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$\DeclareMathOperator\holim{holim}\DeclareMathOperator\hocolim{hocolim}$Let $\mathcal{F}$ be a simplicial (pre)sheaf on some site $\mathcal{C}$ (assume the site has enough stalks; if you like also assume every representable functor on $\mathcal{C}$ is a sheaf). Suppose $\mathcal{G}$ is a (pre)sheaf of groups acting on $\mathcal{F}$. Then we can form a homotopy quotient $\mathcal{F}_{h\mathcal{G}}$ and a homotopy fixed point sheaf $\mathcal{F}^{h\mathcal{G}}$ via a section wise prescription (this is my naive 'on the spot' construction - I don't have a reference and would like one if it exists for the 'correct' construction):

$\mathcal{F}_{h\mathcal{G}}(U) = \mathcal{F}(U)_{h\mathcal{G}} := \hocolim_{\mathcal{G}(U)} F(U)$,

$\mathcal{F}^{h\mathcal{G}}(U) = \mathcal{F}(U)^{h\mathcal{G}} := \holim_{\mathcal{G}(U)} F(U)$,

for $U \in \mathcal{C}$.

If $\mathcal{F}_x$ is a stalk, then we can also form:

$(\mathcal{F}_x)_{h\mathcal{G}} := \hocolim_{\mathcal{G}_x}\mathcal{F}_x$,

$(\mathcal{F}_x)^{h\mathcal{G}} := \holim_{\mathcal{G}_x}\mathcal{F}_x$.

Question 1: Is $(\mathcal{F}_x)_{h\mathcal{G}} \simeq (\mathcal{F}_{h\mathcal{G}})_x$?

Question 2: Is $(\mathcal{F}_x)^{h\mathcal{G}} \simeq (\mathcal{F}^{h\mathcal{G}})_x$?

In both cases I am worried about commuting colimits (albeit filtrant) with homotopy limits/colimits.

Naively, 1) seems to be ok to me: using the explicit Borel model for the homotopy colimit, it's just the diagonal for a bisimplicial complex and stalks are more or less by definition taken level wise. Regardless, I worry I am missing some subtlety, and am quite lost with the dual homotopy fixed point model in terms of a totalization of a cosimplicial set.

Question 3: Is there a decent reference for these constructions for a neophyte for simplicial (pre)sheaves (I am not particularly versed with the subtleties of simplicial methods - my training is in representation theory)? I have Jardine's 'Local Homotopy Theory' which is a godsend for a lot of stuff, but seems to not quite have much along these lines.

Added later: In my questions I am implicitly assuming that stalks are given by a filtrant colimit. This is unnecessary for Question 1, but probably is required for Question 2 to have any hope of having an affirmative answer — along with the group being finite (see Dmitri's answer and the comments underneath it).

Added even later: Question 1 has been sorted in the affirmative by Dmitri's answer below. Question 2 also has an affirmative answer assuming that the diagram is finite (so say $G$ is a constant presheaf with stalk a finite group). This is Lemma 1.20 in Morel-Voevodsky "$\mathbb{A}^1$-homotopy of schemes". They state it without proof, so I give an elementary one here (leaving the details of a final check out).

Please see Maxime Ramzi's counterexample in the comments to the accepted answer. Either I am missing something, or the claim in Morel-Voevodsky needs refinement.

Do note, it has absolutely zilch to do with filtrant limits.:

The homotopy limit over a finite diagram can be expressed as a limit, over a different, but still finite diagram. Stalks, by definition, commute with finite limits. Hence, all that remains to be done is to check that when we do this commutation the resulting limit (in simplicial sets) is precisely the homotopy limit of the stalk (or that the canonical map between the two simplicial sets obtained is a weak equivalence).


In fact, essentially the same proof should also work for an arbitrary site using pullback to a Boolean localization (but I don’t really understand those, so…).

P.S. In all honesty I have not checked the last `check’. However, at this point I am impatient enough to give Morel-Voevodsky the benefit of the doubt (given that they use the cited result everywhere and it is a landmark paper).

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Taking stalks always commutes with taking homotopy orbits, since filtered colimits of simplicial sets are also filtered homotopy colimits, and homotopy colimits commute with homotopy colimits.

Taking stalks commutes with taking homotopy fixed points with respect to a finite group only if additional fibrancy conditions are satisfied, as explained in another answer. Without additional conditions, there are counterexamples.

Concerning references, there are not so many accessible ones. Perhaps Dugger's A primer on homotopy colimits may be helpful. Bousfield and Kan in their book “Homotopy Limits, Completions and Localizations” prove in §XII.3.5 that filtered colimits of simplicial sets are homotopy colimits.

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    $\begingroup$ Many thanks Dmitri. This is exactly the kind of statement I was looking for (the finite group statement also matches with my intuition). Off the top of your head do you know a convenient reference (accessible or not) for these facts (filtered colimits in sSet = filtered hocolims + commutation)? $\endgroup$
    – rvk
    Commented Sep 30, 2021 at 14:05
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    $\begingroup$ Homotopy fixed points along a finite group actions are homotopy limits along $BG$, so they are never finite limits unless $G= *$. In particular, $(-)^{hG}$ usually does not preserve filtered colimits $\endgroup$ Commented Sep 30, 2021 at 14:11
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    $\begingroup$ @rvk: For filtered colimits in sSet = filtered hocolimits, there is a bunch of references listed here: mathoverflow.net/questions/361769/…. For a concrete reference, see Lemma 2.2(iii) in arxiv.org/abs/1510.04969v3. $\endgroup$ Commented Oct 1, 2021 at 0:20
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    $\begingroup$ @rvk: For the commutativity of homotopy colimits and homotopy colimits see, for example, the reference given in mathoverflow.net/questions/33556/…. $\endgroup$ Commented Oct 1, 2021 at 0:23
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    $\begingroup$ Yes, just look at the fixed points of $BG$ under the trivial action. This is $map(BG, BG)$ which is different from $colim_n map(BG, BG^{(n)})$ as the identity doesn't factor through any skeleton of $BG$ $\endgroup$ Commented Oct 1, 2021 at 6:03

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