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Aug 15, 2023 at 8:53 comment added Pietro Majer If $h:K\to K$ is bijective and $f:K\to K$ has fixed points, so does $h\circ f \circ h^{-1}$. Non continuous bijection may still be reasonable objects to work with, but I have no idea about how to prove that a given $g:K\to K$ is conjugated to a continuous $f$.
Jun 15, 2020 at 7:27 history edited CommunityBot
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Dec 1, 2013 at 1:26 history edited Gil Kalai CC BY-SA 3.0
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Nov 18, 2013 at 1:39 history edited Gil Kalai CC BY-SA 3.0
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Nov 10, 2013 at 22:49 comment added domotorp Another class comes from defining a complete lattice on K and taking a function that is monotone w/r/t this lattice. This has a fixed point as shown by the Knaster–Tarski theorem. Many functions can be obtained this way, but e.g. rotations cannot be.
Nov 9, 2013 at 13:39 history edited Gil Kalai
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Jul 6, 2013 at 18:46 history asked Gil Kalai CC BY-SA 3.0