= Solution
Here use the <left lifting property against monomorphisms> as the definition of “strong”; epimorphicity will be established separately. In a lifting square write $g:C\to D$, $m:A\to B$, $k:C\to A$ and $l:D\to B$, with $mk=lg$ and $m$ a <monomorphism>. Any two lifts agree because $mt=mt'=l$.
First let $g$ be the <coequalizer> of $r,s:E\rightrightarrows C$. Every <coequalizer> is an <epimorphism>: if $ug=vg$, the uniqueness clause for the <coequalizer> applied to this common composite gives $u=v$. Moreover,
$$
mkr=lgr=lgs=mks,
$$
so $kr=ks$ by <monomorphism> cancellation. The <coequalizer> therefore supplies $t:D\to A$ with $tg=k$. Then $mtg=mk=lg$, and <epimorphism> cancellation gives $mt=l$. This proves that <regular epimorphisms are strong epimorphisms>.
If $g:C\to D$ is also a <monomorphism>, take $m=g$, $k=1_C$ and $l=1_D$. Its lift satisfies $tg=1_C$ and $gt=1_D$. Hence \b[<monic lifting-only strong morphisms are invertible>].
Next suppose $gf:A\to C$ has the <left lifting property against monomorphisms>. Given a lifting square for $g:B\to C$, with $mk=lg$, precompose its top arrow with $f$. A lift for $gf$ gives $t:C\to X$ with $mt=l$ and $tgf=kf$. Crucially, one does not cancel $f$: instead $mtg=lg=mk$, and the <monomorphism> $m$ gives $tg=k$. Thus \b[the right factor of a strong composite is strong], proving <right-factor cancellation for lifting-only strong morphisms>.
Finally, in $u=iv$ with $u$ strong and $i$ monic, the preceding result makes $i$ strong. The monic-strong argument then makes \b[$i$ an isomorphism]. None of these arguments assumed that a lifting-only strong morphism was already epic.
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