I'm currently trying to show that the critical points of a rational map occur iff their deficiency is non-zero. Just for context, I'm taking the deficiency of a point under a rational map $R$ as $\delta (z) =$ deg($R) - \vert \{R^{-1} (z) \} \vert $
As a result, I want to show that for a rational map $R$, upon translating the problem to the origin:
deg($R) > \vert R'(0) \vert \iff R'(z)=0$ for some $z \in \{R^{-1} (0) \} $
I've already proved the $\implies$ direction. But I'm struggling a bit with the reverse direction. This is what I feel the best approach is:
By contraposition, assume that $R'(z) \neq 0$ for any $z \in \{R^{-1} (0) \}$. By the inverse function theorem, $z$ is locally injective.
$R$ is rational, so can be written in the form $\frac{P(z)}{Q(z)}$ for P,Q coprime polynomials. The degree of $R$ is the $\max\{deg(P),deg(Q)\}$.
$\therefore R(z)=0 \iff P(z) = 0$ which has precisely $deg(P)$ solutions (counting multiplicity). Therefore $\vert R^{-1} (0) \vert \leq$ deg($P) \leq $ deg($R$)... however I don't want the inequality this direction in the contrapositive, I want it the other way; so I've made some mistake somewhere, right?
My guess is that I need to use this injectivity property somewhere in my argument. Any help would be greatly appreciated.