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Stream: theory: category theory

Topic: can one define an adjunction starting from the co/monads?

Matteo Capucci (he/him) (Feb 11 2023 at 15:14):

It is well-known every adjunction $L \dashv R : C \leftrightarrows D$ induces a monad on $D$ and a comonad on $C$. Is the opposite true?
That is, suppose I have a functors $R:C \to D$ and $L:D \to C$ such that $L;R$ is a comonad on $C$ and $R;L$ is a monad on $D$. Can I deduce $L \dashv R$?

Matteo Capucci (he/him) (Feb 11 2023 at 15:16):

The two co/monads obviously give us the unit and counit. But then when it come to prove the triangle identities... I'm stuck

Jacques Carette (Feb 11 2023 at 15:44):

I think you should be stuck, as the monad and comonad may not 'belong' to the same adjunction! Granted, the exceptions would be somewhat weird, but the (seeming) failure of the triangle identity is probably a strong hint of where to look for a counter-example.

Ralph Sarkis (Feb 11 2023 at 15:45):

It feels weird that the (co)multiplications don't play a role in the adjunction. Are there two (co)monads with the same endofunctor and (co)unit but not the same (co)multiplication?

Max New (Feb 11 2023 at 18:03):

The (co)multiplication aren't used in the definition of an adjunction because they are induced by the (co)unit. For instance the multiplication GFGF -> GF is induced by the co unit epsilon : FG -> 1 as G epsilon F

Ralph Sarkis (Feb 11 2023 at 19:26):

Sure, but then you have to ensure that the mutliplication of the monad is actually related to the counit of comonad in this way. Which comes back to what Jacques said.

Matteo Capucci (he/him) (Feb 11 2023 at 19:26):

Jacques Carette said:

I think you should be stuck, as the monad and comonad may not 'belong' to the same adjunction! Granted, the exceptions would be somewhat weird, but the (seeming) failure of the triangle identity is probably a strong hint of where to look for a counter-example.

Indeed... the co/monad _structures_ are not tied to the functors in any way, whereas you'd like the co/multiplications to be related

Matteo Capucci (he/him) (Feb 11 2023 at 19:27):

Once you ask that you end up just stating the triangle laws basically, so... nothing to see here I guess

dusko (Feb 13 2023 at 11:14):

Matteo Capucci (he/him) said:

Jacques Carette said:

I think you should be stuck, as the monad and comonad may not 'belong' to the same adjunction! Granted, the exceptions would be somewhat weird, but the (seeming) failure of the triangle identity is probably a strong hint of where to look for a counter-example.

Indeed... the co/monad _structures_ are not tied to the functors in any way, whereas you'd like the co/multiplications to be related

depending on the example that led to the question, it may be useful to note that a monad and a comonad can be tied to one another by much less than by giving the adjunction that brings them both about. e.g., if we are given a monad $RL$ and a comonad $LR$, and only the object parts of the functors $R$ and $L$, then as far as i can see, the arrow parts are determined by $R\varepsilon$ and $L\eta$ alone, and the assumption that one is an algebra and the other a coalgebra. (i only checked half of this claim, but it looks like both halves are true, which should be easy to check if the claim is of any use.)

if "only the object part" sounds evil, we can say "only on idempotents", and extend the arrow mapings from that. in any case, a correspondence between algebras and coalgebras seems to determine a correspondence on the carriers.