Milnor fundamental theorem of algebra : proof that $f: S^2 \rightarrow S^2$ is smooth In the book of J.Milnor : "Topology from the differentiable viewpoint", in the chapter 1, p.8, he prooves the Fundamental Theorem of Algebra:

Every complex polynomial $p: \mathbb{C} \rightarrow \mathbb{C}$, not constant ($k > 0$),
$$ 
  p(z) = \sum_{k=0}^{N} a_z z^{k}, \quad a_z \in \mathbb{C} \quad 
$$
has at least one zero in $\mathbb{C}$.

To proove that, because $\mathbb{C}$ is not compact, he identifies the complex plane $\mathbb{C}$ with the Riemann sphere $S^2$ via the bijection $h_{+}$ and $h_{-}$ that are the stereographic projection of the north pole and south pole on the plane $\mathbb{R}^2 \times \{0\}$.
This allows us to extend $p: \mathbb{C} \rightarrow \mathbb{C}$ to $\hat{p}: \mathbb{C} \cup \{+\infty\} \rightarrow \mathbb{C} \cup \{+\infty\}$ and define an application,
$$
  f: S^2 \rightarrow S^2
$$
where,
$$
  f(u, v, w) = 
\begin{cases}
  h_{+}^{-1} \circ p \circ h_{+}(u,v,w), &\quad (u,v,w) \neq (0, 0, 1) \\
  (0, 0, 1), &\quad (u,v,w) = (0, 0, 1)
\end{cases}
$$
Now, Milnor claims that $f$ is smooth.
Seeing that f is smooth in $S^2 \setminus \{(0, 0, 1)\}$ is easy but for the north pole (0, 0, 1) I am not sure to understand the proof.
We define,
$$
Q(z) := h_{-} \circ f \circ (h_{-}(z))^{-1}
$$
After some algebra, we can see that $Q(z)$ is a quotient of polynomial and that it is defined in $z = 0$ so that $Q(z)$ is smooth in $z = 0$.
So then $f \circ (h_{-}(0))^{-1} = f(0, 0, 1)$ is smooth at $(0, 0, 1)$ by composition with smooth functions $h_{-}$ and $h_{-}^{-1}$?
 A: Milnor's proof is a little imprecise, but it can easily be repaired.
Let us write $n = (0,0,1)$ = north pole of $S^2$, $s = (0,0,-1)$ = south pole of $S^2$, $ S^2_+ = S^2 \setminus \{n\}$, $S^2_- = S^2 \setminus \{s\}$ and $D = S^2_+ \cap S^2_-$. Note that $D =  S^2_+ \setminus \{s\} =  S^2_- \setminus \{n\}$.
Milnor says that he identifies $\mathbb{R}^2 \times \{0\}$ with $\mathbb C$. Stereographic projection gives two diffeomorphisms $h_+ : S^2_+  \to \mathbb C$ and $h_- : S^2_- \to \mathbb C$. Note that $h_+(s) = h_-(n) = 0$ and thus $h_+(D) = h_-(D) = \mathbb C \setminus  \{0\} = \mathbb C^*$. We have $h_+(h^{-1}_-(z)) =  1/ \bar z$ for $z  \in \mathbb C^*$.
Milnor then defines
$$f(\xi) = \begin{cases} (h_+^{-1} \circ p \circ h_+)(\xi) & \xi \in S^2_+ \\ n  & \xi = n  \end{cases}$$
It seems that Milnor wants to define the map $Q$ by
$$Q = h_- \circ f \circ h^{-1}_- :\mathbb C \stackrel{h_-^{-1}}{\longrightarrow} S^2_- \stackrel{f}{\longrightarrow} S^2_- \stackrel{h_-}{\longrightarrow} \mathbb C .$$
But this only works if
$$f(S^2_-) \subset  S^2_- . \tag{1}$$
Since $f(n) = n \in S^2_- $ this is equivalent to
$$f(D) = (h_+)^{-1}(p(h_+(D))) \subset S^2_- . \tag{2}$$
Since $(h_+)^{-1}$ maps into $S^2_+$ and $h_+(D) = \mathbb C^*$, this means
$$p(\mathbb C^*) \subset h_+(D) = \mathbb C^* .\tag{3}$$
Thus

*

*either Milnor has to assume that $p$ has no zero except possibly at $z = 0$ (which would lead to a proof by contradiction)

*or he has to consider
$$Q :U \stackrel{h_-^{-1}}{\longrightarrow} h_-^{-1}(U) \stackrel{f}{\longrightarrow} S^2_- \stackrel{h_-}{\longrightarrow} \mathbb C $$
with a sufficiently small open neigborhood of $0$ in $\mathbb C$. In fact, $f(h_-^{-1}(0)) = f(n) = n \in S^2_-$. Since $f \circ h_-^{-1} :  \mathbb C \to S^2$ is continuous and $S^2_-$ is open in $S^2$, we find an open neigborhood of $0$ in $\mathbb C$ such that $(f \circ h_-^{-1})(U) \subset S^2_-$.

In both cases we may considers $h_-^{-1} \circ Q \circ h_-$ which is defined in some open neighborhood of $n$, agrees with $f$ on this neighborhood and is smooth because $Q$ is smooth and $h_-$ is a diffeomorphism.
