If $g \circ f$ is injective, so is $g$
I don't think this is true. I think that $f$ has to be surjective.
So I am going to try to prove that:
If $g \circ f$ is injective, and $f$ is surjective, then $g$ is injective.
First off, $g \circ f$ means that $(g \circ f)(a) = (g \circ f)(b)$ then $a=b$.
And $f$ being surjective means that $\forall b \in B, \exists a\in A$ such that $f(a)=b$
Suppose that $g$ was not injective.
$g(f(a))= g(f'(a))$ and $f(a)\ne f'(a)$
But $g(f(a)) = g(f(b)) \to a=b$ $g(f(a))= g(f(b)) \to f(a)=f(b) \to a=b$ Hence contradiction, since $g$ is not injective.
So $g$ is injective.
Is this an acceptable proof, I think my logic is iffy around the second last paragraph.