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. $\square$