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

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

• 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. – Sangchul Lee Mar 24 at 18:34
• @SangchulLee so yes, it is only valid in that case? – Max0815 Mar 24 at 18:35
• 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. – Sangchul Lee Mar 24 at 18:38
• 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. – coffeemath Mar 24 at 18:39
• @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? – Max0815 Mar 24 at 18:39

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.

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

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

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