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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)

1. We begin with a $1$ -category $\Lambda_\infty$ with a $B \Bbb Z $-action. We want to show $$|\Lambda_1| \simeq K(\Bbb Z, 2)$$

So I am trying to understand why this means.

Firstly, which category does this take place in? $B\Bbb Z=A$ lives in com. alg. obj. of $Cat$. I shall use $A$ instead of $B\Bbb Z$ since what $B\Bbb Z$ is seems to be irrelevant for most of the argument.

Hence we can consider $Mod_{A}(Cat)$ of left module objects over $B\Bbb Z$. So the claim is saying that $$\Lambda_\infty \in Mod_{A}(Cat)$$

2. 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 $$\Lambda_1 \simeq colim _{BA} \Lambda_\infty$$

3. 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 $BB\Bbb Z= K(\Bbb Z,2)$.

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)

1. We begin with a $1$ -category $\Lambda_\infty$ with a $B \Bbb Z $-action. We want to show $$|\Lambda_1| \simeq K(\Bbb Z, 2)$$

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.

2. 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 $$\Lambda_1 \simeq colim _{BA} \Lambda_\infty$$

3. 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 $BB\Bbb Z= K(\Bbb Z,2)$.

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).

The problem:

1. We begin with category $C$ with a free $\Bbb Z$-action which fixes the objects of $C$. I.e. Each $g \in \Bbb Z$ induces a functor $$g:C \rightarrow C $$ that fixes the objects. We can thus regard $C$ as an object in $Fun(B\Bbb Z, Cat)$.

2. We construct the quotient category $C':=C/B\Bbb Z$ by quotienting the morphism space by the $\Bbb Z$ action. This is equivalent to taking the colimit of $C$ as an object in $Fun(B\Bbb Z, Cat)$.

3. We wish to compute $|NC'|$. What we do know is that $|NC|\simeq *$.

The proof in the paper goes as follow.

$$|NC'| \simeq |NC|/B\Bbb Z \simeq B(B\Bbb Z) $$

I am rather confused by both the 1st. and 2nd. equivalence. Any comment/elaborations would be appreciated.

My confusions.

For the 1st. It seems to me the author claims we have a commuting diagram $$ Fun(B\Bbb Z,Cat) \rightarrow Cat $$ $$Fun(B\Bbb Z, Spc) \rightarrow Spc $$ where we take colimits horizontally and geometric realization $|N(-)|$ vertically.

For the 2nd. As far as I could find in the literature, we know the case when we have a constant diagram, of which the colimit is given by $BG$.

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)

1. We begin with a $1$ -category $\Lambda_\infty$ with a $B \Bbb Z $-action. We want to show $$|\Lambda_1| \simeq K(\Bbb Z, 2)$$

So I am trying to understand why this means.

Firstly, which category does this take place in? $B\Bbb Z=A$ lives in com. alg. obj. of $Cat$. I shall use $A$ instead of $B\Bbb Z$ since what $B\Bbb Z$ is seems to be irrelevant for most of the argument.

Hence we can consider $Mod_{A}(Cat)$ of left module objects over $B\Bbb Z$. So the claim is saying that $$\Lambda_\infty \in Mod_{A}(Cat)$$

2. 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 $$\Lambda_1 \simeq colim _{BA} \Lambda_\infty$$

3. 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 $BB\Bbb Z= K(\Bbb Z,2)$.

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).

The problem:

1. We begin with category $C$ with a free $\Bbb Z$-action which fixes the objects of $C$. I.e. Each $g \in \Bbb Z$ induces a functor $$g:C \rightarrow C $$ that fixes the objects. We can thus regard $C$ as an object in $Fun(B\Bbb Z, Cat)$.

2. We construct the quotient category $C':=C/B\Bbb Z$ by quotienting the morphism space by the $\Bbb Z$ action. This is equivalent to taking the colimit of $C$ as an object in $Fun(B\Bbb Z, Cat)$.

3. We wish to compute $|NC'|$. What we do know is that $|NC|\simeq *$.

The proof in the paper goes as follow.

$$|NC'| \simeq |NC|/B\Bbb Z \simeq B(B\Bbb Z) $$

I am rather confused by both the 1st. and 2nd. equivalence. Any comment/reference would be appreciated.

My confusions.

For the 1st. It seems to me the author claims we have a commuting diagram $$ Fun(B\Bbb Z,Cat) \rightarrow Cat $$ $$Fun(B\Bbb Z, Spc) \rightarrow Spc $$ where we take colimits horizontally and geometric realization $|N(-)|$ vertically.

For the 2nd. As far as I could find in the literature, we know the case when we have a constant diagram, of which the colimit is given by $BG$.

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).

The problem:

1. We begin with category $C$ with a free $\Bbb Z$-action which fixes the objects of $C$. I.e. Each $g \in \Bbb Z$ induces a functor $$g:C \rightarrow C $$ that fixes the objects. We can thus regard $C$ as an object in $Fun(B\Bbb Z, Cat)$.

2. We construct the quotient category $C':=C/B\Bbb Z$ by quotienting the morphism space by the $\Bbb Z$ action. This is equivalent to taking the colimit of $C$ as an object in $Fun(B\Bbb Z, Cat)$.

3. We wish to compute $|NC'|$. What we do know is that $|NC|\simeq *$.

The proof in the paper goes as follow.

$$|NC'| \simeq |NC|/B\Bbb Z \simeq B(B\Bbb Z) $$

I am rather confused by both the 1st. and 2nd. equivalence. Any comment/elaborations would be appreciated.

My confusions.

For the 1st. It seems to me the author claims we have a commuting diagram $$ Fun(B\Bbb Z,Cat) \rightarrow Cat $$ $$Fun(B\Bbb Z, Spc) \rightarrow Spc $$ where we take colimits horizontally and geometric realization $|N(-)|$ vertically.

For the 2nd. As far as I could find in the literature, we know the case when we have a constant diagram, of which the colimit is given by $BG$.