Let $S^3$ and $S^1$ be the 3-sphere in $\mathbb{C}^2$ and the circle in $\mathbb{C}$, respectively. Is there a continuous map $f: S^3 \to S^1$ where $-f(z_1,z_2) = f(-z_1, -z_2)$ for all $(z_1,z_2) \in S^3$?
I've tried quite a bit to solve this but have gotten nowhere. I figure we can restrict the function to the circle $S^1$ in $\mathbb{C}^2$ and then take advantage of the homotopy properties of functions from $S^1$ to $S^1$, but that hasn't lead me anywhere. Any help is much appreciated.
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Take a point $p$ in $\mathbb S^3$ and consider an oriented great circle $C$ connecting $p$ to $-p$. Now, suppose there is a odd continuous map $f:\mathbb S^3 \to \mathbb S^1$. Let $C_1$ be a semicircle in $C$ connecting $p$ to $-p$ and let $C_2$ be the other semicircle connecting $-p$ to $p$. We have $f(C_1)$ is a curve connecting $f(p)$ to $f(-p)$. Notice $f(C_2) = -f(C_1)$. So, if $f(C_1)$ goes aroung the circle $n+1/2$ times, then $f(C_2)$ also goes around the circle $n+1/2$. So, $f(C)$ goes around the circle $2n+1$ times. Since this number is odd and therefore non-zero we have $f(C)$ is a non-contractible loop, you cannot deform this loop to a point.
On the other hand, $C$ is contractible in $\mathbb S^3$.
Therefore, $f$ take a contractible loop in a non contractible loop, which is impossible. So, this odd function cannot exist.
No, this follows immediately from the (stronger) Borsuk-Ulam theorem. but here is a short proof.
Such an odd function induces a map $\Bbb{RP}^3 \to \Bbb{RP}^1$ which is non-trivial on $\pi_1$: if it were trivial, there would be a lift to the double cover, and hence the map $f: S^3 \to S^1$ would factor through $\Bbb{RP}^3$. But that would mean $f$ is even, not odd. So the map on $\pi_1$ must be non-trivial.
But there are no non-trivial homomorphisms $\Bbb Z/2 \to \Bbb Z$. Contradiction.