1. Interpret the functions correctly

We have $f,g: \mathbb{N} \to \mathbb{N}$ defined by

Notice that both cases are the same as

Also,

Now compute $g(n)$ separately for odd/even $n$:

• If $n$ is even, $(-1)^n=1$, so $g(n)=n-1.$
• If $n$ is odd, $(-1)^n=-1$, so $g(n)=n+1.$

Thus,

2. Find $f\circ g$

We need

Case 1: $n$ is odd

Then

which is even. Therefore,

Case 2: $n$ is even

Then

which is odd. Therefore,

So,

This is exactly the same as $f(n)$. Hence,

3. Check whether $f\circ g$ is one-one

Since $f\circ g=f$, test injectivity of $f$.

Compute a few values:

so two different inputs give the same output. Therefore $f$ is not one-one.

Hence $f\circ g$ is not one-one.

4. Check whether $f\circ g$ is onto

We must see whether every natural number $m\in \mathbb N$ has some $n\in \mathbb N$ such that

But $f\circ g=f$, and for any $m\in\mathbb N$,

• choose $n=2m$, then $f(2m)=\frac{2m}{2}=m.$

So every natural number is attained. Hence $f$ is onto, and therefore $f\circ g$ is onto.

5. Conclusion

$f\circ g$ is

• onto,
• not one-one.

Therefore the correct option is

$\boxed{\text{B: onto but not one-one}}.$

6. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.