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It is a classic result of Kan that the homotopy categories (with appropriate model structures) of simplicial sets and of topological spaces (in fact, one could only care about CW-complexes) are equivalent. One could ask if there is a similar result for G-equivariant homotopy category of spaces, where G is a topological group (usually a Lie group) and a morphism of G-spaces is a map, strictly commuting with G-action: $$f(gx) = g\cdot f(x)$$ Since the G-equivariant category admits a very nice theory of cell decompositions and in fact is equivalent to a homotopy category of space presheaves on the category of G-orbits (Elmendorf's theorem), one could guess that such a result should be true. However, I would like to see it not on the level of $\mathrm{SSet}$-presheaves on orbits, but rather directly in the standard definitions. I see 2 problems. Firstly, while cellular and simplicial descriptions of spaces are very similar, they are not equivalent in any sense I could see (besides giving the same homotopy category). The second problem is that if we pass from topological spaces to simplicial sets, we should also change a topological group into a simplicial group. While there is an obvious way to go back and forth (singular complex $\mathrm{Sing}: \mathrm{Top} \to \mathrm{SSet}$ and geometric realization $|\cdot|: \mathrm{SSet} \to \mathrm{Top}$), it is very unobvious that this gives the correct correspondence, since $X \to |\mathrm{Sing}X|$ is just a weak homotopy equivalence, and the equivariant category doesn't respect homotopy equivalences in general, both for spaces and groups. A contractible space can have different non-equivalent G-structures (e.g. free and trivial), and homotopy equivalent (as spaces) groups can give different equivariant categories.

So the question is, can we generalize Kan's theorem to the equivariant setting? If we do, then how does this correspondence look like and how far is it from the intuition? For example, are there finite-dimensional simplicial groups G, such that their G-homotopy categories are not equivalent to equivariant categories of Lie groups? Can we associate with a space $X$ some simplicial group $G_X$, such that the $G_X$-equivariant category is the same as fibrations over $X$?

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    $\begingroup$ Cyclic sets model $S^1$-equivariant homotopy theory. $\endgroup$ – Fernando Muro Jan 30 '14 at 20:37
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Perhaps you should look at the series of papers by Dwyer and Kan, starting with W. G. Dwyer and D. M. Kan, Singular functors and realization functors, Nederl. Akad. Wetensch. Indag. Math., 87, (1984), 147 – 153.

I do not think you will find all the answers that you require there, as your reluctance to see things in terms of simplicial presheaves is a hinderance from their point of view, but their discussion (and in later papers by others) may help. In fact I seem to remember that Dwyer and Kan's viewpoint (taken up by Dror later on) implies that there is a variant of the singular complex / geometric realisation adjointness that does the job. The standard simplicial models are more or less replaced by the $G/H\times \Delta^n$.

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    $\begingroup$ Can I ask why you do not want the presheaves on orbits idea? I can see that this may be awkward if $G$ is a simplicial group, but then the way forward may be to replace presheaves by $SSet/\overline{W}G$, and work with the Quillen m.c. structure on that. $\endgroup$ – Tim Porter Jan 31 '14 at 11:38
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    $\begingroup$ You say:it doesn't really explain what a G-space is, simplicially? It may do as then it is probably built up from various models such as that one by a colimiting process. $\endgroup$ – Tim Porter Jan 31 '14 at 11:41
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    $\begingroup$ Another way forward is to encode things as being homotopy coherent. Think of $G$ (a simplicial group) as a simplicially entriched groupoid with one object, and then a $G$-simplicial set is just a functor from $G$ to $SSet$ and go on from there. $\endgroup$ – Tim Porter Jan 31 '14 at 11:43
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    $\begingroup$ (I avoid $BG$ and prefer $G[1]$ for the corresponding groupoid.) I was meaning simplicially enriched functors (were you)? The actions that one gets are sensible in their interpretations. If you then go towards the questions you asked in the original post, then there may be a need to use the homotopy coherent nerve of $G[1]$ and to have homotopy actions to get something sensible. $\endgroup$ – Tim Porter Jan 31 '14 at 12:33
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    $\begingroup$ @AntonFetisov The identification of equal things is not really a problem (after all, what are equal things if not identical?); rather it is that equality is too coarse in homotopy type theory for that purpose: a group inside HoTT will be more like some kind of $A_\infty$-group, actions will only be up to coherent homotopy, etc. $\endgroup$ – Zhen Lin Jan 31 '14 at 14:57

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