Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 18 / 1 / b / 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 $mt=m{t}^{\prime }=l$.

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,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 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 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 $i$ an isomorphism. None of these arguments assumed that a lifting-only strong morphism was already epic.