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The bare question:

Let $\mathcal{C}$ be an $\infty$-topos, and let $\tau_{\leq 0}\mathcal{C}$ be the subcategory of 0-truncated objects (which is the nerve of an ordinary Grothendieck topos: see HTT 6.4.1.3).

Does the inclusion $\tau_{\leq 0}\mathcal{C} \hookrightarrow \mathcal{C}$ preserve filtered (or at least directed) ($\infty$-)colimits?

Motivation:

Let $Aff_\mathbb{C}$ the Grothendieck site of complex affine schemes. We can then consider the topos of sheaves, $Shv(Aff_\mathbb{C})$, and the $\infty$-topos of $\infty$-stacks, $Shv_\infty(Aff_\mathbb{C})$. The nerve of the first is equivalent to the subcategory of 0-truncated objects in the second.

Given a scheme $X$ and a closed subscheme $Y$ in it defined by a sheaf of ideals $\mathcal{I}$, we can construct the so-called formal completion of $X$ along $Y$ as the directed colimit $X_Y^{\mbox{^}} = \mathrm{colim}\:V(\mathcal{I}^n)$. Typically this is done in $Shv(Aff_\mathbb{C})$. Since the homotopy theory in the latter is trivial, it is also the homotopy colimit of the same diagram. But how does this play with the inclusion $Shv(Aff_\mathbb{C}) \hookrightarrow Shv_\infty(Aff_\mathbb{C})$? Is $X_Y^{\mbox{^}}$ still the homotopy colimit of the same diagram in $Shv_\infty(Aff_\mathbb{C})$?

An affirmative answer to my last question would be enough for me, but I suppose it is a natural question to ask whether this holds for general formal schemes —i.e., sheaves that are locally formal completions as above.

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I believe the answer is YES and, more generally, that $\tau_{\leq n}\mathcal{C}\subset\mathcal{C}$ preserves filtered colimits for any $\infty$-topos $\mathcal{C}$. For the $\infty$-topos of $\infty$-groupoids this is well-known. This implies the result in any presheaf $\infty$-topos since colimits and truncations are computed objectwise. Finally, if the result is true in $\mathcal{C}$ and $\mathcal{D}\subset\mathcal{C}$ is a left exact localization, then the result is true in $\mathcal{D}$ as well because left exact functors preserve $n$-truncated objects (HTT, Prop. 5.5.6.16).

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  • $\begingroup$ Hi Marc, how does the fact that left exact functors preserve n-truncated objects imply the result for $D$. Maybe I am being slow. $\endgroup$
    – David Carchedi
    Commented Mar 7, 2013 at 3:10
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    $\begingroup$ @David #1: You need to apply this fact to both the inclusion $i:\mathcal{D}\subset\mathcal{C}$ and its left exact left adjoint $a:\mathcal{C}\to\mathcal{D}$. Let $(X_k)$ be a filtered diagram in $\mathcal{D}$. If each $X_k$ is $n$-truncated, so is each $iX_k$, whence also $colim (iX_k)$ by the assumption on $\mathcal{C}$, whence $a(colim (iX_k))=colim (X_k)$. $\endgroup$
    – Marc Hoyois
    Commented Mar 7, 2013 at 3:26
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    $\begingroup$ David: We can compute the colimit by computing in $\mathcal{C}$ and then localizing back down to $\mathcal{D}$. So... truncating a filtered colimit of stuff in $\mathcal{D}$ is the same as truncating the localization of a filtered colimit of stuff in $\mathcal{C}$, which is the same as localizing the truncation of a filtered colimit of stuff in $\mathcal{C}$, which is the same as localizing the filtered colimit of truncations, which is the same taking the filtered colimit of truncations. (Is this what it felt like when people used to write out equations in words?) $\endgroup$
    – Dylan Wilson
    Commented Mar 7, 2013 at 3:29
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    $\begingroup$ @David #2: You're right that for sifted colimits you get different things, but sifted colimits are much more general than filtered colimits. Sifted colimits, unlike filtered ones, do not preserve truncated objects: any $\infty$-groupoid whatsoever is a sifted colimit of discrete ones (e.g. a simplicial set representing it). Another difference with the filtered case is that, for a $1$-category, being sifted as an $\infty$-category is strictly stronger than being sifted in the usual sense (e.g. a reflexive coequalizer is $1$-sifted but not $\infty$-sifted). $\endgroup$
    – Marc Hoyois
    Commented Mar 7, 2013 at 3:39
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    $\begingroup$ @Alberto: I can't think of a reference but here's why it's true. Any filtered colimit of $\infty$-groupoids can be computed as the $1$-colimit of a filtered diagram of simplicial sets (this uses HTT 5.3.1.16 and the fact that filtered colimits of simplicial sets are homotopy colimits). Then just use that the functors $\pi_i(-,x)$ preserve filtered colimits. $\endgroup$
    – Marc Hoyois
    Commented Mar 7, 2013 at 15:11

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