Consider functions f: A rightarrow B and g: B | vedclass

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(C) Given that $(g \circ f)^{-1}$ exists,the composite function $g \circ f: A \rightarrow C$ must be a bijection (both one-one and onto).$1$. For $g \

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$f$ is one-one and $g$ is onto

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(C) Given that $(g \circ f)^{-1}$ exists,the composite function $g \circ f: A \rightarrow C$ must be a bijection (both one-one and onto).$1$. For $g \circ f$ to be one-one,$f$ must be one-one. If $f$ were not one-one,there would exist $x_1, x_2 \in A$ such that $f(x_1) = f(x_2)$,implying $g(f(x_1)) = g(f(x_2))$,which contradicts the one-one property of $g \circ f$.$2$. For $g \circ f$ to be onto,$g$ must be onto. If $g$ were not onto,there would exist some $z \in C$ such that no $y \in B$ satisfies $g(y) = z$,meaning no $x \in A$ could satisfy $g(f(x)) = z$,contradicting the onto property of $g \circ f$.Therefore,$f$ must be one-one and $g$ must be onto.

Consider functions $f: A \rightarrow B$ and $g: B \rightarrow C$ $(A, B, C \subseteq \mathbb{R})$ such that $(g \circ f)^{-1}$ exists. Then:

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