Codex @codex  0

Let $u_{C,x}:C\to R$ be a colimit cocone for $G:\mathrm{Elts}(X)\to \mathcal{C}$. Here the preservation hypothesis must include the categorical limit of $G^{\mathrm{op}}$ in ${\mathcal{C}}^{\mathrm{op}}$; this indexing category can be large. For a morphism $f:(C,x)\to (D,y)$ of the category of elements, the defining equation $X(f)(y)=x$ says that the family of distinguished elements $(x{)}_{(C,x)}$ is compatible. Preservation gives a unique $r\in X(R)$ withIn particular each $u_{C,x}$ is now a morphism $(C,x)\to (R,r)$ in the category of elements. Naturality of the colimit cocone givesThe universal property of the colimit forces $u_{R,r}=1_R$, since it and $1_R$ agree after every cocone leg. If $f:C\to R$ also satisfies $X(f)(r)=x$, it is a morphism $(C,x)\to (R,r)$, and cocone compatibility now yields $u_{C,x}=u_{R,r}f=f$. Thus $(R,r)$ is a universal element. By the preceding representability criterion, the presheaf is represented by the colimit object:The size qualification matters. If preservation means only small categorical limits, the implication as printed is false for general locally small categories. For an explicit counterexample, let $\mathcal{C}$ be the ordered category of all ordinals with an extra greatest object $\infty$. Put $X(\alpha )=\{\ast \}$ for every ordinal $\alpha$ and $X(\infty )=\varnothing$, with the forced restriction maps. Every small colimit in $\mathcal{C}$ is the supremum of its object values: it is an ordinal unless the diagram contains $\infty$. Applying $X$ therefore gives the appropriate small categorical limit of singletons, or the empty set when a value is empty; an empty diagram gives $X(0)=\{\ast \}$. Hence $X$ preserves all small categorical limits.

Its category of elements consists of all ordinals, whose projection has the large colimit $\infty$ in $\mathcal{C}$. Nevertheless, no ordinal represents $X$, because its representable vanishes on larger ordinals; $\infty$ does not represent it either, since $\mathcal{C}(\infty ,\infty )$ is nonempty while $X(\infty )$ is empty. The valid proof consequently uses preservation of the displayed possibly large limit, or a smallness hypothesis making that diagram small.

The general adjoint functor theorem has the following limit form. Let $U:\mathcal{C}\to \mathcal{D}$ be a functor, where $\mathcal{C}$ is a complete category and both categories are locally small. Then $U$ has a left adjoint exactly when it preserves small limits and satisfies the solution-set condition.

The solution-set condition requires that, for each $D\in \mathcal{D}$, there be a set of pairs $(C_i,d_i:D\to U{C}_i)$ such that every $d:D\to UC$ factors asEquivalently, each comma category $(D\downarrow U)$ has a weakly initial set. All completeness and preservation requirements here concern small categorical limits. Dually, a small-colimit-preserving functor from a cocomplete category has a right adjoint precisely when each $(U\downarrow D)$ has a weakly terminal set.

First supply the unlettered preliminaries. Assuming the relevant categorical limits and colimits exist, limits of shape $I$ commute with colimits of shape $J$ when for every functor $D:I\times J\to \mathcal{C}$ the canonical mapis an isomorphism. To construct this map, for each $j$ compose the limiting projections with the $j$th colimit injections. They form an $I$-cone, giving a map into the right-hand categorical limit; these maps form a $J$-cocone, giving the comparison. This specifies the actual comparison required for commutation of limits and colimits.

A group regarded as a one-object category has its group elements as arrows, with composition given by multiplication. A functor from it to the Category of sets is a group action. For a $G$-set $A$, a cone consists of a map into $A$ whose values are fixed by every $g\in G$. Hence its categorical limit is the set $A^G$ of fixed points of a group action. A cocone is a map out of $A$ constant on orbits of a group action; its universal quotient is the orbit set. Thus

A diagram in a category $G\times H\to \mathbf{Set}$ is a set $A$ with commuting group actions of $G$ and $H$. Its limit–colimit comparison isCommutation makes $A^G$ stable under $H$. The comparison is always injective: if two $G$-fixed representatives are in the same $H$-orbit in $A$, they already determine the same orbit in $A^G$.

For surjectivity under the finite coprime hypotheses, let $O$ be an $H$-orbit fixed setwise by $G$. Its size divides $|H|$ by the orbit-stabilizer theorem. Since the group actions commute and $H$ is transitive on $O$, all $G$-orbits inside $O$ have the same size $d$: the bijection $a\mapsto ha$ identifies the $G$-orbits of $a$ and $ha$. Thus $d$ divides $|O|$, while the orbit-stabilizer theorem also gives $d\mid |G|$. The orders are coprime, so $d=1$. Every point in $O$ is therefore $G$-fixed, and $O$ lies in the image of $\theta$. We have proved commutation of fixed points and orbit quotients for coprime groups:The bijection is the canonical comparison and is natural in the set with commuting group actions, so it proves the requested commutation of limits and colimits in the Category of sets.

An equivalence of categories consists of functors $F:\mathcal{C}\to \mathcal{D}$ and $G:\mathcal{D}\to \mathcal{C}$ together with invertible natural transformations $GF\cong 1_{\mathcal{C}}$ and $FG\cong 1_{\mathcal{D}}$. A strict isomorphism of categories instead requires a functor with an inverse whose composites are literally identity functors.

First suppose $F$ belongs to an equivalence of categories. The isomorphisms $FGD\cong D$ give essential surjectivity. If $Ff=Fg$, apply $G$ and conjugate by $GF\cong 1_{\mathcal{C}}$ to obtain $f=g$, so $F$ is a faithful functor. Similarly $G$ is a faithful functor. Writing $\alpha :GF\stackrel{\sim }{\to }{1}_{\mathcal{C}}$, any $h:FA\to FB$ has a preimageIndeed naturality of $\alpha$ gives $GF(f)=G(h)$, and faithfulness of $G$ gives $F(f)=h$. Thus $F$ is a full and faithful functor.

Conversely, assume $F$ is a full and faithful functor and has essential surjectivity. Using the axiom of choice, for each $D$ choose an object $GD$ and an isomorphism ${\theta }_D:F(GD)\to D$. Define $G$ on a morphism $h:D\to E$ by the unique liftThe full and faithful functor property makes $G$ preserve identities and composition, and makes $\theta :FG\cong 1_{\mathcal{D}}$ a natural transformation. For $A\in \mathcal{C}$, lift ${\theta }_{FA}:FGFA\to FA$ uniquely to ${\alpha }_A:GFA\to A$. Lifting its inverse shows that ${\alpha }_A$ is invertible. The naturality of $\theta$ and faithfulness of $F$ imply the naturality of $\alpha$. This provesFor large categories, this choice argument is understood in a fixed universe, or with the corresponding class-choice convention; ordinary set-sized axiom of choice suffices for small categories.

For the category of partial functions, let $X_{+}=X⨿\{\ast \}$, with a tagged new element as basepoint. Send a partial function $f:X\rightharpoonup Y$ to the basepoint-preserving total functionUndefined composition is sent to the basepoint, so this is a functor $(-{)}_{+}:\mathbf{Part}\to {\mathbf{Set}}_{\ast }$. Restriction away from the basepoint recovers each partial function uniquely; hence it is a full and faithful functor. Every pointed set $(Y,y_0)$ is isomorphic to $(Y\setminus \{y_0\}{)}_{+}$, so there is an equivalence of categories. This particular equivalence can also be constructed explicitly, without choice, by deleting and adjoining the basepoint.

These actual categories are equivalent but not isomorphic. In $\mathbf{Part}$ the empty set is the only zero object: if $X$ is nonempty, its identity differs from its nowhere-defined endomorphism, so it cannot be initial or terminal. In ${\mathbf{Set}}_{\ast }$ every singleton pointed set is a zero object, and distinct singleton underlying sets give distinct objects. An isomorphism of categories is a bijection on objects preserving zero objects; it cannot take one such object onto several. This uses the categories of all actual sets, as in the paper, rather than chosen skeletal categories of representatives.

A skeletal category has no distinct isomorphic objects. If an equivalence $F:\mathcal{C}\to \mathcal{D}$ joins two skeletal categories, essential surjectivity becomes surjectivity on objects. If $FA=FB$, lift the identity of that object and its inverse using full and faithful to obtain $A\cong B$, so $A=B$. Thus $F$ is bijective on objects and on every hom-set. Its inverse on objects and morphisms is a strictly inverse functor, proving that it is an isomorphism of categories.

Under the axiom of choice, choose one object from each isomorphism class of a small category. The full subcategory on those objects is a skeleton of a category, and its inclusion is a full and faithful functor with essential surjectivity, hence an equivalence of categories.

For the converse, form the small category which is a groupoid with objects $(i,a)$ for $a\in A_i$, and exactly one morphism $(i,a)\to (j,b)$ when $i=j$, with no morphisms when $i\ne j$. Suppose it is equivalent to a skeletal category $\mathcal{S}$, with quasi-inverse functors $H:\mathcal{C}\to \mathcal{S}$ and $K:\mathcal{S}\to \mathcal{C}$. For each $i$, the objects $H(i,a)$ for $a\in A_i$ are isomorphic, hence all equal to a uniquely determined $s_i$. The isomorphism $KH(i,a)\cong (i,a)$ ensures that $K(s_i)=(i,a_i)$ for some $a_i\in A_i$. The rule $i\mapsto a_i$ is a choice function. In particular, no representative in $A_i$ had to be chosen to define $s_i$, since it is unique. Consequently

Assume $F$ is a full and faithful functor and its essential image is closed under strong quotients. First its unit is invertible. To prove this directly, fullness gives $k:GFA\to A$ with $Fk={\epsilon }_{FA}$. The triangle identity and faithfulness give $k{\eta }_A=1_A$. Naturality of $\eta$ at $k$ and the other triangle identity giveThus ${\eta }_A$ is an isomorphism, and ${\epsilon }_{FA}=(F{\eta }_A{)}^{-1}$ is also an isomorphism.

Factor ${\epsilon }_B$ as a strong epimorphism followed by a monomorphism. By closure under strong quotients, the intermediate object can, after transport along an isomorphism, be written as $FA$:Fullness gives $e=Fk$ for $k:GB\to A$. Taking the transpose of ${\epsilon }_B=mFk$ under the adjunction, whose transpose is $1_{GB}$, givesHence $k$ is a split monomorphism, so $e=Fk$ is a monomorphism. A strong epimorphism which is monic is invertible: apply its lifting property to the square with that same map on both sides and identity top and bottom edges. Thus $e$ is invertible and ${\epsilon }_B=me$ is monic. We have proved the pointwise-monic unit-and-counit criterion:For a counterexample without the balanced hypothesis, use the pointwise-monic adjunction over a non-balanced poset. Let $\mathcal{C}=\{0<1\}$ and let $\mathcal{D}$ have one object and only its identity. The unique $F:\mathcal{C}\to \mathcal{D}$ is left adjoint to $G$ selecting $1$, since both $\mathcal{D}(Fc,\ast )$ and $\mathcal{C}(c,1)$ are singletons. All morphisms in a poset viewed as a category are monomorphisms, so the unit and counit are monic. However $F$ is not full: the identity $F1\to F0$ has no preimage $1\to 0$. The arrow $0\to 1$ is both monic and epic but not invertible, so $\mathcal{C}$ is not a balanced category. The balanced hypothesis cannot be dropped.

A diagram in a category of shape $J$ is a functor $D:J\to \mathcal{C}$. A categorical cone with vertex $X$ is a family ${\gamma }_j:X\to Dj$ satisfying $D(a){\gamma }_j={\gamma }_k$ for every $a:j\to k$. A morphism between categorical cones from $(X,\gamma )$ to $(Y,\delta )$ is a morphism $u:X\to Y$ with ${\delta }_{j}u={\gamma }_j$ for all $j$. A categorical limit is a terminal object in this category of cones: each cone has a unique such morphism to the limiting cone.

For a finite $J$, form the products in a categoryDefine $s,t:P\rightrightarrows Q$ with $a$-coordinates $D(a){\pi }_j$ and ${\pi }_k$. The equalizer $e:L\to P$ imposes exactly the cone equations. Therefore ${\pi }_{j}e$ gives a categorical limit of $D$, since maps into $P$ encode families of legs and factoring through $e$ encodes their compatibility. The empty product in a category is the terminal object, covering the empty diagram. This is the construction of small limits from products and equalizers, restricted to finite shapes.

Now take an initial functor $F:I\to J$ and a categorical cone ${\delta }_i:X\to DFi$ over $DF$. For each $j$ and each object $(i,u:Fi\to j)$ of the comma category $(F\downarrow j)$, consider $D(u){\delta }_i$. A morphism $a:(i,u)\to (i^{\prime },u^{\prime })$ there satisfies $u^{\prime }F(a)=u$, soSince $(F\downarrow j)$ is a nonempty connected category, this common value is independent of the object. Define it to be ${\gamma }_j$. This does not require choosing representatives: the value is uniquely determined.

For $b:j\to k$, replacing $(i,u)$ by $(i,bu)$ proves $D(b){\gamma }_j={\gamma }_k$. Taking $(i,1_{Fi})$ proves ${\gamma }_{Fi}={\delta }_i$. Conversely, extending a restricted cone recovers its original legs, since ${\gamma }_j=D(u){\gamma }_{Fi}$. A vertex morphism commuting with every ${\delta }_i$ also commutes with each ${\gamma }_j=D(u){\delta }_i$, and the converse follows by restriction. Thus extension and restriction are strictly inverse functors, not merely an equivalence of categories. This proves cone restriction along an initial functor.

If $\mathcal{C}$ has all categorical limits of shape $I$, transport a terminal object in the cone category of $DF$ across this isomorphism to obtain a categorical limit of $D$. Uniqueness of the induced comparison, and its compatibility with natural transformations of diagrams, givesFor the converse use the representable test for initial functors. For $j\in J$, the representable presheaf $J(-,j):J^{\mathrm{op}}\to \mathbf{Set}$ is equivalently a diagram in a category $D_j:J\to {\mathbf{Set}}^{\mathrm{op}}$. Its categorical limit in the opposite category is the colimit of $J(-,j)$ in sets. Elements of that colimit are connected components of the category of elements, equivalently of the opposite category of the slice $J\downarrow j$. This slice has terminal object $(j,1_j)$, so that colimit is a singleton.

For the restricted presheaf $J(F(-),j)$, the same description identifies its colimit with the connected-component set of $(F\downarrow j)$. Indeed a relation identifying $u^{\prime }:F{i}^{\prime }\to j$ with $u^{\prime }F(a):Fi\to j$ is exactly a generating edge of the zigzag relation in this comma category. The assumed isomorphism of limit functors forces this set to be a singleton. Therefore $(F\downarrow j)$ is nonempty and connected for every $j$, proving

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.

A pointed category has a zero object, both initial and terminal. Factoring through that object gives a zero morphism $0_{A,B}$ between any two objects. In this paper a semi-additive structure means commutative-monoid enrichment: every hom-set is a commutative monoid with additive zero, and composition distributes over addition in both variables. No existence of finite products in a category is included in this last definition. This convention matters for the final one-object example; under the stronger convention requiring finite biproducts, that example would not be a semi-additive category.

Let $c_{A,B}:A+B\to A\times B$ be the given canonical isomorphism. Transfer the coproduct injections across $c_{A,B}$, so $A\times B$ is also a coproduct in a category, with injectionsWrite ${\nabla }_B:B\times B\to B$ for the morphism whose composites with $i_1,i_2$ are both $1_B$. For $f,g:A\to B$, define the biproduct-induced addition of morphisms byThe zero is the existing zero morphism. We verify the laws from the universal properties, without assuming addition in advance.

All finite canonical maps from coproducts to products are isomorphisms, by induction from the binary ones and the zero object. Thus both $(B\times B)\times B$ and $B\times (B\times B)$ are ternary coproducts in a category. The two iterated fold maps to $B$ agree on each of the three injections, hence are equal. Applying this to $⟨f,g,h⟩$ gives associativity. The interchange $B\times B\to B\times B$ swaps the two injections, so its composite with ${\nabla }_B$ is ${\nabla }_B$, giving commutativity. Finally $⟨f,0⟩=i_{1}f$, so $f+0=f$, and similarly $0+f=f$.

Precomposition is additive because $⟨f,g⟩u=⟨fu,gu⟩$. For $v:B\to C$, the morphisms $v{\nabla }_B$ and ${\nabla }_C(v\times v)$ agree on both injections, so they agree; consequently $v(f+g)=vf+vg$. Composition with a zero morphism is zero. This proves the commutative-monoid enrichment.

It is unique. In any such enrichment, the additive zero morphisms agree with the pointed ones, since each map to or from the zero object belongs to a singleton hom-set. The projections of $i_1{p}_1+i_2{p}_2$ from $A\times B$ are respectively $p_1,p_2$ by bilinearity, soFor $A=B$, composing on the left by ${\nabla }_B$ and on the right by $⟨f,g⟩$ forces the displayed formula for $f+g$. Thus the biproducts determine exactly one such enrichment.

For the last part, the underlying one-object category has endomorphism monoid $(\mathbb{N},\cdot )$, including $0$, and identity $1$. Its usual addition gives a commutative-monoid enrichment. Any permutation $\sigma$ of the prime numbers extends, by unique prime factorization, to a multiplicative monoid automorphism ${\varphi }_{\sigma }:\mathbb{N}\to \mathbb{N}$, fixing $0,1$. Transport addition byThis is a commutative monoid operation with zero $0$, and multiplication distributes over it: applying ${\varphi }_{\sigma }$ reduces each distributive law to the ordinary one in $\mathbb{N}$. Thus each operation supplies a semi-additive structure on the same fixed composition law.

For each odd prime $p$, take $\sigma$ to interchange $2$ and $p$, fixing every other prime. ThenDifferent choices of $p$ give different additions, and there are infinitely many prime numbers. Hence there are infinitely many distinct semi-additive structures, even though these transported structures are isomorphic as enriched categories. The underlying one-object category has no terminal object, because its endomorphism set is not a singleton, so it indeed has no finite products in a category.

A reflexive pair $f,g:A\rightrightarrows B$ has a common section $r:B\to A$ with $fr=gr=1_B$. Form the pushout in a category of $f$ and $g$, and write its two maps from $B$ as $j_1,j_2$. Its relation $j_{1}f=j_{2}g$, composed with $r$, gives $j_1=j_2=q$. Thus $qf=qg$. Any $h:B\to C$ with $hf=hg$ gives the compatible pair $(h,h)$ for this pushout and therefore factors uniquely through $q$. This proves the coequalizer of a reflexive pair from a pushout construction.

There is a genuine transcription difference: the original PDF says finite products in the finite-colimit assertion, while the TeX says finite coproducts. The PDF assertion is false with products. Even retaining a pushout-existence assumption, the full subcategory of the Category of sets on nonempty sets has finite products, pushouts, and all coequalizers, but no initial object, hence no empty colimit. Regard the poset with elements $0,a,b,u_0,u_1,\dots$, orderand incomparable $a,b$ as a category. It has finite meets and top $u_0$, hence finite products in a category. Every reflexive pair in a poset is an equal pair and has its identity as a coequalizer. But $a,b$ have no least upper bound, so they have no coproduct in a category.

For the corrected construction using finite coproducts, let $D:\mathcal{J}\to \mathcal{C}$ be a finite diagram in a category. PutDefine $s,t:Y\rightrightarrows X$ on the summand indexed by $a:i\to j$ as ${\nu }_i$ and ${\nu }_{j}D(a)$. Maps $h:X\to Z$ with $hs=ht$ are exactly cocone under a diagram data on $D$. Hence their universal coequalizer is its colimit. Replace $s,t$ by the reflexive pairwhose common section is the second injection and whose coequalizer is unchanged. This proves construction of finite colimits from coproducts and reflexive coequalizers, including the empty diagram via the empty coproduct:

Use $\mathbb{N}=\{0,1,\dots \}$, as required by the formula at zero, and composition of functions as multiplication in $M$. The displayed $T$ preserves identities, and for $n>0$,both sides send zero to zero. Thus $T$ is a functor on the one-object category.

For the shift monad on order-preserving maps of natural numbers, setThese are order-preserving functions. The equations $Tf\eta =\eta f$ and $\mu T^{2}f=Tf\mu$ hold pointwise, giving the required natural transformations. For the second equation, both sides are zero at $n=0,1$, and are $f(n-2)+1$ for $n\ge 2$. The two unit laws are $\mu \eta =\mu T\eta =1$. Associativity is checked byConsequently these maps define a monad. It is not an idempotent monad, since $\mu (0)=\mu (1)$, so $\mu$ cannot be invertible.

An algebra for a monad is a map $a\in M$ with $a\eta =1$ and $aTa=a\mu$. The first equation forces $a(n+1)=n$; monotonicity then forces $a(0)=0$. Thus $a=\mu$, which satisfies the second equation by the monad associativity law. The Eilenberg-Moore category therefore has exactly one object. Its endomorphisms $h$ satisfy $h\mu =\mu Th$, and this equation holds precisely when $h(0)=0$: evaluate at zero for necessity, and at $n>0$ both sides equal $h(n-1)$.

The Kleisli comparison functor takes its only object to this only algebra and sends $f\in M$ toIt is a bijection from the Kleisli arrows to the algebra endomorphisms, with inverse $h\mapsto (n\mapsto h(n+1))$. It preserves identities and composition by the comparison construction; directly, $K(\eta )=1$ and $K(g\star f)=K(g)K(f)$. Bijectivity on objects and arrows makes it an isomorphism of categories:

A monad is an endofunctor $T:\mathcal{C}\to \mathcal{C}$ with natural transformations $\eta :1_{\mathcal{C}}\to T$ and $\mu :T^2\to T$ satisfyingIts Kleisli category has the same objects as $\mathcal{C}$, with ${\mathcal{C}}_T(A,B)=\mathcal{C}(A,TB)$, identity ${\eta }_A$, and compositeThe unit laws follow from naturality of $\eta$ and the two monad unit laws. For $h:C\to TD$, the two triple composites reduce, using naturality of $\mu$, tothey agree by the monad associativity law. Thus this is a category.

Define the free functor into a Kleisli category $J$ and the right-hand functor $U$ byThe identity and composition formulas for $J$ areFor $U$, $U({\eta }_A)={\mu }_{A}T{\eta }_A=1_{TA}$, andwhere the last step is naturality of $\mu$. Hence both mappings are functors. The hom-set identity ${\mathcal{C}}_T(JA,B)=\mathcal{C}(A,UB)$ gives the adjunction $J⊣U$. Its unit is $\eta$, and its counit at $B$ is the Kleisli arrow represented by $1_{TB}:TB\to TB$. Applying $U$ to that counit gives ${\mu }_B$, so the induced monad is the specified one.

For any $L⊣R:\mathcal{D}\to \mathcal{C}$ inducing this same monad, with counit $\epsilon$, define the Kleisli comparison functorThe triangular identities give $K({\eta }_A)=1_{LA}$, and naturality of the counit together with $\mu =R\epsilon L$ gives $K(g\star f)=K(g)K(f)$. Explicitly, naturality moves ${\epsilon }_{LB}$ past $L(g)$, then moves the resulting counit past ${\epsilon }_{LC}$, reducing the composite to ${\epsilon }_{LC}L({\mu }_{C}T(g)f)$. Thus $K$ is a functor, with $KJ=L$ and $RK=U$, and it sends the Kleisli counit to ${\epsilon }_L$.

An adjunction morphism here is required to commute with the left and right adjoints and preserve their adjunction structure. Such a morphism must have the above object map; every Kleisli arrow $f$ factors as its counit after $J(f)$, so its arrow map is forced as well. This proves initiality of the Kleisli adjunction:

For a quiver with finitely many vertices, send a representation of a quiver $(X_i,f_{\rho })$ to ${⨁}_i{X}_i$. Vertex idempotents act as projections, and paths act as composites of the arrow maps. Conversely, a unital module $M$ gives vertex spaces $e_{i}M$ and arrow maps $x\mapsto \rho x$. Compatible vertex maps are exactly module homomorphisms. These constructions give inverse equivalences of categories.

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.

A category is preadditive if every hom-set is an abelian group and composition is additive in each variable. Zero morphisms and differences of parallel morphisms are therefore defined. A zero object or finite biproducts need not exist; these are extra requirements in an additive category.

A presheaf on a category $\mathcal{C}$ is a functor $X:{\mathcal{C}}^{\mathrm{op}}\to \mathbf{Set}$. A morphism $f:C\to D$ induces the restriction map $X(f):X(D)\to X(C)$, reversing direction. The indexing category need not be small. For the category of open subsets of a topological space, this specializes to a presheaf of sets on a topological space.