If f : A `rarr` B is a bijective function, then `f^(-1) of` is equal to

Options

A. `fof^(-1)`
B. `f`
C. `f^(-1)`
D. `I_(A)` (the identity map of the set A)

Explanation

To solve the problem, we need to determine what \( f^{-1}(f(x)) \) is when \( f: A \to B \) is a bijective function.

### Step-by-Step Solution:

1. **Understanding Bijective Functions**:
A function \( f \) is bijective if it is both injective (one-to-one) and surjective (onto). This means that for every element \( y \) in set \( B \), there is exactly one element \( x \) in set \( A \) such that \( f(x) = y \).

**Hint**: Recall the definitions of injective and surjective functions.

2. **Applying the Function**:
Let \( y = f(x) \). Since \( f \) is a bijective function, we can find a unique \( x \) for every \( y \) in \( B \).

**Hint**: Think about how the function maps elements from set \( A \) to set \( B \).

3. **Finding the Inverse**:
Since \( f \) is bijective, it has an inverse function \( f^{-1} \). By definition of the inverse function, we have:
\[ f^{-1}(y) = x \]
Therefore, if we apply \( f^{-1} \) to both sides of \( y = f(x) \), we get:
\[ f^{-1}(f(x)) = x \]

**Hint**: Remember that applying the inverse function undoes the effect of the original function.

4. **Conclusion**:
Thus, we conclude that:
\[ f^{-1}(f(x)) = x \]
This means that \( f^{-1}(f(x)) \) is the identity function on set \( A \).

**Hint**: The identity function maps every element to itself.

5. **Final Result**:
Therefore, the answer is that \( f^{-1}(f) \) is equal to the identity map on set \( A \).

### Summary:
In conclusion, if \( f: A \to B \) is a bijective function, then:
\[ f^{-1}(f(x)) = x \quad \text{for all } x \in A \]
This confirms that \( f^{-1}(f) \) is the identity function on set \( A \).