@Danu You here?

@BalarkaSen Yes

Also I was thinking for a sec... what if the charts are not balls lol

What if they're disconnected or something

Remind me, wasn't there a condition on $H_{n-1}$ is my manifold wasn't orientable? Sthing like contains a $\Bbb Z/2$ factor?

I don't know

I didn't study orientation in any detail

Darn.

Thanks though.

Can I always produce an atlas with the same or lower number of charts that are all contractible?

I am rusty on this bit. Now that I am reading smooth orientation from G&P, I am trying to simultaneously prove things in the topological setting.

I should have a look at Hatcher, or try to reconstruct it myself.

@Danu Produce an atlas on what?

@BalarkaSen The manifold

I wanted to bound the number of charts from below

what if the charts are not homeo's to contractible subsets

I mean, if you take the definition of chart to be "a subset homeomorphic to the total manifold", that's 1 chart. I don't get the question.

No

I mean

Maybe I could cover e.g. $T^3$ by 3 charts that are homeo to annuli

not balls

(and in fact I probably can...)

Why not. You want to get a bound on that?

Yeah

I can cover $T^2$ by two charts that are annuli of course

so you have to restrict to balls

I dunno how to do this in a general setting if you replace "charts" by "subsets homeo to some fixed topological space $A$".

Well.

Dude

The cup product structure on $H^*(X, A)$ is relevant then, no?

charts are subsets homeo to some open subset of $\Bbb R^n$ to me

what are they to you?

Just things homeomorphic to $\Bbb R^n$ :) Those two are equivalent anyway, so I don't bother with the subset thing.

What do you mean by those two are equivalent?

Not in the setup of this proposed proposition, they're not, hah

The two definition of manifolds, with those two definition of charts.

Of course not.

So okay

you need some condition on the chart domains

that's a shame

If $A$ and $B$ are subspaces of $X$ which are homeomorphic, $H^*(X, A)$ and $H^*(X, B)$ are homeomorphic, yep? Five lemma.

Yeah

Exactly

So if you want to count # of charts (charts being subsets homeomorphic to $A$), then you want to compute the cup product structure on $H^*(X, A)$.

(by cup product on $H^*(X, A)$, I mean the cup product on $H^*(X)$ restricted to $H^*(X, A)$ of course, as $H^*(X, A)$ is not a ring with that structure)

Yeah, too hard :D

But thanks for the input

Not doing that either. No problem.

@BalarkaSen I mean isomorphic here. Typo.

You're talking to yourself a lot now haha

I think stuff out loud a lot in real life actually. It helps.

@Danu No, I am not really convinced about that thing I said. Take $S^1 \vee S^2$. Pinch the wedged circle to get $S^2$ and pinch the equator in $S^2$ to get $S^1 \vee S^2 \vee S^2$. Clearly they have different homology...

Five lemma may not even apply.

You want the little diagram of the two arrows $H^*(X) \to H^*(A)$ and $H^*(X) \to H^*(B)$ to commute.

@BalarkaSen No idea what you mean by pinch?

@BalarkaSen Five lemma definitely solves this.

Collapse. Quotient.

Half of my alg. top. course was based on this

$H^*(X, A) \cong H^*(X/A)$ if $(X, A)$ is CW pair.

I just gave you a counterexample where this doesn't apply.

It must be wrong :P

I mean it's very easy

Just take the short exact sequence for the two pairs

two out of the three are homeo

pass to long exact sequence, done.

Why does that homeo make things commute?

That's just the five lemma

You want the whole diagram to commute for five lemma to apply, is my point.

@Danu Huh?

Wait, which homeo?

You just have a homeo $A \to B$.

Yeah, and $X\to X$

The identity map.

that's two out of the three in the short exact sequence

Why does the diagram induced by those two homeos make the diagram of long exact sequences commute?

So what is your issue?

Hi @Danu. Glad to see you didn't wilt away from starvation :)

@TedShifrin I ate so much!

@BalarkaSen Something something functoriality

I did stay up for two more hours though (?!)

Yes, which is garbage. It's not true in general, because I gave you a counterexample.

No idea why/how

was reading more about Grothendieck

@BalarkaSen I don't think it's a counterexample---I'm willing to bet a lot on that. But I must admit that I don't feel like going through my notes to find how this exactly works right now

So sorry for not being helpful.

Ted, maybe you can enlighten us?

For the 5 lemma, you need 2 on either side of the one you want?

$H^1(S^2 \vee S^1, x_0 \vee S^1)$ and $H^1(S^2 \vee S^1, C)$ are not isomorphic, where $C$ is the equator of $S^2 \vee x_0$.

@TedShifrin Yesh, which you get in the long exact sequence

If you have 2 out of 3 in the short exact sequence

I just came in in the middle of the punchline.

@Danu Your note proves it for a map of pairs $(X, A) \to (X, B)$.

There's no such map here.

Give me a one-sentence statement of where we are here.

Or don't. I have to work on (yet another) letter of recommendation.

True or not?

I am claiming it's not true ^

counterexample.

Hmm ... I think I have a simple counterexample. Take $A$ and $B$ to be points in the closed disk.

You mean isomorphic cohomology groups, anyhow, not homeomorphic.

Ok, so what's the issue? It fails if there is no map of pairs?

Take $A$ to be an interior point, $B$ to be a boundary point.

@Danu Yes.

Yeah, I see it Ted

@BalarkaSen OK, thanks.

that's good to remember

Glad I could stop the fight. :P

:D

Not a fight, come on

I was not fighting.

:)

Mike corrected me on a really idiotic statement in a comment on a question earlier. I told him privately I was momentarily acting like Balarka :P

Also, what does it me to say "a bundle isomorphism covers the identity of the base manifold"?

@Danu was betting on I being wrong about me being wrong. He owes me money.

(from my notes)

@BalarkaSen hahaha :D

It means it maps fiber over $p$ to fiber over $p$.

A lot of reputation on it!

@TedShifrin D'oh. Why not just say it preserves fibers?

In general some authors allow a bundle isomorphism to cover an arbitrary homeomorphism (diffeomorphism) of the base.

@TedShifrin Eh.

So this notion is just... when you have a map of total space you induce one on base?

And then you say the total map "covers" the base map?

Right, @Danu. In general, fibers must map to fibers. Just not necessarily the same one.

For example, you might look at the induced map on the tangent bundle coming from a diffeomorphism of the base manifold.

Now, who starred that, come on!

Teehee

I hardly ever star anything.

You are a star

It seems there are way too many copies of me over there >>>> ... so stop it.

twinkle twinkle little star, show us where math answers are

I guess I could remove those stars---they're not really "useful" for anyone either way.

Danu, what have you been drinking/smoking?

@Danu Anyhow, so it's not clear how to do your counting of weird charts thing anymore, now that I proved I was wrong.

I was not really offended by that particular star. But removing the real fight stars are good idea.

I concur.

There ya go

Lol Balarka, you starred it now?

Damn, I had to speak, didn't I?

You like showing off your power :)

@TedShifrin LOVE IT

MOAR POWERRRR

@Danu Why would I

It diminishes my already diminished reputation

Unless Alessandro is lurking, Danu must have re-starred it himself.

I didn't.

Here, I'll star it to prove it

:)

Well, I sure as hell didn't.

hahaha

Balarka, nice try

Anyhow, enough childishness.

PS Ted you can't star your own messages

@Danu I didn't. Here, I'll star it.

BTW, @Danu, my little almond genoise cakes are yummy. I just tried one to make sure I wouldn't poison my guests this evening.

I'm lurking but not starring from the shadows

Balarka: You have my advanced problems to be working on.

So, @Alessandro, you still loving discrete math and logic? All the stuff I'm no good at?

@TedShifrin Nom nom nom

@TedShifrin Yup. I finally understood orientations yesterday, as in, worked some problems out.

I do, I'm trying to work my way through Shoenfield's "mathematical logic" now that I don't have lectures

There were two math courses I dropped in college, @Alessandro. One was a course in mathematical logic. I loved the discussion of model theory, the compactness theorem, and non-standard analysis. But when he was starting the fourth week of Turing machines to prove Gödel incompleteness, I quit.

By the way @BalarkaSen the proof still works (for general acyclic, open subsets)

Interesting I still remember that from 1973 or so.

Sure, H(X, U) is isom to H(X) if U is acyclic.

@TedShifrin You quit to miss Goedel incompleteness?

It was boring and painful, yes.

I never regretted it.

Denial right there ;)

To me logic is separate from mathematics. :P

I have never seen a Turing machine proof.

Hehehe

Not the basics we all use, but ...

I'd love a course in mathematical logic, but I can't take one this academic year... in 2018 I should be able to though

So you're not at all interested in philosophical underpinnings, Ted?

In the meantime you can learn some analysis and geometry, @Alessandro :P

what was the other course you quit if you don't mind me asking @tedshifrin?

Truly not, @Danu.

@TedShifrin What about $(\infty, 1)$-categories?

I do have courses in both analysis and geometry starting in September

A very off-the-deep end course in algebraic geometry was the other one, although I audited it to the end and took notes (which I only threw out a year ago).

Awesome, @Alessandro. I'm sure you won't need me :)

@TedShifrin Hmm, interesting.

I really don't care if someone doesn't believe in the Axiom of Choice, either, @Danu. Although ... to the best of my knowledge I've never used it in my research. But when it's come up in courses I've taught, I've only even addressed the issue if the course was a point-set topology course (or the one time one of the graduate students in diff geo challenged me because he didn't believe in it).

What? You never needed bases for vector spaces?!

Uncountably-dimensional ones? Only for qualifying exams :)

Damn

I used AoC in Bott & Tu yesterday.

I like finite-looking objects.

I don't remember where it came up in the manifolds course years ago with that graduate student, but it did. Possible partitions of unity for a noncompact manifold, although still that is just countable stuff ... Hmm ...

Out of interest, where, @0celo?

I wouldn't be so sure about that @Tedshifrin but apart from topology the topics of the geometry course are just names I have no idea about to me at the moment so I can't really judge (introduction to homotopy and fundamental groups?)

So what is $S^\infty$?

I hear it's contractible

@TedShifrin Page 43.

@Danu Yeah.

that's serious topology, @Alessandro, not geometry :) When you learn differential or projective geometry, then it's more geometry.

Contract each $S^{n-1}$ in $S^n$ by sliding to the north pole.

I don't know what $S^\infty$ is

$H^q(U\cup V)\cong \mathrm{ker}(r)\oplus\mathrm{im}(r)$ requires AoC if the vector spaces are not finite.

Hmm, @0celo, since manifolds are second countable, that's an at-most-countable AoC, which most people don't think of as AoC.

well I don't know why the course is called "geometry" then!

@TedShifrin What do manifolds have to do with it? It's on the level of vector spaces.

Oh, @Danu. It's a cell complex obtained from gluing two $n$-cells to each $S^{n-1}$, inductively.

I did some projective geometry this year though

Oh, I looked quickly at the page, @0celo, and there were suitable covers ...

@BalarkaSen Nice

Are the cohomology groups countable?

Well $\Bbb R$ isn't countable.

But you mean countably generated?

...countable dimension, I think.

For compact manifolds, they are even finitely generated.

What about non-compact manifolds, @Balarka?

and I can take a course in differential geometry in the third year, if I want to (which I do)

Certainly fundamental group needn't be finitely generated, but what about homology?

@TedShifrin Take uncountably many torii.

Take connected sum.

That's still a countable vector space, @Balarka.

Oh, uncountably many? That's not a manifold.

Manifolds are second countable :D

Oh, second countable assumption mucks stuff up.

hehehe

Cool, @Alessandro. Well, I hope I'm still around to chat with you about it :)

Need those partitions of unity brah

@TedShifrin I asked a main site question about it, and Asaf said you need AoC.

Asaf set theory propaganda ;D

@0celo, if you have a vector space with uncountable dimension, sure.

There might be some way of circumventing it...

@TedShifrin Do the de Rham groups have countable dimension?

Ted's point is that your vector spaces are at most countable

Oh, why?

manifolds

(I don't know a proof)

I think so, @0celo, but I'm not swearing to it yet.

@Danu Nice one line proof :D

I don't know how to get from the second countability assumption to countable homology.

Guess you have to do Cech stuff.

Take a nice cover (i.e., intersections are all contractible), do the nerve theorem. Something.

We can't puncture the plane at uncountably many points and have a manifold, right?

That's because the plane is second-countable.

@BalarkaSen Can you prove that every manifold has a nice cover?

How about instead of intersections contractible, intersections diffeomorphic to $\Bbb R^n$?

@0celo7 Not off the top of my head, but Hatcher's appendix has a proof.

I haven't thought about these subtle set topological points.

I hope that too, September 2018 isn't that far away! @Tedshifrin

It is to me, @Alessandro ... :D I'm old :D

@BalarkaSen By making it "diffeomorphic to $\Bbb R^n$" it becomes a differential topology question

But I think the point is that by second-countability we can reduce everything to countable open covers, so I believe at most countable dimensional homology/cohomology. I need to ponder later.

For smooth manifolds, I think that's easier because they are triangulable.