# Zero in colimit of sheaves category

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

Prerequisition

Let $$\mathscr{F} \in \mathbf{Sheaves} (X)$$ and let $$F = \underset{\rightarrow U}{\mathrm{Lim}} \mathscr{F}(U)$$ where $$U$$ is taken over all open sets of $$X$$. Now let me point out what $$F$$ is. Let $$\mathfrak{S}$$ be the set of 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.

• Your notation $F = \operatorname{colim}_{\to} \mathscr F$ does not make sense. One side is an abelian group, and the other side a sheaf (although already the right hand side does not make sense by itself, because it is unclear what colimit you are taking). Could you please clarify? – R. van Dobben de Bruyn Oct 28 '19 at 15:11
• @R.vanDobbendeBruyn A sheaf is a functor from the category of open sets of $X$ with morphism inclusion to the category of an Abelian category (i.e. Abelian group category in this question). Since the category of Abelian group is cocomplete, there is a colimit for every sheaf $\mathscr{F}$. – XT Chen Oct 28 '19 at 15:18
• @R.vanDobbendeBruyn I hope I have clarified exactly. – XT Chen Oct 28 '19 at 15:22
• Ah, so you mean $\operatorname{colim}_{U} \mathscr F(U)$; that makes more sense. Thanks. Could you please edit the question? The way it's written now suggests you are taking the colimit over some system of sheaves, which is not the case. – R. van Dobben de Bruyn Oct 28 '19 at 17:28
• @R.vanDobbendeBruyn Sure. I've edited. – XT Chen Oct 28 '19 at 17:32