If `f: A->B and g: B->C` are onto functions show that `gof` is an onto function.

Explanation

Let there is an element `x` such that `x in C`.

Since `g` is onto, we can find an element `b` where `b in B` such that `g(b)=x`.

But `f` is onto, so we can also find an element `a` where `a in A` such that `f(a)=b.`

Thus, `(g(f(a))=g(b)=x`, and so `gof` is onto.