The article

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
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    @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

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
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    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 : 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

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