Let $f: R^2 \rightarrow R^2 $ defined by $f(x,y) = (x+y,xy).$
Claim : Inverse image of each point in $R^2$ under f has at most two elements.
My Claim :
Suppose $f(x,y) = (x+y,xy)= (p,q).$ We have to find suitable x and y.
By solving the equations I get,
$x = \dfrac {p \pm \sqrt{p^2 - 4q}}{2q}, y = p-\dfrac {p \pm \sqrt{p^2 - 4q}}{2q}.$
My doubt is that for those $(p,q)$ such that $p^2 < 4q .$ Inverses doesn't exist.
How come inverse exist for all point in $R^2$ ?
On
Given $(p,q)\in{\mathbb R}^2$ you are interested in the set of all pairs $(x,y)$ satisfying $x+y=p$, $xy=q$. By Vieta's theorem $\{x,y\}$ then is the solution set of the quadratic equation $$t^2-pt+q=0\ .\tag{1}$$ Therefore we can say the following: If $p^2-4q>0$ then $(1)$ has two different real solutions $t_1$, $t_2$, giving rise to the two pairs $(t_1,t_2)$ and $(t_2,t_1)$. It follows that $\#f^{-1}(p,q)=2$ in this case. If $p^2-4q=0$ then $(1)$ has one real solution $t_0$, giving rise to the single pair $(t_0,t_0)$. It follows that $\#f^{-1}(p,q)=1$ in this case. If $p^2-4q<0$ then $f^{-1}(p,q)=\emptyset$.
They don't. For example, if $p = 0, q = 1$ then $x + y = p = 0$ implies $x = -y$ but then $xy = -x^2 = q = 1$ has no solution.