# Colimit in the category of (all) simply transitive group actions

Let $\mathcal{C}$ be the category of all group actions, i.e. :

• the objects are the pairs $(G,F)$ where $G$ is a group and $F$ is a functor $F\colon G\to\mathbf{Sets}$
• a morphism between $(G_1,F_1)$ and $(G_2,F_2)$ is a pair $(L,\lambda)$, where $L$ is a functor $L \colon G_1 \to G_2$ and $\lambda$ is a natural transformation from $F_1$ to $F_2 \circ L$.

Consider the subcategory $\mathcal{C}_r$, where the functor $F$ of any object $(G,F) \in \mathcal{C}_r$ is representable.

The general questions are :

• Does $\mathcal{C}$ have all colimits ? If so, how are they constructed ?
• Does $\mathcal{C}_r$ have all colimits ?

For some specific examples: consider $G_1=G_2=\mathbb{Z}_2$, acting on two elements sets $\{p_1,p_2\}$ and $\{q_1,q_2\}$ (where the action of the non-trivial element sends $p_1$ to $p_2$, etc.). What are the colimits in $\mathcal{C}$ and in $\mathcal{C}_r$ if they exist ?

Edit: I've edited the question so that the questions are explicitly stated.

• I don't think $\mathcal{C}_r$ even has an initial object. – Jeremy Rickard Dec 12 '14 at 13:29

## 1 Answer

The category $\mathcal{C}$ of all group actions is complete and cocomplete. First, observe that there is an evident forgetful functor $U : \mathcal{C} \to \mathbf{Set} \times \mathbf{Set}$ that preserves and creates limits. You can also check that $U : \mathcal{C} \to \mathbf{Set} \times \mathbf{Set}$ preserves and creates filtered colimits. With further work, you can eventually show that $\mathcal{C}$ is finitely accessible and hence locally finitely presentable, so cocomplete a fortiori.

On the other hand, the full subcategory $\mathcal{C}_r$ of simply transitive group actions is neither complete nor cocomplete. Indeed, it is easy to check that $\mathcal{C}_r$ is closed under products and filtered colimits in $\mathcal{C}$. It is even a finitely accessible category. On the other hand, as Jeremy Rickard observed, $\mathcal{C}_r$ does not even have an initial object. (Notice that, for $(G, X)$ and $(H, Y)$ in $\mathcal{C}_r$, there is a (non-canonical) bijection between the set of morphisms $(G, X) \to (H, Y)$ and the set $\mathrm{Hom} (G, H) \times Y$, so there is no possible initial object.) It follows that $\mathcal{C}_r$ cannot be complete either – if it did, then it would be locally finitely presentable and hence cocomplete.

• This doesn't affect the main argument, but I don't think the forgetful functor $\mathcal{C} \to \mathbf{Grp} \times \mathbf{Set}$ creates colimits. There are three morphisms from the regular action of the trivial group to the regular action of a cyclic group of order three. Consider the coequalizer of two of these. – Jeremy Rickard Dec 16 '14 at 18:23
• @Zhen Lin: thank you for your answer. I had assumed without thinking much about it that $\mathcal{C}_r$ was complete. May I ask for a simple counterexample ? – OliverX1 Dec 17 '14 at 21:04
• Well, products are not a problem, so it must equalisers. And sure enough, if you think about parallel pairs in $\mathcal{C}$ whose equaliser is empty, you will see that they have no equaliser in $\mathcal{C}_r$. – Zhen Lin Dec 17 '14 at 21:43
• Oh, I see... thanks. – OliverX1 Dec 18 '14 at 11:19