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I'm looking at the following problem from Katok & Hasselblatt (Modern Theory of Dynamical Systems, Problem $11.2.4$ if anyone cares):

Let $f:S^1\to S^1$ be an orientation-reversing homeomorphism of the circle. Show that $f$ has exactly two fixed points, and the rotation number of $f$ is zero.

Now, to start off with I use an easy consequence of the Lefschetz fixed point theorem, which says $f:S^n\to S^n$ has a fixed point if $\deg f\neq(-1)^{n+1}$. Since in our case, $\deg f=-1$, this certainly applies, so $f$ has at least one fixed point. Also, any map with a fixed point has zero rotation number, so $\tau(f)=0$ immediately. However, I'm stuck trying to show the existence of a second fixed point.

Can anybody throw a hint my direction?

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3 Answers 3

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Much much simpler, no need for Lefschtz.

Consider a lift $F$. Notice that $F(0)-0=(F(1)-1)+2$ (make the computations, it is orientation reversing). So, the $2$ gives you the answer!

Presumably you mean that $f^2$ has zero rotation number (makes no sense otherwise according to the definitions, in particular in KH). But indeed it follows as you say.

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  • $\begingroup$ Sorry to ask you to revisit this, but I guess I don't really understand what you mean by "the $2$ gives you the answer". I get that the calculation is done by writing $F(0)=F(1-1)=F(1)+1=(F(1)-1)+2$, but I don't see how writing it in this way shows there are exactly two fixed points. $\endgroup$ Feb 27, 2017 at 17:41
  • $\begingroup$ No problem. Draw the graph of $F$ and look at the points that give fixed points. Indeed, all comes from the graph. $\endgroup$
    – John B
    Feb 27, 2017 at 17:49
  • $\begingroup$ It took some time but I got it now! Thank you. $\endgroup$ Feb 27, 2017 at 19:43
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    $\begingroup$ I can not still understand @AlexMathers could you draw the graph? $\endgroup$ Dec 5, 2020 at 21:29
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Since the above answer was too terse for me to understand until I managed to solve a related problem myself, I'll elaborate on that reasoning, as people are asking in the comments. This leads to a (slightly) weaker result that $f$ must have at least two distinct fixed points.

Let $f$ be the homomorphism in question and $F$ an arbitrary lift. Then define $G(x) = F(x) - x$. Note that this is continuous because $f$ is a homomorphism, and continuous itself (and so $F$, its lift, is too). See then that a fixed point of $f$ means $fx = x \implies F(x) = x + k \implies G(x) \in \mathbb{Z}$. So it suffices to show that $G$ is valued at least two different integers.

Then see, using that $F(x + k) = F(x) + k$ (easily proven as an exercise, by induction, making use of the fact that $f$ preserves orientation) that $G(1) = F(1) - 1 = F(0) - 2 = G(0) - 2$. This means that $G$, a continuous function, increases by $2$ between the inputs $0$ and $1$. This means it must take on two integer values in between, by the Intermediate Value Theorem (graph $G$ against $x$ if you're not convinced). And therefore, these two distinct values correspond to two distinct fixed points of our original function $f$, as required.

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for a different approach: work over $\mathbb C$ and use winding numbers (with the prefered definition as a total continuous change in angle as defined in chp 7 of Beardon's Complex Analysis or chp 3 of Fulton's Algebraic Topology)

with $\gamma(t)=\exp\big(2\pi i \cdot t\big)$ for $t\in[0,1]$
$n\big(f\circ\gamma, 0\big)=-1$
because $f$ is an orientation reversing homeomorphism

with the inversion map $h:z\mapsto z^{-1}$, consider the curve $\sigma$ given by
$\sigma(t) := \big(h\circ \gamma(t)\big) \cdot \big(f\circ\gamma(t)\big)$
$\implies n\big(\sigma,0\big)=n\big(h\circ\gamma, 0\big)+n\big(f\circ\gamma, 0\big)=-1+-1=-2$
$\implies$ at least two distinct values $t',t^*\in \big[0,1\big)$ such that $\sigma(t')=1=\sigma(t^*)$
That is: $\frac{1}{\gamma(t')}\cdot f\big(\gamma(t')\big)=1\implies f\big(\gamma(t')\big)=\gamma(t')$, so $\gamma(t')$ is a fixed point, and the same holds for $\gamma(t^*)$.

justification: if $\sigma^{-1}(1)$ had only one value in $[0,1)$, then via re-parameterization we may assume $\sigma(0)=1=\sigma(1)$, which implies for all $\delta_k:=2^{-k}$, for $k\in \mathbb N$ that $1\gt \big \vert n\big(\sigma_{\big\vert [\delta_k,1-\delta_k]},0\big)\big \vert$ since the curve restricted to $[\delta_k,1-\delta_k]$ does not meet the positive real line but this implies $1\geq \big \vert n\big(\sigma,0\big)\big \vert=2$ which is impossible.

Finally, if $f$ had 3 fixed points, we'd conclude $f\circ \gamma$ is a counter clock-wise curve like $\gamma$ --since $f$ is a homeomorphism there can be no 'back-tracking'. But since $f$ is orientation reversing, this is impossible.

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