# Milnor's proof of the fundamental theorem of algebra (Topology from the Differentiable Viewpoint)

I am studying the proof of the fundamental theorem of algebra out of John Milnor's book Topology from the Differentiable Viewpoint, located on page 8 here:

My questions concern the last line of the proof. I cannot understand why $f$ being zero nowhere implies that it is an onto function - is it because it is an injective map from two finite sets? And why does $f$ being a bijection imply that the polynomial $P$ has a zero? I can see that $f$ is a composition of functions including $P$, but there seems to be a leap here that I am missing.

• where is that PDF? Nov 30, 2014 at 22:42
• Added now, sorry! Nov 30, 2014 at 22:43
• Since $\#f^{-1}(y) > 0$ for all $y$, the function $f$ is onto. Nov 30, 2014 at 22:53

He has proven that the function $$y\to\sharp f^{-1}(y)$$ is constant on the set of regular values. Assume now that $$f$$ is not surjective, i.e. there is some $$y_0$$ with $$\sharp f^{-1}(y_0)=0.$$ This $$y_0$$ is then a regular value. (To check this look at the definition of "regular value".) Thus $$\sharp f^{-1}(y)=0$$ for all regular values $$y$$, so every point is a critical value of $$f$$. But this is of course not possible: a polynomial of degree $$d$$ has at most $$d-1$$ critical points, hence at most $$d-1$$ critical values, so not all points can be critical values of $$f$$.