The Stacks project

Proof. Let $\mathcal{F} \to \mathcal{G}$ be an injective map of abelian sheaves. Suppose $\varphi : \mathcal{F} \to \mathcal{J}$ is a morphism. Arguing as in the proof of More on Algebra, Lemma 15.54.1 we see that it suffices to prove that if $\mathcal{F} \not= \mathcal{G}$, then we can find an abelian sheaf $\mathcal{F}'$, $\mathcal{F} \subset \mathcal{F}' \subset \mathcal{G}$ such that (a) the inclusion $\mathcal{F} \subset \mathcal{F}'$ is strict, and (b) $\varphi $ can be extended to $\mathcal{F}'$. To find $\mathcal{F}'$, let $X$ be an object of $\mathcal{C}$ such that the inclusion $\mathcal{F}(X) \subset \mathcal{G}(X)$ is strict. Pick $s \in \mathcal{G}(X)$, $s \not\in \mathcal{F}(X)$. Let $\psi : \mathbf{Z}_ X^\# \to \mathcal{G}$ be the morphism corresponding to the section $s$ via (19.7.2.1). Set $\mathcal{S} = \psi ^{-1}(\mathcal{F})$. By assumption the morphism

can be extended to a morphism $\varphi ' : \mathbf{Z}_ X^\# \to \mathcal{J}$. Note that $\varphi '$ annihilates the kernel of $\psi $ (as this is true for $\varphi $). Thus $\varphi '$ gives rise to a morphism $\varphi '' : \mathop{\mathrm{Im}}(\psi ) \to \mathcal{J}$ which agrees with $\varphi $ on the intersection $\mathcal{F} \cap \mathop{\mathrm{Im}}(\psi )$ by construction. Thus $\varphi $ and $\varphi ''$ glue to give an extension of $\varphi $ to the strictly bigger subsheaf $\mathcal{F}' = \mathcal{F} + \mathop{\mathrm{Im}}(\psi )$. $\square$