# What is the right notion of morphisms between strict monoidal categories? [Awodey, Exercise 4.5.9]

In Awodey's book "Category Theory", a strict monoidal category is defined as (Definition 4.9):

A strict monoidal category is a category $$\textbf{C}$$ equipped with a binary operation $$\otimes: \textbf{C} \times \textbf{C} \to \textbf{C}$$ which is functorial and associative, $$A \otimes (B \otimes C) = (A \otimes B) \otimes C$$, together with a distinguished object $$I$$ that acts as a unit, $$I \otimes C = C = C \otimes I$$.

As noted in Strict monoidal categories, the definition., this condition will need to hold both for objects and morphisms, so let's assume that for now.

Awodey goes on to define the category $$\textbf{Ord}_\text{fin}$$, the objects of which are the finite ordinal numbers, represented in the usual way as sets via $$0 = \emptyset, n+1 = \{ 0, \ldots, n \}$$, with $$+$$ as the binary operation and $$0$$ as the distinguished object, and the of the category arrows are all functions between the ordinals represented as sets. It is easy to see that the objects form a set satisfying the Peano axioms, and that $$+$$ and $$0$$ form a commutative monoid. It's also easy to verify that $$+$$ is functorial, by setting $$((f: m \to n) + (g: k \to \ell))(x) = \begin{cases} f(x) & x < m \\ g(x - m) + n & \text{otherwise}\end{cases}$$.

Exercise 4.5.9 then asks me to prove that $$\textbf{Ord}_\text{fin}$$ is the free monoidal category on one object.

My assumption is that we will have to show the following universal mapping property: Let $$U = \{ * \}$$ be a set with one element. Then (i) there is a map $$i: U \to (\textbf{Ord}_\text{fin})_0$$ picking an object, and (ii) let $$(\textbf{C}, \otimes, I)$$ be a monoidal category and $$f: U \to \textbf{C}_0$$ a function from $$U$$ to the objects of $$\textbf{C}$$ (i.e., we pick an object of $$\textbf{C}$$). Then there is an arrow $$\overline{f}: \textbf{Ord}_\text{fin} \to \textbf{C}$$ such that $$|\overline{f}|$$ (the part of $$\overline{f}$$ that operates on objects) satisfies $$f = |\overline{f}| \circ i$$, in analogy with the UMP of free monoids (for instance).

From my reading of the definitions, I would assume that an arrow of monoidal categories would, at the very least, be a functor such that $$f(m + n) = f(m) \otimes f(n)$$ and $$f(0) = I$$.

With this assumption, defining $$i$$ is straightforward (pick $$i(*) := 1$$), and it is also easy to define $$|\overline{f}|(n) = \otimes_{i=1}^n f(*)$$. My problem is that I don't see how to define $$\overline{f}$$ such that it is actually a functor: it seems that for any two finite ordinals $$m, n$$, there is at least one map from $$m$$ to $$n$$, except if $$m \neq 0$$, $$n=0$$ -- indeed, we can always take the map $$z(x) = 0$$

But for a given strict monoidal category, I don't see why there should always be a morphism from, say, $$A \otimes A$$ to $$A$$ for some object $$A$$ at all. But that would mean that I would not be able to define $$\overline{f}(z: 2 \to 1)$$, meaning that I wouldn't be able to make $$\overline{f}$$ a functor.

What mistake am I making? Am I using the wrong notion of arrow, or is there in fact a way to prove that there is always an arrow between $$\bigotimes^m A$$ and $$\bigotimes^n A$$, where $$\bigotimes^k A = \otimes_{i=1}^k A$$?

• What edition are you using? There is no exercise 4.5.9 in my edition. (Generally speaking, always specify the edition when you quote a book.) Apr 8, 2022 at 20:49
• Oh right - I'm using the second edition. Apr 8, 2022 at 21:09

This is not an answer to the update, rather it addresses more the original post. $$\def\Mon{\mathrm{Mon}} \def\SMonCat{\mathrm{StrictMonCat}} \def\Set{\mathrm{Set}}$$

Let $$\SMonCat$$ denote the category of (say small) strict monoidal categories together with strict monoidal functors. The left adjoint of the functor $$(C\mapsto C_0):\SMonCat\to\Set$$ is given by simply taking the free monoid over a set and regarding it as a discrete monoidal category. You can verify this directly.

Other way to see it is to observe that $$\SMonCat\to\Set$$ factors through the category of monoids, and check that the functor $$\SMonCat\to\Mon$$ has as left adjoint the functor that regards a monoid as a discrete category (so a monoid is the free strict monoidal category over itself). Then the left adjoint of $$\SMonCat\to\Set$$ is first go from sets to monoids and then include into $$\SMonCat$$.

There may be a mistake in the book can't find exercise 4.5.9 in my second edition, either. I haven't considered lax monoidal functors instead of strict monoidal ones, but I may check what happens later. It is weird, nevertheless, because at least in my edition there is no mention of monoidal functors of any kind.

• I see - the left adjoint gives the actual free functor here, so it's clear that the ordinals are not, in fact, free. Thank you! Apr 9, 2022 at 20:51

Partial progress: it seems that the construction of the functor as outlined above faces an insurmountable obstruction.

Consider the category $$\textbf{M}$$ whose objects are words on $$\{0, 1\}$$, and $$u \to v$$ exactly when $$u$$ is a subword of $$v$$. Take $$\otimes$$ to be concatenation, and $$I$$ to be the empty word. It is easy check that $$\textbf{M}$$ is a strict monoidal category - as a category, it's a poset, $$I \otimes x = x \otimes I = x$$ is clear, and if $$u$$ is a subword of $$u'$$ and $$v$$ a subword of $$v'$$, $$uv$$ is a subword of $$u'v'$$.

Now, note that by the construction outlined above, with $$f(*) = 0$$, we have $$\overline{f}(n) = 0^n$$. If $$f$$ was a functor, there would have to be an arrow $$\overline{f}(z: 2 \to 1) = \overline{f}(2) \to \overline{f}(1) = 0^2 \to 0^1$$, i.e., $$00$$ would have to be a subword of $$0$$, which is clearly not the case.

This still leaves open the question if there is another construction for $$\overline{f}$$ that makes it a functor, which is why I'm leaving the question open.