The Stacks project

Proof. Denote $g : \mathop{\mathit{Sh}}\nolimits (\mathcal{X}_{flat, fppf}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{X}_{fppf})$ the morphism of topoi discussed in Section 104.3. Let $x$ be an object of $\mathcal{X}$ lying over a scheme $U$ such that $U \to \mathcal{X}$ is flat, i.e., $x$ is an object of $\mathcal{X}_{flat, fppf}$. By Lemma 104.4.2 part (2)(b) we have $R\Gamma (x, K) = R\Gamma (\mathcal{X}_{flat, fppf}/x, g^{-1}K)$. However, our assumption means that the cohomology sheaves of the object $g^{-1}K$ of $D(\mathcal{X}_{flat, fppf})$ are zero, see Cohomology of Stacks, Definition 103.9.1. Hence $g^{-1}K = 0$ and we win. $\square$

Proof. Denote $g : \mathop{\mathit{Sh}}\nolimits (\mathcal{X}_{flat, fppf}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{X}_{fppf})$ the morphism of topoi discussed in Section 104.3. By Lemma 104.4.2 part (2)(b) our assumption means that $g^{-1}K$ has vanishing cohomology over every object of $\mathcal{X}_{flat, fppf}$ which lies over an affine scheme. Since every object $x$ of $\mathcal{X}_{flat, fppf}$ has a covering by such objects, we conclude that $g^{-1}K$ has vanishing cohomology sheaves, i.e., we conclude $g^{-1}K = 0$. Then of course $R\Gamma (\mathcal{X}_{flat, fppf}, g^{-1}K) = 0$ which in turn implies what we want by Lemma 104.4.2 part (2)(a). $\square$

Proof. By Lemma 104.5.3 part (2)(c) the object $Lg_!K$ is in $D_{\textit{LQCoh}^{fbc}}(\mathcal{O}_\mathcal {X})$. It follows readily from this that the map displayed in the lemma is an isomorphism if $\mathcal{O}(x') \to \mathcal{O}(x)$ is a flat ring map; we omit the details.

In this paragraph we argue that the question is local for the étale topology. Let $x \to x'$ be a general morphism of $\mathcal{X}_{affine}$. Let $\{ x'_ i \to x'\} $ be a covering in $\mathcal{X}_{affine, {\acute{e}tale}}$. Set $x_ i = x \times _{x'} x'_ i$ so that $\{ x_ i \to x\} $ is a covering of $\mathcal{X}_{affine, {\acute{e}tale}}$ too. Then $\mathcal{O}(x') \to \prod \mathcal{O}(x'_ i)$ is a faithfully flat étale ring map and

Thus a simple algebra argument we omit shows that it suffices to prove the result in the statement of the lemma holds for each of the morphisms $x_ i \to x'_ i$ in $\mathcal{X}_{affine}$. In other words, the problem is local in the étale topology.

Choose a scheme $X$ and a surjective smooth morphism $f : X \to \mathcal{X}$. We may view $f$ as an object of $\mathcal{X}$ (by our abuse of notation) and then $(\mathit{Sch}/X)_{fppf} = \mathcal{X}/f$, see Sheaves on Stacks, Section 96.9. By Sheaves on Stacks, Lemma 96.19.10 for example, there exist an étale covering $\{ x'_ i \to x'\} $ such that $x'_ i : U'_ i = p(x'_ i) \to \mathcal{X}$ factors through $f$. By the result of the previous paragraph, we may assume that $x \to x'$ is a morphism which is the image of a morphism $U \to U'$ of $(\textit{Aff}/X)_{fppf}$ by the functor $(\mathit{Sch}/X)_{fppf} \to \mathcal{X}$. At this point we see use that the restriction to $(\mathit{Sch}/X)_{fppf}$ of $Lg_!K$ is equal to $f^*Lg_!K = L(g')_!(f')^*K$ by Lemma 104.3.2. This reduces us to the case discussed in the next paragraph.

Assume $\mathcal{X} = (\mathit{Sch}/X)_{fppf}$ and $x \to x'$ corresponds to the morphism of affine schemes $U \to U'$. We may still work étale (or Zariski) locally on $U'$ and hence we may assume $U' \to X$ factors through some affine open of $X$. This reduces us to the case discussed in the next paragraph.

Assume $\mathcal{X} = (\mathit{Sch}/X)_{fppf}$ where $X = \mathop{\mathrm{Spec}}(R)$ is an affine scheme and $x \to x'$ corresponds to the morphism of affine schemes $U \to U'$. Let $M^\bullet $ be a complex of $R$-modules representing $R\Gamma (X, K)$. By the construction in More on Algebra, Lemma 15.59.10 we may assume $M^\bullet = \mathop{\mathrm{colim}}\nolimits P_ n^\bullet $ where each $P_ n^\bullet $ is a bounded above complex of free $R$-modules. Details omitted; see also More on Algebra, Remark 15.59.11. Consider the complex of modules $M^\bullet _{flat, fppf}$ on $X_{flat, fppf} = (\mathit{Sch}/X)_{flat, fppf}$ given by the rule

This is a complex of sheaves by the discussion in Descent, Section 35.8. There is a canonical map $M^\bullet _{flat, fppf} \to K$ which by our initial remarks of the proof produces an isomorphism on sections over the affine objects of $X_{flat, fppf}$. Since every object of $X_{flat, fppf}$ has a covering by affine objects we see that $M^\bullet _{flat, fppf}$ agrees with $K$.

Let $M^\bullet _{fppf}$ be the complex of modules on $X_{fppf}$ given by the same formula as displayed above. Recall that $Lg_!\mathcal{O} = g_!\mathcal{O} = \mathcal{O}$. Since $Lg_!$ is the left derived functor of $g_!$ we conclude that $Lg_!P_{n, flat, fppf}^\bullet = P_{n, fppf}^\bullet $. Since the functor $Lg_!$ commutes with homotopy colimits (or by its construction in Cohomology on Sites, Lemma 21.37.2) and since $M^\bullet = \mathop{\mathrm{colim}}\nolimits P_ n^\bullet $ we conclude that $Lg_!M^\bullet _{flat, fppf} = M^\bullet _{fppf}$. Say $U = \mathop{\mathrm{Spec}}(A)$, $U' = \mathop{\mathrm{Spec}}(A')$ and $U \to U'$ corresponds to the ring map $A' \to A$. From the above we see that

Since $M^\bullet $ is a K-flat complex of $R$-modules, by transitivity of tensor product it follows that

is a quasi-isomorphism as desired. $\square$

Proof. By Sheaves on Stacks, Lemma 96.26.6 pullback by the comparison morphism $\epsilon : \mathcal{X}_{affine, fppf} \to \mathcal{X}_{affine}$ identifies $\mathit{QC}(\mathcal{X})$ with a full subcategory $Q_\mathcal {X} \subset D(\mathcal{X}_{affine, fppf}, \mathcal{O})$. Using the equivalence of ringed topoi in Sheaves on Stacks, Equation (96.24.3.1) we may and do view $Q_\mathcal {X}$ as a full subcategory of $D(\mathcal{X}_{fppf}, \mathcal{O})$.

Similarly by Lemma 104.5.4 and Remark 104.5.5 we find that $D_\mathit{QCoh}(\mathcal{O}_\mathcal {X})$ may be viewed as the left orthogonal $\mathcal{A}$ of the left admissible subcategory $D_{\textit{Parasitic} \cap \textit{LQCoh}^{fbc}}(\mathcal{O}_\mathcal {X})$ of $D_{\textit{LQCoh}^{fbc}}(\mathcal{O}_\mathcal {X})$.

To finish we will show that $Q_\mathcal {X}$ is equal to $\mathcal{A}$ as subcategories of $D(\mathcal{X}_{fppf}, \mathcal{O})$.

Step 1: $Q_\mathcal {X}$ is contained in $D_{\textit{LQCoh}^{fbc}}(\mathcal{O}_\mathcal {X})$. An object $K$ of $Q_\mathcal {X}$ is characterized by the property that $K$, viewed as an object of $D(\mathcal{X}_{affine, fppf}, \mathcal{O})$ satisfies $R\epsilon _*K$ is an object of $\mathit{QC}(\mathcal{X}_{affine}, \mathcal{O})$. This in turn means exactly that for all morphisms $x \to x'$ of $\mathcal{X}_{affine}$ the map

is an isomorphism, see footnote in statement of Cohomology on Sites, Lemma 21.43.12. Now, if $x' \to x$ lies over a flat morphism of affine schemes, then this means that

This clearly means that $H^ i(K)$ is a sheaf for the étale topology (Sheaves on Stacks, Lemma 96.25.1) and that it has the flat base change property (small detail omitted).

Step 2: $Q_\mathcal {X}$ is contained in $\mathcal{A}$. To see this it suffices to show that for $K$ in $Q_\mathcal {X}$ we have $\mathop{\mathrm{Hom}}\nolimits (K, P) = 0$ for all $P$ in $D_{\textit{Parasitic} \cap \textit{LQCoh}^{fbc}}(\mathcal{O}_\mathcal {X})$. Consider the object

Let $x$ be an object of $\mathcal{X}$ which lies over an affine scheme $U = p(x)$. By Cohomology on Sites, Lemma 21.35.1 we have the first equality in

The second equality stems from the fact that the topos of the site $\mathcal{X}/x$ is equivalent to the topos of the site $\mathcal{X}_{affine}/x$, see Sheaves on Stacks, Equation (96.24.3.1). We may write $K = \epsilon ^*N$ for some $N$ in $\mathit{QC}(\mathcal{O})$. Then by Cohomology on Sites, Lemma 21.43.13 we see that

By Lemma 104.8.1 we see that $R\Gamma (x, P) = 0$ if $U \to \mathcal{X}$ is flat and hence $R\Gamma (x, H) = 0$ under the same hypothesis. By Lemma 104.8.2 we conclude that $R\Gamma (\mathcal{X}, H) = 0$ and therefore $\mathop{\mathrm{Hom}}\nolimits (K, P) = 0$.

Step 3: $\mathcal{A}$ is contained in $Q_\mathcal {X}$. Let $K$ be an object of $\mathcal{A}$ and let $x \to x'$ be a morphism of $\mathcal{X}_{affine}$. We have to show that

is a quasi-isomorphism, see footnote in statement of Cohomology on Sites, Lemma 21.43.12. By the proof of Lemma 104.5.4 and the discussion in Remark 104.5.5 we see that $\mathcal{A}$ is the image of the restriction of $Lg_!$ to $D_\mathit{QCoh}(\mathcal{O}_{\mathcal{X}_{flat, fppf}})$. Thus we may assume $K = Lg_!M$ for some $M$ in $D_\mathit{QCoh}(\mathcal{O}_{\mathcal{X}_{flat, fppf}})$. Then the desired equality follow from Lemma 104.8.3. $\square$