# How the category of continuous G-sets can be seen as the comonad algebra?

The article

https://ncatlab.org/nlab/show/category+of+G-sets

about the category of continuous G-sets previously suggested that the category could be characterized as a category of coalgebras for a comonad. Does this come from the comonad induced by the adjuntion mentioned there? I.e. is the forgetful functor from continuous G-sets to G-sets comonadic? And what is the description of its purported right adjoint?

Does anyone have a reference to a paper?

• It feels weird because to me it seems as though the category of $G$-sets is the category of algebras for the monad $X\mapsto G\times X$ with obvious unit and multiplication – Max Oct 3 at 20:44
• @Max it is, but the question seems to be about whether the forgetful functor from continuous $\mathbf{G}$-Sets to $\mathbf{Set^G}$ is comonadic. – Colin Oct 3 at 21:00
• @Colin it seems like that indeed, I noticed the title later (in the body of the question, "continuous" is not mentioned) – Max Oct 3 at 21:18

## 1 Answer

This is just a combination of three basic results. For one, the nLab entry on the category of $$G$$-sets shows that the adjoint pair $$i_G\dashv r_G$$ is a geometric morphism between toposes, where $$i_G$$ is the inclusion of $$G$$-sets into $$G_{disc}$$-sets and where $$r_G$$ is its right-adjoint. Now, a geometric morphism is surjective if and only if it is comonadic if and only if the left adjoint $$i_G$$ is faithful, which is easy to see (and also follows from the stuff proven on the nLab page). Thus, $$(G\text{-Set})$$ is naturally equivalent to the category of coalgebras for the comonad $$i_G\circ r_G\colon(G_{disc}\text{-Set})\to(G_{disc}\text{-Set})$$.

• Do you know how to describe $r_G$ concretely? (It must exist because $i_G$ is a colimit-preserving functor between Grothendieck topoi.) – Colin Oct 12 at 13:48
• I thought it would be taking the sub-set of points with open stabilisers, but I have never checked the details, could be horribly wrong. When I find the time, I will think about it again.. – Ben Oct 12 at 19:13
• Dear @Colin, indeed, it's what I thought. This basically boils down to the following lemma, which can be proven similarly to the proof of Proposition 2.1 in ncatlab.org/nlab/show/G-set#ContinuousCharacterization : If $f\colon X\to Y$ is a $G$-equivariant map where $X$ is a $G$-set and $Y$ is a $G_{disc}$-set, then for each $x\in X$ the stabiliser of $f(x)$ is open in $G$. – Ben 2 days ago
• Hi @Ben thanks for taking another look. Alas, I am still not seeing the trick here. Could you be more explicit about how you proved the lemma? – Colin 2 days ago
• Dear @Colin, if $X$ is a $G$-set, $Y$ is a $G_{disc}$-set and $f\colon X\to Y$ is $G$-equivariant, then, since $gf(x) = f(gx)$,$$\mathrm{Stab}_G(f(x)) = \{g\mid f(gx) = f(x)\} = \{g\mid gx\in f^{-1}f(x)\} = \bigcup_{x'\in f^{-1}f(x)}\{g\mid gx = x'\}$$ is open since the sets $\{g\mid gx' = x\}$ are, depending on $x'$, either empty or a translate of an orbit, hence open. Thus, $f$ takes values in $r_G(Y)$. This proves at once that $r_G$ is a functor and provides the adjunction isomorphism. – Ben yesterday