The Stacks project

Proof. The existence and uniqueness of the extension is a general fact, see Categories, Lemma 4.26.2. To see that the lemma applies observe that coherent modules on locally Noetherian algebraic stacks are by definition modules of finite presentation, see Cohomology of Stacks, Definition 103.17.2. Hence a coherent module on $\mathcal{X}$ is a categorically compact object of $\mathit{QCoh}(\mathcal{O}_\mathcal {X})$ by Cohomology of Stacks, Lemma 103.13.5. Finally, every quasi-coherent module is a filtered colimit of its coherent submodules by Cohomology of Stacks, Lemma 103.18.1.

Since $F$ is additive, also the extension of $F$ is additive (details omitted). Since $F$ is a tensor functor and since colimits of modules commute with taking tensor products, also the extension of $F$ is a tensor functor (details omitted).

In this paragraph we show the extension commutes with arbitrary direct sums. If $\mathcal{F} = \bigoplus _{j \in J} \mathcal{H}_ j$ with $\mathcal{H}_ j$ quasi-coherent, then $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits _{J' \subset J\text{ finite}} \bigoplus _{j \in J'} \mathcal{H}_ j$. Denoting the extension of $F$ also by $F$ we obtain

Thus $F$ commutes with arbitrary direct sums.

In this paragraph we show that the extension is right exact. Suppose $0 \to \mathcal{F} \to \mathcal{F}' \to \mathcal{F}'' \to 0$ is a short exact sequence of quasi-coherent $\mathcal{O}_\mathcal {X}$-modules. Then we write $\mathcal{F}' = \bigcup \mathcal{F}'_ i$ as the union of its coherent submodules (see reference given above). Denote $\mathcal{F}''_ i \subset \mathcal{F}''$ the image of $\mathcal{F}'_ i$ and denote $\mathcal{F}_ i = \mathcal{F} \cap \mathcal{F}'_ i = \mathop{\mathrm{Ker}}(\mathcal{F}'_ i \to \mathcal{F}''_ i)$. Then it is clear that $\mathcal{F} = \bigcup \mathcal{F}_ i$ and $\mathcal{F}'' = \bigcup \mathcal{F}''_ i$ and that we have short exact sequences

Since the extension commutes with filtered colimits we have $F(\mathcal{F}) = \mathop{\mathrm{colim}}\nolimits _{i \in I} F(\mathcal{F}_ i)$, $F(\mathcal{F}') = \mathop{\mathrm{colim}}\nolimits _{i \in I} F(\mathcal{F}'_ i)$, and $F(\mathcal{F}'') = \mathop{\mathrm{colim}}\nolimits _{i \in I} F(\mathcal{F}''_ i)$. Since filtered colimits of sheaves of modules is exact we conclude that the extension of $F$ is right exact.

The proof is finished as a right exact functor which commutes with all coproducts commutes with all colimits, see Categories, Lemma 4.14.12. $\square$

Proof. We note that $g$ is quasi-compact and quasi-separated, see Morphisms of Stacks, Lemma 101.7.8. In Cohomology of Stacks, Proposition 103.11.1 we have constructed the functor $g_{\mathit{QCoh}, *} : \mathit{QCoh}(\mathcal{O}_ U) \to \mathit{QCoh}(\mathcal{O}_\mathcal {X})$. By Cohomology of Stacks, Remarks 103.11.3 and 103.10.6 we obtain a multiplication

which turns $g_{\mathit{QCoh}, *}\mathcal{O}_ U$ into a commutative $\mathcal{O}_\mathcal {X}$-algebra. Hence $C = F(g_{\mathit{QCoh}, *}\mathcal{O}_ U)$ is a commutative algebra object in $\text{Mod}_ B$, in other words, $C$ is a commutative $B$-algebra. Observe that we have a map $\kappa : A \to \text{End}_{\mathcal{O}_\mathcal {X}}(g_{\mathit{QCoh}, *}\mathcal{O}_ U)$ such that for any $a \in A$ the diagram

commutes. It follows that we get a map $\kappa ' = F(\kappa ) : A \to \text{End}_ B(C)$ such that $\kappa '(a)(c) c' = \kappa '(a)(cc')$ and of course this means that $a \mapsto \kappa '(a)(1)$ is a ring map $A \to C$.

The morphism $g : U \to \mathcal{X}$ is affine, see Morphisms of Stacks, Lemma 101.9.4. Hence $g_{\mathit{QCoh}, *}$ is exact and commutes with direct sums by Cohomology of Stacks, Lemma 103.13.4. Thus $F \circ g_{\mathit{QCoh}, *} : \text{Mod}_ A \to \text{Mod}_ B$ is a right exact functor which commutes with direct sums and which sends $A$ to $C$. By Functors and Morphisms, Lemma 56.3.1 we see that the functor $F \circ g_{\mathit{QCoh}, *}$ sends an $A$-module $M$ to $M \otimes _ A C$ viewed as a $B$-module. $\square$

Proof. Consider the natural map $1 : \mathcal{O}_\mathcal {X} \to g_{\mathit{QCoh}, *}\mathcal{O}_ U$ in $\mathit{QCoh}(\mathcal{O}_\mathcal {X})$. Pulling back to $U$ and using adjunction we find that the composition

is the identity in $\mathit{QCoh}(\mathcal{O}_ U)$. Write $g_{\mathit{QCoh}, *}\mathcal{O}_ U = \mathop{\mathrm{colim}}\nolimits \mathcal{F}_ i$ as a filtered colimit of coherent $\mathcal{O}_\mathcal {X}$-modules, see Cohomology of Stacks, Lemma 103.18.1. For $i$ large enough the map $1 : \mathcal{O}_\mathcal {X} \to g_{\mathit{QCoh}, *}\mathcal{O}_ U$ factors through $\mathcal{F}_ i$, see Cohomology of Stacks, Lemma 103.13.5. Say $s : \mathcal{O}_\mathcal {X} \to \mathcal{F}_ i$ is the factorization. Then

is the identity. In other words, we see that $s$ becomes the inclusion of a direct summand upon pullback to $U$. Set $\mathcal{F}_ i^\vee = hom(\mathcal{F}_ i, \mathcal{O}_\mathcal {X})$ with notation as in Cohomology of Stacks, Lemma 103.10.8. In particular there is an evaluation map $ev : \mathcal{F}_ i \otimes _{\mathcal{O}_\mathcal {X}} \mathcal{F}_ i^\vee \to \mathcal{O}_\mathcal {X}$. Evaluation at $s$ defines a map $s^\vee : \mathcal{F}_ i^\vee \to \mathcal{O}_\mathcal {X}$. Dual to the statement about $s$ we see that $g^*(s^\vee )$ is surjective, see see Cohomology of Stacks, Section 103.12 for compatibility of $hom$ and $\otimes $ with restriction to $U$. Since $g$ is surjective and flat, we conclude that $s^\vee $ is surjective (see locus citatus). Since $F$ is right exact, we conclude that $F(\mathcal{F}_ i^\vee ) \to F(\mathcal{O}_\mathcal {X}) = B$ is surjective. Choose $\lambda \in F(\mathcal{F}_ i^\vee )$ mapping to $1 \in B$. Denote $e = F(s)(1) \in F(\mathcal{F}_ i)$ the image of $1$ by the map $F(s) : B = F(\mathcal{O}_\mathcal {X}) \to F(\mathcal{F}_ i)$. Then the map

sends $e \otimes \lambda $ to $1$ by construction. Hence the map $B \to F(\mathcal{F}_ i)$, $b \mapsto be$ is universally injective because we have the one-sided inverse $F(\mathcal{F}_ i) \to B$, $\xi \mapsto F(ev)(\xi \otimes \lambda )$. Since this is true for all $i$ large enough we conclude. $\square$

Proof. The map $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(B)$ is surjective as $B \to C$ is universally injective. Thus it suffices to show that $B \to C$ is flat which follows from Descent, Theorem 35.4.25. $\square$

Proof. We abbreviate $\mathcal{P} = \mathcal{Y} \times _\mathcal {X} \mathcal{Z}$. Since $a \circ \text{pr}_1 = f$ and $b \circ \text{pr}_2 = f$ we obtain maps $a_*\mathcal{O}_\mathcal {Y} \to f_*\mathcal{O}_\mathcal {P}$ and $b_*\mathcal{O}_\mathcal {Z} \to f_*\mathcal{O}_\mathcal {P}$ (using relative pullback maps, see Sites, Section 7.45). Hence we obtain a relative cup product

Applying $Q$ and its compatibility with tensor products (Cohomology of Stacks, Remark 103.10.6) we obtain an arrow $Q(\mu ) : a_{\mathit{QCoh}, *}\mathcal{O}_\mathcal {Y} \otimes _{\mathcal{O}_\mathcal {X}} b_{\mathit{QCoh}, *}\mathcal{O}_\mathcal {Z} \to f_{\mathit{QCoh}, *}\mathcal{O}_{\mathcal{Y} \times _\mathcal {X} \mathcal{Z}}$ in $\mathit{QCoh}(\mathcal{O}_\mathcal {X})$. Next, choose a scheme $U$ and a surjective smooth morphism $U \to \mathcal{X}$. It suffices to prove the restriction of $Q(\mu )$ to $U_{\acute{e}tale}$ is an isomorphism, see Cohomology of Stacks, Section 103.12. In turn, by the material in the same section, it suffices to prove the restriction of $\mu $ to $U_{\acute{e}tale}$ is an isomorphism (this uses that the source and target of $\mu $ are locally quasi-coherent modules with the base change property). Moreover, we may compute pushforwards in the étale topology, see Cohomology of Stacks, Proposition 103.8.1. Then finally, we see that $a_*\mathcal{O}_\mathcal {Y}|_{U_{\acute{e}tale}} = (V \to U)_{small, *}\mathcal{O}_ V$ where $V = U \times _\mathcal {X} \mathcal{Y}$. Similarly for $b_*$ and $f_*$. Thus the result follows from the Künneth formula for flat, quasi-compact, quasi-separated morphisms of schemes, see Derived Categories of Schemes, Lemma 36.23.1. $\square$

Proof. Choose an affine scheme $U = \mathop{\mathrm{Spec}}(A)$ and a surjective smooth morphism $g : U \to \mathcal{X}$, see Properties of Stacks, Lemma 100.6.2. Since the diagonal of $\mathcal{X}$ is affine, we see that $U_ i = \mathop{\mathrm{Spec}}(B) \times _{f_ i, \mathcal{X}, g} U$ is affine. Say $U_ i = \mathop{\mathrm{Spec}}(C_ i)$. Then $C_ i$ is the $B$-algebra endowed with ring map $A \to C_ i$ constructed in Lemma 106.16.2 using the functor $F = f_ i^*$. Therefore $t$ induces an isomorphism $C_1 \to C_2$ of $B$-algebras, compatible with the ring maps $A \to C_1$ and $A \to C_2$. In other words, we have a commutative diagrams

This already shows that the objects $f_1$ and $f_2$ of $\mathcal{X}$ over $\mathop{\mathrm{Spec}}(B)$ become isomorphic after the smooth covering $\{ U_1 \to \mathop{\mathrm{Spec}}(B)\} $. To show that this descends to an isomorphism of $f_1$ and $f_2$ over $\mathop{\mathrm{Spec}}(B)$, we have to show that our isomorphism (which comes from the commutative diagrams above) is compatible with the descent data over $U_1 \times _{\mathop{\mathrm{Spec}}(B)} U_1$. For this we observe that $U \times _\mathcal {X} U$ is affine too, that we have the morphism $g' : U \times _\mathcal {X} U \to \mathcal{X}$, and that

It follows that the isomorphism $C_1 \otimes _ B C_1 \to C_2 \otimes _ B C_2$ coming from the isomorphism $C_1 \to C_2$ is compatible with the morphisms $U_ i \times _{\mathop{\mathrm{Spec}}(B)} U_ i \to U \times _\mathcal {X} U$. Some details omitted. $\square$

Proof. Choose a surjective smooth morphism $g : U \to \mathcal{X}$ with $U = \mathop{\mathrm{Spec}}(A)$ affine, see Properties of Stacks, Lemma 100.6.2. Apply Lemma 106.16.2 to get the finite type commutative $B$-algebra $C = F(g_{\mathit{QCoh}, *}\mathcal{O}_ U)$ and the ring map $A \to C$. By Lemma 106.16.3 the ring map $B \to C$ is universally injective. Consider the algebra

Since $g$ is flat, quasi-compact, and quasi-separated by Lemma 106.16.5 we have the first equality in

where $f : U \times _\mathcal {X} U \to \mathcal{X}$ is the obvious morphism and $\text{pr}_2 : U \times _\mathcal {X} U \to U$ is the second projection. The second equality follows from Cohomology of Stacks, Lemma 103.11.5 and $f = g \circ \text{pr}_2$. Since the diagonal of $\mathcal{X}$ is affine, we see that $U \times _\mathcal {X} U = \mathop{\mathrm{Spec}}(R)$ is affine. Let us use $\text{pr}_2 : A \to R$ to view $R$ as an $A$-algebra. All in all we obtain

where the final equality follows from the final statement of Lemma 106.16.2. Since $A \to R$ is flat (because $\text{pr}_2$ is flat as a base change of $U \to \mathcal{X}$), we conclude that $C \otimes _ B C$ is flat over $C$. By Lemma 106.16.4 we conclude that $B \to C$ is faithfully flat.

We claim there is a solid commutative diagram

The arrow $\mathop{\mathrm{Spec}}(C) \to U = \mathop{\mathrm{Spec}}(A)$ comes from the ring map $A \to C$ in the statement of Lemma 106.16.2. The arrow $\mathop{\mathrm{Spec}}(C \otimes _ B C) \to U \times _\mathcal {X} U$ simlarly comes from the ring map $R \to C \otimes _ B C$. To verify the top square commutes use Lemma 106.16.6; details omitted. We conclude we get the dotted arrow $\mathop{\mathrm{Spec}}(B) \to \mathcal{X}$ by Proposition 106.15.1.

The statement that $F$ is the functor corresponding to pullback by the dotted arrow is also clear from this and the corresponding statement in Lemma 106.16.2. Details omitted. $\square$

Proof. By Lemma 106.16.1 we can extend $F$ uniquely to a right exact tensor functor $F : \mathit{QCoh}(\mathcal{O}_\mathcal {X}) \to \text{Mod}_ B$ commuting with all direct susms. Then we can apply Lemma 106.16.7. $\square$