# Uses of stalks of sheaves and germs

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?

• Stalks are very natural. When you construct the étale space you use stalks so that they can be made "explicit". Actually looking at small neighborhoods provides more information than looking at the stalk. For instance locally constant sheaves have isomorphic stalk but globally they're different. About practical theorems, there's the analytic continuation of holomorphic functions. It's useful to know that in the case of sheaves (and not pre sheaves or mono pre sheaves) a morphism between sheaves that is stalkwise an isomorphism is, in fact, an isomorphism. – user40276 Jun 25 '15 at 6:44

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 $B$.
• Thank you for your reply. Sorry for this simple question, but is there some nice geometric reason why one should intuitively appreciate surjectivity on the stalks despite failure of surjectivity on the sections of $\mathcal{F}$ over $U$? (I know specific examples exist, but geometrically is there a nice interpretation of what properties the space has for this to happen?) – Supersingularity Jun 25 '15 at 6:03
• @Supersingularity In some sense, it captures the global gemetry of the open set on which you take sections. For instance, with your example of the complex plane with the sheaf of analytical functions, take the map defined by differentiating. Then it is surjective on the whole space, and on any open disc, but on the plane punctured at the origin, $\Bbb C \setminus \{0\}$, there is no function that differentiates to $\frac{1}{x}$. In fact, the lack of surjectivity is exactly what is measured by sheaf cohomology (in the words of my lecturer: this lack is what makes algebraic geometry interesting) – Arthur Jun 25 '15 at 6:09