$\mathscr{I}$ is injective iff for every $U \subseteq X$, $R \subseteq \mathbb{Z}_U$, and map $f: R \to \mathscr{I}$ extends to $\mathbb{Z}_U$.

algebraic-geometryhomology-cohomologysheaf-theory

This is Hartshorne III.2.6. I'm very stuck on how to proceed since the solutions online I usually refer to have an error here. Here $X$ is a noetherian topological space and $j: U \hookrightarrow X$ is an arbitrary open subset. I denote $\mathbb{Z}_U = j_!j^* \mathbb{Z}$ where $\mathbb{Z}$ is the constant sheaf. Below $R, \mathscr{F}, \mathscr{I}$ and $\mathscr{G}$ are sheaves of abelian groups. We assume that $\mathscr{I}$ satisfies the property that 'for all open sets $U$ and $R \subseteq \mathbb{Z}_U$ and all $f: R \to \mathscr{I}$ we have an extension $\mathbb{Z}_U \to \mathscr{I}.$

To prove $\mathscr{I}$ is injective, we must prove that $\operatorname{Hom}(-, \mathscr{I})$ is exact. In particular, we must show that given $\mathscr{F} \subseteq \mathscr{G}$ the resulting map $\operatorname{Hom}(\mathscr{G}, \mathscr{I}) \to \operatorname{Hom}(\mathscr{F}, \mathscr{I})$ is surjective. Equivalently, we must show that any morphism $\phi: \mathscr{F} \to \mathscr{I}$ can be extended to a morphism $\mathscr{G} \to \mathscr{I}$. The way the other solutions go is this.

Using Zorn's lemma, we find a maximal $\mathscr{F} \subset H \subset \mathscr{G}$ with the property that $H$ extends $\phi$. Now we let $s \in \mathscr{G}(U)$ be a local section which is not in $H$. Let $\langle s \rangle$ denote the subsheaf of $\mathscr{G}$ generated by $s$. Now, most online solutions claim that $\mathbb{Z}_U \cong \langle s \rangle$ and then apply the property to $H \cap \langle s \rangle$. This is incorrect, though, since $s$ might be torsion!

Can this approach be salvaged? One idea is to use the kernel of the map $\mathbb{Z}_U \to \langle s \rangle$, but I am not seeing how to map it to $\mathscr{I}$ in a nice way.

Thanks!

Best Answer

The correct way to do this is to begin by considering the equalizer $\mathcal{K}$ of the morphisms $H\to \mathcal{G}$ and $\Bbb Z_U\to \mathcal{G}$ where the first map is the injection and the second map is the map sending $1\mapsto s$. Since $H\to\mathcal{G}$ is injective, $\mathcal{K}$ may be viewed as a subsheaf of $\Bbb Z_U$ and we can use the property in the question statement to get an extension of $\mathcal{K}\to \mathcal{I}$ to a map $\Bbb Z_U\to \mathcal{I}$. Then $\operatorname{Im}(H\oplus\Bbb Z_U\to \mathcal{G})$ is a subsheaf of $\mathcal{G}$ to which $\phi$ extends and which contains both $H$ and the section $s$, contradicting maximality of $H$. So $H$ must actually have been all of $\mathcal{G}$ to start with, implying that our morphism $\phi$ extends the whole way.

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