The degree of antipodal map, composition of reflections? Here is a bit from Hatcher's book:

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.
 A: 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 $S^n$.
A: 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
$$
R_u(x_1, \dots, x_{i-1}, x_i, x_{i+1}, \dots, x_{n+1}) = (x_1, \dots, x_{i-1}, -x_i, x_{i+1}, \dots, x_{n+1}).
$$
