If f: A→B and g:B→C are onto , then gof:A→C is:
a many-one and onto function
Let c∈C
g:B→C is onto
⇒c∈Chas a pre image
⇒There exists a b ∈B such that g(b) = c
b ∈B and f: A→B is onto
⇒ b∈ B has a pre image
⇒There exists an a ∈A such that f(a) = b
c = g(b) = g[f(a)] = gof (a)
So, c∈Chas a pre image a∈A , such that gof(a) = c
→gof is an onto function