# The Stacks Project

## Tag 006U

Remark 6.7.2. There is always a bit of confusion as to whether it is necessary to say something about the set of sections of a sheaf over the empty set $\emptyset \subset X$. It is necessary, and we already did if you read the definition right. Namely, note that the empty set is covered by the empty open covering, and hence the ''collection of section $s_i$'' from the definition above actually form an element of the empty product which is the final object of the category the sheaf has values in. In other words, if you read the definition right you automatically deduce that $\mathcal{F}(\emptyset) = \textit{a final object}$, which in the case of a sheaf of sets is a singleton. If you do not like this argument, then you can just require that $\mathcal{F}(\emptyset) = \{*\}$.

In particular, this condition will then ensure that if $U, V \subset X$ are open and disjoint then $$\mathcal{F}(U \cup V) = \mathcal{F}(U) \times \mathcal{F}(V).$$ (Because the fibre product over a final object is a product.)

The code snippet corresponding to this tag is a part of the file sheaves.tex and is located in lines 527–551 (see updates for more information).

\begin{remark}
\label{remark-confusion}
There is always a bit of confusion as to whether it is
necessary to say something about the set of sections of
a sheaf over the empty set $\emptyset \subset X$.
definition right. Namely, note that the empty set is
covered by the empty open covering, and hence the collection
of section $s_i$'' from the definition above actually form
an element of the empty product which is the final object
of the category the sheaf has values in. In other words,
if you read the definition right you automatically deduce
that $\mathcal{F}(\emptyset) = \textit{a final object}$,
which in the case of a sheaf of sets is a singleton.
If you do not like this argument, then you can just require
that $\mathcal{F}(\emptyset) = \{*\}$.

\medskip\noindent
In particular, this condition will then ensure that if
$U, V \subset X$ are open and {\it disjoint} then
$$\mathcal{F}(U \cup V) = \mathcal{F}(U) \times \mathcal{F}(V).$$
(Because the fibre product over a final object is a product.)
\end{remark}

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