Let $f:\mathbb{R}^2\to\mathbb{R}^2$ given by $f(x,y)=((x+2)^2+y, 2x-3y-1)$
how do I prove that this function is not injective and thus not globally invertible?
I tried with $f(x,y)=f(u,v)$ but the system is quite impossible/hard for me.. is there another way?
On
Lets solve the equation $f(x,y)=f(u,v)$ parametrically. First note that this gives us two simultaneous equations $(x+2)^2-(u+2)^2=(x-y)(x+y+4)=v-y$ and $2(x-u)=3(y-v).$ Take $x-u=6t$ for some $t\in\mathbb{R},$ then we have three equations $$x-u=6t,\,\,\,\,\,\, x+u=-\dfrac{14}3,\,\,\,\,\,\,y-v=4t$$ of four unknowns. It is clear that we can solve first two equations without introducing a new parameter, but both $y, v$ contained in a single equation and therefore we have to introduce a new parameter for its solution. Lets simply take $v=T$ for some $T\in\mathbb{R}.$ Then we get $$x=-\dfrac73+3t,\,\,\,\,\,\,y=4t+T$$ and $$u=-\dfrac73-3t,\,\,\,\,\,\,v=T$$ for two parameters $t, T$ which is the complete parametric solution for the equation $f(x,y)=f(u,v).$ Now substitute any two $t\neq0$ and $T$ real values to obtain two different pair of pints those have the same image under your function. In other words, $f$ is not injective.
On
Find a case where $((x+2)^2 +y,2x -3y -1) = ((w+2)^2+ u, 2w-3u-1)$ but $(x,y)\ne (w,u)$.
Just spitballing:
If $x =0$ then we need
$(0+2)^2 + y = (w+2)^2 + u$ or
$y+4 = w^2 + 4w + 4 + u$ or $y = w^2 + 4w+u$
And $-3y-1 = 2w-3u-1$ so
$y = u-\frac 23w = w^2 + 4w + u$ so
$w^2 + \frac {14}3w=0$ If $w\ne 0$ then
$w=-\frac {14}3$.
$y = \frac {28}9 + u$ and $u$ can be anything.
So $f(0,\frac {28}9) = (2^2+\frac {28}9, -\frac {28}3-1) = (\frac {64}9,-\frac {31}3)$
And $f(-\frac {14}3,0) = ((-\frac {8}3)^2 + 0, -\frac {28}2-1)=(\frac {64}9, -\frac {31}3)$.
Thus it is not injective.
Probably more efficient ways to solve.
The system of equations
$$ (x+2)^2 + y=0\\ 2x-3y-1=0 $$
has the solutions
$$ x= -11/3, y=-25/9\\ x= -1, y= -1 $$
So both points give f(x,y) = (0,0), which constitutes a counterexample to injectivity.