The Stacks project

Lemma 102.4.1. If $f : \mathcal{X} \to \mathcal{Y}$ is a flat morphism of algebraic stacks then $f^* : \mathit{QCoh}(\mathcal{O}_\mathcal {Y}) \to \mathit{QCoh}(\mathcal{O}_\mathcal {X})$ is an exact functor.

Proof. Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Choose a scheme $U$ and a surjective smooth morphism $U \to V \times _\mathcal {Y} \mathcal{X}$. Then $U \to \mathcal{X}$ is still smooth and surjective as a composition of two such morphisms. From the commutative diagram

\[ \xymatrix{ U \ar[d] \ar[r]_{f'} & V \ar[d] \\ \mathcal{X} \ar[r]^ f & \mathcal{Y} } \]

we obtain a commutative diagram

\[ \xymatrix{ \mathit{QCoh}(\mathcal{O}_ U) & \mathit{QCoh}(\mathcal{O}_ V) \ar[l] \\ \mathit{QCoh}(\mathcal{O}_\mathcal {X}) \ar[u] & \mathit{QCoh}(\mathcal{O}_\mathcal {Y}) \ar[l] \ar[u] } \]

of abelian categories. Our proof that the bottom two categories in this diagram are abelian showed that the vertical functors are faithful exact functors (see proof of Sheaves on Stacks, Lemma 95.15.1). Since $f'$ is a flat morphism of schemes (by our definition of flat morphisms of algebraic stacks) we see that $(f')^*$ is an exact functor on quasi-coherent sheaves on $V$. Thus we win. $\square$

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Your email address will not be published.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 076X. Beware of the difference between the letter 'O' and the digit '0'.