Let `A={x,y,z}=B={u,v,w) and f:A to B` be defined by f `(x)=u, f(y)=v, f(z)=w`. Then, f is

Options

A. surjective but not injective
B. injective but not surjective
C. bijective
D. none of these

Answer

To determine the nature of the function \( f: A \to B \) defined by \( f(x) = u, f(y) = v, f(z) = w \), we need to analyze whether it is injective (one-to-one), surjective (onto), or both (bijective).

### Step-by-step Solution:

1. **Identify the Sets and Function**:
- Let \( A = \{x, y, z\} \)
- Let \( B = \{u, v, w\} \)
- The function \( f \) is defined as:
- \( f(x) = u \)
- \( f(y) = v \)
- \( f(z) = w \)

2. **Check for Injectivity**:
- A function is injective if different elements in the domain map to different elements in the codomain.
- Here, we see:
- \( f(x) = u \)
- \( f(y) = v \)
- \( f(z) = w \)
- Since \( x, y, z \) are distinct and map to distinct \( u, v, w \), the function is injective.

3. **Check for Surjectivity**:
- A function is surjective if every element in the codomain has a pre-image in the domain.
- The codomain \( B \) consists of \( u, v, w \).
- From the function definitions:
- \( u \) has pre-image \( x \)
- \( v \) has pre-image \( y \)
- \( w \) has pre-image \( z \)
- Since every element in \( B \) is covered by the mapping from \( A \), the function is surjective.

4. **Conclusion**:
- Since the function \( f \) is both injective and surjective, it is bijective.

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