Since gof is onto, g is onto. Therefore S-2 is correct. To see that S-1 is correct, we observe that b ∊ B ⇒ g(b) = c ∊ C ⇒ g(b) = (gof) (a) for some a ∊ A ⇒ b = f(a) (since g is one-one) Thus we need gof onto and g one-one for the validity of S-1, but not only g onto.