Determining the image of a function $\psi:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ Determining the image of a function $\psi:\mathbb{R}^2 \rightarrow \mathbb{R}^2$, $\psi(x,y) = (x^2 - y^2, x^2 + y^2)$
I made some observations about $\psi$:
$\psi$ isn't injective, since $\psi(-x,-y) = \psi(x,y)$.
The restriction $\psi_{|D}$, where $D = \{(x,y) \in \mathbb{R}^2 | x,y>0\}$ makes $\psi$ injective, since:
$(x_{1}^2-y_{1}^2,x_{1}^2+y_{1}^2) = (x_{2}^2-y_{2}^2,x_{2}^2+y_{2}^2)$ occurs when $x_{1}^2-y_{1}^2=x_{2}^2-y_{2}^2$ and $x_{1}^2+y_{1}^2=x_{2}^2+y_{2}^2$, which will lead to $(x_1,y_1)=(x_2,y_2)$.
How do I determine the image of  $\psi_{|D}$?  (My objective here is to create a bijection)
 A: Let $(a,b)$ be arbitrary in $\mathbb{R}^2$ and consider $\phi(x,y) = (a,b)$, so:
$$\left\{\begin{array}{rcl}
x^2-y^2 & = & a \\
x^2+y^2 & = & b 
\end{array}\right. \iff 
\left\{\begin{array}{rcl}
x^2 & = & \frac{a+b}{2} \\
y^2 & = & \frac{b-a}{2} 
\end{array}\right.$$
This is only possible when $a+b \ge 0 \iff b \ge -a$ and when $b-a \ge 0 \iff b \ge a$ which can be combined into the condition $b \ge |a|$. In that case, you can solve for $x$ and $y$.
This condition is satisfied by points in the 'infinite triangle' in the $xy$-plane where $y \ge |x|$; top in the origin and extending above along the lines $y=x$ and $y=-x$.
A: Take any $(a,b)\in \mathbb{R}^2$ and try to find a preimage:
We want to find $(x,y)\in D$ such that $x^2-y^2=a$ and $x^2+y^2=b$.
Hence, you have: $2x^2=a+b$ and $2y^2=b-a$ so $\psi(\sqrt{\frac{a+b}2},\sqrt{\frac{b-a}2})=(a,b)$
However, you have to make sure that both expressions for $x$ and $y$ makes sense i.e. that $a+b\geq 0$ and $b-a\geq0$. That's the image of your map
