1. First determine the form of $f$

Given f(n+1)=f(n)+f(1) \forall n\in \mathbb N. Let $f(1)=a\in \mathbb N$.

We compute a few values: $f(2)=f(1)+a=2a,$ $f(3)=f(2)+a=3a,$ and by induction, $f(n)=na\quad \forall n\in \mathbb N.$

So $f$ must be of the form $f(n)=an,\qquad a=f(1)\in \mathbb N.$

2. Check option B: $f$ is one-one

Since $f(n)=an$ with $a\in \mathbb N$, if $f(m)=f(n),$ then $am=an.$ As $a\ge 1$, this gives $m=n.$ Hence $f$ is injective.

So B is true.

3. Check option C: If $f$ is onto, then $f(n)=n\ \forall n\in\mathbb N$

Now $f(n)=an$. For $f$ to be onto from $\mathbb N$ to $\mathbb N$, every natural number must be of the form $an$. That is possible only when $a=1,$ because if $a>1$, numbers like $1$ (and many others) are not in the range.

Thus onto implies $f(n)=n\quad \forall n\in\mathbb N.$

So C is true.

4. Check option D: If $f\circ g$ is one-one, then $g$ is one-one

Suppose $f\circ g$ is one-one. Take $x,y\in\mathbb N$ such that $g(x)=g(y).$ Then $f(g(x))=f(g(y)).$ Since $f\circ g$ is one-one, we get $x=y.$ Therefore $g$ is one-one.

So D is true.

5. Check option A: If $g$ is onto, then $f\circ g$ is one-one

We need to see whether this is always true.

Since $g$ is onto, for every $u\in\mathbb N$ there exists $x\in\mathbb N$ with $g(x)=u$. To conclude $f\circ g$ is one-one, we would need injectivity of $g$ as well, but onto alone does not guarantee that.

A counterexample is enough.

Take $f(n)=n$ (which satisfies the recurrence), and define an onto function $g:\mathbb N\to\mathbb N$ by $g(1)=1,\quad g(2)=1,\quad g(n)=n-1\ \text{for } n\ge 3.$ This $g$ is onto, because:

• $1$ is attained at $1$ and $2$,
• for any $k\ge 2$, $g(k+1)=k$.

But then