14
$\begingroup$

This question is about two definitions of enriched monoidal categories I have:

Let $\mathcal{V}$ be a symmetric monoidal closed category.

The first definition: a $\mathcal{V}$-enriched category $\mathcal{C}$ is a pseudomonoid object in the Day-convolution monoidal category $(\mathcal{V}\text{-}\mathbf{Cat}, \otimes_{\mathrm{Day}}, y(1))$. This is a straightforward generalization of the definition of monoidal categories but everything has to be worked externally and I don't really feel like to work in this definition (for example, I am pretty sure that a kind of coherence theorem holds but I don't know if I will be able to write down the proof).

The second definition can be found here (Bar constructions for topological operads and the Goodwillie derivatives of the identity, Definition 1.10): https://projecteuclid.org/download/pdf_1/euclid.gt/1513799607
In this paper, an enriched monoidal category $(\mathcal{C}, \overline{\wedge}, S)$ over $(\mathcal{V}, \wedge, I)$ is a $\mathcal{V}$-enriched category $\mathcal{C}$ which is tensored (denoted by $\otimes$) and cotensored, together with a monoidal structure on the underlying category $\mathcal{C}_0$ and the distribution natural transformation $d: (X\wedge Y)\otimes (C \overline{\wedge} D)\to (X\otimes C)\overline{\wedge}(Y\otimes D)$ satisfying certain associativity and unitality condition, which is a natural requirement for bar constructions.

So my question is: What is the relation between these two? It will be pleasing to be able to replace the first definition by the second one in other situations but I could not see any immediate implication from one to another or confluence under some conditions.

$\endgroup$

1 Answer 1

9
$\begingroup$

The two definitions are equivalent, for monoidal structures on $\mathcal{V}$-categories that are tensored over $\mathcal{V}$. I'll describe how the tensor product corresponds to the distributivity map:

The tensor product for a monoidal $\mathcal{V}$-category $\mathcal{C}$ in the first sense is given by specifying a tensor product on objects, and compatible maps $\mathcal{C}(C,D) \wedge \mathcal{C}(C',D') \to \mathcal{C}(C\overline{\wedge}C', D\overline{\wedge} D')$ in $\mathcal{V}$. So to show the second definition implies the first, we need to construct these maps. If $\mathcal{C}$ is tensored over $\mathcal{V}$, we can use the tensor product in $\mathcal{C}_0$ to get a natural map of sets $$ \mathcal{V}(X, \mathcal{C}(C,D)) \times \mathcal{V}(Y, \mathcal{C}(C',D')) \cong \mathcal{C}_0(X \otimes C, D) \times \mathcal{C}_0(Y \otimes C', D') \to \mathcal{C}_0((X \otimes C) \overline{\wedge} (Y \otimes C'), D \overline{\wedge} D').$$ Using the distributivity transformation we get a map $$ \mathcal{C}_0((X \otimes C) \overline{\wedge} (Y \otimes C'), D \overline{\wedge} D') \to \mathcal{C}_0((X \wedge Y) \otimes (C \overline{\wedge} C'), D \overline{\wedge} D') \cong \mathcal{V}(X \wedge Y, \mathcal{C}(C \overline{\wedge} C', D \overline{\wedge} D').$$ Now set $X = \mathcal{C}(C,D)$ and $Y = \mathcal{C}(C',D')$, then the pair of identity maps gets sent to the required morphism $\mathcal{C}(C,D) \wedge \mathcal{C}(C',D') \to \mathcal{C}(C\overline{\wedge}C', D\overline{\wedge} D')$.

On the other hand, if we have these maps then for $X,Y \in \mathcal{V}$, $C,D \in \mathcal{C}$, we can form the composite $$ X \wedge Y \to \mathcal{C}(C, X \otimes C) \wedge \mathcal{C}(D, Y \otimes D) \to \mathcal{C}(C \overline{\wedge} D, (X \otimes C) \overline{\wedge} (Y \otimes D)),$$ where the first map is the tensor product of unit maps for the cotensoring adjunction. This is then adjoint to the distributivity morphism $(X \wedge Y) \otimes (C \overline{\wedge} D) \to (X \otimes C) \overline{\wedge} (Y \otimes D))$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.