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?

  • 1
    $\begingroup$ 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. $\endgroup$ Feb 5, 2019 at 21:22

1 Answer 1


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.

  • $\begingroup$ How do we know that $\cal{F}(\{0,1\})$ (with the maps downwards and to the right) is universal? $\endgroup$
    – whetham
    Feb 6, 2019 at 16:27
  • $\begingroup$ @whetham The diagram in the answer is essentially a particular instance of the sheaf condition. A pullback can be formulated as an equalizer of products. $\endgroup$ Feb 6, 2019 at 19:27

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