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This question already has an answer here:

Does there exist a continuous function

$f\colon \Bbb R\rightarrow \Bbb R$ such that $f(f(x))=-x$ for all $x\in\Bbb R$?

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marked as duplicate by Jyrki Lahtonen, Najib Idrissi, Hanul Jeon, Hurkyl, TZakrevskiy Mar 5 '14 at 15:09

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  • $\begingroup$ actually trying to disprove $\endgroup$ – user128956 Mar 5 '14 at 14:46
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Hint: If $f(f(x))=-x$ then $f$ is a bijection and because $f$ is continuous it must also be either order preserving or order reversing.

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    $\begingroup$ not getting.........elaborate please $\endgroup$ – user128956 Mar 5 '14 at 14:48
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    $\begingroup$ I think the hint should be enough for you to get to the answer. Write down the definition of an order preserving/reversing function and see what conclusions you can make from the above. $\endgroup$ – Dan Rust Mar 5 '14 at 14:53
  • $\begingroup$ but why is $f$ a bijection? $\endgroup$ – user2345215 Mar 5 '14 at 14:54
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    $\begingroup$ $f\circ f$ is a bijection, and if $f$ was not a bijection then either $f\circ f$ would not be injective, or it would not be surjective - hence $f$ is bijective. $\endgroup$ – Dan Rust Mar 5 '14 at 14:55
  • $\begingroup$ $f\circ f\circ f\circ f$ is the identity, hence a bijection. Thus $f$ has to be a bijection, too. $\endgroup$ – Jyrki Lahtonen Mar 5 '14 at 14:55

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