Context. I am trying to understand the argument in B.4 of Thomas Nikolaus, Peter Scholze, *On topological cyclic homology*, arXiv:[1707.01799](https://arxiv.org/abs/1707.01799) (on p147).




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I am still lost. But from Maxime's helpful comment and replies, let me list out my concerns - which are listed as (X),(Y),(Z).




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**The proof of B.4 spelt out in steps: (read the numericals for the main steps)**


1. We begin with a $1$ -category $\Lambda\_\infty$ with a $B \Bbb Z $-action. We want to show
$$|\Lambda\_1| \simeq K(\Bbb Z, 2)$$




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So I am trying to understand why this means.


Firstly, which category does this take place in? From answer below, I'd like to understand more how $$ \Lambda\_\infty \in Fun(BB\Bbb Z, Cat)$$ From the construction given.



> 
> (X') So as in comments
> $$object \in Fun(BB\Bbb Z.Cat) \simeq Map(B\Bbb Z, Fun(C,C)^{\simeq}) \simeq Map( \Bbb Z, \Omega (Fun(C,C)^{\simeq}, id)$$
> 
> 
> 


Where I have omitted the subscript category. It would be helpful elaboration what adjunction where are using to obtain such equivalence.
As I am still rather unclear why we have these equivalence.




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2. We construct a new category, $\Lambda\_1:= \Lambda\_\infty/B\Bbb Z= \Lambda\_\infty/A$.




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Now I don't understand what $(-)/B\Bbb Z$ means. i.e. What kind of colimit are we taking?



> 
> (X) for each $A \in CAlg(Cat)$ some object $BA \in Cat$,  
> 
> **$$Mod\_{A}(Cat) \simeq Fun(BA, Cat) $$**
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> 
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Hence
$$\Lambda\_1 \simeq colim \_{BA} \Lambda\_\infty$$




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3. We wish to compute $|N\Lambda\_1|$. Then as $|\quad|$ is left adjoint.


$$ |\Lambda\_1| \simeq colim\_A |\Lambda\_\infty| \simeq |BA| $$




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The second equivalence requires the fact that



> 
> (Y) $Spc^{|BA|} \rightarrow Spc$ is conservative. Does this follow from that $Mod\_{|BA|}(Spc) \rightarrow Spc$ is conservative? .
> 
> 
> 



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> (Z) An explicit formula for $BA$. It doesn't seem clear to me why we would now have $BB\Bbb Z= K(\Bbb Z,2)$.
> 
> 
> 



Computing homotopy colimit of a space with free $S^1$-action
Note that in their context, $C$ has an action of $B\mathbb Z$, not of $\mathbb Z$ ! (Otherwise $C/B\mathbb Z$ wouldn't make sense)
This amounts essentially to a self natural transformation of the identity functor


For the first claim and the commutative square, this is true as geometric realization is a left adjoint $Cat\_\infty \to \mathsf{Spaces}$ so if you're taking the homotopy colimit, it is preserved by the left adjoint (and then they explain why their colimit is a homotopy colimit)


For the second claim, in $Fun(BG, \mathsf{Spaces})$, $\*$ is terminal, and the forgetful functor to $\mathsf{Spaces}$ is conservative, in particular any space with $G$-action (note that this is different from what is often called a genuine $G$-space) whose underlying space is contractible is equivalent, in that category, to $\*$ with the trivial action, which indeed has homotopy colimit $BG$.


Let me adress your new X,Y,Z concerns.


(X) : no, it's not the case that such a $BA$ exists for any $A$, it's specific to the fact that $A$ here is a group (specifically, $S^1$ or $B\mathbb Z$). In fact, I don't think they're claiming that equivalence, I assume that they're taking the right hand side as a *definition* of a category with $G$-action.


More generally, if you have *any* $\infty$-category $D$, $Fun(BG,D)$ is what we define to be "$D$-objects with $G$-action" (note that $G$ doesn't have to be "in" $D$, whatever that would even mean in such generality). It so happens that in the cases $D= Cat\_\infty$ or $\mathsf{Spaces}$, the left hand side also has a meaning, and they happen to agree, but you don't need to know or use that in the proof.


(Y) : For any $\infty$-categories $C,D$, the restriction functor $Fun(C,D)\to Fun(Ob(C),D)$ is conservative, this is just saying that a natural transformation is invertible if and only if each of its components is invertible (which is obvious $1$-categorically, and requires some work $\infty$-categorically, but is not too hard). You then specialize this to $C= BG$ which as only one object, and $D$ whatever. This means that an equivalence between objects with $G$-action is just a $G$-equivariant map which is an equivalence on the underlying objects.


(Z) : $BB\mathbb Z= K(\mathbb Z,2)$ is a classical fact from algebraic topology.


For any (nice) topological group $G$, $\Omega BG\simeq G$, so $\Omega^2BB\mathbb Z \simeq \mathbb Z$, and $BB\mathbb Z$ is simply-connected (essentially by definition of $B(-)$), so it follows that it's a $K(\mathbb Z,2)$.


