Consider `f:{1,\ 2,\ 3}->{a ,\ b ,\ c}` and `g:{a ,\ b ,\ c}->` {apple, ball, cat} defined as `f(1)=a ,\ \ f(2)=b ,\ \ f(3)=c ,\ \ g(a)=` apple, `g(b)=` ball and `g(c)=` cat. Show that `f,\ g\ a n d\ gof` are invertible. Find `f^(-1),\ g^(-1)` and `(gof)^(-1)` and show that `(gof)^(-1)=f^(-1)o\ g^(-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`