Mapping a general curve onto a bijection.

[alex_vs_gmail]

Homework Statement

Denote the xy plane by P. Let C be some general curve in P defined by the equation f ( x , y ) = 0
where f ( x , y ) is some algebraic expression involving x and y.

Verify carefully that if B : P -> P is any bijection then B( C ) is defined by the equation

f ( B^-1 (x , y)) = 0

I have no idea where to even start. Like, I really don't want an answer to it, I'd much rather be given a nudge in the right direction as none of my textbooks mention any explanation as to how these problems could/would be proven.

Thanks!

 

[Stephen Tashi]

I have no idea where to even start.

Start by making yourself an example.
Let C be the curve defined by F(x,y) = x^2 + y^2 - 1 = 0
Let B(x,y) = (x+3.y+3)
B maps the curve C to another curve. What is the equation for the other curve?

 

[alex_vs_gmail]

LOL I'm not even sure how to do that. Ill give it a shot, in F do I swap the function around so I have an expression for x and y, then substitute in the value for x and the value for y into B... if so would it look something like this...

B(F(x,y))= sqrt(1-y^2) +3*sqrt(1-x^2) +3=0 ?

 

[Stephen Tashi]

would it look something like this...

B(F(x,y))= sqrt(1-y^2) +3*sqrt(1-x^2) +3=0 ?

No, "it" wouldn't look like that, no matter which "it" in the problem we are talking about. You'll have to learn the definition of an inverse mapping and the notation for it. Figure out what B^-1(x,y) is.

(There is no B(F(x,y)) mentioned in the problem, so I don't know why you attempted to compute that. Furthermore the notation B(F(x,y)) doesn't make sense because B is a mapping from a 2D point to another 2D point, so B needs two numbers for its argument. F(x,y) is a single number.)

 

[alex_vs_gmail]

ohhhhhhh. wtf. i don't get why they have put this in an assignment if they haven't even taught us it and it isn't in the textbook.

 

[ybcut4?]

Hey I am having trouble with this question as well... can you be a bit more obvious with providing the answer please? :) Stephen Tashi