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Given a pointed $(\infty,n)$-category $\mathcal{C}$, one can define the suspension of $\mathcal{C}$, $\Sigma\mathcal{C}$, via the homotopy pushout of $$\ast\leftarrow \mathcal{C}\rightarrow \ast.$$ Dually one can define $\Omega\mathcal{C}$. Can one explicitly identify these $(\infty,n)$-categories in terms of $\mathcal{C}$?

My vague intuition, based on the case $n=0$, says that the $\Omega\mathcal{C}$ should be the endomorphisms of the distinguished object and $\Sigma\mathcal{C}$ should be what you get when you take the free monoidal $(\infty,n)$-category on $\mathcal{C}$, regard it as an $(\infty,n+1)$-category with one object and then invert the $n+1$-morphisms. However, my understanding of (homotopy) limits and colimits in this setting is pretty poor.

Feel free to use any model you wish.

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    $\begingroup$ If by "homotopy" you mean the usual notion then $\Omega C$ is the automorphisms, not the endomorphisms. Directed delooping in $n$-category theory with $n > (\infty,1)$ requires using "lax" generalizations of homotopy pullbacks, such as "comma objects" ncatlab.org/nlab/show/comma+object $\endgroup$
    – Urs Schreiber
    Commented Sep 5, 2013 at 10:01
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    $\begingroup$ By homotopy pullback/pushout, I will take any construction that is equivalent to the derived pullback/pushout constructed using the projective/injective model structures on diagrams in a combinatorial model category modelling $(\infty,n)$-categories. $\endgroup$
    – Justin Noel
    Commented Sep 5, 2013 at 10:40
  • $\begingroup$ Do you have a proof that $\Omega\mathcal{C}$ is the automorphisms? In particular, why is it always an $(\infty,0)$-category? $\endgroup$
    – Justin Noel
    Commented Sep 5, 2013 at 10:41
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    $\begingroup$ $\Omega \mathcal{C}$ is the $(\infty,n)$-category of automorphisms not the space of automorphisms (i.e. objects are automorphisms but we allow arbitrary morphisms between them). The reason why these are not all endomorphisms is that $\Delta[1]$ is not an interval object but $E[1]$ (the nerve of the contractible groupoid with two objects) is (e.g. in the Joyal model structure or in the Rezk model structure for $\Theta_n$-spaces). $\endgroup$
    – Karol Szumiło
    Commented Sep 5, 2013 at 11:19
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    $\begingroup$ The pushout of $* \sqcup * \to E[1]$ and $* \sqcup * \to *$ is a homotopy pushout too since $* \sqcup * \to E[1]$ is a cofibration. However, I neglected the fact that the analogous pushout of categories is not preserved by the nerve functor. But that's even better, maps $E[1] \to \mathcal{C}$ classify equivalences in $\mathcal{C}$ by definition so maps $E[1] / (* \sqcup *) \to \mathcal{C}$ classify equivalences with the same source and target i.e. automorphisms. $\endgroup$
    – Karol Szumiło
    Commented Sep 5, 2013 at 13:55

1 Answer 1

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First let me thank Urs, Karol, and Rune Haugseng for helpful comments.

Now note that the inclusion, $i$, of $\infty$-groupoids into $(\infty,n)$-categories has an $\infty$-categorical left adjoint, $L$ (for lack of a better name), and a right adjoint $(-)^\prime$.

Given a pointed $(\infty,n)$-category $\mathcal{C}$, the loop category $\Omega \mathcal{C}$ is defined by the following (homotopy) pullback diagram:

\begin{array}{ccc} \Omega\mathcal{C} & \rightarrow & \ast\\ \downarrow & & \downarrow \\ \ast & \rightarrow & \mathcal{C} \end{array}

Now $\mathcal{C}^{\prime}$ is the maximal sub-$\infty$-groupoid of $\mathcal{C}$ (the core). Since $\ast$ is an $\infty$-groupoid, the inclusion $\ast\rightarrow \mathcal{C}$ factors canonically through $\mathcal{C}^\prime$. By a standard finality argument we see that $\Omega \mathcal{C}$ is equivalent to the homotopy pullback: \begin{array}{ccc} \Omega\mathcal{C} \simeq \Omega^{Top}\mathcal{C}^\prime& \rightarrow & \ast\\ \downarrow & & \downarrow \\ \ast & \rightarrow & \mathcal{C}^\prime \end{array}

Regarding $\mathcal{C}^\prime$ as a space (since it is an $\infty$-groupoid), we see that $\Omega\mathcal{C}$ is equivalent to the space of topological (based) loops on $\mathcal{C}^\prime$ since $i$ preserves (homotopy) limits.

Unraveling this a bit, we see that $\Omega(-)$ is naturally equivalent to $i\Omega^{Top}(-)^\prime$ which is a composite of right adjoints. It follows that the left adjoint, $\Sigma(-)$, is naturally equivalent to $i\Sigma^{Top}L(-)$.

As a consequence, an $(\infty,n)$-category which is a loop category is necessarily a loop space. This shows that the inclusion of spectra objects in $(\infty,0)$-categories (i.e., spectra) into spectra objects in $(\infty,n)$-categories is an equivalence. So the two categories have the same stailizations.

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  • $\begingroup$ Could you say a bit more about the "standard finality argument"? $\endgroup$
    – Marc Hoyois
    Commented Sep 27, 2013 at 15:14
  • $\begingroup$ Oh I see: putting $C'$ instead of $C$ in the lower right corner simply doesn't change the universal property of the pullback in the $(\infty,1)$-category of $(\infty,n)$-categories. $\endgroup$
    – Marc Hoyois
    Commented Sep 28, 2013 at 18:31
  • $\begingroup$ Exactly, or one could check that the appropriate comma category is weakly contractible. $\endgroup$
    – Justin Noel
    Commented Oct 1, 2013 at 8:31

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