Given : gof = I_A

In order to prove that f is one-one, it is sufficient to prove that f(x)=f(y) ⇒ x=y ∀ x,y ∈ A .

Let x,y ∈ A such that f(x) = f(y). Then,

f(x) = f(y)

⇒ g(f(x)) = g(f(y))

⇒ gof(x) = gof(y)

⇒ I_A(x) = I_A(y) [∵ gof = I_A is given ]

⇒ x = y [by definition of Identity function, I(x) = x]

Thus, f is one-one.

Now, in order to prove that g : B → A is onto, it is sufficient to prove that each element in A has pre-image in B.

Let x ∈ A.

Also, f : A → B is a function ∴ f(x) ∈ B

Now,

Let f(x) = y

⇒ g(f(x)) = g(y)

⇒ gof(x) = g(y)

⇒ I_A(x) = g(y) [∵ gof = I_A is given ]

⇒ x = g(y) [by definition of Identity function, I(x) = x]

Thus, for every x ∈ A there exists y ∈ B such that g(y) = x.

⇒ g is onto.