Lemma 59.105.3. Let $X$ be a scheme and let $Z \subset X$ be a closed subscheme cut out by a quasi-coherent ideal of finite type. Consider the corresponding blow up square

\[ \xymatrix{ E \ar[d]_\pi \ar[r]_ j & X' \ar[d]^ b \\ Z \ar[r]^ i & X } \]

Suppose given

an object $K'$ of $D^+(X'_{\acute{e}tale})$ with torsion cohomology sheaves,

an object $L$ of $D^+(Z_{\acute{e}tale})$ with torsion cohomology sheaves, and

an isomorphism $\gamma : K'|_ E \to L|_ E$.

Then there exists an object $K$ of $D^+(X_{\acute{e}tale})$ and isomorphisms $f : K|_{X'} \to K'$, $g : K|_ Z \to L$ such that $\gamma = g|_ E \circ f^{-1}|_ E$. Moreover, given

an object $M$ of $D^+(X_{\acute{e}tale})$ with torsion cohomology sheaves,

a morphism $\alpha : K' \to M|_{X'}$ of $D(X'_{\acute{e}tale})$,

a morphism $\beta : L \to M|_ Z$ of $D(Z_{\acute{e}tale})$,

such that

\[ \alpha |_ E = \beta |_ E \circ \gamma . \]

Then there exists a morphism $M \to K$ in $D(X_{\acute{e}tale})$ whose restriction to $X'$ is $a \circ f$ and whose restriction to $Z$ is $b \circ g$.

**Proof.**
If $K$ exists, then Lemma 59.105.1 tells us a distinguished triangle that it fits in. Thus we simply choose a distinguished triangle

\[ K \to Ri_*(L) \oplus Rb_*(K') \to Rc_*(L|_ E) \to K[1] \]

where $c = i \circ \pi = b \circ j$. Here the map $Ri_*(L) \to Rc_*(L|_ E)$ is $Ri_*$ applied to the adjunction mapping $E \to R\pi _*(L|_ E)$. The map $Rb_*(K') \to Rc_*(L|_ E)$ is the composition of the canonical map $Rb_*(K') \to Rc_*(K'|_ E)) = R$ and $Rc_*(\gamma )$. The maps $g$ and $f$ of the statement of the lemma are the adjoints of these maps. If we restrict this distinguished triangle to $Z$ then the map $Rb_*(K) \to Rc_*(L|_ E)$ becomes an isomorphism by the base change theorem (Lemma 59.91.12) and hence the map $g : K|_ Z \to L$ is an isomorphism. Looking at the distinguished triangle we see that $f : K|_{X'} \to K'$ is an isomorphism over $X' \setminus E = X \setminus Z$. Moreover, we have $\gamma \circ f|_ E = g|_ E$ by construction. Then since $\gamma $ and $g$ are isomorphisms we conclude that $f$ induces isomorphisms on stalks at geometric points of $E$ as well. Thus $f$ is an isomorphism.

For the final statement, we may replace $K'$ by $K|_{X'}$, $L$ by $K|_ Z$, and $\gamma $ by the canonical identification. Observe that $\alpha $ and $\beta $ induce a commutative square

\[ \xymatrix{ K \ar[r] \ar@{..>}[d] & Ri_*(K|_ Z) \oplus Rb_*(K|_{X'}) \ar[r] \ar[d]_{\beta \oplus \alpha } & Rc_*(K|_ E) \ar[r] \ar[d]_{\alpha |_ E} & K[1] \ar@{..>}[d] \\ M \ar[r] & Ri_*(M|_ Z) \oplus Rb_*(M|_{X'}) \ar[r] & Rc_*(M|_ E) \ar[r] & M[1] } \]

Thus by the axioms of a derived category we get a dotted arrow producing a morphism of distinguished triangles.
$\square$

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