Proving that orientation-reversing homeomorphisms of the circle have exactly two fixed points using the lift.

Question

I am a Chemistry student who is self-studying mathematics. In particular, I am looking at lecture notes on discrete dynamical systems and am desperately stuck on the following question:

Prove that any orientation-reversing homeomorphism of the circle has exactly two fixed points.

The given hint asks me to use the concept of lifts. I have found the following stack exchange question but do not understand the answer: An orientation-reversing homeomorphism of the circle has two fixed points and rotation number $0$.

I have been able to solve the question using the definition of an orientation-reversing homeomorphism and by considering open sets but do not see how Lifts relate.

Any help would be appreciated.

Answer 1

Since we have $S^1 \sim [0,1]/~$ where ~ identifies 0 and 1, we know that a homeomorphism $h$ of $S^1$ lifts to a homeomorphism from $[0,1] \to [0,1]$. Any homeomorphism from [0,1] to [0,1] has a fixed point (you can use Brouwer’s FPT but really this is just the intermediate value theorem here). Let us call $h(p) = p$ this fixed point. Now we can graph homeomorphisms $S^1 \to S^1$ in the square $[0,1] \times [0,1]$ as a curve, keeping in mind the four corners get mapped to $x$. An orientation-reversing homeomorphism corresponds to a monotonically decreasing map from $[0,1] \to [0,1]$. Since the corners get mapped to $p$, for this curve, the endpoints are one of the corners. We choose to lift $h(0)$ to 1, so that we can draw this homeomorphism as a monotonically decreasing curve. This curve falls to 0 as $x$ equals $1$. Now, the number of fixed points equals the numbers of intersections with the curve $y=x$. There is at least one more intersection (due to the intermediate value theorem again, using the axes adjusted to accommodate for the first fixed point to occur at the corners), and there is exactly one due to monotonicity. Thus, we have a total of two fixed points. An illustration is provided below.

Topological aside: If you've studied any further algebraic topology, note that our graph is in $S^1 \times S^1 = T^2$, and we're essentially calculating the intersection number of the knot $y=x$ (which is $T_{1,1}$) and the knot represented by an orientation-reversing homeomorphism (which is $T_{1,-1}$). Sure enough, their intersection number is $2$, as this calculation shows.

Answer 2

To really understand what a lift is you need to know the concept of a covering map. It is impossible to explain that in all details here, but have a look in the book

Allen Hatcher "Algebraic Topology", https://pi.math.cornell.edu/~hatcher/AT/AT.pdf.

A standard example of a covering map is $$e : \mathbb R \to S^1 = \{ z \in \mathbb C \mid \lvert z \rvert = 1\}, e(t) = e^{it} = \cos(t) + i \sin(t). $$

Note that $e$ maps the closed interval $[0,2\pi] \subset \mathbb R$ surjectively onto $S^1$, where $e(0) = e(2\pi)$ and $e(t) \ne e(t')$ for $t, t' \in [0,2\pi]$ with $\lvert t - t' \rvert < 2\pi$.

Now let $h : S^1 \to S^1$ be any map. Using the theory of covering maps, we see that there exists a lift $H : [0,2\pi] \to \mathbb R$ of $h \circ e \mid_{[0,2\pi]}$. "Lift" means that $e \circ H = h \circ e \mid_{[0,2\pi]}$. Note that lifts are not unique. In fact, for each $k \in \mathbb Z$ the map $H_k(t) = H(t) + 2k\pi$ is also a lift, and each lift of $h$ has the form $H_k$ for some $k$. Let us write $H' : [0,2\pi] \stackrel{H}{\to} J = [a,b] = H([0,2\pi])$. Note that $H([0,2\pi])$ is a compact connected subset of $\mathbb R$, i.e. a closed interval.

If $h$ is a homeomorphism, also $H'$ must be one and $J$ must have length $2\pi$, i.e. $J = [a,a+2\pi]$. Since $H'$ must map the boundary points of $[0,2\pi]$ to the boundary points of $[a,a+2\pi]$, we either have $H'(0) = a, H'(2\pi) = a + 2\pi$ or $H'(0) = a+2\pi, H'(2\pi) = a$. In the first case $h$ is orientation preserving which means that $H'$ is strictly increasing, in the second case $h$ is orientation reversing which means that $H'$ is strictly decreasing. By the way, this can be seen also directly by elementary arguments without using the full theory of covering maps.

Now let us assume that $h$ is orientation reversing. W.l.o.g. we may assume that $H'(0) = a +2\pi \in [0,2\pi)$. i.e. $a \in [-2\pi,0)$ (otherwise replace $H'$ by a suitable $H'_k$). By the IVT there exists a unique $\xi \in [0,2\pi]$ such that $H(\xi) = 0$. We have $\xi = 0$ if $a = -2\pi$. The (possibly degenerate) interval $[0,\xi]$ is mapped by $H'$ homeomorphically (and strictly decreasing) onto $[0,a+2\pi]$ and $[\xi,2\pi]$ homeomorphically (and strictly decreasing) onto $[a,0]$.

Write $z \in S^1$ as $z = e(t)$ with a unique $t \in [0,2\pi)$. Then $h(z) = z$ iff $e(H'(t)) = h(e(t)) = h(z) = z = e(t)$. The latter holds true iff $H'(t) = t$ or $H'(t) = t -2\pi$. Note that $e(H'(t)) = e(t)$ iff $H'(t) = t + 2k\pi$, and only for $k = 0,-1$ we can have $t + 2k\pi \in [a,a+2\pi]$ since $t + 2k\pi \in [2k\pi,2(k+1)\pi)$.

$H'(t) = t$ occurs exactly once for $t \in [0,\xi]$. This is trivial for $\xi = 0$ (which means $a = -2\pi)$). For $\xi > 0$, i.e. $a \in (-2\pi,0)$, consider the function $g(t) = H'(t)- t$. We have $g(0) = H'(0) - 0 = a + 2\pi > 0$ and $g'(\xi) = H'(\xi) - \xi = -\xi < 0$, thus by the IVT we find a unique $t \in (0,\xi)$ such that $g(t) = 0$ which means $H'(t) = t$.

A similar argument shows that $H'(t) = t - 2\pi$ occurs exactly once for $t \in [\xi,2\pi]$.

Thus $h$ has exactly two fixed points.