Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
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
add comment

2 Answers

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

share|improve this answer
add comment

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.

share|improve this answer
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
show 2 more comments

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.