Milnor fundamental theorem of algebra : proof that $f: S^2 \rightarrow S^2$ is smooth

Solution 1:

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.