Let `f: A -> B and g : B -> C` be two functions and `gof: A -> C` is define statement(s) is true?

Options

A. If gof is one-one, then f anf g both are one-one
B. if gof is one-one, then f is one-one
C. If gof is a bijection, then f is one-one and g is onto
D. If f and g are both one-one, then gof is one-one.

Answer

To solve the problem regarding the functions \( f: A \to B \) and \( g: B \to C \), and the composition \( g \circ f: A \to C \), we need to analyze the properties of these functions and determine which statements about them are true.

### Step-by-Step Solution:

1. **Understanding the Functions**:
- We have two functions: \( f \) maps elements from set \( A \) to set \( B \) and \( g \) maps elements from set \( B \) to set \( C \).
- The composition \( g \circ f \) means that for any element \( a \in A \), \( g(f(a)) \) will give us an element in set \( C \).

2. **Analyzing the Statements**:
- We need to evaluate the truth of various statements regarding the functions \( f \) and \( g \) based on their properties (one-one, onto, and bijective).

3. **Statement Analysis**:
- **Statement 1**: If \( g \circ f \) is one-one, then both \( f \) and \( g \) must be one-one.
- This statement is **not necessarily true**. \( g \) could be one-one, and \( f \) could map different elements of \( A \) to the same element in \( B \), making \( g \circ f \) one-one without both being one-one.

- **Statement 2**: If \( g \) is one-one, then \( f \) must also be one-one.
- This statement is **true**. If \( g \) is one-one, then for \( g(f(a)) = g(f(a')) \) to hold, it must be that \( f(a) = f(a') \), which implies \( a = a' \), thus \( f \) must also be one-one.

- **Statement 3**: If \( g \circ f \) is bijective, then \( f \) is one-one and \( g \) is onto.
- This statement is **true**. For \( g \circ f \) to be bijective, \( f \) must be one-one (to ensure distinct inputs map to distinct outputs) and \( g \) must be onto (to ensure every element in \( C \) is covered).

- **Statement 4**: If both \( f \) and \( g \) are one-one, then \( g \circ f \) is one-one.
- This statement is **true**. If both functions are one-one, then the composition will also be one-one since distinct inputs in \( A \) will lead to distinct outputs in \( C \).

4. **Conclusion**:
- The only statement that is definitively true based on the analysis is **Statement 2**: If \( g \) is one-one, then \( f \) must also be one-one.

### Final Answer:
The true statement is: If \( g \) is one-one, then \( f \) is also one-one.