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?

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 '17 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 '17 at 17:49