Let N denote the set of all natural numbers, and Z denote the set of all integers. Consider the functions f : N → Z and g : Z → N defined by

f(n) = { (n + 1)/2 , if n is odd
             (4 − n)/2 , if n is even }

g(n) = { 3 + 2n , if n ≥ 0
            −2n , if n < 0 }

(g ∘ f)(n) = g(f(n)) for all n ∈ ℕ

(f ∘ g)(n) = f(g(n)) for all n ∈ ℤ

Then which of the following statements is (are) TRUE?

The correct statements are:
g ∘ f is NOT one-one and g ∘ f is NOT onto
f is NOT one-one but f is onto

Let us analyze each function and their compositions step-by-step.

1. Analysis of function $f:\mathbb{N}\to \mathbb{Z}$:
The function is defined as:
$f\left(n\right)=\{\begin{array}{ll}\frac{n+1}{2},& \text{if}n\text{is odd}\\ \frac{4-n}{2},& \text{if}n\text{is even}\end{array}$

Let's evaluate $f\left(n\right)$ for the first few natural numbers:
For $n=1$ (odd): $f\left(1\right)=\frac{1+1}{2}=1$
For $n=2$ (even): $f\left(2\right)=\frac{4-2}{2}=1$
Since $f\left(1\right)=f\left(2\right)=1$ for distinct domain elements $1\ne 2$, the function $f$ is NOT one-one.

Now, let's examine the range of $f$:
For odd natural numbers $n\in \{1,3,5,...\}$:
$f\left(n\right)\in \{1,2,3,...\}={\mathbb{Z}}^{+}$
For even natural numbers $n\in \{2,4,6,8,...\}$:
$f\left(n\right)���\{1,0,-1,-2,...\}={\mathbb{Z}}_{\le 1}^{}$
Combining both cases, the range of $f$ covers all integers $\mathbb{Z}$. Therefore, $f$ is onto.

2. Analysis of function $g:\mathbb{Z}\to \mathbb{N}$:
The function is defined as:
$g\left(n\right)=\{\begin{array}{ll}3+2n,& \text{if}n\ge 0\\ -2n,& \text{if}n<0\end{array}$

Let's check the range of $g$:
For $n\ge 0$: $g\left(n\right)\in \{3,5,7,...\}$ (all odd natural numbers starting from 3)
For $n<0$: $g\left(n\right)\in \{2,4,6,...\}$ (all even natural numbers)
Thus, the range of $g$ is $\mathbb{N}-\left\{1\right\}$. Since the element $1\in \mathbb{N}$ has no pre-image in $\mathbb{Z}$, $g$ is NOT onto.

3. Analysis of composition $g\circ f:\mathbb{N}\to \mathbb{N}$:
For $n=1$: $(g\circ f)\left(1\right)=g\left(f\right(1\left)\right)=g\left(1\right)=3+2\left(1\right)=5$
For $n=2$: $(g\circ f)\left(2\right)=g\left(f\right(2\left)\right)=g\left(1\right)=5$
Since $(g\circ f)\left(1\right)=(g\circ f)\left(2\right)=5$, $g\circ f$ is NOT one-one.
Furthermore, the range of $g\circ f$ must be a subset of the range of $g$, which is $\mathbb{N}-\left\{1\right\}$. Thus, the element $1\in \mathbb{N}$ has no pre-image under $g\circ f$. Therefore, $g\circ f$ is NOT onto.

4. Analysis of composition $f\circ g:\mathbb{Z}\to \mathbb{Z}$:
If $n\ge 0$, then $g\left(n\right)=2n+3$ which is odd. Hence,
$(f\circ g)\left(n\right)=f(2n+3)=\frac{(2n+3)+1}{2}=n+2$
If $n<0$, then $g\left(n\right)=-2n$ which is even. Hence,
$(f\circ g)\left(n\right)=f(-2n)=\frac{4-(-2n)}{2}=n+2$
Thus, for all $n\in \mathbb{Z}$, we have $(f\circ g)\left(n\right)=n+2$, which is a bijective function (both one-one and onto).