For ` f : A to A and A = {1, 2, 3, 4} , f = {(1 ,2), (2,3),(3,4),(4,1)}` is

Options

A. injective only
B. surjective only
C. Bijective
D. Neither injective nor subjective

Answer

To solve the problem, we need to analyze the function \( f: A \to A \) where \( A = \{1, 2, 3, 4\} \) and \( f = \{(1, 2), (2, 3), (3, 4), (4, 1)\} \).

### Step-by-Step Solution:

1. **Identify the Function and Its Mapping**:
The function \( f \) is defined as:
- \( f(1) = 2 \)
- \( f(2) = 3 \)
- \( f(3) = 4 \)
- \( f(4) = 1 \)

2. **Determine the Domain and Codomain**:
The domain of the function \( f \) is the set \( A = \{1, 2, 3, 4\} \).
The codomain is also the set \( A = \{1, 2, 3, 4\} \).

3. **Check if the Function is Injective (One-to-One)**:
A function is injective if different elements in the domain map to different elements in the codomain.
- We can see that:
- \( 1 \) maps to \( 2 \)
- \( 2 \) maps to \( 3 \)
- \( 3 \) maps to \( 4 \)
- \( 4 \) maps to \( 1 \)
- Since all outputs are unique, the function is injective.

4. **Check if the Function is Surjective (Onto)**:
A function is surjective if every element in the codomain is mapped to by at least one element in the domain.
- The outputs of the function are \( \{2, 3, 4, 1\} \), which are exactly the elements of the codomain \( A = \{1, 2, 3, 4\} \).
- Since every element in the codomain has a corresponding element in the domain, the function is surjective.

5. **Conclusion - Determine if the Function is Bijective**:
A function is bijective if it is both injective and surjective.
- Since we have established that \( f \) is both injective and surjective, we conclude that \( f \) is bijective.

### Final Answer:
The function \( f \) is bijective.