Can one characterize those sheaves which have Hausdorff etale spaces?

Question

Given a sheaf of sets $F$ on a space $X,$ under the equivalence of categories between etale spaces over $X$ and sheaves over $X,$ $F$ is associated to a local homeomorphism $$E\left(F\right) \to X$$ whose sections are $F$. Is it well known when the space $E\left(F\right)$ is Hausdorff? Note, I do not want to (necessarily) assume that $X$ is Hausdorff, but I seem to remember that there might be a simpler answer in this case.

Is there a way to express this in terms of properties of $F$ on the site of opens of $X$?

Is there a way to express this as a nice categorical property of $F$ in terms of the topos $Sh(X)$. I.e., is there a way to abstractly characterize those sheaves $F \in Sh(X),$ which are Hausdorff? I am not looking for a reformulation, e.g. saying that the diagonal map of $Sh(X)/F$ should be a proper map of topoi.

Answer 1

I don't feel like this is a full answer(and it only adds anything to your first question), but I can't yet comment.

The tautological condition(e.g. if and only if) is that if $\mathscr{F}$ is a sheaf on $X$ and if we have two distinct stalk elements $f \in \mathscr{F}_x$ and $g \in \mathscr{F}_y$, for $x,y \in X$ then there should be two opens $U,V \subseteq X$ with $x \in U$ and $y \in V$ such that $\exists f' \in \mathscr{F}(U)$ with $f'_x = f$, and the same for $g$(mutatis mutandis), such that for any $z \in U \cap V$ we have $f'_z \neq g'_z$.

We can impose a few conditions I think to relate when this can happen to the topology on $X$. If $X$ is not sober this condition can be unsatisfiable. I like to think of sobriety as being broken up into two conditions:

1) Every irreducible closed subset is the closure of some generic point

2) If an irreducible closed subset $K$ is the closure of a generic point, that point is unique in the sense that no other point has closure $K$.

If $X$ fails (2) then no sheaves have Hausdorff etale space. We find $x,y$ with the same closure(failing (2) necessitates the existence of two such points with my wording) and look at the stalks over $x,y$ and pick an element $f$ in them coming from the same open set $W$ containing $x$ and $y$, so on every open subset of the open set corresponding to $f$ of the etale space the open sets have intersection at least at the points $f \in \pi^{-1}(\mathscr{F}_x)$ and $f \in \pi^{-1}(\mathscr{F}_y)$

Answer 2

The answer of the first question is that for all $U$ open in $X$, you need that for all $f,g \in F(U)$ the subset $\left(x \in U| germ_xf \ne germ_xg\right) \subset U$ is open.

If this holds, then for any two points $\tilde x$ and $\tilde y$ in $E(F),$ one may take two opens $U$ of $x$ and $V$ of $y$ respectively, where $x$ and $y$ are their images in $X,$ such that there exists $f \in F(U)$ and $g \in F(V)$ such that $$germ_xf=\tilde x$$ and $$germ_yg=\tilde y.$$ Then the subset $W$ of $U \cap V$ on which $f|_{U\cap V}$ and $g|_{U \cap V}$ agree is closed, and then one may define the open sets $\tilde U:=U - W$ and $\tilde V:V -W.$ One then has that $f(\tilde U)$ and $g(\tilde V)$ are disjoint opens of $\tilde x$ and $\tilde y$..

EDIT: This is wrong, as Mike points out, but the proof can easily be adapted to work when $X$ is Hausdorff:

For the first half above, when $X$ is Hausdorff, if $x$ and $y$ are not equal, one may choose small enough disjoint opens in $X,$ $U$ and $V$ over which there exists sections $$f \in F(U)$$ and $$g \in F(V)$$ such that $\tilde x=germ_x f$ and $\tilde y=germ_y g.$ Then $f(U)$ and $g(V)$ are necessarily disjoint and contain $\tilde x$ and $\tilde y$ respectively. If $x=y,$ then one has that there exists an open $U$ containing $x$ and $$f,g\in F(U)$$ such that $\tilde x=germ_x f$ and $\tilde y=germ_x g.$ Since $\tilde x \ne \tilde y,$ one has that $x \in \left(z \in U| germ_zf \ne germ_z g\right)=:W_x$ which is open. Hence, $f(W_x)$ and $g(W_x)$ are disjoint neighborhoods of $\tilde x$ and $\tilde y.$

(Notice that the converse still holds as stated, without assuming $X$ Hausdorff, so, it follows that the condition is still necessarily when $X$ is not Hausdorff, but not necessarily sufficient.)

Conversely, suppose that $E(F)$ is Hausdorff, and let $U$ be an open of $X.$ Consider for all $f,g \in F(U)$ the subset $Q:=\left(x \in U| germ_xf \ne germ_xg\right) \subset U$. Let $$z_1=germ_x f \ne z_2=germ_x g.$$ There exists disjoint neighborhoods $V_1$ and $V_2$ respectively. Let $$O_x:=f^{-1}(V_1) \cap g^{-1}(V_2).$$ This is a neighborhood of $x.$ $f(O_x) \subset V_1$ and $g(O_x) \subset V_2$ are now also disjoint, hence $O_x \subset Q,$ so $Q$ is open.

Answer 3

$\def\sF{\mathcal{F}}$This is a partial answer, coming from Analysis. It didn't fit as a comment and nobody had pointed it out before, so I am posting it as an answer.

Definition. We say that that a sheaf $\sF$ over $X$ satifies the identity principle if for every open and connected subset $U\subset X$ it holds that if $s,t\in\sF(U)$ are sections such that $s_x=t_x$ for some $x\in U$, then $s=t$.

If $\sF$ is a sheaf whose étale space is Hausdorff, then it must satisfy the identity principle: for an open and connected subset $U\subset X$, one considers the set $$ S=\{x\in U\mid s_x=t_x\}. $$ Then $S$ is open and, using that $|\sF|$ is Hausdorff, $S$ is closed. So $S\neq\varnothing$ is equivalent to $S=U$.

Thus, any candidate for a possible characterization of the Hausdorff condition on the étale space must imply the identity principle.

One can show that if $X$ is locally connected Hausdorff then the converse is also true ($\sF$ satifies id. principle implies $|\sF|$ Hausdorff). This is actually how one proves the Hausdorff condition on the étale space of the sheaf of holomorphic functions on $\mathbb{C}^n$.