Solution:

If we define $g:\{a, b, c\} \rightarrow\{1,2,3\}$ as $g(a)=1, g(b)=2, g(c)=3$, then we have:

$\begin{aligned} &(f \circ g)(a)=f(g(a))=f(1)=a \\\\ &(f \circ g)(b)=f(g(b))=f(2)=b \\\\ &(f \circ g)(c)=f(g(c))=f(3)=c \\ \end{aligned}$

And,

$\begin{aligned} &(g \circ f)(1)=g(f(1))=g(a)=1 \\\\ &(g \circ f)(2)=g(f(2))=g(b)=2 \\\\ &(g \circ f)(3)=g(f(3))=g(c)=3 \\ \end{aligned}$

$\therefore g \circ f=\mathrm{I}_{X} \\ $

and

$f \circ g=\mathrm{I}_{Y} \\ $

where

$X=\{1,2,3\} \\ $

and

$Y=\{a, b, c\} \\ $.

Thus, the inverse of f exists and $f^{-1}=g$.

$\therefore f^{-1}:\{a, b, c\} \rightarrow\{1,2,3\}$ is given by,

Let us now find the inverse of $f^{-1}$ i.e., find the inverse of g.

If we define $h:\{1,2,3\} \rightarrow\{a, b, c\}$ as

$h(1)=a, h(2)=b, h(3)=c$, then we have:

$\begin{aligned} &(g \circ h)(1)=g(h(1))=g(a)=1 \\\\ &(g \circ h)(2)=g(h(2))=g(b)=2 \\\\ &(g \circ h)(3)=g(h(3))=g(c)=3 \\ \end{aligned}$

And,

$\begin{aligned} &(h \circ g)(a)=h(g(a))=h(1)=a \\\\ &(h \circ g)(b)=h(g(b))=h(2)=b \\\\ &(h \circ g)(c)=h(g(c))=h(3)=c \\ \end{aligned}$

$\therefore g \mathrm{o} h=\mathrm{I}_{X}$ and $h \mathrm{og}=\mathrm{I}_{\gamma}$, where $X=\{1,2,3\}$ and $Y=\{a, b, c\}$.

Thus, the inverse of g exists and $g^{-1}=h \Rightarrow\left(f^{-1}\right)^{-1}=h$.

It can be noted that $h=f$.

Hence,

$\left(f^{-1}\right)^{-1}=f \\ $.