= Solution
Let $a(x)=-x$ be the <antipodal map>, whose degree is $(-1)^{n+1}$. The invariance condition is $f\circ a=f$. Multiplicativity of degree gives $\deg(f)(-1)^{n+1}=\deg(f)$, so if $n$ is even,
$$
\boxed{\deg(f)=0}.
$$
For odd $n$, factor the invariant map uniquely through the quotient $q:S^n\to\mathbb{RP}^n$, writing $f=gq$. The standard orientability and homology facts for <Real projective space> are that $\mathbb{RP}^n$ is orientable in odd dimension and has top integral homology $\mathbb Z$. With compatible <orientation> the double cover $q$ has degree two: both local sheets preserve <orientation> because the <antipodal map> does. Thus $g_*$ is multiplication by an integer $k$ and $\deg(f)=2k$.
Every such integer occurs. In the oriented projective space, choose a small oriented closed ball and collapse its exterior to a point. The resulting map $c:\mathbb{RP}^n\to S^n$ has degree one, because a point in the interior image has one <orientation>-preserving preimage. Composing it with a degree-$k$ <sphere> map from part (b) gives
$$
f=\alpha_k\circ c\circ q,\qquad\boxed{\deg(f)=2k}.
$$
Here $\alpha_k$ denotes the dimension-$n$ map constructed by suspension. The factorization through $q$ makes the resulting map antipodally invariant.
Back to article page
