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Let $n\ge2,\ f: S^{n} \to \mathbb{R}$ a continuous function. $A = \{t\in f(S^{n})|\ f^{-1}(t)\text{ is finite}\}$. Then $A$ has cardinality at most $2$.

This exercise is from the chapter about connectedness, so there must be an elementary solution without algebraic topology.

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    Thank you Matt for the information, I didn't know it2011-12-20

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$S^n$ is connected and compact, so its image under $f$ is a closed and bounded interval $[a, b]$ in $\mathbf R$. I claim that $a$ and $b$ are the only possible members of $A$. Pick any point $x \in (a, b)$, and suppose $f^{-1}(x)$ is finite. I think you can argue that $S^n - f^{-1}(x)$ is still connected [$n = 1$ would be bad here]. Why is this a problem?

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    Cool, that was my thought as well.2011-12-20