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