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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!

Daniel
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1 Answers1

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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.

KReiser
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    Let me also caution that Hartshorne's use of "finitely generated" in this exercise is also a bit slippery - $\Bbb Z_U$ is not finitely generated for most $U$ in the standard sense because there are no maps $\Bbb Z^{\oplus n}\to \Bbb Z_U$ since the latter generally has no global sections. Instead, the goal here is to think about sheaves generated by a finite number of local sections. (Maybe this is pedantic, but as long as we're on the topic, I feel compelled to mention it.) – KReiser Jan 19 '22 at 04:29
  • Thank you again! I'm sure I will run into that issue upon moving to the next part of the problem. For now I have been assuming $\mathscr{G}$ to be 'finitely generated' to mean there is a surjection $\bigoplus_i \mathbb{Z}_{U_i} \to \mathscr{G}$ for some finite collection of open sets $U_i$. Hopefully that's the intended notion. – Daniel Jan 19 '22 at 04:41
  • I think you mean an extension of $\mathcal{K} \to \mathscr{I}$ to $\mathbb{Z}U \to \mathscr{I}$ since a function $\mathbb{Z}_U \to \mathcal{G}$ was already defined. Also, is $\mathcal{K} = H \times{\mathcal{G}} \mathbb{Z}_U$ what you meant by setting it equal to the equilizer? It seems to do the trick either way; this was a very enlightening answer. – Daniel Jan 19 '22 at 06:56
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  • Your assumption is a good one. 2) Yes, that was a typo, thanks for pointing that out. 3) Yes, $H\times_\mathcal{G} \Bbb Z_U$ is another way to write down the construction I am describing - originally I was going to write pullback but I thought that might potentially be misleading since there's also the sheaf pullback.
    • – KReiser Jan 19 '22 at 07:04