If the function `f:A to B` is one-one onto and `g:B to A `, is the inverse of f, then fog =?

Options

A. f
B. g
C. `I_(A)`
D. `I_(B)`

Answer

To solve the problem, we need to find the composition of two functions, \( f \) and \( g \), where \( f: A \to B \) is a one-one and onto function, and \( g: B \to A \) is the inverse of \( f \).

### Step-by-Step Solution:

1. **Understanding the Functions**:
- We have a function \( f: A \to B \) which is both one-one (injective) and onto (surjective). This means every element in \( B \) is mapped to by exactly one element in \( A \).
- The function \( g: B \to A \) is defined as the inverse of \( f \). This means that for every \( b \in B \), \( g(b) \) gives us the unique \( a \in A \) such that \( f(a) = b \).

2. **Expressing the Composition**:
- We need to find \( f \circ g \), which means we want to compute \( f(g(b)) \) for any \( b \in B \).

3. **Using the Definition of Inverse**:
- Since \( g \) is the inverse of \( f \), we have:
\[ g(b) = a \quad \text{such that} \quad f(a) = b \]
- Therefore, when we apply \( f \) to \( g(b) \):
\[ f(g(b)) = f(a) = b \]

4. **Conclusion**:
- Thus, we find that:
\[ f \circ g(b) = b \]
- This means that the composition \( f \circ g \) acts as the identity function on the set \( B \). Hence, we can write:
\[ f \circ g = I_B \]
- where \( I_B \) is the identity function on set \( B \).

### Final Answer:
\[ f \circ g = I_B \]