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Several months ago I was browsing through a question posted here ("Applications of Brouwer’s fixed point theorem"); amongst the comments attached it was mentioned that one could derive both Borsuk's Antipodal Theorem as well as the Fundamental Theorem of Algebra from Brouwer's Fixed-Point Theorem. However, to date I haven't seen (correct) proofs of these assertions. Could somebody kindly indicate where such proofs may be found ?

Thanks in advance,


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I think the book by Guillemin and Pollack, "Differential Topology", has it, but I am not certain. – Claudio Gorodski Apr 10 '12 at 18:15
Could you provide a link to the thread you refer to? – Ryan Budney Apr 10 '12 at 18:16
Sorry, I seem to have forgotten how to add links ... I simply copied the title of the thread, but the hypertext marking isn't preserved ... :-( Some advice would be appreciated ! Kind regards, St. – Stephan F. Kroneck Apr 10 '12 at 18:20
Ryan: Google books gives some of the book. – Claudio Gorodski Apr 10 '12 at 18:24
Stephan: I am quite sure the three results are there, but not that the two implications you want are. Nevertheless, I think those proofs are very nice, based on elementary tranversality arguments. Maybe it is worth having a look. You could try Google books. – Claudio Gorodski Apr 10 '12 at 18:27

As far as I'm aware, there isn't a compelling direct argument from Brouwer fixed point to imply the fundamental theorem of algebra. Such an argument isn't impossible -- I can imagine some fairly contrived proofs but I don't know of a very natural one. The references like Guillemin and Pollack don't derive FTOA from Brouwer, they derive both FTOA and Brouwer from degree/intersection theory. In particular they only use mod-2 degree theory for Brouwer but oriented degree theory for FTOA.

I had an argument written down here previously that I thought might work but now I realize it can't work. Oh, but it's fixable.

EDIT I've managed to repair the argument. The downside is it's not as simple.

A polynomial without roots produces a polynomial without fixed points. Specifically, $p(z) \neq 0$ for all $z \in \mathbb C$ means $q(z) = p(z)+z$ has no fixed points in $\mathbb C$. So what? Think of $q(z)$ as a map of the Riemann sphere. Now take the real oriented blow-up of the Riemann sphere at infinity (i.e. replace the point at infinity by its unit normal bundle in the sense of smooth real manifolds). This is a disc. So $q(z)$ becomes a smooth map of the disc, denote it $\hat q$, and identify the blow-up with $D^2$, the unit disc in $\mathbb C$ centred at the origin.

If $q(z) = z^n + a_{n-1}z^{n-1} + \cdots + a_1 z + a_0$ and if we conjugate by $z \longmapsto 1/z$ so

$$\frac{1}{q(1/z)} = \frac{z^n}{1+a_{n-1}z + \cdots + a_0z^n}$$

So when you restrict $\hat q$ to the boundary circle, it becomes $z \longmapsto z^n$.

$z \longmapsto z^n$ has fixed points $z^{n-1}=1$, the $(n-1)$-th roots of unity. So we can not directly appeal to Brouwer, since Brouwer's fixed point theorem might give you a pre-existing fixed point on the boundary.

Consider the vector field $v(z) = z-\hat q(z)$ on $D^2$. It is inward-pointing on the boundary circle with the sole exception of $z^{n-1}=1$, the $(n-1)$ roots of unity. But if we remove a small neighbourhood of $\partial D^2$ from $D^2$, the vector field $v$ restricts to an inward pointing vector field. So you could appeal to Poincare-Hopf and say there has to be a zero in the interior, or you could talk about the flow of the vector field, and Brouwer's fixed-point theorem would then tell you the vector field must have a zero in the interior.

So its not a slick proof, but it can be done.

A suitable identification between $D^2$ and the blow-up of the Riemann sphere at infinity is done by the map $X : D^2 \to \hat{\mathbb C}$ given by $X(z) = \frac{1}{1-|z|^2} z$. So $\hat q$ is the unique continuous extension of $X \circ q \circ X^{-1}$ to $D^2$.

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Ryan Budney: Thank you for your interesting answer ! I wouldn't have thought of that trick; I've been trying all sorts of other maps related to the original polynomial, without success. Still I must agree with you; I would also appreciate a simpler argument. Kind regards ! St. – Stephan F. Kroneck Apr 10 '12 at 19:03
Another version can be found here: . I like these proofs. – Deane Yang Apr 10 '12 at 20:12
@Deane Yang: Thanks for your comment, but I believe the article you are referring to is the faulty one I mentionend above; I think there is a follow-up erratum in the AMM by Arnol'd and Niven ... Kind regards ! St. – Stephan F. Kroneck Apr 10 '12 at 20:22
Stephen, oops. Sorry about that. – Deane Yang Apr 10 '12 at 21:32
Stephen, is the error the same as the one in Ryan's argument? – Deane Yang Apr 10 '12 at 21:55

Since Ryan has discussed the relationship between Brouwer and FTOA, I would like to say something about Brouwer and Borsuk-Ulam(I prefer this name). What I know is there are series of classical theorems on this direction, all closely related:

Thm.1(Brouwer) Continuous map $f:B^{n+1} \to B^{n+1}$ has at least one fixed point.

Thm.2(Hirsch) $\partial B^{n+1}$ is not a deformation retract of $B^{n+1}$.

Thm.3 $\mathrm{id}:S^n \to S^n$ is not homotopic to a constant map.

Thm.4 Continuous map $g:S^n \to S^n$ sending antipodal points to antipodal points is not homotopic to a constant map.

Thm.5 There exists continuous map $h:S^n \to S^m$ sending antipodal points to antipodal points if and only if $n \leq m$.

Thm.6(Borsuk-Ulam)For continuous map $k:S^n \to \mathbb{R}^n$,there exists $x \in S^n$ such that $k(x)=k(-x)$.

Let me explain. Equivalence of 1 and 2 is quite standard, for example, see

Milnor Topology from the differentiable viewpoint

By the way, it also contains a proof of FTOA, using the degree of maps.

To see the relation of 2 and 3, simply glue the boundary to a point.

4 is an obvious generalization of 3. You may prove it by studying $H_*(\mathbb{R}P^n)$. This study also leads to a proof of 5.

Equivalence of 6 and (the "only if" part of) 5 is also not hard, you may take it as a exercise:)

All in all, what I have shown is some knowledge of $H_*(\mathbb{R}P^n)$ is sufficient for both Brouwer and Borsuk-Ulam. Hope this helps.

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@Zhang Xiao: Thank you very much for your comments ! Basically, though, I believe most of the above is well-known (and is contained for instance in Granas-Dugundji II, §§ 5,7). Naturally, using the homological machinery one can prove Borsuk etc., but this seems like "overkill" in view of elementary proofs of Brouwer (cp. and of the FTA (cp. In all sources I can find, though, Borsuk seems to be the stronger statement (than Brouwer); I'm still lacking substantiation of the claims in the other thread ... Kind regards ! St. – Stephan F. Kroneck Apr 10 '12 at 22:07
@Stephan: You are correct. Although homology helps to make a unified picture, it contains strictly more information than Brouwer and Borsuk, hence of no use in setting the equivalence between two. – Zhang Xiao Apr 10 '12 at 23:44

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