Lemma 6.18.2. Let $X$ be a topological space. Let $\mathcal{F}$ be an abelian presheaf on $X$. Then there exists a unique structure of abelian sheaf on $\mathcal{F}^\#$ such that $\mathcal{F} \to \mathcal{F}^\#$ is a morphism of abelian presheaves. Moreover, the following adjointness property holds

$\mathop{Mor}\nolimits _{\textit{PAb}(X)}(\mathcal{F}, i(\mathcal{G})) = \mathop{Mor}\nolimits _{\textit{Ab}(X)}(\mathcal{F}^\# , \mathcal{G}).$

Proof. Recall the sheaf of sets $\Pi (\mathcal{F})$ defined in Section 6.17. All the stalks $\mathcal{F}_ x$ are abelian groups, see Lemma 6.12.1. Hence $\Pi (\mathcal{F})$ is a sheaf of abelian groups by Example 6.15.6. Also, it is clear that the map $\mathcal{F} \to \Pi (\mathcal{F})$ is a morphism of abelian presheaves. If we show that condition $(*)$ of Section 6.17 defines a subgroup of $\Pi (\mathcal{F})(U)$ for all open subsets $U \subset X$, then $\mathcal{F}^\#$ canonically inherits the structure of abelian sheaf. This is quite easy to do by hand, and we leave it to the reader to find a good simple argument. The argument we use here, which generalizes to presheaves of algebraic structures is the following: Lemma 6.18.1 show that $\mathcal{F}^\# (U)$ is the fibre product of a diagram of abelian groups. Thus $\mathcal{F}^\#$ is an abelian subgroup as desired.

Note that at this point $\mathcal{F}^\# _ x$ is an abelian group by Lemma 6.12.1 and that $\mathcal{F}_ x \to \mathcal{F}^\# _ x$ is a bijection (Lemma 6.17.2) and a homomorphism of abelian groups. Hence $\mathcal{F}_ x \to \mathcal{F}^\# _ x$ is an isomorphism of abelian groups. This will be used below without further mention.

To prove the adjointness property we use the adjointness property of sheafification of presheaves of sets. For example if $\psi : \mathcal{F} \to i(\mathcal{G})$ is morphism of presheaves then we obtain a morphism of sheaves $\psi ' : \mathcal{F}^\# \to \mathcal{G}$. What we have to do is to check that this is a morphism of abelian sheaves. We may do this for example by noting that it is true on stalks, by Lemma 6.17.2, and then using Lemma 6.16.4 above. $\square$

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