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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.

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.

In a locally small capital regular category, $\Gamma$ is a regular functor and reflects terminal objects. A well-supported object $A$ has a point: otherwise $A\times A$ is well-supported with no points, so its diagonal is invertible by well-pointedness; then $A$ is subterminal and its regular epimorphism to $1$ is invertible, a contradiction. Pulling a regular epimorphism back along a point gives a well-supported object, so the point lifts. This proves preservation of regular epimorphisms, while representability gives limit preservation. If $\Gamma X$ is a singleton, its unique point $1\to X$ is monic and contains all points; $X$ is well-supported, so that point is invertible. Global sections need not reflect arbitrary isomorphisms between objects of proper support.

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.

For a small category that is a regular category, use a sort for each object, an operation for each morphism, and regular theory axioms describing finite limits and regular epimorphisms. Its models are regular functors into the Category of sets. For a proper subobject $m:S\hookrightarrow A$, the sequent asserting $\forall x:A\exists y:S,m(y)=x$ cannot be derivable, since its categorical interpretation would make $m$ a cover and hence invertible. The chase completeness for regular logic supplies a model where it fails. It is essential to allow empty sorts.

The support of an object $A$ in a regular category is the subterminal object appearing as the image of a morphism in a regular category of $A\to 1$. A regular functor preserves this construction. In the Category of sets, the support is the empty set if $A$ is empty and a singleton otherwise.