For the unit sphere $S^n \subset \mathbb{R}^{n+1}$ let $f : S^n \to S^n$ be the map reversing the signs of all but one coordinate, $$f(x_0, x_1, \dots, x_n) = (x_0, -x_1, \dots, -x_n):$$ Compute the Lefschetz number $L(f)$.
So I am actually wondering, that if I should use the sterographic projection, or I can just use the local parametrization $\psi: \mathbb{R}^m \mapsto S^m \subseteq \mathbb{R}^{m+1}$ that $$\psi(x_1, \dots, x_n) = (\sqrt{1 - x_1^2 - \dots - x_n^2}, x_1, x_2, \dots, x_n)?$$
I am not too convinced with myself that the intersection can be treated locally, hence sterographic projection is not necessary, so I can just drop the first coordinate.
Because $(x_0, x_1, \dots, x_n)$ sits on the unit sphere, $x_0 = \sqrt{1 - x_1^2 - \dots - x_n^2}$. Since $x_0$ is dependent of $x_1, \dots, x_n$, I can simply drop it as rewrite $f$ as: $$\tilde f(\tilde x_1, \dots, \tilde x_n) = (-\tilde x_1, \dots, - \tilde x_n).$$
Then $$df_x - I = \begin{pmatrix} -2 & 0 & \cdots & 0 \\ 0 & -2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & -2 \end{pmatrix}.$$ Hence, $$\det(df_x - I) = (-2)^n.$$ Since the sign of $L_x(f)$ equals the sign of $\det(df_x - I)$, we have $L_x(f) = 1$ if $n$ is even, and $L_x(f) = -1$ is $n$ is odd.
But since $df_x$ is independent of $x$, we have $L(f) = 2L_x(f)$, since clearly, there are two fixed points $(1,0,\dots, 0)$ and $(-1,0,\dots, 0)$. Hence $L(f) = 2$ if $n$ is even, and $L(f) = -2$ is $n$ is odd.