If f: A rightarrow B and g: B rightarrow C ar | vedclass

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(B) Given that the composite function $g \circ f: A \rightarrow C$ is onto. By definition,for every element $z \in C$,there exists an element $x \in A

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$g$ is onto

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(B) Given that the composite function $g \circ f: A \rightarrow C$ is onto. By definition,for every element $z \in C$,there exists an element $x \in A$ such that $(g \circ f)(x) = z$. This can be written as $g(f(x)) = z$. Since $f(x) = y$ for some $y \in B$,we have $g(y) = z$. This implies that for every $z \in C$,there exists at least one $y \in B$ such that $g(y) = z$. Therefore,$g$ must be an onto (surjective) function. Thus,the necessary condition is that $g$ is onto.

If $f: A \rightarrow B$ and $g: B \rightarrow C$ are functions such that $g \circ f: A \rightarrow C$ is onto,then a necessary condition is:

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