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James Wood (Jun 13 2021 at 10:34):Having convinced myself earlier that there are no interesting comonoids in (Set_part, ⊗), I'm curious about Rel. The question: are there any non-trivial bimonoids in (Rel, ⊗) (that is, the monoidal product which is × on objects)? I consider one to be trivial when either the monoid or the comonoid is the diagonal.
Robin Piedeleu (Jun 13 2021 at 10:38):Consider, for example, addition over the naturals (as a relation) and pair it with its converse. This should form a bimonoid that satisfies your requirement.
Jules Hedges (Jun 13 2021 at 11:26):Does every monoid in Set lift to a bimonoid in Rel by taking the comonoid to be the relational converse of the monoid? Does that always satisfy the bialgebra law?
Robin Piedeleu (Jun 13 2021 at 11:56):No, if you consider addition over the integers (instead of the naturals), then it forms a Frobenius algebra when paired with its relational converse. Note that addition and co-addition also satisfy the bimonoid law---the problem lies with the unit laws, e.g. two integers summing to zero are not necessarily both zero.
Robin Piedeleu (Jun 13 2021 at 11:57):So the requirement of positivity played a role in (addition, zero, co-addition, co-zero) satisfying the bimonoid laws.
Mike Shulman (Jun 13 2021 at 15:02):What I believe you can say is that every monoid in Set lifts to a "lax Frobenius algebra" in Rel, where the Frobenius laws hold as inequalities (inclusions) between relations.
JS PL (he/him) (Jun 13 2021 at 17:58):This paper by Hassei seems relevant: "Bialgebras in REL"
https://www.kurims.kyoto-u.ac.jp/~hassei/papers/mfps2010.pdf
JS PL (he/him) (Jun 13 2021 at 18:00):Jules Hedges said:
Does every monoid in Set lift to a bimonoid in Rel by taking the comonoid to be the relational converse of the monoid? Does that always satisfy the bialgebra law?
Every (commutative) monoid in SET induces a (bicommutative) bimonoid in REL where the comultiplication is the copy relation
JS PL (he/him) (Jun 13 2021 at 18:01):James Wood said:
I consider one to be trivial when either the monoid or the comonoid is the diagonal.
Ooops I didn't see this requirement. So my above comment is just the trivial one haha
JS PL (he/him) (Jun 13 2021 at 18:03):A non-trivial example: for every set X, the free (commutative) monoid over X is a (bicommutative) bimonoid in REL which is non-trivial. The comultiplication is given by the dual relation of the multiplicaiton.
JS PL (he/him) (Jun 13 2021 at 18:04):When , the free monoid is the naturals and you get back Robin's example from above.
JS PL (he/him) (Jun 13 2021 at 18:06):The commutative case is important in linear logic. And the fact that it gives bimonoid is important in differential linear logic.