This question is about (not necessarily symmetric) monoidal categories enriched over a symmetric monoidal category $\mathcal{V}$. Assume that $\mathcal{V}$ is closed. You may also assume that $\mathcal{V}$ is (co)complete if you wish.

If $k$ is a commutative ring, a $k$ algebra can be defined in two ways. Either as a $k$-module $R$ together with morphisms $k\rightarrow R$ and $R\otimes_{k}R\rightarrow R$ satisfying the well-known laws, or as a ring homomorphism to the center $k\rightarrow Z(R)$.

Let's see what happens in the categorical context.

The tensor product of $\mathcal{V}$-enriched categories can be straightforwardly defined, see Kelly's book. Then one can define what a monoidal $\mathcal{V}$-category is by reproducing the classical definition in the enriched context.

Assume now that $\mathcal{C}$ is an ordinary monoidal category. I believe that the braided center $Z(\mathcal{C})$ as defined by Joyal and Street is a well known construction. Suppose that we have a strong braided monoidal functor $z : \mathcal{V}\rightarrow Z(\mathcal{C})$ such that the functor $z(-)\otimes Y : \mathcal{V}\rightarrow \mathcal{C}$ has a right adjoint ${Hom}_{\mathcal{C}}(Y,-) : \mathcal{C}\rightarrow\mathcal{V}$ for any object $Y$ in $\mathcal{C}$. The counit is an evaluation morphism in $\mathcal{C}$,

$ev: z( {Hom}_{\mathcal{C}}(Y,Z))\otimes Y\longrightarrow Z$

One can define composition morphisms in $\mathcal{V}$

${Hom}(Y,Z)\otimes {Hom}(X,Y)\longrightarrow {Hom}_{\mathcal{C}}(X,Z) $

as the adjoint of

$z({Hom}(Y,Z)\otimes {Hom}(X,Y))\otimes X \cong z({Hom}(Y,Z))\otimes z({Hom}(X,Y))\otimes X \stackrel{id \otimes ev}\longrightarrow z({Hom}(Y,Z))\otimes Y \stackrel{ev}\longrightarrow Z $

I think it's pretty obvious that $\mathcal{C}$ becomes $\mathcal{V}$-enriched in this way. Moreover, one can also enrich the tensor product in $\mathcal{C}$ in a similar way.

Do you guys agree? Do you know of any reference where this is checked with some detail? Is it even more obvious than I think?

Any comment is welcome!


4 Answers 4


There is a theorem in category theory, generally regarded as folklore, which says that for a symmetric monoidal closed category $V$, the following structures are equivalent:

  1. a category $C$ with an action $V\times C\to C$ of the monoidal category $V$ on $C$, which we may write as $(v,c)\mapsto v*c$, for which $-*c:V\to C$ has a right adjoint for each $c\in C$ (here the action amounts to a strong monoidal functor $V\to [C,C]$.
  2. a $V$-category $C$ for which the $V$-functor $C(c,-):C\to V$ has a left adjoint for each $c\in C$. (such a $V$-category is said to be "tensored'' or "copowered'')

You can see this, for example, in the appendix to this paper.

In your case, unless I've misunderstood, the centre $Z(C)$ plays little role. The point is that your functor $z:V\to C$ induces an action via $v*c=z(v)\otimes c$, and $-*c$ has a right adjoint by assumption, so you get the $V$-enrichment.

(There is an analogous characterization of $V$-categories $C$ which are cotensored/powered: this means that each $C(-,d):C^{op}\to V$ has a left adjoint.)

  • $\begingroup$ Tanks for your answer, Steve. Your comment actually answers many of my questions. Let me explain you the role of the center, maybe you can give me further references then. I'll do it below because the number of characters are limited in these comments. $\endgroup$ Jan 14, 2011 at 0:03
  • $\begingroup$ Sorry, I missed the bit of the question about the monoidal structure. It is known (not sure of a reference) that the monoidal structure lifts in the case where C is symmetric. But I can't recall having seen the situation you describe before. $\endgroup$
    – Steve Lack
    Jan 15, 2011 at 3:17

My "answer" is somewhere between "answer" and "comment", but is much too long to fit in the comment box, so I'll put it here. The point, though, is that I don't really answer the original question. I'm pretty sure that yes, your construction does turn any monoidal category $\mathcal C$ with a sufficiently nice functor $\mathcal V \to Z(\mathcal C)$ into an enriched category. I haven't checked the details, but they don't look hard. I don't know a good reference.

However, you will need more than just enriched monoidal categories in order to build an equivalence between the words "Monoidal category $\mathcal C$ enriched over $\mathcal V$", and "Monoidal category $\mathcal C$ with a braided monoidal functor $\mathcal V \to Z(\mathcal C)$". For example, let $\mathcal C$ be the category of finite-dimensional vector spaces, and $\mathcal V$ the category of all vector spaces, both with their usual tensor structures. Then $\mathcal C$ is certainly enriched over $\mathcal V$, but the corresponding functor $\mathcal V \to Z(\mathcal C)$ does not exist, because $\mathcal V$ is so much bigger than $\mathcal C$.

I do know of at least one situation where something like this should work. It might be broadly known, but I don't think so; we (re?)construct it in current joint-work-in-progress with A. Chirvasitu.

We propose that the correct notion of "2-abelian group" is locally presentable category. The correct 2-category with objects the 2-abelian groups is the one where a 1-morphism $A \to B$ consists of an adjoint pair, a left adjoint $f: A\to B$ and its right adjoint $A \leftarrow B$ (and all the extra stuff). The 2-morphisms are natural transformations of adjunctions. Since a left adjoint determines its right adjoint up to unique isomorphism, in fact we set the 1-morphisms to be precisely those functors that are left adjoints. (Recall that a functor between locally presentable categories is a left adjoint iff it is cocontinuous, and a right adjoint iff it is continuous and commutes with $\kappa$-filtered colimits for sufficiently large cardinals $\kappa$; so the 1,2-opposite category is the one whose 1-morphisms are the continuous $\kappa$-filtered-colimit-preserving functors.)

The category of 2-abelian groups is symmetric monoidal with the "tensor product" defined in the obvious way: it is straightforward to prove that for any two 2-abelian groups $A,B$, there is a 2-abelian group representing the category-valued 2-functor $\hom(A,\hom(B,-))$. (Note that for 2-abelian groups $B,C$, the category $\hom(B,C) = \operatorname{cocontinuous}(B \to C)$ is again a 2-abelian group.) Thus there is a natural notion of "2-ring" and "commutative 2-ring" and the morphisms between them, their modules, etc.

In any case, in this setting given a "commutative 2-ring" $A$ (i.e. a locally presentable category with a symmetric monoidal structure that is cocontinuous in each variable), there is an equivalence of 2-categories between: {commutative monoid objects in $A\text{-mod}$}, and {commutative 2-rings with a distinguished morphism from $A$}. (Note that these 2-categories are somewhat subtle. For example, just like {symmetric monoidal categories} is not full in {monoidal categories}, similarly "module of a commutative 2-ring" requires more data than "module of a 2-ring that happens to be commutative"; also the tensor structure in $A\text{-mod}$ requires some work.)

We don't consider non-commutative 2-algebra, so I'd have to go over the arguments again to see about relaxing the commutativity constraints; but it should work.


Let me clarify the role of the center in my question. The braided functor to the center is needed to extend the ordinary monoidal category structure on $\mathcal{C}$ to a monoidal $\mathcal{V}$-category structure.

A strong braided monoidal functor $z\colon \mathcal{V}\rightarrow z(\mathcal{C})$ is the same as a strong monoidal functor $z\colon \mathcal{V}\rightarrow \mathcal{C}$ together with natural isomorphisms $\zeta(X,Y) : z(X)\otimes Y \cong Y\otimes z(X)$ satisfying some coherence laws. We need these isomorphisms to define the $\mathcal{V}$-enrichment of the tensor product in $\mathcal{C}$: $$\otimes : Hom_{\mathcal{C}}(W,X)\otimes Hom_{\mathcal{C}}(Y,Z)\longrightarrow Hom_{\mathcal{C}}(W\otimes Y, X\otimes Z).$$ This morphism must be the adjoint of:

$\qquad z(Hom_{\mathcal{C}}(W,X)\otimes Hom_{\mathcal{C}}(Y,Z))\otimes W\otimes Y$

$\cong z(Hom_{\mathcal{C}}(W,X))\otimes z(Hom_{\mathcal{C}}(Y,Z))\otimes W\otimes Y$

$\stackrel{\zeta}\cong z(Hom_{\mathcal{C}}(W,X))\otimes W\otimes z(Hom_{\mathcal{C}}(Y,Z))\otimes Y\stackrel{ev \otimes ev}\longrightarrow X \otimes Z.$


Since I do have a reputation smaller then $50$ I have to write this comment as an answer. This is an old question, but I cannot help myself to remark that you do not need $\mathcal{V}$ to be closed in order to define the tensor product for $\mathcal{V}$-categories. The only thing you need is that $\mathcal{V}$ is symmetric. You can find this in Kelly's book, on page $12$ of the $2005$ reprint.

  • $\begingroup$ In fact I think braided monoidal is enough. $\endgroup$
    – Todd Trimble
    Nov 6, 2013 at 3:19

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