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$\DeclareMathOperator\sAn{\mathit{sAn}}\DeclareMathOperator\An{\mathit{An}}$For the category of simpicial animas (simplicial $\infty$-groupoids if you like) $\sAn$, we have the evaluation functor $\mathrm{ev}_n:\sAn\rightarrow \An$ with a left adjoint $\operatorname{const}_n$ and the realization functor $\lvert\ \rvert:\sAn\rightarrow \An$ with a right adjoint being the Rezk nerve.

I wonder why $\Omega \lvert\operatorname{const}_1 X\rvert$ is $X$. This is used in the (4.1.27) in the Lecture notes: Lecture Notes on Algebraic K-Theory.

I can't even give a direct description for $\operatorname{const}_1X$. Is this something concerning the theory of Segal spaces…?

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I don't think this is true as written, for example there are anima $X$ which are not loop spaces of anything (e.g. $X=S^2$). One can also directly see that $\mathrm{ev}_n$ is corepresented by $\Delta^n$, so I think the left adjoint must be given by $X\mapsto \Delta^n\otimes X$ (i.e. the object of $\mathrm{sAn}$ which takes $[m]$ to $\mathrm{Hom}_{\Delta}([m],[n])\otimes X$). This has realisation just equivalent to $X$.

I also don't see the claim you're asking about in the linked notes. In 4.1.27 they rather seem to argue that $\mathrm{const}_1\mathcal{C}$ realizes to $\mathcal{C}^\simeq$ (where they probably mean to first pass from simplicial categories to simplicial anima by discarding noninvertible morphisms levelwise, this then would fit with the above).

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  • $\begingroup$ Thanks for your answer. In 4.1.27 $const_1(C)\rightarrow Q(C)$ only implies $|const_1(C^{\simeq})|\rightarrow |Q(C)^{\simeq}|$, but $K(C)=\Omega |Q(C)^{\simeq}|$(as in definition 4.1.24) so it seems that there may be some natural morphism $C^{\simeq}\rightarrow \Omega|const_1(C^{\simeq})|$. That's why I expected something about $\Omega|const_1(C^{\simeq})|$ to be told. $\endgroup$
    – XiaYu
    Commented Aug 7 at 1:35
  • $\begingroup$ And I totally agree with you that this shouldn't be a equivalence. $\endgroup$
    – XiaYu
    Commented Aug 7 at 1:39
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    $\begingroup$ AH, I think I know what's going on: When they write $\mathrm{const}_1(\mathcal{C})$, they don't mean your $\mathrm{const}_1$, instead they mean the simplicial diagram left Kan extended from the functor $\Delta_{\leq 1}\to \mathrm{Cat}$ taking $[0]\mapsto 0$ and $[1]\mapsto \mathcal{C}$. This should realize to the suspension of $\mathcal{C}^\simeq$, and a map $\Sigma\mathcal{C}^\simeq \to |Q(\mathcal{C})|$ corresponds to a map $\mathcal{C}^\simeq \to \Omega|Q(\mathcal{C})|$ by adjunction. $\endgroup$
    – Achim Krause
    Commented Aug 7 at 9:17

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