Continuous antipodal map $f: S^2 \to S^1$

Question

I want to show that there is no continuous map $f: S^2 \to S^1$ with property $f(-x) = -f(x)$ for all $ x \in S^2$.

My ideas: By asuming that there exist such map $f$ and the fact that the fundamental group $\pi(S^2) =0$ vanishes, $f$ must factorise through universal covering $\mathbb{R}$ of $S^1$. I suppose that the key might be considering paths $\gamma:[0,1] \to S^2$ with some properties like $\gamma(0)= -\gamma(1)$ or something similar, mapping it via $f$ to $S^1$ and lifting them.

But I don't see how to derive here a contradiction to the assumption. Intuitively I guess that one maybe can break here the uniqueness lifting therorem, but I don't find the concrete examples.

Answer 1

Hint: take any great circle $C \subseteq \mathbb{S}^2$ and show that $\deg f \upharpoonright C$ is odd while $f \upharpoonright C$ is clearly null-homotopic.

Answer 2

You have the right idea. Let $\gamma \colon [0,1] \to S^2$ be a constant speed parametrization of a great circle, so that $\gamma(t + \tfrac12) = -\gamma(x)$ for $t \in [0,\tfrac12]$; then the same equation holds for $h = f\gamma$. Lifting $h$ to $\tilde h \colon [0,1] \to \mathbb{R}$, this means that $\tilde h(t + \tfrac12) = \tilde h(t) + n + \tfrac12$ for some integer $n$, which is independent of $t$ (why?). But that means $\tilde h(1) - \tilde h(0) = 2n + 1$ is odd, so $h$ represents a nonzero element of $\pi_1(S^1)$. But this contradicts the fact that $h$ must be nullhomotopic (since $\gamma$ is).

This is most of the proof of the Borsuk-Ulam theorem for $S^2$ (see e.g. Theorem 1.10 in Hatcher).

Answer 3

By Reductio ab absurdum.

Let $f:S^2 \to S^1$ a continous map such that $\space f(-x)=-f(x)$ $ \space \forall \space x \in S^2$.

And let $S^1=\{z \in \mathbb C : |z|=1 \}$.

Consider now the map $h:S^1 \to S^1$ such that $h(z)=z^2$. This map is continous and is a covering map. Let $\alpha : I \to S^2 $ a loop that has as extremal points two antipodal points. Then $hf\alpha:I\to S^1$ is a closed path that is not banal in $\pi_1(S^1)$. (Check this). If we consider $\beta:I\to S^2$ the path $\beta(t)=-\alpha(t)$ then $hf\beta=hf\alpha$, and the loop $\alpha *\beta$ is homotopically trivial in $S^2$. Finally we have :$$0=[hf(\alpha*\beta)]=[hf\alpha][hf\beta]=[hf\alpha][hf\alpha].$$ This is a contraddiction because the foundamental group of $S^1$ is $\mathbb Z$.

Answer 4

This would descend to a map $f_*:\pi_1(\mathbb RP^2) \to \pi_1(\mathbb RP^1)=\mathbb Z$, so it must be the zero map, and hence lifts to $\mathbb R$, making it nullhomotopic.

On the other hand, this cannot be so, see here for example