Solution
= Solution
For each $c\in C$, <surjectivity> of the <function composition> gives $a\in A$ such that $g(f(a))=c$. Taking $b=f(a)\in B$ supplies a preimage of $c$ under $g$. Thus \b[$g$ is a <surjective function>].
= Solution
For each $c\in C$, <surjectivity> of the <function composition> gives $a\in A$ such that $g(f(a))=c$. Taking $b=f(a)\in B$ supplies a preimage of $c$ under $g$. Thus \b[$g$ is a <surjective function>].