Not signed in (Sign In)

Start a new discussion

Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

  • Sign in using OpenID

Discussion Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

Welcome to nForum
If you want to take part in these discussions either sign in now (if you have an account), apply for one now (if you don't).
    • CommentRowNumber1.
    • CommentAuthorZhen Lin
    • CommentTimeAug 11th 2012

    The Eilenberg–Moore construction is essentially a 2-functor from the “2-category of monads on locally small categories” to the “2-category of locally small categories”, modulo certain size issues. In the reverse direction, it is known that any functor Φ:𝒞 𝕊𝒟 𝕋\Phi : \mathcal{C}^\mathbb{S} \to \mathcal{D}^\mathbb{T} such that U 𝕋Φ=FU 𝕊U^\mathbb{T} \Phi = F U^\mathbb{S} for some functor F:𝒞𝒟F : \mathcal{C} \to \mathcal{D}, where U 𝕊:𝒞 𝕊𝒞U^\mathbb{S} : \mathcal{C}^\mathbb{S} \to \mathcal{C} and U 𝕋:𝒟 𝕋𝒟U^\mathbb{T} : \mathcal{D}^\mathbb{T} \to \mathcal{D} are the respective forgetful functors, must come from a unique morphism of monads 𝕊𝕋\mathbb{S} \to \mathbb{T}. (This is basically a stronger version of Theorem 6.3 in Toposes, triples and theories.) It is also not hard to come up with functors 𝒞 𝕊𝒟 𝕋\mathcal{C}^\mathbb{S} \to \mathcal{D}^\mathbb{T} that do not arise in this fashion.

    But what about the 2-cells? Is every natural transformation between functors induced from a morphism of monads also induced by a 2-cell between the monad morphisms? Clearly, the answer is no – any natural transformation that doesn’t factor through the forgetful functor U 𝕊U^\mathbb{S} can’t be induced by a 2-cell between monad morphisms. But I haven’t been able to come up with a proof or counterexample when I assume that the natural transformation does factor through U 𝕊U^\mathbb{S}.

    • CommentRowNumber2.
    • CommentAuthorTodd_Trimble
    • CommentTimeAug 11th 2012
    • (edited Aug 11th 2012)

    Is every natural transformation between functors induced from a morphism of monads also induced by a 2-cell between the monad morphisms? Clearly, the answer is no – any natural transformation that doesn’t factor through the forgetful functor U 𝕊U^\mathbb{S} can’t be induced by a 2-cell between monad morphisms. But I haven’t been able to come up with a proof or counterexample when I assume that the natural transformation does factor through U 𝕊U^\mathbb{S}.

    Good question – the short answer is ’yes’.

    Just to lay all the cards on the table, I assume that by a monad morphism from a monad (C,S,m S:SSS,u S:1 CS)(C, S, m_S: S S \to S, u_S: 1_C \to S) to a monad (D,T,m T:TTT,u T:1 DT)(D, T, m_T: T T \to T, u_T: 1_D \to T), you mean a pair (F:CD,ϕ F:TFFS)(F: C \to D, \phi_F: T F \to F S) where the natural transformation ϕ F\phi_F satisfies an obvious compatibility with the monad multiplications and the monad units (taking the shape of a pentagon and a unit, much as in the case of a distributive law, which is actually a special case). (I point this out because if you don’t think about it too hard, one could guess the opposite direction for ϕ T\phi_T!)

    By a 2-cell from a monad morphism (F,ϕ F)(F, \phi_F) to a monad morphism (G,ϕ G)(G, \phi_G), you must mean a transformation η:FG\eta: F \to G satisfying an obvious compatibility with the ϕ\phi’s.

    Okay, suppose functors F,G:CDF, G: C \to D have lifts to functors Φ,Ψ:C SD T\Phi, \Psi: C^S \to D^T; then, as you say, FF and GG become endowed with appropriate transformations ϕ F\phi_F, ϕ G\phi_G, making them monad morphisms. Now suppose we have a transformation θ:ΦΨ\theta: \Phi \to \Psi which “descends” to a transformation η:FG\eta: F \to G, in the sense that ηU S=U Tθ\eta U_S = U_T \theta, where U SU_S, U TU_T are the forgetful functors from Eilenberg-Moore categories. We want to show η\eta is a transformation between the monad morphisms.

    Here is the critical diagram you need to stare at:

    TF Tu TF TTF Tϕ F TFS ϕ FS FSS Fm S FS Tη TηS ηS TG Tu TG TTG Tϕ G TGS ϕ GS GSS Gm S GS\array{ T F & \stackrel{T u_T F}{\to} & T T F & \stackrel{T\phi_F}{\to} & T F S & \stackrel{\phi_F S}{\to} & F S S & \stackrel{F m_S}{\to} & F S \\ ^\mathllap{T \eta} \downarrow & & & & \downarrow^\mathrlap{T\eta S} & & & & \downarrow^\mathrlap{\eta S} \\ T G & \underset{T u_T G}{\to} & T T G & \underset{T\phi_G}{\to} & T G S & \underset{\phi_G S}{\to} & G S S & \underset{G m_S}{\to} & G S }

    It’s not too hard to see, using the fact that (F,ϕ F)(F, \phi_F) is a monad morphism and one of the unit laws for a monad, that the long top horizontal composite is ϕ F\phi_F; similarly, the long bottom horizontal composite is ϕ G\phi_G. So we’re done if we show the entire diagram commutes. Now the right-hand rectangle commutes essentially because η\eta lifts to a transformation θ:ΦΨ\theta: \Phi \to \Psi. The left rectangle commutes because it’s TT applied to a naturality square for η\eta in disguise – one has to rewrite ϕ Fu TF=Fu S\phi_F \circ u_T F = F u_S and ϕ Gu TG=Gu S\phi_G \circ u_T G = G u_S, again using the fact that (F,ϕ F)(F, \phi_F) and (G,ϕ G)(G, \phi_G) are monad morphisms, to remove the disguise.

    • CommentRowNumber3.
    • CommentAuthorZhen Lin
    • CommentTimeAug 12th 2012

    Perfect, thanks! I got stuck because I subdivided that diagram too much…

    • CommentRowNumber4.
    • CommentAuthorTodd_Trimble
    • CommentTimeAug 12th 2012

    Yup! The same thing happened to me :-)

Add your comments
  • Please log in or leave your comment as a "guest post". If commenting as a "guest", please include your name in the message as a courtesy. Note: only certain categories allow guest posts.
  • To produce a hyperlink to an nLab entry, simply put double square brackets around its name, e.g. [[category]]. To use (La)TeX mathematics in your post, make sure Markdown+Itex is selected below and put your mathematics between dollar signs as usual. Only a subset of the usual TeX math commands are accepted: see here for a list.

  • (Help)