The Stacks project

Proof. The notation $K|_{X'}$ stands for $b_{small}^{-1}K$. Choose a bounded below complex $\mathcal{F}^\bullet$ of abelian sheaves representing $K$. Observe that $i_*(\mathcal{F}^\bullet |_ Z)$ represents $Ri_*(K|_ Z)$ because $i_*$ is exact (Proposition 59.55.2). Choose a quasi-isomorphism $b_{small}^{-1}\mathcal{F}^\bullet \to \mathcal{I}^\bullet$ where $\mathcal{I}^\bullet$ is a bounded below complex of injective abelian sheaves on $X'_{\acute{e}tale}$. This map is adjoint to a map $\mathcal{F}^\bullet \to b_*(\mathcal{I}^\bullet )$ and $b_*(\mathcal{I}^\bullet )$ represents $Rb_*(K|_{X'})$. We have $\pi _*(\mathcal{I}^\bullet |_ E) = (b_*\mathcal{I}^\bullet )|_ Z$ by Lemma 59.91.5 and by Lemma 59.91.12 this complex represents $R\pi _*(K|_ E)$. Hence the map

is represented by the surjective map of bounded below complexes

To get our distinguished triangle it suffices to show that the canonical map $\mathcal{F}^\bullet \to i_*(\mathcal{F}^\bullet |_ Z) \oplus b_*(\mathcal{I}^\bullet )$ maps quasi-isomorphically onto the kernel of the map of complexes displayed above (namely a short exact sequence of complexes determines a distinguished triangle in the derived category, see Derived Categories, Section 13.12). We may check this on stalks at a geometric point $\overline{x}$ of $X$. If $\overline{x}$ is not in $Z$, then $X' \to X$ is an isomorphism over an open neighbourhood of $\overline{x}$. Thus, if $\overline{x}'$ denotes the corresponding geometric point of $X'$ in this case, then we have to show that

is a quasi-isomorphism. This is true by our choice of $\mathcal{I}^\bullet$. If $\overline{x}$ is in $Z$, then $b_(\mathcal{I}^\bullet )_{\overline{x}} \to i_*\left(b_*(\mathcal{I}^\bullet )|_ Z\right)_{\overline{x}}$ is an isomorphism of complexes of abelian groups. Hence the kernel is equal to $i_*(\mathcal{F}^\bullet |_ Z)_{\overline{x}} = \mathcal{F}^\bullet _{\overline{x}}$ as desired. $\square$

First proof. This follows immediately from Lemma 59.105.1 and the fact that

(see Cohomology on Sites, Section 21.14) and similarly for the others. $\square$

Second proof. By Lemma 59.102.7 these cohomology groups are the cohomology of $X, X', E, Z$ with values in some complex of abelian sheaves on the site $(\mathit{Sch}/X)_{ph}$. (Namely, the object $a_ X^{-1}K$ of the derived category, see Lemma 59.102.1 above and recall that $K|_{X'} = b_{small}^{-1}K$.) By More on Flatness, Lemma 38.36.1 the ph sheafification of the diagram of representable presheaves is cocartesian. Thus the lemma follows from the very general Cohomology on Sites, Lemma 21.26.3 applied to the site $(\mathit{Sch}/X)_{ph}$ and the commutative diagram of the lemma. $\square$

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

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

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