Functions Part 1

Given, f(n+1)=f(n)+f(1),∀n∈N
⇒‌‌f(n+1)−f(n)=f(1)
It is an AP with common difference =f(1)
Also, general term =Tn=f(1)+(n−1)f(1)=nf(1)
⇒‌‌f(n)=nf(1)
Clearly, f(n) is one-one.
For fog to be one-one, g must be one-one.
For f to be onto, f(n) should take all the values of natural numbers.
As, f(x) is increasing, f(1)=1
⇒‌‌f(n)=n If g is many-one, then fog is many one.
So, if g is onto, then fog is onto.