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In several places in a segment on cohomology (for example, here (PDF)) in John Baez's online lecture notes for a course in 2007 on quantum gravity, much is made of the fact that the simplex category $\Delta$ (category of finite ordinals) is the "free monoidal category on a monoid". Supposedly this is the reason that this category is the initial object in the category of monoidal categories with monoid object.

But what does this mean? If it's defined in those lecture notes, I can't find it. Under some mild assumptions on the tensor product in a monoidal category $C$, we have a free monoid functor $F\colon C\to Mon(C)$, the adjoint of the forgetful functor, which takes objects $X\mapsto \coprod^\infty_{n=0} X^{\otimes n}.$ One says that $F(X)$ is the free monoid object generated by $X$. Can this be what Baez means? Not really, this is generated by an object, not a monoid. What if we take $X$ to be the underlying object of a monoid object? Still not right, the fact that the object generating the free monoidal object had its own monoidal structure seems to have no effect on the resulting free monoid object. And anyway we're looking for a monoidal category, but we only got a monoid object.

What if we take $C=\mathbf{Cat}$, the category of categories? Then we have $Mon(\mathbf{Cat})=\mathbf{MonCat}$ is the category of monoidal categories, and the free functor $F\colon \mathbf{Cat}\to \mathbf{MonCat}$ does give us a monoidal category. Can this be what Baez means? If $C$ is a category, then $F(C)$ is a category whose objects (resp. morphisms) are tuples of objects (resp. morphisms) in $C$. Now it makes better sense why we can take the free monoidal category generated by a monoid: in this context, a monoid $M$ is just a category with one object, a monoid object in $\mathbf{Set}$, and an object of $\mathbf{Cat}.$ So a tuple of a single object is just a natural number. This is starting to look like $\Delta$, at least the objects are right.

But what are the morphisms? They are tuples of morphisms in the original category, which was a category with one object, but apparently no restrictions on the morphisms. If $M$ was a nontrivial monoid, say $x,y\in M$ with $x\neq y$, then the object 1 in the free monoidal category $F(M)$ will have two distinct endomorphisms. However in $\Delta$, there is only one morphism $1\to 1.$ So $\Delta$ is not the free monoidal category generated by a monoid, unless that monoid is the trivial monoid. Is that what Baez means here? $\Delta$ is the free monoidal category generated by the trivial monoid viewed as a category with one object (and only the trivial arrow)?

He develops the idea also in his This Week's Finds column in 2001. There he says "free monoidal category on a monoid object", and the lack of specificity makes me think that not only does it not matter what monoid you start with, it doesn't even matter what monoidal category you take your monoid object from: you'll still get $\Delta$. But that's just not what I'm seeing. Can someone set me straight?

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I think you are interpreting it as "(free monoidal category) on a monoid", while it's really "free (monoidal category on a monoid)", or rather "free (monoidal category with a monoid)".

Now to construct it you have have to find which morphisms should exists as a consequences of the axioms, i.e. which morphisms you are sure to see whenever you have a monoid in a monoidal category. Clearly, if $X$ is an object in a monoidal category $C$, the only morphisms between $X^{\otimes n}$ and $X^{\otimes m}$ that exists whatever $(C,X)$ are, are the identity when $m=n$. So the free monoidal category is just $\mathbb{N}$ whith $\hom(m,n)=${id} if $m=n$, and is empty otherwise.

But if $X$ is a monoid, you get by definition a morphism $X \otimes X \rightarrow X$ and a morphism $I \rightarrow X$ where $I$ is the identity object. These morphisms clearly extends to morphisms $\delta_i:X^{\otimes n+1}\rightarrow X^{\otimes n}$ and $\sigma_i: X^{\otimes n}\rightarrow X^{\otimes n+1}$ just by tensoring them with identity morphisms, and then you can compose them to get maps $X^{\otimes m}\rightarrow X^{\otimes n}$. Finally, it's easy to see that the commutation relation between these morphisms imposed by the axioms of a monoid are precisely those satisfied by the simplicial maps in $\Delta$.

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It seems to me that the answer to your question is at the end of your first paragraph: "the initial object in the category of monoidal categories with monoid object". A monoidal category with monoid object is a pair $(C,M)$ where $C$ is a monoidal category and $M$ is a monoid-object of $C$ (i.e., $M$ consists of an object of $C$, also called $M$, with a multiplication morphism $M\otimes M\to M$ and unit $I\to M$ satisfying the usual laws for monoids). A morphism from $(C,M)$ to $(C',M')$ is a monoidal functor $F:C\to C'$ such that $F(M)=M'$. That defines the category of monoidal categories with monoid object, and what you want is the initial object there.

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So you've defined "initial object in the category of monoidal categories with monoid object". Do you propose to also take that as the definition of "free monoidal category on a monoid object"? I guessed "free monoidal category" meant something in terms of an adjoint to a forgetful functor, as is usual with free monoids and related. –  Joe Hannon Dec 5 '12 at 17:59
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@ziggurism: yes, that is precisely what Andreas proposes (technically, "initial" should be "2-initial"; roughly speaking, initial up to unique isomorphism), and is what John Baez meant. Sometimes among category theorists, "free" is used as a synonym for "initial". –  Todd Trimble Dec 5 '12 at 18:30
    
Following on from what Todd says, let me give an example to show how "free" and "initial" come to be used interchangeably: the free group on one generator is initial among all (groups equipped with an element). That is, in the category of pairs $(G,g)$ where $G$ is a group and $g\in G$, the initial object is the free group on one generator together with its generator. –  Tom Leinster Dec 5 '12 at 23:58
    
@Todd: It sounds like you, Andreas, and Adrien are all saying the same thing. It's "free" in name only. It's free because initial objects are like free objects. By analogy, the initial object in the category of monoids with maps from set $S$ is $F(S)$, the free monoid of words in $S$. I find this answer somewhat dissatisfying, it seems somehow circular. How about this: there's a forgetful functor $\mathbf{MonCat}\times Mon(C)\to Mon(C)$... hmm no that doesn't make sense. I guess we have the "evaluation at 1" functor $C^{\Delta}\to C$, for $C$ monoidal... no, that's not going to work. Hmmm –  Joe Hannon Dec 6 '12 at 3:07
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It's not that "initial objects are like free objects"; rather, free objects are a special case of initial objects. To use an example that you apparently understand, the free monoid on a set $S$ is the initial object in the category of (monoids with a map from $S$ into their underlying sets). Similarly for other left adjoint functors. The left adjoint (if it exists) of a functor $U:C\to D$ sends an object $S$ of $D$ to the initial object in the category of (objects $X$ of $C$ with a morphism from $S$ to $U(X)$). –  Andreas Blass Dec 6 '12 at 4:01
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