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When one defines the etale space of a presheaf $\mathscr F$ on a topological space $X$, would be assumed that $X$ is a $T0$-space (i.e. for every $x$, $y$ in $X$ exists an open set containg one of them but not the other point)?

If $X$ is not $T0$, I'm not sure that the stalks of $\mathscr F_x$ are disjoint each other for all $x\in X$.

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You take the disjoint union of the stalks. – Fredrik Meyer May 15 '12 at 11:28
@FredrikMeyer is correct, and then you're supposed to glue together compatible germs. In your situation, you'd have isomorphic stalks at $x$ and $y$ for any sheaf. This is perhaps kind of worrying, but if you're willing to think about T0-spaces, then almost certainly you should be willing to allow such behavior. – Aaron Mazel-Gee May 15 '12 at 11:31
Ok, I will take the disjoint union of the stalks if the space $X$ is $T0$. But if $X$ is not $T0$, do you conferm that stalks taken in different point are disjoint (i.e. two germs $f_x$ and $g_y$ can't be the same)? – Dubious May 15 '12 at 12:24
up vote 0 down vote accepted

In order to define the étalé space associated to the presheaf $\mathcal F$ you start with the set $E(\mathcal F)$ of all triples $(x,U,\sigma )$ where:
$\bullet$ $x\in X$ is a point of the space $X$.
$\bullet$ $U$ is an open neighbourhood of $x$.
$\bullet$ $\sigma\in \mathcal F(U)$ is a section of $\mathcal F$ on $U$.
You then introduce the equivalence relation on $\mathcal F$ defined by requiring $(x,U,\sigma )\cong (y,F,\tau )\iff x=y $ and there exists an open neighbourhood $x\in W\subset U\cap V$ such that $\sigma \mid W=\tau \mid W \in \mathcal F(W)$.
The étalé space associated to $\mathcal F$ then has as underlying set $Et(\mathcal F)=E(\mathcal F)/\cong$

An element $[x,U,\sigma] \in Et(\mathcal F)$ is the equivalence class of $(x,U,\sigma) \in E(\mathcal F)$.
The definition of the equivalence relation $\cong $ on $E(\mathcal F)$ forces the implication $$ [x,U,\sigma]=[y,V,\tau] \implies x=y $$
so that the map $$\pi: Et(\mathcal F)\to X:[x,U,\sigma]\mapsto x$$ is well defined, independently of any condition on the topology of $X$.
The stalks $Et(\mathcal F)_x=\pi^{-1}(x)$ are thus always disjoint and that $X$ is or is not $T_0$ plays no role.

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Very well, it's clear. – Dubious May 15 '12 at 13:03

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