https://file.aiquickdraw.com/mathbot/file/1730145499065v5po4jni.jpg

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.

Bestimmen Sie für jede der gegebenen Abbildungen, ob sie injektiv/surjektiv/bijektiv ist.

Analyze given functions to determine if they are injective, surjective, or bijective. This solution covers fundamental concepts in set theory and function properties, with detailed explanations on function types for advanced mathematics students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation