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Stream: learning: questions

Topic: checking my understanding of (co)monoids in monoidal cats?

André Muricy Santos (May 25 2023 at 10:49):

everyone knows $\mathbb{Z}$ is a monoid with $+$ and $0$, and with $*$ and $1$! but im trying to understand better how is this couched in the categorical setting, by looking at the idea in https://ncatlab.org/nlab/show/monoid+in+a+monoidal+category

so what objects in what monoidal categories the examples give? $\mathbb{Z}$ is a set, so it has to be (Set,_, _) (seven sketches-style notation). + and * take two integers so the bifunctor has to be $\mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}$. The unit object then has to be $1$ since the unit arrow has to select the elements 0 and 1 for the two monoids respectively. so the monoidal category in which both of these are objects is (Set, $\times$, $1$).

but then i often hear talk about COmonoids in monoidal categories. reverse the arrows of the unit arrow and the multiply arrow. in this category what you get is _some piece of data_ (NOT an object in the category itself; what is it?) that contains the two arrows
$1 \rightarrow M$
$M \rightarrow M \times M$
a comonoid IN A MONOIDAL CATEGORY still relies on the monoidality of the category, despite comonoids being dual to monoids. its counit arrow is an arrow from the monoidal category's unit object to a (in this case) set, and its comultiplication relies on the multiplication bifunctor. Normally the co- structures are described in a very cleanly separated manner, one thinks of them (at least when learning) almost independently of the original structure, but with monoids this gets complicated for me at least. But this is the correct understanding, right?

Morgan Rogers (he/him) (May 25 2023 at 11:06):

The counit map for a comonoid should be $M \to 1$.

Beware that in a cartesian monoidal category (where the monoidal product is the categorical product), every object can be equipped with a comonoid structure in a unique way: the only choices for the counit and comultiplication maps which satisfy the axioms (dual to the monoid axioms) are the unique map $M \to 1$ and the diagonal map $M \to M \times M$. So you won't find interesting comonoids in $(\mathrm{Set},\times,1)$!

André Muricy Santos (May 25 2023 at 11:19):

Hah, nice. I figured from looking at the type signatures! Thanks Morgan :+1: