Let $C$ be a pointed $\infty$-category which admits finite limits.

Let $Sp(C)$ denote the $\infty$ category of spectrum objects. One way to define, i.e., is by taking the homotopy limit in $Cat_\infty$, the $\infty$-category of categories. $$Sp(C):= \varprojlim \left( \cdots \xrightarrow {\Omega} C \xrightarrow {\Omega} C \right) $$

Let us denote $\Omega^\infty: Sp(C) \rightarrow C$ as the projection onto the last component.

I would like to understand what categorical properties of $\Omega^\infty$ satisfy. My question is

If each $\Omega$ commute with $I$-indexed limit does this imply $\Omega^\infty$ does too?

The reason I am concerned with this question: It is claimed in C.1.4.1, that

if $C$ is a prestable and presentable $\infty$-category and $\Omega:C \rightarrow C$ commutes with filtered colimits then $\Omega^\infty$ commutes with filtered colimits.

A prestable $\infty$-category by definition can be intriscally characterized, C.1.2.1 as a category which satisfies the following conditions

  • pointed and admits finite colimits.
  • suspension is fully fiathful
  • every morphism $Y \rightarrow \Sigma Z$ lies in a pullback square with top right part $X \rightarrow Y \rightarrow \Sigma Z$ and bottom left $0$. Further, the sequence $X \rightarrow Y \rightarrow \Sigma Z$ is a cofiber sequence.

I have recorded my thoughts below, which one may safely ignore.

Both strategies I know do not really apply - these are based on the case $C=S_*$, the $\infty$-cat of pointed spaces.

Strategy 1. $\Omega^\infty: Sp(S) \rightarrow S_*$. $\Omega^\infty$ is corepresented by $\mathbb{S}=\Sigma^\infty S^0$, the sphere spectrum, where we $\Sigma^\infty$ is left adjoint of $\Omega^\infty$. Now by noting that that $S^0$ is a compact object in $S_*$ result follows.

Strategy 2. Consider the $\infty$-cat $Pr^\omega$ of compactly generated, in the sense of, $\infty$-categories with right adjoints. We prove that $S\in Pr^w$ and that $Pr^w \hookrightarrow Cat_\infty$ reflects (filtered) limits.


The result is true, more generally, if you take a class of diagrams $\mathcal K$ and the $\infty$-category $\widehat{Cat_\infty}^\mathcal K$ of $\infty$-categories that have all $\mathcal K$-indexed colimits, and functors between them that preserve those, then the forgetful functor $\widehat{Cat_\infty}^\mathcal K\to \widehat{Cat_\infty}$ preserves all limits, in fact it has a left adjoint.

This is stated as Corollary in Lurie's Higher Topos Theory (with his notations, $\mathcal K' =$ my $\mathcal K$, and his $\mathcal K= \emptyset$).

From this, your result follows, as if $\Omega$ preserves $I$-indexed colimits, then your diagram lives in $\widehat{Cat_\infty}^{\{I\}}$, so its limit does too, and the projection functors too, in particular $\Omega^\infty: Sp(C)\to C$ is one of those projection functors, so it preserves $I$-indexed colimits (this is, of course, assuming that $C$ has all $I$-indexed colimits - which is the case in the statement you refer to, as of course a presentable $\infty$-category has all filtered colimits)

Your strategy 1 is in this sense somehow misguided, as proving that $\mathbb S$ is compact essentially uses that $\Omega^\infty$ preserves filtered colimits.

Actually, a less general, but perhaps easier proof works in the special case of $Sp(C)$ and filtered colimits: $Sp(C)$ can be seen as a certain full subcategory of $Fun(\mathbb{Z\times Z},C)$ (such a functor is a grid, $Sp(C)$ is the full subcategory on those grids that are only $0$ objects off the diagonal, and such that certain squares are pullbacks), and $\Omega^\infty$ is then simply the restriction to this subcategory of the evaluation at $0$.

Now if $C$ has all filtered colimits, and $\Omega$ commutes with those, then $Sp(C)\subset Fun(\mathbb{Z\times Z},C)$ is closed under filtered colimits (the only pullbacks appearing in its definition are pullbacks defining $\Omega$), so that, as in functor categories in general, filtered colimits in $Sp(C)$ are computed pointwise; and so in particular $\Omega^\infty$ commutes with those.

This second proof is less general, but it's easier and gets you what you want- and perhaps it allows for a better understanding in this specific context ?


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