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There is a nice elementary topology problem (proposition) that is often missing from the introductory books on the topic.

PROBLEM. Let $\varphi:\mathbb{S}^{1}\rightarrow\mathbb{S}^{1}$ be a continuous self-map of the circle of degree $\deg(\varphi)=d$. Then $\varphi$ has at least $|d-1|$ fixed points. (For example, if $\varphi$ is an orientation reversing homeomorphism, then it has at least 2 fixed points - the 'two monks walking in opposite directions' problem.)

Its place should be just after the notion of degree, fundamental group etc. In my opinion, it is a very good exercise, as it combines basic notions, such as degree, fixed point, $\pi_{1}(\mathbb{S}^{1})$ and has useful applications. Paradoxically, I don't see it in the appropriate place ("degree", "fundamental group"), but in the more heavy advanced context of Nielsen theory. Nielsen theory, in its turn, is often missing from elementary topology books. The available to me ones are dealing with almost one and the same list of problems (nice, indeed), but this one seems not to be present there.

So my question is: Does anyone know a good elementary proof of this problem/proposition (without referring to advanced things such as Nielsen index theory or so). Any references are welcome as well. Thanks in advance.

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Lift $\varphi$ to a map $f: \mathbb{R} \to \mathbb{R}$ and look at its graph. The condition on the degree forces $f(x+1) = f(x) + d$. But this implies that the graph must intersect at least $d-1$ among the graphs $y = x + k$ with $k \in \mathbb{Z}$ by the mean value theorem. I let you flesh out the details, but that's what I call completely elementary. –  t.b. May 22 '11 at 13:21
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@Qiaochu Yuan: Not exactly, the Lefschetz fixed point theorem implies only that the sum of the fixed points indexes equals |d-1|. Here it is claimed that |d-1| distinct fixed points exist. Anyway, Lefschetz fixed point theorem is an advanced topic as well. I am asking about an 'elementary' proof. –  t22 May 22 '11 at 13:24
    
you're right. My apologies. –  Qiaochu Yuan May 22 '11 at 13:31

1 Answer 1

up vote 13 down vote accepted

I thought I'd better make my comment into an answer:

Lift $\varphi$ to a continuous map $f: \mathbb{R} \to \mathbb{R}$ and look at its graph. The condition on the degree forces $f(x+1) = f(x) + d$ for all $x \in \mathbb{R}$. But this implies that the graph over $[0,1)$ must intersect at least $d-1$ among the graphs $y = x + k$ with $k \in \mathbb{Z}$ by the mean value theorem (to be specific, the ones with $k$ between $[f(0), f(0)+d)$ of which there are at worst $d-1$). I let you flesh out the details, but that's what I call completely elementary.

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Thank you very much, Theo,the nice proof doesn't even need formal arguments, just a drawing. The condition $f(x+1)=f(x)+d$ is crucial here. This may be told to undergraduares even. Now I'm speculating about how to replace the circle by an annulus, but this seems more complicated. –  t22 May 22 '11 at 13:58
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@symbo'leon: You're welcome. To be honest, I don't see how to extend this argument immediately to an annulus, but maybe a bit more thinking could give it (my intuition is very bad in $4$ dimensions). By the way: I learned such arguments from A. Katok in the context of circle mappings in a course on dynamical systems. I'm pretty sure that you should be able to find it in the book Katok-Hasselblatt, Introduction to the modern theory of dynamical systems. –  t.b. May 22 '11 at 14:16

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