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?
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.