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Map two disjoint walking arrows to the two mutually inverse arrows of the walking isomorphism. Their images generate the entire target, so the functor is an extremal epimorphism. Its kernel pair only identifies the matching endpoint objects, not the inverse-composition relations. Mapping both arrows to a nonidentity idempotent in a one-object category equalizes that kernel pair but cannot factor through the target groupoid. Thus the functor is not a regular epimorphism, although the category of small categories has finite limits.

A regular category has finite limits, factorizations of every arrow as a regular epimorphism followed by a monomorphism, and pullback-stable regular epimorphisms. Equivalently one may require coequalizers of kernel pairs and pullback stability of the resulting quotients. A regular epimorphism is a coequalizer of some parallel pair; an extremal epimorphism is an epimorphism $e$ such that $e=mh$, with $m$ monic, forces $m$ to be invertible.

First, every regular epimorphism is extremal without assuming regularity of the whole category. Suppose $e:A\to B$ coequalizes $r,s$ and $e=mh$. Since $m$ is monic, $hr=hs$, so the universal property gives $k:B\to \mathrm{dom}(m)$ with $ke=h$. Then $mke=e$; cancellation of the epimorphism $e$ gives $mk=1_B$. Monicity of $m$ now gives $km=1$, so $m$ is an isomorphism. Conversely, in a regular category, factor an extremal $e$ as $mq$ with $q$ regular epic and $m$ monic. Extremality makes $m$ invertible, so $e$ is regular epic. Thus covers and regular epimorphisms coincide.

For a counterexample, work in the category of small categories, which has finite limits constructed on objects and arrows. Let $C$ be two disjoint walking arrows $a\stackrel{u}{\to }b$ and $c\stackrel{v}{\to }d$. Let $D$ be the walking isomorphism, with two objects $0,1$ and inverse arrows $p:0\to 1$, $q:1\to 0$. Define $F:C\to D$ by $a,d\mapsto 0$, $b,c\mapsto 1$, $u\mapsto p$, $v\mapsto q$. These images generate all of $D$, so two functors agreeing after $F$ agree everywhere; thus $F$ is epic.

A monic functor is injective on objects and faithful, as tested by functors from the terminal category and from the walking arrow. If $F$ factors through such a functor $m:D^{\prime }\to D$, its image contains both objects and both arrows $p,q$. Faithfulness forces their lifted composites to be the identities, so the image contains every arrow of $D$ and $m$ is an isomorphism. Hence $F$ is extremal.

It is not regular. Its kernel pair identifies the endpoint pairs $a,d$ and $b,c$ but has no relation forcing the composites of the images of $u,v$ to be identities. Concretely, send all four objects to the one-object category with endomorphism monoid $\{1,e\}$, $e^2=e\ne 1$, and send both $u,v$ to $e$. This functor equalizes the kernel pair: corresponding identities agree, and the only nonidentity arrow pairs in the kernel pair are an arrow paired with itself. It cannot factor through $D$, because images of $p,q$ would have to be inverse, whereas $e^2\ne 1$. In a category with kernel pairs, every regular epimorphism is the coequalizer of its kernel pair: its original coequalized pair factors through that kernel pair, giving the universal property. Therefore $F$ is an extremal but nonregular epimorphism, the promised nonregular extremal epimorphism in the category of categories.