Let $P(x):= a_nz^n +a_{n-1} z^{n-1} + ... + a_0$ be a polynomial with degree $deg(P) \ge 1$. One know that the map $P: \mathbb{C} \to \mathbb{C}, x \mapsto P(x)$ induced by this polynomial can be exitended to a map $f: \mathbb{PC}^1 \to \mathbb{PC}^1, (z:1) \mapsto (p(z):1)$ by setting $ \infty := (1:0) \mapsto (1:0)$.
Since $\mathbb{PC}^1$ is a differentiable manifold it make sence to talk about critical points of $f$. On the chart $U_1:= \{(z:w) \vert w \neq 0\}$ we know that $f$ "bahaves" like $P$ and therefore the critical points with respect to this chart are exactly the zeros of the derivative $P'(x)$ but what about the infinity point $\infty \in U_2:= \{(z:w) \vert z \neq 0\}$?
I heard that $\infty$ is critical iff $deg(P) \ge 1$? How to show it formally?
By definition on the $U_2$ the map $f$ behaves like
$$f(1:w) =f(\frac{1}{w}:1)= \frac{1}{p(1/w)}= \frac{z^n}{a_n + ... + a_0z^n} =: g(z)$$
But if we try to derive $g(z)$ then we get using quotient rule $g'(z) \vert_{z=0} \neq 0$ if $n=deg(P)=1$ since $a_n \neq 0$.
But this would imply that for $deg(P)=1$ the infinity point $(1:0)$ isn't a zero of $g'(z)$ and therefore isn't a critical point.
Where is the error in my reasoning?
