This is rather specific B.5 of Thomas Nikolaus, Peter Scholze, *On topological cyclic homology*, arXiv:[1707.01799](https://mathoverflow.net/questions/376490/computing-homotopy-colimit-of-a-space-with-free-s1-action) (on last line p147), which I am having fundamental confusion.




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We have the categories $\Lambda:=\Lambda\_\infty/B\Bbb Z, \Lambda\_\infty$ explained in my previous [question](https://mathoverflow.net/questions/376490/computing-homotopy-colimit-of-a-space-with-free-s1-action).


In B.5, the authors describes a functor given by composition
$$ Fun(\Lambda^{op}, C) \rightarrow Fun^{B\Bbb Z}(\Lambda\_\infty^{op}, C) \rightarrow Fun^{B\Bbb Z}(pt, C) = C^{BB\Bbb Z} = C^{B\Bbb T} $$




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I have fundamental confusion in the first 2 arrows. Edit:11/24/20.




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**Q1: What exactly is the cateogry $Fun^{B\Bbb Z}(\Lambda\_\infty^{op}, C)$**. I understand it is to be understood as $B\Bbb Z$ equivariant maps.


But how is this made precise? Irregardless, I'd expect
$$ Fun^{B\Bbb Z}(\Lambda\_\infty, C) \subset Fun(\Lambda\_\infty^{op}, C )$$


But without a concrete meaning, I could not make sense of the following two.


**Q1a: What is the relationship $Map\_{Fun(B\Bbb Z, Cat)}(\Lambda\_\infty,C)$ and $Fun^{B\Bbb Z}(\Lambda\_\infty, C)$?**


In fact, are there general result about symmetric monoidal cateogires $C$ which is enriched over itself and the mapping spaces of its objects?


**Q2 why does taking collimit preserve $B\Bbb Z$-equivariance?**


**Q3: How do we show $Fun^{B\Bbb Z}(pt, C)=C^{BB\Bbb Z}$?**



Equivariant colimit and equivariant functors
Q1: For any $C,D\in Fun(BG,Cat\_\infty)$, $Fun(C,D)$ acquires a $G$-action too. Informally, this is described as $F\mapsto gF(g^{-1}-)$, and this is in fact an accurate description if $G$ is a discrete group and $C,D$ are $1$-categories; but more generally, formally you can see it as an internal hom in $Fun(BG,Cat\_\infty)$.


Indeed, on $Fun(BG,Cat\_\infty)$, $C\times-$ preserves arbitrary colimits (as these are computed pointwise, and it does in $Cat\_\infty$), so that it admits a right adjoint, which we can denote by $Fun(C,-)$. Indeed, one may check (by abstract nonsense) that the underlying object of this $\infty$-category with $G$-action is $Fun(C,D)$.


In particular, taking the $G$-fixed points of $Fun(C,D)$ makes sense; but now we really want the homotopy fixed points even if $G$ is discrete and $C,D$ are $1$-categories: you don't want $gF(g^{-1}-) = F$, but you want the data of an isomorphism $\rho\_g: F\to gF(g^{-1}-)$ so that the various composites are compatible (in a homotopy coherent fashion)


These homotopy fixed points are $Fun^G(C,D)$. In your situation, $G=B\mathbb Z$ and $C=\Lambda\_\infty$ with its $B\mathbb Z$-action.


In particular, note that it is note quite a subcategory, although it does come with a "forgetful" functor $Fun^{B\mathbb Z}(\Lambda\_\infty,D)\to Fun(\Lambda\_\infty, D)$


Q2: The authors claim that $\mathrm{colim}: Fun(\Lambda\_\infty,C)\to C$ is equivariant, which gives their construction by taking $B\mathbb Z$-fixed points.


To see why this functor is actually equivariant takes some work as far as I can tell (although maybe there are simpler solutions) : start by noting that the right adjoint, given by the "diagonal" $C\to Fun(\Lambda\_\infty,C)$ is itself equivariant. This is clear from the definition of $Fun(\Lambda\_\infty,C)$, as this is an internal hom, so it suffices to check that the projection $C\times \Lambda\_\infty\to C$ is equivariant but this is obvious.


So we have a left adjoint whose right adjoint is equivariant. Now the rest of this works in this generality so let's just write it that way : we have an adjunction $L\dashv R$, $L:D\to E$ between $\infty$-categories with $G$-action, where $R:E\to D$ can lives in $Fun(BG,Cat\_\infty)$ (and $L$ only in $Cat\_\infty$ a priori).


But now one may consider $Adj\_R$, the $\infty$-category of $\infty$-categories and right adjoints between them. It follows easily that $R: E\to D$ can be seen as an arrow in $Fun(BG,Adj\_R)$. Now $Adj\_R\simeq Adj\_L^{op}$ in the obvious manner (I write "obvious" but in fact this requires some work to set up $\infty$-categorically), so that we can see $L: D\to E$ as an arrow in $Fun((BG)^{op},Adj\_L)$. Composing with the canonical equivalence $(BG)^{op}\simeq BG$, we see that $L: D\to E$ canonically acquires a $G$-equivariant structure.


There's some detail I'm slipping under the rug here: that $D,E$ have the correct $G$-action when you pass from $Adj\_R$ to $Adj\_L$ and then from $(BG)^{op}$ to $BG$. This follows essentially from the fact that $g$ and $g^{-1}$ act as inverses, and hence are adjoint to one another, essentially uniquely (given that the unit and co-unit are determined).


Again, there may be a simpler way to see that $\mathrm{colim}$ has an equivariant structure, but I'm not sure what that would be (there are certainly other ways to *phrase* what I wrote, and some of them might actually be simpler)


Q3: $Fun^G(pt,C)$ is $(Fun(pt, C))^{hG}$, but $Fun(pt,C) \simeq C$. One may now check that if $C$ has a trivial $G$-action, then so does $Fun(pt,C)$, and the equivalence above is a trivial-action equivalence. It then follows that in this case, $Fun^G(pt,C) = C^{hG}$.


But now, whenever $C$ is an $\infty$-category with trivial $G$-action, its homotopy fixed points are just $Fun(BG,C)$, so in the case $G=B\mathbb Z$, you get $Fun(BB\mathbb Z,C)= C^{BB\mathbb Z}$.


