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Stream: theory: category theory

Topic: Isomorphism in algebraic presentation of monads

Aloïs Rosset (Oct 18 2021 at 13:57):

Hello! I'm reading an article ( https://arxiv.org/pdf/1808.00923.pdf ) by Bonchi, Sokolova and Vignudelli, and there is a paragraph on page 11 that puzzles me. It talks about monads and algebraic presentation of Set-monads:

Since $(Σ, E)$-algebras define the monad $T_{Σ,E}$, it follows from [Mac71, Theorem VI.8.1] that $(Σ, E)$ is a presentation of the monad $T_{Σ,E}$, i.e., $(Σ, E)$-algebras are (isomorphic to) the Eilenberg-Moore algebras of $T_{Σ,E}$. Furthermore, if $(Σ, E)$ is a presentation of a monad $M$, then $M$ is isomorphic to the monad $T_{Σ,E}$, i.e., there is an isomorphism monad map between them. This follows from the fact that every monad is defined by its Eilenberg-Moore algebras, see e.g. [Mac71, Theorem VI.2.1]. In particular, recall that the free $M$-algebra generated by $X$ is the algebra $(MX, µ)$. Hence, if $(Σ, E)$ is a presentation of $M$ then $MX$ is isomorphic to the set of $Σ$-terms with variables in $X$ modulo $E$-equations.

The situation is as follows:

image.png

• The monad $T_{\Sigma,E}$ is defined as $U^\Sigma F^\Sigma$.
• The isomorphism $H$ comes from Beck's monadicity theorem, so it is the comparison functor and thus it commutes with the free and forgetful functors $F^\Sigma, F, U^\Sigma$, and $U$, i.e., the triangle on the right commutes.
• The isomorphism $G$ is there to signify that $(\Sigma,E)$ presents $M$, but it no way does that imply that the triangle on the left commutes.

Hence my question: how can we conclude that $M$ and $T_{\Sigma,E}$ are ismorphic monads, or that the free objects, once considered in $\mathsf{Set}$, are in bijections? To have the isomorphism of monads, we would need a monad map $M => T_{\Sigma,E}$, but I see no way of defining such a natural transformation. The article invokes the theorem that says that a monad is equal to the composition of both adjoint functor to it's EM category. But since the triangle on the left does not commute, to me it looks like the reasoning quickly comes to an end.

Could it be possible that in $\mathsf{Set}$, an isomorphism such as $G$ can always be chosen so that it is a concrete isomorphism?

Thank you in advance for any help!

Nathanael Arkor (Oct 18 2021 at 15:32):

If you are happy that the category of algebras for $T_{(\Sigma, E)}$ is (concretely) equivalent to the category of algebras for $M$ (by definition of "being a presentation"), then it follows that the two monads are isomorphic. You can prove this directly, or deduce it abstractly.

Nathanael Arkor (Oct 18 2021 at 15:38):

@Dylan McDermott just pointed out to me that Definition 3.5 of that paper seems wrong. A presentation should require that the two categories of algebras are concretely isomorphic, i.e. isomorphic over $\mathrm{Set}$. I'm not convinced that with their definition the result holds.

Aloïs Rosset (Oct 18 2021 at 15:43):

I have always seen the definition of presentation as only requiring an isomorphism and not a concrete isomorphism, even though sometimes it would feel more fitting.

Nathanael Arkor (Oct 18 2021 at 15:46):

I don't think I've seen a definition of presentation before that involves the Eilenberg–Moore category like this (rather than the monad directly, or something equivalent, like an algebraic theory).

Nathanael Arkor (Oct 18 2021 at 15:48):

If the two monads are isomorphic, then there is a concrete isomorphism of EM categories, so this stronger definition is not incorrect. But I would like to see a proof that the weaker definition implies the stronger one if that definition is the one being used.

Aloïs Rosset (Oct 18 2021 at 15:51):

Exactly! I've been looking in the literature for what you just said, that having an isomorphism between the categories of algebras imply an isomorphism of monads, but without success so far

Nathanael Arkor (Oct 18 2021 at 15:54):

I would imagine it is simply an (easy) oversight on the behalf of the authors. Perhaps @Filippo Bonchi could comment?

Mike Shulman (Oct 18 2021 at 17:43):

It's certainly not true for monads on an arbitrary category that an isomorphism between categories of algebras implies an isomorphism of monads. For instance, the lattice of subsets of $\{0,1\}$ has two different reflective sublattices that are abstractly isomorphic. I suppose it's possible that something sneaky happens in the special case of monads on $\rm Set$, but even if so, I think it would be bad form to rely on that.