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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.

Take two copies of the category of rings, identify their terminal zero rings, and adjoin a strict initial object $\perp$. Same-copy finite categorical limits are ordinary ring limits; products of nonzero objects in different copies are $\perp$. Ring surjections and $1_{\perp }$ are precisely the strong epimorphisms. They are stable under pullback along monomorphisms, so image factorizations exist and Frobenius reciprocity for subobjects holds. But pulling $\mathbb{Z}$ in one copy $\to 0$ back along $\mathbb{Z}[x]$ in the other copy $\to 0$ gives $\perp \to \mathbb{Z}[x]$. This is not epic: evaluations $x\mapsto 0,1$ agree after precomposing with it. Thus the category is not regular, since strong epimorphisms would then be regular and pullback-stable.

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 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.