TITLE: Meaning of the expression "orientation preserving" homeomorphism
QUESTION [12 upvotes]: 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!

REPLY [4 votes]: 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 $S^1$.