Prove that there exists $p\in \mathbb{S}^1$, that $f(p)=f(-p)$, where $f:\mathbb{S}^1\rightarrow \mathbb{R}$ be the continuous projection

Question

Let $f:\mathbb{S}^1\rightarrow \mathbb{R}$ be the continuous mapping of the circle $\{(x,y)\in \mathbb{R}^2\;|\;x^2+y^2=1\}$ to the set of real numbers. Prove that there exists $p\in \mathbb{S}^1$, that $f(p)=f(-p)$.

So I need help with this kind of proof, I don't know how to yet. Any will be highly appreciated.

Answer 1

Define $F: \mathbb S^1 \longrightarrow \mathbb{R}$ as $$F(x) = f(x) - f( -x)$$

$F$ is continuous. Your aim is to show that $F$ has a zero.

Fix any point $p \in \mathbb S^1$. If $ F(p)=0$, then you are done.

If $F(p) >0$, then $F(-p) <0$. Since $F( \mathbb S^1 )$ is connected, necessarily $0 \in F( \mathbb S^1 )$, and you are done.

The same argument applies if $F(p) <0$. In any case we ha ve a zero: this concludes the proof.