Proving bijection between a region

[Gekko]

Homework Statement

Prove bijection between the regions
0<x<1, 0<y<1, 0<u, 0<v, u+v<pi/2

Homework Equations

x=sinu/cosv y = sinv/cosu

The Attempt at a Solution

We need to show that an inverse function exists to prove the bijection so obviously, (u,v) maps to one and only one (x,y) for the above. But what about the other way around? What is the best approach
Do we need to calculate:
u=arcsin(xcosv), v=arccos(sinu/x), u=arccos(sinv/y), v=arcsin(ycosu) and then look at each individually? Or could we divide one by the other and obtain tan(u)tan(v)=xy so that u=arctan(xy/tan(v) and v=arctan(xy/tan(u))?

 

[Mark44]

Homework Statement

Prove bijection between the regions
0<x<1, 0<y<1, 0<u, 0<v, u+v<pi/2

Homework Equations

x=sinu/cosv y = sinv/cosu

The Attempt at a Solution

We need to show that an inverse function exists to prove the bijection so obviously, (u,v) maps to one and only one (x,y) for the above. But what about the other way around? What is the best approach
Do we need to calculate:
u=arcsin(xcosv), v=arccos(sinu/x), u=arccos(sinv/y), v=arcsin(ycosu) and then look at each individually? Or could we divide one by the other and obtain tan(u)tan(v)=xy so that u=arctan(xy/tan(v) and v=arctan(xy/tan(u))?

I'm not sure that you can solve for u and v from your equations for x and y. Maybe it's possible, but I haven't come up with anything. Your attempts seem like good ideas at first, but you need u in terms of x and y alone, and the same for v. If you have learned about the Inverse Function Theorem (see http://en.wikipedia.org/wiki/Inverse_function_theorem), you can use it to show that there is an inverse mapping from a point (x, y) to a point (u, v).

 

[Gekko]

Thanks Mark. Yes, I too tried to put u and v in terms of x and y only but didnt see how

Suppose the way then is to prove by showing injectivity and surjectivity

This actually appears in "Proofs from the book", Pi^2/6 here

http://books.google.com/books?id=Cu...resnum=2&ved=0CBcQ6AEwAQ#v=onepage&q&f=false"

 

[Office_Shredder]

The u and v one paragraph above is the inverse transformation

 

[Gekko]

Of course! Thanks