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Let f : $S^1$ → R be continuous, where $S^1$ is the unit circle in $R^2$. (a) Show that there is a point z ∈ $S^1$ such that f(z) = f(−z). [z = (x; y), −z = (−x;−y)]\ (b) Show that f is not surjective.

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What did you try? Where are you stuck? –  Pedro Tamaroff Apr 10 '13 at 21:04
    
Did you try looking at $F(x)=f(x)-f(-x)$? It is continuous. –  Pedro Tamaroff Apr 10 '13 at 21:07
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Did you solve the first 776 exercises? –  Euler....IS_ALIVE Apr 10 '13 at 21:33
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closed as off-topic by Bruno Joyal, Stefan4024, Daniel Fischer, Macavity, dfeuer Nov 13 '13 at 19:27

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2 Answers

HINTS: $S^1$ is compact and connected. The map $S^1\to\Bbb R:z\mapsto f(z)-f(-z)$ is useful.

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Hints:

(a) Let $g(z)=f(z)-f(-z)$ and apply the Intermediate Value Theorem.

(b) You know that $S^1$ is compact. What do you know about the image of a compact space under a continuous map?

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Compact...and connected (even path-connected)... –  DonAntonio Apr 11 '13 at 23:31
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