Question

Let S = {a, b, c} and T = {1, 2, 3}. Find F −1 of the following functions F from S to T, if it exists. (i) F = {(a, 3), (b, 2), (c, 1)} (ii) F = {(a, 2), (b, 1), (c, 1)}

Answer 1

Given, $S=\{a,b,c\}\:\:and\:\:T=\{1,2,3\}$
(i) $F=\{(a,3),(b,2),(c,1)\}$
e.g., F(a) = 3 , F(b) = 2 and F(c) = 1
As each element of S is having different image of T for function F . and it is one - one . also it is onto . because co - domain = range = set T
so, f is inversible.
$\implies F^{-1}(3)=a,F^{-1}(2)=b,F^{-1}(1)=c$
so, $F^{-1}=\{(3,a),(2,b),(1,c)\}$

(ii) here F(b) = 1 = F(c) but b ≠ c
hence, it is not one - one function.
but we know, any function is inversible only when function is one one and onto.
hence, F is not inversible.