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

Explanation

`f: {1, 2, 3} to {a, b, c}`
and `f(1) = a, f(2) = b and f(3) = c `
Let, in `g: {a, b, c} to { 1, 2, 3}, g(a)=1, g(b) = 2 and g(c ) = 3`
`therefore " " (fog) (a) = f{g(a)} = f(1) =a`
`" " (fog) (b) = f{g(b)} = f(2) = b`
`" " (fog) (c ) = f{g(c)} = f(3) = c`
and `" " (gof) (1) = g{f(1)} = g(a) =1 `
`" " (gof)(2) = g{f(2)}= g(b) = 2`
`" "(gof) (3) = g{f(3)} = g(c) = 3`
`therefore " " gof = I_X and fog = I_Y`
where `X = { 1, 2, 3} and Y = {a, b, c}`
`therefore " " f^(-1) = g `
and `f^(-1) : {a, b, c} to {1, 2, 3 }`
and `f^(-1)(a) = 1, f^(-1) (b) = 2, f^(-1) (c) = 3`
Again let, in `h: {1, 2, 3} to {a, b, c}`
`" " h(1) = a, h(2) = b, h(3) = c`
`" " (goh) (1) = g{h(1)} = g(a)= 1`
`" " (goh)(2) = g{h(2)} = g(b) = 2 `
`" " (goh)(3) = g{h(3)}= g(c) = 3`
and `" " (hog)(a) = h{g(a)} = h(1) =a`
`" " (hog )(b) = h{g(b)} = h(2) = b`
`" " (hog) (c) = h{g(c)} = h(3) =c `
`therefore " " goh = I_X and hog = I_Y`
where `" " X= {1, 2, 3} and y = {a, b, c}`
`thererfore " "g^(-1) = h`
`rArr " " (f^(-1)) = f`