point-set-topology

[urban zinc]

^

[abstract saffron]

Munkres all the way

[quick bough]

any other honorable mentions?

maybe its good to have more literature

[abstract saffron]

Euhm, I guess yes? There are many other proofs for that fact, it's a standard exercise

[quick bough]

(also do you mean Elements Of Algebraic Topology by munkres?)

[abstract saffron]

Yes, that one

I'm also doing it rn

[quick bough]

oh damn

[abstract saffron]

I tried many, but no other really stuck

[quick bough]

what about hatcher?

also, do you have the book as a pdf or smth? haha

[urban zinc]

I'm reading hatcher rn... it's really confusing

[quick bough]

oh, i see

[urban zinc]

I'm gonna power through it though

it does have nice pictures, I'll give it that

[abstract saffron]

Don't

Just. Don't.

[quick bough]

okay hahah

[abstract saffron]

Yes, I always have pdfs

[quick bough]

you mind sharing? 👁️

[abstract saffron]

For munkres you don't even need to pirate

[quick bough]

oh rlly

[abstract saffron]

It's there on Google

[quick bough]

lemme search

[urban zinc]

Wait how is that not pirating

[plain raven]

It's an equivalent statement.

[hidden crag]

if you're new to alg top Rotman is good imo (freely available online), Hatcher has good treatments of certain topics and some good examples/exercises. I think his treatment of covering stuff is a good supplement to other ressources

[plain raven]

@urban zinc I strongly recommend the book by Spanier, especially the first 3 chapters.

[urban zinc]

I'm in a Hatcher reading group and I dunno if I can handle more than one textbook but I'll give it a look, thanks!

[quick bough]

thank you guys, ill look into those

[plain raven]

After that it basically turns into a comprehensive reference and becomes like, unreadable because of the scope and generality of the results.

But you can still use it for clear understandable proofs if you want a good proof of the result

[hidden crag]

Yeah hatcher refers to spanier several times i think

[quick bough]

😭

[abstract saffron]

I guess it's a matter of taste. Munkres went the chronological way, built everything from simplicial complex before deriving Eilenberg-Steenrod axioms. It goes slowly, and very gently. Singular homology becomes almost immediate because of that, as opposed to Hatcher.

[plain raven]

Actually i have some caveats

[abstract saffron]

That's why I said not Hatcher

[quick bough]

i see i see

[plain raven]

Chapters 1-3 (on introductory alg top, fundamental group, fibrations, simplicial complexes) are pretty readable
Chapter 7 (introduction to homotopy theory) is readable
Chapters 4-6 (homology and cohomology) are an encyclopedia and should only be read if you have somebody telling you exactly what results you actually need to read and understand. Also the important results are listed at the beginning of each section so you could just read those and save a lot of time

[abstract saffron]

Also Munkres focused only on Homology for the majority of the book.

Hatched wrote about almost everything. You have to endure quite a lot before you see some homology.

[coarse night]

spanier has too much content imo

[plain raven]

yes.

a lot of the results are of only technical interest.

[urban zinc]

anyone have any good exposition suggestions on CW complexes?

trying to wrap my head around them

[coarse night]

if you want to get the idea, it's probably not a good first read. Maybe comeback to it later when you know which results you really want to see

[quick bough]

which one would you suggest, ryu

[abstract saffron]

Munkres

[plain raven]

They are in Spanier, in Ch 7.6 i think?
They're also in Strom's Modern classical homotopy theory

[quick bough]

munkres for everything

[urban zinc]

hatcher's chapter zero is so painful

[abstract saffron]

Chapter 4 of Munkres is on Singular and Cellular homology

[quick bough]

lmao, even chapter 0 is painful?

[abstract saffron]

hatcher is painful

I told y'all so, many times

[coarse night]

depends on your taste, if you like hatcher's treatment then hatcher. If you want rigour then Rotman. If details then Spanier. If diff topo and geo viewpoint then Bredon and if you're insane then May

[abstract saffron]

May isn't too bad. At least it's concise

[hidden crag]

It's unreadable

if you're not already familiar with the content

[coarse night]

if you have no idea what those things are then yes. I went back to ch0 few months back and I liked it

(I used to hate hatcher btw)

[abstract saffron]

I hate hatcher with passion

[hidden crag]

It's not worse than may

[abstract saffron]

This is equivalent to saying it's not worse than Lang

[abstract saffron]

If you are not so bad, you might wanna try SGA.

Just to see how others live

[quick bough]

will do, thank you:)

[abstract saffron]

It's the best piece of Alg Topo, wdym

It's commutative, it's computable, it's nice, it's cute

As opposed to that horrible piece of homotopy. It can go fuck itself.

[unreal stratus]

Bruh

Cohomology is kinda cuter etc tho

[tiny ridge]

Hatcher is a good book for people who’ll end up doing geometric topology

[plain raven]

yes

[unreal stratus]

Found chap 0 a bit disorganised lol

[tiny ridge]

It has the right cultural context. The standard of proof and discussion in Hatcher is essentially that of modern geometric topology.

[abstract saffron]

Well, it has a nice little ring on it

Turns out it's quite cool

[plain raven]

it's the worst chapter because it's meant as a handwavy introduction. i mean the book is handwavy as a whole but the introduction is extra handwavy. if you try to read it rigorously it's painful

[tiny ridge]

I like Chapter 0 a bunch, it has all the necessary theorems to make everything precise

[warm quiver]

I like Chapter 0 as a hand-wavey intro, but some of the exercises at the end feel completely unjustified lol

[urban zinc]

imma just watch some videos on cw complexes

[warm quiver]

Like he wants you to practice handwaving also

[tiny ridge]

Do you remember a specific example

[abstract saffron]

Italian school?

It's all fine and good to develop intuition. I study mostly with my intuition these days. But topology is a dangerous land.

[warm quiver]

I remember hating this exercise when I first read through

[urban zinc]

wtf the torus is a 2-cell attached to the figure 8

[tiny ridge]

Hatcher has a high starting bar, I didnt really understand anything before reading Munkres part 2

[tiny ridge]

Hah, yeah, that’s a funny one

[abstract saffron]

Lmaoooo

What is wedge product again? I completely forgot

[urban zinc]

disjoint union, except identify a point

[abstract saffron]

Ahhh

[warm quiver]

He just doesnt construct any explicit homotopies in the chapter, so it's really more like playing with imaginary playdoh

[abstract saffron]

I always think of wedge in tensor algebra.

[hidden crag]

different wedge

[tiny ridge]

Proposition 0.15 basically makes all the arguments he makes precise,

[urban zinc]

how does one visualize adding a 2-cell to a figure 8 to make a torus

is this something y'all can visualize

[abstract saffron]

Yeah

[tiny ridge]

Think very hard about the square picture for a torus

It took me some getting used to, the first time I learnt CW complexes

[warm quiver]

I agree, but it's hard to see that when you are the naive self-studyer

[urban zinc]

okay lemme think more

[tiny ridge]

Oh absolutely, it clicked way later for me, essentially in Chapter 2

[unreal stratus]

Or think of the torus and take two interlocking circles like the big and small one

that is the 1-skeleton of the torus

[abstract saffron]

Was about to say that

[unreal stratus]

But yeah it's still a bit weird visualising

[abstract saffron]

But the final attaching the 2-cell isn't that easy to grasp

[tiny ridge]

That’s cheating, because you are not visualizing how the disk is physically being attached

That’s the most interesting part IMO

[unreal stratus]

Oh sure I mean I was more thinking about giving a decomposition

[tiny ridge]

I gotchu

[unreal stratus]

which i kinda feel is the ssame if you think about the square picture

[abstract saffron]

I mean, I can visualise it in my head how the 2-cell got stretched to cover the tours, but I don't have a blackboard with me

😄

[tiny ridge]

@abstract saffron I have an exercise for you

Do you know what an immersion is

[abstract saffron]

I heard the term before 😄

but keep going

[tiny ridge]

It’s a local embedding, but not necessarily an embedding.

Prove that there’s an immersion from the punctured torus to R^2

Or rather, construct such a thing.

[urban zinc]

Oh I watched a video and now I get the definition of a CW complex at least

That was really straightforward

It looked a lot scarier in Hatcher

[tiny ridge]

Now construct the CW structure on the torus, or more generally on the surface of genus g, in complete gory details.

[abstract saffron]

Might as well go with handles at that point

[tiny ridge]

Quite, handlebody decompositions are exactly like CW structures

[abstract saffron]

If I am not mistaken, an immersion from torus to R^2 is trivial. The punctured point makes things complicated a bit

[tiny ridge]

By trivial you mean there’s none, I think

That’s correct (why?)

[abstract saffron]

But I really never remember immersion, imbedding, embedding, and all of those things

[tiny ridge]

Immersion = local embedding, thats all

[abstract saffron]

Believe it or not, i don't know embedding either 😄

I just know intuitively what it is

[tiny ridge]

You know point-set topology?

[abstract saffron]

I do, but up to basic differentiable manifolds

[tiny ridge]

An embedding is just homeomorphism onto image.

[abstract saffron]

I remember reading about these things in Sternberg, but I keep forgetting them

[tiny ridge]

Don’t need differentiable manifolds for this

Just ol’e Munkres knowledge is sufficient

[abstract saffron]

i never read Munkres' Topology either, if that's what you mean 😄

But ok, gimme a sec, I think my intuition can give something

[tiny ridge]

Oh, where did you learn point set topology then?

[abstract saffron]

We had a course last semester, up to IVT and some small applications, e.g. Larange's multipliers

[tiny ridge]

That’s calculus, not point-set topology, though.

[abstract saffron]

After that I'm taking a course on Diff Geo, and studying Diff Geo, Alg Topo, and Knot theory by myself

[tiny ridge]

You should go through Hatcher’s 50 page notes

[abstract saffron]

Not Hatcher again

[tiny ridge]

Quick and dirty intro to point-set top

Well, Munkres is too long. I wouldnt recommend it if you want to pick up point-set backgrounds for AT

[hidden crag]

||deform and project||

roughly speaking

[tiny ridge]

Most books on point set topology are unnecessarily long IMO

[hidden crag]

is that what you're referring to

[gritty widget]

lukewarm take the only bad thing about hatcher is the typesetting

i cannot read this book. my eyes just glaze over the paragraphs

[tiny ridge]

Seems unclear that your construction gives a local embedding.

Unless I’m misunderstanding

DM me more details?

[quiet thorn]

What is much of the content of point set actually good for? Outside of to learn point set

Maybe that was a stupid question

[tiny ridge]

Not much. Just that we topologists need to write in that language, otherwise the rest of the mathematicians will crucify us. We already get too much flak for being imprecise

[hidden crag]

no i just haven't given it detailed thoughts, maybe it's even wrong

[abstract saffron]

Ohhh, those embedding theorems

[tiny ridge]

Else, I’d start sentences with like “Let X be a playdough, we squish the subdough Y…”

[abstract saffron]

But that only explains why torus has no immersion onto R^2

[warm quiver]

The typesetting might be my favorite part of the book ngl

[hidden crag]

||basically doing the deformation to S^1 v S^1 but leaving it "thickened" s.t. it's still a diffeo and not just a homotopy and "flatten" it that way s.t. we can project it into the plane||

[tiny ridge]

This is exactly correct

[warm quiver]

I also glaze over but I think that's just cause Hatcher loves to talk and talk and keep talking

[abstract saffron]

Wait, this makes little sense. I keep thinking about torus as a differentiable manifold, and the atlast gives an immersion, or am I tripping?

[hidden crag]

you might be

[tiny ridge]

Each individual chart in the atlas gives a local embedding, but these do not patch up to an immersion

You need a map T^2 -> R^2

@hidden crag ||Reinterpretation: the Pochhammer contour is nullhomologous in the twice punctured plane||

[coarse night]

btw is there any stronger result of embed-ability of smooth manifolds inside Rⁿ. I know one constraint can be given using SW class. Is there more restrictive one?

immersability*

One I saw in Lee saying closes manifold of dim n can be immersed in 2n-k where k is the number of 1's in the binary expansion of n

like wtf?

[tiny ridge]

Any characteristic class that detects stable triviality should do the job. I don’t know of stronger invariants per se

[tiny ridge]

lol

[hidden crag]

||reinterpretation: you can hang a picture onto two nails with a string such that it falls off when you remove one||

[tiny ridge]

||Very nice!||

[coarse night]

sorry what?

let me look up the buzz words

[abstract saffron]

Can you give me a hint why?

[novel acorn]

analysis

[abstract saffron]

The very nice structure of atlast should make it possible, although my intuition says no

[novel acorn]

A lot of topology is used in stuff like functional analysis

[tiny ridge]

The charts do not agree on the intersections, only upto a transition function they agree.

[novel acorn]

not to mention that like analysis on compact abelian topological groups is used a lot in Tate's thesis for example

[tiny ridge]

But for the chart maps to glue up you need literal agreement

Coz gluing lemma etc

Tate’s thesis to justify point set topology

bruh

[novel acorn]

bruh give me a break I'm trying 😭

[tiny ridge]

peak algebraist

[novel acorn]

I literally am not even gonna joke abt that lmao

Idk if you want some better use you can always cite some stupid technical lemma in AT

[quiet thorn]

I was more so referring to the endless pile of definitions they make you keep track of in a point set class

[coarse night]

btw how did you give the immersion of T²\pt

[quiet thorn]

Like all the T1234567 spaces

[tiny ridge]

exercise

[abstract saffron]

That was to avoid getting crucified

[novel acorn]

Tbh later on I found these kinda cool (because they're used in cool stuff like topological groups lol (I like topological groups a lot))

[tiny ridge]

i will email subject board to ask you that specifically in the oral exam, @coarse night

[novel acorn]

Tho realistically I only know what T0,1,2,3,3 1/2 and 4 are

[abstract saffron]

My guess is immersing it to S^2. Removing a point is removing a point from S^2, which gives R^2. I don't see another way to work around that stupid point

[coarse night]

I need to worry about CA first

[quiet thorn]

Topological groups

[novel acorn]

completely regular spaces my beloved

[hidden crag]

see my spoilered text for a description

or is that not what you meant?

[tiny ridge]

me and timo leave it as an exercise to you to interpret our cryptic spoiler messages

[hidden crag]

||This one is just to waste your time||

[quiet thorn]

I remember forgetting definitions on my point set final and not being able to do the problems and just sitting there for an hour staring out the window

[hidden crag]

the only point set exam i ever had was oral

that was convenient

[quiet thorn]

I hope I never have to read munkres again

[abstract saffron]

Shit, I can't think of anything. Torus is a bit weird of an object.

TIL I need to study point-set rigorously again

[tiny ridge]

youll figure it out, im confident

need to think about the definitions a little bit, and the rest is visualization tricks — up your alley

[abstract saffron]

The thing is, im my head, I pick a point, then zoom in to find a neighborhood that just locally flat-ish, and immerse that to R^2. This same trick would also show T^2 is immersible to R^2, which is wrong (for a completely stupid reason in my head, one is compact and the other is not).

I'm having hard time rigorously arguing where the fault lies

[hidden crag]

the sentence after "this same trick" is wrong

you said that already

i should read the whole msg before replying

[abstract saffron]

I know, I know nothing is correct in my reasoning 😄

[hidden crag]

you're trying to patch together a global map out of local diffeos

there's no reason that should work

[gritty widget]

whitney and induct

[abstract saffron]

I don't see why it shouldn't either

Transition maps are just too nice to make it fail

[obtuse meteor]

No?

They have to agree on overlap

That’s a quite strong condition

If this worked you could build@anything you wanted to In difftop

And it would be a meaningless subject

[gritty widget]

"and it would be a meaningless subject" is one of my favorite ways to tell someone their argument is wrong lmao

[obtuse meteor]

becoming sheafpilled

[obtuse meteor]

It’s a good way

[abstract saffron]

As I said

[gritty widget]

all symplectic manifolds are globally R^2n sorry ibsen

[hidden crag]

with which form tterra........

[obtuse meteor]

Like

That does it

[hidden crag]

spoiler

[obtuse meteor]

Oop

Timo did you already post that

[hidden crag]

.

[obtuse meteor]

Oopsy

[hidden crag]

yeah i think we mean the same thing

[unreal stratus]

Pain

[obtuse meteor]

Yes yeah

[gritty widget]

the form

you know like, the one

[unreal stratus]

Yeah.

[abstract saffron]

Ok, I guess this will work: I will take a leap of faith and say a punctured torus is just an annulus

[unreal stratus]

So like automorphic forms

[abstract saffron]

Then it all makes sense

[obtuse meteor]

||Once you do the deformation retract like “halfway” you just get two strands vibing in space and you can boop it”||

||is this not an embedding into R^2 though I feel like I’m missing something||

[abstract saffron]

But there are still so many gaps

[unreal stratus]

immersion ting bruv

skrrt

[abstract saffron]

I'm feeling like feather rn 😄

[obtuse meteor]

This is just false

[hidden crag]

Jumped into the mud with that one

[obtuse meteor]

It is an identification space of an Annulus tho which is maybe a useful way to think of it

[unreal stratus]

punctured torus is dangerous though

You don't want that jam leaking out!

[obtuse meteor]

||the flattening step may mess up this as an embedding but I can’t quite see it. Thoughts Timo?||

[unreal stratus]

Or the coffee, as the case may be

Hopefully you punctured the handle

[abstract saffron]

I guess that's what I meant? I saw a picture today about Morse theory and attaching handles

[unreal stratus]

I don't believe in Morse theory lol it's just a theory

[abstract saffron]

That's my last guess actually, I'm out of moves now

[gritty widget]

have you ever even seen a nondegenerate critical point in the wild? they're too good at hiding

[obtuse meteor]

A good idea for this problem is to think of the fundamental polygon

It’s really useful also to like

Have you seen that torus cut a point is homotopy equiv to a wedge of two circles

[abstract saffron]

Is this a Sard joke?

[obtuse meteor]

I don’t think I could do this proof unless I had seen that before

[gritty widget]

it's a joke about the DN category of manifolds

[unreal stratus]

DN you say?

[abstract saffron]

CW complex thing-y? Yes

[obtuse meteor]

Ok but like do you know the visual proof

[tiny ridge]

Ah, but which R^2n? There are exotic symplectic R^2n s!!!

[gritty widget]

you're an exotic symplectic R^2n

[abstract saffron]

Yes. That's how I remember how to construct CW complex of torus 😄

[gritty widget]

a rare and very nice find

[tiny ridge]

||it wont be an embedding, youre right: just an immersion. the point being, the nbhd of the 1 skeleton of the torus isnt an annulus with two holes in R^2, even though they look kinda same||

[obtuse meteor]

||is the essential problem flattening the juncture point like I said? It feels like you’d have to like. Twist both strips of paper in a way that would “shear” and tear something at the center to flatten||

[tiny ridge]

fuck i was beaten to this joke

[obtuse meteor]

||this is very informal but that is my sense of the issue||

[tiny ridge]

||exactly, but you can do this without twisting badly even though thats the natural instinct. draw the joint like a usual cross on the plane, a thickening of the letter X, and do the noninjective rigamarole elsewhere||

[obtuse meteor]

Yeah this fits

Cool

[obtuse meteor]

I wish I had this sort of intuition for higher dimensional objects

[plain raven]

T1234567 is an extremely strong separation axiom.

[abstract saffron]

Well, there's another proof with the square identifying edges. Removing a point leaves only the four edges, which up to identification is wedge of two circles

[tiny ridge]

how high do you want to be

5 or <= 4

[abstract saffron]

I guess that's what you meant?

[obtuse meteor]

This is not true though

That’s true up to homotopy

Which might be what you meant

[obtuse meteor]

Curious you removed 5 here

[tiny ridge]

im bad with inequalities

[obtuse meteor]

Usually if I were going to remove a dimension it would be 3 or 4

[tiny ridge]

you get what i meam

[obtuse meteor]

All :)

[tiny ridge]

more than or equal to 5, or less

more than or equal to 5 is uhm

messy

3 = 2 + 1 but personally i find 4 easier since 4 = 2 * 2

[hidden crag]

most analytically gifted topologist

[tiny ridge]

i draw all 4 dimensional handlebodies just like 2

[abstract saffron]

Ok, I'm out 😄

[tiny ridge]

the attaching maps are framed links and thats the only 3 d intuition you need

Kirby calculus is quite goated for this stuff

[abstract saffron]

I keep mixing different parts of topology

[hidden crag]

cya megumi

[abstract saffron]

Nah, I'm out mean I'm done with this exercise

I'll stay here tho

[tiny ridge]

Go Megumi! But stay here!

[obtuse meteor]

I need to learn so many more things

[coarse night]

what's the H^*(Spec Z[x])

[quiet thorn]

Unfortunately I was unable to remember their definitions

[coarse night]

[abstract saffron]

Me neither

[quiet thorn]

I only remember 0 1 and one of them is hausdorff

[tiny ridge]

too much to read / too little time / soon we all die

[abstract saffron]

T3?

T4 is quite strong, def not Hausdorff

[hidden crag]

T2 is hausdorff

[abstract saffron]

One of them is Kolmogorov. Never remember what is what

[tiny ridge]

T1000 is inseparable. No I mean like literally, they tried cracking it open after dipping it in liquid nitrogen

[hidden crag]

wait until you hear about 1001

[tiny ridge]

soon you have a movie franchise about separation axioms

Let M be a noncompact manifold with trivial tangent bundle, of dimension n

Then M immerses in R^n

[gritty widget]

proof: whitney and induct

[tiny ridge]

Nearly there, but just one thing

[abstract saffron]

me, not knowing what a bundle is: yeah, sure, why not

[tiny ridge]

alternative to “trivial tangent bundle”: there are vector fields X1, …, Xn on M which span every tangent space at every point

n = dim M

[abstract saffron]

Why are y'all surprised 😄 I told y'all I didn't know much

[tiny ridge]

proof: triangulate M, then push the triangles by fingers from infinity until they all deformation retract to their faces, which gives a deformation retract of M to a codimension 1 skeleton, wiggle waggle the skeleton, then undeformation retract it back

by which I mean of course, apply the Smale-Hirsch immersion theorem

wish i could write papers like this

[abstract saffron]

[hidden crag]

I was just surprised because you talked about smooth scructures etc earlier, i didn't mean for that to seem like i'm mocking you

[abstract saffron]

I just spent time on this ? ain't no way I could find that

[tiny ridge]

oh but its easier for punctured torus

[abstract saffron]

I mean, I know a lot of stuff, but in no way systematic. I only know well what I was interested in at some point.

[tiny ridge]

i can assure you that you can 100% do it for punctured torus, with some amount of patience

[abstract saffron]

There are always a lot of holes and gaps like this

[hidden crag]

bro is not contractible

[tiny ridge]

Could be interesting!

[abstract saffron]

The amount of math we had in uni is abysmal :sadge:

[tiny ridge]

theres an indie horror game which came out last year which, when you move your POV around and point at various objects, shows you in huge font sizes on the screen “Could be interesting!”

to get you to like interact with it

ill start irl all-caps that whenever i point to something interesting from now on

[unreal stratus]

Mathematics

[abstract saffron]

Today I learned of a very cool proof that R^4 admits uncountably many smooth structures

But don't test me on that yet, I haven't grasped all the details 😄

[tiny ridge]

isnt that a super hard theorem lol

[gritty widget]

any positive-dimensional topological manifold which admits one smooth structure admits uncountably many

[tiny ridge]

i barely know a tenth of the details

[tiny ridge]

lol

[gritty widget]

:^)

[tiny ridge]

ok, you got me

[gritty widget]

[abstract saffron]

It is. But there's a nice series of argument from knot theory

Basically every topologically but not smoothly slice knot gives rise to such a structure.

[gritty widget]

megumi you may want to distinguish between distinct and nonisomorphic smooth structures

[tiny ridge]

i think that implication involves seiberg witten theory

because why should that be true right

[abstract saffron]

Ugh...

😄 Well

As I said, don't test me on the details

[tiny ridge]

i agree construction of those knots is nice tho

[coarse night]

knot interested

jk

[tiny ridge]

how does it go again? bunch of hooks?

it necessarily has to be a wild knot

[abstract saffron]

Simpler I think

[tiny ridge]

oh im not just saying some parts of that implication maybe harder than others. but again, indont know much, this is a guess

[abstract saffron]

Any knot with trivial Alexander polynomial is topologically slice

[tiny ridge]

Oh wow but is that obvious

[abstract saffron]

No, it's hell deep

😄

[tiny ridge]

I know smoothly slice implies its f(t) f(t)^-1

[abstract saffron]

Showing the non-smooth slice can be done with Rasmussen's s-invariant

[tiny ridge]

Oh damn ok

[abstract saffron]

That's why I said don't test me on the details. This single series of argument by itself can be an honour thesis

[tiny ridge]

learn the deets and explain it to me

[abstract saffron]

Even that is hell deep I think 😄

[tiny ridge]

Much easier than everything else you said

[abstract saffron]

Hmm, let's see if I can figure out a proof

[warm quiver]

strategic rwby reference with the boop

[obtuse meteor]

Yes precisely

[urban zinc]

Eric Working Through AT

[chrome ridge]

Trying to compute the homology groups of this space

is it $H_0(X) = Z, H_1(X) = Z^{n+m}, H_2(X) = Z$ ?

[gentle osprey]

ru0xffian

[abstract saffron]

n+m doesn't feel right to me

Idk, have to check that bit

[chrome ridge]

n+m 1-cells and the cellular boundary map is zero

agrees with homology of Torus when u plugin n = m = 1

[novel acorn]

I believe this to be correct

There's one 0-cell n+m 1-cells and one 2-cell

Then the d_0 map is 0 since there's only one zero cell

And d_1 is 0 since the attaching map for the 2-cell is a^mb^na^-mb^-n which disappears when abelianized

[chrome ridge]

yeah this agrees with my reasoning

[tiny ridge]

Where are you getting these m+n 1-cells from

I don’t understand your cell structure

[novel acorn]

the identification of the circles gives you a wedge of m+n circles

[tiny ridge]

How?

[unreal stratus]

Doesn't that just turn the circle into another circle lol

But changes behaviour around it

[tiny ridge]

Yeah

Its the same homology as the torus

On the meridian youre just doing S^1/(Z_m)

[novel acorn]

ah
I understood it as you identify the points that are 2pi/m away from the basepoint

[urban zinc]

Is there a particular reason why AT textbooks tend to use \overline for the inverse path instead of ^-1

[tiny ridge]

Ahhhh. But then you should get Z^(m+n+2)

[novel acorn]

tho now that I read it again this makes more sense

[unreal stratus]

overline kinda arguably more sense i think like

it's not a true inverse

[urban zinc]

Hmm but it is, modulo path homotopy right?

[unreal stratus]

But yeah I mean i don't think ^-1 is gonna lead to bad habits lol

Indeed

[novel acorn]

me when cos^-1

[urban zinc]

Okay lol just checking ty

[unreal stratus]

Lol

[tiny ridge]

i write inverse

[unreal stratus]

I write 1/γ

[tiny ridge]

goated

[unreal stratus]

Okay this is making me realise I need to do more explicit calculations with (co)homology lol

[urban zinc]

Hey John!

[coral pivot]

i hate using overline bc its so over used!

hey

[unreal stratus]

Usually they are more geometrically understandable spaces than this kek

[coral pivot]

but also inverse is less than ideal here

[abstract saffron]

We should make a Discord Bot, that will give us some simple structure and we can practice computing (co)homology

[coral pivot]

I usually say $[\gamma]^{-1}$, the [] implies its an equivalence class so its implicitly part of the group

[gentle osprey]

JohnDS

[coral pivot]

but ofc this doesnt cover all cases 😭

[unreal stratus]

Lol

[unreal stratus]

Lol wonder what the homotopy groups are

[abstract saffron]

Integration Bee is outdated. Now we have Homology Bee

[coral pivot]

time to use only H_1 so i can call it -\gamma

[urban zinc]

Hmmm I see

This makes sense

[unreal stratus]

Wait actually lol silly question

Uhhhh

Nvm

I am just struggling to visualise that space lol

[novel acorn]

I would unironically watch this

[unreal stratus]

but i imagine you kinda can't

[tiny ridge]

Its gluing a disk to a wedge of circles by a^n b^m a^-n b^-m. Bit of a mess

[coral pivot]

oh yeah the space is kinda wierd

[unreal stratus]

Yeah sure

[tiny ridge]

But thats the simplest I can describe it

[coral pivot]

but its easy to write a cell complex

right

[unreal stratus]

Ibsen just did

lel

[abstract saffron]

You can even go hardcore and write simplicial complex

[tiny ridge]

Who’s running the relative homotopy LES?

[unreal stratus]

Me

[abstract saffron]

LES?

[unreal stratus]

long exact sequence

[abstract saffron]

Bruh

[tiny ridge]

Conjecture, this has contractible universal cover

[unreal stratus]

ye lol

hence pi_i = 0 for i > 1

peng

[abstract saffron]

No shit

[tiny ridge]

seems like easy

itll be a bunch of squares pasted weirdly

lots of flaps

[unreal stratus]

this is kinda an octagon with side identifications right lol

[abstract saffron]

Indeed lots of flaps

[unreal stratus]

So we can probably construct a universal cover by pasting those together

[abstract saffron]

I think more of a huge grid

No?

[unreal stratus]

No I mean the space

Not octagon, sorry

Some big n-gon

not n because n has already been used but ye

[abstract saffron]

Huh? But we only have the torus. Lots of identifications, sure, but still from a torus

I don't see why it would make some n-gon

[coral pivot]

hmm you can get a nm cover that is like

[tiny ridge]

take an m x n grid, take mnZ^2 many of these. paste top and bottom along a 1/m subedge

[coral pivot]

a bunch of torus connected glued together

[tiny ridge]

and left and right along a 1/n subedge

[coral pivot]

and yeah it should be what ibsen describes i think

[abstract saffron]

Yeah, so where does n-gon come from? Or am I missing something 😄

[tiny ridge]

m x n grid = mn-gon essentially

[abstract saffron]

Oh bruh

Well, sure, you can say that 😄 no objections to that

[coral pivot]

yeah the wierd thing is that the edges of this grid will really only be of lenght one (we will identify things beyond that), but R^2 should still be a cover I think

or hmm

yeah it should be

[tiny ridge]

seems right lol

very flaring R^2

[unreal stratus]

can't wait for exams to be over and to do more topology

[tiny ridge]

more tautology

[abstract saffron]

I can't wait to get better at topology

:sadge:

[unreal stratus]

lol just realised i got sent topology resources on a zoom call so i no longer have access to them

hope i can remember / find them lol

[urban zinc]

The fact that path composition is in the opposite order as function composition is tripping me up so much lol

[tiny ridge]

i find the path stuff much easier. a first then b, so ab

to this day i confuse which order functions go in

[urban zinc]

It's a good thing I don't have to think about permutations

Having two different conventions is the worst

[urban zinc]

The world if mathematical notation were logical:

https://tenor.com/view/car-flying-car-coche-automobil-coche-volando-gif-19204586

[pseudo coral]

Same I have to picture it out each time and@make@sure where I’m starting at

[urban zinc]

Wow wait this problem is cool, didn't expect the results of c and d even though the proof is really simple

[tiny ridge]

this is a good one

[urban zinc]

Wow the AT chapters of Munkres Topology are a lot more readable than Hatcher 😭

[chrome ridge]

yeah I had this problem on some pset before; really cool result

[urban zinc]

Seems to go a good bit slower though

Which I guess is the tradeoff

[plain raven]

7c+d is called the Eckmann-Hilton argument

[tiny ridge]

heres a way of doing this which is visually entertaining. i will write \otimes vertically and * horizontally. then, f g is equal to
f e
e g
which is equal to
e f
g e
which is equal to g f

[urban zinc]

Neat! so this is a version of the interchange law

[tiny ridge]

theres actually some geometry behind this interpretation

[novel acorn]

you can also show that the homology ring of a topological group has an algebra structure

[urban zinc]

I guess I'll learn what a homology ring is soon enough

[novel acorn]

(also that the action of G of this ring factors through pi_0(G))

[plain raven]

The most general version of this theorem I know of is:

Let $M$ be a set, and let $\cdot$ and $\otimes$ be binary operations on $M$. Let $e\in M$ such that $e$ is a left and right unit for both $\otimes$ and $\cdot$, so $e\otimes x=x=x\otimes e$ and $e\cdot x = x = x\cdot e$.

Suppose that we have $(a\cdot b)\otimes (c\cdot d)=(a\otimes c)\cdot (b\otimes d)$.

Then $\otimes=\cdot$ and $M$ is an Abelian monoid.

[gentle osprey]

diligentClerk

[plain raven]

In particular, these operations are associative.

[novel acorn]

Unrelated but I just found this

[coral pivot]

group object in grp is abelain group.

[urban zinc]

That's really cool

[plain raven]

A useful application of this is when $\otimes$ is known to be a group, but $\cdot$ is only known to be a monoid. After applying the theorem we know that $\cdot$ has inverses.

[gentle osprey]

diligentClerk

[tiny ridge]

Let Omega(X, x0) denote the space of maps S^1 -> X sending 1 to x0. Equip it with the compact-open topology (or if X is a metric space, put the sup metric). Consider the product defined by taking two such maps f, g and letting fg = f * g. This is not a topological group, because path composition isnt a group operation, but it is an “H space” ie upto homotopy it is a group. Use this to show pi_1 Omega(X, x0) is abelian anyway

[coral pivot]

this follows from eckmon hilton

you can use this to show that since pi1 preserves products, it preserves group objects

so pi1 of top group is abelian!

[plain raven]

oh that's nice.

[tiny ridge]

Incidentally, pi_1 Omega(X, x0) is known as pi_2(X, x0)

Omega is called the loopspace

Higher homotopy groups are defined by iteratively taking loopspaces

Define Omega^k(X, x0) as the k-fold loopspace Omega Omega Omega … Omega X. Reinterpret pi_1 Omega^k(X, x0) as homotopy classes of maps from S^k+1 to X

—end of exercise—

[white oxide]

Doesn't it follow from group objects having an inversion morphism?

[tiny ridge]

G x G -> G needs to be a homomorphism

that basically means G abelian

also inverse yeah

[white oxide]

Yup

[plain raven]

The theorem holds for monoids though.

So it can't entirely be about inverses.

A monoid object in the category of monoids is an Abelian monoid.

[white oxide]

Yeah, but for the purposes of the abstract nonsense proof it works without it, afaict

[plain raven]

What are you saying?

"it works without it"

what works without what

[white oxide]

The argument for showing that the fundamental group of topological groups is abelian using the fact that as a functor pi preserves product? So it maps group objects (which topological groups are) to group group objects, which are abelian.

[coral pivot]

hmm right the extra inverse condition in grp object of group might allow you to circumvent eckmon hilton maybe? idk the proof I know uses eckmon hilton

[plain raven]

Since pi preserves products, it sends group objects to group objects.
That much does not require eckmann hilton.
But then to say that group objects in Grp are Abelian requires Eckmann-Hilton.
You said "Doesn't it follow from group objects having an inversion morphism?"
The argument does require the existence of inverses at all. It is a theorem about monoids, which happens to be true for groups without change.
I don't know what you can get using inverses because "inverse" only makes sense when you have a product and unit.

[coral pivot]

the inverse condition only tells you the two inverses coincide I think?

[uncut fulcrum]

Could someone explain this definition to me? K is a field with nontrivial nonarchimedian norm.

[novel acorn]

I feel like there's a lot of context missing

Also the moment you said nonarhimedean valuation i knew you were talking Abt number theory lol so maybe ask in #advanced-number-theory

[unreal stratus]

also lol are there any good sources for like interesting kinda geometric fundamental group questions

maybe i should just spam some hatcher

[pseudo coral]

Hatcher makes me cringe 😬

[urban zinc]

Oh wow Munkres's proof of lifting to the covering space using the Lebesgue number lemma is nice

[coarse night]

how else do you prove it

[abstract saffron]

Is it the thing with the degree of the covering map?

[unreal stratus]

Hm I don't think the proof I've seen use Lebesgue number lemma lol

Do you mean lifting homotopies

[tiny ridge]

You need to chop up the domain into little pieces so that they map into evenly covered open sets

And then lift

This requires the Lebesgue number lemma

[plain raven]

Oh, that's nice.

I didn't know there was a name for that result

[unreal stratus]

Oh sure

Ye

[urban zinc]

hatcher does it without mentioning the name of the result, it feels handwavy

munkres's proof is like.. I think it's shorter? and also feels more logical

maybe I don't have enough background for hatcher

[tiny ridge]

youll get used to it

no big deal

[coarse night]

If you get the idea of what's happening, you'll be able to come up with the definition

I'll suggest you look up fibrations from Spanier since you've seen lifting

[bitter smelt]

hatcher reads a bit handwavy at times

[abstract saffron]

Just a bit?

[violet summit]

Could someone verify what I have for this is correct?

Let (B) be decomposed into disjoint clopen sets (B_1,B_2).
The sets (B_1\cap A) and (B_2\cap A) are disjoint open sets
that have union (A). Assume (B_1\cap A=A) (since (A) is
connected). This implies (A \subseteq B_1). With respect to
(\ol{A})'s topology, (B_1=C\cap B) for
some closed set (C) in (\ol{A}). Since (C) is a closed
set that contains (A), it must be all of (\ol{A}).
Therefore (B_1=B), and (B) is connected.

[gentle osprey]

pramana

[novel acorn]

looks good

maybe just add in the last line since C is a closed set in A-bar that contains A

this is justified by the fact that A-bar is the smallest closed set containing A

[pseudo coral]

Yep

[willow viper]

Hello how can I prove here that the disjoint union topology is the unique topology?

[muted raptor]

Hi, I hope you're doing well. Could someone verify an exercise I did? I just need to make sure if it's correct

[worldly bobcat]

hey, I have a question. Does anyone know what 'manifold' is in french? I've looked it up online and its probably either 'collecteur' or 'variete'. Does anyone know what the correct french translation is?

[tiny ridge]

they call it a variety i think yeah

[worldly bobcat]

really? thanks.

[unreal stratus]

The spice of life.

[fickle hamlet]

If S is a subset of X^w, consisting of sequences which are eventually 0, what is the closure of S?

I think it should be X^w.

[coarse night]

what's X?

[fickle hamlet]

this is R

given product topology

[novel acorn]

Is X^w a countable product of Rs?

[fickle hamlet]

ya i guess, since they are talkin about sequences

cuz it doesnt say

My thinking is, if x is not in S, then any open set O=O1x...xOnxXxX... containing x will have an element of S s_a=x_a, for a<=n and s_a=0 for a>n, which is contained in O

and this is not x since x is not in S

if x is in S then its in the closure

[stark fog]

using the distance from a point to a set and then using extreme value theorem right?

it is a nice proof btw

[fervent root]

complex general linear group is a 2n^2 manifold since general linear group is n^2 and complex numbers have two components right

[fickle hamlet]

If instead of the product topology, it had the topology induced by the metric p=sup{min{d(x_a,y_a),1}| a in J}, would the closure of S be the set of sequences which converge to 0?

[rancid umbra]

what did you try?

[rancid umbra]

just ask

[coarse night]

sorta yes

[fervent root]

is there a better interpretation

[coarse night]

M_2(C) is 2n² and GL is an open subset

it's the same

[fervent root]

I don’t understand how to apply the concept of a neighborhood. A sequence converges is for every neighborhood U of q there is an N such that q_i in Y for all i >= N, but what neighborhood do i pick? can i pick?

[coarse night]

you can pick any, you'll get a corresponding N depending on your nbd

[fervent root]

so i need to just pick one and then give an N that makes it work

[coarse night]

you have show N can be choose for all nbd U

[fervent root]

in the trivial topological space theirs only 3 choices right

[coarse night]

yes

but empty set is not a nbd of anything

also 3?

[fervent root]

oh Y is included in itself yeah

so just two

[coarse night]

empty set doesn't count as nbd of anything

[fervent root]

so i need to show Y converges to Y? just pick N = 1

[coarse night]

Y converges? Y is not a sequence, it's a set

[fervent root]

so picking a nbd is the same as picking a sequence then?

[coarse night]

sorry?

[fervent root]

i’m lost if you couldn’t tell already

i don’t understand how to apply the definition of a sequence converging

[white oxide]

What topological space are you in?

In general top spaces there is no single choice, you have to "verify the existence of each N for each NBD of the point". This is usually impossible as there are infinitely many such points but in sufficiently nice spaces you can infer it.

[fervent root]

the trivial space

[white oxide]

For example, take a complete metric space and a Cauchy sequence. This sequence has a limit you can verify that it also converges by the abstract topological definition (Why? What characterizes open sets in a metric space?)

The empty set?

[fervent root]

and the space itself yes

[white oxide]

What?

[fervent root]

an open set only has interior points

[white oxide]

Okay, take any random sequence from your trivial topological space and some point y that you want to verify the sequence converges to, alr?

For every single open set that contains y you have to find some N so that x_n with n>N lie in that open set.

Luckily, there aren't that many open sets to check here

[fervent root]

well we determined that the only set to check was Y

since the empty set isn’t a neighborhood of anything

[white oxide]

Yup

So, does there exist some N such that the rest of the sequence lies in Y?

[fervent root]

it just has to be 1 right

[white oxide]

Any number works, aye.

[fervent root]

what is the sequence though that’s what i’m not getting

[white oxide]

So your sequence converges to y. Not only that it converges to every single point in that space.

Any sequence of choice

[fervent root]

well the only points in the space is the whole space

so that’s to be expected i hope

[white oxide]

Points are elements of the set Y

You could also view the set of real numbers with the trivial topology for the exact same argument.

[fervent root]

now that i’m going to have to map open sets to open sets, i think i actually like epsilon delta

[white oxide]

Think of coarser topologies as being really blurry, sequences are unable to distinguish points since everything looks the same. If you had a super fine topology you can see whether or not any sequence actually reaches the limit or not.

For something in-between like the reals with the standard topology you have that sequences approaching infinitely close to some point look like they actually reach it.

[floral bear]

how does a torus have two 1-dimensional holes

it cant be the center and the interior, bc the interior is 2d

[cedar pebble]

here are the two "holes," in red and pink respectively

[languid patrol]

It is in fact the center and the interior tho

[floral bear]

yeah ik those are the basis cycles for the homology group

but like i dont see how those correspond to holes

[languid patrol]

Yes also for the fundamental group

[floral bear]

are "holes" in the homology group not the same as "holes" in the geometry?

[cedar pebble]

I mean "holes" isn't really a precise thing I think until you define it with homology or something similar

[floral bear]

doesn't that kinda defeat the point of homology, which was (initially) supposed to be a formalization of hole-counting?

[cedar pebble]

but like here's another way you could see this

let's remove a single point from the torus, then the punctured torus is homotopy equivalent to a wedge sum of two circles, and this definitely feels more like it has two holes...

I guess this is a little deceptive since this kind of reasoning doesn't really work in general

[floral bear]

so is the interior both 1- and 2-dimensional then?

[floral bear]

also how is a punctured torus homotopy equivalent to a wedge sum of circles (you dont have to answer this its just confusing me)

(idrc if i dont understand that rn but if its easy to explain i'll listen)

[willow viper]

I don't really know how to start showing this.

[lime sable]

https://upload.wikimedia.org/wikipedia/commons/thumb/f/f2/TorusAsSquare.svg/1280px-TorusAsSquare.svg.png

glue together the sides as instructed

you get a torus

puncturing it in the middle, you can imagine widening this hole until it's just the edges left to get a homotopy equivalent shape

[cedar pebble]

in general you can do this sort of thing

[lime sable]

that is just the wedge of two circles

[gritty widget]

S'VS'VS'VS'VS'V

[tiny ridge]

I hate the holes description for homology

A better way to say it is that you need two independent cuts to make the torus into a disk

This is how Betti originally defined Betti numbers, actually

Also interesting: “How many ways can conversation of energy fail on your manifold?”

[nimble portal]

Conversation of energy is my favorite

When different forms of energy talk to each other

Just the best

:hehe:

[tiny ridge]

Oh funny Freudian slip

[rancid umbra]

take any topology on the disjoint union with that property and show it must be equal to the disjoint union topology

[abstract saffron]

Yes, it's a strange notion. I find it more natural for homotopy theory, a hole being a non-contractible cycle.

[unreal stratus]

Or head in the sand axiomatics

[umbral panther]

Do you mean for H1, or something more general?

[tiny ridge]

@umbral panther It might be a stretch, but I sometimes think of integral of a k-form on a k-submanifold as some sort of a work done while something was tracing out the k-submanifold, against the “force field” provided by the k-form.

This is only literal for k = 1

But with that charitable generalization of “work”, b_k = 0 is equivalent to conservation of energy.

[nimble portal]

Um

[umbral panther]

Actually, i like the first step, but balk at the conservation of energy

[nimble portal]

I have this map that sends every point on S^n to upper half space

[abstract saffron]

Yes?

I've been waiting for your questions

[tiny ridge]

Interesting

[nimble portal]

What it does is set the x^(n+1)th coordinate to 0 and then omits the nth coordinate and shifts everything down

So (x^1, ..., x^(n+1)) becomes (x^1, ..., x^(n-1), 0)

[abstract saffron]

Huh, interesting

[nimble portal]

So ig my question is

Is the image open in H^n?

Because technically the image is just the boundary of H^n right?

[abstract saffron]

H^n?

[nimble portal]

Upper half space

[abstract saffron]

Poincare's half space?

Oh, that half space

[abstract saffron]

I don't see why not

Unless I'm missing something

Wait, hmm

[nimble portal]

I'm just unsure because it sends every point on the sphere to the boundary of H^n

[abstract saffron]

It is closed, no?

[nimble portal]

What is?

The image?

[abstract saffron]

The image

Hence can't be open

But wait, it's not in R^n. Gotta check for H^n, hmm

Yeah, I think it's not open

[nimble portal]

I'm trying to just visualize it in R^2

[abstract saffron]

R^2 is shit in this case. Try R^3, it helps better

[nimble portal]

Yeah I don't think it's open either

[abstract saffron]

Basically my intuition tells me the image is the closed disc D^(n-1)

Or at least up to some projection/inclusion/bla bla

One day my intuition will bite my back. But until then, Imma keep abusing it

[nimble portal]

Hmm

I think

Okay let me add a bit more context

So my map is something like u = (x^1, ..., x^n) -> (2x^1, ..., 2x^n, |u|^2 - 1)

So if u is in S^n then |u| = 1 so x^(n+1)th component is 0

THen when we remove the x^nth component and shift everything down, the new x^nth component is zero, so it lies in the boundary

But if u was not in S^n then it's in B^n, so |u| < 1, so the last coordinate won't be 0 (in fact it'll be less than 0)

Ahhh but then it wouldn't be in upper halfspace, it'd be in lower :(

But isn't that basically the same thing? As far as being in half-space in R^n goes

[abstract saffron]

Depends on what you wanna do

But yes, I guess?

[nimble portal]

This is gonna be really ugly I'm just trying to sketch it out

So here's the problem

[abstract saffron]

You can always switch some signs

[nimble portal]

This is some scratch work I've got

I'm trying to get a chart that takes points on S^(n-1) to the boundary of M and points on B^n to the interior of M

And this does do that but the issue is phi(S^(n-1)) isn't open in H^n, so it couldn't be a chart (ignoring any issues with bijectivity, I haven't checked those yet)

[abstract saffron]

Chapter 2 of Milnor's book on Differential Geometry covers this, but we can forget that

[nimble portal]

But if I change it so it's not S^(n-1) and just a set with nonempty intersection with S^(n-1)

Then I think it's A ok?

I'm speaking gibberish but I understand what I'm trying to say :^) let me clean it up a bit and see

(And eys I'm aware phi doesn't map R^n -> R^n)

[abstract saffron]

I guess at least in this case, you can work more geometrically

I will recommend trying to work explicitly for n=3 first. You have geometry in your favour for this problem.

Also, hint: ||have you ever tried to flatten orange's peel?||

[nimble portal]

Silly question but

An open subset of a closed set can contain only points from the interior or points from the interior + points on the boundary (but not the entire boundary), but cannot contain only points from the boundary right?

[abstract saffron]

Yucc... first thing: what is the topology here?

Because we do have quite a few subset topologies, and one set can be open in one but not in another

[nimble portal]

Uh

Subset topology of R^n

Basically I want to be able to say that

Since M is the union of B^n and S^(n-1), a point in U is either in B^n or S^(n-1)

[gritty widget]

set theory...

[nimble portal]

So I'm trying to check if that's even correct LOL

[gritty widget]

there's absolutely nothing topological about what you are worrying about

a point in U is also a point in M, which is the union of...

[abstract saffron]

Well, that's totaulogy

[nimble portal]

Well I'm gonna take the image of U so I need to make sure it's open in M

I'm so not confident in what I know versus what I don't know

Constantly second guessing myself

[abstract saffron]

remember: you also have the notion of closed set in subset topology

[nimble portal]

It's closed iff it's not open

I guess not

😭

[gritty widget]

have you ever used a door before?

[nimble portal]

Maybe once or twice, yeah

Why?

[abstract saffron]

R is both closed and open, remember?

[nimble portal]

It's closed IF it's not open

._.

[abstract saffron]

Should have done point-set topo

[nimble portal]

I know :hehe:

[abstract saffron]

That's not the def, certainly not

[gritty widget]

Q is not open in R but it's definitely not closed

[nimble portal]

In my defense I haven't done math in a couple days...? 😭

Open and closed sets are gonna be the death of me, I always get confused when what's what

Let me take a look

[gritty widget]

closed sets are complements of open sets

[abstract saffron]

It's closed if the complement is open

[nimble portal]

Yes

I remember that

OH

lol

[abstract saffron]

Bruh

[nimble portal]

Okay yes

[abstract saffron]

Back to this

I'm sure you can verify yourself if what you have is closed or open

[nimble portal]

Well I know if it lies entirely in the interior it's open

So I just gotta check if the complement of a set containing part of the interior and part of the boundary is open

[abstract saffron]

... do you know definition of subset topo?

[nimble portal]

Yes?

I'm scared to say yes to anything

[abstract saffron]

Can you quickly recall what it is?

[nimble portal]

Yeah it's the topology generated by the intersection of the subset with the topology of the larger set right?

[abstract saffron]

Yes

So an open set in subset topo O_X is open if it is of the form U n X for some U open.

Guess what? You can play the same game for closed sets

Now I let you choose some smart closed/open sets to suit your need

[nimble portal]

I'm literally negative brain cells

I don't understand 😭 I'm just doubting everything I think I know

[tiny ridge]

What was the original q

Like the cosmogonic egg

Is there a bite sized way of asking it

[nimble portal]

Um

Oh

Wait

I think I answered my question

LOL

[tiny ridge]

Perfect

Glad to be of no help, youre welcome

[abstract saffron]

Did you just take a bath or smth 😄

[nimble portal]

If a subset U of a closed set M is open in M, then either U is disjoint with the boundary of M or it contains some of the boundary of M

[abstract saffron]

Except one edge case: if U = M

You beat me to it 😄

[nimble portal]

But then it wouldn't be open since M is closed

[abstract saffron]

Ahhh

But you see, M is open wrt to M, even if M is closed wrt to R^n

[nimble portal]

And we can't have U = bd(M) since that's closed in M

[abstract saffron]

That's why I asked what topology we're dealing with

[nimble portal]

My brain hurts

[abstract saffron]

There are two ways to see this fact

Ugly drawings and intution

Or going back to definition

[abstract saffron]

That's good. It means you're learning something new.

[nimble portal]

I'm not

I'm just so lost in definitions lmao

[gritty widget]

or you have a headache. drink some water

[nimble portal]

:hehe:

[quiet thorn]

and possibly eat some food

[abstract saffron]

and watch a movie too

[nimble portal]

I think I've answered my question just not really proved it to ymself lol

[quiet thorn]

intuitive understanding is good too!

[abstract saffron]

You know, the God said himself: there are two ways to crack a nut

You use force, crack it hard, very hard, then it might open.

Or you soak it, for a long time, multiple times, then it'll open on its own

[tiny ridge]

alexander tadokoro

[abstract saffron]

Do something else, then come back. Thinking too long makes it less productive

Another God liked to do something else when he did Math. He believed that mind still worked on the problem unconsciously

[tiny ridge]

grothendieck, when he wasnt doing math, was a pick up artist

poincare was sleeping instead

[abstract saffron]

What/who did he pick up?

Girls?

[tiny ridge]

‘It is also the case that the most totally consuming ambition is powerless to make or to demonstrate the simplest mathematical discovery - even as it is powerless ( for example) to “score” ( in the vulgar sense). Whether one is male or female, that which allows one to ‘score’ is not ambition, the desire to shine, to exhibit one’s prowess, sexual in this case. Quite the contrary!’ - actual quote from Grothendieck

refer to Recoltes et Semailles for more pick up advice

[abstract saffron]

I forgot he was also a philosopher

[tiny ridge]

*pick up artist

[fickle hamlet]

Can some1 check this? I think the closure of sequences that are eventually 0 is R^w in the product topology but in the second topology, it should be I think the sequences which converge to 0

[quick bough]

im rlly struggling to understand why this already implies that f x id_K is a quotient map, this should follow from the universal property of the final topology apparently, but i dont quite see it

(this is theorem 2.38)

[untold lily]

well, you are trying to prove (b) for the map (f x id_K) right?

[quick bough]

yeah pretty much

[untold lily]

so you are showing this map is the map that induces the final topology on Y x K, I am assuming that is how the quotient map is defined in your book

or maybe there's some equivalence that has been proven

[quick bough]

quotient maps f : X -> Y are continuous surjections that satisfy U is in Y open if and only if f^{-1}(U) is open

for my case i guess

[untold lily]

has nothing been proven about how quotient maps induce the final topology on Y with respect to the map?

[quick bough]

is our definition of final topology i guess

(in this case, we would have Y standing for Y x K)

[untold lily]

right, and I'm asking if there's no prior discussion in the book about the relation between final topologies and quotient maps

to be clear, it's not hard to figure out on your own, but it does come out of nowhere in the proof

[quick bough]

not rlly tbh

[untold lily]

if you have a quotient map pi : X -> Y this is essentially saying Y has the final topology induced by pi, try to show that

[quick bough]

can you elaborate what you mean by Y having the final topology induced by pi?

so the topology is just defined by the open sets U in Y such that pi^{-1}(U) is open in X?

[untold lily]

imagine Y as just a set

X is a top space

pi : X -> Y is a map

there is the final topology that is induced on Y from this one map pi

[quick bough]

so this?

just to be clear

[untold lily]

and this will be the same topology as the original one

yes

wait

not open sets

any set in Y is defined to be open if its preimage under pi is open

[quick bough]

yeah

isnt that what i said

hm, i find this a bit weird because essentially, the way we defined a quotient map is to be between top spaces

if we consider Y to be just a set and we have a quotient map from X to Y

and induce the final topology this way

it seems like we're going a different way if that makes sense

[untold lily]

yes

in summary

[quick bough]

the proof would literally just be the "definition" of the quotient map

[untold lily]

I'm saying, if you have a quotient map X -> Y where Y is a top space, this is the same as saying Y has the final topology induced by pi

like these two concepts coincide

[quick bough]

ah, okay, i got it

yeah

[quick bough]

okay, wait, so for this

how do we get that f x id_K is a quotient map then

[untold lily]

you are trying to show (b) holds in that earlier thing

[untold lily]

here

[quick bough]

yeah

after ive shown it

how does that imply that f is a quotient map

[untold lily]

that means you will have shown the target space has the final topology induced by f x id

by our earlier discussion, that makes f x id quotient map

[quick bough]

ah, i see, okok, thank you

you're also indirectly using (c) with uniqueness

[untold lily]

yes

[quick bough]

got it, thanks

[untold lily]

is this a book btw? it seems nice

[quick bough]

haha, no, those are lecture notes

i can send those to you if you want

[untold lily]

I would appreciate it

[quick bough]

sent it in dms

[untold lily]

thanks

[fickle hamlet]

can someone help me understand why part b is true?

I'm assuming everything after n is just X for U

[urban zinc]

Give a value of x which has no neighborhood in U, according to the uniform metric

What was your solution for part a?

[fickle hamlet]

For part a, wouldnt every point just work cuz eps<1, so U would be X after some N but in uniform topology, even after that N, it would continue to be smaller than eps

[nimble portal]

How are we supposed to have phi(U) open in H^n but have nonempty intersection with bd(H^n)

I mean I see how

Just struggling to put it into words zz

[urban zinc]

what do you mean X after some N?

[fickle hamlet]

the points (x_N+1-eps,x_N+1+eps) x ... and so on, it wouldnt be eps exactly but it would be smaller than eps

Im assuming that U is X after the N right

or R in this case

[urban zinc]

I believe (U(\mathbf x,\epsilon)) here is meant to be defined to be
[\prod_{i \in \bN} (x_i-\epsilon, x_i+\epsilon)]

[gentle osprey]

boolean_satisfiERIC

[urban zinc]

I don't think it stops at the n

[fickle hamlet]

but why would they show n

arbitrarily

[urban zinc]

Just so you know what the general term looks like

[fickle hamlet]

ok i see

that changes things

its the box topology then

I was thinking product topology

ok ill come back after thinkin about it

[urban zinc]

U doesn't determine the topology

It's just a set

[fickle hamlet]

I know but the point is to compare topologies

[urban zinc]

Yeah it's open in the box topology

[fickle hamlet]

its a set but it would open in the box topology

ya

[urban zinc]

open in H^n means that it's the intersection of an open set of R^n with H^n

It's the subspace topology

[nimble portal]

Would this be open in H^n then?

It's open in R^n

And I believe it's the intersection of R^n with...that set...

[gritty widget]

no it's not open in R^n

[nimble portal]

I just ChatGPT'd whether it was or not fuck 😭

[gritty widget]

[nimble portal]

I never thought this open-closed stuff would be this tuff

[gritty widget]

it's gonna be hard if you're using chatgpt

R^{n-1} x (0, 1) is open in R^n, for example

[quiet thorn]

chatgpt for math is pretty terrible

[gritty widget]

R^{n-1} x (0, 1] is not. no point whose last coordinate is 1 has any neighborhood contained in the set

[nimble portal]

Hm

[urban zinc]

Don't use ChatGPT for this lol, especially if you're unsure about something

If you can BS-check it and you're just using it for a dirty computation for a subject you've already learned, ig that's different