Enriched Categories — Examples

The hom-categories, horizontal composition, and identity 1-morphisms of $\mathcal{K}$ are exactly the data that [L1] identifies with the hom-objects, enriched composition, and enriched identities of a $\mathbf{C}\mathbf{a}\mathbf{t}$-enriched category.

Therefore every such strict 2-category is read directly as a $\mathbf{C}\mathbf{a}\mathbf{t}$-enriched category.

The displayed hom-object assignment is the standard construction used in [L1]. Reflexivity gives the enriched identities and transitivity gives enriched composition.

Hence the preorder $(P,\le )$ is an example of a $\mathbf{2}$-enriched category.

Because the unit of the base is $0$, the identity axiom for [L2] is exactly the statement $0\ge d(x,x)$, which holds by [L1].

The composition axiom in the reversed-order preorder is $d(y,z)+d(x,y)\ge d(x,z)$, which is exactly the triangle inequality from [L1].

Therefore every metric space is a Lawvere-style enriched category over the base $([0,\mathrm{\infty }],\ge ,+,0)$.

By [L1], $R$ is the endomorphism ring of a one-object preadditive category.

Applying [L2] to that preadditive category turns it into a one-object $\mathbf{A}\mathbf{b}$-enriched category, with the additive group of $R$ as hom-object and ring multiplication as enriched composition.

Read $\mathcal{K}$ as a $\mathbf{C}\mathbf{a}\mathbf{t}$-enriched category using [L1].

Then [L2] says that the hom-set in the underlying category is the set of objects of the corresponding hom-category of $\mathcal{K}$. Those objects are the 1-morphisms, while the 2-morphisms are morphisms inside the hom-category and are forgotten.

The universal property in [L2] is exactly the $\mathbf{S}\mathbf{e}\mathbf{t}$-instance of the cotensor formula from [L1].

Therefore the cotensor of $C$ by $X$ in $\mathbf{S}\mathbf{e}\mathbf{t}$ is the power $C^X$.

In the witness of [L1], the sole hom-category is the one-object category $E$ with endomorphism monoid $\mathbb{N}$, and the weight $X$ is the discrete two-object category. The candidate is the sole enriched object. Its underlying hom-set and the underlying set of $[X,E]$ are both singletons, so the required underlying bijection is present and is natural in the only test object.

The enriched cotensor property would require $E\cong [X,E]\cong E\times E$. This is impossible because the endomorphism monoids $\mathbb{N}$ and ${\mathbb{N}}^2$ are not isomorphic. Thus the candidate has the expected natural underlying hom-set bijection but is not a cotensor.

Apply [L1] with base functor $U:\mathbf{A}\mathbf{b}\to \mathbf{S}\mathbf{e}\mathbf{t}$.

The result keeps the same objects and replaces each abelian-group hom-object by its underlying set, so one recovers the ordinary category obtained by forgetting addition on homs.

If the statement were true, then conical enriched limits and underlying ordinary limits would coincide.

A concrete failure occurs for $\mathcal{V}=\mathbf{C}\mathbf{a}\mathbf{t}$. Let $\mathcal{M}$ be the strict $2$-category freely generated by objects $a,b$, a $1$-cell $f:a\to b$, and a nonidentity $2$-cell $\alpha :f\Rightarrow f$. Its underlying category is the walking-arrow category, so $b$ is the ordinary product $b\times b$. But a conical product would require $\mathcal{M}(a,b)\cong \mathcal{M}(a,b)\times \mathcal{M}(a,b)$ as categories. Here $\mathcal{M}(a,b)$ is the one-object category with endomorphism monoid $\mathbb{N}$, which is not isomorphic to its square. Thus this ordinary product is not conical.

Hence the converse of [L2] fails: an underlying ordinary limit need not satisfy the enriched conical universal property. The displayed statement is false.

The theorem [L1] gives an enriched category whose underlying ordinary constant functor has no enriched lift.

That witness contradicts the universal assertion in the statement. Hence the statement is false.

By [L1], distinct enriched hom-objects can induce the same underlying ordinary hom-set data.

Therefore the underlying ordinary category does not determine all enriched structure, so the statement is false.

The theorem [L1] states the precise equivalence and includes the smallness hypotheses as part of the claim.

Removing those hypotheses enlarges the statement beyond [L1], so the displayed unrestricted claim is false.

The theorem-page remark [L1] already identifies the failure: the component data may look ordinary, but the enriched naturality square asks for more.

So the statement that enriched naturality is only ordinary naturality is false.

The theorem [L1] proves only the existence of one end attached to one representable functor.

By [L2], that does not construct the whole enriched functor category for a large source, and [L3] does not change that size issue. Therefore the statement is false.

On $\mathbb{Z}\mathrm{/}2$-graded vector spaces, the ordinary swap $x\otimes y\mapsto y\otimes x$ and the Koszul swap $x\otimes y\mapsto (-1{)}^{\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }y\mathrm{\mid }}y\otimes x$ are two natural symmetries of the same tensor product. Because the characteristic is not $2$, these two maps differ on odd-degree simple tensors.

Since these two symmetries differ on odd-degree simple tensors, the monoidal category has more than one symmetry. Therefore the statement is false.