Map from circle to real line

I am asked to show that, for any continuous $\phi:\;S^1\to\mathbb{R}$ where $S^1=\{ \|\mathbf{x}\|=1,\;\mathbf{x}\in\mathbb{R}^2\}$, there exists $\mathbf{z}\neq 0$ such that: $$\phi(\mathbf{z})=\phi(-\mathbf{z})$$ It is suggest that I use connectedness.

I know both sets are connected, and that a continuous map preserves connectedness, but I can't see how this helps. I thought of considering arcs from $\mathbf{z}$ to $-\mathbf{z}$, but again I cannot see how to argue that there must be an arc such that the image of its endpoints are collapsed to a single point in $\mathbb{R}$.

Help?

• Hint: consider the function $f({\bf z}):=\phi({\bf z})-\phi(-{\bf z})$. – Akiva Weinberger Feb 5 '16 at 17:27
• N.B. : this is known as the Borsuk-Ulam theorem. – Watson Feb 5 '16 at 17:34
• @Watson Well, the generalization is. It's much harder for $n\ge2$. – Akiva Weinberger Feb 5 '16 at 17:38

Can you use continuity and the Intermediate Value Theorem for a Connected domain to show that the function $$\phi(z)-\phi(-z)=0$$ for some $z$? If so, you are done.
Since it is suggested that connectedness be explicitly used, I might phrase it like this: The image of the connected domain of the function $\mathbf z \mapsto \phi(\mathbf z) - \phi(-\mathbf z)$ must be a connected subset of $\mathbb R$. If it is not everywhere $0$, then for any point $\mathbf z_0$ where it is not $0$, the function changes signs as $\mathbf z$ goes from $\mathbf z_0$ to $-\mathbf z_0$. Thus the image includes both positive and negative numbers. All connected subsets of $\mathbb R$ that contain both positive and negative numbers contain $0$.