# 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. – Dimitri Chikhladze Apr 17 '14 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. – Dimitri Chikhladze Apr 17 '14 at 0:17
• But I am not sure what you mean by Mod(T). – Dimitri Chikhladze Apr 17 '14 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. – Mariano Suárez-Álvarez Apr 17 '14 at 9:10
• @MartinBrandenburg, that shows that Mod(T) has coproducts, but not that $\tilde{G}$ preserves them. – Mike Shulman Apr 19 '14 at 21:55

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.