Lemma 19.7.3. Suppose $\mathcal{J}$ is a sheaf of abelian groups with the following property: For all $X\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, for any abelian subsheaf $\mathcal{S} \subset \mathbf{Z}_ X^\#$ and any morphism $\varphi : \mathcal{S} \to \mathcal{J}$, there exists a morphism $\mathbf{Z}_ X^\# \to \mathcal{J}$ extending $\varphi$. Then $\mathcal{J}$ is an injective sheaf of abelian groups.

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

$\mathcal{S} \xrightarrow {\psi } \mathcal{F} \xrightarrow {\varphi } \mathcal{J}$

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 $\mathcal{F}' = \mathcal{F} + \mathop{\mathrm{Im}}(\psi )$. $\square$

Comment #5029 by Remy on

This is also a special case of Tag 19.11.6 (which appears later). Maybe slightly more natural is to formulate the latter for a generating set instead of a single generator, but it doesn't really matter here since $\mathcal C$ is small so you can take the sum of all $\mathbf Z_X$.

Comment #5263 by on

Dear Remy, yes. The reason for the formulation in 19.11.6 is that when we talk of Grothendieck abelian categories, it is natural to talk about a generator because the very definition of Grothendieck abelian categories is that they should have one.

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