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

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

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

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

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

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

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.
$\square$

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