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Let $f\in C([0,1]^n,\mathbb R^n) $ with $[0,1]^n \subset f([0,1]^n)$

Is it true that $\exists x \in [0,1]^n, f(x) =x$?

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This is a simple counterexample a counterexample.

You use two disjoint parts of the square to cover the square.

In one dimension you can prove this using the intermediate value theorem.

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    $\begingroup$ Beautifully illustrated. $\endgroup$ – Reinstate Monica Jan 26 at 18:35
  • $\begingroup$ I disagree but thanks:) $\endgroup$ – Thomas Rot Jan 26 at 20:26

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