Determining that a certain diffeomorphism of $\Bbb R^n-\{0\}$ is orientation preserving or not Consider the diffeomorphism $f:\Bbb R^n-\{0\} \to \Bbb R^n-\{0\}$ (whose inverse is itself) given by $x\mapsto x/|x|^2$.
How can we determine that $f$ is orientation preserving? For $n=1$ it is clearly orientation reversing, and also for $n=2$, but it seems not easy to compute its Jacobian determinant for large $n$, so I think there should be another method. Can I get a hint?
 A: Note that the set $U:= \Bbb{R}^n \setminus\{0\}$ is open and connected, and the map $f:U \to U$ you defined is $C^{\infty}$. So, the function $\phi:U \to \Bbb{R}$ given as
\begin{align}
\phi(x) := \det Df_x = (\det \circ Df)(x)
\end{align}
is a composition of smooth maps hence smooth (in particular continuous). Also, since $f$ is a diffeomorphism, we know that $\phi$ is no-where vanishing. Thus, by the intermediate value theorem, the image of $\phi$ lies either in $(-\infty,0)$ or in $(0, \infty)$. To figure out which, all we have to do is calculate $\phi$ at a simple point, for example $\phi(e_1)$, where $e_1 = (1,0, \dots, 0)$. Now, note that the matrix of $Df_x$ consists of the various partial derivatives $\dfrac{\partial f_i}{\partial x_j}$. These are easy enough to calculate (quotient rule):
\begin{align}
\dfrac{\partial f_i}{\partial x_j}\bigg|_x &= \dfrac{\lVert x\rVert^2 \delta_{ij} - x_i\left( 2x_j\right)}{\lVert x\rVert^4}
\end{align}
In particular, at $x = e_1 = (1, 0, \dots, 0)$, this reduces to
\begin{align}
\dfrac{\partial f_i}{\partial x_j}\bigg|_{x = (1, 0, \dots, 0)} &= \delta_{ij} - 2 \delta_{i1} \delta_{j1}
\end{align}
If you assemble this into a matrix, it is the following block diagonal matrix:
\begin{align}
[Df_x] &= 
\begin{pmatrix}
-1 & 0 \\
0 & I_{n-1}
\end{pmatrix}
\end{align}
So, what is the determinant?
A: You have to calculate the derivative in one point only, since the sign of the Jacobian cannot change on the connected set on which the map is defined (because otherwise it would have a zero somewhere).
Now note that the restrtiction of the map to the unit sphere is just the identity, and that it's derivative, in $x$ with $|x|=1$, will map $x$ to $-x$.
From this you can easily get a diagonal representation of the derivative in $x \in  S^{n-1}$, which allows you to directly read off the sign of the determinant.
