Question

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 and gof are invertible. Find out f –1, g–1 and (gof)–1 and show that(gof)–1 = f –1o g–1.​

Answer 1

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Solution :Note that by definition, f and g are bijective functions. Letf –1: {a, b, c} → (1, 2, 3} and g

–1 : {apple, ball, cat} → {a, b, c} be defined as

f –1{a} = 1, f –1{b} = 2, f –1{c} = 3, g–1{apple} = a, g

–1{ball} = b and g–1{cat} = c.It is easy to verify that f –1o f = I{1, 2, 3}, f o f –1 = I{a, b, c}, g–1og = I{a, b, c} and go g–1 = IDwhere, D = {apple, ball, cat}. Now, gof : {1, 2, 3} → {apple, ball, cat} is given by

gof(1) = apple, gof(2) = ball, gof(3) = cat. We can define(gof)

–1 : {apple, ball, cat} → {1, 2, 3} by (gof)

–1 (apple) = 1,(gof)

–1 (ball) = 2 and(g of)

–1 (cat) = 3. It is easy to see that (g o f)

–1 o (g o f) = I{1, 2, 3} and(gof) o (gof)–1 = ID

. Thus, we have seen that f, g and gof are invertible.Now, f–1og–1 (apple)= f –1(g–1(apple)) = f –1(a) = 1 = (gof)–1 (apple)f

–1og

–1 (ball) = f –1(g

–1(ball)) = f –1(b) = 2 = (gof)

–1 (ball) andf

–1og

–1 (cat) = f –1(g

–1(cat)) = f –1(c) = 3 = (gof)

–1 (cat).

Hence (gof)–1 = f –1og–1

.The above result is true in general situation also.