# Universal property of module categories over monads

Let $T$ be a monad on a cocomplete category $\mathcal{C}$. Let's assume that $T$ preserves reflexive coequalizers (or something weaker?). Then the category of $T$-modules $\mathsf{Mod}(T)$ is cocomplete (Linton), and I think that we have the following universal property:

The category of cocontinuous functors $\mathsf{Mod}(T) \to \mathcal{D}$, where $\mathcal{D}$ is a cocomplete category, is equivalent to the category of cocontinuous functors $G : \mathcal{C} \to \mathcal{D}$ equipped with a right action $GT \to G$.

Sketch of proof: The free functor $F : \mathcal{C} \to \mathsf{Mod}(T)$ is cocontinuous and carries a right action $FT \to F$, induced by $\mu : T^2 \to T$. If $G : \mathcal{C} \to \mathcal{D}$ with $GT \to G$ is given, since every $T$-module $(X,a)$ has a canonical presentation $F(T(M)) \rightrightarrows F(M) \to (X,a)$ where the parallel arrows are given by the action $FT \to F$ and $F(a)$, and the right arrow is $a$, we have to define $\tilde{G} : \mathsf{Mod}(T) \to \mathcal{D}$ by $\tilde{G}(M,a):=$ coequalizer of $G(T(M)) \rightrightarrows G(M)$. Then $\tilde{G} F \cong G$ and $\tilde{G}$ preserves reflexive coequalizers (this is where I need that $T$ preserves reflexive coequalizers), so that $\tilde{G}$ is cocontinuous. QED

Actually I think there is also a version for cocomplete tensor categories and cocontinuous tensor functors; here $T$ should be a symmetric monoidal monad. This is the setting I'm actually interested in.

I don't really know much literature about category theory and in particular monad theory, but I suspect that this is known or even well-known. Is there any reference? (Again I need this in my dissertation and don't want to spam it with proofs of known facts.)

• Forgetting colimits the category with that universal property is the Kleisli category for the monad. One can define a Kleisli object in any 2-category. If your T is cocontinuous, then the universal category that you are after would be the Kleisli object of T in the 2-category of cocomplete categories and cocontinuous functors. This is purely formal statement of course. Commented Apr 17, 2014 at 0:08
• One way to actually construct the Kleisli object in the 2-category of cocomplete categories and cocontinuous functors I imagine is to somehow make the Kleisli category of T cocomplete. I dont't know how this would work. Commented Apr 17, 2014 at 0:17
• But I am not sure what you mean by Mod(T). Commented Apr 17, 2014 at 0:18
• @MartinBrandenburg, well, modules are algbras in the sense of universal algebra (with one unary operator per element in the ring, satisfyign the relations you know) That is where the name $T$-algebra comes from, I guess. Commented Apr 17, 2014 at 9:10
• @MartinBrandenburg, that shows that Mod(T) has coproducts, but not that $\tilde{G}$ preserves them. Commented Apr 19, 2014 at 21:55

## 1 Answer

Here is a partial answer with some literature pointers.

It's a general fact that in any bicategory with local reflexive coequalizers (i.e. reflexive coequalizers in each hom-category, preserved by composition on either side), Eilenberg-Moore objects coincide with Kleisli objects. I believe this was first observed by R. J. Wood, "Proarrows II" (Corollary 23). A different proof is in Carboni, Kasangian, and Walters, "An axiomatics for bicategories of modules", Remark 2.6(ii). In 1301.3191, Richard Garner and I observed (Corollary 15.16) that this is a bicategorical version of a "Cauchy colimit", akin to the fact that finite products and coproducts coincide in any category enriched over abelian monoids.

This general observation implies the following facts, which are close to your claim.

1. Working in the 2-category of categories with reflexive coequalizers and functors preserving them, we see that if $T$ preserves reflexive coequalizers, then it has your universal property but with "cocontinuous" replaced everywhere by "reflexive-coequalizer preserving".

2. Working in the 2-category of cocomplete categories and cocontinuous functors, we see that if $T$ is cocontinuous, then it has your universal property.

Your claim is a bit mismatched, assuming only that $T$ preserves reflexive coequalizers, but obtaining a universal property about cocontinuous functors. But it might be possible to deduce this from (1) using Linton's construction.