Lemma 95.4.1. A morphism $(f, \varphi ) : (Y, \mathcal{G}) \to (X, \mathcal{F})$ of $\mathcal{QC}\! \mathit{oh}$ is strongly cartesian if and only if the map $\varphi $ induces an isomorphism $f^*\mathcal{F} \to \mathcal{G}$.
95.4 Quasi-coherent sheaves
We define a category $\mathcal{QC}\! \mathit{oh}$ as follows:
An object of $\mathcal{QC}\! \mathit{oh}$ is a pair $(X, \mathcal{F})$, where $X/S$ is an object of $(\mathit{Sch}/S)_{fppf}$, and $\mathcal{F}$ is a quasi-coherent $\mathcal{O}_ X$-module, and
a morphism $(f, \varphi ) : (Y, \mathcal{G}) \to (X, \mathcal{F})$ is a pair consisting of a morphism $f : Y \to X$ of schemes over $S$ and an $f$-map (see Sheaves, Section 6.26) $\varphi : \mathcal{F} \to \mathcal{G}$.
The composition of morphisms
\[ (Z, \mathcal{H}) \xrightarrow {(g, \psi )} (Y, \mathcal{G}) \xrightarrow {(f, \phi )} (X, \mathcal{F}) \]is $(f \circ g, \psi \circ \phi )$ where $\psi \circ \phi $ is the composition of $f$-maps.
Thus $\mathcal{QC}\! \mathit{oh}$ is a category and
\[ p : \mathcal{QC}\! \mathit{oh}\to (\mathit{Sch}/S)_{fppf}, \quad (X, \mathcal{F}) \mapsto X \]
is a functor. Note that the fibre category of $\mathcal{QC}\! \mathit{oh}$ over a scheme $X$ is the opposite of the category $\mathit{QCoh}(\mathcal{O}_ X)$ of quasi-coherent $\mathcal{O}_ X$-modules. We remark for later use that given $(X, \mathcal{F}), (Y, \mathcal{G}) \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{QC}\! \mathit{oh})$ we have
95.4.0.1
\begin{equation} \label{examples-stacks-equation-morphisms-qcoh} \mathop{\mathrm{Mor}}\nolimits _{\mathcal{QC}\! \mathit{oh}}((Y, \mathcal{G}), (X, \mathcal{F})) = \coprod \nolimits _{f \in \mathop{\mathrm{Mor}}\nolimits _ S(Y, X)} \mathop{\mathrm{Mor}}\nolimits _{\mathit{QCoh}(\mathcal{O}_ Y)}(f^*\mathcal{F}, \mathcal{G}) \end{equation}
See the discussion on $f$-maps of modules in Sheaves, Section 6.26.
The category $\mathcal{QC}\! \mathit{oh}$ is not a stack over $(\mathit{Sch}/S)_{fppf}$ because its collection of objects is a proper class. On the other hand we will see that it does satisfy all the axioms of a stack. We will get around the set theoretical issue in Section 95.5.
Proof. Let $(X, \mathcal{F}) \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{QC}\! \mathit{oh})$. Let $f : Y \to X$ be a morphism of $(\mathit{Sch}/S)_{fppf}$. Note that there is a canonical $f$-map $c : \mathcal{F} \to f^*\mathcal{F}$ and hence we get a morphism $(f, c) : (Y, f^*\mathcal{F}) \to (X, \mathcal{F})$. We claim that $(f, c)$ is strongly cartesian. Namely, for any object $(Z, \mathcal{H})$ of $\mathcal{QC}\! \mathit{oh}$ we have
\begin{align*} \mathop{\mathrm{Mor}}\nolimits _{\mathcal{QC}\! \mathit{oh}}((Z, \mathcal{H}), (Y, f^*\mathcal{F})) & = \coprod \nolimits _{g \in \mathop{\mathrm{Mor}}\nolimits _ S(Z, Y)} \mathop{\mathrm{Mor}}\nolimits _{\mathit{QCoh}(\mathcal{O}_ Z)}(g^*f^*\mathcal{F}, \mathcal{H}) \\ & = \coprod \nolimits _{g \in \mathop{\mathrm{Mor}}\nolimits _ S(Z, Y)} \mathop{\mathrm{Mor}}\nolimits _{\mathit{QCoh}(\mathcal{O}_ Z)}((f \circ g)^*\mathcal{F}, \mathcal{H}) \\ & = \mathop{\mathrm{Mor}}\nolimits _{\mathcal{QC}\! \mathit{oh}}((Z, \mathcal{H}), (X, \mathcal{F})) \times _{\mathop{\mathrm{Mor}}\nolimits _ S(Z, X)} \mathop{\mathrm{Mor}}\nolimits _ S(Z, Y) \end{align*}
where we have used Equation (95.4.0.1) twice. This proves that the condition of Categories, Definition 4.33.1 holds for $(f, c)$, and hence our claim is true. Now by Categories, Lemma 4.33.2 we see that isomorphisms are strongly cartesian and compositions of strongly cartesian morphisms are strongly cartesian which proves the “if” part of the lemma. For the converse, note that given $(X, \mathcal{F})$ and $f : Y \to X$, if there exists a strongly cartesian morphism lifting $f$ with target $(X, \mathcal{F})$ then it has to be isomorphic to $(f, c)$ (see discussion following Categories, Definition 4.33.1). Hence the "only if" part of the lemma holds. $\square$
Lemma 95.4.2. The functor $p : \mathcal{QC}\! \mathit{oh}\to (\mathit{Sch}/S)_{fppf}$ satisfies conditions (1), (2) and (3) of Stacks, Definition 8.4.1.
Proof. It is clear from Lemma 95.4.1 that $\mathcal{QC}\! \mathit{oh}$ is a fibred category over $(\mathit{Sch}/S)_{fppf}$. Given covering $\mathcal{U} = \{ X_ i \to X\} _{i \in I}$ of $(\mathit{Sch}/S)_{fppf}$ the functor
\[ \mathit{QCoh}(\mathcal{O}_ X) \longrightarrow DD(\mathcal{U}) \]
is fully faithful and essentially surjective, see Descent, Proposition 35.5.2. Hence Stacks, Lemma 8.4.2 applies to show that $\mathcal{QC}\! \mathit{oh}$ satisfies all the axioms of a stack. $\square$
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