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I was reading John Milnor's Topology from the Differentiable Viewpoint and there's a proof of the fundamental theorem of algebra at the end of the first chapter that I don't fully understand. I can post a picture if anyone asks for it, it's less than a page, but the book can be easily googled.

What I don't get is the final parragraph. He defines $f$ by $$f(z)=h_+^{-1}Ph_+$$

Being $h_+$ the north stereographic projection, and $P$ a non constant polynomial of arbitrary degree.

He then says:

Observe that $f$ has only a finite number of critical points, for $P$ fails to be a local diffeomorphism only at the zeros of the derivative $P'=\sum a_{n-j}jz^{j-1}$ and there are only finitely many zeros since $P'$ is not identically zero.

Ok, so for $f$ to have a finite number of critical points, $df=P'$ must vanish in a finite number of points. I see that $P'$ must vanish in a finite number of points, but I don't really see why he talks about it failing to be a local diffeomorphism.

Then it follows:

Therefore the set of all regular values on the sphere is connected. Hence the locally constant function $\#f^{-1}(y)$ must be constant on the full set. Since $\#f^{-1}$ can't be zero everywhere, then it's zero nowhere. Thus $f$ is an onto maping, and $P$ must have a zero.

The second sentence bugs me probably from a lack of knowledge in general topology. If a function is locally constant, and we have a connected set, must it be constant on the set? I can intuitively see that and actually prove it for some cases.

MyUserIsThis
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  • Yes, a locally constant function on a connected space is constant. For every possible value $v$, the preimage $c^{-1}(v)$ under a locally constant function $c$ is open. Thus you have a partition of the space into disjoint open sets. If the space is connected, only one of the partition sets can be nonempty. – Daniel Fischer Jun 24 '14 at 11:34
  • Assume that $f$ is locally constant and $f(x)=c$ in some $x$. Define $C$ to be the set of all points $y$ where $f(y)=c$. You can then easilly show that $C$ is both closed an open, so it must be the whole conencted space. – Peter Franek Jun 24 '14 at 11:45
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    Thank you both, it was a lot simpler than I thought. – MyUserIsThis Jun 24 '14 at 11:47

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Firstly, at the zeros of the derivative it is not invertible and therefore not a local diffeomorphism.

Secondly, if a function on a connected set is locally constant if and only if it is constant. See here for a similar question

Locally Constant Functions on Connected Spaces are Constant

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  • I see your "firstly", I just don't see why it's neccesary in the argument. The fact that there are a finite number of points in which $df$ is singular comes from the expression of $P'$, right? – MyUserIsThis Jun 24 '14 at 11:56