show that equivalence classes of ker f are concentric circles

1. What does it mean for two equivalence classes to be concentric circles?

Concentric circles refer to a set of circles that share the same center point. In the context of equivalence classes of a function, it means that all elements within each class have the same image (or output) under the function.

2. How can we show that equivalence classes of ker f are concentric circles?

In order to show that equivalence classes of ker f are concentric circles, we need to prove that all elements within each class have the same image under the function f. This can be done by showing that the elements within each class are related by the same equivalence relation.

3. What is the significance of proving that equivalence classes of ker f are concentric circles?

Proving that equivalence classes of ker f are concentric circles helps us understand the structure of the kernel of the function f. It also allows us to group elements with similar outputs together, which can be useful in further analysis of the function.

4. Can equivalence classes of ker f be any other shape besides concentric circles?

Yes, equivalence classes of ker f can be any shape as long as all elements within each class have the same image under the function f. However, concentric circles are a common and intuitive way to illustrate this concept.

5. How does proving that equivalence classes of ker f are concentric circles relate to the concept of equivalence relations?

Proving that equivalence classes of ker f are concentric circles demonstrates the transitive, symmetric, and reflexive properties of the equivalence relation. This relation is necessary in order to group elements into equivalence classes, which are represented by the concentric circles in this scenario.

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