# If $f \circ g$ is a bijection, then $f$ is a surjection and $g$ is an injection [Help with Understanding]

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".

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 $$a = a'$$.
$$(f\circ g)(a)$$ 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.