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Show that if `f : A ->B`and `g : B ->C`are one-one, then `gof : A ->C`is also one-one.

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To prove that if \( f: A \to B \) and \( g: B \to C \) are one-one functions, then the composition \( g \circ f: A \to C \) is also one-one, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding One-One Function**: - A function \( f \) is called one-one (or injective) if for any two elements \( x_1, x_2 \in A \), \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \). Similarly, \( g \) is one-one if \( g(y_1) = g(y_2) \) implies \( y_1 = y_2 \) for any \( y_1, y_2 \in B \). 2. **Assume \( g \circ f \) is not one-one**: ...
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