Homotopy fixed points of involutive automorphisms of discrete groups

Question

$\DeclareMathOperator\Map{Map}$Precis: I am looking for a reference/explanation of a (general) computation of homotopy fixed points $(BG)^{h\Gamma}$ of a finite group $\Gamma$ acting on a discrete (but not necessarily finite) group $G$.

I am primary interested in the case of $\Gamma = \mathbb{Z}/2\mathbb{Z}$, but don't see why the story shouldn't be completely general. Ideally, someone will explain precisely why (or why not) this is a purely algebraic computation (morally this is true because there is no topological input involved involved apart from the general gluing of simplices in the nerve construction).

Note: I can rephrase this question as one about computing a particular type of homotopy limit (homotopy fixed points, finite group) of a homotopy colimit (the classifying space $BG$) and maybe that is the "correct" way to look at this.

In more detail: Suppose given an involutive automorphism of a discrete group $G$. I will write the automorphism as $ g\mapsto \bar g$

For any (discrete) group $G$, I will write $\mathbb{B}G$ for the groupoid with one object and morphisms given by $G$. Let $\Gamma = \mathbb{Z}/2\mathbb{Z}$ and write $\mathbb{E}\Gamma$ for the groupoid with objects as well as morphisms given by $\Gamma$ (I hope it is clear in this context what the morphisms are doing and where I am going with this).

Then $\Gamma$ acts on these groupoids (on $\mathbb{B}G$ via the involution) in the evident way (in the sense that it acts on the set of objects and morphisms in a way that all structure maps are equivariant).

Let $ \Map_{\Gamma}(\mathbb{E}\Gamma, \mathbb{B}G) $ denote the groupoid whose objects are $\Gamma$-equivariant functors between the mentioned groupoids. Then just playing around with the definitions it is straightforward to show (hopefully, I didn't mess this up):

$$ \Map_{\Gamma}(\mathbb{E}\Gamma, \mathbb{B}G) \cong \bigsqcup_{[\sigma] \in H^1(\Gamma; G)} \mathbb{B}K_{\sigma} $$

where

$$ K_{\sigma} = \{ g\in G \mid \bar g = \bar{\sigma}g\sigma \} $$

and $H^1(\Gamma; G)$ is the set

$$Z^1(\Gamma; G) = \{ \sigma\in G \mid \sigma\bar \sigma = 1\}.$$

modulo the relation $\sigma \sim g\sigma\bar g^{-1}$ for all $g\in G$.

Of course, now my desire is to say that the above formula holds topologically. There is certainly a canonical map from the geometric realization of the groupoid above to $(BG)^{h\Gamma}$ but it isn't obvious to me that it is (or isn't) a weak equivalence.

Added later: I think I may have just reverse engineered the Bousfield-Kan model of the homotopy limit here (that's what I get for trying to first understand it via its formal properties before looking at an explicit model). I am staring at Ch XI, Section 3.2 of the monograph "Homotopy limits, completions and localizations", and unless I am misreading/misunderstanding, the explicit construction of a model for a homotopy limit given there is precisely the description above in terms of the groupoid of functors (in this particular situation).

Hopefully, someone with expertise can confirm or explain.

Added later: For anyone interested in this, Bertram Arnold's comment below is essentially the correct answer, as far as I can tell (it took me a while to decipher it though).

Answer 1

$\DeclareMathOperator\Map{Map}$ In case some other novice like me comes across this, let me provide a (generalized) answer to what is going on.

Write

$$B\colon \text{Groupoids} \to \text{simplicial sets}$$ for the nerve functor. Then, amongst other nice features, $B$ commutes with limits (it has a left adjoint).

Let $\Gamma$ be a group acting on another group $G$ via automorphisms. As in the original question, write $\Map_{\Gamma}(\mathbb{E}\Gamma, \mathbb{B}G)$ for the groupoid of $\Gamma$-equivariant functors and equivariant natural transformations. As before, $\mathbb{B}G$ is the groupoid with single object and automorphisms $G$, etc.

More or less from the definitions and the aforementioned commutation with limits (in particular products), it follows that the obvious canonical map

$$ B\Map_{\Gamma}(\mathbb{E}\Gamma, \mathbb{B}G) \to \Map_{\Gamma}(E\Gamma, BG) $$

is an isomorphism. Here, the right hand side is the (equivariant) simplicial function complex, and $E\Gamma = B\mathbb{E}\Gamma$, etc. As $\mathbb{B}G$ is a groupoid, we have that $BG$ is a Kan complex. In other words, my notation isn't misleading - the right hand side is, as far as homotopy theory is concerned, the usual topological homotopy fixed point space.

The cute upside of this is that working with groupoids and just sitting down and writing explict formulae gives that:

$$\Map_{\Gamma}(E\Gamma, BG) \simeq \bigsqcup_{[\sigma]\in H^1(\Gamma; G)} BK_{\sigma}, $$

where $H^1(\Gamma; G)$ is the first (non-abelian) cohomology of $\Gamma$ with coefficients in $G$, and $K_{\sigma} \subset G$ is the centralizer subgroup

$$ K_{\sigma} = \{ \alpha\in G \mid \sigma(g)\cdot g\alpha\cdot \sigma(g)^{-1} = \alpha \text{ for all $g\in\Gamma$}\}. $$

A pedantic comment: that decomposition $\bigsqcup K_{\sigma}$ can't be picked canonically, except for in trivial situations (you need to pick representatives for $[\sigma]\in H^1(\Gamma; G)$). If you want to keep things canonical and not run into making mistakes with diagrams not commuting, you need to replace the decomposition with the homotopy/Borel orbit space of the 1-cocycles $Z^1(\Gamma; G)$ under the twisted conjugation action (that at the level of components gives $H^1(\Gamma; G)$.

Further, note that this works precisely because there is no topology on the groups involved. If $G$ was a Lie group, then the (enriched) nerve takes the topology into account (we are really dealing with simplicial groupoids then), and the canonical map to the function complex being an isomorphism doesn't hold. Although, surprisingly to me, even in that case if $\Gamma$ is a $p$-group, $G$ is a compact connected Lie group and the $\Gamma$-action is trivial, the isomorphism does hold after localizing at $p$ (there's a very nice paper of Dwyer-Zabrodsky whose main result is essentially this). However, this uses Miller's theorem/the Sullivan conjecture.