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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$\begingroup$ What did you try? Where are you stuck? $\endgroup$– Pedro ♦Apr 10, 2013 at 21:04
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$\begingroup$ Did you try looking at $F(x)=f(x)-f(-x)$? It is continuous. $\endgroup$– Pedro ♦Apr 10, 2013 at 21:07
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3$\begingroup$ Did you solve the first 776 exercises? $\endgroup$– Euler....IS_ALIVEApr 10, 2013 at 21:33
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2 Answers
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HINTS: $S^1$ is compact and connected. The map $S^1\to\Bbb R:z\mapsto f(z)-f(-z)$ is useful.
answered Apr 10, 2013 at 21:07
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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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$\begingroup$ Compact...and connected (even path-connected)... $\endgroup$ Apr 11, 2013 at 23:31