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

f$= \{(1, a), (2, 6), (3, c)\}$. Clearly/is one-one and onto.

Also, ${f^{ - 1}} = \{(0, 4), (6, 2), (c, 3)\}$

$\Rightarrow$ ${f^{ - 1}}(a) = 1,{f^{ - 1}}(b) = 2,{f^{ - 1}}(c) = 3$

Also, ${({f^{ - 1}})^1} = \{ (1,a),(2,b),(3,c)\} = f$

Hence, the result ${({f^{ - 1}})^{ - 1}} = f$ .