This question is motivated by showing that the category $\mathbf{Sheaves} (X)$ from the non empty open subset category to the category of abelian group $\mathbf{Ab}$ has enough injective objects.


Let $\mathscr{F} \in \mathbf{Sheaves} (X)$ and let $F = \underset{\rightarrow U}{\operatorname{colimit}} \mathscr{F}(U)$. Now let me point out what $F$ is. Let $\mathfrak{S}$ be the set of non empty open sets of $X$. Then by the knowledge of category theory, we can take coproduct then coequalizer to get colimits. See This. So $F = \bigoplus_\limits{S \in \mathfrak{S}} \mathscr{F}(S) / A$ and $\mathscr{F}(S) \rightarrow F$ is the composition of caronical injection and quotient, where $A$ is generated by $$(a)_S - (f(a))_\mathrm{Codom\{f\}}$$ The subscript means which the summand coordinate the element is in. $a \in \mathscr{F}(S), S \in \mathfrak{S}$ and $f$ is a morphism induced by an inclusion in $\mathfrak{S}$.

Then my question is

How to show that $a \in \mathscr{F}(S) \mapsto 0 \in F$ through the map in the universal cone of the colimit iff there are $a'\in \mathscr{F}(S')$ and $f_1 $, $f_2$ induced by inclusions such that $$f_1(a') = a$$ and $$f_2(a') = 0$$

The if part is easy since $(a)_S = [(a')_{S'} - (f_2(a'))] - [ (a')_{S'} - (f_1(a'))_S] \in A$. To show the only if part, maybe we should show that if $(a)_S$ can be decomposed to sum of $(a')_{S'} - (f(a'))_\mathrm{Codom\{f\}}$ then it can be decomposed to

$$[(a')_{S'} - (f_2(a'))] - [ (a')_{S'} - (f_1(a'))_S]$$

Or more specifically, let $\mathfrak{S}_x$ be the set of all subsets containing $x \in X$. Show that $a \in \mathscr{F}(S_x)$ maps to zero in $\underset{\rightarrow}{\mathrm{Lim}} \mathscr{F}(\mathfrak{S}_x)$ iff $f(a) = 0$ for some $f$ induced by an inclusion. This proposition is more closed to my original purpose.

  • $\begingroup$ Colimit of sheaves can also be computed as the sheafification of the colimit of the underlying presheaf and colimits of presheaves are computed pointwise. Maybe this other description may help. $\endgroup$ Oct 30 '19 at 16:46
  • $\begingroup$ @RobinCarlier Good point! I would think about it later cuz I am busy now. $\endgroup$
    – XT Chen
    Oct 30 '19 at 16:47
  • $\begingroup$ Actually, don't mind my comment, I thought you were taking a colimit of sheaves but you are taking a colimit of abelian groups. If I understand your notations, you are taking the colimit of the functor $\mathscr{F}$ over the category of all opens sets of $X$, right? $\endgroup$ Oct 30 '19 at 17:09
  • $\begingroup$ Now that I think about it: since $\mathscr{F}$ is a sheaf, one must have $\mathscr{F}(\varnothing) = \{0\}$. For any $S$, one has a map $\mathscr{F}(S) \to \mathscr{F}(\varnothing)$ induced by the inclusion $\varnothing \to S$. This map can be the $f_2$ you are looking for. But this seems a little silly and I feel like this is not what you are looking for. $\endgroup$ Nov 1 '19 at 0:20
  • $\begingroup$ @RobinCarlier I don’t think empty set is included in a sheaf. Otherwise for every sheaf the colimit of it (deem as a functional from the open set category to Abelian category) is zero. $\endgroup$
    – XT Chen
    Nov 1 '19 at 0:27

The property you are trying to prove is actually false. I will give a counterexample as well as a few explanations.

Let's begin with a counter-example. Let $X = \{1,2\}$ be the set with two elements, with its discrete topology, and let $\mathscr{F}$ be the presheaf defined by $\mathscr{F}(\{1\}) = \mathbb{Z}/2\mathbb{Z}$, $\mathscr{F}(\{2\}) = \mathbb{Z}/3\mathbb{Z}$. If we want the sheaf condition to hold, one must have $\mathscr{F}(\{1,2\}) = \mathbb{Z}/2\mathbb{Z} \oplus\mathbb{Z}/3\mathbb{Z}$. We define the maps $\pi_1: \mathscr{F}(\{1,2\}) \to \mathscr{F}(\{1\})$ and $\pi_2: \mathscr{F}(\{1,2\}) \to \mathscr{F}(\{2\})$ to be the natural projections on the relevant factor.

The category of non-empty open subsets of $X$ is simple enough to be described. In this situation, it looks like the category of the diagram of a pullback. Since $\mathscr{F}$ is contravariant, the diagram we get for computing $\mathrm{colim}\mathscr{F}$ will be that of a pushout, i.e a diagram $\mathbb{Z}/2\mathbb{Z} \leftarrow \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/3\mathbb{Z} \to \mathbb{Z}/3\mathbb{Z}$.

Explicitly, the colimit will be $\left((\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/3\mathbb{Z}) \oplus \mathbb{Z}/2\mathbb{Z} \oplus\mathbb{Z}/3\mathbb{Z}\right)/A$ where $A$ is generated by elements of the form $i(x) - \pi_1(x)$ and $i(x) - \pi_2(x)$ for $x$ in the first factor where $i$ denote the inclusion of the first factor $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/3\mathbb{Z}$.

The element $(1,1)$ is sent to $0$ in the colimit, indeed $i((1, 1)) = i((0, 1)) - \pi_1((0,1)) + i((1,0)) - \pi_2((1,0))$ since $\pi_1((0,1)) = \pi_2((1,0)) = 0$. Yet $(1,1)$ is not sent to zero by $\pi_1$ or by $\pi_2$. This provides a counterexample: in your notations, $f_1$ must be the identity (since $\{1,2\}$ is the whole space, there is no bigger open), and $f_2$ must be either $\pi_1$ or $\pi_2$, since these are the only maps, but none of these maps send $(1,1)$ to $0$.

This counterexample is essentially the one from there. The key point, here, is that the property you are trying to prove is only true in general when you are taking colimit of filtered diagrams, in this case, what you want to prove is exactly this lemma. However, when you remove the empty set from it, the posetal category of open subsets (or rather, its opposite) is not filtered anymore, and the property does not hold in general, as the counterexample show.

However, when you only consider open subsets that contains a given point $x$, the diagram become filtered again (the intersection of two open sets containing $x$ is an open set that contains $x$). In other words, what you say at the end of your question turns out to be true: a section $s$ has $s_x = 0$ (where $s_x$ is the stalk at $x$, i.e the class of $s$ in the colimit) if and only if $s$ restricts to the zero section on some open set containing $x$.

Finally, since you mentionnend your goal was to prove that the category of sheaves of abelian groups have enough injective, let me point you to this reference for a direct proof, and to this section for a way more general (although way harder to understand) criterion that ensure that an abelian category has enough injectives.


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