Let $p$ be a non-constant polynomial in one variable in complex coefficients. Let $G:\mathbf C\to \mathbf CP^1$ be the map defined as $G(z)=[z:1]$.
Problem. To show that there is a unique smooth map $\tilde p:\mathbf CP^1\to \mathbf CP^1$ such that following diagram commutes. $$ \begin{array}[rrr] \ \mathbf C & \xrightarrow{G} & \mathbf CP^1\\ \downarrow{} & & \downarrow\\ \mathbf C & \xrightarrow{G}& \mathbf CP^1 \end{array} $$ where the left vertical arrow is $p$ and the right vertical arrow is $\tilde p$.
The continuity of $\tilde p$ alone forces that $\tilde p([1:0])= [1:0]$. So, by passing to the standard chart about $[1:0]$, the smoothness of $\tilde p$ is equivalent to the smoothness of the map $f:U\to \mathbf C$ defined as
$$ f(z) = \left\{ \begin{array}[rr] \ 1/p(1/z) & \text{if }z\neq 0\\ 0 & \text{if } z=0 \end{array} \right. $$ where $U\subseteq \mathbf C$ is a sufficiently small neighborhood of the origin, so that $p(1/z)$ is not zero whenever $z\in U$.
It so happens that it is easy to show that $f$ is complex differentiable, and hence smooth.
But is there a way to show the smoothness of $f$ without going into complex analysis? Thanks.