Proving T is a Bijection: D* to D

[ironman2]

Homework Statement

D* = {(u,v) | u>0, v>0, u + v < pi/2} and D = {(x,y) | 0<x<1, 0<y<1}

The map T : R² -> R² is given by (x,y) = T(u,v) = (sin u/cos v, sin v/cos u).

To prove that T is a bijection

Homework Equations

and

The Attempt at a Solution

I'm kinda stuck finding the limits for u and v and get the region for D*. Once I can nail down the region for D* I can break into different paths and get the solution. I've done squares to squares and stuff before, but can't get my head around the triangular region for D*.

Any ideas guys?

 

[lanedance]

so if u is horizontal, v is vertical axis

D* is the positive quadrant bounded on the upper right by the line that passses through (0,pi/2) and (pi/2,0)

 

[ironman2]

Ok, so I've figured out that I can prove that T is one-to-one if the determination of the transformation matrix A is not zero. How do I find the transformation matrix given that sin and cos are involved? Also, I'm guessing that if I can build an inverse function that expresses (u,v) as some (inverse trig?) function of x & y, I can prove bijection... but I'm stuck.. help?!

 

[lanedance]

thats only for linear maps, this not a linear map...

do you know about the Inverse Function Theorem?

 

[ironman2]

Oh my bad, ofcourse this isn't a linear map, so the determinants methods is ruled out.

No, I'm not familiar with Inverse Function Theorem... can you tell me what it is and how can I use it here?

 

[lanedance]

http://en.wikipedia.org/wiki/Inverse_function_theorem

it basically says a function has an inverse on a neighbourhood if the determininant of the Jacobian ins non-zero

this is similar to saying the function can be approximated by a linear function on a small enough neighbourhood eg. taylor series

 

[ironman2]

Ok, so if we take f1 = sin u/cosv and f2 = sin v/cos u, the determinant of the Jacobian becomes 1 - tan²vsin²u... which is a mess. I don't get the bigger picture from here. What do I from the Jacobian?

 

[lanedance]

its not much of a mess, plus all you need to know is whether it can be zero on the given inetrval... the jacobian shows the function is locally 1:1

then you need to show it is globally 1:1 and onto (each x,y has a unique corresponding u,v)

 

[ironman2]

So, the function is locally 1:1 if the Jacobian is zero on the given interval?

Can you give me some more hints on how to proceed to prove that it is globally 1:1?

 

[lanedance]

the function is locally 1:1 if the Jacobian is non-zero on the given interval