Antipodal mapping $f:S^n\to S^1$

Question

I know that " There does not exist an antipodal mapping $f:S^{n}\to S^{n-1}$" by Borsuk-Ulam. But, ¿There does not exist an antipodal mapping $f:S^{n}\to S^{1}$ (continous)?

Answer 1

No, embed $S^1 \hookrightarrow S^{m-1}$ equatorially so that $(x_1,x_2) \mapsto (x_1, x_2,0,\dots,0) \in S^{m-1}$ and note that such a map would provide an odd map $S^n \to S^{m-1}$.

I guess if you wanted to do this from scratch:

Take $f:S^n \to S^1$ an odd map. So, we can embed $i:S^1 \hookrightarrow S^n$ equatorially, and note that the composition $f \circ i:S^1 \to S^1$ is nontrivial since its degree is $\pm 1$, while it factors through $S^n$ which has trivial $\pi_1$.

Answer 2

Just to be precise, by an antipodal map $f$ between spheres you mean one satisfying $f(-x) = -f(x)$ for all $x$.

Obviously antipodal maps exist for $n = 1$ (e.g. the identity on $S^1$).

If there would exist an antipodal map for $n > 1$, then you would get one for $n = 2$ because $S^2$ embeds antipodally into $S^n$. Now have a look at

Map $f:S^2 \to S^1$ with $f(-x) = -f(x)$ .