Given f : A → A and g : A → A are two bijections. So, both f and g are one-one and onto functions.

We know (fog)(x) = f(g(x))

Thus, fog is also defined from A to A.

(i) First, we will prove that fog is an interjection.

Let x₁, x₂ϵ A (domain) such that (fog)(x₁) = (fog)(x₂)

⇒ f(g(x₁)) = f(g(x₂))

⇒ g(x₁) = g(x₂) [since f is one-one]

∴ x₁ = x₂ [since g is one-one]

So, we have (fog)(x₁) = (fog)(x₂) ⇒ x₁ = x₂.

Thus, function fog is an interjection.

(ii) Now, we will prove that fog is a surjection.

Let z ϵ A, the co-domain of fog.

As f is onto, we have y ϵ A (domain of f) such that f(y) = z.

However, as g is also onto and y belongs to the co-domain of g, we have x ϵ A (domain of g) such that g(x) = y.

Hence, (fog)(x) = f(g(x)) = f(y) = z.

Here, x belongs to the domain of fog (A) and z belongs to the co-domain of fog (A).

Thus, function fog is a surjection.