Lemma 19.4.1. Let $X$ be a topological space. The category of abelian sheaves on $X$ has enough injectives. In fact it has functorial injective embeddings.

Proof. For an abelian group $A$ we denote $j : A \to J(A)$ the functorial injective embedding constructed in More on Algebra, Section 15.55. Let $\mathcal{F}$ be an abelian sheaf on $X$. By Sheaves, Example 6.7.5 the assignment

$\mathcal{I} : U \mapsto \mathcal{I}(U) = \prod \nolimits _{x\in U} J(\mathcal{F}_ x)$

is an abelian sheaf. There is a canonical map $\mathcal{F} \to \mathcal{I}$ given by mapping $s \in \mathcal{F}(U)$ to $\prod _{x \in U} j(s_ x)$ where $s_ x \in \mathcal{F}_ x$ denotes the germ of $s$ at $x$. This map is injective, see Sheaves, Lemma 6.11.1 for example.

It remains to prove the following: Given a rule $x \mapsto I_ x$ which assigns to each point $x \in X$ an injective abelian group the sheaf $\mathcal{I} : U \mapsto \prod _{x \in U} I_ x$ is injective. Note that

$\mathcal{I} = \prod \nolimits _{x \in X} i_{x, *}I_ x$

is the product of the skyscraper sheaves $i_{x, *}I_ x$ (see Sheaves, Section 6.27 for notation.) We have

$\mathop{\mathrm{Mor}}\nolimits _{\textit{Ab}}(\mathcal{F}_ x, I_ x) = \mathop{\mathrm{Mor}}\nolimits _{\textit{Ab}(X)}(\mathcal{F}, i_{x, *}I_ x).$

see Sheaves, Lemma 6.27.3. Hence it is clear that each $i_{x, *}I_ x$ is injective. Hence the injectivity of $\mathcal{I}$ follows from Homology, Lemma 12.27.3. $\square$

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