Let  $f,g:N\to N$such that $f\left(n+1\right)=f\left(n\right)+f\left(1\right)\forall n\in N$  and  be any arbitrary function. Which of the following statements is NOT true?

$\begin{array}{l}f:N\to N,g:N\to N\\ f(n+1)=f\left(n\right)+f\left(1\right)\forall n\in N\\ f\left(2\right)=2f\left(1\right),f\left(3\right)=3f\left(1\right),f\left(4\right)=4f\left(1\right),f\left(n\right)=nf\left(1\right)\end{array}$

it gives the function $f\left(x\right)$ is one to one

it gives $fog\left(x\right)$ is one to one only when $g\left(x\right)$ is one to one

If g is onto,  $fog\left(x\right)$ is need not to be one to one