The Stacks project

Lemma 105.16.1. Let $\mathcal{X}$ and $\mathcal{Y}$ be Noetherian algebraic stacks. Any right exact tensor functor $F : \textit{Coh}(\mathcal{O}_\mathcal {X}) \to \textit{Coh}(\mathcal{O}_\mathcal {Y})$ extends uniquely to a right exact tensor functor $F : \mathit{QCoh}(\mathcal{O}_\mathcal {X}) \to \mathit{QCoh}(\mathcal{O}_\mathcal {Y})$ commuting with all colimits.

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 102.17.2. Hence a coherent module on $\mathcal{X}$ is a categorically compact object of $\mathit{QCoh}(\mathcal{O}_\mathcal {X})$ by Cohomology of Stacks, Lemma 102.13.5. Finally, every quasi-coherent module is a filtered colimit of its coherent submodules by Cohomology of Stacks, Lemma 102.18.1.

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

\begin{align*} F(\mathcal{F}) & = \mathop{\mathrm{colim}}\nolimits _{J' \subset J\text{ finite}} F(\bigoplus \nolimits _{j \in J'} \mathcal{H}_ j) \\ & = \mathop{\mathrm{colim}}\nolimits _{J' \subset J\text{ finite}} \bigoplus \nolimits _{j \in J'} F(\mathcal{H}_ j) \\ & = \bigoplus \nolimits _{j \in J} F(\mathcal{H}_ j) \end{align*}

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

\[ 0 \to \mathcal{F}_ i \to \mathcal{F}_ i' \to \mathcal{F}_ i'' \to 0 \]

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. $\square$


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