Question
If $f : A \rightarrow B$ is a one-one onto function, then prove that :$
\left(f^{-1}\right)^{-1}=f
$

Answer

$\because f: A \rightarrow B$, is a one-one onto function.
$\therefore$ there is inverse function of this funciton $f^{-1}: B$
$\rightarrow A$ and $f^{-1}$ will be also one-one onto function.
suppose $f^{-1}=g$
then $\quad\left(f^{-1}\right)^{-1}=g^{-1}$
Again suppose that $g^{-1}(x)=y$
then $x=g(y) \Rightarrow x=f^{-1}(y) \Rightarrow f(x)=y$$
\begin{array}{lll}
\Rightarrow & f(x)=g^{-1}(x) & {[\text { from equation (2)] }} \\
\Rightarrow & & f=g^{-1} \Rightarrow f=\left(f^{-1}\right)^{-1} \\
& & {[\text { from equation (1)] }}
\end{array}
$

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