Check the injecti

It is given that f : N → N given by f (x) = x³

We can see that for x, y ϵ N,

f(x) = f(y)

⇒ x³ = y³

⇒ x = y

⇒ f is injective.

Now, let 2 ϵ N. But, we can see that there does not exists any x in N such that

f(x) = x³ = 2

⇒ f is not surjective.

Therefore, function f is injective but not surjective.