Injectivity and Surjectivity of Composite Functions on Sets of Integers and Naturals

To determine which of the given statements are true, let us analyze the functions $f: \mathbb{N} \to \mathbb{Z}$ and $g: \mathbb{Z} \to \mathbb{N}$, where $\mathbb{N} = \{1, 2, 3, \dots\}$.

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1. Analysis of $f: \mathbb{N} \to \mathbb{Z}$

The function $f$ is defined by: $f(n) = \begin{cases} \dfrac{n + 1}{2}, & \text{if } n \text{ is odd} \\ \dfrac{4 - n}{2}, & \text{if } n \text{ is even} \end{cases}$

• Injectivity (One-One):

  • For $n = 1$ (odd): $f(1) = \dfrac{1 + 1}{2} = 1$.
  • For $n = 2$ (even): $f(2) = \dfrac{4 - 2}{2} = 1$.

  Since $f(1) = f(2) = 1$ for $1 \neq 2$, the function $f$ is NOT one-one.

• Surjectivity (Onto):

  • If $n$ is odd, let $n = 2k - 1$ where $k \in \mathbb{N}$. Then: $f(2k - 1) = \frac{(2k - 1) + 1}{2} = k \in \{1, 2, 3, \dots\} = \mathbb{N}$
  • If $n$ is even, let $n = 2k$ where $k \in \mathbb{N}$. Then: $f(2k) = \frac{4 - 2k}{2} = 2 - k \in \{1, 0, -1, -2, \dots\}$

  The union of these two sets is: $\text{Range}(f) = \{1, 2, 3, \dots\} \cup \{1, 0, -1, -2, \dots\} = \mathbb{Z}$ Since the range equals the codomain $\mathbb{Z}$, $f$ is onto.

Thus, statement (D) is TRUE.

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2. Analysis of $g: \mathbb{Z} \to \mathbb{N}$

The function $g$ is defined by: $g(n) = \begin{cases} 3 + 2n, & \text{if } n \ge 0 \\ -2n, & \text{if } n < 0 \end{cases}$

• Surjectivity (Onto):
  • For $n \ge 0$, $g(n) \in \{3, 5, 7, 9, \dots\}$ (odd integers $\ge 3$).
  • For $n < 0$, $g(n) \in \{2, 4, 6, 8, \dots\}$ (all positive even integers).
  • The range of $g$ is: $\text{Range}(g) = \{2, 3, 4, 5, \dots\} = \mathbb{N} \setminus \{1\}$ Since $1 \notin \text{Range}(g)$, $g$ is NOT onto.

Thus, statement (C) is FALSE.

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3. Analysis of $g \circ f: \mathbb{N} \to \mathbb{N}$

• Injectivity: $(g \circ f)(1) = g(f(1)) = g(1) = 3 + 2(1) = 5$ $(g \circ f)(2) = g(f(2)) = g(1) = 3 + 2(1) = 5$ Since $(g \circ f)(1) = (g \circ f)(2) = 5$ for $1 \neq 2$, $g \circ f$ is NOT one-one.

• Surjectivity: Since $\text{Range}(g \circ f) \subseteq \text{Range}(g) = \mathbb{N} \setminus \{1\}$, the value $1 \in \mathbb{N}$ is never attained. Therefore, $g \circ f$ is NOT onto.

Thus, statement (A) is TRUE.

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4. Analysis of $f \circ g: \mathbb{Z} \to \mathbb{Z}$

• For $n \ge 0$, $g(n) = 3 + 2n$, which is an odd integer. Thus: $(f \circ g)(n) = f(3 + 2n) = \frac{(3 + 2n) + 1}{2} = n + 2$
• For $n < 0$, $g(n) = -2n$, which is an even integer. Thus: $(f \circ g)(n) = f(-2n) = \frac{4 - (-2n)}{2} = n + 2$

Hence, for all $n \in \mathbb{Z}$: $(f \circ g)(n) = n + 2$ The function $h(n) = n + 2$ on $\mathbb{Z}$ is both one-one and onto (a bijection).

Thus, statement (B) is FALSE.

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Conclusion:

The correct statements are (A) and (D).