# Thread: Prove a mapping is 1 to 1 and onto

1. ## Prove a mapping is 1 to 1 and onto

The mapping T:R^2 -> R^2 is (x,y)=T(u,v)=(sinu/cosv, sinv/cosu).

I was able to prove that the mapping was 1 to 1 by setting T(u,v)=T(u',v') and showing that u=u' and v=v'.

I'm not sure how to show its onto. I think since I have x and y in terms of u and v, if i write u and v in terms of x and y, this will prove onto. The only problem is I am having trouble doing this.
I was given the hint to square both x(u,v) and y(u,v) and use sinx^2 +cosx^2=1 but i cant see how this helps.

thanks

2. Originally Posted by BrianMath
The mapping T:R^2 -> R^2 is (x,y)=T(u,v)=(sinu/cosv, sinv/cosu).

I was able to prove that the mapping was 1 to 1 by setting T(u,v)=T(u',v') and showing that u=u' and v=v'.
Maybe I am misreading or there is something awfully wrong with it.
Is this defined

Is this true

3. Sorry, forgot the domain. Its a projection from the triangle u>0, v>0, u+v=pi/2 to the unit square (0<x<1, 0<y<1). this eliminates your issue with it not being defined

still need help with solving for u and v in terms of x and y

4. Originally Posted by BrianMath
Sorry, forgot the domain. Its a projection from the triangle u>0, v>0, u+v=pi/2 to the unit square (0<x<1, 0<y<1). this eliminates your issue with it not being defined

still need help with solving for u and v in terms of x and y
I'm not certain this will actually help you, but I can show you how to solve the problem. Here is a sketch:

$sin(u) = x~cos(v) \text{ and }cos(u) = \frac{sin(v)}{y}$

Then use

$sin^2(u) + cos^2(u) = 1 \implies x^2~cos^2(v) + \frac{sin^2(v)}{y^2} = 1$

Replace this with sin^2(v) = 1 - cos^2(v). Now solve for cos(v).

-Dan

5. ok thanks!. am i correct in assuming that if i can solve for u and v in terms of x and y, this will show that that mapping is onto? ie. for all values of x and y s.t. (xy)^2 does not equal one, we can find a value for u and v?