An equivalence between function $f(x)$ and $f^{-1}(x)$.

Question

Recently on this answer to one of my questions user farruhota replied that

Alternatively, note the property of inverse function: $$f(f^{-1}(x))=f^{-1}(f(x))=x$$ Hence: $$f(f(x))=x \iff f(x)=f^{-1}(x)$$

How is $f(f(x))=x \iff f(x)=f^{-1}(x)$ derived from the equation $f(f^{-1}(x))=f^{-1}(f(x))=x$?

Is this "$f(f^{-1}(x))=f^{-1}(f(x))=x$" thing only valid when the function is $f(f(x))=x$?

Thanks,
Max0815

Answer 1

Suppose $f: X \to X$ has an inverse $f^{-1}: X \to X$ (in particular, $f$ is a bijection)

If $f(f(x)) = x$, apply $f^{-1}$ to get $f(x) = f^{-1}(x)$.

If $f(x) = f^{-1}(x)$ apply $f$ to get $f(f(x)) = x$. Very simple.

Answer 2

Ok. So to generalize, I have

Suppose $f: X \to X$ has an inverse $f^{-1}: X \to X$ (in particular, $f$ is a bijection)

If $f(f(x)) = x$, apply $f^{-1}$ to get $f(x) = f^{-1}(x)$.

If $f(x) = f^{-1}(x)$ apply $f$ to get $f(f(x)) = x$. Very simple.

user Mariah

as the answer to my first question.

For my second question,

If $f:X\rightarrow X$ satisfies $f(f(x))=x$ for all $x\in X$, then $f$ has an inverse $f^{-1}$ and in fact, $f^{-1}=f$

user SangChul Lee

This relation $f^{-1}=f$ is valid when functions are involutions, when a function $f:X\rightarrow X$ maps a number $x$ to $y$, and another application of the same function maps $y$ to $x$. However,

this property holds whenever a function $f:X\rightarrow X$ has an inverse.

user SangChul Lee

Also,

Note that the domain and codomain must be equal (and $f$ invertible) in order for $f(f^{−1}(x))=f^{−1}(f(x))$ to even make sense. That each is $x$ also holds, but still only when domain same as codomain.

user coffeemath