This question is truly stupid, but is driving me crazy. I just need an outside viewpoint to sort out what's going on.
- In my textbook: "Show that every linear fractional (LF) transformation of $\hat{\mathbb{C}}$ is orientation preserving.
This is completely baffling. $f(z) = \frac{1}{z}$ is certainly LF, but it's equal to $\frac{\overline{z}}{|z|^2}$. I can't think of a more not orientation preserving map than complex conjugation followed by scaling.
- From Wikipedia: The set of LF transformations is isomorphic to the orientation-preserving isometries of $\mathbb{H}^3$.
What is the isomorphism here, explicitly? Realize $\mathbb{H}^3$ as the upper-half space in $E^3$. It can't be Poincaré extension, since we can just take the map $f$ from above and extend it to get reflection through a sphere, which is not orientation preserving. What orientation-preserving map of $\mathbb{H}^3$ will be the image of $f$?
Holomorphic functions, including Möbius transformations, are locally orientation preserving. This means that a small circle $t\mapsto z_0+r e^{it}$ $\ (0\leq t\leq 2\pi)$ going counterclockwise around a point $z_0$ where $f'(z_0)\ne0$ is mapped onto a small circlelike closed curve going counterclockwise around the point $f(z_0)$. This has to do with the fact that the Jacobian determinant at $z_0$ of the underlying real map ${\bf f}:\ {\bf z}\to{\bf f}({\bf z})$ is given by $|f'(z_0)|^2>0$.
Draw a figure and convince yourself that this property is manifest also in the case of the map $f:\ z\mapsto {1\over z}$ wherever it is defined.