Show $x \mapsto \frac{x}{\Vert x\Vert^2}$ is orientation reversing. I wanted to show that the map $\mathbb{R}^n- \{0\} → \mathbb{R}^n- \{0\},\ x\mapsto \frac{x}{\Vert x\Vert^2}$ is orientation reversing. But i don't know how to tackle it appropriately.
The solution proposed the following:

At $p = (1,0,...,0)$ the differential of the given map is the identity
on $0 \oplus \mathbb{R}^{n-1} \subset T_p\mathbb{R}^n$ while it is
multiplication by $-1$ on $\mathbb{R}  \oplus 0 \subset
 T_p\mathbb{R}^{n}$.

Then they justify it by connectedness of $\mathbb{R}^{n}$ for $n$ greater than $1$.
I am not sure whether I fully understand the solution. For simplicity let's assume $n = 2$, then at $(1,0)$ the differential of the map is $(-1,1)$, similarly $(-1,1,1...,1)$ for arbitrary $n$. How is it the identity on $0 \oplus \mathbb{R}^{n-1} \subset T_p\mathbb{R}^n$?
Could someone elaborate? I tried to look up similar problems in my books but couldn't find any. I would really like to understand this.
Thanks for any help!
 A: Calculate the matrix representing the differential in standard coordinates in $\mathbb{R}^n$, that is, the Jacobian of the given map: The derivative of the $j$-th component of $\frac{x}{ \| x \|^2}$ with respect to $x^i$, is $\frac{ \delta_i^j \| x \|^2 - 2 x_i x^j }{ \| x \|^4}$.  Evaluated at the point $(1, 0, \dots, 0)$, this gives zero if $i \ne j$, gives $-1$ if $i=j=1$, and $1$ if $i = j \ne 1$, so the Jacobian is block diagonal with the upper left $1\times 1$ block being $-1$ and the lower right $(n-1)\times (n-1)$ block being the identity, which is what the solution you quote is saying in a nicer way.
Makes sense geometrically too, as the map is an inversion in the unit sphere in $\mathbb{R}^n$, and at any point $p$, any tangent vectors perpendicular to $p$ are simply rescaled by $1/\|p\|^2$, whereas the ones parallel to $p$ switch direction.
A: Define $\vec{f}(x)=\frac{\vec{x}}{x\cdot x}$, which coordinate-wise is
$$
f_j(x)=\frac{x_j}{x\cdot x}\tag1
$$
The Jacobian of $f$ is
$$
\partial_if_j(x)=\frac{\delta_{ij}x\cdot x-2x_ix_j}{(x\cdot x)^2}\tag2
$$
Written as matrices, this is
$$
\frac{I-2\frac{xx^T}{x^Tx}}{x^Tx}\tag3
$$
The matrix $I-2\frac{xx^T}{x^Tx}$ is a reflection in the direction of $x$; that is, locally, the map flips in the radial direction and scales by the inverse of the square of the distance from the origin.
The reflection reverses the orientation.
