# Proving Zariski descent

I want to understand why the functor $$\mathscr{D}$$ sending an affine scheme to its associated derived $$\infty$$-category satisfies Zariski descent. My understanding is that one has to show that given a covering $$\mathfrak{U}$$ of some open $$U$$, there is an equivalence of categories $$\mathscr{D}(U)\simeq\mathscr{D}\left(\check{C}\mathfrak{U}\right)$$. Here, $$\check{C}\mathfrak{U}$$ denotes the Čech nerve associated to $$\mathfrak{U}$$.

In the mathoverflow post Why stable $\infty$-categories?, the author of the answer states that it suffices to show that squares of the form $$\require{AMScd} \begin{CD} \mathscr{D}(R) @>>> \mathscr{D}(R[1/f])\\ @VVV @VVV\\ \mathscr{D}(R[1/g]) @>>> \mathscr{D}(R[1/fg])\end{CD}$$ are cartesian. I do not understand this reduction step: What happened to the Čech nerve?

Many thanks!

• The Zariski topology is generated (as a Grothendieck topology) by covering families with two elements of the form indicated in your question, together with the empty covering family of the empty scheme (should be added to your statement). A proof can be found in Stacks Project, for example, and there is also a survey by Dugger. Therefore, the descent condition can be specified only for such generating families. (And for the empty family, one must say that D(∅) is the terminal ∞-category.) Commented Apr 27, 2023 at 15:15
• @DmitriPavlov "Therefore, the descent condition can be specified only for such generating families." I do not understand this: How does this work, exactly? And, again, what happened to the Čech nerve? Commented Apr 27, 2023 at 15:39
• The best way to define sheaves is not via the cech nerve, but rather via sieves. (This is how it's done in HTT for sheaves with values in an $\infty$-category.) For a covering, one can now express the associated sieve as geometric realisation of the Cech nerve, this leads to the sheaf condition in terms of the Cech nerve. But for a covering consisting of a pair of subobjects, another way to write the covering sieve is as a pushout. This leads to the pullback squares above. And one can show that those sieves generate the Zariski topology. Commented Apr 27, 2023 at 17:31
• @user141099: If you write down the whole Čech nerve diagram, its terms correspond to finite intersections of elements of the open cover. For an open cover with two elements, there is only one nontrivial finite intersection, which is in the bottom right corner of the square. Thus, the whole Čech diagram can be replaced by the bottom and right legs of the square, and the homotopy limit of the Čech diagram can be computed (using the fact that the relevant inclusion of indexing categories is homotopy initial) as the homotopy pullback of the bottom and right leg of the square. Commented Apr 27, 2023 at 19:45
• @DmitriPavlov A reference question: which statement in Stacks Project proves that covering families of two objects (aka. affine Zariski excision) generate the Zariski topology? I find it in [Asok–Hoyois–Wendt, Prop 2.1.3], but not in Stacks Project.
– Z. M
Commented Apr 29, 2023 at 11:30

Ok, here we go. For a covering $$U_1,\ldots,U_n\to U$$ in the Zariski topology, let us denote by $$S(U_1,\ldots,U_n;U)$$ the subpresheaf of the Yoneda embedding $$\underline{U}$$ which assigns to each $$V$$ the subset of all maps $$V\to U$$ which factor through one of the $$U_i$$. An $$F\in \operatorname{PSh}(\operatorname{Aff};\mathcal{C})$$ is then a sheaf if the canonical map $$\operatorname{Map}_{\operatorname{PSh}}(\underline{U}, F) \to \operatorname{Map}_{\operatorname{PSh}}(S(U_1,\ldots,U_n;U), F)$$ is an equivalence, for each covering $$U_1,\ldots,U_n\to U$$. Here we view $$\underline{U}$$ and $$S(U_1,\ldots,U_n;U)$$ as (discrete) space-valued presheaves, and assume that $$\mathcal{C}$$ has enough limits to be copowered over spaces for the mapping spaces above to make sense.
So the sheaves are by definition precisely those presheaves which are local with respect to the inclusions $$S(U_1,\ldots,U_n;U)\to \underline{U}$$. We now claim that it suffices to have this condition for binary covers, i.e. $$n=2$$. So assume $$F$$ is local with respect to all binary covers. Inductively, assume we already know $$F$$ to be local with respect to all $$n$$-ary covers, and let $$U_1,\ldots,U_{n+1}\to U$$ be a cover by $$n+1$$ opens. Let $$V_1,\ldots,V_n$$ be defined by $$V_i=U_i\cap U_{n+1}$$. Then one checks $$S(U_1,\ldots,U_{n+1};U) = S(U_1,\ldots,U_n; \bigcup_{i=1}^n U_i) \amalg_{S(V_1,\ldots,V_n;\bigcup_{i=1}^n V_i)} \underline{U_{n+1}}$$ and $$S(\bigcup_{i=1}^n U_i, U_{n+1}; U) = \underline{\bigcup_{i=1}^n U_i} \amalg_{\underline{\bigcup_{i=1}^n V_i}} \underline{U_{n+1}}$$ so $$F$$ is local with respect to both of the maps $$S(U_1,\ldots,U_{n+1};U)\to S(\bigcup_{i=1}^n U_i, U_{n+1}; U) \to \underline{U}$$ by the inductive assumption. These pushout descriptions take place in space-valued presheaves, so to check them you just check them pointwise, but since everything is discrete and the maps are injective no funny derived business appears.
Also note that the pushout description of $$S(U,V;U\cup V)$$ for binary covers that we used in the proof above directly tells you how to check whether $$F$$ is local with respect to $$S(U,V;U\cup V)\to \underline{U\cup V}$$: $$\operatorname{Map}(\underline{U\cup V},F) = F(U\cup V)$$ by Yoneda, and $$\operatorname{Map}(S(U,V;U\cup V), F) = F(U)\times_{F(U\cap V)} F(V)$$ by the pushout description, so descent now reduces to the pullback squares that you asked about.
• I am not sure how the desired descent property follows from this formal argument. Note that the union of two basic affine open subsets is not necessarily affine (let alone being an affine open subset), and I think that this is a special property of the Zariski topology: say, if $x_1+\dots+x_n=1$ for $n>2$, then $D(x_1)$ and $D(x_2+\dots+x_n)$ covers the whole space, and apply the induction hypothesis to $D(x_2+\dots+x_n)$.