= Solution
A <regular epimorphism> $e:A\to B$ is the <coequalizer> of some parallel pair $r,s:R\rightrightarrows A$. A <strong epimorphism> has the left lifting property against every <monomorphism>: in every commutative square
$$
\begin{array}{ccc}
A&\xrightarrow{u}&C\\
e\downarrow&&\downarrow m\\
B&\xrightarrow{v}&D
\end{array}
$$
with $m$ monic, there is $d:B\to C$ satisfying $de=u$ and $md=v$.
If $e$ coequalizes $r,s$, then $mur=ver=ves=mus$. Monicity gives $ur=us$, so the <coequalizer> property gives a unique $d$ with $de=u$. Now $mde=mu=ve$; every coequalizer is epic, hence $md=v$. This proves that every <regular epimorphism> has the lifting property of a <strong epimorphism>.
For an arbitrary $f:A\to B$, form its <kernel pair> $r,s:R=A\times_BA\rightrightarrows A$ and let $e:A\to I$ be its coequalizer. There is a unique $m:I\to B$ with $f=me$. To show $m$ monic, suppose $mu=mv$ for $u,v:X\rightrightarrows I$. Pull back $e$ along $u$, obtaining an epic $p:P\to X$ and $a:P\to A$ with $ea=up$. Pull back $e$ along $vp$, obtaining an epic $q:Q\to P$ and $b:Q\to A$ with $eb=vpq$. Then
$$
f(aq)=mupq=mvpq=f(b),
$$
so $(aq,b)$ factors through the kernel pair of $f$. Since $e$ coequalizes that pair, $eaq=eb$, and hence $upq=vpq$. The composite $pq$ is epic, so $u=v$. This proves the <regular-epimorphism-monomorphism factorization from kernel pairs>.
If $f$ is strong, apply its lifting property to $f=me$:
$$
\begin{array}{ccc}
A&\xrightarrow{e}&I\\
f\downarrow&&\downarrow m\\
B&\xrightarrow{1_B}&B.
\end{array}
$$
A diagonal $d:B\to I$ satisfies $md=1_B$. Since $m$ is also monic, $dm=1_I$, so $m$ is an <isomorphism> and $f$ is regular.
For the category of small categories, let $\mathcal A$ have objects $0,1,1',2$, nonidentity arrows $a:0\to1$ and $b:1'\to2$, and no other nonidentity generators. Send it to the ordinal category $[2]=(0\to1\to2)$ by identifying $1$ with $1'$ and sending $a,b$ to the two generating arrows. The image generates $[2]$, so this is a <strong epimorphism in the category of small categories>, but it is not full because the composite $0\to2$ has no preimage; therefore it is not regular.
Every functor factors through the subcategory of its codomain generated by its image. The first functor is strong epic and the inclusion is monic, so strong-epi--mono factorizations always exist in $\mathbf{Cat}$. Regular-epi--mono factorizations do not always exist: the displayed strong nonregular functor would make its monic factor an isomorphism by strongness, forcing the original functor to be regular.
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