Inverse of $\psi(t,s)=(e^t,se^{2t})$

Question

I am trying to solve a problem involving finding the inverse of this map

$$\psi:\mathbb{R}^2\to \mathbb{R}^2$$ $$\psi(t,s)=(e^t,se^{2t})$$

Here is what I am thinking

Let $x=e^t$ and $y=se^{2t}$.

Exchange $x,y$ and $s,t$.

$t=e^x$ and $s=ye^{2x}$.

Solving for $x$ and $y$,

$x=\ln (t)$ and $y=e^{s-2\ln (\ln(t))}$

Hence

$\psi^{-1}(x,y)=(\ln (x), e^{y-2\ln (\ln(x))})$

I computed $\psi\circ \psi^{-1} $ but didn’t get the identity, so I guess there is a mistake. Any help would be appreciated. Thank you.

Answer 1

$\psi$ is not a bijection since the range is contained in $\{(x,y): x >0\}$

If you restrict the codomain to this set you can find the inverse as follows: $t=\ln x$ and $s=ye^{-2t}=ye^{-2\ln x}=y/x^{2}$

So $\psi^{-1}(x,y)=(\ln x , yx^{-2})$ on $\{(x,y): x>0, y \in \mathbb R\}$.