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$\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\Fun{Fun}\newcommand\Cat{\text{Cat}}\DeclareMathOperator\CoInd{CoInd}\newcommand\Spaces{\text{Spaces}}$It is stated in line 10, p76, Thomas Nikolaus, Peter Scholze, On topological cyclic homology, arXiv:1707.01799.

The functor $(-)^{tC_p}:\Sp^{BC_p}\rightarrow \Sp$ is $BC_p$ equivariant, by Theorem I.4.1.

I have two confusions here

Q1. How are $\Sp^{BC_p}$ and $\Sp$ regarded as $BC_p$ equivariant objects?

In the spirit of Maxime's answer to How does one rigorously lift a map $Sp \rightarrow Sp$ of spectra to equivariant spectra? (which I clearly haven't internalized), my understanding is that we should have an adjunction $$ \Fun(B(BC_p), \Cat) \xrightarrow{U} \Cat$$ with right adjoint $\CoInd$, such that $\CoInd(\Sp) \simeq \Sp^{BC_p}$.

Q2. How does the result followm from I.4.1?

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    $\begingroup$ presumably $BC_p$ acts on itself by translation and hence on the functor category $\mathsf{Fun}(\mathrm{B}C_p, \mathsf{Sp})$, and it acts trivially on $\mathsf{Sp}$. (Be careful not to confuse a BC_p action for a C_p action!) for Q2, I'd guess they're pointing to I.4.1 since it gives a statement that is manifestly functorial in the Kan complex S. $\endgroup$
    – Dylan Wilson
    Commented Nov 23, 2020 at 23:43
  • $\begingroup$ Thanks Dylan, for Q1, i edited slightly - is what I comment what we should expect? For Q2, Im confused with your comment, I hope you don't mind elaborating. $\endgroup$
    – Bryan Shih
    Commented Nov 24, 2020 at 0:45
  • $\begingroup$ for your edit: a BC_p-equivariant object in Cat would be a functor from B(BC_p) to Cat, not from BC_p to Cat. $\endgroup$
    – Dylan Wilson
    Commented Nov 24, 2020 at 13:10
  • $\begingroup$ for Q2: Let S be a Kan complex. Then the diagonal p^*: Sp-->Sp^S is certainly Aut(S)-equivariant for the trivial action on the source. It follows that the right adjoint p_* inherits an Aut(S)-action. Now, there is a localization functor Fun(Sp^S, Sp)-->Fun(Sp^S,S) (modulo set theory) which approximates a functor by one which annihilates the 'induced' objects; so this functor then induces a functor on Aut(S)-equivariant objects. The value of the functor on p_* is p_*^T. Now apply the discussion to the case S=BC_p and restrict the Aut(BC_p) action to BC_p, which acts on itself by translation $\endgroup$
    – Dylan Wilson
    Commented Nov 24, 2020 at 13:16
  • $\begingroup$ I'm still lost by "so this functor then induces a functor on Aut(S)-equivariant objects. " - but I should think about this for some more time then come back. And yes! I had a typo on Q1. $\endgroup$
    – Bryan Shih
    Commented Nov 24, 2020 at 15:18

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