The Stacks project

Proof. Let $\mathcal{K}$ be an object of $\mathit{QCoh}(\mathcal{O}_{X \times _ R Y})$ and $F_\mathcal {K}$ the functor (56.5.1.1). By the discussion in Example 56.5.1 we already know that $F$ is $R$-linear and commutes with arbitrary direct sums. Since $\text{pr}_2 : X \times _ R Y \to Y$ is affine (Morphisms, Lemma 29.11.8) the functor $\text{pr}_{2, *}$ is exact, see Cohomology of Schemes, Lemma 30.2.3. Hence $F$ is right exact as well, in other words $F$ is as in (1).

Let $F$ be as in (1). Say $X = \mathop{\mathrm{Spec}}(A)$. Consider the quasi-coherent $\mathcal{O}_ Y$-module $\mathcal{G} = F(\mathcal{O}_ X)$. The functor $F$ induces an $R$-linear map $A \to \text{End}_{\mathcal{O}_ Y}(\mathcal{G})$, $a \mapsto F(a \cdot \text{id})$. Thus $\mathcal{G}$ is a sheaf of modules over

By Morphisms, Lemma 29.11.6 we find that there is a unique quasi-coherent module $\mathcal{K}$ on $X \times _ R Y$ such that $F(\mathcal{O}_ X) = \mathcal{G} = \text{pr}_{2, *}\mathcal{K}$ compatible with action of $A$ and $\mathcal{O}_ Y$. Denote $F_\mathcal {K}$ the functor given by (56.5.1.1). There is an equivalence $\text{Mod}_ A \to \mathit{QCoh}(\mathcal{O}_ X)$ sending $A$ to $\mathcal{O}_ X$, see Schemes, Lemma 26.7.5. Hence we find an isomorphism $F \cong F_\mathcal {K}$ by Lemma 56.2.6 because we have an isomorphism $F(\mathcal{O}_ X) \cong F_\mathcal {K}(\mathcal{O}_ X)$ compatible with $A$-action by construction.

This shows that the functor sending $\mathcal{K}$ to $F_\mathcal {K}$ is essentially surjective. We omit the verification of fully faithfulness. $\square$

Proof. Proof of (1). Consider the commutative diagram

Let $\mathcal{F}'$ be a quasi-coherent module on $X'$. We have

Here the first equality is affine base change for the left hand square in the diagram, see Cohomology of Schemes, Lemma 30.5.1. The second equality hold by Remark 56.5.3. The third equality is functoriality of pushforwards for modules. This proves (1).

Proof of (2). Consider the commutative diagram

We have

The first equality by flat base change for the square in the diagram, see Cohomology of Schemes, Lemma 30.5.2. The second equality by functoriality of pullback and the fact that a pullback of tensor products it the tensor product of the pullbacks.

Part (3) is a special case of (2). $\square$

Proof. Consider a morphism $f' : X' \to X$ with $X'$ affine. Since the diagonal of $X$ is affine, we see that $f'$ is an affine morphism (Morphisms, Lemma 29.11.11). Thus $f'_* : \mathit{QCoh}(\mathcal{O}_{X'}) \to \mathit{QCoh}(\mathcal{O}_ X)$ is an $R$-linear exact functor (Cohomology of Schemes, Lemma 30.2.3) which commutes with direct sums (Cohomology of Schemes, Lemma 30.6.1). Thus $F \circ f'_*$ is an $R$-linear, right exact functor which commutes with arbitrary direct sums. Whence $F \circ f'_* = F_{\mathcal{K}'}$ for some $\mathcal{K}'$ on $X' \times _ R Y$ by Lemma 56.5.2. Moreover, given a morphism $f'' : X'' \to X'$ with $X''$ affine we obtain a canonical identification $(f'' \times \text{id}_ Y)^*\mathcal{K}' = \mathcal{K}''$ by the references already given combined with Lemma 56.5.4. These identifications satisfy a cocycle condition given another morphism $f''' : X''' \to X''$ which we leave it to the reader to spell out.

Choose an affine open covering $X = \bigcup _{i = 1, \ldots , n} U_ i$. Since the diagonal of $X$ is affine, we see that the intersections $U_{i_0 \ldots i_ p} = U_{i_0} \cap \ldots \cap U_{i_ p}$ are affine. As above the inclusion morphisms $j_{i_0 \ldots i_ p} : U_{i_0 \ldots i_ p} \to X$ are affine. Denote $\mathcal{K}_{i_0 \ldots i_ p}$ the quasi-coherent module on $U_{i_0 \ldots i_ p} \times _ R Y$ corresponding to $F \circ j_{i_0 \ldots i_ p *}$ as above. By the above we obtain identifications

which satisfy the usual compatibilites for glueing. In other words, we obtain a unique quasi-coherent module $\mathcal{K}$ on $X \times _ R Y$ whose restriction to $U_{i_0 \ldots i_ p} \times _ R Y$ is $\mathcal{K}_{i_0 \ldots i_ p}$ compatible with the displayed identifications.

Next, we construct the transformation $t$. Given a quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F}$ denote $\mathcal{F}_{i_0 \ldots i_ p}$ the restriction of $\mathcal{F}$ to $U_{i_0 \ldots i_ p}$ and denote $(\text{pr}_1^*\mathcal{F} \otimes \mathcal{K})_{i_0 \ldots i_ p}$ the restriction of $\text{pr}_1^*\mathcal{F} \otimes \mathcal{K}$ to $U_{i_0 \ldots i_ p} \times _ R Y$. Observe that

where $\text{pr}_{i_0 \ldots i_ p, 2} : U_{i_0 \ldots i_ p} \times _ R Y \to Y$ is the projection and similarly for the other projection. Moreover, these identifications are compatible with the displayed identifications in the previous paragraph. Recall, from Cohomology of Schemes, Lemma 30.7.1 that the relative Čech complex

computes $R\text{pr}_{2, *}(\text{pr}_1^*\mathcal{F} \otimes \mathcal{K})$. Hence the cohomology sheaf in degree $0$ is $F_\mathcal {K}(\mathcal{F})$. Thus we obtain the desired map $t : F(\mathcal{F}) \to F_\mathcal {K}(\mathcal{F})$ by contemplating the following commutative diagram

We obtain the top row by applying $F$ to the (exact) complex $0 \to \mathcal{F} \to \bigoplus j_{i_0*}\mathcal{F}_{i_0} \to \bigoplus j_{i_0i_1*}\mathcal{F}_{i_0i_1}$ (but since $F$ is not exact, the top row is just a complex and not necessarily exact). The solid vertical arrows are the identifications above. This does indeed define the dotted arrow as desired. The arrow is functorial in $\mathcal{F}$; we omit the details.

We still have to prove the final assertion. Let $f : X' \to X$ be as in the statement of the lemma and let $\mathcal{K}'$ be the quasi-coherent module on $X' \times _ R Y$ constructed in the first paragraph of the proof. If the morphism $f : X' \to X$ maps into one of the opens $U_ i$, then the result follows from Lemma 56.5.4 because in this case we know that $\mathcal{K}_ i = \mathcal{K}|_{U_ i \times _ R Y}$ pulls back to $\mathcal{K}$. In general, we obtain an affine open covering $X' = \bigcup U'_ i$ with $U'_ i = f^{-1}(U_ i)$ and we obtain isomorphisms $\mathcal{K}'|_{U'_ i} = f_ i^*\mathcal{K}_ i$ where $f_ i : U'_ i \to U_ i$ is the induced morphism. These morphisms satisfy the compatibility conditions needed to glue to an isomorphism $\mathcal{K}' = f^*\mathcal{K}$ and we conclude. Some details omitted. $\square$

Proof. We may assume $X$ is affine, so we are in the case of Lemma 56.5.2. By Lemma 56.5.4 we may assume $Y$ is affine. In the affine case the statement translates into Remark 56.3.5. $\square$

Proof. Let $\mathcal{K}$ be as in (2). The functor $F$ in (56.5.1.1) commutes with direct sums. Since by (1) (a) the modules $\mathcal{K}$ is $X$-flat, we see that given a short exact sequence $0 \to \mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3 \to 0$ we obtain a short exact sequence

Since by (2)(b) the higher direct image $R^1\text{pr}_{2, *}$ on the first term is zero, we conclude that $0 \to F(\mathcal{F}_1) \to F(\mathcal{F}_2) \to F(\mathcal{F}_3) \to 0$ is exact and we see that $F$ is as in (1).

Let $F$ be as in (1). Let $\mathcal{K}$ and $t : F \to F_\mathcal {K}$ be as in Lemma 56.5.5. By Lemma 56.5.6 we see that $\mathcal{K}$ is flat over $X$. To finish the proof we have to show that $t$ is an isomorphism and the statement on higher direct images. Both of these follow from the fact that the relative Čech complex

computes $R\text{pr}_{2, *}(\text{pr}_1^*\mathcal{F} \otimes \mathcal{K})$. Please see proof of Lemma 56.5.5 for notation and for the reason why this is so. In the proof of Lemma 56.5.5 we also found that this complex is equal to $F$ applied to the complex

This complex is exact except in degree zero with cohomology sheaf equal to $\mathcal{F}$. Hence since $F$ is an exact functor we conclude $F = F_\mathcal {K}$ and that (2)(b) holds.

We omit the proof that the construction that sends $F$ to $\mathcal{K}$ is functorial and a quasi-inverse to the functor sending $\mathcal{K}$ to the functor $F_\mathcal {K}$ determined by (56.5.1.1). $\square$

Proof. Since $G \circ F : \mathit{QCoh}(\mathcal{O}_ X) \to \mathit{QCoh}(\mathcal{O}_ Z)$ is $R$-linear, exact, and commutes with arbitrary direct sums, we find by Lemma 56.5.7 that there exists an $\mathcal{M}$ in $\mathit{QCoh}(\mathcal{O}_{X \times _ R Z})$ corresponding to $G \circ F$. On the other hand, denote $\mathcal{E} = \text{pr}_{13, *}(\text{pr}_{12}^*\mathcal{K} \otimes \text{pr}_{23}^*\mathcal{L})$. Here and in the rest of the proof we omit the subscript from the tensor products. Let $U \subset X$ and $W \subset Z$ be affine open subschemes. To prove the lemma, we will construct an isomorphism

compatible with restriction mappings for varying $U$ and $W$.

First, we observe that

by construction. Thus we have to show that the same thing is true for $\mathcal{M}$.

Write $U = \mathop{\mathrm{Spec}}(A)$ and denote $j : U \to X$ the inclusion morphism. Recall from the construction of $\mathcal{M}$ in the proof of Lemma 56.5.2 that

where the $A$-module action on the right hand side is given by the action of $A$ on $\mathcal{O}_ U$. The correspondence between $F$ and $\mathcal{K}$ tells us that $F(j_*\mathcal{O}_ U) = b_*(a^*j_*\mathcal{O}_ U \otimes \mathcal{K})$ where $a : X \times _ R Y \to X$ and $b : X \times _ R Y \to Y$ are the projection morphisms. Since $j$ is an affine morphism, we have $a^*j_*\mathcal{O}_ U = (j \times \text{id}_ Y)_*\mathcal{O}_{U \times _ R Y}$ by Cohomology of Schemes, Lemma 30.5.1. Next, we have $(j \times \text{id}_ Y)_*\mathcal{O}_{U \times _ R Y} \otimes \mathcal{K} = (j \times \text{id}_ Y)_*\mathcal{K}|_{U \times _ R Y}$ by Remark 56.5.3 for example. Putting what we have found together we find

with obvious $A$-action. (This formula is implicit in the proof of Lemma 56.5.2.) Applying the functor $G$ we obtain

where $s : Y \times _ R Z \to Y$ and $t : Y \times _ R Z \to Z$ are the projection morphisms. Again using affine base change (Cohomology of Schemes, Lemma 30.5.1) but this time for the square

we obtain

Using Remark 56.5.3 again we find

Applying the functor $\Gamma (W, t_*(-)) = \Gamma (Y \times _ R W, -)$ to this we obtain

as desired. We omit the verification that these isomorphisms are compatible with restriction mappings. $\square$

Proof. The question is local on $T$ hence we may assume $T$ is affine. In this case we can consider the diagram

whose vertical arrows are affine. In particular the pushforward along $T \times _ R Y \to Y$ is faithful and exact (Cohomology of Schemes, Lemma 30.2.3 and Morphisms, Lemma 29.11.6). Chasing around in the diagram using that higher direct images along affine morphisms vanish (see reference above) we see that it suffices to prove

is zero which is true by assumption on $\mathcal{K}$. The equality holds by Remark 56.5.3. $\square$

Proof. Let $\Delta : X \to X \times _ R X$ be the diagonal morphism and set $\mathcal{O}_\Delta = \Delta _*\mathcal{O}_ X$. Since $\Delta $ is a closed immersion have a short exact sequence

Since $\mathcal{K}$ is flat over $X$, the pullback $\text{pr}_{23}^*\mathcal{K}$ to $X \times _ R X \times _ R Y$ is flat over $X \times _ R X$. We obtain a short exact sequence

on $X \times _ R X \times _ R Y$, see Modules, Lemma 17.20.4. Thus, by Lemma 56.5.10 we obtain a surjection

By flat base change (Cohomology of Schemes, Lemma 30.5.2) the source of this arrow is equal to $\text{pr}_2^*\text{pr}_{2, *}\mathcal{K} = \mathcal{O}_ X \boxtimes F(\mathcal{O}_ X)$. On the other hand the target is equal to

which finishes the proof. The first equality holds for example by Cohomology, Lemma 20.54.4 and the fact that $\text{pr}_{12}^*\mathcal{O}_\Delta = (\Delta \times \text{id}_ Y)_*\mathcal{O}_{X \times _ R Y}$. $\square$