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Given any monoidal category $(\mathcal{C}, \otimes, I)$ we have an associated multicategory $M_{\mathcal{C}}$ with underlying category $\mathcal{C}$, and $k$-morphisms $M_{\mathcal{C}}(A_1,\ldots,A_k; B) = \mathcal{C}(A_1\otimes\cdots\otimes A_k, B)$. Given any (lax) monoidal functor $F\colon \mathcal{C}\rightarrow \mathcal{D}$ we get a multifunctor $M_\mathcal{C}\rightarrow M_\mathcal{D}$.

Now suppose that we have two monoidal categories $(\mathcal{C}, \otimes, I)$ and $(\mathcal{D}, \boxtimes, I')$, and a multifunctor $F\colon M_\mathcal{C}\rightarrow M_\mathcal{D}$. On the underlying categories $\mathcal{C}$ and $\mathcal{D}$, is the functor $F$ monoidal?

In order for $F$ to be monoidal we need two types of structure maps. It's clear how to get the structure map $F(A)\boxtimes F(B) \rightarrow F(A\otimes B)$: just look at the image under $F$ of the identity in $M_\mathcal{C}(A,B;A\otimes B)$. But we also need a morphism $I'\rightarrow F(I)$ (which is nicely coherent), and I don't see where in the structure of the multicategory that could be encoded.

So does this mean that there are $F$'s that don't come from monoidal functors, or is there some other structure that guarantees that there is an underlying monoidal functor?

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3 Answers 3

up vote 6 down vote accepted

I guess that you can take the image under $F$ of the identity in $M_C(;I)$ (the arity 0 is allowed).

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See Leinster's "Higher Operads, Higher Categories" Example 2.1.10. Maps of multicategories which come from monoidal categories are precisely the same as lax monoidal functors.

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14  
Thanks for the precise reference, David. I was looking in that book, and cursing it because I couldn't locate this fact. –  Tom Leinster Feb 13 '12 at 0:45

Lax monoidal functors $\mathcal{C} \to \mathcal{D}$ correspond one-to-one with multifunctors $M_\mathcal{C} \to M_\mathcal{D}$. Indeed, the notions of transformation are compatible too, so that you get an isomorphism of categories. As Guillaume says, there's nothing special about the arity $0$.

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