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Exactly under what conditions would the equality $f(x)=f^{-1}(x)$ hold? The proof attached below considers a special case when $f$ is strictly increasing. The theorem then says that the set of solutions to $f(x)=x$ will always be a superset of the set of solutions to $f(x)=f^{-1}(x)$. Any suggestions on how to improve this proof will be much appreciated.

However, most high school texts seem to assume in all cases to solve $f(x)=f^{-1}(x)$, simply solve $f(x)= x$. We know that this method, if it works at all, is at best a sufficient but not necessary condition. So, is there a precise set of conditions such that our desired equality will hold?

It's a superset

Edit: to avoid any miscommunications, my question is not related to the concept of involution functions. Involution functions are functions that are their own inverses. On the other hand, my question doesn't require $f^{-1}$ to be equal to $f$. I am merely interested in finding under what conditions will the equality $f=f^{-1}$ holds. And yes, just to clarify, I am looking at invertible functions. Thank you to whoever who pointed that out.

mrtaurho
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Hello
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    To be clear, you are asking that given a particular invertible function $f$ under what conditions for $x$ will $f(x)=f^{-1}(x)$? You are not asking under what conditions for $f$ will $f(x)=f^{-1}(x)$ for all $x$? – JMoravitz Apr 27 '19 at 13:51
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    Great question, but unfortunately a duplicate of 'Functions that are their own inversion'. – Toby Mak Apr 27 '19 at 13:52
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    I am voting to repoen because this is not a duplicate of the linked question. Crucially, the OP is asking for particular $x$s in the domain of $f$ so that $f$ and $f^{-1}$ coincide, and not requiring that it is true for all $x$. While I doubt this will have a satisfactory general solution, it is not a duplicate. – YiFan Tey Apr 28 '19 at 02:20

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