Let f : A → B and g : B → C be the bijective functions. Then (g o f)^–1 is

Given that, f : A → B and g : B → C be the bijective functions.

Let A = {1,3,4}, B ={2,5,1} and C = {3,1,2}

f : A → B is bijective function.

∴ f = {(1, 2), (3, 5), (4, 1)

f^-1 = {(2,1),(5,3),(1,4)}

g : B → C is bijective function.

∴ g = {(2, 3), (5, 1), (1, 4)}

g^-1 ={(3,2),(1,5),(4,1)}

Now,

gof (1) = g(f(1)) = g(2) = 3

gof (3) = g(f(3)) = g(5) = 1

gof (4) = g(f(4)) = g(1) = 4

∴ gof = {(1,3),(3,1),(4,4)} (1)

(gof)^-1 = {(3,1),(1,3),(4,4)} (2)

fog (2) = f(g(2)) = f(3) = 5

fog (5) = f(g(5)) = f(1) = 2

fog (1) = f(g(1)) = f(4) = 1

∴ fog = {(2,5),(5,2),(1,1)} (3)

(fog)^-1 = {(5,2),(2,5),(1,1)} (4)

f^-1og^-1 (3) = f^-1(g^-1(3)) = f^-1(2) = 1

f^-1og^-1 (1) = f^-1(g^-1(1)) = f^-1(5) = 3

f^-1og^-1 (4) = f^-1(g^-1(4)) = f^-1(1) = 4

∴ f^-1og^-1 = {(3,1),(1,3),(4,4)} (5)

g^-1of^-1 (2) = g^-1(f^-1(2)) = g^-1(1) = 5

g^-1of^-1 (5) = g^-1(f^-1(5)) = g^-1(3) = 2

g^-1of^-1 (1) = g^-1(f^-1(1)) = g^-1(4) = 1

∴ g^-1of^-1 = {(2,5),(5,2),(1,1)} (6)

On comparing 1,2,3,4,5 and 6 we observe that 2 and 5 are same.

i.e (g o f)^–1 = f^-1og^-1