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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

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  • $\begingroup$ For the second question, it is a direct consequence of the definition of inverse function. Recall that $g : Y \to X$ is called an inverse function of $f : X \to Y$ if $$g(f(x)) = x \text{ for all } x \in X, \qquad f(g(y)) = y \text{ for all } y \in Y.$$ Now the property in the second question is simply the case when $X = Y$ so that $f : X \to Y$ and its inverse $f^{-1} : X \to X$ live on the same set. $\endgroup$ – Sangchul Lee Mar 24 at 18:34
  • $\begingroup$ @SangchulLee so yes, it is only valid in that case? $\endgroup$ – Max0815 Mar 24 at 18:35
  • $\begingroup$ The second property holds whenever a function $f : X \to X$ has an inverse. It need not be an involution (i.e. $f(f(x)) = x$) to satisfy the property, although any involution will certainly do. $\endgroup$ – Sangchul Lee Mar 24 at 18:38
  • $\begingroup$ 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. $\endgroup$ – coffeemath Mar 24 at 18:39
  • $\begingroup$ @SangchulLee so you mean that if $f(x)$ has an inverse and satisfies that $f(f(x))=x$, then $f(x)=f^{-1}(x)$ is true? $\endgroup$ – Max0815 Mar 24 at 18:39
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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.

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  • $\begingroup$ A question on your notation in the first sentence. You mean that this is only true when $f(x)=x$? $\endgroup$ – Max0815 Mar 24 at 18:37
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    $\begingroup$ @Max0815 no, we only assumed $f$ is a function from $X$ to $X$ with an inverse. $\endgroup$ – Mariah Mar 24 at 18:38
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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

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  • $\begingroup$ If I understand you correctly, you still seem to think that if $f$ has an inverse then it is an involution - this is wrong. $\endgroup$ – Mariah Mar 25 at 2:38
  • $\begingroup$ @Mariah no I think that if f has inverse and has an involution, then the property f=f-1 is definitely true $\endgroup$ – Max0815 Mar 25 at 15:10

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