1. Understand the function $g$

The map $g: \mathbb N \to \mathbb N$ is defined in cycles of length $3$:

for all $n\ge 0$.

So on each block $\{3n+1,3n+2,3n+3\}$, the function $g$ acts as

Thus $g$ is a permutation of $\mathbb N$ made of disjoint 3-cycles.

Hence,

That is,

2. Check option C: $g\circ g\circ g = g$

But we just found that

not $g$.

Since $g \ne I$, we have

So Option C is false.

3. Check option A: There exists an onto function $f: \mathbb N \to \mathbb N$ such that $f\circ g = f$

Condition $f\circ g=f$ means

Since $g$ cycles elements in each triple,

for every $n\ge 0$.

So $f$ must be constant on each block $\{3n+1,3n+2,3n+3\}$.

Now define

Then:

• each block maps to one natural number,
• every natural number $m$ is attained by the block corresponding to $n=m-1$.

So $f$ is onto.

Also, since $f$ is constant on each 3-cycle, clearly

for all $x$.

Hence Option A is true.

4. Check option B: There exists a one-one function $f: \mathbb N \to \mathbb N$ such that $f\circ g=f$

If $f\circ g=f$, then as above,

for every $n$.

In particular,

so $f$ cannot be injective.

Therefore Option B is false.

5. Check option D: There exists a function $f: \mathbb N \to \mathbb N$ such that $g\circ f=f$

Condition $g\circ f=f$ means

So every value in the image of $f$ must be a fixed point of $g$.

Let us see whether $g$ has any fixed point.

Suppose $g(k)=k$. Then $k$ must be one of the forms:

• $k=3n+1$, then $g(k)=3n+2\ne k$,
• $k=3n+2$, then $g(k)=3n+3\ne k$,
• $k=3n+3$, then $g(k)=3n+1\ne k$.

So $g$ has no fixed point.

Hence there is no $y\in \mathbb N$ such that $g(y)=y$. Therefore no function $f$ can satisfy $g\circ f=f$.

So Option D is false.

6. Conclusion

Only Option A is true.