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.
Best Answer
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 $S^n$.