A map of even degree must identify a pair of antipodal points

Question

I'm stuck on this problem in Armstrong's Basic Topology Chapter 9, P206.

Prove that if $f:S^n\to S^n$ has even degree, it must identify a pair of antipodal points of $S^n$.

I know some similar conclusions such as "if $f$ preserves antipodal points, then $f$ has odd degree", "if f has odd degree, it must carry some pair of antipodal points into a pair of antipodal points".
I tried a similar skill used in solving the second problem: consider a homotopy $F(x,t)=\frac{tf(x)+(1-t)f(-x)}{||tf(x)+(1-t)f(-x)||}$ and the degree of $F(x,1/2)$. I let $F(x,t)=\frac{tf(x)-(1-t)f(-x)}{||tf(x)-(1-t)f(-x)||}$ in this problem to look for contradiction. But I can get nothing about the degree of these $F(x,t)$ for certain $t$s. And since $deg(-f(-x))=deg(f)$ holds for any $f$, I 've found no contradiction so far.

Is this a wrong approach? Any small tip would be appreciated.

Answer 1

Your approach is good. Hint: use one of the homotopies you've mentioned (the one made possible by the assumption $f(x) \not = f(-x)$) and the fact that an odd map has odd degree.

For $F(x,t)=\frac{tf(x)-(1-t)f(-x)}{||tf(x)-(1-t)f(-x)||},$ setting $t = 1/2$ gives you a map whose degree has to equal $\deg (f)$ and yet has to be odd.