Enriched Categories

By [L1], the lozenge compares two composites from $\mathcal{A}(A,B)$ to $\mathcal{B}(TA,SB)$ obtained by first applying $S_{A,B}$ or $T_{A,B}$ and then composing with the component of $\alpha$ at the source or target. The canonical maps $\mathcal{B}({\alpha }_A,1)$ and $\mathcal{B}(1,{\alpha }_B)$ are precisely those postcomposition and precomposition maps written without the unitors.

After identifying $\mathbf{1}\otimes \mathcal{A}(A,B)$ and $\mathcal{A}(A,B)\otimes \mathbf{1}$ with $\mathcal{A}(A,B)$ by the unitors, the two composites in the lozenge become exactly $\mathcal{B}({\alpha }_A,1)\circ S_{A,B}$ and $\mathcal{B}(1,{\alpha }_B)\circ T_{A,B}$. Hence the lozenge equation holds if and only if those two morphisms are equal, which is exactly the commutativity of the displayed square.

Therefore the lozenge and compact-square formulations are equivalent.

Take objects to be the set-object $\mathcal{V}$-categories. For fixed $\mathcal{A},\mathcal{B}$, let the objects of the hom-category $\mathcal{V}\text{-}\mathbf{C}\mathbf{a}\mathbf{t}(\mathcal{A},\mathcal{B})$ be the $\mathcal{V}$-functors $\mathcal{A}\to \mathcal{B}$ from [L2], and let the morphisms be the $\mathcal{V}$-natural transformations from [L3]. The identity 2-cell on a functor $T$ has components $j_{TA}:\mathbf{1}\to \mathcal{B}(TA,TA)$ from [L1], and vertical composition of $\alpha :T\Rightarrow S$ and $\beta :S\Rightarrow R$ is defined componentwise by $\mathbf{1}\cong \mathbf{1}\otimes \mathbf{1}\stackrel{{\beta }_A\otimes {\alpha }_A}{\to }\mathcal{B}(SA,RA)\otimes \mathcal{B}(TA,SA)\stackrel{M^{\mathcal{B}}}{\to }\mathcal{B}(TA,RA)$. Associativity and the unit laws for this vertical composition follow directly from the associativity and unit diagrams of [L1].

Horizontal composition of 1-morphisms is ordinary composition of $\mathcal{V}$-functors: if $T:\mathcal{A}\to \mathcal{B}$ and $U:\mathcal{B}\to \mathcal{C}$, then $(UT{)}_{A,B}=U_{TA,TB}\circ T_{A,B}$ and the compatibility axioms follow from those in [L2]. Whiskering of 2-cells is also componentwise: $(U\alpha {)}_A:=U_{TA,SA}\circ {\alpha }_A$ and $(\alpha H{)}_A:={\alpha }_{HA}$. These formulas preserve naturality because [L2] and [L3] are already written against enriched composition.

Step 1.1 gives a category of 1-morphisms and 2-morphisms for every ordered pair of objects, and step 1.2 gives horizontal composition functors between those hom-categories. The associativity and unit laws are strict because they are literal equalities of composed functions and of componentwise composites. Thus [L4] applies.

On objects, define $T_0:{\mathcal{A}}_0\to {\mathcal{B}}_0$ by the same object map as $T$. On a morphism $f:\mathbf{1}\to \mathcal{A}(A,B)$, define $T_0(f):=T_{A,B}\circ f:\mathbf{1}\to \mathcal{B}(TA,TB)$. Because [L2] preserves enriched composition and identities, [L1] shows that $T_0$ preserves ordinary composition and identity morphisms.

On 2-cells, send $\alpha :T\Rightarrow S$ to the ordinary natural transformation ${\alpha }_0:T_0\Rightarrow S_0$ whose component at $A$ is exactly the same morphism ${\alpha }_A:\mathbf{1}\to \mathcal{B}(TA,SA)$ from [L3]. The enriched naturality equation of [L3] implies ordinary naturality after reading the hom-objects through [L1].

Identity functors, composite functors, identity 2-cells, and vertical and horizontal composites are preserved strictly because the definitions in steps 1.1 and 1.2 forget no object-level data and simply reuse the same component maps. Therefore the assignment is a strict 2-functor.

Assume first that $\mathcal{A}$ is enriched in $\mathbf{C}\mathbf{a}\mathbf{t}$. By [L1], each hom-object $\mathcal{A}(A,B)$ is a small category, and the object set of $\mathcal{A}$ is already a set. Because the monoidal product in [L3] is the cartesian product, the enriched composition morphism is a functor $\mathcal{A}(B,C)\times \mathcal{A}(A,B)\to \mathcal{A}(A,C)$, which is exactly horizontal composition on 1-morphisms and 2-morphisms. The identity morphism $\mathbf{1}\to \mathcal{A}(A,A)$ picks out an object of the hom-category, hence an identity 1-morphism. The enriched associativity and unit diagrams are therefore exactly the strict 2-category axioms of [L2].

Conversely, let $\mathcal{K}$ be a strict 2-category with a set of objects and small hom-categories. Use the hom-categories $\mathcal{K}(A,B)$ as the hom-objects. The horizontal-composition functor of [L2] supplies $\mathcal{K}(B,C)\times \mathcal{K}(A,B)\to \mathcal{K}(A,C)$, and each identity 1-morphism gives a functor $\mathbf{1}\to \mathcal{K}(A,A)$. Since the tensor product in [L3] is cartesian product, these data satisfy the definition of [L1].

Steps 1.1 and 1.2 are inverse unpackings of the same data, so the two notions agree exactly.

Suppose the base $\mathcal{V}$ is thin. Then for any objects $A,B$ of $\mathcal{A}$, the hom-set of the underlying ordinary category is $\mathcal{V}(\mathbf{1},\mathcal{A}(A,B))$. Because $\mathcal{V}$ has at most one morphism between any two objects, this hom-set has at most one element. The enriched identities and composition from [L3] become ordinary identities and composition, so the underlying category is a category all of whose hom-sets are subsingletons. Defining $A\le B$ when such a morphism exists gives a reflexive and transitive relation by [L2], hence a preorder by [L1].

Now take $\mathcal{V}=\mathbf{S}\mathbf{e}\mathbf{t}$ with cartesian product and singleton unit. Then a hom-object of [L3] is literally a set of morphisms, an identity map $\{\ast \}\to \mathcal{A}(A,A)$ chooses an identity element, and the composition morphism $\mathcal{A}(B,C)\times \mathcal{A}(A,B)\to \mathcal{A}(A,C)$ is ordinary composition of elements. The associativity and unit diagrams of [L3] are exactly the axioms of [L2]. Conversely, every small ordinary category gives such data by taking its hom-sets as the enriched hom-objects; its object collection is a set as required by [L3].

Therefore thin-base enrichment recovers a preorder on the object set, and Set-enrichment recovers precisely a small ordinary category.

Assume $\mathcal{A}$ is $\mathbf{A}\mathbf{b}$-enriched. By [L1], each hom-object $\mathcal{A}(A,B)$ is an abelian group. The composition morphism is a morphism in $\mathbf{A}\mathbf{b}$ out of the tensor product, so by [L3] it is exactly a bilinear map $\mathcal{A}(B,C)\times \mathcal{A}(A,B)\to \mathcal{A}(A,C)$. The identity map $\mathbb{Z}\to \mathcal{A}(A,A)$ selects the identity element in the endomorphism group. Thus the underlying ordinary category has abelian-group homs and bilinear composition, so [L2] makes it preadditive.

Conversely, let $\mathcal{C}$ be preadditive. By [L4], each hom-set is an abelian group and the two-variable hom-assignment is additive in each variable. So for every triple $A,B,C$, the bilinear composition map $\mathcal{C}(B,C)\times \mathcal{C}(A,B)\to \mathcal{C}(A,C)$ transposes uniquely, by [L3], to a group homomorphism $\mathcal{C}(B,C){\otimes }_{\mathbb{Z}}\mathcal{C}(A,B)\to \mathcal{C}(A,C)$. Sending $1\in \mathbb{Z}$ to the identity morphism of $A$ gives the required unit map $\mathbb{Z}\to \mathcal{C}(A,A)$. The ordinary associativity and unit laws are exactly the enriched ones after this transposition.

Steps 1.1 and 1.2 show that the two notions encode the same data.

By [L1], $\mathcal{A}$ is preadditive and has finite biproducts; in particular it has a zero object as part of that finite-biproduct structure.

The preadditivity from step 1.1 is exactly the hypothesis of [L2], so $\mathcal{A}$ is $\mathbf{A}\mathbf{b}$-enriched. The zero object and finite biproducts remain part of the underlying ordinary category because [L2] only rephrases the hom-structure.

Therefore every additive category has the stated enriched form.

By [L1], every ordered pair $(X,Y)$ has an internal hom-object $[X,Y]$. Take that object to be the enriched hom-object $\underset{‾}{\mathcal{V}}(X,Y)$.

Use the composition morphism of [L2] as the enriched composition map $\underset{‾}{\mathcal{V}}(Y,Z)\otimes \underset{‾}{\mathcal{V}}(X,Y)\to \underset{‾}{\mathcal{V}}(X,Z)$. For each $X$, use the unit morphism $\mathbf{1}\to [X,X]$ from [L2] as the enriched identity. The associativity and unit axioms required by Enriched category over a monoidal base are exactly the associativity and unit laws already proved in [L2].

Thus the data of steps 1.1 and 1.2 satisfy the definition of a $\mathcal{V}$-category. The bijection of [L3] explains why the global elements of the enriched hom-object recover the ordinary morphisms of $\mathcal{V}$.

Let $\mathcal{A}$ be $\mathbf{2}$-enriched. Define a relation on its object set by $A\le B⟺\mathcal{A}(A,B)=1$. Because the unit object of the base is $1$, the identity morphism of [L2] forces $\mathcal{A}(A,A)=1$ for every $A$, so the relation is reflexive. Since composition in the base is $\wedge$, the composition morphism $\mathcal{A}(B,C)\wedge \mathcal{A}(A,B)\to \mathcal{A}(A,C)$ implies that if both hom-objects on the left are $1$, then $\mathcal{A}(A,C)=1$ on the right. So the relation is transitive. By [L1], it is a preorder.

Conversely, given a preorder $(P,\le )$, put the object set equal to $P$ and define the hom-object by $\mathcal{A}(x,y)=1$ when $x\le y$ and $\mathcal{A}(x,y)=0$ otherwise. Reflexivity gives the identity maps, and transitivity gives the composition morphism because $1\wedge 1=1$ exactly in the composable case. Thus the preorder data satisfy [L2].

The two constructions are inverse restatements of the same information, so $\mathbf{2}$-enrichment and preorder structure are equivalent.

Let $\alpha :\mathcal{A}(K,-)\Rightarrow F$ be $\mathcal{V}$-natural. By [L4], its component at $K$ corresponds to an ordinary morphism ${\bar{\alpha }}_K:\mathcal{A}(K,K)\to FK$. Define $x_{\alpha }:={\bar{\alpha }}_K\circ j_K:\mathbf{1}\to FK$, where $j_K$ is the enriched identity of $K$. This is the evaluation map from the statement.

Conversely, let $x:\mathbf{1}\to FK$. For each object $A$, the functor structure of $F$ gives a morphism $F_{K,A}:\mathcal{A}(K,A)\to [FK,FA]$, and [L3] turns it into $F_{K,A}^{\mathrm{♯}}:\mathcal{A}(K,A)\otimes FK\to FA$. Compose with $1\otimes x$ and the right unitor to obtain ${\bar{\alpha }}_A^x:\mathcal{A}(K,A)\to FA$. By [L4], this determines a unique component ${\alpha }_A^x:\mathbf{1}\to [\mathcal{A}(K,A),FA]$.

The square criterion of [L2] for ${\alpha }^x$ is exactly the compatibility of $F_{K,-}$ with enriched composition: both sides become the same composite $\mathcal{A}(A,B)\otimes \mathcal{A}(K,A)\otimes FK\to FB$ after evaluating the internal homs and inserting $x$. So ${\alpha }^x$ is $\mathcal{V}$-natural.

Starting from $\alpha$, step 1.2 applied to $x_{\alpha }$ reconstructs the same family of components because the naturality square at $(K,A)$ sends the identity element $j_K$ to the value of ${\alpha }_A$ on $x_{\alpha }$. Starting from $x$, step 1.1 evaluates ${\alpha }_K^x$ at $j_K$ and recovers $x$ by definition. Hence the two constructions are inverse bijections.

Therefore evaluation at the enriched identity yields the stated bijection.

Fix an object $X$ of $\mathcal{V}$. An enriched wedge from $X$ to the displayed enriched end is, by [L2] and [L3], the same data as a family of morphisms $X\otimes \mathcal{A}(K,A)\to FA$ satisfying the enriched dinaturality equations. After transposing by [L2], these are exactly the components and enriched naturality equations of a $\mathcal{V}$-natural transformation $\mathcal{A}(K,-)\Rightarrow [X,F-]$, where $[X,F-]$ is obtained by applying $[X,-]$ objectwise in the self-enrichment of [L4].

Apply [L1] to the functor $G=[X,F-]$. This gives a bijection between the wedge data of step 1.1 and $\mathcal{V}(\mathbf{1},[X,FK])\cong \mathcal{V}(X,FK)$, the second bijection coming from [L2]. The correspondence is natural in $X$.

Step 2.1 says exactly that morphisms $X\to FK$ are in natural bijection with wedges from $X$ to the displayed diagram. By [L3], that is the universal property of the end, so $FK$ is the end ${\int }_A[\mathcal{A}(K,A),FA]$.

Therefore the particular end exists and is naturally isomorphic to $FK$.

By [L1], the hom-object from the contravariant representable $\mathcal{C}(-,A)$ to the contravariant representable $\mathcal{C}(-,B)$ is the end ${\int }_X[\mathcal{C}(X,A),\mathcal{C}(X,B)]\mathrm{.}$

Apply [L2] to the $\mathcal{V}$-category ${\mathcal{C}}^{\mathrm{o}\mathrm{p}}$, the object $A$, and the $\mathcal{V}$-functor $\mathcal{C}(-,B):{\mathcal{C}}^{\mathrm{o}\mathrm{p}}\to \mathcal{V}$. Since ${\mathcal{C}}^{\mathrm{o}\mathrm{p}}(A,X)=\mathcal{C}(X,A)$, it identifies the end of step 1.1 with $(\mathcal{C}(-,B))(A)=\mathcal{C}(A,B)$.

Therefore each hom-object map of the enriched Yoneda assignment is an isomorphism, which is exactly enriched full faithfulness.

Give $E$ its strict monoidal structure induced by addition: it has one object, both composition and tensor of endomorphisms are addition in $\mathbb{N}$, and commutativity gives the interchange law. Hence there is a one-object $\mathbf{C}\mathbf{a}\mathbf{t}$-enriched category $\mathcal{B}$ with sole object $C$, hom-category $\mathcal{B}(C,C)=E$, and enriched composition given by this tensor.

The underlying category ${\mathcal{B}}_0$ has one object and one morphism, since the objects of $E$ form a singleton. Thus, for its only test object $B=C$, there is a bijection ${\mathcal{B}}_0(B,C)\cong {\mathbf{C}\mathbf{a}\mathbf{t}}_0(X,\mathcal{B}(B,C)):$ both sides are singletons, because a functor from the discrete two-object category $X$ to the one-object category $E$ is unique on objects and identities. This bijection is automatically natural in the one-object category ${\mathcal{B}}_0$.

If $C$ were its own cotensor by $X$, [L1] would require an isomorphism of categories $E=\mathcal{B}(C,C)\cong [X,E]\cong E\times E\mathrm{.}$ But the endomorphism monoids of the unique objects are respectively $\mathbb{N}$ and ${\mathbb{N}}^2$, which are not isomorphic: the former has one indecomposable nonzero generator and the latter has two. Hence the natural underlying hom-set bijection of step 2.1 does not exhibit a cotensor.

Define ${\mathcal{L}}_{\mathcal{V}}$ to have the same objects as $\mathcal{L}$ and hom-object ${\coprod }_{\mathcal{L}(A,B)}\mathbf{1}$ from $A$ to $B$. The summand indexed by $f:A\to B$ is the enriched name of the ordinary arrow $f$, the identity map of $\mathcal{L}$ names the enriched identity, and ordinary composition in $\mathcal{L}$ induces the enriched composition morphisms by the coproduct universal property. Thus [L2] gives a $\mathcal{V}$-category.

Conversely, let $F:\mathcal{L}\to {\mathcal{B}}_0$ be an ordinary functor. On objects keep the same map. On each hom-object ${\coprod }_{\mathcal{L}(A,B)}\mathbf{1}$, define the corresponding map into $\mathcal{B}(FA,FB)$ by naming the ordinary morphism $F(f)$ on the summand indexed by $f$. The functoriality of $F$ makes these maps preserve identities and composition, so [L2] gives a $\mathcal{V}$-functor ${\mathcal{L}}_{\mathcal{V}}\to \mathcal{B}$.

Let $T:{\mathcal{L}}_{\mathcal{V}}\to \mathcal{B}$ be a $\mathcal{V}$-functor. For every ordinary arrow $f:A\to B$ in $\mathcal{L}$, the corresponding coproduct summand $\mathbf{1}\to {\mathcal{L}}_{\mathcal{V}}(A,B)$ followed by the hom-map of $T$ gives a global element $\mathbf{1}\to \mathcal{B}(TA,TB)$, hence by [L1] an ordinary morphism $TA\to TB$ in ${\mathcal{B}}_0$. Compatibility of $T$ with identities and composition makes these assignments an ordinary functor $\mathcal{L}\to {\mathcal{B}}_0$.

The same correspondence acts on $2$-cells. A $\mathcal{V}$-natural transformation $\alpha :T\Rightarrow S$ has components $\mathbf{1}\to \mathcal{B}(TA,SA)$, hence ordinary components $TA\to SA$ in ${\mathcal{B}}_0$ by [L1]. Its enriched naturality equation on the coproduct hom-object of ${\mathcal{L}}_{\mathcal{V}}$ holds exactly when the ordinary naturality square holds on every summand indexed by an arrow of $\mathcal{L}$. Thus enriched natural transformations correspond bijectively to ordinary natural transformations, compatibly with identities and composition.

Steps 1.2 and 2.1 give mutually inverse correspondences on functors, while step 2.2 gives the corresponding isomorphism on morphisms. They therefore define a natural isomorphism of hom-categories $[{\mathcal{L}}_{\mathcal{V}},\mathcal{B}{]}_0\cong [\mathcal{L},{\mathcal{B}}_0],$ which is the claimed $2$-adjunction.

Let $\mathcal{I}$ be the one-object $\mathbf{A}\mathbf{b}$-category with hom-object $\mathbb{Z}$, so its unique object has endomorphism object the tensor unit. Let $\mathcal{T}$ be the one-object $\mathbf{A}\mathbf{b}$-category with hom-object $0$, the terminal object of $\mathbf{A}\mathbf{b}$; the identity map $\mathbb{Z}\to 0$ and the zero composition make this a valid $\mathbf{A}\mathbf{b}$-category by [L3].

The underlying ordinary category ${\mathcal{T}}_0$ has one object and one morphism, because $\mathbf{A}\mathbf{b}(\mathbb{Z},0)$ is a singleton by [L2]. The underlying ordinary category ${\mathcal{I}}_0$ also has one object, with morphism set $\mathbf{A}\mathbf{b}(\mathbb{Z},\mathbb{Z})$. Sending the unique object of ${\mathcal{T}}_0$ to the unique object of ${\mathcal{I}}_0$ and its identity to $1_{\mathbb{Z}}$ therefore defines an ordinary constant functor ${\mathcal{T}}_0\to {\mathcal{I}}_0$.

A $\mathbf{A}\mathbf{b}$-enriched functor $\mathcal{T}\to \mathcal{I}$ would need a hom-object map $0\to \mathbb{Z}$ preserving the enriched identity. But the identity in $\mathcal{T}$ is the unique map $\mathbb{Z}\to 0$, and the identity in $\mathcal{I}$ is $1_{\mathbb{Z}}:\mathbb{Z}\to \mathbb{Z}$; preserving identities would force the composite $\mathbb{Z}\to 0\to \mathbb{Z}$ to be $1_{\mathbb{Z}}$, impossible because the composite through $0$ is the zero homomorphism. This contradicts [L1].

Hence the ordinary constant functor of step 2.1 has no enriched lift, so constant enriched functors need not exist.

By [L1], a conical limit uses the constant weight at the tensor unit. By [L2], weighted limits already include the one-object weights given by arbitrary objects $X\in \mathcal{V}$; these are the cotensors $X⋔C$. Thus conical weights constitute only the tensor-unit case of this family.

For $\mathcal{V}=\mathbf{A}\mathbf{b}$, the tensor unit is $\mathbb{Z}$, while $X=\mathbb{Z}\mathrm{/}2$ is a legitimate nonunit weight. Its weighted-limit universal property is $\mathcal{B}(B,(\mathbb{Z}\mathrm{/}2)⋔C)\cong [\mathbb{Z}\mathrm{/}2,\mathcal{B}(B,C)],$ whereas the conical weight on the one-object free enriched category is the constant weight $\mathbb{Z}$. Since $\mathbb{Z}\mathrm{/}2\cong ̸\mathbb{Z}$, these are different weights and different specified universal properties.

Thus the class of enriched weights is strictly larger than the class of conical weights, already over $\mathbf{A}\mathbf{b}$. A particular cotensor may nevertheless be computable from conical limits—for example, when it exists, $(\mathbb{Z}\mathrm{/}2)⋔C$ can be the conical equalizer of $2_C$ and $0_C$. The distinction proved here is between the weighted problems, not a prohibition on alternative constructions of their representing objects.

If $L$ is a conical enriched limit, then the enriched universal morphism gives, after applying the underlying-hom functor to each hom-object, exactly the ordinary cone bijection in the underlying category. So every conical enriched limit is an ordinary limit after forgetting enrichment.

So the conical enriched universal property is stronger than the underlying ordinary one: once the former exists, the latter follows by forgetting to global elements of the enriched hom-objects.