# general topology exercise 777 [closed]

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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## closed as off-topic by Bruno Joyal, Stefan4024, Daniel Fischer♦, Macavity, dfeuerNov 13 '13 at 19:27

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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
Did you solve the first 776 exercises? – Euler....IS_ALIVE Apr 10 '13 at 21:33

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