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Here is a bit from Hatcher's book:

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I don't understand part (f); why is the antipodal map the composition of $n+1$ reflections? Even if I accept that, I still don't know why does it have degree $(-1)^{n+1}$. What would that mean for $S^2$? We fix the equator $S^1$ and take every other point $x$ to $-x$? And then we are doing this for $S^1$? That sounds a little pointless. In general if $\operatorname{deg}f=x$ and $\operatorname{deg}g=y$ then what is $\operatorname{deg}f\circ g$? It's $xy$ I guess.

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    $\begingroup$ On $S^2$. You take the map $f_x \colon S^2 \to S^2 \,\, f_x(x,y,z) = (-x,y,z)$ and define $f_y$ and $f_z$ in a similar way. Then observe that the antipodal map $-\mathbb{1}$ is just $-\mathbb{1} = f_x \circ f_y \circ f_z$. By (e), each of the $f_i$ has degree $-1$, so by (d) you have $\mathrm{deg}\, (-\mathbb{1}) = (-1)(-1)(-1) = -1$. Just do the same things on $S^n$. $\endgroup$
    – Ivo
    Oct 16, 2013 at 13:12
  • $\begingroup$ possible duplicate of The degree of antipodal map. $\endgroup$
    – Noah Snyder
    Dec 5, 2014 at 4:06
  • $\begingroup$ which page of Hatcher? $\endgroup$
    – Idonotknow
    Nov 3, 2018 at 18:23

2 Answers 2

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Consider $S^n$ as a subspace of $\mathbb{R}^{n+1}$. Then a point $x$ in $S^n$ is given by an $(n+1)$--tuple $(x_0,\dotsc,x_n)$, hence the antipodal map is given by $$ (x_0,\dotsc,x_n)\mapsto (-x_0,\dotsc,-x_n) $$ we can get this map by composing all $n+1$ reflections with respect to the hyperplanesplanes orthogonal to the coordinate axes. (I.e. the maps $$ (x_0,\dotsc, x_k,\dotsc,x_n)\mapsto (x_0,\dotsc,-x_k,\dotsc,x_n) $$ Furthermore, since homology is functorial, we have $H_n(f\circ g,\mathbb{Z}) = H_n(f,\mathbb{Z})\circ H_n(g,\mathbb{Z})$. Thus given $f,g\colon S^n\to S^n$ with $$ H_n(f,\mathbb{Z})=f_*\colon H_n(S^n,\mathbb{Z})\to H_n(S^n,\mathbb{Z}) $$ $$x\mapsto\alpha x$$ and $$ H_n(g,\mathbb{Z})=g_*\colon H_n(S^n,\mathbb{Z})\to H_n(S^n,\mathbb{Z}) $$

$$x\mapsto\beta x $$ we have $(f\circ g)_*(x) = f_*\circ g_*(x) = f_*(\beta x) = \alpha\beta x$, so as you assumed $\operatorname{deg}(f\circ g) = \operatorname{deg}(f)\operatorname{deg}(g)$. And combining both results, we get the degree $(-1)^{n+1}$ for the antipodal map on $S^n$.

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  • $\begingroup$ If it is $S^\infty$ how to compute the deg ? For example, $E$ is infinity dimensional Banach space,$\Omega={x\in E:||x||=1}$,$L:\Omega\rightarrow\Omega ,L(x)=-x$ $\endgroup$
    – Farmer
    Mar 2, 2016 at 2:01
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If $u$ is a unit vector in $\mathbf{R}^{n+1}$ and $H = u^\perp$ is the hyperplane orthogonal to $u$, then reflection in $H$ is the linear transformation $R_u(v) = v - 2\langle u, v\rangle u$.

For example, if $u = \mathbf{e}_i$ is the $i$th standard basis vector (so $H$ is the hyperplane $\{x_i = 0\}$), then $$ R_u(x_1, \dots, x_{i-1}, x_i, x_{i+1}, \dots, x_{n+1}) = (x_1, \dots, x_{i-1}, -x_i, x_{i+1}, \dots, x_{n+1}). $$

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