# Can one characterize those sheaves which have Hausdorff etale spaces?

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.

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I would like to suggest "Can one characterize Hausdorff étale spaces" as a title. – Martin Brandenburg Jul 17 '12 at 12:12
Changed accordingly. – David Carchedi Jul 17 '12 at 13:57
For what it's worth, in case anyone else is confused, here is an example of a well-supported sheaf on a non-Hausdorff space whose total space is Hausdorff. Let $X$ be the real line with a doubled origin, and let $E(F) = \mathbb{R} + \mathbb{R}$ with the two maps to $X$ being two "copies of the identity" one going through each copy of the origin. Then $E(F)$ is certainly Hausdorff and its map to $X$ is surjective, and it's easy to see that it is also a local homeomorphism. – Mike Shulman Jul 19 '12 at 17:31
Another comment along these lines: $E(F)$ can only be Hausdorff if $X$ is locally Hausdorff. Also, if $X$ is locally Hausdorff, then one can choose a covering by Hausdorff neighborhoods, and the canonical projection to $X$ from the disjoint union of the elements of this cover is a local homeomorphism, hence a sheaf. This generalizes Mike's example. – David Carchedi Jul 19 '12 at 19:11

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 $f \in \pi^{-1}(\mathscr{F}_y)$

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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.

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This can't possibly be right unless $X$ is Hausdorff, since the terminal sheaf always satisfies your condition, but its etale space is just $X$. I think the flaw in your proof is that $W$ might contain $x$ or $y$, so that $f(\tilde{U})$ and $g(\tilde{V})$ need not contain $\tilde{x}$ and $\tilde{y}$ respectively. – Mike Shulman Jul 17 '12 at 20:19
And actually, doesn't your condition basically amount to saying that the diagonal $E(F) \to E(F) \times_X E(F)$ is closed, i.e. that the geometric morphism $Sh(X)/F \to Sh(X)$ is separated? Which, given that $Sh(X)$ is a separated topos, is equivalent to $Sh(X)/F$ being separated by B3.2.25. – Mike Shulman Jul 17 '12 at 20:21
Thanks Mike. I'll have a look. I wrote this up rather hastily because my collaborator is visiting me. I'll have a think and revise this answer. – David Carchedi Jul 17 '12 at 23:14
Actually, as Serre observes in FAC, chapter 1, section 1 it might happen that $E(F)$ is not separated although $X$ is. – Filippo Alberto Edoardo Jul 18 '12 at 3:14
@Filippo: That is why I am asking though. Sometimes a sheaf is Hausdorff, even when the underlying space isn't. I'd like to understand precisely when this happens. – David Carchedi Jul 18 '12 at 12:13