Let f : N → N, be a function given by f(1) = f(2) = 1 and f(x) = x-1 for every x > 2.

Since, f(1) = 1 = f(2)

⇒ 1 and 2 does not have unique image.

Thus, f is not one-one.

Let f(x) = y

⇒ y = x-1

⇒ x = y+1

⇒ for each y ∈ R there exists x ∈ N such that f(x) = y.

Thus, f is onto.