Prove $f(f^{-1}(x))=x$

Question

If we have a problem $f(x)=x$ it's mean $x=f^{-1}(x)$. Hence if $f(x)=x$, then $f(x)=f^{-1}(x)$

But behind the problem, we know that we must take the inverse for both sides. And it leaves $f^{-1}(f(x))=f(f^{-1}(x))=x$

How do i prove $f(f^{-1}(x))=x$ ? I can't found it on any books. Sorry for the stupid question....

I know that all of us already know about this. But if someone asks me "how do you prove this?" I don't know how to answer this.

Answer 1

''If we have a problem f(x)=x it's mean x=f−1(x). Hence if f(x)=x, then f(x)=f−1(x).''

Well, if $f(x)=x$ for all $x$, then $f$ is the identity map whose inverse is the identity map as well.

''But behind the problem, we know that we must take the inverse for both sides. And it leaves f−1(f(x))=f(f−1(x))=x.''

Well, a function $f:A\rightarrow A$ is invertible if there is a function $g:A\rightarrow A$ such that $fg = id =gf$ where $id$ is the identity mapping.

One can show that the invertible is uniquely determined so that we can write $f^{-1}$ for the inverse of $f$. Then we have $f^{-1}f = id = ff^{-1}$. Moreover, a function is invertible iff it is both injective and surjective.

Answer 2

Say $f$ is injective. If $\color{red}{y=f(x)}$ then $f^{-1}(\color{red}{y})=x$ so $f^{-1}(\color{red}{f(x)})=x$

Answer 3

Function $f:A\to B$ has an inverse function if and only there are functions $g,h:B\to A$ such that $f\circ g$ and $h\circ f$ are identities.

If such functions exists then it can be shown that the functions $g,h$ are unique and coincide.

This opens the possibility to speak of the inverse of $f$ and to denote it without ambiguity as $f^{-1}$.

So the characteristic properties of $f^{-1}$ are $f\circ f^{-1}=\mathsf{id}_B$ and $f^{-1}\circ f=\mathsf{id}_A$, which means that by definition we have $f(f^{-1}(x))=x$ for every $x\in B$ and $f^{-1}(f(x))=x$ for every $x\in A$.

Also it well known that $f$ has an inverse function if and only if $f$ is bijective.