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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeMay 1st 2012
    • (edited May 1st 2012)
    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTimeMay 1st 2012
    • (edited May 1st 2012)

    added to the Definition-section also the formulation “the diagonal is proper”

    • CommentRowNumber3.
    • CommentAuthorzskoda
    • CommentTimeMay 7th 2012

    So, how do you include separated scheme as a special case of a Hausdorff topos (the paragraph claims it is a special case) ? The separated scheme XX has a closed diagonal where the diagonal is in X×XX\times X which itself does NOT have a product topology.

    • CommentRowNumber4.
    • CommentAuthorUrs
    • CommentTimeMay 7th 2012

    The diagonal is compact.

    • CommentRowNumber5.
    • CommentAuthorzskoda
    • CommentTimeMay 7th 2012
    • (edited May 7th 2012)

    I am saying that I do not see how you consider a topology of X×XX\times X in scheme world ANYHOW related to a topology of X×XX\times X in the world of topoi (even if I understood how you meant to consider a scheme in general as simply a topos). So the first condition can not be a special case of the second condition. The fact that we are using word “diagonal” for two different things can not unify the subject as a special case, but only an analogy.

    • CommentRowNumber6.
    • CommentAuthorUrs
    • CommentTimeMay 7th 2012

    I don’t quite understand what you are saying.

    even if I understood how you meant to consider a scheme in general as simply a topos

    The topos corresponding of a scheme is the sheaf topos over the site of the scheme.

    The statement must be in EGA or somewhere. You will know this better than I do.

    • CommentRowNumber7.
    • CommentAuthorZhen Lin
    • CommentTimeMay 7th 2012

    The unadorned sheaf topos over a scheme (qua locally ringed space) doesn’t see the scheme structure, and it is the scheme structure that controls what X× YXX \times_Y X looks like, not the underlying space. So there’s no chance of Sh(X)× Sh(Y)Sh(X)\mathbf{Sh}(X) \times_{\mathbf{Sh}(Y)} \mathbf{Sh}(X) giving the right answer… whatever that expression even means! (For a concrete example: let Y=SpecY = Spec \mathbb{Q}, and let X=Spec¯X = Spec \overline{\mathbb{Q}}. Then the canonical morphism XYX \to Y is a homeomorphism, but the scheme X× YXX \times_Y X has rather a lot more than one point.) I wonder if the topos over a larger Zariski site might give the right answer though.

    • CommentRowNumber8.
    • CommentAuthorzskoda
    • CommentTimeMay 7th 2012
    • (edited May 7th 2012)

    Urs, the Zariski topology on scheme X×XX\times X differs significantly from the product topology on X×XX\times X for XX given by its Zariski topology, so any statement about the topological (Tihonov) product (and the topos product is such) says nothing about the scheme theoretic topology of the product. Hence the topological product of a diagonal in scheme theoretic sense has nothing to do with topological properties of diagonal in Tihonov product sense (what is reflected in topos picture as well). Am I speaking Spanish ? – I mean this is easy to understand and I do not see you are addressing this…

    • CommentRowNumber9.
    • CommentAuthorUrs
    • CommentTimeMay 7th 2012
    • (edited May 7th 2012)

    If I understand correctly you have in mind (the topos over) the underlying topological space. But the “site of the scheme” XX is Aff/XAff/X.

    • CommentRowNumber10.
    • CommentAuthorzskoda
    • CommentTimeMay 7th 2012
    • (edited May 7th 2012)

    If you take the Zariski Grothendieck topology on Aff/XAff/X it is ultimately induced by the usual Zariski topology on XX. Similarly, the Zariski topology on Aff/X×XAff/X\times X. The latter is very different from what one would expect from the product topology and the topos on the product: restrict for example the site to the embeddings of Zariski open subsets into XX and into X×XX\times X. Can’t you explain which magic makes the huge difference between the two topologies on X×XX\times X disappear once one takes a topos of sheaves of sets on the induced topology on Aff/X×XAff/X\times X ?? Maybe it does, but I have not heard any argument here so far. The problem does not disappear after replacing a site by topos or any other change of level of abstraction.

    • CommentRowNumber11.
    • CommentAuthorzskoda
    • CommentTimeMay 7th 2012
    • (edited May 7th 2012)

    Let me repeat it precisely: the statement about a Hausdorff topos is a relation about a morphism of topoi 𝒳𝒳×𝒳\mathcal{X}\to\mathcal{X}\times\mathcal{X}. If 𝒳=Sh(Aff/X,Zariski)\mathcal{X} = Sh(Aff/X, Zariski) why do you think that 𝒳×𝒳\mathcal{X}\times\mathcal{X} would have anything to do with 𝒴=Sh(Aff/X×X,Zariski)\mathcal{Y} = Sh(Aff/X\times X, Zariski) ? (Maybe you are right but I did not hear the argument so far)

    • CommentRowNumber12.
    • CommentAuthorUrs
    • CommentTimeMay 8th 2012
    • (edited May 8th 2012)

    Because pullback of toposes (discussed here) is computed by pushout of the underlying “2-frames”, the underlying sites as categories with finite limits.

    For slice sites C/aC/a and C/bC/b with a,bCa, b \in C their coproduct in Cat lexCat^{lex} is indeed C/(a×b)C/(a \times b). To see this, observe that for any two lex functors F 1:C/aDF_1 : C/a \to D and F 2:C/bDF_2 : C/b \to D the initial cocone morphism is

    C/a C/(a×b) C/b F 1 η F 2 D, \array{ C/a &\to& C/(a \times b) &\leftarrow& C/b \\ & {}_{\mathllap{F_1}}\searrow & \downarrow^\eta & \swarrow_{\mathrlap{F_2}} \\ && D } \,,

    where the inclusion on the left takes [Uϕa][U \stackrel{\phi}{\to} a] to [U×b(ϕp 1,p 2)a×b][U \times b \stackrel{(\phi\circ p_1, p_2)}{\to} a \times b] and similarly for the inclusion on the right, and where η\eta takes [U(ϕ 1,ϕ 2)a×b][U \stackrel{(\phi_1, \phi_2)}{\to} a \times b] to F 1(ϕ 1)×F 2(ϕ 2)F_1(\phi_1) \times F_2(\phi_2) in DD.

    Do you agree with that?

    • CommentRowNumber13.
    • CommentAuthorzskoda
    • CommentTimeMay 8th 2012
    • (edited May 8th 2012)

    Sounds as correct start! So Lurie says that for some sites giving the given topos the morphisms are just the lex functors and that in terms of those the product in the underlying category whose slice is taken are used for the coproduct. This convinces me that the statement that the topos induced by the scheme X×XX\times X is probably correct codomain. Now one of course needs to have more precise form – as the original sites are not of the given form (that the morphisms of sites are precisely the lex functors), and one should still see the crucial part: the statement that the diagonal of XX×XX\to X\times X is closed in Zariski topology as locally ringed space (what is in the definition of the separatedness) should somehow induce that the diagonal of the corresponding topoi (up to isomorphism we just discussed) is closed in the topological sense, and (likely harder) the converse.

    So, Urs, I now believe that the topos 𝒳×𝒳\mathcal{X}\times\mathcal{X} is (up to equivalence) the correct codomain but do not see that the closedness of the diagonal in the sense of schemes as locally ringed spaces is equivalent to the closedness of the diagonal in the sense of associated topoi. The latter, as you kindly reminded me, sees the full Grothendieck topology on AffAff (or on SchSch) while the first sees by the very definition just the topology of X×XX\times X as a locally ringed space. So how are the closedness conditions equivalent ?

    • CommentRowNumber14.
    • CommentAuthorUrs
    • CommentTimeMay 8th 2012
    • (edited May 8th 2012)

    Okay, so for the last bit:

    by the discussion at compact topos a slice topos is compact (relative to a base) precisely if the object that it is sliced over is compact in the sense (“quasi-compact”) that every cover (relative to the base) has a finite subcover.

    So the geometric morphism of toposes Sh(Sch/X)Sh(Sch/X)×Sh(Sch/X)Sh(Sch/X×X)Sh(Sch/X) \to Sh(Sch/X) \times Sh(Sch/X) \simeq Sh(Sch/X \times X) is separated in that it exhibits its domain as a compact topos over its codomain if XX×XX \to X \times X exhibits XX as (“quasi”-)compact over X×XX \times X, so if XX is separated.

    • CommentRowNumber15.
    • CommentAuthorzskoda
    • CommentTimeMay 8th 2012

    Thanks, Urs. If a scheme is separated then the diagonal is quasi-compact. The converse does not hold in general, there are essential conditions for that (e.g. expressed using valuations). So, the direction which I said should be harder does not seem to follow: the topos condition seems weaker.

    • CommentRowNumber16.
    • CommentAuthorZhen Lin
    • CommentTimeMay 9th 2012

    Fortunately, those algebraic geometers think of everything: the condition that the diagonal be quasi-compact is called “quasi-seperated”.

    • CommentRowNumber17.
    • CommentAuthorzskoda
    • CommentTimeMay 9th 2012

    Right, we have the entry quasi-separated morphism in the nLab. Yet weaker property is semiseparatedness after Thomason and Trobaugh, cf. semiseparated scheme.

    • CommentRowNumber18.
    • CommentAuthorUrs
    • CommentTimeMar 17th 2017

    I gave the Properties-section at Hausdorff topological space more structure. Also added the bare statement of Hausdorff reflection

    • CommentRowNumber19.
    • CommentAuthorUrs
    • CommentTimeApr 10th 2017

    Added the details of Hausdorff reflection (here).

    • CommentRowNumber20.
    • CommentAuthorTodd_Trimble
    • CommentTimeApr 11th 2017

    Isn’t the equivalence relation \sim also describable as the topological closure of the equality relation Δ:XX×X\Delta: X \to X \times X?

    • CommentRowNumber21.
    • CommentAuthorUrs
    • CommentTimeApr 11th 2017

    This is good, thanks. I have added this here

    • CommentRowNumber22.
    • CommentAuthorTodd_Trimble
    • CommentTimeApr 16th 2017

    Oh, bother. I’m sorry; the topological closure of Δ\Delta need not be transitive, so my suggestion in #20 was wrong. One might think \sim is instead describable as the minimal closed equivalence relation, i.e., the intersection of all closed equivalence relations, but that might not be right either! (It doesn’t seem straightforward to prove X/X/\sim is actually Hausdorff. There are traps lurking.)

    So perhaps that material should be excised until such problems are fixed.

    • CommentRowNumber23.
    • CommentAuthorUrs
    • CommentTimeApr 16th 2017

    There is a careful account of how to do this in section 4 here.

    • CommentRowNumber24.
    • CommentAuthorTodd_Trimble
    • CommentTimeApr 16th 2017

    Yes. By now I had seen all that (it’s discussed at MO as well), but that transfinite construction it is not to my aesthetic liking. I like what you had written before I wrote #20 much more.

    Of course, all this can also be seen as coming about via an adjoint functor theorem. :-) The solution set condition is easy (consider all possible Hausdorff topologies on quotient sets, which form at most a set).

    • CommentRowNumber25.
    • CommentAuthorTodd_Trimble
    • CommentTimeApr 16th 2017
    • (edited Apr 16th 2017)

    I see you’ve now included the transfinite construction – thanks for that addition.

    However, I don’t really agree that it’s “more direct” than that given by the first proof, which is much simpler. I suppose that some readers might feel uneasy about appealing to a proper class of Hausdorff spaces YY to describe the equivalence relation on XX, but that objection is spurious because each continuous f:XYf: X \to Y factors through a quotient set f(X)f(X) equipped with a Hausdorff topology (the subspace topology inherited from YY). There are only set-many such quotient sets of XX equipped with Hausdorff topologies, and you can restrict the YY to just those.

    It reminds me of a canard that the adjoint functor theorem doesn’t provide explicit, direct constructions. Of course that’s wrong: it gives perfectly explicit constructions, as long as you have explicit solution sets in your hands.

    • CommentRowNumber26.
    • CommentAuthorUrs
    • CommentTimeApr 16th 2017
    • (edited Apr 17th 2017)

    I like what you had written before I wrote #20 much more.

    That’s still in the entry here. But it has the disadvantage that it only gives the existence of the Hausdorff reflection, not a practical way to compute it.

    On the other hand, that argument has the advantage that it evidently generalizes also to T 0T_0 and T 1T_1. For T 0T_0 then there is of course an easy explicit construction of the reflection, the Kolmogorov quotient. But for T 1T_1 I suppose the explicit construction again needs a transfinite induction?

    • CommentRowNumber27.
    • CommentAuthorTodd_Trimble
    • CommentTimeApr 16th 2017
    • (edited Apr 16th 2017)

    That’s still in the entry here.

    I know.

    But it has the disadvantage that it only gives the existence of the Hausdorff reflection, not a practical way to compute it.

    No, not just existence, but an actual construction.

    I don’t consider the transfinite construction any more “practical”, or “explicit”. And the first method doesn’t require any of the infrastructure of ordinals (the method of transfinite constructions is usually justified by appeal to the replacement axiom, when you get down to it).

    • CommentRowNumber28.
    • CommentAuthorTodd_Trimble
    • CommentTimeApr 16th 2017

    I took the liberty of changing to “Some readers may find the following a more direct way of constructing the Hausdorff reflection:”, which I think we can both agree on. :-)

    • CommentRowNumber29.
    • CommentAuthorUrs
    • CommentTimeApr 16th 2017
    • (edited Apr 16th 2017)

    which I think we can both agree on.

    Sure, I don’t have a strong opinion on this.

    But I’d like to understand your attitude. I seems to me that we just saw by our own example that it can be tricky to determine what concretely the equivalence relation conditioned on the set of all surjective continuous functions to some T nT_n space divides out. For T 0T_0 the “evident” answer that it’s those pairs of points which are not properly separated is correct, for T 2T_2 it turns out more subtle. Lemma 4.13 in van Munster’s note, which says that the naive answer is correct after all if there are only finitely many not properly separated points, gives a prescription that seems efficient for concrete examples.

    • CommentRowNumber30.
    • CommentAuthorTodd_Trimble
    • CommentTimeApr 16th 2017

    It might be good then to reinforce the point that naive ideas (like mine in #20, or the one in #22 that was floated and rejected in the same sentence) are liable not to work. (In my defense, analogous constructions for more structured objects do work; for example, for topological groups, one mods out by the closure of the identity, which is a normal subgroup, to get the Hausdorff reflection.) Thus one could read the more elaborate transfinite construction partly as a cautionary tale – and the discussion in the article could be rounded out by citing some counterexamples.

    I mean, the method of the second proof is “venerable” (if at first you don’t succeed, try, try again – infinitely many times) and I agree it’s good to mention it, as surely it will appeal to many readers. My own aesthetic sensibility is that when available, I usually prefer one-shot top-down descriptions, even if seemingly abstract, over ordinal recursive iterations (from the bottom up, so to speak). For example, the proof that Pataraia gave of his eponymous fixed-point theorem is simply masterful, and perfectly constructive, and to me much nicer than any ordinal iterative proof (which as I said earlier does require ordinal infrastructure, which can be tricky in constructive settings, as well as requiring replacement) – and I certainly don’t find the latter more practical or explicit. So it was only the assertion “more direct” that I was arguing with. :-) I don’t happen to agree with that myself.

    Paul Taylor is another who seems to share my aesthetic preference in these matters.

    Maybe I’m wrong, but lemma 4.13 strikes me as dealing with somewhat exceptional hypotheses; I’m not wowed by it. Somehow I expect there would be more pleasant hypotheses that ensure that the closure of the diagonal is already transitive (so that one doesn’t even need the second half of h 1h^1, which is taking the transitive closure of the topological closure). But I haven’t thought about this.

    I think there are probably many situations where ordinal iterations are needed and a one-shot description isn’t available. An example might be the coreflection of monads into idempotent monads, among many others (small object arguments to name another?).

    • CommentRowNumber31.
    • CommentAuthorUrs
    • CommentTimeApr 17th 2017
    • (edited Apr 17th 2017)

    Maybe I am not expressing myself well. If we run into an example of a non-Hausdorff space we should be interested in having a good grasp about which points exactly need to be identified to make it a Hausdorff space. That lemma 4.13 achieves this for a class of examples that one will encounter, certainly for the first class of examples that one will discuss as examples. I didn’t think of this as being about aesthetics or wow-factors, but just about the matter-of-fact statements of the kind: here is an explicit description of a topological space, and here is an explicit description of its Hausdorff reflection.

    But anyway, I am fine with your changes of the wording. Sorry for making you do it twice, this was a copy-and-pasting mistake of mine.

    • CommentRowNumber32.
    • CommentAuthorTodd_Trimble
    • CommentTimeApr 17th 2017

    Oh, I see. So I think the case you are making is that one method is “more concrete” than the other. I think you’d have a point there (I don’t deny that the other method of proof appears more abstract). Thanks for your patience, and sorry for not taking your meaning sooner.

    • CommentRowNumber33.
    • CommentAuthorTobyBartels
    • CommentTimeApr 17th 2017

    @Urs #26:

    Bad link, should be

    That’s still in the entry here.

    or

    That’s still in the entry here.

    • CommentRowNumber34.
    • CommentAuthorUrs
    • CommentTimeApr 17th 2017

    Thanks for catching this. I have fixed it.

    • CommentRowNumber35.
    • CommentAuthorTobyBartels
    • CommentTimeApr 17th 2017

    Possibly we should have (in addition to or as part of Proposition 4.3) this variation, which incorporates Todd's point from #25):

    Let (X,τ)(X,\tau) be a topological space and consider the equivalence relation \sim on the underlying set XX for which xyx \sim y precisely if for every equivalence relation \approx such that the quotient topology makes X/X /{\approx} into a Hausdorff topological space we have xyx \approx y.

    Then the set of equivalence classes

    HXX/ H X \coloneqq X /{\sim}

    equipped with the quotient topology is a Hausdorff topological space and the quotient map h X:XX/h_X \;\colon\; X \to X/{\sim} exhibits the Hausdorff reflection of XX, according to prop. 4.1.

    This is still impredicative, but the impredicativity is of the size 2 2 |X|2^{2^{|X|}} (the size of the solution set in the adjoint functor theorem) rather than |Ord||Ord| (the size of a proper class), so may seem more concrete.

    (By the way, that link to Proposition 4.1 is another case where a section and a theorem have the same name, although in this case the theorem is the first item in the section, so it's not really a problem.)

    • CommentRowNumber36.
    • CommentAuthorUrs
    • CommentTimeApr 17th 2017

    rather than |Ord||Ord| (the size of a proper class)

    as stated in the entry with respect to surjective maps into any Hausdorff spaces, the quantification is indeed over a small set.

    (By the way, that link to Proposition 4.1 is another case where a section and a theorem have the same name,

    Thanks for catching this. I have fixed it now.

    although in this case the theorem is the first item in the section, so it’s not really a problem.)

    In fact it was a problem. The pointer to prop. 4.1 came out as “prop. 2”. (That the numbering breaks is not surprising, but that it breaks in such a seemingly erratic way is what makes debugging the duplicated anchors so hard, for long entries.)

    • CommentRowNumber37.
    • CommentAuthorTobyBartels
    • CommentTimeApr 18th 2017

    as stated in the entry with respect to surjective maps into any Hausdorff spaces, the quantification is indeed over a small set.

    I don't see where it's stated. And strictly speaking, at least from a material perspective, it is not over a small set; there are largely many topological spaces homeomorphic to any given inhabited space and so largely many surjective maps to a Hausdorff space from any given inhabited space1. Of course, it is an essentially small number of surjective maps, which is the number that really matters (and the only number that makes sense from a structural perspective); even so, the way to prove this is to note that surjective maps are parametrized by equivalence relations.


    1. As soon as the inhabited space has at least one surjective map to an inhabited Hausdorff space, which it has, say to the point. 

    • CommentRowNumber38.
    • CommentAuthorUrs
    • CommentTimeApr 19th 2017
    • (edited Apr 19th 2017)

    I don’t see where it’s stated.

    the proposition (presently 4.3) has "…for which xyx \sim y precisely if for every surjective continuous function…"

    But don’t read that as saying that your suggestion in #35 should not be implemented. Please do implement it!

    • CommentRowNumber39.
    • CommentAuthorTobyBartels
    • CommentTimeApr 20th 2017

    If you read ‘for every surjective continuous function [from a given topological space]’ and think that this says that the quantification is over a small set, then I guess that you've internalized the structural perspective!

    I'll implement #35 when I have time and decide where to put it ….

    • CommentRowNumber40.
    • CommentAuthorTobyBartels
    • CommentTimeApr 20th 2017
    • (edited Apr 20th 2017)

    @Urs #36:

    Of course, the other thing that is (potentially) of size |Ord||Ord| is the ordinal-inductive construction. But again, it is actually bounded by 2 2 |X|2^{2^{|X|}} (Lemma 4.11 in van Munster), and again this is how one proves that the construction works.