Let's examine each function to determine if it is injective, surjective, or bijective.

Part (a): $f_1 : \mathbb{Z} \rightarrow \mathbb{Z}, \, f_1(x) = 1 - x$

1. Injectivity: To check if $f_1$ is injective, assume $f_1(x_1) = f_1(x_2)$. Then: $1 - x_1 = 1 - x_2$ which implies $x_1 = x_2$. Therefore, $f_1$ is injective.

2. Surjectivity: Since for any integer $y \in \mathbb{Z}$, we can find an $x = 1 - y$ such that $f_1(x) = y$, $f_1$ is surjective.

3. Conclusion: $f_1$ is both injective and surjective, so it is bijective.

Part (b): $f_2 : \mathbb{N} \rightarrow \mathbb{N}, \, f_2(x) = 1 + x$

1. Injectivity: Assume $f_2(x_1) = f_2(x_2)$. Then: $1 + x_1 = 1 + x_2$ which implies $x_1 = x_2$. Therefore, $f_2$ is injective.

2. Surjectivity: To be surjective, every natural number $y \in \mathbb{N}$ must have a preimage in $\mathbb{N}$. However, $f_2(x) = 1 + x$ cannot produce $y = 1$, as the minimum value of $f_2(x)$ is 2 when $x = 1$. Therefore, $f_2$ is not surjective.

3. Conclusion: $f_2$ is injective but not surjective.

Part (c): $f_3 : \mathbb{R} \rightarrow \mathbb{R}_{\geq 0}, \, f_3(x) = (x + 1)^2$

1. Injectivity: Assume $f_3(x_1) = f_3(x_2)$. Then: $(x_1 + 1)^2 = (x_2 + 1)^2$ This implies $x_1 + 1 = x_2 + 1$ or $x_1 + 1 = -(x_2 + 1)$. Therefore, $x_1 = x_2$ or $x_1 = -x_2 - 2$, meaning $f_3$ is not injective.

2. Surjectivity: For any $y \in \mathbb{R}_{\geq 0}$, we can solve $y = (x + 1)^2$, yielding $x = \pm \sqrt{y} - 1$. Thus, every non-negative real number $y$ has a preimage, so $f_3$ is surjective.

3. Conclusion: $f_3$ is surjective but not injective.

Part (d): $f_4 : \mathcal{P}(\mathbb{R}) \rightarrow \mathcal{P}(\mathbb{N}), \, f_4(M) = M \cap \mathbb{N}$

1. Injectivity: Suppose $f_4(M_1) = f_4(M_2)$, which implies $M_1 \cap \mathbb{N} = M_2 \cap \mathbb{N}$. However, $M_1$ and $M_2$ could differ outside of $\mathbb{N}$ while still having the same intersection with $\mathbb{N}$. Thus, $f_4$ is not injective.

2. Surjectivity: For any subset $A \subseteq \mathbb{N}$, we can take $M = A \subseteq \mathbb{R}$. Then $f_4(M) = M \cap \mathbb{N} = A$, so $f_4$ is surjective.

3. Conclusion: $f_4$ is surjective but not injective.

Summary of Results:

• (a) $f_1$ is bijective.
• (b) $f_2$ is injective but not surjective.
• (c) $f_3$ is surjective but not injective.
• (d) $f_4$ is surjective but not injective.

Would you like further details on any part?

Related Questions:

1. What is the formal definition of an injective function?
2. How can we determine the range of a given function?
3. What are the differences between injective, surjective, and bijective mappings?
4. How do domain restrictions affect injectivity and surjectivity?
5. Can a function from $\mathbb{R}$ to $\mathbb{Z}$ be bijective?

Tip:

For functions, proving injectivity typically involves showing that $f(x_1) = f(x_2)$ implies $x_1 = x_2$, while proving surjectivity often requires finding a preimage for every element in the codomain.