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Proof: Let \(f: A \rightarrow B\) and \(g: B \rightarrow C\) be functions. If \(g \circ f\) is bijective, then \(f\) is injective and \(g\) is surjective.
in Mathematics by Platinum (130,522 points) | 155 views

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Proof. Suppose the statement does not hold, so \(f\) is not injective or \(g\) is not surjective. Let us consider both cases:
- \(\mathrm{f}\) is not injective, which means that \(\exists a, a^{\prime} \in A\) such that \(f(a)=f\left(a^{\prime}\right)\).
Now,
\[
(g \circ f)(a)=g(f(a))=g\left(f\left(a^{\prime}\right)\right)
\]
so we have \((g \circ f)(a)=(g \circ f)\left(a^{\prime}\right)\) but \(a \neq a^{\prime}\). So \(g \circ f\) is not injective, hence it is not bijective.
- g is not surjective, which means that \(\exists c \in C\) such that for all \(b \in B, g(b) \neq c .\) Moreover, \(g \circ f\) is surjective, so \(\exists a \in A\) such that \((g \circ f)(a)=c .\) Now, if \(b=f(a)\), then \(g(b)=c\), which is a contradiction!

Both cases lead us to the contradiction, hence we may conclude that if \(g \circ f\) is bijective, then \(f\) is injective and \(g\) is surjective.
by Platinum (130,522 points)

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