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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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Hey stef. People around here like it when you accept some of their answers from time to time. To do so you click on the tick symbol next to the answer you'd like to accept. – Rudy the Reindeer Dec 20 '11 at 8:08
Thank you Matt for the information, I didn't know it – WLOG Dec 20 '11 at 8:12
up vote 2 down vote accepted

$S^n$ 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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Sorry Dylan, but I don't understand why |A| is at most 2. – WLOG Dec 20 '11 at 8:14
@stef Maybe I should make this clearer in the hints above, but the argument is that any point in the interior of the interval has an infinite preimage under $f$. How many points are left? – Dylan Moreland Dec 20 '11 at 8:19
Now it's clear ! – WLOG Dec 20 '11 at 8:28
@stef I'm glad! Maybe the only loose end here is proving that the $n$-sphere $(n \geq 2)$ less a finite set of points is connected. Are you good with this? – Dylan Moreland Dec 20 '11 at 8:31
I think we can use stereographic projection and then move to R^n – WLOG Dec 20 '11 at 8:37

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