# Free monoidal category over a set

From nLab's article on coherence theorems, there seems to be a notion of free monoidal category over a set $S$.

I guess this corresponds to the left adjoint to the functor $Ob : MonCat \to Set$ which given a monoidal category $C$, it gives back the set of objects of $C$. The free monoidal category over $S$ would be a monoidal category $C$ together with a function $i : S \to Ob C$ such that for any other monoidal category $D$ and function $f : S \to Ob D$, I get a unique monoidal functor $\bar{f} : C \to D$ such that $Ob \bar{f} \circ i = f$.

After thinking of that definition, I thought that the objects of the free monoidal category over $\{ * \}$ would be $I$, $\{ * \}$, $\{ * \}^2$, etc.

Then, I thought that a similar presentation can be made for a free strict monoidal category over a set $S$, by replacing "monoidal" by "strict monoidal".

However, I started to wonder, wouldn't the free strict monoidal category over $\{ * \}$ satisfy the condition of being the free (non-necessarily strict) monoidal category over $\{ * \}$? If that was the case, I think the nLab page wouldn't make much sense. What's wrong here? Is my definition of free monoidal category?

The construction you describe is exactly similar to the free-strict-monoidal category but it is not the free-monoidal category: the difference being that in a strict monoidal category associators and left and right units are required to be the identities of the respective objects, a requirement that you did not add in your construction.

The free-monoidal-(biased)-category $F(S)$ over $S$ should have:

• as set of object the set of all the formal expressions you can build from the set of symbols in $S$ with the operators $\otimes$ and $1$
• as set of morphisms is generated by associators, left and right unit (and their inverses) through application of compositions and monoidal products, quotienting by the various equations needed for a monoidal categories (associativity and identities for compositions, functoriality of the monoidal product and most importantly coherence conditions).

One of the interesting things is that coherence conditions give you the ability to describe more explicitly the set of morphisms in this category: for each pair of objects in $F(S)$ you have exactly one morphism between two objects if and only if they are equivalent up to associativity and unit.

This data give you a category with the wished property: namely that for every monoidal category $\mathbf C$ and a function $f \colon S \to Ob(\mathbf C)$ there is a monoidal functor $\bar f \colon F(S) \to \mathbf C$ such that the equation $f=Ob(\bar f) \circ i$ holds, where $i \colon S \to Ob(F(S))$ is the obvious embedding sending every element of $S$ in itself seen as an element of $F(S)$.

P.s.: I realize that what I've described is not a fully formal description of what the free monoidal category over the set $S$ should be, but I hope that is enough to get the idea.

P.p.s.: above I've called $F(S)$ to distinguish it from the free-unbiased-monoidal categories described in Leinster's book Higher Operads, Higher categories.

• But doesn't the free strict monoidal category (the one I described, not all expressions as objects but only "normalized" ones) also satisfy this definition of free monoidal category? For which $\mathbb{C}$ and $f : S \to Ob(\mathbb{C})$ it doesn't hold? – user329838 Apr 8 '16 at 20:16
• Not if you are working in the $1$-category of monoidal categories and functor between them. To understand why try to answer this question: if your category were initial it should exist a functor into my category, what should be the image of an object $x_1 x_2 x_3$ through this functor? Should it be $(x_1 x_2)x_3$ or $x_1 (x_2 x_3)$? – Giorgio Mossa Apr 8 '16 at 20:22
• If I can choose any of the two, there is more than one $\bar{f}$ possible? So it is essential that $\bar{f}$ is unique? – user329838 Apr 8 '16 at 20:30
• If you work with $1$-categories I guess so. Nonetheless if you instead want to consider the $2$-category of monoidal categories and monoidal functor between them.... well I think the story could be different: I think (if not entirely sure) that in that case the two categories should be equivalent, but then you shouldn't have a unique functor $\bar f$, only one unique up to natural isomorphism. – Giorgio Mossa Apr 8 '16 at 20:35
• Just a speculation. – Giorgio Mossa Apr 8 '16 at 20:36