Consider $f:\{1,2,3\} \rightarrow\{a, b, c\}$ given by $f(1)=a, \,f(2)=b$ and $f(3)=c .$ Find $f^{-1}$ and show that $\left(f^{-1}\right)^{-1}=f$.

Function $f :\{1,2,3\} \rightarrow\{ a ,\, b , \,c \}$ is given by $f (1)= a ,\, f (2)= b ,$ and $f (3)= c$

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

We have

$(f og)(a)=f(g(a))=f(1)=a$

$(f o g)(b)=f(g(b))=f(2)=b$

$(f og)(c)=f(g(c))=f(3)=c$

and

$(\operatorname{gof})(1)=g(f(1))=f(a)=1$

$(\operatorname{gof})(2)=g(f(2))=f(b)=2$

$(\operatorname{gof})(3)=g(f(3))=f(c)=3$

$\therefore$ $gof$ $= I_X$ and $fog =$ $I_Y$, 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 $f ^{-1}( a )=1,\, f ^{-1}( b )=2,\, f ^{-1}( c )=3$

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 $h (1)= a , \,h (2)= b , \,h (3)= c$

We have

$(\operatorname{goh})(1)=g(h(1))=g(a)=1$

$(\operatorname{goh})(2)=g(h(2))=g(b)=2$

$(\operatorname{goh})(3)=g(h(3))=g(c)=3$

and

$(\operatorname{hog})(a)=h(g(a))=h(1)=a$

$(\operatorname{hog})(b)=h(g(b))=h(2)=b$

$(\operatorname{hog})(c)=h(g(c))=h(3)=c$

$\therefore \operatorname{goh}=$ $I_X$ and $hog$ $=$ $I_Y$, 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, $\left(f^{-1}\right)^{-1}=f$