I am trying to understand the motivation behind defining stalks of sheaves, but I suppose my complex geometry is a little weak. I know they are meant to represent germs of holomorphic functions at a point, but generalized to any sheaf.
Here is what I have in mind currently: holomorphic functions are complex differentiable locally, so it doesn't make sense to talk about holomorphic functions at a point $z \in \mathbb{C}$. However, we can talk about the germs of holomorphic functions at $z$ (so identify all those functions that agree on some arbitrarily small neighborhood of $z$). In this way, it seems we can try to talk about complex functions "holomorphic at a point" by just using direct limits.
So my question is, why do we use germs/stalks at all? How can they reveal (geometric/local) structure that cannot be deduced by simply looking at small neighborhoods (which is what I think of when I think local). Are there any theorems that become easy to prove using germs (or maybe almost impossible/unclear without using germs)? Also, why do we want morphisms of sheaves to induce maps on the stalks? Is this just a convenient bookkeeping mechanism, or is there some genuine geometric meaning to it all?
Best Answer
Sheaves have a very local nature. This is reflected in the definition (the part that separates them from presheafs), and in many constructions. For instance, in Hartshorne, the structure sheaf on an affine scheme, the (quasi)coherent sheaf associated to a module, and the sheaf associated to a presheaf are all constructed by specifying what the stalks should be, and how they're glued together.
A map of sheaves that is surjective might not be surjective on all sections. But it is surjective on all stalks. In fact, a sequence of sheaves and morphisms between them is exact iff it induces exact sequences on all stalks.
Also, the fact that a map of locally ringed spaces should induce local maps on the stalks is needed for the one-to-one correspondence between homomorphisms $A\to B$ and scheme-morphisms $\operatorname{Spec} B\to \operatorname{Spec} A$ for rings $A$ and $B$.