3
$\begingroup$

$\DeclareMathOperator\sAn{\mathit{sAn}}\DeclareMathOperator\An{\mathit{An}}$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…?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

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).

Improve this answer
$\endgroup$
3
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.