Solution
= Solution
For any $c\in C$, the <surjectivity> of $g\circ f$ supplies $a\in A$ with $g(f(a))=c$. Therefore $f(a)\in B$ is a preimage of $c$ under $g$, proving <surjectivity> of $g$.
For the requested counterexample, take $A=C=\{0\}$, $B=\{0,1\}$, $f(0)=0$, and $g(0)=g(1)=0$. The <function composition> is the identity <bijection> of the singleton, but $f$ is not <surjective> and $g$ is not <injective>.