The Stacks project

Lemma 56.6.4. Let $X$ be a quasi-compact and quasi-separated scheme. Let $Z \subset X$ be a closed subset such that $U = X \setminus Z$ is quasi-compact. Let $\mathcal{A} \subset \mathit{QCoh}(\mathcal{O}_ X)$ be the full subcategory whose objects are the quasi-coherent modules supported on $Z$. Then the restriction functor $\mathit{QCoh}(\mathcal{O}_ X) \to \mathit{QCoh}(\mathcal{O}_ U)$ induces an equivalence $\mathit{QCoh}(\mathcal{O}_ X)/\mathcal{A} \cong \mathit{QCoh}(\mathcal{O}_ U)$.

Proof. By the universal property of the quotient construction (Homology, Lemma 12.10.6) we certainly obtain an induced functor $\mathit{QCoh}(\mathcal{O}_ X)/\mathcal{A} \cong \mathit{QCoh}(\mathcal{O}_ U)$. Denote $j : U \to X$ the inclusion morphism. Since $j$ is quasi-compact and quasi-separated we obtain a functor $j_* : \mathit{QCoh}(\mathcal{O}_ U) \to \mathit{QCoh}(\mathcal{O}_ X)$. The reader shows that this defines a quasi-inverse; details omitted. $\square$

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Your email address will not be published.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