Let $X$ be a topological space and $\cal F$ the sheaf of continuous real valued functions on it.
Somewhere I read the following definition:
Let $x\in X$. Then the 'stalk' of $\cal F$ at $x$ is $\Gamma(X,{\cal F})/A$ where $A$ is the set of $f\in\Gamma(X,\cal F)$ such that $f\rvert_U=0$ for some open neighborhood $U$ of $x$.
(I am not sure where I read this and if it was really a definition instead of an equivalent characterization)
As this is clearly not the standard definition I will denote this 'global stalk' by ${\cal F}^x$. Note that we always have a canonical injective map ${\cal F}^x\hookrightarrow{\cal F}_x$ where ${\cal F}_x$ denotes the usual stalk. Via this map we can identify ${\cal F}^x$ with the image of $\Gamma(X,\cal F)$ under the map $\Gamma(X,{\cal F})\to{\cal F}_x$.
My question: What would be an example of a space $X$ where ${\cal F}^x\ne{\cal F}_x$ or is there a characterization of such spaces? I.e. when is the sheaf of continuous real valued functions generated by global sections? (I think 'generated by global sections' usually means something little different, but maybe the term makes sense here, too.)
Some notes:
If we consider other sheaves it is very easy to come up with examples, take e.g. $X=\Bbb P^n_k=\operatorname{Proj} k[X_0,\dots,X_n]$ or a compact complex analytic manifold with their usual structure sheaves (dimension $>0$).
We have ${\cal F}^x={\cal F}_x$ if $X$ satisfies the following two conditions:
- Any continuous function on a closed subset of $X$ admits an extension to $X$ ($\Leftrightarrow X$ is normal)
- For each open neighborhood $U$ of $x$ there exists a closed (in $X$) subset $V$ of $U$ such that $V$ contains another open neighborhood of $x$.
Indeed in this case let $f\in\Gamma(U,\cal F)$ where $U$ is any neighborhood of $x$ and let $V$ be as in 2. Then by 1. there is an extension $g\in\Gamma(U,\cal F)$ such that $g\rvert_V=f\rvert_V$ and thus $f_x=g_x$.
These conditions are not necessary: Consider an infinite set $X$ together with the cofinite topology. Clearly $\Gamma(U,\cal F)=\Bbb R$ for all open sets $U\ne\emptyset$ (because $X$ is irreducible), so ${\cal F}^x=\Bbb R={\cal F}_x$