About the etale space 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 $x\in X$.
 A: 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.
