Is $f$ globally invertible on $R^2$?

Question

$f(x,y) = (x^2y + 2xy, xy^2 + xy)$

Q. Is $f$ globally invertible on $R^2$?

I found that $f_1(-2,1) = f_1(0,0)$. But, $f_2(-2,1) \not = f_2(-2,1).$

Is the case for $f_1$ enough to show $f$ is not injective? or should I find $(x_0, y_0)$ and $(x_1,y_1)$ such that both $f_1$ and $f_2$ have the case for equality?

Answer 1

No, it is not enough. But noticing that $f(0,0)=f(0,1)$ is.