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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.

A category specifies objects of a category, categorical morphisms between them, and composition in a category. An arrow $f:A\to B$ can be followed by $g:B\to C$, giving $g\circ f:A\to C$. Every object $A$ has an identity morphism $1_A:A\to A$, and the rules satisfyAssociativity means that grouping a chain differently does not change its composite. It does not assert commutativity, and the sources and targets must match before a composite is defined.

In the Category of sets, arrows are functions. Take two functions on the real numbers, $f(x)=x+1$ and $g(x)=2x$. Then $(g\circ f)(x)=2x+2$, whereas $(f\circ g)(x)=2x+1$. At $x=3$, the two orders give 8 and 7, respectively. Both orders are defined here because these maps share their source and target; for arbitrary arrows the reverse composite need not exist.

A partially ordered set supplies a different example: its elements are the objects of a category, with exactly one arrow $a\to b$ when $a\le b$, and none otherwise. Transitivity gives composition in a category, and reflexivity gives identity morphisms. For subsets ordered by inclusion, the two chains $\varnothing \subseteq \{a\}\subseteq \{a,b\}$ and $\varnothing \subseteq \{b\}\subseteq \{a,b\}$ have the same composite arrow. Thus arrows need not be maps of elements: they can record a relation.

For commuting actions of finite groups $G,H$ with coprime orders, the canonical map $A^G/H\to (A/H{)}^G$ is a bijection. It is always injective. On any $G$-stable $H$-orbit, all $G$-orbits have the same size, dividing both $|G|$ and the $H$-orbit size, hence dividing $|H|$. Coprimality forces that size to be one, proving surjectivity. Thus categorical limits indexed by $G$ commute with colimits indexed by $H$ in the Category of sets.

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.

A countable coproduct is indexed by the natural numbers. In the Category of sets it is the disjoint union ${\coprod }_{n\in \mathbb{N}}{A}_n$.

An epimorphism $e:A\to B$ is right-cancellative: $fe=ge$ implies $f=g$. In the Category of sets and in abelian groups, epimorphisms are precisely the surjective homomorphisms.

A filtered colimit in a category is a colimit whose indexing category is a small filtered category. In the Category of sets, filtered colimits commute with finite limits in sets. A filtered colimit of modules is an example of this general categorical construction.

In the Category of sets, filtered colimits commute with finite limits. Elements in a finite limiting diagram involve only finitely many representatives and finitely many equalities, all of which can be realized at one common stage of a filtered diagram.

A fixed point of a group action is an element fixed by every group element. The set $A^G=\{a:g\cdot a=a\text{ for all }g\in G\}$ is the categorical limit of the action viewed as a functor from the one-object category associated with $G$ to the Category of sets.

A functor assigns an object of a category $F(A)$ to each object $A$ and a categorical morphism $F(f):F(A)\to F(B)$ to each arrow $f:A\to B$. It preserves the rules for composing arrows:For example, the forgetful functor from the category of groups to the Category of sets keeps the underlying sets and maps of elements while forgetting the group operations. A group homomorphism remains a function, and composing the underlying functions gives the underlying map of the composite group homomorphism.

Another example sends a set $X$ to its power set $\mathcal{P}(X)$. It sends a function $f:X\to Y$ to the function $\mathcal{P}(f):\mathcal{P}(X)\to \mathcal{P}(Y)$ defined by $S\mapsto f(S)$, the image of the subset $S$. For $f(x)=x+1$ and $g(x)=2x$, the subset $\{0,2\}$ goes to $\{1,3\}$ and then $\{2,6\}$; applying the composite $g\circ f$ directly also gives $\{2,6\}$. The equality of these routes expresses preservation of composition in a category.

An initial object has a unique morphism to every object.

For small categories and a set-valued functor $Q$, each displayed colimit exists in the Category of sets. A representative $(c,u:Fc\to d,x\in Qc)$ maps under a natural transformation $\eta :Q\to GF$ to $G(u){\eta }_c(x)$. The colimit identifications make this well-defined, proving the adjunction between left Kan extension and precomposition.

On the Category of sets, the list monad sends $X$ to all finite ordered lists of elements of $X$, including the empty list. Its unit inserts a singleton list and its multiplication concatenates a list of lists. Functoriality is entrywise application of functions. Ordered flattening proves the monad laws. This is the free-monoid monad, so order must not be discarded.

An object of ${\mathcal{C}}_n$ is a set with a finite list of partial unary operations. The first is total; the domain of each later operation consists exactly of the defined fixed points of the immediately preceding operation. Morphisms preserve every defined operation. Take ${\mathcal{C}}_0$ to be the Category of sets. This tower gives examples of arbitrary finite monadic length even though its long free–forgetful composites induce the same first monad.

For the Category of sets, take the monoidal tensor product to be the Cartesian product, $A\otimes B=A\times B$, with unit a singleton $1$. On morphisms use $(f\times g)(a,b)=(f(a),g(b))$. Define the associator by $((a,b),c)\mapsto (a,(b,c))$ and the unitors by $(\ast ,a)\mapsto a$ and $(a,\ast )\mapsto a$. These are natural isomorphisms, with inverses obtained by rebracketing or adjoining $\ast$. Both paths in the pentagon send $(((a,b),c),d)$ to $(a,(b,(c,d)))$, and both paths in the triangle send $((a,\ast ),b)$ to $(a,b)$. This proves the axioms elementwise.

There is a second structure: use disjoint union $A⨿B$ and unit $\varnothing$. Define the associator by preserving each element and its summand while changing the nested tags, and define the unitors by removing the empty summand. These are natural and invertible. In the pentagon, either path takes an element in any of the four summands to that same element in the same summand of $A⨿(B⨿(C⨿D))$; the triangle removes an empty intermediate summand on either path. Thus both axioms hold.

These product and coproduct monoidal structures on sets have different units, $1$ and $\varnothing$, which are not isomorphic. They are even inequivalent as monoidal categories: an equivalence of the underlying Category of sets preserves terminal objects, so it cannot take the singleton unit of the product structure to the empty unit of the coproduct structure. Hence the monoidal structure on sets is not unique.

The Category of sets is monoidal under Cartesian product, with any singleton as unit. The tensor of maps is $(f\times g)(x,y)=(f(x),g(y))$, the associator sends $((x,y),z)$ to $(x,(y,z))$, and the left and right unitors discard the singleton coordinate. These are natural bijections. Both pentagon paths send a nested quadruple to the same rebracketed quadruple, and both triangle paths discard the same singleton coordinate, verifying the axioms directly.

This is not the only structure. Disjoint union $X⨿Y$ is another tensor bifunctor, with empty-set unit. Its associator changes parentheses while retaining the tag indicating the original summand, and its unitors remove an empty summand. Every coherence path preserves both the original tag and the element, so again the pentagon and triangle hold.

These product and coproduct monoidal structures on sets are distinct even up to monoidal equivalence. Any equivalence of the underlying category preserves initial and terminal objects up to isomorphism. It therefore cannot carry the empty-set unit of the coproduct structure to the singleton unit of the product structure. Hence Set has at least these two inequivalent monoidal structures.

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.

Suppose every component ${\alpha }_A$ is a monomorphism in the Category of sets, hence an injective function. If $\beta ,\gamma :H\to F$ are natural transformations with $\alpha \beta =\alpha \gamma$, then ${\alpha }_A{\beta }_A={\alpha }_A{\gamma }_A$ at every object. Componentwise cancellation gives ${\beta }_A={\gamma }_A$, so $\beta =\gamma$. Thus $\alpha$ is a monomorphism in the functor category.

Conversely, suppose $\alpha$ is a monomorphism and $x,y\in F(A)$ satisfy ${\alpha }_A(x)={\alpha }_A(y)$. The Yoneda lemma supplies natural transformations ${\theta }^x,{\theta }^y:Y(A)\to F$. Its naturality in $F$ identifies $\alpha {\theta }^x$ and $\alpha {\theta }^y$ with the equal elements ${\alpha }_A(x)$ and ${\alpha }_A(y)$. Cancellation by $\alpha$ gives ${\theta }^x={\theta }^y$; evaluating at $1_A$ gives $x=y$. Therefore a natural transformation is a monomorphism exactly when all its components are injective functions. This is the pointwise monomorphism in a set-valued functor category criterion.

In the Category of sets, an epimorphism is exactly a surjective function. A surjective function is right-cancellable. If $q:X\to Y$ misses $y\in Y$, the constant-zero function and the function that is one at $y$ and zero elsewhere are distinct maps $Y\to \{0,1\}$ with equal composites with $q$.

If every ${\alpha }_A$ is surjective, equality $\beta \alpha =\gamma \alpha$ for natural transformations $\beta ,\gamma :G\Rightarrow H$ gives ${\beta }_A={\gamma }_A$ for every $A$. Thus $\alpha$ is an epimorphism in the functor category.

For the converse, construct the pointwise amalgamated doublewhere exactly the two copies of each element of $\mathrm{im}{\alpha }_A$ are identified; different elements of $G(A)$ remain different. Define $H(u)[y,i]=[G(u)y,i]$. Naturality of $\alpha$ implies that $G(u)$ takes its image into the image at the target, so this formula is well-defined and gives a functor. The maps $j_{i,A}(y)=[y,i]$ form natural transformations $j_0,j_1:G\Rightarrow H$, with $j_0\alpha =j_1\alpha$.

If $\alpha$ is an epimorphism, $j_0=j_1$. Since the two copies of an element outside $\mathrm{im}{\alpha }_A$ would be distinct, every element must lie in that image. Hence $\alpha$ is epic if and only if every component is epic. This proves the pointwise epimorphism in a functor category criterion directly, including the needed existence and naturality of the separating functor.