Orientation of stereographic projection I was reading Lee and was wondering if the stereographic projection of the north pole and the south pole are in the same oriented smooth atlas. To be more precise, we have $\sigma_N: S^n \to \mathbb{R}^n$ defined by:
$$
\sigma_N(x^1, \ldots, x^{n+1}) = \frac{(x^1, \ldots, x^n)}{1-x^{n+1}}
$$ 
and: 
$$
\sigma_N^{-1}(u^1, \ldots, u^n) = \frac{(2u^1, \ldots, 2u^n, |u|^2 - 1)}{|u|^2+1}
$$
Then the projection from the south pole is given by: $\sigma_S(x) = -\sigma_N(x)$. The composition of these two maps yields:
$$
\sigma_S \circ \sigma_N^{-1} (x) = \frac{x}{|x|^2}
$$
If the determinant of the total derivative of this map is negative, then these two functions do not belong to the same oriented smooth atlas. I have been able to show this in two and in three dimensions with some laborious computations in Mathematica, but I don't see how I can can compute that in the $n$-dimensional case. Is there an easier way to show that these two maps do not belong to the same oriented smooth atlas?  
 A: It is not true that $\sigma_S(x)= −\sigma_N(x)$. In fact, we have $\sigma_S(x) = \sigma_N(r(x))$, where $r : \mathbb R^{n+1} \to \mathbb R^{n+1}, r(x_1,\ldots,x_{n+1})=(x_1,\ldots,x_n,−x_{n+1})$. Nevertheless we correctly get
$$ \sigma_S \circ \sigma_N^{-1} (x) = f(x) :=\frac{x}{|x|^2} .$$
Note that the transition map $f$ is defined on $\mathbb R^n \setminus \{ 0\}$. The Jacobian matrix $Jf(x)$ of $f$ at $x$ has entries
$$\dfrac{\partial f_i}{\partial x_j}(x) = \begin{cases} \dfrac{-2x_ix_j}{\lvert x \rvert ^4} & i \ne j \\ \dfrac{\lvert x \rvert^2 -2x_i^2}{\lvert x \rvert^4} & i = j \end{cases}$$
Now consider the point $\xi = (1,0,\ldots,0)$. The Jacobian $Jf(\xi)$ is a diagonal matrix with entries $d_{11} = -1$, $d_{ii} = 1$ for $i > 1$. Its determinant is $-1 < 0$. Therefore $\sigma_N, \sigma_S$ cannot be contained in the same oriented smooth atlas.
Remark:
Consider the function
$$d : \mathbb R^n \setminus \{ 0\} \to \mathbb R \setminus \{ 0\}, d(x) = \det Jf(x) .$$
This is a continuous function. For $n = 1$ we get $d(x) = -\dfrac{1}{x^2} < 0$ for all $x$. For $n > 1$ it is difficult to explicitly compute $d(x)$ for a general $x$, but it is easy to see that $d(x) < 0$ for all $x$. Since $\mathbb R^n \setminus \{ 0\}$ is connected, the image of $d$ is a connected subset of $\mathbb R \setminus \{ 0\}$ which contains $d(\xi) = -1$. Thus the image is contained in $(-\infty,0)$. Actually the image is $(-\infty,0)$ because $d(r\xi) = -\dfrac{1}{r^{2n}}$.
