TITLE: Monoid under Day convolution and lax monoidal functor: Is strength necessary?
QUESTION [13 upvotes]: nCatLab states that a monoid in the Day-convolution monoidal category is equivalent to a lax monoidal functor. In the Haskell community this result is used to explain the construction of the applicative functor. Rivas and Jaskelioff show that a monoid in the endofunctor category $[Set, Set]$ with Day convolution is equivalent to a strong lax monoidal functor. They need strength to get the natural transformation $Id \to F$ from the (lax) preservation of the unit $1 \to F(1)$. I'm trying to understand why they need strength even though the nCatLab result doesn't. Granted, nCatLab deals with enriched functors, and self-enrichment is related to strength, but functors in $Set$ are both strong and self-enriched.
Instead of starting from $[Set Set]$, I'd like to start from a monoidal category $C$ and work with the functor category $[C, Set]$ with Day convolution. A monoid is a functor $F$ in this category, together with two natural transformations. I'm interested in one of them.
The unit under Day convolution is $C(I, -)$ where $I$ is the unit object in $C$. Consider the set of natural transformations from $C(I, -)$ to $F$:
$[C, Set](C(I), -), F) = \int_a C(I, a) \to F(a)$
By Yoneda, this is equal to $F(I)$. 
A function from $1$, the singleton set, will pick up a natural transformation in $[C, Set](C(I, -), F)$ and an element of $F(I)$:
$1 \to F(I)$
This is exactly one of the morphisms defining the lax monoidal functor. It seems like, in this formulation, strength is not needed. (It's not needed in the rest of the proof either.) What am I missing?

REPLY [2 votes]: You are correct. 
The most straightforward structure that you have on $[C, Set]$ is that of a multicategory (https://ncatlab.org/nlab/show/multicategory). Given functors $F_1, F_2, ..., F_n$ (n can be 0) and $F$ the set of multimorphisms $(F_1, F_2, ..., F_n) \rightarrow F$ is defined as the set
$$Nat(F_1(X_1)\otimes  ... \otimes F_n(X_n), F(X_1\otimes ... \otimes X_n)).$$
The multicategory structure is obvious.
The notion of monoid makes sense in any multicategory. Particularly, a monoid in the multicategory $[C, Set]$ is exactly a lax monoidal functor (not necessarily strong).
Furthermore, via Day convolution $[C, Set]$ happens to be a representable multicategory. More concretely, pretty much by (one of) the definitions of the Day convolution we have universal natural transformations
$$F_1(X_1)\otimes  ... \otimes F_n(X_n) \rightarrow (F_1\otimes_{Day}...\otimes_{Day}F_n)(X_1\otimes ... \otimes X_n)$$
which is the property guaranteeing representability (See C. Hermida, Representable multicategories). Specifically, the multicategory structure on $[C, Set]$ is represented by the Day monoidal structure on $[C, Set]$, which of course just means that
$$[C, Set]_{Mult}((F_1, F_2, ..., F_n), F) \cong [C, Set](F_1\otimes_{Day}...\otimes_{Day}F_n, F).$$ 
This implies that a monoid in the multicategory [C, Set], i.e. a lax monoidal functor as noted before, is the same as a monoid in the Day-convolution monoidal category [C, Set]. There is no need of further conditions.