Showing a function is one-to-one

Question

So I know for one-to-one I need to show Ψ(u1,v1)=Ψ(u2,v2) but I am unsure how to go to about it for this function...

Answer 1

Suppose that $\Psi(u,v)=\Psi(x,y)$, that is, that$$(u\cosh v,u\sinh v,u^2)=(x\cosh y,x\sinh y,x^2).$$Then $u^2=x^2$ and therefore $u=\pm x$. Suppose that that $u=-x$. Since $u\neq 0$ we have then that $-x\cosh v=x\cosh y$, and therefore $-\cosh v=\cosh y$. This is impossible, because $\cosh v,\cosh y>0$. So, $u=x$.

Now, we know that $u\sinh v=u\sinh y$. But then $\sinh v=\sinh y$ and therefore $v=y$, since $\sinh$ is one-to-one.