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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

$$\psi(x_1, \dots, x_n) = (\sqrt{1 – x_1^2 – \dots – x_n^2}, x_1, x_2, \dots, x_n)?$$

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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.

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