The Stacks project

Lemma 103.10.1. Let $\mathcal{X}$ be an algebraic stack. Let $\textit{LQCoh}^{fbc}(\mathcal{O}_\mathcal {X})$ be the category of locally quasi-coherent modules with the flat base change property, see Section 103.8. The inclusion functor $i : \mathit{QCoh}(\mathcal{O}_\mathcal {X}) \to \textit{LQCoh}^{fbc}(\mathcal{O}_\mathcal {X})$ has a right adjoint

\[ Q : \textit{LQCoh}^{fbc}(\mathcal{O}_\mathcal {X}) \to \mathit{QCoh}(\mathcal{O}_\mathcal {X}) \]

such that $Q \circ i$ is the identity functor.

Proof. Choose a scheme $U$ and a surjective smooth morphism $f : U \to \mathcal{X}$. Set $R = U \times _\mathcal {X} U$ so that we obtain a smooth groupoid $(U, R, s, t, c)$ in algebraic spaces with the property that $\mathcal{X} = [U/R]$, see Algebraic Stacks, Lemma 94.16.2. We may and do replace $\mathcal{X}$ by $[U/R]$. By Sheaves on Stacks, Proposition 96.14.3 there is an equivalence

\[ q_1 : \mathit{QCoh}(U, R, s, t, c) \longrightarrow \mathit{QCoh}(\mathcal{O}_\mathcal {X}) \]

Let us construct a functor

\[ q_2 : \textit{LQCoh}^{fbc}(\mathcal{O}_\mathcal {X}) \longrightarrow \mathit{QCoh}(U, R, s, t, c) \]

by the following rule: if $\mathcal{F}$ is an object of $\textit{LQCoh}^{fbc}(\mathcal{O}_\mathcal {X})$ then we set

\[ q_2(\mathcal{F}) = (f^*\mathcal{F}|_{U_{\acute{e}tale}}, \alpha ) \]

where $\alpha $ is the isomorphism

\[ t_{small}^*(f^*\mathcal{F}|_{U_{\acute{e}tale}}) \to t^*f^*\mathcal{F}|_{R_{\acute{e}tale}} \to s^*f^*\mathcal{F}|_{R_{\acute{e}tale}} \to s_{small}^*(f^*\mathcal{F}|_{U_{\acute{e}tale}}) \]

where the outer two morphisms are the comparison maps. Note that $q_2(\mathcal{F})$ is quasi-coherent precisely because $\mathcal{F}$ is locally quasi-coherent and that we used (and needed) the flat base change property in the construction of the descent datum $\alpha $. We omit the verification that the cocycle condition (see Groupoids in Spaces, Definition 78.12.1) holds. Looking at the proof of Sheaves on Stacks, Proposition 96.14.3 we see that $q_2 \circ i$ is the quasi-inverse to $q_1$. We define $Q = q_1 \circ q_2$. Let $\mathcal{F}$ be an object of $\textit{LQCoh}^{fbc}(\mathcal{O}_\mathcal {X})$ and let $\mathcal{G}$ be an object of $\mathit{QCoh}(\mathcal{O}_\mathcal {X})$. We have

\begin{align*} \mathop{\mathrm{Mor}}\nolimits _{\textit{LQCoh}^{fbc}(\mathcal{O}_\mathcal {X})} (i(\mathcal{G}), \mathcal{F}) & = \mathop{\mathrm{Mor}}\nolimits _{\mathit{QCoh}(U, R, s, t, c)}(q_2(i(\mathcal{G})), q_2(\mathcal{F})) \\ & = \mathop{\mathrm{Mor}}\nolimits _{\mathit{QCoh}(\mathcal{O}_\mathcal {X})}(\mathcal{G}, Q(\mathcal{F})) \end{align*}

where the first equality is Sheaves on Stacks, Lemma 96.14.4 and the second equality holds because $q_1 \circ i$ and $q_2$ are quasi-inverse equivalences of categories. The assertion $Q \circ i \cong \text{id}$ is a formal consequence of the fact that $i$ is fully faithful. $\square$


Comments (0)


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 0778. Beware of the difference between the letter 'O' and the digit '0'.