Consider `f : {1, 2, 3} ->{a , b , c}`and `g : {a , b , c} ->{a p p l e , b a l l , c a t}`defined as `f (1) = a`, `f (2) = b`, `f (3) = c`, `g(a) = a p p l e`,`g(b) = b a l l and g(c) = cat`. Show that `f , g and gof` are invertible .Find out `f^-1 , g^-1` and `(gof)^-1` and show that `(gof)^-1=f^-1og^-1`

Explanation

Given as `f = {(1, a)`, `(2, b), (c, 3)}` and `g` = {(a, apple), `(b, ball), (c, cat)}`.
Clearly, `f` and `g` are bijections.
∴ `f` and `g` are invertible.
Then, `f^-1 = {(a ,1) , (b , 2) , (3,c)}` and `g^-1` = {(apple, a), `(ball , b), (cat , c)}`

∴ `f^-1og^-1`= {apple, 1), `(ball, 2), (cat, 3)}` …(1)
`f: {1,2,3,} →{a, b, c}` and `g: {a, b, c}` → {apple, ball, cat}
∴ `gof: {1, 2, 3}` → {apple, ball, cat}

⇒ `(gof(1)) = g(f(1)) = g(a)` = apple
`(gof)(2) = g(f(2)) = g(b)` = ball
And `(gof)(3) = g(f(3)) = g(c)` = cat
∴ `gof` = {(1, apple), `(2, ball), (3, cat)}`
Clearly, gof is a bijection.
∴ `gof` is invertible.
`(gof)^-1` = {(apple, 1), `(ball, 2), (cat, 3)}` ...(2)
Form equations (1) and (2), we get
`(gof)^-1=f^-1og^-1`