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Let f: A to B and g: B to C be two funct...

Let `f: A to B` and `g: B to C` be two functions. Then; if gof is onto then g is onto; if gof is one one then f is one-one and if gof is onto and g is one one then f is onto and if gof is one one and f is onto then g is one one.

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To solve the problem regarding the functions \( f: A \to B \) and \( g: B \to C \), we will analyze the implications of the composition of these functions, denoted as \( g \circ f \). We will prove each statement step by step. ### Step 1: If \( g \circ f \) is onto, then \( g \) is onto. **Proof:** 1. Assume \( g \circ f \) is onto. This means that for every element \( c \in C \), there exists an element \( a \in A \) such that \( g(f(a)) = c \). 2. Since \( g(f(a)) = c \), for every \( c \in C \), there exists \( b = f(a) \in B \) such that \( g(b) = c \). 3. This shows that every element \( c \in C \) is covered by \( g \), hence \( g \) is onto. ...
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