Lemma 85.12.10. In Situation 85.3.3 let $\mathcal{O}$ be a sheaf of rings on $\mathcal{C}_{total}$. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}$-modules. Then $\mathcal{F}$ is quasi-coherent in the sense of Modules on Sites, Definition 18.23.1 if and only if $\mathcal{F}$ is cartesian and $\mathcal{F}_ n$ is a quasi-coherent $\mathcal{O}_ n$-module for all $n$.

Proof. Assume $\mathcal{F}$ is quasi-coherent. Since pullbacks of quasi-coherent modules are quasi-coherent (Modules on Sites, Lemma 18.23.4) we see that $\mathcal{F}_ n$ is a quasi-coherent $\mathcal{O}_ n$-module for all $n$. To show that $\mathcal{F}$ is cartesian, let $U$ be an object of $\mathcal{C}_ n$ for some $n$. Let us view $U$ as an object of $\mathcal{C}_{total}$. Because $\mathcal{F}$ is quasi-coherent there exists a covering $\{ U_ i \to U\}$ and for each $i$ a presentation

$\bigoplus \nolimits _{j \in J_ i} \mathcal{O}_{\mathcal{C}_{total}/U_ i} \to \bigoplus \nolimits _{k \in K_ i} \mathcal{O}_{\mathcal{C}_{total}/U_ i} \to \mathcal{F}|_{\mathcal{C}_{total}/U_ i} \to 0$

Observe that $\{ U_ i \to U\}$ is a covering of $\mathcal{C}_ n$ by the construction of the site $\mathcal{C}_{total}$. Next, let $V$ be an object of $\mathcal{C}_ m$ for some $m$ and let $V \to U$ be a morphism of $\mathcal{C}_{total}$ lying over $\varphi : [n] \to [m]$. The fibre products $V_ i = V \times _ U U_ i$ exist and we get an induced covering $\{ V_ i \to V\}$ in $\mathcal{C}_ m$. Restricting the presentation above to the sites $\mathcal{C}_ n/U_ i$ and $\mathcal{C}_ m/V_ i$ we obtain presentations

$\bigoplus \nolimits _{j \in J_ i} \mathcal{O}_{\mathcal{C}_ m/U_ i} \to \bigoplus \nolimits _{k \in K_ i} \mathcal{O}_{\mathcal{C}_ m/U_ i} \to \mathcal{F}_ n|_{\mathcal{C}_ n/U_ i} \to 0$

and

$\bigoplus \nolimits _{j \in J_ i} \mathcal{O}_{\mathcal{C}_ m/V_ i} \to \bigoplus \nolimits _{k \in K_ i} \mathcal{O}_{\mathcal{C}_ m/V_ i} \to \mathcal{F}_ m|_{\mathcal{C}_ m/V_ i} \to 0$

These presentations are compatible with the map $\mathcal{F}(\varphi ) : f_\varphi ^*\mathcal{F}_ n \to \mathcal{F}_ m$ (as this map is defined using the restriction maps of $\mathcal{F}$ along morphisms of $\mathcal{C}_{total}$ lying over $\varphi$). We conclude that $\mathcal{F}(\varphi )|_{\mathcal{C}_ m/V_ i}$ is an isomorphism. As $\{ V_ i \to V\}$ is a covering we conclude $\mathcal{F}(\varphi )|_{\mathcal{C}_ m/V}$ is an isomorphism. Since $V$ and $U$ were arbitrary this proves that $\mathcal{F}$ is cartesian. (In case A use Sites, Lemma 7.14.10.)

Conversely, assume $\mathcal{F}_ n$ is quasi-coherent for all $n$ and that $\mathcal{F}$ is cartesian. Then for any $n$ and object $U$ of $\mathcal{C}_ n$ we can choose a covering $\{ U_ i \to U\}$ of $\mathcal{C}_ n$ and for each $i$ a presentation

$\bigoplus \nolimits _{j \in J_ i} \mathcal{O}_{\mathcal{C}_ m/U_ i} \to \bigoplus \nolimits _{k \in K_ i} \mathcal{O}_{\mathcal{C}_ m/U_ i} \to \mathcal{F}_ n|_{\mathcal{C}_ n/U_ i} \to 0$

Pulling back to $\mathcal{C}_{total}/U_ i$ we obtain complexes

$\bigoplus \nolimits _{j \in J_ i} \mathcal{O}_{\mathcal{C}_{total}/U_ i} \to \bigoplus \nolimits _{k \in K_ i} \mathcal{O}_{\mathcal{C}_{total}/U_ i} \to \mathcal{F}|_{\mathcal{C}_{total}/U_ i} \to 0$

of modules on $\mathcal{C}_{total}/U_ i$. Then the property that $\mathcal{F}$ is cartesian implies that this is exact. We omit the details. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0D7M. Beware of the difference between the letter 'O' and the digit '0'.