Example 6.11.5. Let $X$ be a topological space. Let $A_ x$ be a set for each $x \in X$. Consider the sheaf $\mathcal{F} : U \mapsto \prod _{x\in U} A_ x$ of Example 6.7.5. We would just like to point out here that the stalk $\mathcal{F}_ x$ of $\mathcal{F}$ at $x$ is in general not equal to the set $A_ x$. Of course there is a map $\mathcal{F}_ x \to A_ x$, but that is in general the best you can say. For example, suppose $x = \mathop{\mathrm{lim}}\nolimits x_ n$ with $x_ n \not= x_ m$ for all $n \not= m$ and suppose that $A_ y = \{ 0, 1\}$ for all $y \in X$. Then $\mathcal{F}_ x$ maps onto the (infinite) set of tails of sequences of $0$s and $1$s. Namely, every open neighbourhood of $x$ contains almost all of the $x_ n$. On the other hand, if every neighbourhood of $x$ contains a point $y$ such that $A_ y = \emptyset$, then $\mathcal{F}_ x = \emptyset$.

## Comments (3)

Comment #429 by JuanPablo on

"if each neighbourhood of $x$ has infinitely many points, and each $A_{x'}$ has exactly two elements, then $\mathcal{F}_x$ has infinitely many elements"

I think this statement is false.

Take for ease of notation $A_x=\{0,1\}$. Take $X$ an infinite set and take $\tau$ a non-principal ultrafilter of $X$, and take the open sets of $X$ to be exactly those sets which are in $\tau$. Then for any point $x\in X$, any neighbourhood of $x$ contains infinitely many points.

But $\mathcal{F}_x$ has four points, one is given by the section $s\in\mathcal{F}(X)$ constant with $s_y=1$ for all $y\in X$, another is given by $t\in\mathcal{F}(X)$ such that $t_y=1$ iff $y=x$. The other two are by interchanging $1$ and $0$ for this two.

This are the only four because for any $r\in\mathcal{F}(U)$ with $U\in\tau$ we can define $A=\{y\in U: r_y=0\}$ and $B=\{y\in U: r_y=1\}$ so that $A\cup B=U$, and as $\tau$ is an ultrafilter then either $A\in\tau$ or $B\in\tau$, this with $r_x=0$ or $r_x=1$ correspond to the four elements.

(There seems to be a problem in my computer with the subscripts in the preview)

Comment #430 by on

The preview should now be fine, there was a silly mistake in the code. I will let Johan handle the math :).

Comment #431 by on

Yes, very good! How bizarre to get 4 elements! I replaced it with the statement: $x = \lim x_n$ with $x_n \not = x$ and all sets have two elements then the stalks are infinite. The fix is here. Stop by my office if you are in NY and I'll give you a Stacks project mug, if I have't run out by then.

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• 2 comment(s) on Section 6.11: Stalks

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