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Let $g: A \to B$ and $f: B \to C$ be two functions.

Show that if $f \circ g$ is a bijection, then $f$ is a surjection and $g$ is an injection.

I have trouble understanding the proof.

I saw a post on this and the person answered by using Proof by Contradiction. He stated:

"If $g$ is not injective, we have $g(a) = g(a')$ for some $a\not=a'$. But $f \circ g(a)=f \circ g(a')$, so $f \circ g$ is not bijective. Contradiction. So $g$ must be injective.

If $f$ is not surjective, then for some $c \in C$ we have $f(b)\not=c$ for any $b \in B$. So we cannot have $f \circ g(a)=c$. So $f \circ g$ is not bijective. Contradiction. So $f$ must be surjective."

I understand the second part (surjection proof) for this, but I don't understand the first part (injection proof). I just don't know how the user got from the assumption to "But $f \circ g(a)=f \circ g(a')$" and how he concluded that "so $f \circ g$ is not bijective".

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2 Answers 2

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If $g(a) = g(a')$, let $c := g(a)$. Then $f(c) = f(c)$ but $c= g(a)$ and $c= g(a')$, so $f(g(a)) = f(c) = f(c) = f(g(a'))$.

Thus $f(g(a)) = f(g(a'))$, then since $f \circ g$ is a bijection, it is an injection, hence $a = a'$.

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$(f\circ g)(a)$ is $f(g(a))$, which is equal to $f(g(a'))$ as we've postulated that $g(a)=g(a')$. Therefore $f\circ g$ is not injective.

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