Solution
= Solution
Let $c\in C$. Since $g$ is <surjective>, choose $b\in B$ with $g(b)=c$. Since $f$ is <surjective>, choose $a\in A$ with $f(a)=b$. Then $(g\circ f)(a)=g(b)=c$. Thus every element of $C$ has a preimage, proving that $g\circ f$ is <surjective>.