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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:

http://webmath2.unito.it/paginepersonali/sergio.console/Dispense/Milnor%20Topology%20from%20%23681EA.pdf.

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.

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

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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 $f$.

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