Solution
= Solution
\b[True.] Given $a'\in A'$, surjectivity supplies $a\in A$ with $f(a)=g(a')$. Then $(a,a')$ lies in the fibre product and maps to $a'$.
= Solution
\b[True.] Given $a'\in A'$, surjectivity supplies $a\in A$ with $f(a)=g(a')$. Then $(a,a')$ lies in the fibre product and maps to $a'$.