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Stream: learning: questions

Topic: Free pseudomonoid

Asad Saeeduddin (Sep 15 2020 at 04:59):

John Baez said:

Asad Saeeduddin said:

this is probably a question for a different thread, but I was wondering if there is such a thing as a "free pseudofunctor to the bicategory of pseudomonoids" for any monoidal bicategory. if so, since Cat is a monoidal bicategory, does the free pseudofunctor to the bicategory of monoids of Cat give us on 1-cells (functors) the corresponding "free monoidal functor" in some sense?

I'm not sure I understand this. Let's see. In any monoidal bicategory C we can define pseudomonoids, and we get a bicategory (not monoidal!) of pseudomonoids in C - call it Ps(C).

There's always a forgetful functor U: Ps(C) $\to$ C since any pseudomonoid has an underlying object: it's an object together with some other stuff.

I imagine you're asking if U has a left adjoint F: C $\to$ Ps(C) sending any object to the free pseudomonoid on that object.

This clearly won't work unless C has enough 2-colimits!

It's probably best to start one level down: in any monoidal category C we can define monoids, and we get a category Mon(C) and a forgetful functor U: Mon(C) $\to$ C, and we can ask if this has a left adjoint F: C $\to$ Mon(C).

If C = Set with its cartesian monoidal structure, the answer is "yes", and we have

$F(S) = 1 + S + S^2 + S^3 \cdots$

So here we are taking a coproduct. In the category of finite sets this wouldn't work!

Asad Saeeduddin (Sep 15 2020 at 05:00):

So assuming our monoidal bicategory is "cocartesian" (in some sense suitable for bicategories), do we indeed have a free pseudofunctor to its bicategory of pseudomonoids?

Asad Saeeduddin (Sep 15 2020 at 05:01):

does the monoidal structure in question need to be related to the cocartesian structure in some way?

Asad Saeeduddin (Sep 15 2020 at 05:06):

or, to take the analogous specialization of C = Set: is there a 2-functor Cat -> MonCat from the 2-category of small categories and functors to the 2-category of monoidal categories and (lax/strong?) monoidal functors?

John Baez (Sep 15 2020 at 05:07):

There's too much to say about all this. But in the example I gave, of "the free monoid on a set", we took advantage of the fact that Set has countable coproducts, finite products, and products distribute over coproducts. That's what we use to show

$F(S) = 1 + S + S^2 + \cdots$

is a monoid and in fact the free monoid on $S$. So you can generalize this idea in various ways.

John Baez (Sep 15 2020 at 05:08):

Other examples work differently!

John Baez (Sep 15 2020 at 05:09):

Exercise: work out the multiplication in $F(S)$ and show it's associative.

Asad Saeeduddin (Sep 15 2020 at 05:12):

I think I can see how to generalize the concept of products distributing over coproducts via the idea of a distributive monoidal category (i don't think the actual projections of a product are relevant). But I'm not sure what the analog of "countability" for coproducts is

John Baez (Sep 15 2020 at 05:23):

I think a useful level of generality is a monoidal category with countable coproducts where the tensor product distributes over countable coproducts. In any category of this sort, the formula I gave (using tensor powers instead of cartesian powers) gives the free monoid. For example (Vect, $\otimes$) or (AbGp, $\otimes$) or (Set, $\times$) are all like this.... and lots more too.

John Baez (Sep 15 2020 at 06:11):

In fact all these examples have all coproducts, not just countable ones.