Added the papers by Veblen and Whitehead, and also Whitney.

]]>Added link to category SmoothManifolds.

]]>In the cold light of day, perhaps the key point alluded to in the exercise I quote in #44 is that the open covers on manifolds can be generated by an explicitly given class of open covers. This might be helpful to pin down what it means for a sheaf on $CartSp$ to be locally representable.

]]>As the reference is only alluded above here it is (about the “class of open maps”) for the record, in full:

- Andre Joyal, Ieke Moerdijk,
*A completness theorem for open maps*, Annals of Pure and Applied Logic**70**, Issue 1, 18 November 1994, p. 51-86, MR95j:03104, doi

Going back to the points raised in #2 and #3, I read in MacLane&Moerdijk just recently an exercise that says that the usual open cover topology on the category of manifolds (presumably separable ones, and also finite dimensional) is the smallest one containing the covering families

$U_1 = \{(-\infty,1),(-1,\infty)\}$

$U_2 = \{(-n,n)|n\geq 1\}$

Now these are covering families of $\mathbb{R}$ in the category $CartSp_0$ of cartesian spaces and maps preserving zero. They generate covering families on all $\mathbb{R}^n$ by taking products of these.

I’m not sure what this means, I’m just putting ’on paper’ as it were (and it’s late here, for me). Any thoughts?

]]>It is effectively the coverage data, I think

I feel like the “class of open maps” (a term which I much prefer to “admissibility structure,” ugh) is slightly more than giving a coverage (or pretopology) which generates the topology — it’s more like a class of maps which *could appear* in a generating covering family, but which don’t necessarily do so. There is an abstract notion of “class of open maps” which has been studied by I think Joyal and Moerdijk, among others.

Zoran, I think you are right, I was wrong. I need to come back this whole issue here and do things right. I should maybe not have started this without the time to concentrate on it. On the other hand, except that I am making a fool of myself, I am enjoying that we are discussing this! :-)

Here is another thought that I won’t have time to look at in detail right now, but which I’ll mention anyway:

as a slight variant of the definition in *Structured Spaces* with more an emphasis on the big topos $Sh(CartSp)$: maybe it makes sense to say that for $X \in Sh(CartSp)$ and $U_i \in CartSp \hookrightarrow Sh(CartSp)$ we have an effective epi $\coprod_i U_i \to X$ plus the condition that the corresponding morphisms in $Topos/Sh(CartSp)$

sit in the right half of a suitable factorization system. But I still need to understand better why the analogous condition is not needed in def. 2.3.9 of *Structured Spaces* .

Urs, do you have a way out of the “gap” in your Hausdorffness proof I seem to point out in 27 ? I am still not seeing it.

]]>Maybe the issue is more vivid in terms of the classifying topos description: $\mathcal{G}$-structures are classified by geometric morphisms into $Sh(\mathcal{G})$. That does not involve the admissibility structure. Instead, the admissibility structure on $\mathcal{G}$ is equivalently a natural factorization system on $Topos(-,Sh(\mathcal{G}))$, hence affects only the morphisms. The morphisms of $\mathcal{G}$-structured toposes are those landing in the right half of this factorization system. But every equivalence is in there, so it does not affect the notion of equivalence. I’d think.

]]>But doesn’t the “admissibility structure” there essentially encode the same data about which maps are “open”?

Yes. (It is effectively the coverage data, I think.)

I had thought this does not crucially play a role in def 2.3.9, because it doesn’t in saying that $\coprod_i U_i \to *_{\mathcal{X}}$ is a regular epi and not in saying that we have equivalences of structured topoi $Spec U_i \simeq \mathcal{X}/U_i$ (I think, because an equivalence of structure sheaves just as left exact functors should automatically be a morphism that respects the admissibility structure and hence be an equivalence as structure sheaves). Clearly I should think more about this.

]]>There must be a way, as in def. 2.3.9 of Structured Spaces

But doesn’t the “admissibility structure” there essentially encode the same data about which maps are “open”?

]]>Urs, I wasn’t quite able to convince myself some months back that the regular monos in the category of smooth *Hausdorff* manifolds are precisely the closed embeddings, but I still believe that’s what they’ll probably wind up being.

@Urs 33

since we are working with a coverage on CartSp (good open covers), then perhaps we don’t ask for the pullback $\mathbb{R}^n \times_X \coprod U_i$, but a weak pullback, as in the definition of coverage.

We can restate a map (of topological spaces or manifolds) $X \to Y$ being étale as being existence of covers $\coprod U_i =: U \to X, U\to Y$ making the obvious triangle commute (or better, draw it as a square with an equality on one side). Surely we can state the topos theoretic condition as some sort of existence theorem about covers etc.?

]]>There must be a way, as in def. 2.3.9 of Structured Spaces:

Pass to the little ($\infty$-)topos $\mathcal{X}$ over $X$, ask for an effective epi $\coprod_i U_i \to *$ *in* the topos, and ask that each $\mathcal{X}/U_i$ is equivalent, as a ringed topos, to a representable.

If I understand well then the fact that we ask the covering to be a morphism in the little topos makes it étale, so that’s where the topology is taken care of.

]]>It seems unlikely to me that we’ll be able to avoid equipping the site CartSp with some extra structure, specifying which maps are “open,” if we want to be able to define a notion which reduces to manifolds.

]]>the pullback of the cover along $f$ yields a coproduct of representables)?

This cannot quite work: the intersection of two contractible opens is in general not contractible.

But something that sounds similar is that one could ask that each $U_{i } \times_X U_j$ is either empty representable, which is the beginning of saying that $\{U_i \to X\}$ is a *good* open cover. But that does not seem to get around having to invoke the topology on $X$ to get a manifold.

When schemes are defined as locally representable sheaves on the Zariski site, there is also the explicit condition that the cover is open (as on slide 3 here). That’s why I said in #1

smooth manifolds are equivalently the locally representable sheaves on CartSp (more precisely: the $\mathcal{G} = CartSp$-schemes).

only to trick myself into ignoring the “more precisely”-clause later on ;-)

]]>Regarding local representability: how about asking that there is a cover $\coprod U_i \to X$ which is representable (as in, given any map $f:\mathbb{R}^n \to X$, the pullback of the cover along $f$ yields a coproduct of representables)?

]]>I should maybe stop posting to this here while busy with something else. But: what are the regular monos?

]]>Edit: this is to address the concern about Hausdorffness, not local representability

Something like a condition on the diagonal $X \to X\times X$, I reckon. cf algebraic spaces. Maybe require the diagonal is representable and closed….

]]>Let’s see, I want a way to say this that does not make use of the by-hand verifiction that a concrete sheaf has canonically an underlying topological space. A more abstract way. Hm…

]]>Yeah, I am screwing it up. I guess you are right and I will have to require that $X$ is concrete and the $U_i \to X$ are open embeddings.

]]>Say $x$ does not, then we can find an open neighbourhood for $x$ in $U_i$ that does not intersect $U_i \cap U_j$.

Why ? What if $x$ and $y$ are both in the closure of $U_i \cap U_j$ while not in $U_i\cap U_j$ ? I mean I do not see how this proof works for the usual examples of nonHausdorff manifolds.

]]>How does just assuming that each $U_i\hookrightarrow X$ is monic imply that they are *open* subsets? What if I take two copies of $\mathbb{R}$ and glue them together at a point? Then don’t these two copies of $\mathbb{R}$ satisfy your conditions, but aren’t open subsets of the glued space?

So how do you rule out non-Hausdorff manifolds ?

Let $X$ be as in #23. Let $x,y \in X$ be two points. The pojnt $x$ is in the image of some $U_i$, the point $y$ in the image of some $U_j$. If $U_i \cap U_J$ is empty, they serve as open subsets that separate $x$ and $y$. if not, either both $x$ and $y$ sit in the intersection $U_i \cap U_j$, or at least one does not. Say $x$ does not, then we can find an open neighbourhood for $x$ in $U_i$ that does not intersect $U_i \cap U_j$. So again they are separated. If both points are in the interseciton $U_i \cap U_j$, use that this is an open subsets of $U_i$ and hence is Hausdorff.

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