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I have noticed that the function that maps a monad $T : C \to C$ to the Eilenberg-Moore category $C^T$ can easily be extended into a functor $E_C$ from the category of monads $\textbf{Mnd}_C$ to $\textbf{Cat}^\text{op}$.

Questions Is it a well-known functor? How can it be dualized to co-Eilenberg-Moore categories?

My own attempt Le $i$ be the isomorphism between $\textbf{Comnd}_C$ and $(\textbf{Mnd}_{C^\textit{op}})^{\textit{op}}$. The composite $({E_{C^\textit{op}}})^\textit{op} \circ i$ is a functor from $\textbf{Comnd}_C$ to $\textbf{Cat}$. But it is not the right one because its object part maps a comonad $D$ to the $\textit{op}$ of the co-Eilenberg-Moore category $C_D$. The $\textit{op}$ is too much.

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    $\begingroup$ You should write out explicitly the definitions of the category of monads and the category of comonads. Then you will see that your isomorphism is not correct. (You have one too many ${}^\textrm{op}$, as you surmised.) $\endgroup$
    – Zhen Lin
    Feb 18 at 22:45
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    $\begingroup$ @ZhenLin I disagree. I cannot define an isomorphism between $\textbf{Comnd}_C$ and $(\textbf{Mnd}_{C})^{\textit{op}}$ because I would have to map a comonad on $C$ to a monad on $C$. And I cannot define an isomorphism between $\textbf{Comnd}_C$ and $\textbf{Mnd}_{C^\textit{op}}$ because I would have to map a comonad morphism between $D_1$ and $D_2$ to a monad morphism between $i(D_1)$ and $i(D_2)$. Both $\textit{op}$ are necessary. $\endgroup$
    – Bob
    Feb 19 at 9:43
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    $\begingroup$ @ZhenLin I took the monad morphism definition from Borceux's Volume 2 of Handbook of Categorical Algebra. As for a comonad morphism between $(D_1, \eta_1, \mu_1)$ and $(D_2, \eta_2, \mu_2)$, I guess it is a natural transformation $\lambda : D_1 \to D_2$ such that $\eta_2 \circ \lambda = \eta_1$ and $(\lambda \cdot \lambda) \circ \mu_1 = \mu_2 \circ \lambda$. Is it wrong? $\endgroup$
    – Bob
    Feb 19 at 10:35
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    $\begingroup$ My expression was poor. There are many definitions of monad morphism, some more general than others, and appropriate for different situations. I can think of one for which there is a dual pair of functors of the kind you are looking for. The one you have chosen is not such a definition. $\endgroup$
    – Zhen Lin
    Feb 19 at 12:47
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    $\begingroup$ @ZhenLin It is well known how to generalize the (co)monad morphism definitions above so that they can relate (co)monads on different categories. But, except that, I have never heard of any other definitions of (co)monad morphism. This looks very interesting. Can I find those other definitions on nLab or anywhere else? $\endgroup$
    – Bob
    Feb 19 at 13:22

1 Answer 1

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Here is one way of defining "monad".

Definition 1. A monad in a 2-category $\mathcal{K}$ is a lax 2-functor $\mathbf{1} \to \mathcal{K}$, where $\mathbf{1}$ is the 2-category with a single object and only the trivial morphism and 2-morphism.

But there is a general theorem to the effect that lax 2-functors are the same as strict 2-functors, after replacing the domain with a different 2-category (the so-called lax morphism classifier, though in my opinion it should be called a coclassifier rather than a classifier). In this case, we have a very explicit construction.

Let $\mathbf{M}$ be the following 2-category:

  • There is only one object, $*$.

  • The morphisms are $\textrm{id}, t, t^2, t^3, \ldots$.

  • The 2-morphisms $t^n \Rightarrow t^m$ are the monotone maps $\{ 0, \ldots, n - 1 \} \to \{ 0, \ldots, m - 1 \}$.

In other words, $\mathbf{M}$ is the delooping of the augmented simplex category considered as a strict monoidal category. (Beware of differing numbering conventions for the objects in the augmented simplex category.)

Definition 2. A monad in 2-category $\mathcal{K}$ is a strict 2-functor $\mathbf{M} \to \mathcal{K}$.

The two definitions are equivalent in the following sense:

Proposition. There is a lax 2-functor $\mathbf{1} \to \mathbf{M}$ such that, for every lax 2-functor $\mathbf{1} \to \mathcal{K}$, there is a unique strict 2-functor $\mathbf{M} \to \mathcal{K}$ such that the composite $\mathbf{1} \to \mathbf{M} \to \mathcal{K}$ is the given lax 2-functor.

So much for monads. What about morphisms of monads? If we follow definition 1, the obvious notion of morphism is that of a lax natural transformation (of lax 2-functors), but we could also consider icons (= identity-component oplax natural transformations). If we follow definition 2, the obvious notion of morphism is that of a strict natural transformation (of strict 2-functors). But we could also look at lax or oplax or pseudo natural transformations (of strict 2-functors). Every strict natural transformation is pseudo, and every pseudo natural transformation is both lax and oplax. In principle, we might also consider things similar to pseudo natural transformations except that the naturality squares only commute up to an unspecified 2-isomorphism and no coherence axioms are imposed, but they do not seem to be useful.

As I said, what notion of morphism is appropriate depends on the application. For Eilenberg–Moore objects, it seems the right notion is that of a lax natural transformation of strict 2-functors. I described the universal property of the Eilenberg–Moore object of a monad previously, but if we define monad following definition 2, there is a succinct description: it is the lax limit of the strict 2-functor $\mathbb{T} : \mathbf{M} \to \mathcal{K}$. That is, it is an object $A$ in $\mathcal{K}$ together with a universal lax cone $\Delta A \Rightarrow \mathbb{T}$. Universality means that for every object $B$ in $\mathcal{K}$ and every lax cone $\Delta B \Rightarrow \mathbb{T}$ there is a unique morphism $B \to A$ such that the composite $\Delta B \Rightarrow \Delta A \Rightarrow \mathbb{T}$ is the lax cone you started with. If you have another strict 2-functor $\mathbb{T}' : \mathbf{M} \to \mathcal{K}$ and a lax natural transformation $\mathbb{T}' \Rightarrow \mathbb{T}$, you get an induced morphism from the lax limit of $\mathbb{T}'$ to the lax limit of $\mathbb{T}$, i.e. a morphism between the Eilenberg–Moore objects of the respective monads. In fact:

Proposition. Let $[\mathbf{M}, \mathcal{K}]_\textrm{strict, lax}$ be the following 2-category:

  • The objects are the strict 2-functors $\mathbf{M} \to \mathcal{K}$.
  • The morphisms are the lax natural transformations.
  • The 2-morphisms are the modifications.

Assuming $\mathcal{K}$ has Eilenberg–Moore objects for every monad, there is a strict 2-functor $[\mathbf{M}, \mathcal{K}]_\textrm{strict, lax} \to \mathcal{K}$ sending each monad (i.e. strict 2-functor $\mathbf{M} \to \mathcal{K}$) to its Eilenberg–Moore object (i.e. lax limit).

We can easily dualise this to get the corresponding construction for comonads: just replace $\mathcal{K}$ with $\mathcal{K}^\textrm{co}$. If we shuffle the modifiers to get $\mathcal{K}$ to appear without ${}^\textrm{co}$, we get:

Proposition. Let $[\mathbf{M}^\textrm{co}, \mathcal{K}]_\textrm{strict, oplax}$ be the following 2-category:

  • The objects are the strict 2-functors $\mathbf{M}^\textrm{co} \to \mathcal{K}$.
  • The morphisms are the oplax natural transformations.
  • The 2-morphisms are the modifications.

Assuming $\mathcal{K}$ has Eilenberg–Moore objects for every comonad, there is a strict 2-functor $[\mathbf{M}^\textrm{co}, \mathcal{K}]_\textrm{strict, oplax} \to \mathcal{K}$ sending each comonad (i.e. strict 2-functor $\mathbf{M}^\textrm{co} \to \mathcal{K}$) to its Eilenberg–Moore object (i.e. oplax limit).

If you want to focus on monads on a single category you could take the evident full sub-2-category of $[\mathbf{M}, \mathcal{K}]_\textrm{strict, lax}$, or you could cut down the 1-morphisms (as Borceux does) to identity-component lax natural transformations. (Note that an identity-component lax natural transformation $\mathbb{T}' \Rightarrow \mathbb{T}$ will have among its data 2-morphisms $t \Rightarrow t'$. The arrows point in opposite directions, but this is what falls out of the definitions. You can fix this by thinking about icons instead, but that obscures the duality.)

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    $\begingroup$ Thank you very much for this wealth of information. But I will need a few weeks to digest it in order to determine in which way it answers my question. $\endgroup$
    – Bob
    Feb 20 at 7:53
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    $\begingroup$ I find it very bizarre that this answer is attracting downvotes to cancel out the upvotes. It's as if someone is trying to keep the score at zero. $\endgroup$
    – Zhen Lin
    Feb 20 at 22:12
  • $\begingroup$ I have tried hard but I can't see how it answers my questions. From your comment I understand that you propose here an alternative definition of monad morphism "for which there is a dual pair of functors of the kind [I am] looking for". Can you please make explicit this alternative definition in terms on 1-category theory? $\endgroup$
    – Bob
    Feb 24 at 7:17
  • $\begingroup$ It is the one you claimed is well known: a monad morphism from $\mathbb{T}'$ on $C'$ to $\mathbb{T}$ on $C$ consists of a functor $C' \to C$ together with a natural transformation from the composite $C' \to C \to C$ to the composite $C' \to C' \to C$, such that various equations are satisfied. Note the direction of the natural transformation in particular. $\endgroup$
    – Zhen Lin
    Feb 24 at 10:40
  • $\begingroup$ Yes, but in my question $C' = C$. So the natural choice for the functor is the identity and this gives back the usual notion of monad morphism. Do you think of another choice for the functor that would provide an answer to my question? $\endgroup$
    – Bob
    Feb 24 at 11:19

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