# Does inclusion from n-stacks into (n+1)-stacks preserve the sheaf condition?

I'm going to describe two situations that seem to contradict each other, and I'm interested to know precisely what's wrong with this reasoning.

1. Let $$M$$ be a manifold, and consider the presheaf $$C^*(-,\mathbb{Z})$$ on $$M$$ sending $$U$$ to $$C^*(U;\mathbb{Z})$$, the complex of (singular) cochains on $$U$$. Then this association is a homotopy sheaf in the sense that for any $$U$$ and a covering $$\{U_i\}_i$$ of $$U$$, the map $$C^*(U;\mathbb{Z}) \to \mathrm{holim} \, C^*(N(\{U_i\});\mathbb{Z})$$ is an equivalence (e.g., in Lurie's formalism by considering the quasicategory $$\operatorname{Mod}_{\mathbb{Z}}$$ and ordinary limits in that sense, or homotopy limits in the model category of chain complexes). For a covering $$U = U_1 \cup U_2$$, the fact that it's a homotopy sheaf is essentially equivalent to Mayer-Vietoris (and more generally, it corresponds to the Mayer-Vietoris spectral sequence for a general covering).

Here's the crux: when viewed as taking values in chain complexes, the constant sheaf $$\underline{\mathbb{Z}}$$ is NOT a homotopy sheaf (even though it is an ordinary sheaf), and the map $$\underline{\mathbb{Z}} \to C^*(-,\mathbb{Z})$$ is a homotopy sheafification.

What this tells us is: even something that's a sheaf in the ordinary $$1$$-categorical sense is NOT necessarily a homotopy sheaf (e.g. assuming $$M$$ has nontrivial cohomology).

Since the sheaf condition is a condition on limits, this tells us that the inclusion of quasicategories from $$\operatorname{Ab}$$ into $$\operatorname{Mod}_{\mathbb{Z}}$$ does not preserve limits.

Something similar happens with etale cohomology on the etale site.

1. It's well-known that schemes are algebraic stacks. In particular, if I have a representable functor on the category of schemes, then if we view it as a functor from schemes to groupoids (via the inclusion of sets into groupoids), then it's a sheaf (aka stack) for the fppf topology. I assume this means that if I have ANY set-valued sheaf for the fppf topology, then if I consider it as a presheaf valued in groupoids, it remains a sheaf (in the appropriate $$2$$-categorical sense).

More generally, a scheme is an algebraic $$n$$-stack, and an $$n$$-stack is an $$m$$-stack for $$m \ge n$$ (c.f. https://arxiv.org/pdf/alg-geom/9609014.pdf, p.3). There's no discussion of needing to sheafify when you do this.

So what's the difference? Am I wrong about one of these two situations? (and if so, which one and how?) Is it some fundamental difference between the abelian and non-abelian settings? Or something else? I talked to my advisor, who has some expertise in this area, but they were confused by this as well.

You said it yourself in the question! The reason that sheaves of abelian groups are not $$\infty$$-sheaves in general, when considered as presheaves taking values in the $$\infty$$-category $$\mathsf{Mod}_{\mathbb Z}$$, is that the $$\infty$$-functor $$\mathsf {Ab} \longrightarrow \mathsf{Mod}_{\mathbb Z}$$ does not preserve limits. And the reason that sheaves of sets are $$\infty$$-sheaves is that the $$\infty$$-functor $$\mathsf{Set} \longrightarrow \mathsf{Spaces}$$ does preserve limits: it has a left adjoint, given by $$\pi_0$$. The same holds for the inclusion of $$k$$-truncated spaces into $$(k+1)$$-truncated spaces, in which case the adjoint is Postnikov truncation, and this answers your question about higher stacks. More generally the full subcategory of $$k$$-truncated objects in any $$\infty$$-category is closed under all limits.

You wrote: "Is it some fundamental difference between the abelian and non-abelian settings?". Perhaps not. The inclusion of the 1-category of abelian groups into the $$\infty$$-category of nonnegatively graded chain complexes preserves all limits, and this is arguably the "true" abelian analogue of the inclusion of sets into spaces via Dold-Kan.

• To clarify, I meant that my question was why one of those functors preserves limits and the other doesn't. Anyway, it seems that the answer is simply that one consists only of connective objects while the other is stable! Apr 7, 2021 at 21:34
• So does the inclusion from sets into spectra not preserve limits, even though the inclusion from sets into spaces does? Apr 7, 2021 at 21:34
• @DavidCorwin: The inclusion (if you can call it that) of sets into spectra does not preserve homotopy limits. The inclusion of spaces into spectra also does not preserve homotopy limits. Apr 7, 2021 at 23:54
• Yes David, you've got it exactly. So the point is that $\pi_0$ is a left adjoint unstably, but stably it is the composition of two truncation functors, one which kills positive degrees (and is a left adjoint) and another which kills negative degrees (and is a right adjoint). Apr 8, 2021 at 2:09

the constant sheaf Z is NOT a homotopy sheaf (even though it is an ordinary sheaf)

This type of phrasing is ambiguous and is probably responsible for the confusion. In this sentence, Z is used to refer to two completely different presheaves:

• the presheaf Z of abelian groups, which sends U to the set of locally constant Z-valued functions on U;

• the presheaf Z[0] of unbounded chain complexes, which sends U to the unbounded chain complex concentrated in degree 0, where it is given by the abelian group of locally constant Z-valued functions on U.

The presheaf Z is indeed a 1-sheaf and an ∞-sheaf of abelian groups.

The presheaf Z[0] is a 1-sheaf of unbounded chain complexes. It is not an ∞-sheaf of unbounded chain complexes and its ∞-sheafification can be computed as the ∞-sheaf of integral singular cochains.

To summarize, regardless of the site (manifolds or schemes), presheaves of connective objects (i.e., with homotopy groups concentrated in nonnegative degrees), whether sets, abelian groups, simplicial sets, nonnegatively graded chain complexes, or connective spectra, are automatically ∞-sheaves, provided their individual values have vanishing homotopy groups in degree 1 and above, and their π_0 is a sheaf in the ordinary sense.

On the other hand, passing to the nonconnective setting, whether unbounded chain complexes or spectra, almost always destroys the ∞-sheaf property, precisely because higher sheaf cohomology groups can be nonvanishing. (Exceptions exist in special cases, e.g., quasicoherent sheaves on affine schemes or Stein spaces, or representable sheaves of abelian Lie groups on cartesian spaces.)

Indeed, the inclusion of connective objects into nonconnective objects already fails to preserve (say) pullbacks. For example, consider an object A with trivial π_k for all k≥1. (Assume all objects are pointed, for simplicity.) The loop space of A is the homotopy pullback of 1→A←1. In the nonconnective setting, π_{−1} of this homotopy pullback will be nontrivial (namely, A in the case of unbounded chain complexes or spectra), whereas in the connective setting the homotopy pullback is 1, with trivial homotopy groups in all degrees.

• I thought the original question is pretty clear. The question is what happens when we include from the category of abelian groups into the category of complexes. Apr 7, 2021 at 20:50
• @DavidCorwin: You asked “So what's the difference?”, and the answer is that the difference is that in one case Z is a presheaf of abelian groups and in the other case, Z is a presheaf of chain complexes, and the two categories have completely different notions of homotopy limits and ∞-sheaves. In your main post, you already pointed out that the inclusion of abelian groups into chain complexes does not preserve homotopy limits, which means that ∞-sheaves are not preserved. Apr 7, 2021 at 21:04
• I made a slight edit to clarify this. Apr 8, 2021 at 20:30
• @DavidCorwin: I added a few more paragraphs based on the newly edited version of the question. Apr 9, 2021 at 1:43