If the mappings f :` A to B `and g: `B to C` are both bijective, then the mapping gof: A `to` C is also bijective.
A. true B. False

Answer

To determine whether the mapping \( g \circ f: A \to C \) is bijective given that \( f: A \to B \) and \( g: B \to C \) are both bijective, we need to check if \( g \circ f \) is both one-to-one (injective) and onto (surjective).

### Step 1: Check if \( g \circ f \) is One-to-One

1. **Assume \( g(f(x_1)) = g(f(x_2)) \)** for some \( x_1, x_2 \in A \).
2. Since \( g \) is bijective (one-to-one), we can conclude that \( f(x_1) = f(x_2) \).
3. Now, since \( f \) is also bijective (one-to-one), we can conclude that \( x_1 = x_2 \).

Thus, \( g \circ f \) is one-to-one.

### Step 2: Check if \( g \circ f \) is Onto

1. **Take any \( c \in C \)**. Since \( g \) is onto, there exists some \( b \in B \) such that \( g(b) = c \).
2. Since \( f \) is onto, there exists some \( a \in A \) such that \( f(a) = b \).
3. Therefore, \( g(f(a)) = g(b) = c \).

This shows that for every \( c \in C \), there exists an \( a \in A \) such that \( g(f(a)) = c \). Thus, \( g \circ f \) is onto.

### Conclusion

Since \( g \circ f \) is both one-to-one and onto, we conclude that \( g \circ f \) is bijective.

Therefore, the statement is **True**.

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