In Abstract and Concrete Category of Adamek, Herrlich and Stretcker, I'm dealing with exercise 7Dd. It says that a strict monomorphism $f:A\to B$ followed by a section $g:B\to C$ give rise to a a strict monomorphism $fg:A\to C$.
I try to prove it: Let $f'$ such that, for any $r,s:C\to\cdot$,we have $fgr=fgs$ implies $f'r=f's$. We have to show that $f'=af$ for a unique $a$. Let $h$ such that $gh=1$. Then $fu=fv$ implies $fghu=fghv$ which implies $f'hu=f'hv$. Since $f'$ is strict, by assumption, we obtain $f'h=af$ for a unique $f$. I have difficulties to deduce $f'=afg$ from it.