This is from Problem 2 from Milnor's "Topology from a differentiable viewpoint".
Show that every complex polynomial of degree $n$ gives rise to a smooth map from the Gauss sphere $S^2$ to itself of degree $n$.
My candidate for the desired map $f$ is the natural one: We can project $S^2\backslash\{N\}$ ($N$ is the North pole) onto the plane via the stereographic projection from $N$, do the polynomial map, then take the inverse projection. Then define $N\mapsto N$. It is clear that $f$ is smooth everywhere away from $N$.
To show that $f$ is smooth at $N$, I tried to consider the other chart of this map, by projecting $S^2\backslash \{S\}$ from the South pole $S$. If we decompose the polynomial map $P:\mathbb{C}\to \mathbb{C}$ to $(Q,R):\mathbb{R}^2\to \mathbb{R}^2$, then the map $f$ on this chart is $$(x,y)\mapsto \begin{cases} \left(\frac{Q(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2})}{Q(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2})^2+R(\frac{x}{x^2+y^2},\frac{x}{x^2+y^2})^2},\frac{R(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2})}{Q(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2})^2+R(\frac{x}{x^2+y^2},\frac{x}{x^2+y^2})^2}\right) &\text{ if }(x,y)\neq (0,0)\\ (0,0) &\text{ if }(x,y)=(0,0)\end{cases}.$$ Maybe we can discuss the derivatives at $(0,0)$, but I can't see a clean argument in this way. I also think that there maybe something about an analytic continuation to $(0,0)$ since the function defined on $\mathbb{R}^2\setminus (0,0)$ is rational, but I am not familiar with this type of argument.
Any help to find a smooth map/show that my map is smooth at $(0,0)$ is greatly appreciated!
The resulting rational function is much easier to analyze in complex form, where the transition map $\tau$ between the two steriographic projections has the form $\tau(z)=z/|z|^2$, or, more simply $\tau(z)=1/\overline{z}$, where $\overline{z}$ is the complex conjugate of $z$. Let $P(z)=p_0+p_1z+\cdots+p_dz^d$ be the polynomial of interest (with $p_d\neq 0$), and $Q=\tau\circ P\circ\tau$. We then have $$ Q(z)=\frac{1}{\overline{P(1/\overline{z})}}=\frac{1}{\sum_{n=0}^d\overline{p}_n/z^n}=\frac{z^d}{\sum_{n=1}^d\overline{p}_nz^{d-n}} $$ We see that the resulting function is a complex rational, and the constant term in the denominator is $\bar{p}_d\neq 0$. Clearly, then, there is no issue extending $Q$ smoothly to $0$.