Lemma 21.26.3. In the situation above assume

$h_ X^\# = h_ Y^\# \amalg _{h_ E^\# } h_ Z^\# $, and

$h_ E^\# \to h_ Y^\# $ is injective.

Then the construction of Lemma 21.26.1 produces a distinguished triangle

\[ R\Gamma (X, K) \to R\Gamma (Z, K) \oplus R\Gamma (Y, K) \to R\Gamma (E, K) \to R\Gamma (X, K)[1] \]

functorial for $K$ in $D(\mathcal{C})$.

**Proof.**
We can represent $K$ by a K-injective complex whose terms are injective abelian sheaves, see Section 21.19. Thus it suffices to show: if $\mathcal{I}$ is an injective abelian sheaf, then

\[ 0 \to \mathcal{I}(X) \to \mathcal{I}(Z) \oplus \mathcal{I}(Y) \to \mathcal{I}(E) \to 0 \]

is a short exact sequence. The first arrow is injective because by condition (1) the map $h_ Y \amalg h_ Z \to h_ X$ becomes surjective after sheafification, which means that $\{ Y \to X, Z \to X\} $ can be refined by a covering of $X$. The last arrow is surjective because $\mathcal{I}(Y) \to \mathcal{I}(E)$ is surjective. Namely, we have $\mathcal{I}(E) = \mathop{\mathrm{Hom}}\nolimits (\mathbf{Z}_ E^\# , \mathcal{I})$, $\mathcal{I}(Y) = \mathop{\mathrm{Hom}}\nolimits (\mathbf{Z}_ Y^\# , \mathcal{I})$, the map $\mathbf{Z}_ E^\# \to \mathbf{Z}_ Y^\# $ is injective by (2), and $\mathcal{I}$ is an injective abelian sheaf. Please compare with Modules on Sites, Section 18.5. Finally, suppose we have $s \in \mathcal{I}(Y)$ and $t \in \mathcal{F}(Z)$ mapping to the same element of $\mathcal{I}(E)$. Then $s$ and $t$ define a map

\[ s \amalg t : h_ Y^\# \amalg h_ Z^\# \longrightarrow \mathcal{I} \]

which by assumption factors through $h_ Y^\# \amalg _{h_ E^\# } h_ Z^\# $. Thus by assumption (1) we obtain a unique map $h_ X^\# \to \mathcal{I}$ which corresponds to an element of $\mathcal{I}(X)$ restricting to $s$ on $Y$ and $t$ on $Z$.
$\square$

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