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Stream: theory: category theory

Topic: monomial composition comonoids

Jules Hedges (Jul 25 2024 at 16:52):

On the category of endofunctors on Set and natural transformations, there is a monoidal product given by functor composition. Comonoids for that monoidal product are exactly comonads. The monoidal product restricts to the full subcategory of polynomial endofunctors, and then comonoids become exactly categories. It also restricts further to the full subcategory of monomial endofunctors - does anybody know what its comonoids are?

Jules Hedges (Jul 25 2024 at 16:52):

(In case of doubt, a monomial endofunctor is one of the form $F (y) = A \times y^B$, or equivalently it's the coproduct of a single representable with itself set-many times)

Jules Hedges (Jul 25 2024 at 16:54):

My first guess was that they should be monoid actions, where you can view a monoid $M$ acting on a set $X$ as a category whose objects are $X$ and morphisms $X \times M$, where $(x, m) : x \to x \cdot m$. But my second guess is that that's too restrictive

Kevin Carlson (Jul 25 2024 at 17:14):

These should be arbitrary category structures on object set $A$ such that for every $a,$ the morphisms out of $a$ are in given isomorphism with the set $B.$ It doesn't seem like a particularly natural class, category-theoretically.

Kevin Carlson (Jul 25 2024 at 17:14):

For instance, you could have the disjoint union of two arbitrary monoid structures on $B$ and $A=2.$

Jules Hedges (Jul 25 2024 at 17:47):

Yeah, that sounds unpleasant

Jules Hedges (Jul 26 2024 at 16:10):

Here's my attempt to say what this is in the least horrible way I can. Health warning: I'm entirely eyeballing this. It's a 2-sorted algebraic theory given by functions:
(identity) $e : A \to B$
(codomain) $\bullet : A \times B \to A$
(composition) $c : A \times B \times B \to B$
... satisfying
(unit codomain) $x \bullet e(x) = x$
(composition codomain) $(x \bullet f) \bullet g = x \bullet c (x, f, g)$
(left unit composition) $c (x, e(x), f) = f$
(right unit composition) $c (x, f, e (x \bullet f)) = f$
(associativity) $c (x, f, c (x \bullet f, g, h)) = c (x, c (x, f, g), h)$

Kevin Carlson (Jul 26 2024 at 18:44):

That looks mathematically right. (Still ethically questionable though.)