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?



Which functors between multicategories that come from monoidal categories are monoidal?
I guess that you can take the image under $F$ of the identity in $M\_C(;I)$ (the arity 0 is allowed).


