Computing homotopy colimit of a space with free $S^1$-action

Question

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).

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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 $BA \in Cat$,
$$Mod_{A}(Cat) \simeq Fun(BA, Cat) $$

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? .

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

Answer 1

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 $K(\mathbb Z,2)$.