# Understanding sheaves on a $2$-element set

I'm working through the Geometry of Schemes and wanted some clarification for an exercise.

Exercise I-5 considers a two-element set $$X=\{0,1\}$$ with the discrete topology and asks the reader to find the relations between the objects of a sheaf (of abelian groups) on $$X$$.

If we let $$\mathcal{F}$$ be the sheaf, then it's clear that $$\mathcal{F}(\emptyset)$$ is the trivial group and that we have a commutative diagram of restrictions from $$\mathcal{F}(\{0\})$$ to $$\mathcal{F}(\emptyset)$$, from $$\mathcal{F}(\{1\})$$ to $$\mathcal{F}(\emptyset)$$, from $$\mathcal{F}(\{0,1\})$$ to $$\mathcal{F}(\{0\})$$, and from $$\mathcal{F}(\{0,1\})$$ to $$\mathcal{F}(\emptyset)$$.

By the sheaf axiom, it seems like for any $$s\in\mathcal{F}(\{0\})$$ and $$t\in\mathcal{F}({\{1\}})$$, $$s$$ and $$t$$ restricted to the intersection, $$\emptyset$$, must be the same, so there is some unique section in $$\mathcal{F}(\{0,1\})$$ that restricts to $$s$$ and $$t$$.

What does this say about $$\mathcal{F}(\{0,1\})$$? I'm guessing it's related to the fiber product, but I'm not particularly well-versed in category theory. Also, how does this generalize to sheaves over different categories?

• If you haven't already, it is easy to prove that $\mathcal F(\varnothing)$ is always the terminal object for any topological space $X$ for sheaves of sets. You can also prove that if this is true for sheaves of sets, then it's also true for sheaves of any essentially algebraic structure. Feb 5, 2019 at 21:22

In this case, $$\mathcal{F}(\{0,1\})$$ is just the direct product of the groups $$\mathcal{F}(\{0\})$$ and $$\mathcal{F}(\{1\})$$.
This commutative diagram has to be a pullback $$\require{AMScd}$$ $$\begin{CD} \mathcal{F}(\{0,1\}) @>>> \mathcal{F}(\{1\})\\ @V V V @VV V\\ \mathcal{F}(\{0\}) @>>> \mathcal{F}(\emptyset)=\{0\} \end{CD}$$ but as the southeast corner is trivial, the northwest group is the direct product of the other corners, and the nontrivial maps are the projections from a direct product to its factors.
• How do we know that $\cal{F}(\{0,1\})$ (with the maps downwards and to the right) is universal? Feb 6, 2019 at 16:27