= Solution
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 $h_0g=k_0g$, 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
$$
\boxed{\text{regular monomorphism}\ \Longrightarrow\ \text{strict monomorphism}\ \Longrightarrow\ \text{strong monomorphism}.}
$$
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