Given `f: A to B and g: B to A`, such that `gof (x) = x , AA x in B , gof = I_(B)` and `fog = I_(A)`, then `fog(x)=`

Options

A. a. `x `
B. b. `-x`
C. c. `x^(2)`
D. d. None of these

Answer

To solve the problem step by step, we need to analyze the given functions and their properties.

### Step 1: Understand the Given Functions
We have two functions:
- \( f: A \to B \)
- \( g: B \to A \)

We are given the following properties:
1. \( g \circ f(x) = x \) for all \( x \in B \)
2. \( g \circ f = I_B \) (the identity function on set B)
3. \( f \circ g = I_A \) (the identity function on set A)

### Step 2: Analyze the Identity Functions
The identity function \( I_B \) means that for any element \( y \in B \), \( I_B(y) = y \). Similarly, \( I_A \) means that for any element \( z \in A \), \( I_A(z) = z \).

From the property \( g \circ f(x) = x \), we can conclude that \( g \) is a left inverse of \( f \). This means that applying \( f \) followed by \( g \) will return the original element from set B.

### Step 3: Use the Second Property
The second property states that \( g \circ f = I_B \). This confirms that \( g \) undoes the action of \( f \) for every element in \( B \).

### Step 4: Analyze the Third Property
The property \( f \circ g = I_A \) indicates that \( f \) is a left inverse of \( g \). This means that applying \( g \) followed by \( f \) will return the original element from set A.

### Step 5: Find \( f \circ g(x) \)
Now, we need to find \( f \circ g(x) \):
- Since we know that \( f \circ g = I_A \), we can conclude that for any \( x \in A \):
\[ f \circ g(x) = x \]

### Conclusion
Thus, the final answer is:
\[ f \circ g(x) = x \]