The Stacks project

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 $Z$. $\square$