While i was reading through my calculus recitations, i stumbled upon this example: suppose we have $D = \{(x,y) \in \mathbb{R^2}: 0 \leq y,0 \leq x\leq 9, 1 \leq x^2-y^2 \leq 9\}$.
$U=\{(x,y) \in \mathbb{R^2} : 0<x<y\}$
$V=\{(x,y) \in \mathbb{R^2} : 0<x, 0<y\}$
let $T:U\rightarrow V$ and $S:V\rightarrow U$ two transformations, defined as:
\begin{equation} T \begin{pmatrix} x \\\ y \end{pmatrix} =\begin{pmatrix} \sqrt{y} \\\ \sqrt{y-x} \end{pmatrix} \end{equation}
\begin{equation} S \begin{pmatrix} x \\\ y \end{pmatrix} =\begin{pmatrix} x^2 - y^2 \\\ x^2 \end{pmatrix} \end{equation}
we can see that $S=T^{-1}$.
so in the recitations they calculated the set $T^{-1}(D)$ as the following:
$T^{-1}$$ (D)= \{(x,y) \in \mathbb{R^2}: 0 \leq \sqrt{y-x},0 \leq \sqrt{y} \leq 9, 1 \leq \sqrt{y^2}-\sqrt{(y-x)^2} \leq 9\}$
I'm struggling to understand why this is the right set, i see that we applied the transformation $T$ over the endpoints of the intervals of D instead of applying $T^{-1}=S$
