Counterexample of converse of "if $f: A\to B$ and $g: B\to C$ are surjective, then $g\circ f$ is surjective" I managed to prove the statement:

If $f: A\to B$ and $g: B\to C$ are surjective, then $g\circ f$ is surjective.

But now I require a counterexample to the converse of this statement. I am not sure how to formulate the counterexample. Similarly I need a counterexample of the statement being "injective" instead of "surjective".
 A: Converse : "If $g\circ f$ is surjective, then $f$ and $g$ are surjective".
Saying this statement is false means $g\circ f$ is surjective, but either $g$ or $f$ is not. But if $g$ is not surjective, $g\circ f$ can't be either (check it). So your counterexample has to be composed of $f:A\to B$ non surjective, and $g:B\to C$ surjective such that $g\circ f$ is surjective.
For example : $A=\left\{0,1\right\}$, $B=\left\{a,b,c\right\}$, $C=\left\{\alpha,\beta\right\}$. $f(0)=a$, $f(1)=b$, $g(a)=\alpha$, $g(b)=\beta=g(c)$.
You can check $g\circ f$ is surjective, but $f$ is not.
For "injective" instead of "surjective", do the same analysis, and consider the fact that if $f$ is not injective, $g\circ f$ can't be either.
A: Hint: A function $h: A \to \{0\}$ is surjective for any nonempty set $A$.
A: Hints: You may find it useful to use the characterizations
$$f:A\to B \textrm{ surjective }\iff f\circ f^{-1}=\operatorname{Id}_B$$
and
$$f:A\to B \textrm{ injective }\iff f^{-1}\circ f=\operatorname{Id}_A$$
