The Stacks project

Lemma 75.10.8. Let $S$ be a scheme. Let $(U \subset X, V \to X)$ be an elementary distinguished square of algebraic spaces over $S$. Suppose given

  1. an object $A$ of $D(\mathcal{O}_ U)$,

  2. an object $B$ of $D(\mathcal{O}_ V)$, and

  3. an isomorphism $c : A|_{U \times _ X V} \to B|_{U \times _ X V}$.

Then there exists an object $F$ of $D(\mathcal{O}_ X)$ and isomorphisms $f : F|_ U \to A$, $g : F|_ V \to B$ such that $c = g|_{U \times _ X V} \circ f^{-1}|_{U \times _ X V}$. Moreover, given

  1. an object $E$ of $D(\mathcal{O}_ X)$,

  2. a morphism $a : A \to E|_ U$ of $D(\mathcal{O}_ U)$,

  3. a morphism $b : B \to E|_ V$ of $D(\mathcal{O}_ V)$,

such that

\[ a|_{U \times _ X V} = b|_{U \times _ X V} \circ c. \]

Then there exists a morphism $F \to E$ in $D(\mathcal{O}_ X)$ whose restriction to $U$ is $a \circ f$ and whose restriction to $V$ is $b \circ g$.

Proof. Denote $j_ U$, $j_ V$, $j_{U \times _ X V}$ the corresponding morphisms towards $X$. Choose a distinguished triangle

\[ F \to Rj_{U, *}A \oplus Rj_{V, *}B \to Rj_{U \times _ X V, *}(B|_{U \times _ X V}) \to F[1] \]

Here the map $Rj_{V, *}B \to Rj_{U \times _ X V, *}(B|_{U \times _ X V})$ is the obvious one. The map $Rj_{U, *}A \to Rj_{U \times _ X V, *}(B|_{U \times _ X V})$ is the composition of $Rj_{U, *}A \to Rj_{U \times _ X V, *}(A|_{U \times _ X V})$ with $Rj_{U \times _ X V, *}c$. Restricting to $U$ we obtain

\[ F|_ U \to A \oplus (Rj_{V, *}B)|_ U \to (Rj_{U \times _ X V, *}(B|_{U \times _ X V}))|_ U \to F|_ U[1] \]

Denote $j : U \times _ X V \to U$. Compatibility of restriction and total direct image (Lemma 75.3.1) shows that both $(Rj_{V, *}B)|_ U$ and $(Rj_{U \times _ X V, *}(B|_{U \times _ X V}))|_ U$ are canonically isomorphic to $Rj_*(B|_{U \times _ X V})$. Hence the second arrow of the last displayed equation has a section, and we conclude that the morphism $F|_ U \to A$ is an isomorphism.

To see that the morphism $F|_ V \to B$ is an isomorphism we will use a trick. Namely, choose a distinguished triangle

\[ F|_ V \to B \to B' \to F[1]|_ V \]

in $D(\mathcal{O}_ V)$. Since $F|_ U \to A$ is an isomorphism, and since we have the isomorphism $c : A|_{U \times _ X V} \to B|_{U \times _ X V}$ the restriction of $F|_ V \to B$ is an isomorphism over $U \times _ X V$. Thus $B'$ is supported on $j_ V^{-1}(T)$ where $T = |X| \setminus |U|$. On the other hand, there is a morphism of distinguished triangles

\[ \xymatrix{ F \ar[r] \ar[d] & Rj_{U, *}F|_ U \oplus Rj_{V, *}F|_ V \ar[r] \ar[d] & Rj_{U \times _ X V, *}F|_{U \times _ X V} \ar[r] \ar[d] & F[1] \ar[d] \\ F \ar[r] & Rj_{U, *}A \oplus Rj_{V, *}B \ar[r] & Rj_{U \times _ X V, *}(B|_{U \times _ X V}) \ar[r] & F[1] } \]

The all of the vertical maps in this diagram are isomorphisms, except for the map $Rj_{V, *}F|_ V \to Rj_{V, *}B$, hence that is an isomorphism too (Derived Categories, Lemma 13.4.3). This implies that $Rj_{V, *}B' = 0$. Hence $B' = 0$ by Lemma 75.10.7.

The existence of the morphism $F \to E$ follows from the Mayer-Vietoris sequence for $\mathop{\mathrm{Hom}}\nolimits $, see Lemma 75.10.4. $\square$

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 08HB. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 75.10.8 as pdf