Proving Zariski descent

Question

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!

Answer 1

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.