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Suppose that $X$ is a topological space with a sheaf of rings $\mathcal{O}_X$. In general, the stalk at a point $p \in X$ is the direct limit of the rings $\mathcal{O}_X(U)$ for all open sets $U$ containing $p$.

Here are two questions on computing stalks - I think both should be true, since a direct limit should be some sort of "limiting process", but that's far from convincing for me.

Can I compute the stalk of $\mathcal{O}_X$ at a point $p \in X$ by only limiting over basic open sets of $X$ containing $p$?

Can I compute the stalk of $\mathcal{O}_X$ at a point $p \in X$ by excluding some finite number of "large" open sets around $p$, and then limiting over the remaining open sets around $p$?

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This is definitely a question for Math Overflow... mathoverflow.net –  Noldorin Aug 2 '10 at 16:22
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@Noldorin: I disagree. It is a basic question in the theory of sheaves and is definitely appropriate for this site. –  Eric O. Korman Aug 2 '10 at 16:30
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@Noldorin: I definitely agree that this is not undergraduate level. However, many questions on this site are not and I did not think this site was restricted to undergraduate level mathematics. –  Eric O. Korman Aug 2 '10 at 16:33
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@Noldorin et. al. This question is not appropriate for MO, as it would likely not be of interest to professional mathematicians, as many of those using sheaves will know this. However, I think this is an excellent question for this website, precisely because it is useful for those learning sheaf theory. Additionally the question is well asked. We do not wish to discourage this sort of question. If we want to discuss this further we might want to start a meta thread, rather than pollute this fine question. –  BBischof Aug 2 '10 at 17:08
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I've deleted all comments not specifically about this question. Whether or not any particular question is suitable should be discussed on meta. –  Larry Wang Aug 5 '10 at 17:13
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3 Answers

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Yes. The general statement is the following: limit over a poset is equal to limit over its any coinitial subset. Formal proof is easy (hint: construct maps in both directions) and informally it's an analogue of "subsequence has the same limit as a sequence" theorem.

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(Re: Formal proof is easy) ..especially if one use set-theoretic definition of direct limit (described by Akhil in other answer) and not category-theoretic definition (as an adjoint functor aka as something satisfying some universal property). Is the question why this two definitions give the same result? –  Grigory M Aug 2 '10 at 18:20
    
This is fantastic, thanks Grigory. In fact, the isomorphism you describe follows readily from the facts I proved in an Atiyah-Macdonald exercise, and also easily from Akhil's compatible germs definition. As for equivalence, I've proven on my own that the germs definition matches the direct limit in the category of $R$-modules. –  Dalron Aug 3 '10 at 14:05
    
it's my pleasure –  Grigory M Aug 3 '10 at 15:26
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Basically, here's a way to think of the stalk that is more "down-to-earth" than direct limits. An element of the stalk $O_x$ is given by a pair $(f, U)$ where $f$ is a section over the open set $U$ and $U$ contains $x$. Two pairs $(f,U), (g,V)$ are considered equivalent if $f=g$ on a neighborhood $W$ of $x$ (contained in $U \cap V$).

With this definition, it's easy to see that what happens at $x$ doesn't depend on what happens on $F$, where $F$ is any closed set disjoint from $x$. The stalk is a purely local construction.

As for why this is equivalent to the direct limit: that's a direct corollary of how the construction works in most familiar categories with which one might define a sheaf (sets, groups, rings, etc.)

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This is sometimes referred to as the "compatible germs" definition –  BBischof Aug 2 '10 at 18:18
    
Thank you for the good answer - I wish I could accept two. I do switch to the "germs" way of looking at things, which sometimes makes a problem much simpler... –  Dalron Aug 3 '10 at 13:58
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I think there is a missing word in Akhil Mathew's answer: and it's "filtered".

You can do that because stalks are filtered colimits (aka "direct limits").

For filtered colimits, $\varinjlim_i X_i$, you can take representatives of elements $x \in \varinjlim_i X_i$ for some $i$ belonging to the set of indexes $I$ (in our case, the open sets $U$). That is, you can find some $i$ and $x_i \in X_i$ that goes to your $x$ through the universal arrow $X_i \longrightarrow \varinjlim_i X_i$. For instance, every element of the stalk $O_{X,p}$ can be represented by a section $f \in O_X(U)$ for some open set $U$.

But this is not true for other kinds of colimits.

For instance, take the push-out of two arrows $f: A \longrightarrow B$ and $g: A \longrightarrow C$ in the category of, say, abelian groups. Elements of this push out $B \oplus_A C$ are classes of pairs $(b,c) \in B\oplus C$, where you quotient out elements of the form $(f(a), 0) - (0, g(a))$, for all $a\in A$. That is, $(f(a),0) = (0,g(a))$ in $B\oplus_A C$.

Elements of $B\oplus_A C$ cannot be represented, in general, by elements coming from just $B$ or $C$, which are of the form $(b,0)$ or $(0,c)$, respectively: so, for a general $(b,c) \in B\oplus_A C$ there is no $b \in B$, nor $c\in C$, that represents it.

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Good point. I make the mistake of forgetting that not all colimits are filtered too frequently. –  Akhil Mathew Aug 5 '10 at 21:48
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