# continuous map from $S^1\rightarrow \mathbb{R}^1$

well, so far I know there exist no injective map from $S^n\rightarrow R^n$(due to Borsuk-Ulam), so in the case of $3.8$ they are asking are there different point on $S^1$ whic maps to same point in $\mathbb{R}$? so by Borsuk Ulam theorem I can say "Yes", If the $f$ is constant then $A$ is uncounatble, but I have no idea about if the $f$ is non-constant.

for $3.9$ I can say same argument right?

I will be happy about responses. Thank you.

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It might help to consider the function $\phi(x) = f(x)-f(-x)$. Then $\phi(-x) = -\phi(x)$. –  copper.hat Jul 20 '12 at 18:50
@copper.hat clearly by IVT there exist $x_0$ such that $\phi(x_0)=0$ –  La Belle Noiseuse Jul 20 '12 at 18:52

For 3.8, if $f$ is not constant, let $a$ and $b$ be any two values of $f$ and let $u$ and $v$ be any of their preimages. By continuity, any point in $[a,b]$ must have preimages in both of the segments into which $u$ and $v$ divide the circle, and there are uncountably many such points.