Sufficient / necessary conditions for $g \circ f$ being injective, surjective or bijective

What I know about your nice question is as follows. I suppose that $f:A\to B$ and $g:B\to C$.

1. If $g\circ f$ is 1-1 then $f$ is 1-1.

2. If $g\circ f$ is onto then $g$ is onto.

so if $g\circ f$ is bijective then $f$ is 1-1 and $g$ is onto but, the converse is not true.

1. $f(x)=f(x')\Longrightarrow g(f(x))=g(f(x'))\Longrightarrow (g\circ f)(x)=(g\circ f)(x')\Longrightarrow x=x'$

2. if $c\in C$ then $\exists a (a\in A \wedge (g\circ f)(a)=c)\Longrightarrow\exists a(a\in A \wedge (g\big(f(a)\big)=c)$ so if we take $b=f(a)\in B$ then $g(b)=c$.

Now take $A=\{1,2\}, B=\{3,4,5\}, C=\{6,7\}$ and $$f:A\to B\\f(1)=3,f(2)=4$$ $$g:B\to C\\ g(3)=g(4)=6,g(5)=7$$ $f$ is 1-1 and $g$ is onto but $$(g\circ f)(1)=g(3), (g\circ f)(2)=6$$ which means that $g\circ f$ is not a bijective map.