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(B) Given: $f: A \rightarrow B$ and $g: B \rightarrow C$. Also,$g \circ f: A \rightarrow C$ is a bijection. First,we prove $f$ is an injection: Let $x

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$f$ is an injection and $g$ is a surjection

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(B) Given: $f: A \rightarrow B$ and $g: B \rightarrow C$. Also,$g \circ f: A \rightarrow C$ is a bijection. First,we prove $f$ is an injection: Let $x_1, x_2 \in A$ such that $f(x_1) = f(x_2)$. Then $g(f(x_1)) = g(f(x_2))$,which implies $(g \circ f)(x_1) = (g \circ f)(x_2)$. Since $g \circ f$ is a bijection,it is injective,so $x_1 = x_2$. Thus,$f$ is an injection. Next,we prove $g$ is a surjection: Let $z \in C$. Since $g \circ f: A \rightarrow C$ is a bijection,it is surjective. Therefore,there exists $x \in A$ such that $(g \circ f)(x) = z$. This means $g(f(x)) = z$. Since $f(x) \in B$,let $y = f(x)$. Then $g(y) = z$ for some $y \in B$. Thus,$g$ is a surjection. Therefore,$f$ is an injection and $g$ is a surjection.

If $f: A \rightarrow B$ and $g: B \rightarrow C$ are two functions such that $g \circ f: A \rightarrow C$ is a bijection,then which one of the following is always true?

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