Context. I am trying to understand the argument in B.4 of Thomas Nikolaus, Peter Scholze, On topological cyclic homology, arXiv:1707.01799 (on p147).
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).
The proof of B.4 spelt out in steps: (read the numericals for the main steps)
- We begin with a $1$ -category $\Lambda_\infty$ with a $B \Bbb Z $-action. We want to show $$|\Lambda_1| \simeq K(\Bbb Z, 2)$$
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.
- We construct a new category, $\Lambda_1:= \Lambda_\infty/B\Bbb Z= \Lambda_\infty/A$.
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 $BA \in Cat$,
$$Mod_{A}(Cat) \simeq Fun(BA, Cat) $$
Hence $$\Lambda_1 \simeq colim _{BA} \Lambda_\infty$$
- We wish to compute $|N\Lambda_1|$. Then as $|\quad|$ is left adjoint.
$$ |\Lambda_1| \simeq colim_A |\Lambda_\infty| \simeq |BA| $$
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? .
(Z) An explicit formula for $BA$. It doesn't seem clear to me why we would now have $BB\Bbb Z= K(\Bbb Z,2)$.