So I can say that because $f$ is injective $f(g(a)) = f(g(b)) \Rightarrow g(a) =g(b)$,right?
And because $g$ is not surjective there is $g(a) \neq A$.
I'm not sure how to combine these two together and prove that $f(g(a)) \neq A$.
Am I in the right direction with these claims? Any tips?
Yes, you are in the right direction. Suppose that $g$ is a function from $X$ into $Y$. Since $g$ is not surjective, there is some $y\in Y$ which is not of the form $g(x)$, for some $x\in X$. Then $f(y)$ does not belong to the range of $f\circ g$. Indeed, if $f(y)=f\bigl(g(x)\bigr)$ for some $x\in X$, then, since $f$ is injective, $y=g(x)$ .
It's easier to do a direct proof that if $f \circ g$ is surjective and $f$ is injective, then $g$ is surjective.
Indeed, note that whenever $f \circ g$ is surjective, $f$ must also be surjective. If $f$ is also injective, this means that $f$ is a bijection. Therefore, since $f \circ g$ is surjective, so is $g$.
