= Solution
A <monomorphism> $m:A\to B$ satisfies $mu=mv\Rightarrow u=v$ for every parallel pair into $A$. An <epimorphism> $e:X\to Y$ satisfies $ue=ve\Rightarrow u=v$ for every parallel pair out of $Y$. A <strong monomorphism> is a <monomorphism> with the following lifting property: whenever $e:X\to Y$ is an <epimorphism> and $mu=ve$, with $u:X\to A$ and $v:Y\to B$, there is a unique $d:Y\to A$ with $de=u$ and $md=v$. A <regular monomorphism> is a <morphism> occurring as the <equalizer> of two parallel maps. A <balanced category> is one in which every <morphism> that is both a <monomorphism> and an <epimorphism> is an <isomorphism>.
Suppose $m:A\to B$ equalizes $p,q:B\rightrightarrows D$. If $mu=mv$, the common composite has two factorizations through the <equalizer>, so its uniqueness property gives $u=v$. Hence every <regular monomorphism> is indeed a <monomorphism>.
Now take the lifting square just described. The equations $pmu=qmu$ imply $pve=qve$. Cancel the <epimorphism> $e$ to obtain $pv=qv$. The <equalizer> gives a unique $d$ with $md=v$. Since $mde=ve=mu$ and $m$ is a <monomorphism>, $de=u$. Thus \b[every <regular monomorphism> is a <strong monomorphism>]; the existence of the lift uses epic cancellation, and its uniqueness uses the equalizer property.
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