Is there a sensible way to enrich over SymMonCat such that id_X is not the monoidal unit?

SymMonCat is the cartesian 2-category of symmetric monoidal categories, braided monoidal functors, and monoidal natural transformations. The terminal symmetric monoidal category 1 has one object $I$ and $I \otimes I = I$.

A category enriched over a monoidal category $V$ assigns to each pair of objects $X, Y$ an object hom$(X,Y)$ in $V$ and to each object $X$ a morphism $id_X:I \to \mbox{hom}(X,X)$ in $V$.

When $V =$ SymMonCat, the morphism $id_X:1 \to \mbox{hom}(X,X)$ is a braided monoidal functor; since monoidal functors preserve the monoidal unit and tensor product, it must map the unit $I$ in 1 to the unit $I$ in hom$(X,X)$.

Is there a different way of enriching over SymMonCat such that $id_X$ does not pick out the monoidal unit (other than considering it a subcategory of Cat)?

• I've added the ct.category-theory tag. – Finn Lawler Nov 9 '10 at 21:24
• Symmetric monoidal categories are something like commutative rings, and so the natural monoidal structure to put on SymMonCat is the tensor product of categories. If you add the right adjectives and construct the tensor product correctly, then like in the ring case it is the coproduct of symmetric monoidal categories. In particular, you should not expect the kind of "hom-tensor adjunction" that you usually want when writing down a good theory of enriched categories. (continued) – Theo Johnson-Freyd Nov 9 '10 at 22:20
• (continuation) You can try to work with SymMonCat^{op}, which is something like "affine schemes"; but then for a good theory you need to extend to more general "algebraic spaces". (More precisely, symmetric monoidal categories are essentially categories of QCoh sheaves on stacks --- all stacks are "affine" in this sense --- so really I'm talking about more general "nonaffine stacks" in some sense.) But, anyway, before answering your question at the 1/2 level, try it a level lower: what do you want the answer to be for categories enriched over CommutativeRings? – Theo Johnson-Freyd Nov 9 '10 at 22:23
• I think symmetric monoidal categories are more like abelian groups than they are like commutative rings. The analogue of a ring would be a bimonoidal category. For SMCs, the tensor is not the coproduct---but also the unit object is not the terminal category, but rather the free symmetric monoidal category on one generating object, which is the disjoint union of the delooped symmetric groups. Just as the unit for the tensor product structure on abelian groups is the integers, i.e. the free abelian group on one generator. – Mike Shulman Nov 10 '10 at 2:59
• I was thinking of the one defined by Hyland and Power, which acts on strong monoidal functors, since I suspect from your question that you are considering only strong monoidal functors. Schmitt's deals instead with lax monoidal functors. – Mike Shulman Nov 10 '10 at 21:19

[Ignore this first part, I'm just leaving it for the context to the comments below.] It is hard for me to understand why you would want to enrich in symmetric monoidal categories, have an identity, and also want this identity to not be the unit of the symmetric monoidal category.

That said, you can always do away with units altogether and consider "enriched categories without identities". Is this what you are after?

After Mike's example I am now on board. What you probably want to do is enrich over the symmetric monoidal 2-category of symmetric monoidal categories where the monoidal structure is the "tensor product of symmetric monoidal categories". What is this you ask?

The functor category between two symmetric monoidal categories $Fun^\otimes(B,C)$ is naturally equipped with a symmetric monoidal structure (using pointwise multiplication). The tensor product of symmetric monoidal categories is $(-) \otimes B$ is the (weak) left adjoint to the functor $Fun(B, -)$. Thus $A \otimes B$ is a symmetric monoidal category such that symmetirc monoidal functors from it to C are the same as "bilinear" functors $A \times B \to C$. Now the monoidal unit for this tensor product is the free symmetric monoidal category on one object $\mathbb{F}$ (which is the category of finite sets and permutations).

In this way, if you enrich in (SymCat, $\otimes$) you get a unit being a functor $\mathbb{F} \to Hom(a,a)$, which is equivalent to just some element, not necessarily the unit object of $Hom(a,a)$.

The prototypical example is the 2-category of symmetric monoidal categories itself.

• As an example, the category of symmetric monoidal categories is itself enriched over symmetric monoidal categories. The symmetric monoidal structure on SymMonCat(C,D) is pointwise, which means that its monoidal unit is the functor constant at the unit object of D; whereas of course the identity morphism in SymMonCat(C,C) is the identity functor, not the functor constant at the unit object. – Mike Shulman Nov 10 '10 at 21:17
• Ahh. I See, we are not using the cartesian product of symmetric monoidal categories, we are using the tensor product of symmetric monoidal categories. – Chris Schommer-Pries Nov 11 '10 at 22:27
• I see that Mike Shulman has also suggested this same approach in the comments to the main question. – Chris Schommer-Pries Nov 12 '10 at 1:58
• Thanks to both of you! I voted up Mike's comment, but since it wasn't given as an "answer", I'm accepting Chris' answer. – Mike Stay Nov 12 '10 at 16:18

I suppose that the monoidal structure for $SymMonCat$ you mean is the cartesian one, and as braided functor you mean a pseudo-monidal symmetrical (funtors that commutes with the canonical isomorphism of symmetry, and the coherence morphism data are isomorphisms). Then $\mathcal{C}(X, X)\in SymMonCat$ has a internal symmetric monoidal structure and another monoidal structure given by monoidal moltiplication $\mathcal{C}(X, X)\times\mathcal{C}(X, X)\to \mathcal{C}(X, X)$ with the realtive axioms related to a $SymMonCat$-enriched category. Then $\mathcal{C}(X, X)$ is a bimomoidal category, and the two “monidal Identity" for the two monoidal structures are $I_X$ (for the internal monoidal structure) and the (essential) image of the morphism you called $id_X$, and a very reduced example of this (de-categorification) is a $rig$, a algebraic structure by two monoidal law, one abelian. In general the two units are different (trivial example $(\mathbb{N}, +, 0 ; \times , 1)$.

• I've discovered that I don't want the cartesian product, for the same reason that a one-object category enriched over (Ab, $\times$, 1) is just an abelian group, not a ring. You have to enrich over (Ab, $\otimes$, $\mathbb{Z}$) to get a ring. So my question has become how to define $\otimes$ properly on SymMonCat and what its monoidal unit is. – Mike Stay Nov 10 '10 at 19:19

A monoidal functor preserves the unit and tensor only 'laxly', so that $1_X$ is not the local unit but comes with a 2-cell $I \Rightarrow 1_X$.

If you're talking about strong monoidal functors then enriching over SymMonCat will by definition imply that identities are (isomorphic to) local units.

I don't know what you're trying to do exactly, but it seems it's just a matter of picking the right kind of functor, and thus the right category to enrich over.

• Thinking about it some more, I guess I have the same question about Ab-enriched categories, aka ringoids. A one-object Ab-enriched category is a ring; multiplication is composition and addition comes from the abelian group structure. Every object $X$ comes with a group homomorphism from the trivial group to hom$(X,X)$; this preserves the unit and products, so $id_X$ has to pick out the additive identity, not the multiplicative identity. Where am I going wrong? – Mike Stay Nov 9 '10 at 21:27
• Just thought of at least one place I'm going wrong: the monoidal unit in (Ab, $\otimes$) is $\mathbb{Z}$, not the trivial group. Is there a similar tensor product on SymMonCat? – Mike Stay Nov 9 '10 at 22:10
• Many people use the word "monoidal functor" for "strong" monoidal functors, and mark "lax" or "oplax" as needed. Grammatically, this is formally similar to the habit of using the word "ring" for commutative ring, and "noncommutative ring" for something that may or may not be commutative; see also "noncommutative geometry". – Theo Johnson-Freyd Nov 9 '10 at 22:16

Yes, the tensor product I want is the one described by Vincent Schmitt; it satisfies the universal property that any other braided monoidal bifunctor is naturally isomorphic to it.