For $n \geq 2$, there is no $\phi : S^n \to S^1$ such that $\phi(-x) = -\phi(x)$. 
Show that for $n\geq 2$ there is not any function $\phi: S^n \rightarrow S^1$ such that $\phi(-x)=-\phi(x)$.

I have no idea about how to solve this problem. It is quite similar to Borsuk-Ulam Theorem.
Any hint? 
Thanks for help!

The Borsuk-Ulam Theorem states that for any continuous odd function $f : S^n \to \mathbb{R}^n$, there is $p \in S^n$ such that $f(p) = 0$. 
If such a map $\phi$ existed, then $f = i\circ\phi$ would be a continuous odd map $S^n \to \mathbb{R}^n$ (here $i$ denotes the inclusion $S^1 \to \mathbb{R}^n$). By Borsuk-Ulam, there is $p \in S^n$ such that $f(p) = 0$, but $f(p) = i(\phi(p)) = \phi(p) \neq 0$ as $\phi(p) \in S^1$. Therefore there is no such map $\phi$.
