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Is it true that there exist $x\in S^1$ such that $f(x)=x$ where $f:D\rightarrow S^1$ is a continuous map. $D$ is closed unit disk.

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up vote 8 down vote accepted

Yes, by the Brouwer fixed-point theorem (we consider $f$ to be a map from $D$ to $D$, which is OK, as $S^1 \subset D$, and the map remains continuous when we extend the codomain trivially like this). The promised fixed point must lie on $S^1$ (as all values of $f$).

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