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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 monomorphism is a morphism $m:X\to Y$ such that $mu=mv$ implies $u=v$ for every pair $u,v:Z\to X$. Dually, an epimorphism $e:X\to Y$ satisfies $ue=ve\Rightarrow u=v$. A regular epimorphism is a coequalizer of some parallel pair. An isomorphism has a two-sided inverse.

For a product in a category, write ${\Delta }_A:A\to A\times A$ for its categorical diagonal. Since ${\pi }_1{\Delta }_A=1_A$, equality ${\Delta }_{A}u={\Delta }_{A}v$ gives $u={\pi }_1{\Delta }_{A}u={\pi }_1{\Delta }_{A}v=v$. Thus the diagonal is a monomorphism, indeed a split monomorphism.

Use the displayed square's notation $f:A\to B$, $h:A\to C$, $g:B\to D$, $k:C\to D$, with $gf=kh$ and $g$ epic. Form the biproduct and the morphismsThe map $q$ is an epimorphism, since its restriction to $B$ is $g$: equality after $q$ implies equality after $g$. The pullback in a category property says exactly that $j$ is a categorical kernel of $q$. Indeed $qj=gf-kh=0$, and a map into $B\oplus C$ killed by $q$ is a pair $(x,y)$ with $gx=ky$, which factors uniquely through $(f,h)$.

Use the standard abelian category property that every epimorphism is the categorical cokernel of its categorical kernel. If $u:B\to X$ and $v:C\to X$ satisfy $uf=vh$, then $[u,-v]j=0$. Hence there is a unique $w:D\to X$ with $wq=[u,-v]$. Restriction to the two summands gives $wg=u$ and $wk=v$. This is the pushout in a category universal property. Thus a pullback of an epimorphism is a pushout in an abelian category.

In this pushout, $g$ is the pushout of $h$ along $f$. The reflection result of part (b) therefore makes $h$ epic. This proves pullback stability of epimorphisms in an abelian category.

Finally, let $e:E\to B$ be an epimorphism and let $p_1,p_2:R\rightrightarrows E$ be its kernel pair. Their pullback square is a pushout by the result just proved. If $u:E\to X$ satisfies $u{p}_1=u{p}_2$, the two copies of $u$ form a pushout cocone. There is a unique $v:B\to X$ with $ve=u$. Hence epimorphisms in an abelian category are coequalizers of their kernel pairs, so they are regular epimorphisms.

The required condition is the common-refinement condition for nonempty-sieve coverage: for every pair $f:V\to U$, $g:W\to U$, there are arrows $h:T\to V$, $k:T\to W$ withNecessity follows by pulling back the nonempty sieve generated by $f$ along $g$: a member $k$ of the pullback sieve supplies such an $h$. Conversely, this condition makes the pullback of every nonempty sieve on a category nonempty. The maximal sieve is nonempty, and the transitivity axiom holds: if a sieve $S$ is locally covering along every member of a nonempty covering sieve $R$, choose $f\in R$ and then $g\in f^{\ast }S$; their composite is in $S$. Hence the nonempty sieves form a Grothendieck topology, called the atomic topology.

For all functions between nonempty finite sets, the two maps from a singleton to different points of a two-point set have no common refinement. Every potential domain remains nonempty, so the two constant composites cannot agree. The condition fails.

For surjections it holds: $V{\times }_{U}W$ is nonempty and both projections are surjective. Work from now on in a small skeleton $\mathcal{D}$ of nonempty finite sets and surjections. Every morphism $f:P\twoheadrightarrow U$ is a regular epimorphism, with kernel pair $P{\times }_{U}P\rightrightarrows P$, and is the coequalizer of that pair in $\mathcal{D}$.

A matching family in a representable $yV$ on the sieve generated by $f$ is determined by a surjection $t:P\to V$ equalizing that kernel pair. It factors uniquely through a function $U\to V$, which is surjective because $t$ is. This gives the unique amalgamation. A general nonempty covering sieve contains such an $f$; after amalgamating there, common refinements with any other member force agreement on the entire sieve. Therefore every representable is a sheaf, so this atomic site is subcanonical.

For any sheaf $F$, every restriction $F(f)$ is injective: equality after a covering arrow forces equality by the separated part of the sheaf condition. We shall also use descent along any surjection $q:n\twoheadrightarrow k$:is an equalizer of sets. These are the descent identities for the atomic finite-surjection site.

Consider primitive $x\in F(m)$, $y\in F(n)$ with a common restriction along $\alpha :P\twoheadrightarrow m$, $\beta :P\twoheadrightarrow n$. Suppose $a,b\in P$ have $\alpha (a)=\alpha (b)$ but $\beta (a)\ne \beta (b)$. Let $q:n\twoheadrightarrow n-1$ identify just the two points $\beta (a),\beta (b)$. Define the finite nonempty setBoth projections $t_1,t_2:T\twoheadrightarrow P$ are surjective, since $T$ contains every diagonal pair. There is also a surjectionIndeed the target consists of diagonal pairs, which are reached because $\beta$ is surjective, and the two off-diagonal pairs corresponding to $\beta (a),\beta (b)$, reached by $(a,b)$ and $(b,a)$.

Since $\alpha t_1=\alpha t_2$, the common-restriction equality gives $F(\beta t_1)y=F(\beta t_2)y$. If ${\pi }_1,{\pi }_2$ are the target kernel-pair projections, this isInjectivity of $F(h)$ gives the kernel-pair matching condition on $y$. Descent along $q$ then writes $y=F(q)y^{\prime }$, contradicting primitivity. Thus $\ker \alpha \subseteq \ker \beta$; interchange the roles to obtain equality. This is the primitive-element kernel rigidity lemma.

Equal kernels produce a unique bijection $\gamma :m\to n$ with $\beta =\gamma \alpha$. Now $F(\alpha )x=F(\alpha )F(\gamma )y$, and injectivity impliesIn particular equivalent primitive elements have the same cardinality and differ only by transport along a bijection.

Every element $z\in F(p)$ descends to a primitive one: whenever it is not primitive, descend along a surjection reducing the cardinality by one; this process terminates at or before cardinality one. Kernel rigidity shows that all primitive ancestors of $z$ lie in one equivalence class. Let $F_C(P)$ be the elements with primitive-ancestor class $C$. Restriction along a surjection preserves this class, so each $F_C$ is a subfunctor andpointwise. Each $F_C$ is a sheaf. A matching family glues in $F$, and one member along a nonempty covering arrow already determines the primitive class of the glued element; it must be $C$.

Choose a representative primitive $x\in F(m)$ of $C$. The Yoneda map $ym\to F$ named by $x$ has image exactly $F_C$. It reaches all descendants of $x$, and every equivalent primitive ancestor is its transport along a bijection. It is therefore pointwise surjective onto $F_C$ and is epic as a map of sheaves. We obtain the primitive decomposition of an atomic finite-surjection sheafThis includes the empty coproduct for an empty sheaf.

Each nonempty $F_C$ is an atom in a topos. If a sheaf subobject $S\hookrightarrow F_C$ has an element $F(\alpha )x$ at some $P$, membership descends along the covering surjection $\alpha :P\to m$, so $x$ belongs to $S$. All its restrictions then belong to $S$, giving $S=F_C$. Thus every subobject of any $F$ selects entire components of this coproduct, and its complementary selection is again a sheaf subobject. Its characteristic map sends selected components to $\top$ and all others to $\perp$.

The constant two-element presheaf is a sheaf: a matching family on a nonempty sieve has the same value on all its arrows, since any two have a common refinement. The value extends uniquely. It therefore supplies these characteristic maps, with truth the inclusion of the $\top$ value. Equivalently, a J-closed sieve here is either empty or maximal, because every nonempty sieve covers. Hence

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.

Let $P$ be a tiny object of $[\mathcal{C},\mathbf{Set}]$, and let $0$ be the initial object of $\mathcal{C}$. The representable functor $h_0=\mathcal{C}(0,-)$ is the terminal functor, since each of its values is a singleton. The exponential adjunction and the Yoneda lemma giveSince $P$ is tiny, $E_P:F\mapsto F^P$ is itself a left adjoint, so preserves all colimits. Evaluation at $0$ also preserves colimits, because they are pointwise in a set-valued functor category. Therefore $\mathrm{Nat}(P,-)$ preserves the coproduct construction.

It also preserves epimorphisms. Indeed, every such epimorphism is a pointwise epimorphism in a functor category, hence the coequalizer of its kernel pair, as can be checked pointwise in sets. A left adjoint preserves that coequalizer and therefore sends it to a regular epimorphism, which is an epimorphism; evaluation at zero preserves surjectivity. Thus $P$ is an irreducible projective in a set-valued functor category.

A Cauchy-complete category is one in which every idempotent morphism splits. By the characterization allowed in the question, irreducible projectives are representable functors for such an indexing category. Applying it to $P$ proves the tiny covariant functor on a Cauchy-complete category with an initial object is representable conclusion:

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 squarewith $m$ 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 $r,s:R=A{\times }_{B}A\rightrightarrows A$ 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 $eb=vpq$. Thenso $(aq,b)$ 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 $f=me$: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 $0,1,1^{\prime },2$, nonidentity arrows $a:0\to 1$ and $b:1^{\prime }\to 2$, and no other nonidentity generators. Send it to the ordinal category $[2]=(0\to 1\to 2)$ by identifying $1$ with $1^{\prime }$ 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 $0\to 2$ 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.

A regular category has finite limits, every morphism factors through its image of a morphism in a regular category as a regular epimorphism followed by a monomorphism, and regular epimorphisms are stable under every pullback in a category. A cover is a strong epimorphism. Every regular epimorphism is strong: if $e$ is the coequalizer of $r,s$ and a square has $e$ on the left and a monomorphism $m$ on the right, monicity shows that the upper arrow coequalizes $r,s$. It therefore factors through $e$, and the epimorphism property of $e$ shows that this factor is the required diagonal. Conversely, factor a strong epimorphism $f$ as $f=me$ with $e$ regular epic and $m$ monic. The lifting property gives a two-sided inverse to $m$, so $m$ is an isomorphism and $f$ is regular epic. Thus regular epimorphisms and covers coincide.

Let $L:\mathcal{C}\to \mathcal{D}$ be the left-exact reflector and let $m:A^{\prime }\hookrightarrow A$. Since $L$ preserves finite limits, $Lm$ is monic. Define $c_A(A^{\prime })$ as the pullbackNaturality of the unit supplies a map $A^{\prime }\to c_A(A^{\prime })$ over $A$, proving $A^{\prime }\le c_A(A^{\prime })$. A factorization $A^{\prime }\le A^{\prime \prime }$ induces $L{A}^{\prime }\le L{A}^{\prime \prime }$ and therefore $c_A(A^{\prime })\le c_A(A^{\prime \prime })$, so $c_A$ is order-preserving.

Apply $L$ to the defining pullback. Left exactness and the fact that $L{\eta }_A:LA\to LLA$ is an isomorphism identify $L(c_A(A^{\prime }))$ with $L{A}^{\prime }$. Pulling back once more therefore givesFor a map $f:B\to A$, left exactness identifies $L(f^{\ast }{A}^{\prime })$ with $(Lf{)}^{\ast }(L{A}^{\prime })$. Pasting the two pullback squares then yieldsso this closure operation induced by a left-exact reflector commutes with pullback.

Assume $A$ lies in $\mathcal{D}$, so ${\eta }_A$ is an isomorphism. If $A^{\prime }$ also lies in $\mathcal{D}$, its unit is an isomorphism and the defining square gives $c_A(A^{\prime })\cong A^{\prime }$. Conversely, if $A^{\prime }$ is closed, that square expresses $A^{\prime }$ as a finite limit of $A$, $L{A}^{\prime }$, and $LA$, all fixed by $L$. Fixed objects of a left-exact reflective subcategory are closed under finite limits, so $A^{\prime }$ belongs to $\mathcal{D}$.

Finally suppose $\mathcal{C}$ is regular. The fixed objects have finite limits. For $f:A\to B$ in $\mathcal{D}$, factor it in $\mathcal{C}$ asApplying $L$ gives $A\stackrel{Le}{\to }LI\stackrel{Lm}{\to }B$. The map $Le$ is regular epic because a left adjoint preserves the coequalizer presenting $e$, and $Lm$ is monic because $L$ is left exact. Thus $\mathcal{D}$ has image factorizations. Their image subobject is the closure $c_B(I)$. A map in $\mathcal{D}$ is regular epic exactly when this closure is all of its codomain. Images in $\mathcal{C}$ commute with pullback, and the closure operation also commutes with pullback, so this condition is pullback-stable. Hence $\mathcal{D}$ is regular, as stated by the left-exact reflective subcategory of a regular category theorem.

Here $I$ is a set regarded as a discrete category. Finite limits and regular epimorphisms are coordinatewise. A well-supported object is a family of nonempty sets. With the axiom of choice, every coordinate element extends to a global point of its product. Any subobject containing all global points must therefore contain every coordinate element, and be the whole family. Thus the category is capital. Families with some empty coordinates explain why global points need not reflect arbitrary isomorphisms.

Suppose a category has pullbacks and coequalizers and every pullback of a regular epimorphism is epic. Coequalizing the kernel pair of any morphism produces a factorization as a regular epimorphism followed by a monomorphism. Consequently every strong epimorphism is regular.

The coequalizer universal property supplies the diagonal filler against any monomorphism, so every regular epimorphism is strong. In a regular category, every strong epimorphism is regular: factor it as a regular epimorphism followed by a monomorphism and use the lifting property to make the monomorphism an isomorphism.

A functor between regular categories is regular if it preserves finite limits and regular epimorphisms. It then preserves image factorizations, since finite-limit preservation preserves monomorphisms. A representable functor $\mathcal{C}(P,-)$ is regular exactly when $P$ is projective with respect to regular epimorphisms. Regularity of a functor does not alone imply faithfulness or reflection of isomorphisms.

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.

A functor between small categories is a strong epimorphism precisely when its image generates its codomain as a subcategory. It is a regular epimorphism precisely when it is surjective on objects and full. Hence a functor can be strong without being regular when its image arrows generate additional composites that have no single preimage.

In a regular category, a relation $R\hookrightarrow A\times B$ is total if its projection to $A$ is a regular epimorphism, and single-valued if two pairs with the same first component have the same second component. The latter makes that projection a monomorphism. A regular epimorphism that is monic is invertible, so $f={\pi }_B{\pi }_A^{-1}$ has graph exactly $R$. No axiom of choice or set-based element selection is used.