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

Options

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

Answer

To solve the problem, we need to show that if \( g \circ f = I_A \), where \( I_A \) is the identity function on set \( A \), then \( f \) is an injection (one-to-one) and \( g \) is a surjection (onto).

### Step 1: Show that \( f \) is an injection.

1. **Assume \( f(x_1) = f(x_2) \)** for some \( x_1, x_2 \in A \).
2. Apply \( g \) to both sides of the equation:
\[ g(f(x_1)) = g(f(x_2)). \]
3. By the property of composition, we can rewrite this as:
\[ g(f(x_1)) = I_A(x_1) \quad \text{and} \quad g(f(x_2)) = I_A(x_2). \]
4. Since \( g \circ f = I_A \), we have:
\[ I_A(x_1) = x_1 \quad \text{and} \quad I_A(x_2) = x_2. \]
5. Thus, we get:
\[ x_1 = x_2. \]
6. Since \( x_1 \) and \( x_2 \) were arbitrary, this means that \( f \) is injective.

### Step 2: Show that \( g \) is a surjection.

1. **Let \( y \) be any element in \( B \)**.
2. We need to show that there exists an \( x \in A \) such that \( g(y) = x \).
3. Since \( g \circ f = I_A \), we can take \( x = f(x') \) for some \( x' \in A \).
4. Then, we have:
\[ g(f(x')) = I_A(x') = x'. \]
5. This means that for every \( x' \in A \), there exists a \( y \in B \) such that \( g(y) = x' \).
6. Therefore, \( g \) is surjective.

### Conclusion

From the above steps, we conclude that:
- \( f \) is an injection.
- \( g \) is a surjection.

Thus, the statement is verified.