# What does the stalk of a sheaf of discontinuous sections look like?

I'm having some kind of cognitive dissonance here, but I'm having trouble figuring out which of my beliefs is false. Let $\mathscr{F}$ be a sheaf of abelian groups on a topological space $X$ and $x \in X$. Write $\mathscr{G} := \prod_{x \in X} i_\ast(\mathscr{F}_x)$, where $i_\ast(\mathscr{F}_x)$ is the $\mathscr{F}_x$-valued skyscraper sheaf at $x$.

We have $\mathscr{G}(U) := \prod_{x \in X} \Gamma(U, i_\ast(\mathscr{F}_x)) = \prod_{x \in U} \mathscr{F}_x$, which we can regard as the set of functions $s : U \to \bigsqcup_{x \in U} \mathscr{F}_x$ by writing $s(x)$ for the $x$-entry of the direct product. So $\mathscr{G}$ is the sheaf of discontinuous sections of $\mathscr{F}$.

Since taking stalks commutes with taking direct sums products, we have $\mathscr{G}_x = \mathscr{F}_x$ for each $x \in X$.

On the other hand, looking at elements of $\mathscr{G}_x$ directly, they represent functions $s : U \to \bigsqcup_{x \in U} \mathscr{F}_x$ up to equivalence on some open set. But it's easy to construct all sorts of such functions which don't agree on any open neighborhood of $x$ but which have the same value for $s(x)$ -- just send all the neighboring points absolutely anywhere. So $\mathscr{G}_x$ is much, much larger than $\mathscr{F}_x$.

Where is the mistake?

Update: Thanks to both of you for helpful answers; it was tough choosing just one.

• Dear Daniel: taking stalks commutes with direct sums and doesn't necessarily commute with infinite direct products. So your edit modified a correct assertion into an incorrect one! – Georges Elencwajg May 3 '13 at 8:04
• Is the description of $\mathscr{G}_x$ in my second paragraph correct, then? – Daniel McLaury May 3 '13 at 8:10
• The description in your second paragraph is correct provided you remove the equality with $\prod_{x \in X} \Gamma(U, i_\ast(\mathscr{F}_x))$: see the answers below. – Georges Elencwajg May 3 '13 at 11:02

## 2 Answers

First let me note that the asociation $\mathcal F\to\mathcal G=\mathcal G(\mathcal F)$ is an extremely useful functor introduced by Godement in his book and through which he calculates the Grothedieck cohomology of sheaves, the main virtue of his sheaves being that they are flabby, hence acyclic.
So your choice of the letter $\mathcal G$ for this Godementification is excellent!

And you are perfectly right that the stalk $\mathcal G_x$ is much, much larger than the corresponding $\mathcal F_x$: congratulations for not sharing the common misconception that these stalks are equal.

Your difficulty just results from two facts:

1) Your notation $i_\ast \mathcal F_x$ does not make sense: what is $i$?
You might introduce the collection of $i_x$'s (and then take into account Matt's remark) but it is better to directly define $\mathcal G$ by your (and Godement's!) formula $\mathcal{G}(U) := \prod_{x \in U} \mathcal{F}_x$.

2) More importantly, it is completely false that taking stalks commutes with infinite products.
This explains the discrepancy between your two reasonings.

• By $i_\ast(\mathscr{F}_x)$ I mean $(i_x)_\ast(\mathscr{F}_x)$ where $i_x : \{x\} \to X$ is the inclusion. This notation (including omitting the subscript) is what's used in Hartshorne. – Daniel McLaury May 3 '13 at 12:50
• The impression I get, then, is that it would be completely hopeless to try and use this technique to manually construct an injective resolution since you'd have to find an injective that the indescribable objects $\mathscr{G}_x$ inject into. Is this right? – Daniel McLaury May 3 '13 at 12:59
• Dear @Daniel, I know you meant $i_x$ when you wrote $i$ and I was quite surprized to see that Hartshorne abuses language in this fashion. Despite my admiration for him, I think his is a wrong decision: some students, less alert than you, might not notice that there is an abuse of notation. – Georges Elencwajg May 3 '13 at 13:55
• @Daniel(continued) I think Hartshorne in his Proposition 2.2, Chapter III, does something in between: he starts from $\mathcal F$ takes the stalks $\mathcal F_x$, embeds them in injective modules $I_x$ and then takes the sheaf of discontinuous functions associated to these injective modules. I agree that "indescribable" is a good qualification of the whole process, and I find Godement's method more pleasant.The drawback is that Godementification only yields flabby sheaves, not injective ones, and this might be a drawback in some contexts. Buy it suffices for defining cohomology. – Georges Elencwajg May 3 '13 at 14:08
• This is very interesting to read, thanks Georges. – Joachim May 3 '13 at 14:25

In the third line you write that taking stalks commutes with direct sums, but the actual construction of $\mathcal G$ is a direct product, not a direct sum (unless $X$ is finite).

Even if $X$ is finite, it won't be true that $\mathcal F = \mathcal G$ unless all the points $x$ are closed (so that $X$ is actually discrete, in which case there is no difference between continuous and discontinuous sections).

The reason is that if $x$ is not closed, then $i_*(\mathcal F_x)$ has non-zero stalks not just at $x$, but at all the points in the closure of $x$.

• Sorry, of course I meant "direct product" like I wrote in the equations rather than "direct sum" like I wrote in the sentence. Do direct products not commute with taking sheaves? I thought that they commuted with taking sections over an open set, and that that property was stronger. – Daniel McLaury May 3 '13 at 7:51
• Also, I'm not sure I understand your second paragraph. Should it say $\mathscr{F}_x$ and $\mathscr{G}_x$ instead of $\mathcal{F}$ and $\mathcal{G}$? – Daniel McLaury May 3 '13 at 7:51
• Finally, what does $\mathscr{G}_x$ look like? Is the description in my last paragraph correct? – Daniel McLaury May 3 '13 at 7:52
• Dear @Matt, your answer somehow misled Daniel into modifying his correct assertion that taking stalks commutes with direct sums into the incorrect assertion that taking stalks commutes with direct products. Maybe you could modify your (absolutely correct!) answer so as to dispel this misunderstanding. – Georges Elencwajg May 3 '13 at 8:01
• @Daniel: There is a natural map from $\mathcal F$ to $\mathcal G$, so this is an isomorphism if and only if it is an isomorphism on stalks. So $\mathcal F_x = \mathcal G_x$ for all $x$ is equivalent to $\mathcal F = \mathcal G$, if the equalities are to be understood via the natural isomorphism. (This answers one of your questions.) Also, stalks are defined as a direct limit, and while it is reasonable to try to commute direct limits past other direct limits (by taking the "double limit", so to speak), you can't expect to interchange a direct limit and a product in general. ... – Matt E May 3 '13 at 11:29