Why is the empty set assigned the trivial group in the definition of a pre-sheaf?

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

1. $$\mathscr{F}(\emptyset) = 0$$ (the trivial group),
2. $$\rho_{UU} = \mathrm{id}$$ for all open sets $$U$$ in $$X$$,
3. 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?

• I'm no geometer, but this ensures that the functor preserves initial objects. – Randall Sep 9 at 19:49
• @Randall Ah yes, I noticed that. But I am not aware of the geometric significance of that fact, hopefully someone can elaborate if there are any :) – Brahadeesh Sep 9 at 19:53
• I don't think that's standard; but in a sheaf it must be the trivial group. – Lord Shark the Unknown Sep 9 at 20:04

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 $$\emptyset$$.
• Oh, I didn't notice that $\mathscr{F}(\emptyset) = 0$ is automatic for sheaves. In Lang's Algebra, he includes this condition as a part of the definition of a sheaf, and I didn't think too deeply about it. – Brahadeesh Sep 9 at 20:09
• Is the empty cover $\bigcup_{U\in \emptyset}U$ different from the cover by the empty set $(\emptyset)$? – s.harp Sep 9 at 20:38
• @s.harp: That depends on what you mean by "the cover by the empty set". Here I mean the cover which writes $\emptyset$ as the union of no open sets, not the cover which writes $\emptyset$ as the union of one open set (namely, $\emptyset$). – Eric Wofsey Sep 9 at 20:39