A map $f:S^n\rightarrow S^n$ satisfying $f(x)=f(-x)$ for all $x\in S^n$ is said to be an even map. Show that if $f:S^n\rightarrow S^n$ is an even map, then $\deg f=0$ when $n$ is even and $\deg f$ is even when $n$ is odd. Moreover show that when $n$ is odd, there exist even maps of any given even degree.
My Try:
So $f=(-I)\circ f$. Then $\deg f=\deg (-I)\times \deg f$. If $n$ is even $\deg (-I)=-1$. Hence, $2\deg f=0$, so $\deg f=0$. But how do I show the results when $n$ is odd. No clue at all. This problem is in Hatchers book (A very complicated text). There is a hint but I do not understand it. Can somebody help me to proceed?
Best Answer
Note that $f$ factors over $\Bbb RP^n$. Therefore, for $n$ odd, it is enough to show that the $2$-fold covering map $S^n\to \Bbb RP^n$ induces multiplication by $2$ on $n$-th homology. Can you prove that?