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When defining a distributive law between endofunctors, I think one keeps in mind the canonical example of distributing over and says that a 2-cell is a "distributivity" of over , and not the other way round: distributivity of products over sums is indeed a rule to transform a product of sums, , into a sum of products, , so a map . So:
However, this definition has a problem: when trying to extend this to a "distributivity" of over an -ary functor, for example the tensor of a monoidal structure, lax monoidality means that there is a codistributivity of over in the sense that there exists a 2-cell , and instead distributivity captures oplax monoidal functors.
Is this clash of notation just an unfortunate naming? Is it fixable? Am I mistaken somewhere?
(Maybe it's not completely straightforward why I'd like to talk about "distributivity" between F and a tensor functor... the point is that a natural transformation (+ axioms) corresponds, as soon as the ambient category has countable sums, to a distributive law between F and the free monoid monad ... but in the sense that distributes over , i.e. with a map transforming a tensor of F(thing)'s into the F of a tensor of things!
this means essentially that lax monoidality is a codistributivity request. Which is ugly, nomenclature-wise! One would like laxity to be captured by distributivity, and colaxity by codistributivity. However, inverting the initial nomenclature would result in the monad of monoids codistributing over the one of abelian groups, equally ugly because the semiring axioms really are about "multiplication distributive over sum".)
So, am I making a stupid mistake here that I can't see?
No that's exactly how it works unfortunately
It makes sense though, because distirbutivity takes the point of view of one structure (eg monoidal prod or functor application) and says whether the other one preserves it. Thus changing which structure you take as primary you get a different direction.
Check §4 here for instance: https://arxiv.org/abs/2206.06858
Matteo Capucci (he/him) said:
No that's exactly how it works unfortunately
Well, at least now I'm sure I'm not mistaken