6
$\begingroup$

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. 1.4.2.24, 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 5.5.7.1, $\infty$-categories with right adjoints. We prove that $S\in Pr^w$ and that $Pr^w \hookrightarrow Cat_\infty$ reflects (filtered) limits.


$\endgroup$
5
$\begingroup$

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 5.3.6.10. 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 ?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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