does monomorphism of sheaves of abelian groups imply it's injective on the level of open sets?

Question

In Vakil's The Rising Sea: Foundation of Algebraic Geometry, August 29, 2022 version, there is an exercise discussing monomorphism(subsheaf) from the perspective of stalks, quoted as follows, on page 83, source here

2.4.M. EXERCISE. Suppose $\phi: \mathscr{F} \rightarrow \mathscr{G}$ is a morphism of sheaves of sets on a topological space $X$. Show that the following are equivalent.

(a) $\phi$ is a monomorphism in the category of sheaves.

(b) $\phi$ is injective on the level of stalks: $\phi_p: \mathscr{F}_p \rightarrow \mathscr{G}_p$ is injective for all $p \in X$.

(c) $\phi$ is injective on the level of open sets: $\phi(\mathrm{U}): \mathscr{F}(\mathrm{U}) \rightarrow \mathscr{G}(\mathrm{U})$ is injective for all open $\mathrm{U} \subset X$.

(Possible hints: for (b) implies (a), recall that morphisms are determined by stalks, Exercise 2.4.C. For (a) implies (c), use the "indicator sheaf" with one section over every open set contained in $\mathrm{U}$, and no section over any other open set. If these conditions hold, we say that $\mathscr{F}$ is a subsheaf of $\mathscr{G}$ (where the "inclusion" $\phi$ is sometimes left implicit).

(You may later wish to extend your solution to Exercise 2.4.M to show that for any morphism of presheaves, if all maps of sections are injective, then all stalk maps are injective. And furthermore, if $\phi: \mathscr{F} \rightarrow \mathscr{G}$ is a morphism from a separated presheaf to an arbitrary presheaf, then injectivity on the level of stalks implies that $\phi$ is a monomorphism in the category of presheaves. This is useful in some approaches to Exercise 2.6.C.)

Using the hint given by the author, I figured out the proof of this exercise for sheaves of sets. Moreover, since in the following two sections(section 2.6) this exercise is used to prove that sheaves of abelian group (or $\mathscr{O}_X$-modules) form an abelian category, I am trying to prove it for sheaves of abelian groups.

As the hint, the key point is to use the "indicator sheaf". In the case of sheaves of sets, I use $\mathscr{I}_U$ defined by $\mathscr{I}_U(V) = \{e\}$ if $V \subset U$ and $\mathscr{I}_U(V) = \varnothing$ otherwise. I can prove $\mathscr{I}$ is a sheaf of sets and use it to prove that (a) implies (c).

Nevertheless, in the case sheaf of abelian groups, I cannot find such a sheaf. Suppose $\phi(U): \mathscr{F}(U) \to \mathscr{G}(U)$ is not injective for some open set $U \subset X$. Hence there is some $a \in \mathscr{F}(U), a \neq 0$ such that $\phi(U)(a) = 0$. I tried the following constructions, but all failed:

1. $\mathscr{I}(V) = <\operatorname{res}_{U,V} a>$, i.e. the cyclic subgroup of $\mathscr{F}(V)$ generated by $\operatorname{res}_{U,V} a$ if $V \subset U$, $\{0\}$ otherwise. I can only prove it is a presheaf of abelian groups and failed to prove it is a sheaf of abelian groups. (It seems to be not a sheaf of abelian groups.)
2. $\mathscr{I}(V) = \Bbb{Z}$ if $V \subset U$, $\{0\}$ otherwise. It is a construction similar to the constant presheaf. I see it is not a sheaf.
3. $\mathscr{I}(V) = \{\text{all local constant map }s:V \to \Bbb{Z}\}$ if $\varnothing \neq V \subset U$, $\{0\}$ otherwise. It is a construction similar to the constant sheaf. However, I failed to construct a morphism $\mathscr{I} \to \mathscr{F}$, which is essential in the proof.

So how to prove that (a) implies (c) in the case morphism of sheaves of abelian groups? Is it true? (Section 2.6 in Vakil's FOAG seems only using (a) and (b) are equivalent for sheaves of abelian groups)

Sorry for the context being kind of fuzzy. The detail I tried is too tedious to write all of it down (stuffs like checking axioms of being presheaves, being sheaves, being morphism).

Any help is appreciated. Thanks!

Answer 1

Your question is primarily about extending the proof of (a) $\implies$ (c) to sheaves of modules. To do so, it is helpful to first consider the special case of sheaves on $1$, which are just (up to equivalence of categories) ordinary modules. We consider the category of $R$-modules.

The task for (a) $\implies$ (c) in this case is as follows: we must show that if we have $R$-modules $F, G$ and a monomorphism $\phi : F \to G$ in the category of $R$-modules, then $\phi$ is injective.

To prove this, we note that $R$ itself provides a representation of the forgetful functor $R$-mod $\to Set$. In other words, there is a natural bijection $Hom(R, H) \to H$, where $H$ is an $R$-module. The bijection sends $f \in Hom(R, H)$ to $f(1)$. This means that the forgetful functor preserves monomorphisms, as all representable functors do (this is a good exercise - show that all representable functors preserve all limits, and therefore preserve monomorphisms). Thus, $\phi$ is a monomorphism of sets, hence injective.

Rephrasing in the language of sheaves on $1$, we found that the global sections functor $F \mapsto F(1)$ from sheaves of $R$-modules to sets is representable by $R$. So if we want to generalise to sheaves on $X$, we will want to show that the functors $\mathcal{F} \mapsto \mathcal{F}(U)$ are representable for all $U$.

In the general case, we have a sheaf of rings $\mathcal{R}$, and we wish to represent the functor $\mathcal{F} \mapsto \mathcal{F}(U) : $sheaves of $\mathcal{R}$-modules $\to Set$. It turns out that the relevant representing object is given by $C_U = i(D_U)$, where $i : PSh(X) \to Sh(X)$ is the sheafification functor and $D_U$ is defined by

$D(U)(V) = \begin{cases} R(V) & V \subseteq U \\ 0 & otherwise \end{cases}$

You considered $D(U)$ as a possibility in your question (it was option 3), but you didn't come up with the insight that you had to sheafify it.

One can get a natural bijection $Hom_{\text{sheaves of } \mathcal{R} \text{-modules}}(C_U, \mathcal{F}) \cong Hom_{\text{presheaves of } \mathcal{R} \text{-modules}}(D_U, \mathcal{F}) \cong \mathcal{F}(U)$. This is enough to complete the proof, and these are the "indicator sheaves" you are looking for. $\square$

If all you care about is finishing your proof, you can stop reading here. However, you may understandably find this a bit frustrating, since it is not entirely obvious where the magic $C_U$ come from. I will attempt to shed some light on the subject.

Another route to the proof in the case of ordinary modules that monomorphisms are injections goes like this. First, note that given any set $S$, we can explicitly construct $\bigoplus\limits_{s \in S} R$, the free $R$-module on $S$ - the construction is functorial in the obvious way. We can show that this free functor is the left adjoint of the forgetful functor $R$-mod $\to Set$. Therefore, the forgetful functor, being a right adjoint, preserves limits and thus preserves monos. A mono in the category of $R$-modules is therefore a mono in the category of sets.

This construction actually works in any topos with natural numbers object. In particular, it works in the topos of sheaves of sets on a space $X$. Given a sheaf of sets $\mathcal{S}$, we can construct the internal free module $\bigoplus\limits_{s \in \mathcal{S}} \mathcal{R}$. If you are familiar with the internal logic of a topos, this can be done in a fairly straightforward way by carrying out the normal set-theoretic construction within the internal logic (but you must be careful to use a purely constructive approach - defining $\bigoplus\limits_{s \in S} R = \{f \in \prod\limits_{s \in S} R \mid \{s \in S \mid f(s) \neq 0\}$ is finite$\}$ will not work unless every open set in $X$ is also closed). From here, we can show that the internal free module functor is the left adjoint to the forgetful functor which takes a sheaf of modules and outputs its sheaf of sets. A monomorphism in the category of sheaves of modules is therefore a monomorphism in the category of sheaves of sets. We then apply the fact that (a) $\implies$ (c) holds for monomorphisms of sheaves of sets to conclude it also holds for monomorphisms of sheaves of modules.

This approach is very general. It applies not only to modules but also to all kinds of purely equational algebraic structures, and it applies not only to sheaves on a space but also to all sorts of other toposes with NNO (and even $\Pi W$ pretoposes). And it sheds some light on how we came up with $C_U$.

Given any open $U \subseteq X$, we have the yoneda sheaf $y(U) \in Sh(X)$, which is defined by

$y(U)(V) = \begin{cases} 1 & V \subseteq U \\ \emptyset & otherwise \end{cases}$

and has the property that, in the category of sheaves of sets, there is a natural isomorphism $Hom(y(U), \mathcal{F}) \to \mathcal{F}(U)$.

Then it turns out that $C_U$ is the sheaf $\bigoplus\limits_{s \in y(U)} \mathcal{R}$. Thus, $Hom_{\text{sheaves of }\mathcal{R} \text{-modules}}(C_U, \mathcal{F}) \cong Hom_{\text{sheaves of sets}}(y(U), \mathcal{F}) \cong F(U)$. So there was actually no "magic" at all needed to come up with $C_U$, merely some intuition about how the internal direct sum works.

Answer 2

With Mark Saving's remarkable answer, I figure it out how to define a non-zero morphism $\varphi: \mathscr{I}_U \to \mathscr{F}$ such that $\phi \circ \varphi = 0$. In the following text, I will work in the context and use the symbol as in the question, remove some categorical argument (I will try to return to some days later after getting more familiar with the content), and keep it kind of elementary.

The "indicator sheaf" of $U$ is defined by:

$$ \mathscr{I}_U(V) = \begin{cases} \{\text{all local constant map }s:V \to \Bbb{Z}\} & \text{if } \varnothing \neq V \subset U \\ \{0\} & \text{otherwise} \end{cases}. $$

And define $$ \varphi: \mathscr{I}_U \to \mathscr{F} $$ by:

• if $V = \varnothing$ or $V \nsubseteq U$, $\varphi(V)$ defined to be the zero map.
• otherwise:

$\varphi(V): \mathscr{I}_U(V) \to \mathscr{F}(V)$ is defined by taking $1$, the constant map to $\operatorname{res}_{U,V}a$: $$\varphi(V)(1: V \to \Bbb{Z}) = \operatorname{res}_{U,V}a $$

Then for any other local constant map $i: V \to \Bbb{Z}$, since it's local constant，for any $p \in V$, there is some open neighborhood $p \in V_p \subset V$, such that $i(q) = i(p)$ for all $q \in V_p$.

Define $f_p = i(p)\operatorname{res}_{U,V}a \in \mathscr{F}(V_p)$. Then $\{V_p\}$ and $\{f_p\}$ satisfy the condition for the gluablitity axiom. Hence There is some $f \in \mathscr{F}(V)$ such that $\operatorname{res}_{U,V}f = f_p$. Define: $$ \varphi(V)(i)=f. $$

Then $\varphi$ is a morphism (skipped the reality check) from $\mathscr{I}_U$ to $\mathscr{F}$ such that $\phi \circ \varphi = 0 = \phi \circ 0$. Hence $\varphi = 0$. Especially $\varphi(U) = 0$ and $a = 0$. It's a contradiction. $\square$

This argument is also the manual proof for $Hom(C_U, \mathcal{F}) \cong \mathcal{F}(U)$ I comment in Mark Saving's answer. I don't know how to show it category theoretically, although. (Is it some kind related to the Yoneda lemma?)

Answer 3

If $\phi$ is a monomorphism, then the sheaf $\ker\phi=0$, so its sections $\ker\phi(U)=(\ker\phi)(U)=0$. Thus $\phi(U)$ is a monomorphism.