An orientation-reversing homeomorphism of the circle has two fixed points and rotation number $0$.

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?

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.

• 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. Feb 27, 2017 at 17:41
• No problem. Draw the graph of $F$ and look at the points that give fixed points. Indeed, all comes from the graph. Feb 27, 2017 at 17:49
• It took some time but I got it now! Thank you. Feb 27, 2017 at 19:43
• I can not still understand @AlexMathers could you draw the graph? Dec 5, 2020 at 21:29

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.

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.