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The only time that I've heard the term "orientation-preserving map" was in Linear Algebra, but today I read the term orientation-preserving homeomorphism of the circle in the following context: If a homeomorphism $f$ of $S^1$ has a rational rotation number $r$ ($r=\lim_{n\to\infty} \frac{F^n(x)-x}{n}$ where $F$ is a "lift" of $f$ to $\mathbb{R}$) and if $f$ is "orientation-preserving", then every periodic point of $f$ has the same period. My question is simply:

What does it mean (rigorously) for a homeomorphism $f$ of the circle $S^1$ to be "orientation-preserving"?

Thank you in advance!

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  • $\begingroup$ It would help to have some context. Could you tell us where you read this? $\endgroup$ – A.P. Jun 9 '15 at 22:51
  • $\begingroup$ Sure, I've edited the question with some context. $\endgroup$ – u1571372 Jun 9 '15 at 22:56
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Let $p:\Bbb{R} \longrightarrow S^1$ be the fundamental covering map $t \mapsto e^{2\pi i t}$. Then $f$ preserves orientation if there exists some increasing homeomorphism $g: \Bbb{R} \longrightarrow \Bbb{R}$ such that $p \circ g = f \circ p$, i.e. if $f$ can be lifted to an increasing homeomorphism of $\Bbb{R}$.

Note that any homeomorphism of $\Bbb{R}$ is necessarily monotone, so this gives the notion of "preserving / swapping the orientation" for a homeomorphism of $S^1$.

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  • $\begingroup$ Thank you, that's precisely the kind of definition that I expected. $\endgroup$ – u1571372 Jun 9 '15 at 23:05
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I think that in this context orientation-preserving means that $f$ doesn't flip local charts.

Since $S^1$ is a $1$-dimensional manifold, given a point $x$ on $S^1$ you can always find open neighbourhoods $U$ of $x$ and $V$ of $f(x)$ homeomorphic to, say, $(-1/2,1/2)$, where $x$ and $f(x)$ correspond to $0$ (under the respective homeomorphisms). Furthermore, up to substituting $U$ with a smaller neighbourhood of $x$ we can assume that $f(U) \subseteq V$.

These homeomorphisms induce a partial order on $U$ and $V$ and we can say that $f$ is orientation-preserving at $x$ if $f(y) > f(x)$ for every $y \in U$ such that $y > x$. Then $f$ is orientation-preserving on $S^1$ if it is at every point of $S^1$.

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