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A regular category is almost totally supported if every object is a well-supported object or a strict initial object. The case in which every object is well-supported is allowed. There can be at most one strict initial object up to isomorphism. In the Category of sets, these two cases distinguish nonempty sets from the empty set. A meet-semilattice regarded as a category satisfies the condition precisely when it has at most two objects up to isomorphism.

Choose one regular functor witnessing the regular-logic separation of a proper subobject for each proper subobject represented by a monomorphism of the small regular category. Their product is regular and reflects invertibility of monomorphisms. If an arbitrary morphism becomes invertible, its image inclusion becomes invertible and so does the diagonal of its kernel pair. The original morphism is both a regular epimorphism and a monomorphism, hence invertible. The target is a power of sets as a capital regular category. This proves an isomorphism-reflecting regular representation without asserting full faithfulness.

A capital regular category, in the convention of Definitions 1.1–1.3, is a regular category whose well-supported objects are well-pointed objects in a category. It does not require arbitrary objects to be detected by global points. A partially ordered set with finite meets, regarded as a category, is capital: its only well-supported object is the top element. Such a category can have many proper subterminal objects, all with no points. This example distinguishes the definition from the stronger convention that the global-sections functor reflects every isomorphism.

For a regular category $\mathcal{C}$, the category $\mathbf{Rel}(\mathcal{C})$ has the objects of $\mathcal{C}$ and subobjects of $A\times B$ as relations $A⇝B$. Composition takes the image of the pullback expressing existential quantification over the middle object.

For $\mathcal{C}=\mathbf{Set}$ this is the category of relations (sets); the general construction also applies to other regular categories.

The category of unital rings has identity-preserving ring homomorphisms as morphisms. The zero ring, allowing $0=1$, is terminal, and $\mathbb{Z}$ is initial. Nonzero rings admit no maps from the zero ring. Monomorphisms are injective: maps from $\mathbb{Z}[x]$ probe individual elements. It is a regular category, with surjective homomorphisms as regular epimorphisms; these are distinct from arbitrary categorical epimorphisms.

For a regular category that is a small category, a regular functor to the Category of sets that reflects isomorphisms exists exactly when the category has almost total support for a regular category. Necessity follows because a nonempty image forces the object's support to be terminal, while an empty image makes all projections $A\times X\to A$ invertible and makes all equalizers of maps out of $A$ invertible. This proves that $A$ is strict initial. For sufficiency, use the capitalization theorem to map conservatively and regularly into a capital regular category, then take global sections of a capital regular category. Well-supported objects retain points; proper strict initial objects retain empty point sets. A bijection between nonempty point sets makes the image subobject entire and makes the kernel-pair diagonal entire, by well-pointedness. Thus the image arrow is both a regular epimorphism and a monomorphism, and is invertible. The capitalization theorem is a substantive existence input; this argument does not infer arbitrary large-category representability from small-category hypotheses.

An abelian category has finite limits, and every morphism factors through its image as an epimorphism followed by a monomorphism. Every epimorphism is the cokernel of its kernel, hence a regular epimorphism, and epimorphisms in an abelian category are stable under pullback. Therefore every abelian category is a regular category.

An epimorphism $e$ is extremal when a factorization $e=mh$ with $m$ a monomorphism forces $m$ to be an isomorphism. Every regular epimorphism is extremal: its coequalizer property factors $h$ back through $e$, giving an inverse to $m$. In a regular category, the converse follows by applying extremality to the regular-epi/mono image factorization.

An image of $f:A\to B$ is the least subobject $m:I\hookrightarrow B$ through which $f$ factors. In a category with pullback in a category constructions this gives a strong epimorphism–monomorphism factorization $f=me$: pull back any mono in a lifting square for $e$ to $I$; minimality of the image forces the resulting mono to be invertible, producing a diagonal. Conversely, the lifting property of the strong part proves minimality. No stability under pullback is asserted; that is an additional property in a regular category.

In a regular category, the image of $f:A\to B$ is the monomorphism in its essentially unique factorizationas a regular epimorphism followed by a monomorphism. Images are stable under pullback.

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.

Work in ordinary set theory with the axiom of choice. A regular category has finite limits, image factorizations as regular epimorphisms followed by monomorphisms, and pullback-stable regular epimorphisms. A regular functor preserves finite limits and regular epimorphisms.

A capital regular category is one whose well-supported objects are well-pointed objects in a category. Here well-supported means $A\to 1$ is a regular epimorphism; well-pointed means a subobject of $A$ containing every point $1\to A$ must be the whole object. This is the convention of Definition 1.3. It permits objects of proper support with no points. Replacing this condition by global points detecting every isomorphism would change the assertion: a three-element chain is regular but cannot admit a conservative finite-limit-preserving functor to such a category.

Here is a regular-logic proof sketch of capitalization of a small regular category. Use a sort for every object of $\mathcal{C}$, a function symbol for every morphism, and regular theory axioms for composition, finite limiting diagrams and regular-epimorphism lifting. Its set models are precisely regular functors $\mathcal{C}\to \mathbf{Set}$. Empty sorts must be allowed.

The key separation lemma is the regular-logic separation of a proper subobject. For $m:S\hookrightarrow A$ proper, the sequentis not derivable: interpreting a derivation in $\mathcal{C}$ would make the monomorphism $m$ a regular epimorphism, hence an isomorphism. The chase completeness for regular logic supplies a set model in which it fails. Briefly, start with a named antecedent element, adjoin witnesses for the existential axioms, and quotient by required equalities. Every consequent about the original parameter arising at a finite stage has a finite proof; thus a nonderivable consequent remains false. The filtered union is the desired model. This witness construction, rather than an assumption that every sort is inhabited, is what keeps proper supports distinguishable.

Since $\mathcal{C}$ is small, choose a set-indexed family $(M_i{)}_{i\in I}$ of regular functors, one detecting each proper monomorphism, and formIt is regular by coordinatewise finite limits and surjections. It reflects invertibility of monomorphisms. If $P(f)$ is invertible, apply this fact to the image inclusion of $f$ and to the diagonal of its kernel pair. They are invertible, so $f$ is both a regular epimorphism and a monomorphism, hence invertible. Thus $P$ is a conservative functor.

The target is a power of sets as a capital regular category. A well-supported family has every coordinate nonempty, and choice extends any coordinate element to a global tuple. A subobject containing every tuple must consequently contain every coordinate element. Therefore it is capital, giving the required isomorphism-reflecting regular functor. This is the separation aspect of the regular representation results originating in Barr's embedding work; full faithfulness is unnecessary here.

The stronger target of a single set category has a precise restriction: $\mathcal{C}$ admits a conservative regular functor to the Category of sets exactly when it is an almost totally supported regular category. This means every object is well-supported or a strict initial object. Here are necessity and sufficiency.

Suppose $V:\mathcal{C}\to \mathbf{Set}$ is regular and conservative. If $VA$ is nonempty, its support in sets is $1$. Since $V$ preserves images, it sends the support inclusion $\sigma (A)\hookrightarrow 1$ to an isomorphism; conservativity makes $\sigma (A)=1$, so $A$ is well-supported.

If $VA$ is empty, every projection $A\times X\to A$ is sent to the unique bijection between empty sets, hence is invertible. This supplies a morphism $A\to X$ for every $X$. The equalizer of any two such morphisms is also sent to a bijection, hence is invertible, proving uniqueness. Thus $A$ is initial. Every arrow $Y\to A$ forces $VY$ to be empty and is then sent to a bijection; conservativity makes it invertible. So $A$ is strict initial.

Conversely suppose this support condition holds. Use $P$ above and takeProducts preserve finite limits and, by choice, coordinatewise surjections, so $V$ is regular. A well-supported object has all coordinates nonempty. A proper strict initial object has a subterminal image in every coordinate, with at least one empty coordinate: otherwise its arrow to $1$ would be sent by $P$ to an isomorphism, contradicting conservativity.

If $Vf$ is a bijection between well-supported objects, the conservativity of set products on totally supported families makes each $M_i(f)$ bijective: a missing coordinate preimage or a pair of equal-image coordinate elements extends to a tuple witnessing failure of surjectivity or injectivity. Hence $P(f)$ is invertible and so is $f$. An arrow between strict initial objects is already invertible; an arrow from a strict initial object to a well-supported object cannot become a bijection between empty and nonempty sets. An arrow into a strict initial object is invertible by definition. These cases prove conservativity and the conservative regular representation in sets criterion.

A regular category has finite limits, coequalizers of kernel pairs, and regular epimorphisms stable under pullback. Equivalently, every morphism factors as a regular epimorphism followed by a monomorphism, and these factorizations are stable under pullback. A regular functor preserves finite limits and regular epimorphisms, hence also preserves these image factorizations.

For this question, the capital regular category convention is the one in which every well-supported object is a well-pointed object in a category. Here an object $A$ is well-supported if $A\to 1$ is a regular epimorphism. It is well-pointed if a subobject $m:B\hookrightarrow A$ through which every point $1\to A$ factors must be the whole object. This convention is documented in Definitions 1.1–1.3. It imposes no such point-detection requirement on objects whose support is a proper subobject of $1$. In particular, it should not be replaced here by the stronger assertion that $\mathcal{A}(1,-)$ reflects every isomorphism.

First prove two facts about a capital regular category $\mathcal{A}$. Every well-supported object $A$ has a point. Otherwise $A\times A$ is well-supported and has no points, so its diagonal $A\hookrightarrow A\times A$ contains all points and is invertible. That makes $A$ a subterminal object; the regular epimorphism $A\to 1$ is then also monic and hence invertible, contradicting the absence of a point. If $e:X\to Y$ is a regular epimorphism and $y:1\to Y$ is a point, its pullback in a category is a well-supported object over $1$, so it has a point lifting $y$. Thus the representable functorpreserves regular epimorphisms; it already preserves all existing limits. Hence $\Gamma$ is a regular functor. Second, if $\Gamma X$ is a singleton, its unique point $x:1\to X$ is a monomorphism, $X$ is well-supported, and $x$ contains all points of $X$. Well-pointedness makes $x$ invertible. Thus $\Gamma$ reflects terminal objects, even though it need not reflect arbitrary isomorphisms.

Now let $\mathcal{C}$ be a small category that is a regular category. For each object $B$, its slice category $\mathcal{C}/B$ is small and regular: finite limits, images and the stability of regular epimorphisms are computed using the corresponding constructions in $\mathcal{C}$. By the permitted capitalization result choose a conservative functor that is regular,with ${\mathcal{A}}_B$ a small capital regular category. The base-change functor $B^{\ast }:\mathcal{C}\to \mathcal{C}/B$, given by $X\mapsto (X\times B\to B)$, preserves finite limits and regular epimorphisms. So doesTaking these as the coordinates defines a regular functorThe exponent is a set of indices, so this is a power of the Category of sets, with finite limits and regular epimorphisms computed coordinatewise.

To prove reflection of isomorphisms, suppose $R(f)$ is invertible for $f:A\to B$. In $\mathcal{C}/B$, the object $(A,f)$ is the pullback of $B^{\ast }f:(A\times B\to B)\to (B\times B\to B)$ along the point given by the diagonal $B\to B\times B$. Applying ${\Gamma }_B{J}_B$ shows that ${\Gamma }_B{J}_B(A,f)$ is the fibre of the bijection $R_B(f)$ over this diagonal point, and is a singleton. The preceding lemma gives $J_B(A,f)\cong 1$ through its canonical arrow to $1$. Since $J_B$ reflects isomorphisms, $(A,f)\to (B,1_B)$ is invertible. Its underlying morphism is $f$. Therefore

In the small-category setting of this representation result, the elementary condition for replacing the power by one copy of $\mathbf{Set}$ is:This is almost total support for a regular category. “Strict initial” means initial and every arrow into it is invertible; the condition allows the case in which all objects are well-supported and no initial object exists.

For necessity, let $V:\mathcal{C}\to \mathbf{Set}$ be regular and conservative. If $VA\ne \varnothing$, factor $A\to 1$ through its support of an object in a regular category $S\hookrightarrow 1$. Preservation of images gives $VS=1$. Conservativity makes $S\to 1$ invertible, so $A$ is well-supported. If $VA=\varnothing$, then for every $X$ the projection $A\times X\to A$ is sent to the bijection $\varnothing \to \varnothing$, hence is invertible. Its inverse followed by the other projection supplies an arrow $A\to X$. For any two such arrows, their equalizer is sent to a bijection onto $VA$, hence is invertible, proving uniqueness. Thus $A$ is initial. If $X\to A$ exists, then $VX\to \varnothing$ exists, so $VX=\varnothing$ and that arrow is sent to a bijection. Conservativity makes it invertible. Hence $A$ is strict initial.

For sufficiency, choose the permitted conservative regular functor $J:\mathcal{C}\to \mathcal{A}$ into a small capital regular category, and put $V=\Gamma J$. This is regular. A well-supported object of $\mathcal{C}$ is sent to a well-supported object of $\mathcal{A}$, which has a point. A non-well-supported object is strict initial by the condition, and is consequently a proper subterminal object. The functor $J$ preserves subterminality and reflects the noninvertibility of its arrow to $1$; its image is a proper subterminal object and has no points. Thus $VA$ is empty exactly on the non-well-supported objects.

Suppose $Vf$ is a bijection for $f:A\to B$. If both sets are empty, $A$ and $B$ are strict initial objects and $f$ is invertible. Otherwise $A,B$ are well-supported. Factor $Jf$ as a regular epimorphism followed by a monomorphism $m:I\hookrightarrow JB$. Surjectivity of $Vf$ makes every point of $JB$ factor through $m$; capitality of $\mathcal{A}$ makes $m$ invertible. Hence $Jf$ is a regular epimorphism. Its kernel pair $P=JA{\times }_{JB}JA$ is well-supported, because $P\to JA$ and $JA\to 1$ are regular epimorphisms. Injectivity of $Vf$ implies that every point of $P$ has equal coordinates, and therefore factors through the diagonal $JA\hookrightarrow P$. Since $P$ is well-pointed, this diagonal is invertible. Thus $Jf$ is also monic, hence an isomorphism, and conservativity of $J$ makes $f$ invertible. This proves sufficiency with actual point and image arguments, rather than asserting that global sections of every capital regular category is conservative.

In a regular category, interpret existential quantification by the image of the appropriate projection. The relation $f(x)=y$ is the graph of $f$, isomorphic to $A$; its projection to $B$ is $f$ itself. Thus the subobject interpreting the existential predicate is $\mathrm{im}(f)\hookrightarrow B$. Semantic validity of the sequent means this image is the whole object.

The regular image factorization writes $f=me$ with $e$ a regular epimorphism and $m$ its monic image inclusion. Its image is all of $B$ exactly when $m$ is invertible, equivalently when $f$ is itself a regular epimorphism. This is precisely the cover condition in a regular category.

A coherent category is a regular category with finite unions of subobjects, including the empty union, stable under pullback. Thus it has finite limits, pullback-stable regular-epimorphism/monomorphism image factorizations, and pullback-stable finite joins in each subobject lattice.

Write $U=A\vee B\hookrightarrow X$ and $C=A\wedge B=A{\times }_{X}B$. Their inclusions give a commutative squareTo establish the pushout property, take arrows $a:A\to Y$, $b:B\to Y$ agreeing on $C$. Regard their graphs as subobjects of $U\times Y$ and form their union $R$. Internally, this relation says that a point of $U$ is related to its $a$-value when it lies in $A$, or to its $b$-value when it lies in $B$.

The relation is total because every point of $U$ belongs to at least one of $A,B$. It is single-valued: two witnesses in the same summand give the same value by functionality of that map; witnesses in different summands have their shared point in $C$, where the two maps agree. By the relation argument proved in Question 7's unheaded continuation, the projection $R\to U$ is both a regular epimorphism and monic, hence an isomorphism. It consequently gives a map $h:U\to Y$ restricting to $a,b$.

For uniqueness, the equalizer of two possible maps contains $A$ and $B$, hence contains their union $U$ and is all of $U$. Thus the square is a pushout, provingThis is the unions of subobjects are pushouts property; it does not assume arbitrary pushouts already exist in a coherent category.

In a small regular category $\mathcal{C}$, the regular coverage has one-arrow covers: each regular epimorphism $\alpha :B\twoheadrightarrow A$ is a covering family by itself. Identity maps, composites and pullbacks of such arrows are again covers, so this is a basis for a Grothendieck topology. A presheaf is a sheaf precisely when every compatible section over a covering arrow descends uniquely along that arrow.

It is subcanonical. A section of the representable presheaf $yX=\mathcal{C}(-,X)$ over $B$ is a map $g:B\to X$. Its matching condition is $g{\pi }_1=g{\pi }_2$ on the kernel pair $B{\times }_{A}B$. Since a regular epimorphism is the coequalizer of its kernel pair, there is a unique $\bar{g}:A\to X$ with $\bar{g}\alpha =g$. This is exactly the sheaf condition for $yX$.

For a subfunctor $F^{\prime }\subseteq F$ of a regular-coverage sheaf, define local membership byThis is a subfunctor. If $t:C\to A$ and $\alpha$ witnesses membership, pull back $\alpha$ along $t$; the resulting cover of $C$ witnesses membership of $F(t)x$, because $F^{\prime }$ is stable under restriction.

It is a sheaf as well. Let $\beta :B\twoheadrightarrow A$ cover and let $y\in F^{\prime \prime }(B)$ satisfy the kernel-pair matching condition. The sheaf $F$ gives a unique $x\in F(A)$ with $F(\beta )x=y$. Choose a cover $\alpha :C\twoheadrightarrow B$ with $F(\alpha )y\in F^{\prime }(C)$. The composite $\beta \alpha$ is a cover of $A$ and witnesses $x\in F^{\prime \prime }(A)$. Uniqueness comes from $F$. Thus the inclusion $F^{\prime \prime }\hookrightarrow F$ is a mono between sheaves and is closed for the associated local operator.

Moreover, any closed subobject $S\subseteq F$ containing $F^{\prime }$ is itself a sheaf. If $x\in F^{\prime \prime }(A)$, its restriction along a witnessing cover lies in $S(B)$, and the sheaf condition for $S$ descends it to a section in $S(A)$. Its image in $F(A)$ is $x$ by uniqueness. Hence $F^{\prime \prime }$ is the least closed subobject containing $F^{\prime }$, andThis proves the local membership closure for the regular coverage, with a single regular epimorphism as witness.

Suppose $yA$ is the union, in the sheaf topos, of a family of subsheaves $F_i$. The union in presheaves is the pointwise union $F^{\prime }={\bigcup }_i{F}_i$; its closure is the union in sheaves. Thus $1_A\in F^{\prime \prime }(A)$. A witnessing cover $\alpha :B\twoheadrightarrow A$ belongs to $F_i(B)$ for one index $i$. The two restrictions of $\alpha$ to its kernel pair agree. Since $F_i$ is a subsheaf, this matching section descends in $F_i$ to $1_A\in F_i(A)$. Every map $t:C\to A$ is the restriction of $1_A$ along $t$, so $F_i=yA$. Therefore every representable sheaf is irreducible, even with respect to arbitrary unions. There are no empty covering families in this coverage, and $1_A$ also shows that the representable cannot be initial.

Finally, in the regular syntactic category of a regular theory, the context formula $\varphi$ defines an object $A$. Each formula $\varphi \wedge {\psi }_i$ defines a mono $A_i\hookrightarrow A$, and the associated representable subsheaf is its interpretation inside $yA$ in the generic model. A derivation of the coherent disjunction says that these finitely many subsheaves have union $yA$. Irreducibility gives $y{A}_i=yA$ for some $i$. The Yoneda embedding is full and faithful, since the coverage is subcanonical, and therefore reflects this isomorphism. Hence $A_i\hookrightarrow A$ is invertible in the syntactic category: $\varphi$ entails ${\psi }_i$. ThusThe disjunction is an outer coherent conclusion; the theory and all constituent formulas are regular. This disjunction property of regular theories follows from single-arrow descent, rather than from ordinary first-order compactness.

A regular category has finite categorical limits, coequalizers of kernel pairs, and regular epimorphism–monomorphism image factorizations whose regular epimorphisms are stable under every pullback in a category. Equivalently, finite limits, pullback-stable regular epimorphism–monomorphism factorizations suffice. In particular, every strong epimorphism in a regular category is regular.

For the weaker hypotheses here, interpret an image factorization as $f=me$ with $m:I\hookrightarrow B$ the least subobject through which $f$ factors, and $e:A\to I$ a strong epimorphism. With pullback in a category constructions, the least-subobject characterization itself implies that $e$ is strong: in a lifting square against a monomorphism, pull that monomorphism back to $I$; since $e$ factors through this pullback, image minimality forces its mono into $I$ to be invertible, giving the required diagonal. Thus the two descriptions of images agree under the stated hypotheses.

For $a:A^{\prime }\hookrightarrow A$, define ${\exists }_f(A^{\prime })$ as the image factorization subobject of $fa:A^{\prime }\to B$. For $b:B^{\prime }\hookrightarrow B$, the pullback in a category universal property and image minimality giveHenceThis is an adjunction between the posets of subobjects.

For Frobenius reciprocity for subobjects, factor $fa=me$ through its image $m:I\hookrightarrow B$, and put $P=I{\times }_B{B}^{\prime }$. Its mono into $B$ represents ${\exists }_f(A^{\prime })\cap B^{\prime }$. The pullback in a category of $e$ along the monomorphism $P\to I$ has domain canonically $A^{\prime }{\times }_B{B}^{\prime }$, which represents $A^{\prime }\cap f^{\ast }(B^{\prime })$. If strong epimorphisms are stable under pullback along monomorphisms, that pulled-back arrow is strong, so its composite with $P\hookrightarrow B$ is an image factorization. This givesEquality here is equality of subobjects, hence an isomorphism of their representatives.

Conversely, let $e:X\to Y$ be a strong epimorphism and $b:B^{\prime }\hookrightarrow Y$ be monic. Its image is all of $Y$: in any image factorization of $e$, the lifting property makes the mono invertible. Apply Frobenius reciprocity for subobjects to $f=e$ and $A^{\prime }=X$. The induced morphism $p:X{\times }_Y{B}^{\prime }\to B^{\prime }$ has image all of $B^{\prime }$, because the image of $bp$ in $Y$ is $B^{\prime }$. Its image factorization therefore has invertible mono, so $p$ is a strong epimorphism. This proves the converse.

We now use the glued ring categories counterexample to regularity. All rings and ring homomorphisms are unital; the zero ring, where $0=1$, is allowed. It is the terminal object of $\mathbf{Rng}$, and there is no unital ring homomorphism from it to a nonzero ring. Denote the common zero ring by $Z$, and the newly adjoined strict initial object by $\perp$. Nonzero objects in different copies have no morphisms between them. There is one morphism from every object to $Z$ and from $\perp$ to every object, and no morphism into $\perp$ except its identity.

Finite products in a category and equalizers can be described explicitly. The terminal object is $Z$. The product in a category of two nonzero objects in one copy is their ordinary ring product; the product in a category of nonzero objects in different copies is $\perp$, since only $\perp$ can map to both. A product with $Z$ is the other factor, and one with $\perp$ is $\perp$. For parallel morphisms between nonzero objects in the same copy, the ordinary ring equalizer works; it is nonzero because its subring contains distinct $0$ and $1$. All other parallel pairs are equal, and their equalizer is the identity of the domain. Thus the construction of small limits from products and equalizers gives all finite categorical limits.

Within either ring copy, monomorphisms are precisely injective ring homomorphisms: maps from $\mathbb{Z}[x]$ detect unequal elements. The only monos from outside a copy are the maps $\perp \to X$, and the only subobjects of $Z$ are $\perp$ and $Z$. Indeed a nonzero ring $R$ is not subterminal, since $\mathbb{Z}[x]\to R$ can send $x$ to $0$ or to $1$.

Ordinary ring-image factorizations remain image factorizations in the glued category. Their surjective parts remain strong epimorphisms: a lifting square into a nonzero ring stays in one copy; a square into $Z$ with mono $\perp \to Z$ would require a nonexistent map from a noninitial domain to $\perp$. Maps with domain $\perp$ are already monic and have identity strong part. Conversely, a strong epimorphism within a ring copy has surjective ring image, since its mono image part must be invertible. A map $\perp \to X$ with $X\ne \perp$ is monic but noninvertible, and therefore cannot be strong. This classifies the strong epimorphisms as the identities of $\perp$ and the surjective homomorphisms in the copies, including maps to $Z$.

Pullback along a monomorphism within a copy preserves those surjections, using the given regularity of $\mathbf{Rng}$. Pullback along $\perp \to X$ gives $1_{\perp }$, and these exhaust the additional monos, including those into $Z$. Thus strong epimorphisms are stable under pullback along monos, so Frobenius reciprocity for subobjects holds.

However, let $e:{\mathbb{Z}}_{\mathcal{D}}\to Z$ be the surjective ring homomorphism in the first copy, and pull it back along $\mathbb{Z}[x{]}_{\mathcal{E}}\to Z$ in the other copy. The resulting arrow isIt is not even an epimorphism, since the distinct evaluation ring homomorphisms $x\mapsto 0$ and $x\mapsto 1$ to ${\mathbb{Z}}_{\mathcal{E}}$ agree after composing with $\perp \to \mathbb{Z}[x{]}_{\mathcal{E}}$. Since $e$ is strong, it would be a regular epimorphism in a regular category, whose pullbacks must be regular, in particular epic. Therefore the glued category has finite limits and images and satisfies Frobenius reciprocity, but is not regular.