Let `f: A to B; g: B to A` be two functions such that `fog = I_B`. Then; f is a surjection and g is an injection.

Options

A. f and g both are injections
B. f and both are surjections
C. f is and injection and g is a surjection
D. f is a injections and g is a surjection

Answer

To prove that if \( f: A \to B \) and \( g: B \to A \) are two functions such that \( f \circ g = I_B \) (where \( I_B \) is the identity function on set \( B \)), then \( f \) is a surjection and \( g \) is an injection, we can follow these steps:

### Step 1: Prove that \( f \) is a surjection

1. **Consider an arbitrary element in \( B \)**: Let \( y \in B \) be any arbitrary element.

2. **Use the function \( g \)**: Since \( g: B \to A \), we can apply \( g \) to \( y \). Let \( x = g(y) \). Thus, \( x \in A \).

3. **Apply the function \( f \)**: Now, we apply \( f \) to \( x \):
\[ f(x) = f(g(y)) \]

4. **Use the composition property**: From the given information, we know that \( f \circ g = I_B \). Therefore:
\[ f(g(y)) = I_B(y) \]
Since \( I_B \) is the identity function on \( B \), we have:
\[ I_B(y) = y \]
Thus:
\[ f(g(y)) = y \]

5. **Conclusion for surjection**: We have shown that for every \( y \in B \), there exists an \( x \in A \) (specifically \( x = g(y) \)) such that \( f(x) = y \). This means that \( f \) is a surjection.

### Step 2: Prove that \( g \) is an injection

1. **Assume \( g(x_1) = g(x_2) \)**: Let \( x_1, x_2 \in B \) be such that \( g(x_1) = g(x_2) \).

2. **Apply the function \( f \)**: We apply \( f \) to both sides:
\[ f(g(x_1)) = f(g(x_2)) \]

3. **Use the composition property**: Again, using \( f \circ g = I_B \), we have:
\[ f(g(x_1)) = I_B(x_1) \quad \text{and} \quad f(g(x_2)) = I_B(x_2) \]
Therefore:
\[ I_B(x_1) = I_B(x_2) \]

4. **Identity function property**: Since \( I_B(x_1) = x_1 \) and \( I_B(x_2) = x_2 \), we can conclude:
\[ x_1 = x_2 \]

5. **Conclusion for injection**: Since \( g(x_1) = g(x_2) \) implies \( x_1 = x_2 \), we conclude that \( g \) is an injection.

### Final Conclusion
We have shown that \( f \) is a surjection and \( g \) is an injection.

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