Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 18 / 2 / a

A monomorphism $m:A\to B$ satisfies $mu=mv\Rightarrow u=v$, and an epimorphism $e:A\to B$ satisfies $ue=ve\Rightarrow u=v$. A strong monomorphism is a monomorphism with the right lifting property against all epimorphisms: every square $mu=ve$ has a diagonal $d$ satisfying $de=u$ and $md=v$. The diagonal is unique by monicity. A regular monomorphism is an equalizer of some pair $h,k:B\rightrightarrows Z$. It is monic because two equalizer factorizations of the same arrow must agree.

The converted TeX omits the remainder of this subpart. For the printed strict monomorphism condition, an arrow $g:C\to B$ is admissible when, for every pair $h,k$ out of $B$, the implication $hm=km\Rightarrow hg=kg$ holds; strictness says each such $g$ factors uniquely through $m$. If $m$ equalizes $h_0,k_0$, every admissible $g$ satisfies $h_{0}g=k_{0}g$, and the equalizer property supplies its unique factorization. Hence every regular monomorphism is strict.

Strictness itself implies monicity: whenever $mu=mv$, the common composite is admissible, so uniqueness of its factor through $m$ gives $u=v$. Now take a square $mu=ve$ with $e$ epic. Whenever $hm=km$, we have $hve=hmu=kmu=kve$, hence $hv=kv$. Thus $v$ is admissible and has a unique factor $d$ through $m$. Monicity gives $de=u$. We have proved the chain