point-set-topology

[novel acorn]

Ok so can you find the final i sent you i nthe to other chat

[odd flame]

actually wait i have an example

[novel acorn]

And look at P2

[odd flame]

i know it's S2 v S1 v S1

[novel acorn]

Oh god

[novel acorn]

Cool so you already know the homotopy

Look at my problem cuz it's more complicated

Sent it for ease

P2

You have a 3-cell glued along the identified faces as can be seen

(also don't look at my answers lol I think I messed it up cuz I was rushing)

[odd flame]

so one 3 cell

three 2 cells

four 1 cells...?

and idk how many 0

[novel acorn]

yeah something like that

but let's focus on the 3-cell now

the 2-cell and 1-cell are the same thing

for the 3-cell

[odd flame]

like same process right

[novel acorn]

yea
just be careful when quotienting and you'll be good

[odd flame]

[novel acorn]

Now for the 3-cell

we will compute the local degree lol

[odd flame]

not gonna work it all through just wanna know the 3 cell stuff

degree?

i only know degree of like antipodal map lol

[novel acorn]

notice that the 3-cell is attached to the same 2-cell twice

but once it's rotated + mirrored

[odd flame]

i see

[novel acorn]

now rotation doesn't affect the orientation

but mirroring does

[odd flame]

i was gonna say, rotation is the same as the 2 cells to 1 cells right

[novel acorn]

(you can see this if you draw like a loop going one way in the 2-cell to find the orientation)

So what you have

is that your orientation is shifted on the other attaching to the same 2-cell

so when computing the degree

remeber that the degree is just the sum of local degrees

so here your local degree is like 1 and -1

Where the - minues from the orientation reversal

so the total degree is 0

and thus the map is 0

(Here we refer back to that original definition of a cellular boundary map that's horrible but needed)

Also seb isn't it like 6am for you

[odd flame]

,ti

[gentle osprey]

The current time for stμ₂dying is 06:56 AM (EDT) on Wed, 10/05/2023.

[odd flame]

exam at 9

[novel acorn]

have you slept at all???

[odd flame]

i went to bet at like 10 and woke up at 4 yeah

[novel acorn]

lol ok then

[odd flame]

better than staying up late and not remembering anything

it's my first time trying this actually, i usually stay up late but i like this more

but im fucking hungry

[novel acorn]

lol yeah

[arctic socket]

Hello! I have a problem trying to prove this rigorously. Let $X$ be a set and $\mathscr{B} \subseteq 2^X$. Define $\mathscr{B} = {{x} | x \in X}$. Show that the topology generated by $\mathscr{B}$ is equal to $2^X$. The first step that I did was apply the theorem that the topology generated by a basis is just the collection of all arbitrary unions of elements of the basis set. Intuitively, the collection of arbitrary unions of singletons is just the power set but I am having trouble putting this thought rigorously. Can someone help?

[gentle osprey]

kolybelnaya

[novel acorn]

I mean you can do like inclusions

to show equality

[odd flame]

irony

wanna do a fund. group/homology calculation with me super quick

[novel acorn]

sure

[odd flame]

i was considering a few spaces

T # T

T \ a point

S^2 v S^1

[novel acorn]

sure

[odd flame]

which one are you most comfortable with

i just wanna practice MV and SVKT

[novel acorn]

Okay uhhh

I feel like for the latter two you don't have to use either of those lol

for the first one sure

if you want to I can give you some more interesting MV/SvT problems

[odd flame]

i mean i just wanna get the process clear in my head before the exam lol

we can do T # T

or P #P

@novel acorn

[novel acorn]

Mmmm

[odd flame]

can we do P # P if that's good for SVT

[novel acorn]

yeah let's to the T#T

[odd flame]

ok sure

[novel acorn]

idk it doesn't matter really but P#P is uh weird lol so I feel more comfy doing T#T

[novel acorn]

Yeah so take A = T^2\D^2 and B is the same thing

[odd flame]

minus a disc?

[novel acorn]

yea

[odd flame]

i had seen a different way

[novel acorn]

I mean that's how T#T is defined

[odd flame]

oh but i guess they intersect at a disk

[novel acorn]

you cut out a disk

and glue them as the cutout

[novel acorn]

they intersect at a circle
Since it's not filled in

OK so you have that A and B are homotopy equivalent to S^1VS^1

and the intersection is homotopy equivalent to a circle

(Rigorously you have to add a bit more each part to make it open iirc)

[odd flame]

i know fundamental group of that wedge is Z * Z

[novel acorn]

first
mental check
Is the intersection path connected

[odd flame]

yeah

[novel acorn]

good

that important for SvK so don't forget to check that

[odd flame]

oui oui baguette

[novel acorn]

Cool OK so

now using SvK what do we get as our fundamental group
(don't calculate it explicitly just give me a quotient)

[odd flame]

it's this free prod itself quotient the commutator?

free product if funky

[novel acorn]

yea

[odd flame]

[novel acorn]

OK so that's like

<a1,b1,a2,b2>

quotient something

[odd flame]

i know the answer lol i was more worried about like

process

[novel acorn]

yeah lol
This is how I'd do it haha

[odd flame]

but i think coming up with U and V on the spot is up to god

for an arbitrary space

in any case then

[novel acorn]

it's usually pretty easy to see

at least idk

[odd flame]

mayer vietoris

[novel acorn]

that was the easy part for me haha

[novel acorn]

Yeah MV uh I'll let you try now that I went over this
(if you remember how MV goes haha)

[odd flame]

i need to remember first lol

it's basically the same homology is just weird lol

[novel acorn]

I think for this case MV is pretty OK
The examples we had were

[odd flame]

yeah like i said ive gotten by just kinda doing the homeworkss

they take forever but it's gotten me by

plus undergrad curve

[odd flame]

actually that notation

it's come up once

H(X ; Z)

[novel acorn]

It means H with coefficients in Z

[odd flame]

this problem fucked me up lol

but i understand it better now

actually no i dot

this needs tensor stuff right

[novel acorn]

no not really

it can but that would be a hassle

[odd flame]

this was the solution he gave us

[novel acorn]

ignore the tensor part

for you that means

"take the coefficients mod bla"

[odd flame]

oh okay lol

and then just same process as we did for the other cells/chains right

[novel acorn]

yeah
Just always remember that it'd mod 4 or whatever

[novel acorn]

also this is a case where you really have to be careful with the quotients cuz they're p non trivial

[lean knot]

G is a topological group, U, V, W are open sets such that U·V⊆W, then do we have V⊆W?

[novel acorn]

I don't believe that is true generally

Take the group to be R\{0} with multiplication

U = V = (-1,0)

Now W = (0,1)

but neither U nor V is a subset of W

[lean knot]

Thanks!

[odd flame]

ohhhhh shit shit

what are common examples of covering spaces

[hidden crag]

R -> S^1

covering spaces of wedge sums of Circles

[odd flame]

S^n -> P^n just by antipodal identification right

[hidden crag]

yes

[odd flame]

unrelated but

how can we prove that any map S^n -> S^1 is htpic to a constant map

this has to do with the fact that any n dim sphere can't retract to it's boundary right

[languid patrol]

You can compute this using the universal cover.

[hidden crag]

S^n is simply connected

and the map lifts to the universal cover

which is R

[odd flame]

oh simply connected => trivial pi

[languid patrol]

The same proof that shows that paths lift uniquely to the universal cover shows that maps from simply connected spaces lift uniquely up to homotopy once you fix a basepoint.

So if X is simply connected then Hom(X, S^1) \cong Hom(X, R) \cong Hom(X, *) up to homotopy

[languid patrol]

Yes but I think you might be misunderstanding timo's argument

[odd flame]

hmmm

[hidden crag]

how?

[languid patrol]

Well simply connected => trivial \pi doesn't really seem like a full explanation of what's going on

[hidden crag]

fair but it's relevant to my argument so let him cook for now

[languid patrol]

But I don't know, Sebb do you understand the fact that maps from simply connected spaces to a space X lift to the universal cover?

[odd flame]

https://tenor.com/view/woody-woody-toy-story-sora-kh3-kh-gif-27223516

except im done cooking lol

final in 20 min

just quick reminders rn

[hidden crag]

we've been over this fact

[languid patrol]

Okay okay

[hidden crag]

when a lift exists

you can do it

[odd flame]

https://tenor.com/view/gh-gif-22201979

that exam is stuck in there with me

[languid patrol]

You got this

[odd flame]

someone's asshole is getting widened today

and it wont be mine

[coarse night]

[quiet thorn]

Wtf did I just read

[languid patrol]

he's charging up

[odd flame]

that' the realest shit you'll read all day

[quiet thorn]

[languid patrol]

And apparently about to try to sodomize a math test

[odd flame]

try to?

[languid patrol]

err succeed at

[lean knot]

Is continuous preimage of a meagre set also meagre?

[odd flame]

https://tenor.com/view/anime-goku-dbz-dragon-ball-super-saiyan-gif-8246706

see yall later

thanks for all help everyone

[hidden crag]

good luck

[quiet thorn]

Good luck on sodomizing your math exam(?) I guess

[surreal hornet]

How does the proof go for showing that R^n and R^m are homeomorphic iff n = m using homology? Is it something on the lines that H_p(R^n \ {p}) = H_p(R^m \ {p}) and since we know the homology for the sphere we get that n = m?

[coarse night]

your exam over?

[coarse night]

yes

too many p's tho

[lean knot]

τ and σ are two topologies on the same space X, A is a subset of X, A is nowhere dense in τ, then is A nowhere dense in σ?

[languid patrol]

Are you sure you want absolutely no relationship between \tau and \sigma? Because then the answer is "definitely not"

[lean knot]

σ is finer than τ

[languid patrol]

The answer is no, no nonempty set is nowhere dense in the discrete topology.

[lean knot]

X and Y are topological spaces, f is a continuous function from X to Y, A is a meagre set in Y, then is its preimage under f also meagre in X?

[languid patrol]

You should try to figure out the answer given what we've already done

🙂

[abstract saffron]

It's not a hard question, no? No offensive

But unless I'm mistaken, you can follow the def and should find it ok

[lean knot]

I get it now

Sorry, I'm trying to prove an equality in a paper, which isn't given a proof. After trying and trying, my brain.exe malfunctioned😭

[odd flame]

SODOMIZED

https://media.discordapp.net/attachments/738548949488762954/818299625182199878/image0.gif

[nimble portal]

Yeah you fuck that ass Sebbb!

[gaunt linden]

Could you guys find a more family-friendly metaphor here, please?

[coarse night]

how did your exam go?

[surreal hornet]

How can I prove that there does not exist a continuous map f : S^{2n} -> S^{2n} such that f(x) is orthogonal to x for all x in S^{2n} and n >= 1? I think that it's a special case of the hairy ball theorem?

[winged viper]

Yeah I think that’s just the hairy ball theorem since if it’s orthogonal to x it can be identified with a tangent vector at x

[languid patrol]

It’s not “exactly” the hairy ball theorem unless you know that the unit tangent bundle admits a section if and only if the tangent bundle has a non vanishing section. One direction is totally obvious but it is not the helpful one.

The other direction is pretty trivial but in a first manifolds class perhaps it would be good to show it

[unreal stratus]

ye this is a special case

To see equivalence with hairy ball is basically gram schmidt go brr

[lean knot]

G is a topological group, W is an open subset, A is a subset of W, A is nonmeager. How can I prove that there are two open sets U and V , s.t. U · V ⊆ W and there are nonmeagrely many g in V such that there are comeagrely many h in U, such that hg ∈ A?

[novel acorn]

So you said that you're trying to prove some equality from some paper
Can you maybe post what you're looking at?

[surreal hornet]

could someone help me understand the last part of the proof here? http://www.homepages.ucl.ac.uk/~ucahjde/tg/html/pi1-01.html

I don't understand this \gamma_0 and \gamma_R having different winding numbers implying n = 0

[thorny agate]

Currently self studying topology using chapter 1 of Bredon

It's nice so far but sorely lacking examples

anyone have recommendations of some sort of resource with a bunch of examples? doesn't have to be a text, online notes or whatever is fine

Extra bonus points if it has examples & counter examples for stuff

[nocturne basalt]

theres a homotopy between them and since winding number is a homotopy invariant \gamma_R's winding number, n, is the same as \gamma_0's winding number, 0

[round dust]

I'm working through this section in hatcher, and was just wondering why the order of pi tau is flipped to tau pi in the last line, after moving to homology

[gritty widget]

you are passing to cohomology

[round dust]

ah right

for the homology case though?

would it still be flipped or no

ah indeed it would not be flipped it seems

[surreal hornet]

I'm trying to show if H_0(X, A) = 0, then A meets every path-component of X. By the definition H_0(X, A) = ker(∂_0)/im(∂_1) = C_0(X,A)/im(∂_1) = 0 so i think this says that every 0-cycle is a boundary, but I don't know why this means that A meets every path-component?

[nocturne basalt]

homology is a covariant functor, cohomology is a contravariant one

[radiant cedar]

what does it mean to map the northern and southern hemispheres to the southern hemisphere via f?

[gritty widget]

each hemisphere of S^n is homeomorphic to D^n so you can consider f as a map on them

[abstract saffron]

Just read The Rising Sea

Holy hell, Chapter 2 is Sheaves

Is this normal?

[unreal stratus]

wdym

[abstract saffron]

I haven't seen any textbook on Alg Topo with Sheaves before

[unreal stratus]

The Rising Sea isn't an alg topo textbook tho

It's alg geo lol

[abstract saffron]

So it's the worst of both

[unreal stratus]

What lol

[gritty widget]

[coarse night]

[unreal stratus]

Lol do you have a question

[weak vale]

I will come up with one, give me a moment

[abstract saffron]

The Rising Sea

[coarse night]

Why in AT

[bitter smelt]

Yeah putting aside the "why are you AG posting in TAT," there is the greater question or why are you using a version of the book which has been outdated since 2022

[coarse night]

It’s the calm sea now

[abstract saffron]

It's the pdf you would find on Google

I know because I just downloaded that exact version

[bitter smelt]

Reading the first few links of Google search result? Nah, click first link no questions asked, never look back

[abstract saffron]

Well, a book is a book. If I search Herstein, I will find 1st ed, and complain why it's so confusing, but to me initially, 1st and 2nd are the same anyway. Not everyone would know if there's any benefits in finding the most recent version.

[weak vale]

Where is the Alg top in Alg geo? I am not sure how to connect them

[abstract saffron]

Scroll down the Rising sea, and you'll see a bunch of cohomology

[grizzled ibex]

Any good class playlists to alg top?

Im a guy who likes to use both books + classes

@abstract saffron any idea?

[grizzled ibex]

that's something category theory can answer you quickly

but i'll left it for you as an exercise

[novel acorn]

https://www.youtube.com/watch?v=XxFGokyYo6g&list=PLpRLWqLFLVTCL15U6N3o35g4uhMSBVA2b
Also why ping a random person for this question??

[radiant cedar]

can u elaborate this? i know that they are homeomorphic but what do you mean by considering f as a map on them

[languid patrol]

N = northern hemisphere, S = southern

f: D^n \to D^n

define t: N \to S to be the obvious bijection coming from reflection accross the z axis

then g: S^n \to S^n is defined by g(n) = f \circ t if n \in N and g(s) = f(s) if s \in S

this is well defined on the overlap

[radiant cedar]

so t is just swapping the last coordinate x_{n+1} to -x_{n+1}?

[languid patrol]

yep

[radiant cedar]

i'm lagging, how is this a map to S^n when f(t(x)) could be on the interior of the equator disc D^n which isn't on S^n?

[languid patrol]

I'm considering f as a map f: S \to S

[languid patrol]

^

[radiant cedar]

so its essentially just pushing the disc to lay over the southern hemisphere and identifying D^n with S therefore

[languid patrol]

Yeah the specifics of how you do it are not so important, but that's the easiest way

[radiant cedar]

a bit much packed into one word "via". soo g is not surjective and has therefore degree zero, do maps with degree zero neccessarily have fixed points?

[languid patrol]

mathematical maturity will require you to pack more and more information into things like that as time goes by

[languid patrol]

Yes, do you see why? hint: try to construct a homotopy to the identity assuming the map doesn't have a fixed point.

[gritty widget]

tteg did it wonderfully for me

[abstract saffron]

Munkres

Also, why me? I know the least amongst all the people here

[grizzled ibex]

you are active...

[novel acorn]

yeah but that doesn't give you an excuse to randomly ping them lol

[grizzled ibex]

he's gentle

and educated

and i feel good comunicating with him

need any other reasons?

excuse != reason btw

[abstract saffron]

me? educated?

I'm probably the least qualified person

[grizzled ibex]

with me at least..

with educated i mean respectful

😭

[grizzled ibex]

well, thank you for the link anyway

[abstract saffron]

It costs nothing to be a good person 🤷

[abstract saffron]

I won't mind tbh, for the record. But others might not like it so much, idk

[grizzled ibex]

i was going to send a gif

#chill here

[radiant cedar]

if it has no fixed point, then at least we have a homotopy to the antipodal map that has degree (-1)^{n+1}

[languid patrol]

great!

[abstract saffron]

Has anyone seen feather these days?

#point-set-topology feels so strange without those questions

[languid patrol]

i think she's probably busy with exams

[viscid nebula]

How can I prove that a set U is open if and only if U is equal to its interior? Is it because the interior is the union of the open sets in Y and (idk if this is true) the union of open sets has to be open?

[abstract saffron]

Hint: Interior of U is the biggest open set contained within U

[viscid nebula]

I can‘t understand why U can only be open if it is equal to its biggest open set, can’t it additionally have non open sets and still be open??? (Sorry i am very new to this topic)

[abstract saffron]

Ok, if U is open, then U equals to its interior, no problem on that?

[viscid nebula]

I get that if U is equal to its interior then it has to be open, but not the other way around

[viscid nebula]

So this is what i dont understand

[abstract saffron]

Let's say the interior is X. It's the biggest open set that is contained in U, so in particular, if there's another open set Y contained in U, then Y is in X.

The thing is, U is open and is contained in U, so X must also contain U

But U contains X 😄 so they are equal

[viscid nebula]

Ohhhhh

Thank you 😭

[viscid nebula]

Is there a book for topology that you guys would recommend?

[abstract saffron]

Munkres?

He's like, a God

[viscid nebula]

Nice time to pirate

[graceful oriole]

Here is an exercise about the closure properties of sequentially compact subsets under union and intersection in metric spaces. My question is: for (b), do we really need the condition that every set in the family is a subset of some sequentially compact subset K of X? Does it hold for arbitrary intersection (of some non-empty family) of seq. compact subsets? Every sequence in the intersection is also a sequence in each subset, and since each subset is seq. compact, obviously, the sequence has a cluster point in each one of the subsets, and therefore in the intersection

[nimble portal]

Though it is p boring

[nimble portal]

Yesterday was last exam just kinda chillin rn

[fickle hamlet]

Let X and Y be topological spaces; let p : X → Y be a surjective map.
The map p is said to be a quotient map provided a subset U of Y is open in Y if and
only if p−1(U) is open in X. Let X and Y be topological spaces; let f : X → Y be a bijection. If both the function f
and the inverse function are continuous, then f is called a homeomorphism. If the quotient map is also injective then is it a homeomorphism?

[plain raven]

Yes

[urban zinc]

Any finite product of Hausdorff spaces is Hausdorff
Only one of the spaces has to be Hausdorff, right?

Oh wait

I'm dumb nvm lol

Ignore that

[civic verge]

I am afraid of Munkres' book

Next semester topology :''3

[rancid umbra]

munkres imo is a pretty fun read and good intro to topology

[coarse night]

but a little dry

[quiet thorn]

dont tell darq

[quasi steppe]

I liked the book by John McCleary

[coarse night]

Topology and modern analysis by Simmons

[novel acorn]

man for a second there I thought you were talking abt his spectral sequence book lmao

[quasi steppe]

Lol

Never read it

Should I?

[solemn oar]

The spectral sequence book?
If you're gonna be computing stuff it will be helpful to read it at some point.

[gritty widget]

Hey guys. Not sure if this is the right channel, but does somebody know a book or some source where i can find out more about the space of σ-finite measures on IR^d, equipped with the vague topology? Like some properties about it, separable, complete etc

[weak vale]

Do any of you happen to know what a
2-dimensional cell complex consisting of 29 vertices (0-cells) where three or more edges meet at each vertex so that the 51 total 1-cells which are the interiors of the edges connecting these vertices give us 23 total 2-skeletons by removing the 1-skeletons from the entire space, with Euler characteristic 29-51+23=1, can be classified as

^ the same as me asking is the above description a solid definition of the House with Two Rooms (an actual question)

[coarse night]

[nimble portal]

How are these for exercises on Chapter 1 of Lee ISM?

Tried to cover the topics in the chapter but still have things I find interesting to prove

H3 seems a little unrelated to the contents but it looks cool

[coarse night]

have you seen Lie groups? If not then come back to it later otherwise it'll be some ad-hoc thing which you'll forget or won't get the satisfaction

[nimble portal]

I have not, I just chose it because I know they come up later in diffgeo and I wanted to play with them now

[coarse night]

these aren't really that "enlightening" exercise

[nimble portal]

What should I replace it with?

[nimble portal]

All of them or the one on Lie groups specifically?

[coarse night]

I mean showing some set satisfies some properties is never fun

the Lie group ones

other ones are good for practice

[nimble portal]

What should I replace it with?

Maybe I will get shot for this but I don

[coarse night]

1.1 is a counter example to keep in mind

[nimble portal]

t really care about whether something is enlightening or not, just whether it's an interesting result or plays with interesting objects

[coarse night]

okay then you can try to show they are Lie groups

It's on you but apart from that, there won't be any Lie group till chapter 6/7

there might be idk

[nimble portal]

Nah nah idm taking that out I just meant for like what to replace those problems with

[coarse night]

for me, I don't find them to be a good exercise, I never solved them myself

but good to have an idea on how it's done

[nimble portal]

It doesn't seem related to manifolds, I just wanted to do it because I saw "Lie groups" LOL, I can wait until they're actually relevant :3

[coarse night]

No they are integral part of manifolds depending on what you are doing

[nimble portal]

I know

This problem specifically*

[coarse night]

group structure on any space (or category) forces a lot of symmetry

for example S¹ ∨ S¹ is not a topological group, even though it's a topological space. Since points are not the same everywhere

[unreal stratus]

nonabelian fundamental group 🤓

[coarse night]

that's one way

[sand dust]

Helloo

i have a introduction to topology subject this semester

[sand dust]

haha thank you

[nimble portal]

Good luck :)

[sand dust]

was wondering if its related to abstract algerba

[nimble portal]

Absolutely

Look at the title of the channel

Algebraic topology :3

[novel acorn]

ok well let's step back a second lol

[nimble portal]

We can look at topological spaces through their topological invariants, like fundamental groups

[novel acorn]

point set (the stuff covered in intro topology)
Isn't really too connected to algebra
(It can be, just look at topological groups)
But in your course you probably won't see that much algebra
it'll be more similar to analysis if anything

[nimble portal]

Ya good point

[novel acorn]

Realistically at some point you might start doing algebraic topology
And you'll see connections to algebra there
But it isn't immediate

[sand dust]

idk tbh xd
am not so good with abs algebra but i still studied last year so maybe it could help

thank u both !
wish me good luck:')

[nimble portal]

You’ll be okay, don’t stress <3

[abstract saffron]

Ironically I'm doing Munkres

It's like, Hilton and Wylie, but better

My roommate shares my hatred toward Hatcher and Lee. Gotta thank him for the recommendation.

[nimble portal]

Also wait this might be a stupid question but

[gritty widget]

the answer is 2

[coarse night]

No 7

[unreal stratus]

Solve y^3 = x^2 + 118

Just had to do this lol

[nimble portal]

So on manifolds we have this notion of two different smooth atlases defining the same smooth structure

[unreal stratus]

Or at least, find a solution

[coarse night]

,w solve y^3=x^2+118

[gentle osprey]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Ryuzaki#5323

[coarse night]

Great solution

[unreal stratus]

Imma need to learn the cubes and square fr

Anyway ||7^3 = 15^2 + 118||

Wai this is topology

[coarse night]

Yes, more accurately they define the same smooth structure if their charts are inter compatible

[nimble portal]

Yes

That's what I was about to say, we can say they define the same smooth structure if their charts are smoothly compatible

Then we define the smooth structure generated by an atlas as a maximal smooth atlas

But you could also do it through an equivalence relation (A ~ A' if A & A are smoothly compatible)

Could you...quotient...by that equivalence?

[abstract saffron]

Try "Mordell's equation". For all the small cases, the solutions are known. And Mordell proved there can be only finitely many solutions.

[cosmic socket]

why are S^{n-1} -fiber bundles classified by homotopy classes of maps in BGL_1(S^{n-1})?

[coarse night]

What’s BGL?

[nimble portal]

Not sure whatever the use of that would be (or if it even makes sense), it just kinda came to mind

[coarse night]

what are you trying to do here?

[nimble portal]

Nothing

[abstract saffron]

I guess you can define an algebraic structure?

[nimble portal]

I only learned about quotient spaces recently so now when I hear equivalence relation I just think about quotienting under it

[abstract saffron]

Yes, was about to say the same thing

[nimble portal]

So I was just wondering if quotienting by atlases was a thing

[coarse night]

sorry algebraic structure on the set of all smooth structures?

[nimble portal]

Me or Megu?

Mmmm wait I think that doesn't make sense actually

Don't you have to quotient by an equivalence relation on elements in your space

An atlas isn't something in a manifold, it's something attached to a manifold

[coarse night]

oh you meant classifying space

[abstract saffron]

"Let us consider the category of smooth structures..."

[coarse night]

lmao that's the definition of BGL_1

[abstract saffron]

I hear Lang lurking around

[gritty widget]

an atlas is still something in a set and your set has the equivalence relation of "two atlases are equivalent iff..."

"set of atlases for the topological manifold M" or something

[nimble portal]

🤔

[cosmic socket]

Yes classifying space of GL_1(S^{n-1}). I think you can give this space the structure of a topological group

[coarse night]

can you make a moduli space out of it?

[nimble portal]

So we can send every atlas to its equivalence class, all the equivalence classes would be the different unique smooth structures on a manifold

That's what the quotient space would be yeah?

[abstract saffron]

That's the quotient, yes

"Quotient space" has a different meaning

[coarse night]

I mean that is the definition.... what definition of BG are you using?

[nimble portal]

What's the difference?

[abstract saffron]

Quotient space means you have a topological structure

I thought you read Bredon?

[gritty widget]

or algebraic, in particular quotient vector spaces

reserve "space" for when you have some kind of structure

[nimble portal]

Okay

[abstract saffron]

Yes. Anyhow, we're only having equivalence classes for now

[gritty widget]

it doesn't make much sense to call something which you only know as a set a space

[nimble portal]

Yep

[cosmic socket]

No I’m saying why you can take classifying space of GL_1(S^{n-1}).

[nimble portal]

Why isn't it already a space? Doesn't it inherit the quotient topology?

[cosmic socket]

The construction you use for BG doesn’t matter

[languid patrol]

I don't think this is a very interesting notion, the more valuable one is: when are two manifolds with two different smooth structures diffeomorphic

[abstract saffron]

Well, what are the open sets?

[languid patrol]

there are infinitely many equivalence classes of atlases on R for instance

(under this definition)

which all induce the obvious smooth manifold up to diffeomorphism

[gritty widget]

refer to my previous comment

[nimble portal]

Preimages of the smooth structures? Idk lol, more interested in what Tteg said now

[gritty widget]

you do not have a topology on the set of all smooth structures on your space

[abstract saffron]

And this is well-studied 😄

[languid patrol]

yes but i was trying to avoid giving "this is what people decided to look at" as a reason

and try to think of a more satisfying one

[nimble portal]

I'm not sure how to put it formally but in my mind it's something like "if we can go back and forth between the maximal atlases defining the smooth structures without issue", whatever that means?

[languid patrol]

i guess to some extent the whole point of the theory of abstract manifolds is to try not to focus on the coordinates you're putting on the space (otherwise, by whitney embedding, why not just consider them as cut out in some euclidean space by some smooth functions?), and this seems like it's going in the opposite direction

[abstract saffron]

Well, any two atlases induces a homeomorphism between (subsets of) R^n.

[languid patrol]

yes, if there is a diffeomorphism between the two resulting manifolds. there are notable smooth structures (like the milnor example) which have nothing to do with the naive one on a topological space, and give two totally unrelated smooth manifolds with the same topological space

[abstract saffron]

But this is roughly what Topo-girl said above about differential structure

[nimble portal]

I think what I'm thinking is wrong actually

But Tteg said yes so I think I accidentally wrote something correct

[languid patrol]

I don't know I'm just guessing

[nimble portal]

I meant like

[abstract saffron]

Same

[languid patrol]

at what you're trying to say

I think I wouldn't get hung up on things like "we extend these charts to a maximal smooth atlas" it's just something mathematicians say to make themselves feel better, but it doesn't really factor into the theory in practice.

[nimble portal]

I think what I meant was that if you have two different maximal atlases A & B that define different smooth structures, if we can find "a (diffeomorphic?) way" to "go between" the charts in A and the charts in B, then the two manifolds would be diffeomorphic

I'm so bad at words

Like "intermediary maps", I'm hesitant to call them transition maps though because if they're different maximal atlases then doesn't that mean they're not smoothly compatible? Lol

So like a diffeomorphism between the two atlases?

[languid patrol]

We cannot in general

Some examples

the structure on R given by the identity and the one given by x^{2n+1} are not "equivalent" by your definition but they are diffeomorphic in the obvious way

On S^7 there are known to be many non-diffeomorphic smooth structures, but they are hard to construct

on R^4 there are known to be uncountably many maximal atlases which you cannot find a diffeomorphism between.

[umbral panther]

They’re easy to construct! They’re just S^3 bundles over S^4. (Well, maybe that’s just 7, not 28, I’m not sure)

[languid patrol]

Oh I didn't know that! I had never really looked into it but that makes a great deal of sense... now I have to count the nonisomorphic S^3 bundles over S^4....

Aha this is asking about H^1(S^4, O(4)) I see

[abstract saffron]

Me, never read Milnor's original paper:

[abstract saffron]

but iirc, this is the only case, all other dimensions have finitely many smooth structures

[languid patrol]

and this is \pi_3(O(4))

[umbral panther]

Also, there should be a similarly easy to construct example of an infinite family of manifolds that are homotopy equivalent but not diffeo that can be detected much more easily, using just characteristic classes, not the signature theorem

[abstract saffron]

Also, i'm curious: I recall there''s also something similar about decomposing S^3 into a bunch of S^2 (Hopf sphere or something?) Are these related to quaternions and octonions?

[umbral panther]

Maybe S^4 bundles over S^4?

[languid patrol]

Okay \pi_3(SO(4)) seems to be infinite, have I done something stupid?

[umbral panther]

Yes, it’s infinite

[languid patrol]

Ah I am being stupid though

When are these diffeomorphic/homeomorphic to S^7?

[umbral panther]

It’s Z^2. There’s a coset of Z^1 that is homotopy equivalent to S^7

Diffeomorphic is much harder. There’s a map to Z/7 that obstructs diffeo

[languid patrol]

Ah I see, yes this is what I should've said instead of "difficult to construct," difficult to verify that what you get is not equivalent to S^7

[nimble portal]

Is it ok if I interrupt with questions too?

[languid patrol]

that's fine

[nimble portal]

So under this definition, how do I actually show that a space is locally Euclidean? So more specifically I'm trying to show the line with two origins is locally Euclidean, but I'm getting tripped up on "for each p in M we can find a neighborhood containing p". How do I show such a neighborhood exists?

Why can't I just say "consider p \in U \subseteq M" and then define a chart?

[languid patrol]

usually you have to write down a neighborhood explicitly for each point

So for the sphere we noticed that S^n - \infty and S^n - 0 were both just R^n

so that gives us that S^n is locally euclidean

for the line with two origins you can do something similar.

[umbral panther]

Yes. It’s the generalized Hopf fibration
https://en.m.wikipedia.org/wiki/Hopf_fibration

[nimble portal]

Hm

I think I phrased my question wrong?

Like, how do you take a point in your set and show there's a nbhd containing that point?

Since we're in R^2 I could just take an open ball around the point I guess

But like more generally how do you do that?

Lee's example for the sphere doesn't do any of this point business, he just takes a neighborhood in the space where S^n is embedded (R^(n+1)) and takes the intersection of that with S^n for the chart domain

[abstract saffron]

Holy hell, I still remember some stuff 😄

[nimble portal]

I feel like such a dork I don't know wtf I'm even trying to prove for local Euclidity 💀

Like M locally looks like a line except for where x = 0, I know that intuitively, but how tf do I prove that

I need to show that given a point in M there's an open nbhd around that point? Which in the quotient topology means that the preimage of the projection is open in X

[gritty widget]

"euclidity"

[nimble portal]

Local Euclidianism?

[tiny obsidian]

Euclidianness I think

[nimble portal]

Isn't the preimage of any point in M just a doubleton {(x, -1), (x, 1)}? X inherits the subspace topology from R^2, so just take open balls of radius 1/2 around the two points and ggs?

[tiny obsidian]

careful if the origin(s) are within 1/2 of the points

[nimble portal]

Right

Well X only has one origin

[nimble portal]

uh

So a function is injective if f(a) = f(b) implies a = b

Does equivalence count as equality? So if f(a) = f(b) implies a ~ b, then is the function injective?

[gritty widget]

why should it

[nimble portal]

I don't know, but it sounds really nice if that's the case lol. So I'm on some hopium that it is 🙃

[gritty widget]

rather than tell you the answer i am going to suggest you try just a few examples

[nimble portal]

Can I show the context for why I'm asking?

[gritty widget]

sure

[nimble portal]

It's got an example in it

So we can map (0, 1) and (0, -1) to 0 in R

So technically speaking f((0, 1)) = f((0, -1)) does not imply (0, 1) = (0, -1), since, well, (0, 1) isn't (0, -1)

But they are equivalent!

So if that was good enough to call phi injective then I'd have that phi is a homeomorphism and I'm done with the first part I think?

Since projection maps are continuous & open

[gritty widget]

i still strongly suggest this

[nimble portal]

Yeah isn't this an example?

[gritty widget]

too much reading for me

[nimble portal]

That's what I'm asking, if it can only be equality equality then it wouldn't be injective, but if it can just be equivalence then it's good enough

[gritty widget]

where did you find this? i don't see at all why the second last line should be true

seems like nonsense to me

[nimble portal]

...I wrote it

[gritty widget]

yeah what you wrote is nonsense sorry

[nimble portal]

Why?

[gritty widget]

.

also for the love of god learn latex

and i don't mean sit down with an intro to latex how to type a million things. learn it by typing your stuff and looking up what you need

[nimble portal]

If you have a point in M it's either (0,1), (0, -1), or x right? Where [x] is the equivalence class of (x, +-1)

I think what I just wrote is nonsense too lol...

[gritty widget]

every element of M is an equivalence class

every element of M is either [(0, -1)], [(0, 1)], or [(x, 1)] = [(x, -1]) for some non-zero x

[nimble portal]

What are the equivalence classes of (0, -1) and (0, 1)?

[gritty widget]

they are the two origins

[nimble portal]

I think that's where my thinking failed, I never realized there was anything equivalent to them since the equivalence relation said x neq 0

[gritty widget]

{(0, -1)} and {(0, 1)}

the equivalence classes of (0, -1) and (0, 1) are the singletons consisting of these points since they aren't equivalent to anything else

[nimble portal]

I see

[gritty widget]

please do this

it is so false

[nimble portal]

I'm not trying to be ignorant I swear, I legit thought that

s what I was doing by sharing what I wrote

Giving an example and then interpreting it

[gaunt linden]

When the definition says "the equivalence relation (x,-1) ~ (x,1) for x != 0" it means the reflexive transitive closure of the explicitly given related pairs.

[gritty widget]

a few examples not just one

[nimble portal]

But anyways what I was trying to say was that given one of these elements in M, the only things that map to it are (0, 1) or (0, -1) or (x, 1) or (x, -1)

So the preimg of a point of M is a subset of the set of those 4 elements

[gaunt linden]

Wait, do you think x is a fixed real?

[nimble portal]

Well when you pick a point in M doesn't that automatically fix x?

[gritty widget]

||let X and Y be any sets and f: X -> Y be any function and consider the equivalence relation on X making everything equivalent to everything. then f(x) = f(x') always implies x ~ y but f could be the least injective function in the world||

[nimble portal]

Assuming it's not one of the two origins

[gritty widget]

for a less trivial example ||let f: R -> R be given by f(x) = x^2 and consider the equivalence relation on R given by x ~ y iff |x| = |y| ||

[gaunt linden]

M contains both [(2,1)] and [(3,1)], for example -- they're different points.

[nimble portal]

Yeah so even though -2 & 2 are equivalent they're not equal so it's not injective

That's just what I wanted to confirm, if equivalence is good enough instead of equality or if it has to be equality

[nimble portal]

Sure but if you pick a point (say [(2, 1)]) then that fixes x = 2 in X no?

[gritty widget]

you could have confirmed it yourself by doing what people typically do when they don't understand something: check it against examples

[gaunt linden]

What to you mean by "fixes" in that context?

[nimble portal]

I was asking how to interpret the example I had

It was similar to yours

[gritty widget]

my final comment is the following. suppose f: X -> Y is a function of sets and ~ is an equivalence relation on X. suppose f "factors through" the quotient function, i.e., there's a g: X/~ -> Y with g([x]) = f(x) for all x in X. the condition you proposed, that "f(a) = f(b) implies a ~ b", is equivalent to injectivity of g

[gaunt linden]

This sounded like you think there is a single number x such that X = { (0,1), (0,-1), (x,1), (x,-1) }.

[nimble portal]

Well like for [(2, 1)] in M, aren't there only two points in X that map to [(2, 1)]? Being (2, 1) and (2, -1)

So any point [(x, y)] in M (with x nonzero) has the preimage {(x, 1), (x, -1)}

[gaunt linden]

That is right.

[gritty widget]

actually no that's not my final comment i want to say more. if X is a set and ~ is an equivalence relation on X then you have the function p: X -> X/~ sending each element to its equivalence class. note that p(x) = p(y) is quite literally the condition x ~ y. injectivity of p would mean the equivalence relation is trivial: every element is equivalent only to itself

[nimble portal]

Right, so what's wrong with what I said here?

If p = [(0, 1)] then the preimage of p is {(0, 1}}
If p = [(0, -1)] then the preimage of p is {(0, -1)}
If p = [(x, y)], then the preimage of p is {(x, 1), (x, -1)}

[gaunt linden]

This is true.

[nimble portal]

So regardless the preimage is in the set I wrote above

Is it because I said {p}?

The reason I did {p} was because I needed to show that for any point p in M we can find an open subset containing p, so I used {p} and showed that {p} was open in the quotient topology

(Or that was my intent, at least)

[gaunt linden]

Wait, what? {p} is definitely not open.

[nimble portal]

Why not? Its preimage is open

[gaunt linden]

We're still talking about the line with two origins, right?

[nimble portal]

Yes

[gaunt linden]

You listed the possible preimages here. Neither of those sets look open to me.

Are you perhaps confusing the single point (0,1) with the open interval (0,1)?
The former is an element of R^2, the latter is a subset of R.

[nimble portal]

Hmm wait a second

[nimble portal]

I think that is what I was doing

[grizzled ibex]

there shouldn't be a retraction deformation from a solid ball D^n without a point on the boundary to S^(n-1) right?

[gaunt linden]

No, not if the missing point is on the boundary.

[grizzled ibex]

yeah that's what i was wondering

again i was betrayed by imagination

[nimble portal]

Do you see the idea of what I was trying to do Trop?

I know I did it wrong, just want to ask how I can salvage it without straying too far from what I was intending

[gaunt linden]

Not entirely. I got somewhat sideways into the conversation.

[grizzled ibex]

i imagined the ball a with piece removed being "hollowed", then i thought if we coneccted that piece and squished it to the boundary it would be ok

again, betrayed 😭

[nimble portal]

Hehe okay I've got to head out for dinner anyways so I think I'll put this on hold and do some work on paper first then come back with a second attempt

Maybe with some pictures so I can show what I was trying to do

[gritty widget]

there's also like everything i said about the equivalence relation stuff 🙂

[nimble portal]

Yes

I read that

I'll figure out how to do it right promise Tt!

[gaunt linden]

One piece of advice here: Do the case for x != 0 first without trying to handle the two origins at the same time.

[nimble portal]

Will do

I’m assuming it’s those two origins that make it not Hausdorff too

[gaunt linden]

Indeed.

But it seems you need some familiarity with the quotient construction before attacking those, and the points away from the origins is a chance to get at least a little of that.

[viscid nebula]

What would this hexagon look like if we apply the same thing we apply to the unit square to make it a torus? The green and orange sides are undefined.

[novel acorn]

I don't think it will be anything concretely named tbh

[viscid nebula]

Yeah, it’s a question in my homework and I was confused because idk how to describe the resulting shape. Weird.

[hidden crag]

Can you send the exercise

[viscid nebula]

It’s in german

[hidden crag]

ich werds überleben

[viscid nebula]

Lolllll

[hidden crag]

weird, i thought the actual exercise would provide more info but i guess not

[viscid nebula]

[abstract saffron]

When I passed my Lin Alg exam, I thought I would never have to do Gaussian elimination again

Joke on me: I suffered for 1 hour while computing some stupid simplicial homology because I could not bring myself to do Gaussian properly

[unreal stratus]

Lol

Rip

[hidden crag]

can't you just like

feed it to some CAS

maple came in clutch for me last time i had to do that

[abstract saffron]

I hate to rely on computers, especially since it's a simple object in this case

But man, I'm horrible at computation. One day it's gonna bite me back

[novel acorn]

Me: megumi, tell me a prime number
Megumi: 64

[unreal stratus]

57

1

[gaunt linden]

Up to homeomorphism, a torus with a hole cut out, I should think. This is also the same as a disk with one handle.

The intention is probably that in (c) you can glue a sequence of these patches together, yellow edge of one of them to green edge of the next one, to get a surface of arbitrary genus. But it seems at bit ambitious to expect you to generalize the concept of "genus" from that line-and-a-half it it's your first exposure to the concept,

[urban zinc]

Lol are you also doing Lee intro to smooth manifolds

[bitter smelt]

Last I heard she was reading Tu

But any smooth manifolds text will have this example

[nimble portal]

I switched to Lee a while ago!

[tulip bluff]

Is this problem doable using elementary methods? Initially I was thinking about excision, but it did not seem to work. There is some duality theorem that gives this, but I would like to see whether it can be done without it.

[coarse night]

what the things you're allowed to use?

It's just Jordan Brower separation theorem but it's highly nontrivial to prove

[tulip bluff]

I have no restrictions, it's for my own sake. Can you elaborate on how this comes from the separation theorem?

[coarse night]

Literally the statement

[tulip bluff]

The theorem says that for the image of S^n in R^{n+1} under an injective mapping the complement separates R^{n+1} to two connected components.

Okay I see that Hatcher has this exact formulation.

[umbral panther]

If you can prove this, you can prove full fledged Alexander duality

Maybe use the ANR absolute neighborhood retract property?

[cosmic socket]

Does anyone have a reference for desuspension

Of spectra

[viscid nebula]

Thank you for the answer but I don’t really understand what this picture represents

[gaunt linden]

It's the same hexagon, just deformed such that it looks like the rectangle-that-becomes-a-torus with a notch cut into it from one of the corners.

So when you glue the black edges together as indicated, you get a torus with a thin slit missing from it.

[viscid nebula]

Oh wow, makes sense!

Am I allowed to deform it however I want?

[gaunt linden]

Yes, generally.

(As long as your deformations are in fact homeomorphisms).

[viscid nebula]

Thanks again

[nimble portal]

Okay uh

I'm still not really sure what's wrong with what I wrote yesterday ._.

The quotient map preimg of {p} is {(x, -1)} or {(x, 1)}. Both of these are in the subspace topology of X, so they're open, so {p} is open in M no?

[coarse night]

what's the problem statement

[nimble portal]

I swear I'm constantly checking definitions to see the flaw in my logic but I don't see it

A set is open in the quotient topology if its preimage under canonical projection is open

Oops I made a little mistake, the (x, +-1) points should be in one set

But anyways regardless the preimage of {p} has to be one of those three sets (or just one of those one sets if I do as Trop said and ignore the origins)

And those are in the subspace topology of X, because they're the intersection of S with {(x, +-1)} or {(0, +-1)}

(which are open sets in R^2)

So what's the issue so far?

[gaunt linden]

Still no. Single points are not open in X.

[nimble portal]

But they're not points, they're sets

[gaunt linden]

They are sets with one element each.

(x,-1) is one point.

[nimble portal]

Why not? A set is open in X if it's in the subspace topology of X right? And if it's in the subspace topology then that means it's the intersection of X with an open set in R^2

[gaunt linden]

(x,1) is one point.

[nimble portal]

Yeah I meant to put (x, +-1) into one set, sorry

But that doesn't fix the (0, +-1) points

[gaunt linden]

Can you show an open set in R^2 whose intersection with X is {(3,1),(3,-1)}?

[nimble portal]

{(3, 1), (3, -1)}? That's what I was thinking for the other points as well

[gaunt linden]

{(3,1),(3,-1)} is not an open subset of R^2.

[nimble portal]

._.

I see

Oh I'm such an idiot argh

I see

I need an interval don't I

Or rather a rectangle?

[gaunt linden]

Yes.

[nimble portal]

Okay yeah thank you

[nimble portal]

@gaunt linden Do you mind checking over a rough sketch of how I think I can reapproach the problem?

(Just let me know if you don't want me to ping you again)

[urban zinc]

I just did this problem two days ago so I can check over it

Lol

@nimble portal

[nimble portal]

One second

From what I can tell \varphi is a homeomorphism right?

It's essentially just a projection onto the first factor

And the assumption that y = 1 guarantees injectivity

With phi^(-1) given by x -> [(x, 1)]

[urban zinc]

So locally euclidean means every point has a neigborhood which is homeomorphic to R right? What point are you considering in this proof

[nimble portal]

Just take a point in U

[urban zinc]

I'm a little confused by the notation

That's not how this works, I give you a point and you give me a neighborhood of that point and a homeomorphism

[nimble portal]

If you give me a point I pick epsilon so that it's in U (a neighborhood containing the point)

[gritty widget]

you want to put that "for some epsilon..." outside of the {} at the start

[nimble portal]

I believe I showed U was open in M

I think I see why

And then I planned to use that function phi as my homeomorphism (which I think meets the conditions to be a homeomorphism)

[urban zinc]

Oh okay so you're saying my point is [(p,q)] and you're telling me I should consider the neighborhood U of that point (which depends on what epsilon is) which is the set {[(x,y)] such that x is less than epsilon away from p and y is either 1 or -1}?

[gaunt linden]

Your preimage of U shouldn't contain (0,-y), with your latest U.

[nimble portal]

I think, yes. In my head I was picturing: you pick a point [(p, y)] in M, I define a neighborhood about that point consisting of all [(x, y)] where x is less than epsilon away from p. I said suppose y = 1 so y would just be 1

[nimble portal]

I'm sorry, I don't understand what you mean

[urban zinc]

You can't suppose without loss of generality y=1 because [(0,1)] and [(0,-1)] are different

[nimble portal]

Yeah but the elements of M are equivalence classes

Oh you're using the origins

[urban zinc]

I mean they are in different equivalence classes

Are you assuming for this part of the proof that p is nonzero?

[nimble portal]

No, but Trop suggested that

I stubbornly keep wanting to see if I can do it all in one go

[urban zinc]

The origin here is fundamentally different

[nimble portal]

You can just mirror hte proof if y = -1 no?

Just splitting it up into two cases

[urban zinc]

Oh so you're saying suppose the y of the point that I give you is 1? I'm a bit confused because you only ever use y in your definition of U

[gaunt linden]

You can do it all in (almost) one go -- I suggested doing the easy part first because it looked like you had conceptual problem with the quotient construction, and it would be good to get some practice with just that before you get to tackle the double origin.

[nimble portal]

Yes

[urban zinc]

Okay I see now

[nimble portal]

It would help if I specified that first at the start I think

[urban zinc]

So does U only contain points that look like [(x,1)] in that case?

[nimble portal]

Yes! Where x is within epsilon away from p

[urban zinc]

Ah okay

[gaunt linden]

If I get you right you defined U as { [(x,1)] | -eps < x < eps }.
So far so good. But the preimage of this U is not (-eps,eps) × {1} cup (-eps,eps) × {-1}. It almost is, but (0,-1) is not actually in the preimage.

[urban zinc]

^ but this isn't actually a problem because that set is still open in X

[gaunt linden]

Right, it's easily fixed.

[urban zinc]

I still think the proof would be clearer if you considered two separate cases

You should think of proofwriting as explaining why something is true to someone else in your class, basically

[nimble portal]

Why isn't (0, -1) in the preimage? Under the projection it gets mapped to its equivalence class [(0, -1)] ~ [(0, 1)], which is in U

[urban zinc]

(0,-1) and (0,1) are not in the same equivalence class

[gaunt linden]

I suppose there is something to be said for doing just the neighborhood-of-one-of-the-origins case first and then observing, "oh, this happens to work for all the other points too".

[urban zinc]

The preimage of (0,1) is just {(0,1)}

[nimble portal]

Eric have you seen me talk? I speak in pictures and handwaving 💀 that's why this is so hard to me

I see

But the easy fix is just remove the (-epsilon, epsilon) x {-1} from the preimage right?

[urban zinc]

wait

The only thing to remove from the preimage is (0,-1)

[gaunt linden]

All the other points in (-eps,eps) × {-1} are and must be in the preimage.

[urban zinc]

All the other fibers look like {(x,1),(x,-1)}

it's just the x=0 case that is special

[nimble portal]

OH I see what I was getting confused

Okay

I keep wanting to treat the two origins as being equivalent lol, thank you

[urban zinc]

hahaha

So if your set in the quotient space is {[(x,1)] where x is < eps away from p} then the preimage should be {(x,1) where x is < eps away from p} U {(x,-1) where x is < eps away from p} minus (0,-1)

Which is open (why?)

[nimble portal]

You mean minus {(0, -1)} right?

I could not prove to you in symbols/mathspeak why it's open lol, but I can explain why intuitively 🙃

[gaunt linden]

It might be clearer to write it as:
(-eps,eps) × {1} cup (-eps,0) × {-1} cup (0,eps) × {-1}

[nimble portal]

Yeah that's what I was planning to do, just write the negative half as a union with 0 excluded

[urban zinc]

Yeah the reason why I didn't write it like that is that I'm still not clear on if U is supposed to be centered at p or not

[urban zinc]

Yeah

[urban zinc]

Do you remember when things are open in the subspace topology?

[nimble portal]

I suppose it doesn't necessarily need to be centered at p but in my ehad that's how I pictured it, that's an easy fix though (just define at the start what p is)

[nimble portal]

Yeah, a set is open in the subspace topology if it's the intersection of the subspace with an open set in the larger space

[gaunt linden]

In this case it's actually less work to keep it centered on 0.

Then you don't need a special case for "what if 0 is not between our endpoints".

[urban zinc]

Okay yeah that's true

[nimble portal]

I didn't think that'd be an issue, you could just redefine epsilon so 0 is in the set

[nimble portal]

I think I see what you mean though

(That's presently what I'm doing right? Implicitly)

[gaunt linden]

Yes.

[urban zinc]

You could have your point be like [(x,y)] and then your U be like {[(x',y)] for x' between -|x|-1 and |x|+1}

[nimble portal]

WHen does analysis show up in diffgeo?

This feels analysisy, not really since it's just the most basic of inequalities but I'm sure you could go deeper into similar stuff

[urban zinc]

Well we haven't done any calculus yet haha

This is just topology

Without the smooth part yet

[nimble portal]

Oops right lol

I'll be getting to that in the next problem

Okay so I think I fixed all the glaring issues there, are there any counterexamples I'm missing with my projection map (phi)? Or does it pass as a homeomorphism? lol

How do y'all notice the flaws/counterexamples to stuff so quickly? I've noticed that's what always gets me as far as my logic goes, there's always some case I didn't consider

[gaunt linden]

It's a homeomorphism all right. In principle you have more footwork to do to show that, but in the interest of moving on to the interesting parts, there could be a good case for leaving that for later and work on the other parts of the exercise instead.

[novel acorn]

it's something you develop over time

idk sometimes when reading something it just doesn't feel right so you go back and think it over more clearly

or sometimes you just realize something is blatantly false lol

[nimble portal]

I'd have to explicitly prove the bijectivity & continuity, and likewise for hte inverse to do it thoroughly right?

[gaunt linden]

Yeah, if you want all the relevant letters dotted and crossed.

[nimble portal]

I'm too tired to bother 💀 onto second countability

Isn't this just figuring out a basis for U? Since I let U just be basically any nbhd in M

[cosmic socket]

are there any good problems/exercises to get more comfortable with spectra

[cedar pebble]

what is your background?

what sorts of things are you interested in doing with spectra? like you could do explicit computations with the Adams spectral sequence and this would give you some level of familiarity working with examples of spectra, but idk if that sort of example would be useful for most people

[urban zinc]

Note: if you've already proven it's an inverse, then that already shows it's bijective

But you do need to show that both it is continuous and the inverse is continuous (which should be fairly simple if you're comfortable with quotient topologies and subspace topologies and good practice if you're not)

[urban zinc]

Well I just did this exercise haha

But it helps to list out explicitly exactly what statement you need to prove

Mathematics is sort of nice in that if you write out everything perfectly explicitly, then you need no knowledge to be able to check something (in practice, we don't write everything out, but it's good to understand what we're leaving out)

[urban zinc]

Luckily this one is simple

Proving that it's not Hausdorff is also simple but it's probably more interesting

[nimble portal]

I'm stuck on covering every point in U for my basis

[urban zinc]

What is U

Do you mean M

[nimble portal]

My first thought was just to partition the neighborhood U I defined earlier into two parts

[urban zinc]

Maybe use the following definition for basis: B is a basis iff for every x in M and open U containing x, there's a set in B which contains x and is contained in U

[nimble portal]

One consisting of all the equivalence classes "to the left of" the midpoint (epsilon/2) and one with the equivalence classes "to the right of it"

But that doesn't cover the equivalence class x = [(epsilon/2, y)]

[urban zinc]

Do you know the proof that R is second countable?

If you know that, then this should be a relatively simple modification

[nimble portal]

I do not, but I know it has something to do with open intervals. I will take a look at it 👍 thanks!

uh

[nimble portal]

if I just take U as what I defined before (but let epsilon be a positive rational instead of positive real)

And partition the interval (-epsilon, epsilon) into open intervals whose endpoints are rational

Doesn't that work?

Oh I don't think you even need to go through all that

[urban zinc]

U here is given to you

So I give you x and an arbitrary open set U which contains x

You need to give me a set in B which contains x and is contained in U

Open intervals with rational endpoints is the right idea though

[nimble portal]

I see in my head what to do, just struggling to put it into words lol

Let's say you give me the point q = [(p, y)] and an arbitrary open set U that contains q

I can define a set similar to what I did for the first part, let's call it V = {[(x, y)] : a < p < b}, where a is the closest rational approximation to p s.t. [(a, y)] is still in U, and similar for b

That feels overly complicated though

You don't even need "closest rational approximation" lol, just that [(a, y)] and [(b, y)] are still in U

[urban zinc]

sorry do you mean a<x<b instead of a<p<b?

If so I think that works!

[nimble portal]

Yeah sorry a < x < b, but we still need a < p < b alongside

[urban zinc]

Right

[nimble portal]

Once I write it out I'm sure I'll think of some ways to clean it up

[urban zinc]

And so what is your basis for M?

[nimble portal]

B = {V_{a,b}}, where V_{a, b} = {[(x, y)] : a < x < b}, where a, b are rational

[urban zinc]

yup perfect! and y is either 1 or -1

and why is that countable?

[nimble portal]

I don't like the notation V_{a, b} but I'll fix that when I can TeX it, it's countable since each V_{a, b} is defined by open intervals in R with rational endpoints, which we know is a countable set

[urban zinc]

Yeah, so to be more precise, Q is countable, so Q×Q×{-1,1} is countable, so B is countable

And then just make sure you understand why all of the sets in B are open in M and you're done

Also note V depends on a, b, and y so I guess V_{a,b,y} is better haha

[nimble portal]

Well they're open in M by nearly the same logic/process as how I showed my neighborhood in the previous part was open

Basically the same neighborhoods, just different rational endpoints instead of a fixed real epsilon

[nimble portal]

You know

This is a great example of where visual intuition fails

[urban zinc]

Only if your visual intuition is bad

[nimble portal]

I keep visualizing M as a single line but in a sense it's sort of two lines stacked on top of each other (glued together by the quotient)

[urban zinc]

Okay fr though I feel like learning topology is the act of replacing bad visual intuition with good visual intuition

[nimble portal]

I never thought I'd enjoy point set but this is actually cool

[urban zinc]

Many of the arguments will be visual but importantly you need to be able to formalize them as soon as someone asks

[nimble portal]

Yep

That's my issue

[urban zinc]

Yeah, it's like the topology professor who, after a student asks them to explain a visual argument, immediately covers the entire board with a single equation and then moves on

(Source: the topology professor here)

[nimble portal]

Alright I’m away from my computer now so no formality but let’s think about Hausdorff 🤔

So given two distinct points we can always form disjoint open nbhds around them

Can’t you just use the basis for the topology from earlier w the origins as a counterexample

[gritty widget]

write down your two points and explain why there are no disjoint opens containing each

[nimble portal]

Yep I see it

It’s because when we take a nbhd about either origin we’ll always have points within the “line” part of M inside

Even if one nbhd is smaller than the other, the points on the “line” nearest to the origins will be inside the nbhd around the other origin, so they’re never going to have empty intersection

[gritty widget]

sounds like the right idea

got a picture?

[nimble portal]

Sure

Is the line with two origins an example of a space we can’t embed into R^2?

[gritty widget]

you can't embed it into any R^n because it is not hausdorff

all subspaces of hausdorff spaces are hausdorff

[nimble portal]

Oh

That sounds like it leads to Whitney Embedding Thm

Sort of

[gritty widget]

it shows you that hausdorff-ness is necessary for the theorem to hold

[nimble portal]

Here’s the picture, the two origins are supposed to be “infinitely close” to the line though

Obviously the neighborhoods can’t be balls but

You know what I’m getting at I hope

That was fun

I’m scared of the stereographic projection problem though, that looks really hard

But I’ll get to that tomorrow

[urban zinc]

Yeah that's correct

Just make sure you know how to formally write it out and everything

[nimble portal]

Yeah I’m pretty sure I can tom!

[urban zinc]

It's not too bad

(Although I admit I didn't check that it gives the same smooth structure as the other one they gave, but that's also not too bad)

[nimble portal]

The locally Euclidean part definitely seems to be hardest for basically everything lmao

[urban zinc]

Yeah the other properties are usually routine

[nimble portal]

I think that was a hint tbh

I’m gonna bet there’s some kind of circular symmetry you take advantage of

Well ig that’s obvious because it’s a sphere

[abstract saffron]

I have a question: supposed you have an orientable surface whose boundary is homeomorphic to a circle, and you identify the boundary. Can the resulting 2-manifold always be embedded in S^3?

It's an easy No without the "orientable" part, just take Mobius strip

[coarse night]

Identifying boundary gives you a oriented manifold ( I think) but you already have the classification

[abstract saffron]

It seems that the answer is yes

[coarse night]

wait your mani is not necessarily compact, is it?

[abstract saffron]

What I considered initially was compact, but I forgot to include that

But now it seems more interesting without compactness condition

Well, it should be closed, of course

[languid patrol]

@coarse night why do you get something orientable?

[coarse night]

was getting it from this, M is orientable then by definiton H_n(M, ∂M)= ℤ= H_n(M/ ∂M)

Though I am fully sure of the quotient is a manifold

feels like it

[gaunt linden]

There's a neighborhood of the circular boundary that's homeomorphic to { x in R^2 : 1 <= |x| < 1+epsilon }.
Identifying the boundary corresponds to mapping this annulus to { x in R^2 : |x| < 1+epsilon } by multiplying every x by (|x|-1)·(1+epsilon)/epsilon.

[coarse night]

M/ ∂M = M ∪ C ∂M

[gaunt linden]

This lets us continue an orientation of the interior of the original manifold to a neighborhood of the point that represents the collapsed boundary.

[coarse night]

if we go along the definition of nonvanishing forms, we have a non vanishing form on the S¹ and attaching a D² to it is collapsing the boundary, so we can continue the form radially to entire D² giving non vanishing form on M u c(\del M)

only issue is the center

ok we can normalize the form on the boundary

we have a tubular nbd of S¹. Say S¹ and as attach a S¹x[0,1] via the boundary to extend and given an non vanishing form on S¹x{0} we can normalise the form to get a unit valued from on S¹x{1}

[gritty widget]

How to estimate ${p\in\beta\mathbb{N} : p\leq_\text{RF} q}$ where $q\in\beta\mathbb{N}$ and $\leq_\text{RF}$ is the Rudin-Frolik order?

[gentle osprey]

Blitzkrieg

[gritty widget]

I define $p\leq_\text{RF} q$ when there is injective $f:\mathbb{N}\to\beta\mathbb{N}$ with discrete image such that $f^(p) = q$, $f^$ being the continuous extension of $f$ to $\beta\mathbb{N}$

[gentle osprey]

Blitzkrieg

[gritty widget]

the cardinality of*

[radiant cedar]

If X = {(x,y) in S^n x S^n | x \ne y } and f : S^n -> X is a map f(x) = (x, -x) how can I show that f is a homotopy equivalence? Will it work to take g : X -> S^n as the projection g(x,y) = x?

Then we would have g o f = id_{S^n} for free and f o g could be homotoped to id_{X} with H: A x [0,1] -> A, H(x,y,t) = (1-t)(x, -x) + t(x,y)/|(1-t)(x, -x) + t(x,y)|

[worldly panther]

Hey topology people can y’all help me 😭

[bitter smelt]

no, that probably isn't even a knot

0/10

[worldly panther]

This is an extension cord that was wrapped all around inside itself

[solemn oar]

Contract to a point and re-extend to a line segment

[coarse night]

Plug it in and touch the other end

[gritty widget]

I did it already

[coarse night]

Are they?

Doesn’t feel right, take n=1 then X becomes contractible

[urban zinc]

Excuse me read the channel description, this channel is for anime

[worldly panther]

I fixed it and didn’t see this till now, wdym by contract to a point?

[abstract saffron]

yeah, it's not a knot

[gritty widget]

isn't there a book with this name

[languid patrol]

it can't be contractible, it has fundamental group Z

[coarse night]

oh right

[languid patrol]

Maybe an easier way to visualize this: this is an R^n bundle over S^n

[coarse night]

it's just S¹

[languid patrol]

yep!

Up to homotopy anyway

[coarse night]

yeah

[unreal stratus]

Indeed this is the (ordered) configuration space of two points in S^n and looking that up i found this lol

Interestingly if you use three points instead of two you get the unit tangent bundle of S^n, hmmm

[languid patrol]

Yep in general it's just the complement of the big diagonal in (S^n)^m

[unreal stratus]

Well that's what I meant by three points lol

Like ordered conf w 3 points

[languid patrol]

Yeah and I'm saying if you picked more than 3 points that's what you get 🙂

[unreal stratus]

Hm wdym

[languid patrol]

aha i fixed a typo

does it make sense now?

[unreal stratus]

Hm idk just sounds like you said what I said lol

Like I mean the ordered configuration space is by definitij what you said right so idk how what ur saying is non tautological xd

I am going to learn more conf spaces tbh hehe exciting

[languid patrol]

I guess I wasn't sure if you'd realized there was a simple description of Conf(m, S^n), I think from that description it's pretty clear why for m = 3 it retracts onto the unit tangent bundle 🤷‍♀️

[unreal stratus]

Yee sure

Hm did you have any equivalent characterisation of the conf space in mind

[languid patrol]

I mean nothing besides just the definition!

[unreal stratus]

Yeee sure

OwO

[coarse night]

where do these spaces come up?

Any interesting application of configuration spaces?

[quasi steppe]

Does anyone here do topological data analysis?

Or study it?

[languid patrol]

all over the place, maybe the most interesting thing about them is their fundamental groups which have been studied by mathematicians for a long time

[unreal stratus]

Doing a summer project on them and my advisor talked about how it's conjectured that configuration spaces can parametrise certain cohomology operations lol

but also like interesting spaces in themselves, interesting (co)homology w/ links to lie algebra stuffs hehe

[onyx raft]

How do we get these isomorphisms

$H^k(M,\partial M;R)\cong H_c^k(M-\partial M;R)$ and $H_{n-k}(M;R)\cong H_{n-k}(M-\partial M;R)$

[gentle osprey]

Kähler

[onyx raft]

From the fact we have a collared neighborhood we have that through excision

$H^k(M,\partial M;R)\cong H^k(M-\partial M, (0,\epsilon)\times \partial M$ for some $\epsilon>0$ but I am not sure how we'd use this for anything

[gentle osprey]

Kähler

[rapid lagoon]

Hi all, can you guys help me with this problem from Munkres? It is rather elementary, but I am rather stuck on it

The problem asks for a given set X with a family of topologies (with an index set i in I) T_i, find the smallest topology containing all of T_i

Now my answer to this is rather straight forward

All we do is construct a family of topologies $\tau_\beta$ with $\beta$ being an index, such that $\forall \beta, \bigcup T_i \subseteq \tau_\beta$

[gentle osprey]

dulg

[rapid lagoon]

Hence, we have $\bigcup T_i \subseteq \bigcap \tau_\beta$ and that $\tau_\beta$ is the smallest topology that contains all of $T_i$

[gentle osprey]

dulg