I'm currently taking a course on Complex Geometry, where we were introduced to the concept of a pre-sheaf.
A pre-sheaf $\mathscr{F}$ of abelian groups on a topological space $X$ is an assignment to each open set $U$ in $X$ an abelian group $\mathscr{F}(U)$, and for every pair of open sets $V \subset U$ in $X$, a group homomorphism (called the restriction homomorphism) $\rho_{UV} \colon \mathscr{F}(U) \to \mathscr{F}(V)$ such that
- $\mathscr{F}(\emptyset) = 0$ (the trivial group),
- $\rho_{UU} = \mathrm{id}$ for all open sets $U$ in $X$,
- if $W \subset V \subset U$ are open sets in $X$, then $\rho_{UW} = \rho_{VW} \circ \rho_{UV}$.
If I understand the definition correctly, then what we are doing is the following: given a topological space $X$, we have an associated category, say denoted by $\mathscr{C}_X$, whose objects are the open sets in $X$, and such that there is a unique morphism from $V$ to $U$ precisely when $V \subset U$ (given by inclusion, say). Then, a pre-sheaf is just a contravariant functor from $\mathscr{C}_X$ to $\mathbf{Ab}$, the category of abelian groups, such that $\mathscr{F}(\emptyset) = 0$.
Viewing it this way, it is not clear to me why we want the additional condition $\mathscr{F}(\emptyset) = 0$. Is there any advantage to requiring that the empty set be always assigned the trivial group when working with pre-sheaves or sheaves?
Best Answer
This is not the standard definition; the standard definition imposes no restriction on $\mathscr{F}(\emptyset)$. There is no real advantage to this definition (and some minor disadvantages), but mathematicians are often skittish about the empty set and like to impose unnecessary restrictions on it. Note that for sheaves, it is automatic that $\mathscr{F}(\emptyset)$ is trivial, by considering gluing condition for the empty cover of $\emptyset$.