point-set-topology

[prime cobalt]

this is the definition give by chatgpt

[wispy veldt]

ile assume the interior is define as such since its a metric space

x in interior C subset R^3 means we can find a open ball of R^3 of radius r thats completely contained in C

open balls in R^3 must contain non-zero z-value

do you see why this implies your interior is empty?

also i think sharing chatgpt responses is against the server rules , might want to delete it

[novel acorn]

● Do not spam, flood, or post unsolicited advertisements. Do not post responses written by ChatGPT or a similar AI tool.

[prime cobalt]

i have no idea about open ball ,

[wispy veldt]

its defined in the picture you sent

[prime cobalt]

if the R^3 ball split by x-y plane , it may contain some zero z-value ?

[marsh forge]

It seems like you are missing some key definitions

You should make sure you understand the definition here

before trying the exercise

I guess its not an exercise, before reading that proof

[grim knot]

does sombody have a clear distinction and thus intuition of homotopy, homeomorphism, retraction, deformationsretract, strong deformationsretract

I'd love to gain an intuition from it, I do not quite imagine and thus recognize the difference of the above named functions

[hidden crag]

Can you think of an example of a homotopy equivalence that is not a homeomorphism?

And why is one stronger than the other

[grim knot]

maybe shaping the punctured disc to S1 (?)

[grim knot]

one IS a bijection, the others are only homotpy inverse I guess

there is not an equality?

but I can't really imagine what that means

[hidden crag]

Yes, injectivity fails here

[grim knot]

does it mean that we have the identity up to an homotopy?

[hidden crag]

Another good one is contracting a disk to a point

Because there obv is no bijection between a one point space and a disk

[hidden crag]

Yes

[grim knot]

ohhh that makes sense, so we identify the point on the disc as if they were all the same right?

this is why injection fails, cause we have the same value but not the same starting point

can you also give me an intuition on how to imagine an homotopy

[hidden crag]

Homotopies between maps or homotopy equivalences

[grim knot]

homotopy equivalence is what you said before right?

two spaces are considered homotopy equivalent if to every continuous function f:X->Y we can find an homotopy inverse

it means that we can shape to Y an reshape back continuosly right?

[unreal stratus]

this isn't the case

[hidden crag]

There is at least one that has an inverse

[unreal stratus]

Two spaces X,Y are homotopy equivalent if there are maps f: X -> Y, g : Y -> X such that fg and gf are both identity up to homotopy

So you just take the notion of homeomorphism but don't require them to be identities `on the nose' but only up to homotopy

[hidden crag]

You basically weaken homeomorphism up to homotopy

Wow

Sniped

[grim knot]

so I have the image of an homotopy in minda

like shaping things continuosly

[unreal stratus]

I think the like prototypical example is that I = [0,1] is homotopy equivalent to a point ( = contractible) but not homeomorphic to a point

[grim knot]

an homeomorphism is now stronger right? What is an example?

[hidden crag]

We already said that potato

Well every homeomorphism is a htpy equivalence

(Why?)

[unreal stratus]

The closed disk is homeomorphic to a square

[marsh forge]

torus and coffee cup

[grim knot]

I thought this was only an homotopy

[unreal stratus]

(say homotopy equivalence rather than homeomorphism)

[marsh forge]

no

also homotopy equivalence* homotopy refers to a thing between continuous maps

[grim knot]

yes you're right, sorry

[marsh forge]

np

[grim knot]

are two completely different things

so what we "lose" between from an homeomorphism to an homotopy equivalence is the injectivity?

[marsh forge]

You lose a lot, not just injectivity

[unreal stratus]

the existence of a homotopy equivalence between [0,1] and {0} shows that homotopy equivalences can be very far from injective/surjective

[grim knot]

is [0,1] homeomorphic to (0,1)?

[unreal stratus]

No

e.g. the former is compact, the latter ain't

Or if you remove any point from (0,1) it is no longer connected, whilst [0,1] is connected after removing 0 or 1

[grim knot]

oh yes, sorry dumb mistake

[unreal stratus]

Any good place to look at E_{infty} rings etc

[marsh forge]

no

[unreal stratus]

There's obvs classic May texts but idk if should look again at more modern resources

Higher Algebra lol

[marsh forge]

honestly there might genuinely be no good source to learn these from

[unreal stratus]

Oh I thought you were just being v sarcastic lol

[marsh forge]

higher algebra might be my unironic suggestion

no

[unreal stratus]

Sure fair enough lol

Interestingly was gonna look at H_{infty} stuff a bit lol

[marsh forge]

the issue is like

you need to speak operadese

[unreal stratus]

Not sure how much they are worked with nowadays

Sure

Yeah I mean I was gonna learn about operads as part of it

[marsh forge]

people use H_infinity sometimes

[novel acorn]

https://florianadler.github.io/AlgebraBonn/KTheory.pdf
There's this

[marsh forge]

but i don't know of any H_infinity rings that aren't also E_infinity

im sure there are some

[unreal stratus]

Just yeah it was like power operations on Morava E theory

Ah ok

But yeah i mean those are E_{infty} anyway lol

[novel acorn]

oop wait
Those don't cover E_infty rings
Only E_infty groups and monoids

[marsh forge]

this looks pretty good from the TOC anyway

[unreal stratus]

Ooh very nice thanks

[marsh forge]

its under brave new algebra

[unreal stratus]

Okay Max beat me to stuff twice lol

[marsh forge]

i mean theres no reason to work in spectra

but its good nonetheless

[unreal stratus]

what would you work in otherwise?

[marsh forge]

an arbitrary (presentably) symmetric monoidal stable infinity category

[unreal stratus]

Sure yeah

This does talk about it being more general than spectra i think

[marsh forge]

i mean spectra are probably the most important examples to be fair

[grim knot]

Since I'm getting to understand this, what is then the different between a retract, a deformation retract and a strong deformation retract?

[marsh forge]

A retract is any situation in which you have $A\xrightarrow{i} B\xrightarrow{r} C$ and $r\circ i=id$.

[gentle osprey]

themaxj

[marsh forge]

A deformation retract is this relation holding up to homotopy

and a strong deformation retract is this relation holding up to a homotopy that fixes A at all times t

(pretty sure I am remembering these right, haven't used this terminology in awhile)

[grim knot]

I see what retracts are, but do you have an example for the other two?

[tiny ridge]

The canonical example is the topologists's broom which deformation retracts to a point (which point?) but not strongly so. This is somewhere in Hatcher ch 0

I think a more useful example might be EG. For a general group G this does not strongly deformation retract to the 0-simplex given by the identity, but does so weakly.

Try G = Z/2 to see this in action

[marsh forge]

this requires a specific model for EG, no?

[tiny ridge]

Yes, correct. I was going to clarify I meant the Milnor model

[marsh forge]

gotcha

can you do better with a different model? Never really though about this

[tiny ridge]

No clue

Cool q

[grim knot]

guys I have a question on an exercise about seifert van kampen

[opaque scroll]

A deformation retract is a retract that is also a homotopy equivalence.

So the same relation should still hold

[marsh forge]

Oh wait it has to be a homotopy equivalence?

oops

i knew i should've double checked

[marsh forge]

go ahead and post

[grim knot]

The problem is that it is in german

I'm gonna rewrtite it on the ipad gimme a sec

[opaque scroll]

Yeah, you want ri to be the identity and ir to be homotopic to the identity

[marsh forge]

ah

I honestly don't know if I have ever seen a deformation retract come up since my first course in AT

[grim knot]

i_B is defined the other wa arounf, I did a mistake

I hope you can read my writing

in my solution they say that i_A(a)=a'2, but I do not see why

sorry it's in german, but I think it's more clear like this

[marsh forge]

Both i_A and i_B are homomorphisms out of Z, so that it suffices to pick a single generator and see where it goes. Moreover, pi_1(B) is the trivial group, so [fill in]. All that remains is to understand i_A. It might help to draw a picture

[grim knot]

I mean, I_B is just the constant function right?

[marsh forge]

yes

[grim knot]

I don't see why we need a square

[marsh forge]

?

[hidden crag]

Schick mal noch mehr davon

[grim knot]

this sentence confuses me

this is the solution

I understand it besides from why the inclusion is defined like that

[marsh forge]

right thats why i am suggesting the picture

draw a picture

is my answer to that last question

this is the type of thing thats super annoying to explain in words

and is incredibly obvious with the right picture

[opaque scroll]

I guess you wanna ask yourself how you know A and A\cap B are homotopy equivalent to a circle. That will give you the generators for the fundamental groups. Then just observe that the inclusion of A\cap B sends the generator to the square of the one in A.

[marsh forge]

u can use this picture if u want

[unreal stratus]

I bet it's quicker just to adapt the covering space argument to make it doable by hand

[opaque scroll]

Drawing the generators onto MaxJs picture should make it intuitively clear

[marsh forge]

this uses a different presentation of RP2 no?

the one thats S^2 / Z/2

[grim knot]

this is a mobius strip right?

[marsh forge]

no

its RP2

[unreal stratus]

Yeah

[grim knot]

kinda ashamed of it, but I don't know how to draw it, otherwise I had already done it

[marsh forge]

its this haha

The point is that B is the inner disk mod the relation (but the relation is trivial)

and A is the outer border mod the relation (which is no longer trivial)

Now, we will pick a clever representative of the identity element in pi_1 B

it will be the circle of radius slightly less than the border

this is of course contractible

But if we keep pushing this ring to the edge, we eventually get the full black circle

but because of the identification we made, the black circle actually traverses all the way around A twice

[grim knot]

so one question, it is true that pi(A n B) is generated only by one path since it is isomorphi to Z?

[marsh forge]

yes

[grim knot]

and the same holds for pi(A)

[marsh forge]

yeah

[grim knot]

OH MY GOD I GOT IT

It took so long

thank you to all of you who helped me through my slowness

[marsh forge]

np

[urban zinc]

wait this is so neat

[untold lily]

what's the book?

[urban zinc]

ghrist elementary applied topology

he doesn't prove it but I can check the references if you want

[untold lily]

I'll check them myself

interesting book

[unreal stratus]

configuration spaces r the best

[obtuse meteor]

This theorem cracked

[urban zinc]

yeahhh it's so cool

I've never thought about a manifold in that way

but it makes sense

[solemn oar]

If you could post the reference here that would be awesome.

[urban zinc]

tfw one of the references is Kaczynski T

[solemn oar]

Thank you!

[inner rune]

Hello, can anyone tell me if what I explained is correct? I still have doubts. that two continuous functions f and g from a topological space X to a topological space Y will only be considered homotopic if there is a continuous function H: X × [0,1] -> Y, where H(x,0) = f(x ) and H(x,1) = g(x). In this case, H is called a homotopy between f and g.

[solemn oar]

Indeed, this is correct.

[void tapir]

why aren't the inclusions into the free product of the fundamental groups labelled?
in the definition of U_ab do you not need to use that inclusion to view the homotopy classes as words in the free product instead of as homotopy classes in the fundamental groups?

[tough hamlet]

the inclusion is implied

they didn't think it was worth labelling

[void tapir]

how do you know if [a] is a word or a homotopy class then

by virtue of it being in U?

would it not make sense to make it so we know [a] is always a homotopy class and i([a]) is always a word

[tough hamlet]

perhaps but that would involve writing more stuff down

they chose to leave it up to context

[void tapir]

i see, thanks

[tough hamlet]

no tadder

[broken nacelle]

students whose ability to conceptualize geometric dimensions greater than three is not limited:

[abstract saffron]

I'm actually curious how ppl conceptualise higher dim 😄

I can, sort of, but I def can't explain how I do it

[high hill]

expand the product of spaces as defined

S1xS2 is like a copy of S2 at every point of S1.

ig i think about long line in the same way

[thorny agate]

this may be a dumb question but I've genuinely never run into these before (except topology texts assuming I already know what they are)

how do I prove some function f: U -> R^n is C^\infty?

For some functions sure it's easy, like e^x or sin(x) with easy patterns for repeated derivatives

but are there some other conditions?

[trail charm]

induction works i think

[thorny agate]

is that the general way?

or are there some nice conditions that I should keep in mind?

[trail charm]

i believe that is the general way

if you know that it’s analytic, then it’s C infinity (but not vice versa)

[thorny agate]

ah true

[abstract wigeon]

i want to verify that the following forms a topology on R:
the set, T, containing the empty set and R, as well as all half-bounded intervals of the form (a,inf) for some a in R

1. empty set and R are clearly in T

my question is this: for the arbitrary union axiom, can we say that the union of any collection of open sets, {U_i}, in T is either R (if one of the open sets is R), or (min{a_i},inf) for a_i in U_i

[novel acorn]

those are called rays and they do form the regular topology on R iirc

[abstract wigeon]

i just don't know if the min{a_i} breaks down because there are an arbitrary number of a_i's

im pretty sure the same argument works for the intersection property because there are only finitely many open sets to intersect

[novel acorn]

since you defined a_i to be in R I don't see a way this can become pathological

I think it's good tbh

[abstract wigeon]

maybe i am just confusing myself but iirc, the min function only works on a finite set of values

[novel acorn]

no?

I mean if you wanna be pedantic you can call if like the infinum of the set

[abstract wigeon]

oh so i am mistaken

[novel acorn]

Yeah it works in the infinite case
I think for the continuous case you'd usually write inf of a set but for like {1,2,3,...} you can def write like min of this set is 1

[abstract wigeon]

okay infimum is probably the more precise way to put it, accounting for the cont. case

thank you @novel acorn i appreciate it

[queen prism]

min means it’s included in the set

[novel acorn]

inf can also be included in the set

[thorny agate]

At the end

that should rank $(F \circ f \circ G^{-1})$ right?

[gentle osprey]

spamakin

[thorny agate]

not the inverse outside of the parentheses?

[gritty widget]

@thorny agate yes

[abstract wigeon]

this may be a dumb question but for the projective space P^1(R), why must all of it's elements be lines in R^2 that pass through the origin

is it because the line must contain 0

[queen prism]

straight lines through the origin are vector subspaces

[brittle rapids]

otherwise you don't have an equivalence relation

[abstract wigeon]

namely x ~ x' iff (x = x' or x = -x') ?

because one wishes to have points as equivalent iff they are antipodal on S1?

[brittle rapids]

x ~ x' iff (there exists nonzero λ in R such that x = λx')

[abstract wigeon]

okay thanks!

[brittle rapids]

i'm assuming the alternative is like, arbitrary lines

if you're asking why don't we have lines that pass through (1, 1) instead of (0, 0), then ya it's because we want subspaces

[abstract wigeon]

that's right

okay, i am still unclear as to why the line passing through (0,0) forms a subspace and a line not passing through (0,0) doesn't

[brittle rapids]

think about what it means to scale a vector in R^2

[abstract wigeon]

okay

[brittle rapids]

a subspace should contain all scaled copies of its members

so if x is in your subspace, the subspace should also contain the line that passes through x and 0

[abstract wigeon]

ohh okay that makes more sense

thank you

[grim knot]

Hey guys, is the real projective line homeomorphic to D1/~ where u~v : <=> u0v or u = -v for u,v in S0?

[tough hamlet]

well that's the circle

so actually yes

[grim knot]

but how can this then be homeomorphic to S1/~ where u~v : <=> u=v or u = -v for u,v in S1?

I don't see it

I see the homeomorphism to S1

but not to S1/~

does it hold thanks to the universal property?

[tough hamlet]

S^1/~ happens to be homeomorphic to S^1

this fails for n>1

a homeomorphism is induced by the square map z->z^2 on S^1

[grim knot]

okay I get it, thank you!

[tough hamlet]

np

[abstract saffron]

take a closed loop, and join two opposite points, you get a figure-8

lay one half of the figure-8 on the other

that's the equivalence. kinda clear that it's S^1

oh, nvm, I thought the question was between S1 and S1/~

[urban zinc]

For what positive integer values of n can S^n be expressed as the disjoint sum of unit circles?

I know the answer for odd n but even n above 2 are a mystery to me

I have no idea how to even begin conceptualizing it

[tough hamlet]

disjoint sum?

[umbral panther]

Only odd n. If you can partition a space into circles, it is a circle bundle over the space of circles. But the Euler characteristic is multiplicative, so the total space has Euler characteristic zero

[urban zinc]

Sorry disjoint union

[urban zinc]

Ty!!

I def would not have thought about that, I'll look those words up but I think that makes sense

[urban zinc]

rad af

[coarse night]

You meant like Hopf fibration?

[umbral panther]

I guess it’s possible to partition a space into circles without it being locally trivial. For example, partition S3 into circles, find a torus, a circle of circles, and switch which circles you use in that torus

There’s a more interesting situation in which you follow a path of circle and suddenly when you come to the limit, it’s a double cover. If it’s smooth, this is a manifold with a non vanishing vector field, so has Euler characteristic zero. I think it’s zero in general

[urban zinc]

Yeah

[coarse night]

just projective planes then no? (z1, .. zn) → [z1:z2:..:zn]

[umbral panther]

There are free circle actions on spheres with quotient not complex projective space. They are homotopy equivalent, but not diffeomorphic. Indeed, if you have a manifold homotopy equivalent to complex projective space, it has a circle bundle who’s total space is homotopy equivalent to a sphere

[grim knot]

how can the möbius strip be homotopy equivalent to a circle?

by the way, I'm still confusing deformation retract and strong deformation retract, so if somebody is still in the mood of helping me, I'd really appreciate it ❤️

[opaque scroll]

If you shrink the width of a Möbius strip down to 0, you get a circle. The circle is even a strong deformation retract.

[opaque scroll]

A strong deformation retract is when you fix a subspace in place, and the rest of the space deforms until everything is inside the subspace.

A deformation retract is similar, but the subspace doesn't have to be fixed in place as you deform. It can wiggle around as long as it ends up back where it started after deforming

Examples that are deformation retracts, but not strong deformation retracts are usually quite complicated

[grim knot]

oh this explanation is GOOD

[grim knot]

so for example, D2 is a strong deformation retract of R2 right?

[opaque scroll]

Yes, that's right

[grim knot]

what is an example of a deformation retract though?

[opaque scroll]

Here is one example
https://math.stackexchange.com/q/1841883/306319

(Also every strong deformation retract is in particular a deformation retract)

[grim knot]

are there a lot of deformation retracts which are not strong?

I might just learn them by heart

I think we only discuss about the most important cases, which mostly are strong defo retract

[opaque scroll]

I don't think there is any fixed list, but just knowing one example, or even just knowing that an example exists is probably sufficient.

Depends what you need it for I guess

[tawdry widget]

Is there discrete version, like the version on polytopes , of Gauss-Bonnet theorem? I know for a 2-manifolds we have a discrete analog called Descartes formula. But I wonder if there is a general analog for all even dimensions.

[tawdry widget]

Nvm, found an article “Descartes’s theorem in n dimensions” by Branko discussing it.

[arctic relic]

For showing that a vector bundle possesses a Euclidean metric implies that it’s isomorphic to the dual Hom bundle, is that basically Riesz

Or am I supposed to do w/o func analysis

[coarse night]

finite dimensional bundle right?

[arctic relic]

Yeah finite dimensional vs in the fibers

Vs = vector space

[coarse night]

you don't need full force of functional analysis that's just basic LA

[arctic relic]

Never took LA

I am so cool

[coarse night]

you don't need to have a LA background to use it

[arctic relic]

Yeah I took functional analysis but not LA so that’s why my first thought was Riesz vs something simpler

[coarse night]

v ↦⟨_, v⟩ ∈ V* show it's injection and use dimension argument

but Riesz is perfectly fine

it's not something anyone will point you out for

[arctic relic]

Yeah that makes sense

[unreal stratus]

Finite dim riesz

[queen prism]

wait how do you do functional without linear

[gritty widget]

you don't need to know how to solve a system of equations to apply the open mapping theorem / closed graph theorem / uniform boundedness theorem

[tough hamlet]

there's not much nontrivial stuff that carries over

[grim knot]

if we are looking for a deformation retract, do you agree with me that there might be a mistake in the definition of rho? shouldn the last line be for |x| > 2022=

the above is also a strong deformation retract right?

[eager herald]

yeah

[eager herald]

should be

[grim knot]

okay, thank youu

[rough cedar]

stupid question but

what is an explicit description of the weak topology

[ebon galleon]

Weak topology on what? I believe the term can refer to a variety of things

[rough cedar]

in the definition of weakly continuous

[umbral panther]

When topologists use it, they mean fine. When analysts use it, they mean coarse

[rough cedar]

what about algebraists

[ebon galleon]

Yeah that's what I had in mind lol, weak as in initial topology vs weak as in (say) CW complexes

[ebon galleon]

Not 100% sure, but from skimming the definitions online, it should be the weakest (coarsest) topology for which a map (or collection of maps with a common domain) is continuous. I think in certain contexts, there are additional requirements put on the topology (i.e. TVS topology)

[tough hamlet]

I'm pretty sure everyone means coarse

[grim knot]

if I use van Kampen on the Torus, I get that the fundamentalgroup os given by Z*Z /<<aba^-1b^-1>>, why is this isomorphic to ZxZ?

[rough cedar]

yeah but I'm asking about an explicit description

[ebon galleon]

The W in CW complex stands for weak topology. It's the finest topology for which the characteristic maps (disk inclusions) are continuous

So sometimes it can go either way

[ebon galleon]

Ah, that I'm not sure of, sorry 😅

[abstract saffron]

that literally means ab = ba, everything is commutative

[grim knot]

oh okay, I thought it wass something else

thank you

[hidden crag]

Check the Wikipedia article

They give a subbasis for it

Ah nvm that’s only for func Ana contexts

[tough hamlet]

topologists are fighting a losing battle

[hidden crag]

What do you need the explicit description for anyway

[marsh forge]

its the smallest topology containing all of the preimages of open sets of the maps you're taking the weak topology with respect to

sometimes this is easy to describe and other times maybe more annoying

[paper wedge]

its just the smallest topology such that the linear functionals are continous

[hidden crag]

…

[abstract saffron]

joke of you to think it's a battle 😄

[coarse night]

[rough cedar]

so true

[urban zinc]

Trying to grok cohomology is killing my soul

Too many duals

[marsh forge]

do more computations

[urban zinc]

Also, going back and doing some basic exercises for the fun of it: cone over Q works for this right?

Or even, cone over any not locally path connected space?

also wait, are equivalence and isomorphism synonymous?

Okay yknow what I'm just gonna try to do all the exercises in rotman, there aren't that many per chapter

[ebon galleon]

I think so, it seems like it should work

[ebon galleon]

I've never seen this used as a categorical definition of "equivalence", that's just isomorphism

[urban zinc]

Interessting, I guess Rotman's terminology is weird here

[ebon galleon]

There is a general notion of equivalence in 2-categories (i.e adjoint equivalence of categories, homotopy equivalence, etc)

But not this lol

[urban zinc]

Weird

He does define homotopy equivalence right below this as an equivalence in hTop

[marsh forge]

how does he even define hTop without defining homotopy equivalences

[ebon galleon]

Homotopy classes of maps as morphisms, then defining a h.e. to be an iso in hTop maybe?

Technically don't need to define h.e. to verify that is a category

[urban zinc]

yeah ^

[marsh forge]

fair enuf

[novel acorn]

does dense + compact imply closed?

here I'm talking abt some subset of a topological space

[unreal stratus]

{1} as a subset of {1,2} with indiscrete topology

[novel acorn]

hm okay

then it isn't clear what Neukirch is doing exactly

he shows that a map has compact and dense image and then concludes it's surjective

[ebon galleon]

Is the codomain Hausdorff?

That would imply the image is closed since it would be compact

[unreal stratus]

Yeah exactly

[novel acorn]

Ngl
I don't even being to have any idea
idek what topology the codomain has

[hidden crag]

send the problem

[unreal stratus]

Can u show the map

But I assume it is what has been said above lol

[novel acorn]

Here U^(1) is the set of principal units

first image is the general setup
Second image is the map itself

[hidden crag]

is this discrete for some reason

[ebon galleon]

Yeah not sure where topology comes in? Unless there's more context

[unreal stratus]

Ig we gotta look at Iwasawa lol

[hidden crag]

non hausdorff spaces aren't real anyway

[ebon galleon]

LMAO

I think it's safe to conclude whatever topology U^{(1)} has is Hausdorff though, that's pretty much the only way we could conclude that

[novel acorn]

I think it has the cofinite topology

I THINK

with a big I THINK

since we're working with local fields here

[hidden crag]

uuh

that'd be a problem

that would not be hausdorff

[ebon galleon]

Unless it's finite itself lol

[hidden crag]

unless it's finite

[ebon galleon]

But would compact sets be closed?

[novel acorn]

oh wait lol

me dumb

this is what U^(1) is so idk what topology it has

[hidden crag]

you made me think about the cofinite topology

i do not like that

[ebon galleon]

it's not real anyways

not Hausdorff

[hidden crag]

true true

[hidden crag]

no though

[ebon galleon]

didn't think so, but technically compact sets being closed is slightly weaker than Hausdorff

[novel acorn]

uhhhh

[hidden crag]

in the cofinite topology everything is compact

[novel acorn]

I think it might be cofinite

[hidden crag]

but only finite stuff is closed roughly

[novel acorn]

OH WAIT

NO IT'S NOT THE COFINITE TOPOLOGY

me dumb

[ebon galleon]

bruv

[novel acorn]

what is the name

PROFINITE

PROFINITE

that's it lmao

Me an big dumb lol

[hidden crag]

sounds close enough

that is indeed hausdorff

[ebon galleon]

Yeah

Subspace of a product of discrete (Hausdorff) spaces, right?

As a limit

[novel acorn]

WOOOO YEAH
good then now it make sense le tank u

[hidden crag]

so U^(1) mod U^(n+1) is finite right

[unreal stratus]

Profinite

[hidden crag]

otherwise the profinite thing doesn't really make sense or am i smoking

[novel acorn]

ye
it's of cardinality q^n where q is the cardinality of the residue class field

[hidden crag]

okok good

then it's resolved

[dusk heron]

Let $F:(\mathbb{C}\setminus{0})\times\mathbb{C}\to(\mathbb{C}\setminus{0})\times\mathbb{C}$ be the map $F(z_1,z_2)=(1/z_1,z_2)$. Is it true that if I glue a copy of $\mathbb{C}^2$ to another copy of $\mathbb{C}^2$ along the map $F$, then I obtain $\mathbb{C}P^2\setminus{[0:0:1]}$?

[gentle osprey]

gustavn64

[broken nacelle]

valid

[urban zinc]

This message gives me life

[urban zinc]

The fact that the intersection is path-connected isn't necessary, is it...?

[gritty widget]

yeah it's completely unnecessary

if you have points a in A and b in B, and even just one single point x common to A and B, then you can get a path from a to b by going first from a to x, and then x to b

[urban zinc]

Okay cool

also is this not just trivial from the fact that path-connected implies connected?

I don't see why the hint is necessary

[steel glen]

is this rotman still

how did he define lpc

[urban zinc]

Yeah

[steel glen]

is it just a neighborhood at each point that's pc

[urban zinc]

Oh I just realized V doesn't need to be path-connected

I guess that complicates things slightly

That's a weird definition

[steel glen]

i assume the definition of locally connected is similar, in which case i don't see why you need the hint either

[urban zinc]

Okay I mean the exercise is easy, but why does Rotman define it this way..?

Isn't it equivalent to having V be path connected

Oh he proves that it's equivalent

I guess he wanted to start with something weaker and build up

Still, so weird

[steel glen]

yeah i think that's what it is

[urban zinc]

Also am I crazy or do these have the same exact solution lol

Like literally the same formula

[steel glen]

basically the same thing yeah, but you ofc have the rest of R^n+1 to worry about in 1.29

[urban zinc]

So the exercise with more to worry about is first

This exercise design confuses me but I'll roll with it lol

[steel glen]

if you have the homeomorphisms between punctured spheres and R^n then that's probably why 1.29 is first

[gritty widget]

i can't believe they stabbed S^n. poor sphere...

[urban zinc]

Every continuous map is an injection and then a homotopy equivalence

Mapping cylinders my beloved

Also can we drop "is homotopic to" here? Aren't they literally equal?

i being an injection into the mapping cylinder of f and then r just collapsing the mapping cylinder into Y

[steel glen]

how would something like a constant map be equal to the composition of an injective map and a homotopy equivalence, granted the spaces are not simply connected

[empty grove]

Inject as the top face of the mapping cylinder of the map, then homotope the image to the bottom face of the cylinder

This is the general recipe for this factorization, not just for constant maps

[urban zinc]

If f : X -> Y is a constant map at y then let M_f be Y but attach a cone with the pointy end going down to y

i is injecting X at the top of the cone

r is sliding everything in the cone down to y

[steel glen]

ah that makes sense yeah

[urban zinc]

So like we can drop "is homotopic to" right?

I don't see why we need it

[empty grove]

Yeah

[broken nacelle]

Path connected doesn't imply connected hurb

[lime tendon]

Could someone help me in a topology exercice ?

Let X_1, X_2, X_3 be 3 connected spaces such that X_1 \cap X_2 is not empty and X_2 \cap X_3 is not empty.
Show that X_1 \cup X_2 \cup X_3 is connected

[lime tendon]

That's the exercise

[opaque scroll]

Write your space as the disjoint union of two open sets. Can you show that a connected subspace must be fully contained in one of them? Then Why can't X1, X2 and X3 be in different components?

[lime tendon]

One of the 2 open space could be X_1 and the other X_2 \cup X3. That's what you mean ?

[opaque scroll]

Not quite. Say the union of the three equals U\cup V for two disjoint open sets

First thing you should show is that X2 is either entirely contained in U or entirely contained in V

[wispy veldt]

alternatively you can use connected iff no proper clopen sets

[lime tendon]

Will we show that U or V is empty ?

[broken nacelle]

Oh wait it's the opposite lmfao

Mb

[opaque scroll]

That will be the eventual goal, yes.

[wispy veldt]

topologist sine curve my beloved

[lime tendon]

Thanks the fact that we know that X_1 \cap X_2 is not empty and X_2 \cap X_3 is not empty, X1, X2 and X3 can't be in different components

[lime tendon]

But I don't know how to show this

[opaque scroll]

Alright, how is your definition of connected framed?

[lime tendon]

A set is connected if it is not disconnected.
And a set X is disconnected if there exists 2 open sets U, V such that X = U \cup V and U \cap V = empty set

[opaque scroll]

So what do you get if you intersect X2 with U and V?

[lime tendon]

The intersection will not be empty

If X = U \cup V and X_2 is in X

[opaque scroll]

If you intersect U and V with X2, you get a decomposition like the one in the definition of disconnected

[lime tendon]

Yes ok, so 2 open sets

[opaque scroll]

And you know X2 is connected

[lime tendon]

But a connected set can contain 2 open sets 🤔

[ebon galleon]

If you haven't worked it out already, this follows from a slightly simpler statement by induction
Check that if X and Y are connected subspaces such that X \cap Y is not empty, the X \cup Y is connected

[echo oyster]

f : X → f (X) contained in Y, is a homeomorphism
g : Y→g (Y) contained in X, is a homeomorphism
But X & Y is not homeomorphic

X=R & Y=[0,1] works i think ?

[opaque scroll]

Yeah, that works.

[echo oyster]

Ok ty

[unreal stratus]

So like "CBS for topological spaces fails?"

[tough hamlet]

what's cbs

[red yoke]

Cantor Shröder Bernstein?

[hoary breach]

cauchy bunyakovsky schwarz?

[queen prism]

columbia broadcasting system?

[tough hamlet]

thanks besties

[robust fable]

Is f continuous because q and g are continuous and q isn't necessarily constant or am I just tripping?

from the wikipedia page of quotient topology spaces

uniqueness I get, but continuity I'm guessing is just inherited

[ebon galleon]

If U \subseteq Z is open, then g^{-1} U is open in X. By commutativity, this should be saturated w.r.t q I believe, which means qg^{-1} U = f^{-1}U is open in X/~

Or something like that I think?

[robust fable]

yeah that tracks. Thanks!

[urban zinc]

Lol I was about to be like wtf

[balmy field]

This was from a midterm who’s due date is now passed. I did the first one by pulling back to the cantor set and using real analysis techniques (that is, assuming p was a limit of things in the set, norms, and so on). Is there a nice way of doing it without this?

[ebon galleon]

That would be my first thought, show p_\infty is the only limit point of {p_n}

[raw cargo]

Agreed. What I did was that for any point x outside the set, there exist indices m < k such that x_m = 0 and x_k = 1, and a neighborhood U = U_1 X ... where U_m = {0} and U_k = {1}, which it does not intersect, hence x is not a limit point.

[raw cargo]

How do you prove it using the Cantor set?

[urban zinc]

Personally I would do it like this: for any point p not in that set (let's call it S), find an open neighborhood around p that doesn't intersect S

Finding that open neighborhood is not very difficult

[ebon galleon]

wait that should be much easier lol

[urban zinc]

You just need a way of describing the complement of S

[ebon galleon]

||Has a 0 before a 1||

[urban zinc]

Yeah but shh

Spoiler it

[ebon galleon]

my bad lmao

but yeah, should be easier than sequences (though that shouldn't be too terrible either)

[raw cargo]

Am I the only one who can see my message?

[cursive tendon]

let $x \in \mathbb{R}$, what property guarantees that there is some open interval $ (a, b) $ that contains $x$?

[gentle osprey]

Mattuwu

[tough hamlet]

this is true for all of them

[hoary breach]

that the open intervals are a topological basis?

[cursive tendon]

of course you can just construct something like (-|x|-1, |x| + 1)

I'm curious what's the bare minimum

[tough hamlet]

property of what

[cursive tendon]

idk, just an open question, I'm not familiar with topology 😂

the answer I want to hear is something like: in a xxx space, all points have an open neighborhood

[tough hamlet]

this is true in any topological space

the whole space is an open neighbourhood of every point

[cursive tendon]

and is it true that, in R, an open neighborhood V of some x must contain an open interval that also contains x?

[tough hamlet]

yes

[tough hamlet]

.

[cursive tendon]

interesting, thanks!

[tough hamlet]

np

[hoary breach]

any open set in R can be written as an at most countable union of open intervals with rational end points

[cursive tendon]

https://en.m.wikipedia.org/wiki/Base_(topology)
was reading this, is "base" a prefered terminology? it feels like it distinguishes itself from basis as in vector spaces

[next crystal]

Ive seen basis used more often

Normally theres no confusion from context

[hoary breach]

my book calls it topological basis but ig you can use just base, the idea between vector spaces and topo space is the same, want to express a larger space using a smaller collection either by a unique linear combination (in v.s.) or a union (in topo space). there are differences tho in terms of results, for e.g. in a v.s. all bases are equicardinal. in R, the euclidean topology is a (topo) basis for itself and is uncountable (same cardinality as R) while the set of all open intervals with rational end points is a countable (topo) basis

[unreal stratus]

Tbh I kinda wouldn't compare it to bases in algebra since the uniqueness of expression is usually a key part of bases in algebra

Idk I've never found it to be profitable

[ebon galleon]

I mean they're analogous in the sense of "generating" the structure

But in lin alg it's in terms of representation, for a topology in essentially a set that's "almost" a topology. In this sense it aligns more with something like filter bases

[balmy field]

Sorry for the late reply but here’s how I did it:

[balmy field]

This is a super nice argument

[severe dawn]

Hi guys. I need your help.

Recently I came across this space-filling surface. I have been trying to find a name for it or a rigorous way of describing it. I asked 2 professors at my university about this space-filling surface. One professor (a theoretical physicist) told me that he wasn't an expert in topology, and the other (a mathematician) told me that she wasn't an expert in fractals.

These images show a 3D representation of the surface, but the surface can be generalized for higher dimensions. You can interpret these 3D structures as slices of a 4D object. The surfaces may look 2D but, as you zoom in, you notice that every surface is made of several parallel "sub-surfaces" (similar to zooming into the Cantor set).

Do you guys know if there is a name for such structures?

[nimble portal]

shouldn’t e0 be reversed here

[ebon galleon]

no

well, the labeling seems odd in general

I would take this to be the labelling: [\begin{tikzcd}
& 1 \
0 && 2
\arrow["{e_2}", from=2-1, to=1-2]
\arrow["{e_0}", from=1-2, to=2-3]
\arrow["{e_1}"', from=2-1, to=2-3]
\end{tikzcd}]

[gentle osprey]

Ryxiann

[ebon galleon]

from which it's clear the labelling is just order of integers

[nimble portal]

what

that makes no sense to me

why would you not just label them in increasing order so it’s cyclic

[ebon galleon]

maybe this isn't exactly the same thing

[nimble portal]

This is my first time learning about this

no comprende

[ebon galleon]

Actually I think you're right, since it should make a cycle as indicated in the picture you sent, my bad. I don't know much (co)homology, I was thinking of the simplex we use for stuff like simplicial sets, in which the order is induced by a total order on the vertices

[hidden crag]

Can you send the context

(the text above and below)

But yes the induced orientation by integer ordering on the vertices would indeed reverse e_1 in Ryx picture

[hidden crag]

the same works for simplicial complexes

[ebon galleon]

Is there a nice proof that \overline{D} is continuous? Or is it really just something that you have to verify with the definition

Like I've been looking for a diagramatic proof that these conditions imply i:A -->X is a Hurewicz cofibration (so this proof can be abstracted), but I don't think \overline{D} can be induced diagramatically

And idk what else to use instead of \overline{D} otherwise lmao

[unreal stratus]

Should just be pasting lemma right

But yeah doesn't seem to have a nice categorical way of putting it but hm

[ebon galleon]

with what closed sets tho?

Like of course A is closed by A=\phi^{-1}(0), so A \times [0,1] would be a reasonable choice, but what else

[unreal stratus]

The set S of points where t <= phi(x) is closed right

Like preimage of a closed subset of I^2 under a continuous map

[ebon galleon]

Maybe? But it's t < phi(x), not <=
The only problem is when considering neighborhoods of points in A \times {0}, since at (a,0), we are taking D(a,1), and nearby for x \nin A, t=0, we are taking D(x,0)

So somehow relativity comes in

Regardless, I think there's no nice way to check it, but I think there's an algebraic way to get the result I want without this lemma

[unreal stratus]

doesn't matter

since 1 = t/φ(x) when t = φ(x)

that's what the pasting lemma is for anyway innit like show for t <= φ(x) and t >= φ(x) and then compatibility on the intersection i.e. where t = φ(x) is obvious

[ebon galleon]

t=\phi(x)=0 tho?

maybe I'm just being dense for a sec, but that still seems like a problem

[fading vale]

Phi(x) = 0 on A and D is a homotopy rel A so on A x I Dbar is definitionally the identity

[novel acorn]

the ordering is always by index of vertices

it always goes from higher to lower

so e0 goes from 2 to 0

[nimble portal]

I don’t understand

2 to 0? What is 2 and what is 0?

[novel acorn]

your vertices are indexed as v0 v1 and v2

[nimble portal]

Yes

[novel acorn]

so for the 0-simplices you'd write [v0], [v1] and [v2]

for the 1-simplices you'd write [v0v1], [v1v2] and [v0v2]

wait

ur book does something weird

is that v1 or v0 at the top???

bruh

I think they messed up lol

[ebon galleon]

https://math.uchicago.edu/~may/VIGRE/VIGRE2007/REUPapers/FINALFULL/Nadathur.pdf

pretty sure it's this

[novel acorn]

I am so confused

why does the ordering have the extra (-1)^i?

that isn't standard from how simplices are usually defined

[ebon galleon]

Is this standard for topological simplex as well?

[novel acorn]

yes
Books I've read usually define it like so

[ebon galleon]

okay that's what I thought, but that pdf confused me too lmao

[nimble portal]

I’ll just look at it from a proper topology book when I can, I was just curious to learn more about homology and wanted something quick and dirty so I just Googled it and picked the first pdf

[hidden crag]

Feather skipping steps again

…

[raw cargo]

Don't worry. I actually meant to respond to my other message, since the other two guys were discussing about a solution exactly to mine and I thought I had some problem with my Internet connection. This solution is pretty interesting, didn't know it could be done with analysis.

[nimble portal]

I’m on vacation I have an excuse this time 💀

[grim knot]

Hey guys, what are all the possible identifications of a torus?

[steel glen]

what do you mean

[grim knot]

I think there are multiple ways of defining it right?

[steel glen]

yeah, are you looking for a specific construction?

[grim knot]

I'm doing an exercise and they say: "we identify the Torus T2 by R^2/Z^2"

where does this come from? 😐

[steel glen]

have you seen the construction that takes a square and glues the top-bottom and left-right

[grim knot]

yes I did

is that isomorphic to the latter expression?

[steel glen]

well they are all homeomorphic by virtue of describing the same spaces

but yes that's basically what we're doing

a similar thing is S^1 as R/Z

[grim knot]

moreover, I would love to get a better intuition of what a covering really is

[steel glen]

what kind of covering

like a covering space?

[grim knot]

yes covering a space

[cursive tendon]

I was thinking of the following interesting claim.

Let $S$ be a subset of $\mathbb{R}$, let $x \in \mathbb{R}$,

Claim: $\exists (a, b) \ni x\ldotp (a,x) \cup (x,b) \subset S $ if and only if $x \in \operatorname{int} \operatorname{cl} S$

It seems straight forward to show. It's interesting because it's a equivalent definition of the interior of the closure(at least in the R case)

[gentle osprey]

Mattuwu

[steel glen]

do you know the definition?

[cursive tendon]

Does this remind anybody of any thing?

[grim knot]

yes I do, I have it in front of me

[steel glen]

i don't understand your claim

can you say it in words

[grim knot]

but I'd love to get more intution by an easy example

[steel glen]

do you know what the universal cover of S^1 is?

[grim knot]

nope I do not................

[cursive tendon]

context: I was reading the definition of the limit of a function, and realized one of the defintions actually works exactly for points in the closure of the interior of the domain.

[steel glen]

it's R

[grim knot]

which intution can I gain from this?

[steel glen]

the visual being R/Z, or looping R over and over again like this

[grim knot]

like I don't see the meaning of having a cover

[steel glen]

well you should think of having a bunch of "sheets" of your space

[grim knot]

oh this is beautifuk

[steel glen]

basically X has open neighborhoods which are locally homeomorphic to some open neighborhoods in Y

a collection of them

and the projection is like flattening the covering space Y down to X

[steel glen]

is the claim:

(there exists a set (a,b), x in (a,b), such that (a,x) u (x,b) is a subset of S) iff (x is in int cl S)

[cursive tendon]

yep!

[grim knot]

oh okay that makes sense

so for example the covering of S1 -> R flattens R to the unit circle(?)

[steel glen]

R is being flattened to the unit circle yes

[cursive tendon]

the claim seems correct (unless?? 😳 ), it's thus an equivalent description of cl int S given by the existence of punctured intervals. I wonder if there is an analogy in general

[grim knot]

okay, going to work on this, thank you

moreover I have an algebraic question I think(?), but is still in the topology exercise

why can't two space be isomorphic, if the quotient is not?

like Z x 4Z and 2Zx2Z are not isomorphic because the above are not

is there a theorem that implies this, or does it just follow but surjectivity of the projection

[hidden crag]

Context please

Are these homology groups

[grim knot]

you understand german

this is what I'm working with

[hidden crag]

Z/4Z and Z/2Z x Z/2Z are not iso because one has an element of order 4

And the other doesn’t

If that’s your question

[grim knot]

yes, I know that, but why does this imply that the other two spaces are not ?

I do not understand where the last three sentences come from

[steel glen]

which two other spaces

(ZxZ)/(2Z x 2Z) and (ZxZ)/(Z x 4Z)?

[hidden crag]

Im at a tattoo appointment

Can get back to it later

[grim knot]

I do not understand why the fact that those quotient spaces are not isomorphic implies that the isomorphism F can not exist

[steel glen]

what is the isomorphism between

in general, if G and H are not isomoprhic, then there's no isomoprhism between them

by definition

[grim knot]

do you see F in the picture I sent?

[steel glen]

if F is an isomorphism, then F restricted to a subgroup (and the codomain restricted to the image) is still an isomorphism
what would be the converse of that

[grim knot]

is the quotioent space a subgroup(?)

[steel glen]

the converse is that
if F restricted to a subgroup isn't an isomorphism, then F isn't an isomorphism

[grim knot]

I see that, but what I'm asking id if the Quotient space is a subgroup

It is a subset, but is it a subgroup(?)

[steel glen]

it isn't a subgroup

the operation is not the same

i see. what i said above is also not important here i think

i assume (i don't know german) that what they're saying is
suppose F is an automorphism of ZxZ, with F(Z x 4Z) = 2Z x 2Z

this would give (Z x Z)/(Z x 4Z) = Z/4Z, and (Z x Z)/(2Z x 2Z) = Z/2Z x Z/2Z

but in general, if F is an isomorphism of G (to some group G') with normal subgroup H, then G/H is isomorphic to F(G)/F(H)

for us that would mean
(ZxZ)/(Zx4Z) is isomorphic to (ZxZ)/(2Zx2Z)
but that's not the case

@grim knot does that make sense?

[grim knot]

Ah okay, I see that now, thank you @steel glen

sorry for taking so long to understand

I also wanted to ask, how does the number of covering spaces (up to isomorphy) depend on the fundamental group

or a different related question, which pieces of information do the fundamental groups give ?

[hidden crag]

Have a look at the correspondence theorem

Between subgroups of pi_1 up to conjugacy and and covering spaces

(For nice enough spaces)

[hidden crag]

It’s a homotopy invariant

If two spaces have different fundamental groups they are not homotopy equivalent

[hidden crag]

If you’ve seen universal coverings

[grim knot]

I'm looking for it in the notes

you do not mean this right?

[hidden crag]

No that’s just the lifting property

The correspondence is deeper and needs more assumptions

[grim knot]

it's this?

jeez I don't find anything

[hidden crag]

That’s part of it

But still not the big thing

[grim knot]

I do not have more ...

like the script finishes here

[hidden crag]

Will more be uploaded

Or does the course finish with that

[grim knot]

I finished the course

I'm preparing the exam haha

[hidden crag]

„normal divisor“

Lol

German moment

[grim knot]

HAHAHAHAHHAHA

Normalteiler be like

I found this on the internet

is it this?

[hidden crag]

Yes

That’s it

[grim knot]

I do not have this theorem in my notes

[hidden crag]

You can see this by looking at the universal covering that exists iff the space is [insert all the assumptions]

And modding out decktrafos

[grim knot]

dumb question

I do not have a definition of "universelle Überlagerung"

but they mention it in the solution of my exercise

what is this?

[hidden crag]

Eine einfach zusammenhängende Überlagerung

[grim knot]

ah okay top, thanks

[hidden crag]

Die ist dann schon eindeutig bis auf homöomorphie

That’s very weird though

[grim knot]

I feel completely high guys.............

Why the freak is the hawaiian earring not semilocally simply connected?

I see that if we look at 0 and take a neighborhood, we get that there exists a circle completely contained in our U. But why is this a contradiction? I don't see it

[novel acorn]

You can take a loop around the circle and that isn't nullhomotopic

[steel glen]

it’s connected, just not semilocally simply

[grim knot]

If we take a loop at 0 between the circle contained in U and the next circle, why can't I homotopically move it to 0?

I don't see it

Like I get that it is not, but why?

[steel glen]

is S^1 simply connected

[novel acorn]

Take a loop staring at 0
Go around the circle once
And back to 0
This loop isn't contractible

[grim knot]

emh nope

[steel glen]

then visualize this

[grim knot]

oh wait, I was looking at the space

without the circle

[arctic relic]

Why is there no way to define a cup product on homology? I think I have a vague idea but not completely sure. Namely the failure of chain groups to possess a natural evaluation.

[grim knot]

but the space is the union of the circles

jeez thx for the time guys

[novel acorn]

The cup product is defined using the chain sequence C(X) x C(X) -> C(X x X) -> C(X), where the first map is the Künneth map and the second map is the map indices from mapping X into the diagonal of X x X

[hidden crag]

There’s a MO thread on this

[novel acorn]

This can't define a product in homology since the maps of chains would go the other way

[hidden crag]

https://mathoverflow.net/questions/415/does-homology-have-a-coproduct/419#419

[arctic relic]

Yeah that makes sense thank you

[novel acorn]

The cup product is the coproduct on cohomology?
Interesting
What's the categorical way to view the cap product then?

[hidden crag]

I think they just mean the terminology

Product on cohomology, why no „co-product“ on dual structure

[urban zinc]

This might help

[unreal stratus]

Pretty cool when you have an H-space and get a Hopf algebra structure

[fading vale]

I think intuitively the cup product in cohomology is taking low dimensional information and splicing it together into something higher dimensional and theres not really going to be a canonical way of going down from n-simplices to k-simplices for k < n without losing information

this is captured in what eric sent: going from X x X (or X x Y) to X is necessarily going to lose some information

[umbral panther]

Of course it does, since H stands for Hopf

[opaque scroll]

Where does the antipode come from?

[umbral panther]

If the monoid of components is not a group, you don’t get an antipode. But if it is a group, in particular, if it is a connected, you do

[opaque scroll]

Hmmm, so is there like an explicit description using the H-space multiplication or is it more of a "it has to exist" type thing?

[tiny ridge]

The cohomology algebra carries the same information as the homology coalgebra

It's just a historical artifact that we are more used to product than coproduct

We don't write 5 = (1,4)+(2,3)+(3,2)+(4,1)

If we did we'd probably use the homology coalgebra primarily

[umbral panther]

Somewhere in between. There is a shearing map XxX -> XxX, x,y -> x,xy. If the monoid of components is a group, this is a homotopy equivalence, and its inverse can be used to define an up-to-homotopy inverse on X

But also, a graded bialgebra with degree zero just the constants has a unique antipode

[tiny ridge]

Here is an interesting problem. What kind of an algebra is H_*(LM)?

LM = free loop space of a manifold M

There is a fibration Omega M -> LM -> M where the projection map is evaluation at 1

The E2 page of the spectral sequence is H*(M; H*(Omega M))

Homology of Omega M is a Hopf algebra

Homology of a manifold has a product of its own, the intersection product

What is the synthesis of these 50 different products?

[umbral panther]

Have you read the literature on string topology starting from Chas and Sullivan? I think there is a complete answer to this question of how these products interact. This structure only uses the hypothesis to M is a Poincaré duality space. There is hope to use the manifold structure to build more structure

But it’s kind of a dead field, subsumed into symplectic topology. The lagrangian conjecture says (among other things) that the cotangent bundle detects the smooth structure. A piece of the theory of J-holomorphic curves captures everything in string topology, so rather than trying to define more structure on string topology, people study the full theory of J-holomorphic curves applied to the cotangent bundle.

[tiny ridge]

@umbral panther Yes, I am familiar with string topology

[umbral panther]

…but you don’t think it contains a complete answer?

[tiny ridge]

I didnt ask a genuine question, aka I did know the answer

It was more of a puzzle

[umbral panther]

Oh

[tiny ridge]

It is believed (Eliashberg mentions this in a talk), by the way, that J-holomorphic curve techniques may not be enough to detect all of the symplectic topology of cotangent bundles

There has been some fantastic results, for instance by Abouzaid. But its nowhere close to a definitive answer

[umbral panther]

Yeah, it’s not clear how much they should do, it’s just that they subsumed string topology and killed it off. It’s clear that the next place to look for more structure is there

[tiny ridge]

True

What do you actually work on, @umbral panther? You seem to know a decent amount of everything

[umbral panther]

I am spread too thin

[tiny ridge]

Welcome to the club.

[unreal stratus]

What are homotopy quotients in general oop I can't see to find much on them

Like say actions on a manifold

[languid patrol]

They are the correct kinds of quotients from the homotopy theoretic perspective when the group action is not well behaved. Concretely if you have a group G acting on X then (X/G)^h is (X \times EG)/G where G acts diagonally

[unreal stratus]

Sure

Yeah I've seen that much but hm do you know of anywhere nice to learn more

Like I imagine there is a more universal way to think of them too?

As a like homotopy coequaliswe i guess actually would make sense

[languid patrol]

It depends on what you want to know, the point is to construct a space Y so that Y is homotopy equivalent to X but Y \to Y/G is a fibration

[unreal stratus]

Sure

[languid patrol]

So there is a motivation from model category theory

[unreal stratus]

Oh and so if G is for example a covering space action then this is automatic but more generally we need a suitably derived notion ig

[languid patrol]

You can’t expect any good motivation for any particular choice because homotopy quotients are not really spaces as they are homotopy colimits

So they have a very involved universal property

[unreal stratus]

Oh I mean that's why i asked about universal stuff aha ye

[languid patrol]

By particular choice I mean particular space which represents the homotopy quotient

[unreal stratus]

Yeah

[novel acorn]

For risk of sounding like a dumbass is this construction in any way related to moduli spaces

[languid patrol]

You might be best off just reading a little bit about (infinity, 1) categories if you want to understand how they fit into a broader category theoretic framework. Or you could read quillen on model categories for a more old fashioned viewpoint. Or you could read a primer on equivariant cohomology if you want to see homotopy quotients in action

[unreal stratus]

Oh I mean sure I have read about model cats and a bit of infty cats but they didn't come up in that lol

[languid patrol]

@novel acorn very loosely they are connected

[unreal stratus]

But I'll search through those more to see if they have anything specific for thos

Thank

[novel acorn]

I attended a talk Abt certain constructions in moduli spaces that had to do with chow quotients and this sounded vaguely reminiscent of stuff they did there

Tbh the presenter wasn't the best at explaining exactly what was going on so I only have snippets of the talk in my head

[languid patrol]

@novel acorn homotopy quotients definitely can arise to represent certain functors on the category of spaces. But it’s a very general construction

[unreal stratus]

uwu

[ebon galleon]

If you don't mind me asking, what's a good resource to learn about the infinity category-theoretic stuff? I know at least basics of model categories and am curious to learn more along these lines

[novel acorn]

Some people will say HTT (I am one of those people)

[unreal stratus]

Look at HTT, you can read it in an afternoon

[languid patrol]

@ebon galleon there really isn’t one

The Kerodon project is coming along but is still in its infancy

[ebon galleon]

it's 900 pages lmaooo 💀

not quite an afternoon

[languid patrol]

But it aims to be the stacks project for higher category theory

*Kerodon

[novel acorn]

It's comprehensive tho

Crafted by the all knowing Lurie in the ancient forges of the IAS

[languid patrol]

Higher topos theory is good but it’s a bit like reading EGA

[ebon galleon]

ofc ofc, I might read HTT, I was just curious other sources

And it is by Lurie 👀 which is based

[languid patrol]

To read one part of it you have to read another, etc, etc.

[novel acorn]

Have you read EGA lol

Idk french unfortunately :(

[languid patrol]

I read a lot of SGA and EGA when I was younger and had worse taste

[novel acorn]

Lol

[languid patrol]

I do think there was a time in the 90’s when it really did give you a competitive advantage over other people

[novel acorn]

I am in that phase rn I feel like

[languid patrol]

Like Kisin just decided to read SGA because he didn’t have a thesis problem and it went very well

[unreal stratus]

So tteg must be 50 I guess then

jk

[novel acorn]

She read EGA when she was 5 surely

[languid patrol]

If you read older papers from the 70’s-80’s you will definitely end up peeping some EGA. But in general the stacks project has much of the good stuff now

[novel acorn]

Neukirch likes to cite EGA in that arithmetic geo book of his and so I was curious to read it

[languid patrol]

Yes I think people did that for a long time

Just because there wasn’t anything else

[cosmic socket]

Is HTT/HA similar in modern times?

[languid patrol]

When certain people were doing math

@cosmic socket hopefully! I think it will take time for people to fully digest HTT but it’s getting there

[cosmic socket]

There doesn’t seem to be any alternative source on the modern perspective on spectra

Or related stuff

[unreal stratus]

[novel acorn]

Uh oh

Him

[unreal stratus]

Btw bigradedsphere lol what is ur nick from

Or is it like nonsensical

[novel acorn]

Urs started doing some arithmetic topology recently I've heard

[unreal stratus]

Arithmetic top sounds inch resting

[novel acorn]

A friend I know went down that rabbit hole recently
We're probably gonna do a project Abt that soonish?

[unreal stratus]

Btw irony r u like phd student

or end of bach or smth

idk

[empty grove]

Multigraded spheres are like the collection {S^n} but the n is now in ℤ^d

[novel acorn]

I am first year UG

[empty grove]

And S^(i,0,...,0) = S^i in the usual sense

[novel acorn]

[cosmic socket]

Motivic spheres

[unreal stratus]

Damn

[novel acorn]

I'm supposed to be a second year but uh I didn't finish my first year at uni

Something about depression?
So I transferred

I was at UCSD but now I'm EU based

[unreal stratus]

Aw im sorry sure

[alpine oriole]

EU masterrace

[unreal stratus]

eu is pog

Hi bobles

[novel acorn]

I'm at like a uni that does a shitton of Langlands stuff

[alpine oriole]

yo

[unreal stratus]

me too

I think

[novel acorn]

I'm in the nowhere country of the Uni of Zagreb lol

[unreal stratus]

Yeah maybe not the best thing to call masterrace

[tough hamlet]

lmao

[alpine oriole]

i managed to avoid all algebra in my 3rd year

[unreal stratus]

How

[ebon galleon]

Unfortunate

[unreal stratus]

I had 8 modules and like 7 were algebra

Or partly so

[novel acorn]

The way to go

[ebon galleon]

I've got a pretty good split next year lol

[alpine oriole]

analysis >>>

[tough hamlet]

what was the other one

[novel acorn]

We need an analysis out emote

[tough hamlet]

which

[unreal stratus]

Logic

[novel acorn]

Logic is for dumbasses just use ur head

[unreal stratus]

But then the final had a question on vector spaces over a finite field

So lol

[alpine oriole]

3rd year analysis for us is just functional analysis

[novel acorn]

Prove the axiom of choice

[tough hamlet]

fym "for us"

[unreal stratus]

That's set theory

[tough hamlet]

do you not know who I am

[alpine oriole]

as in at my uni

[unreal stratus]

Lol

[alpine oriole]

logic seems boring to me ngl

[novel acorn]

Logic is cool
Set theory is super boring to me tho

[alpine oriole]

then again i havent looked that far into it

[cursive tendon]

is there a concise description of this set in terms of S?

https://media.discordapp.net/attachments/844681108473118750/1136085832382566480/336361466376617986.png

where S is a subset of the reals

it seems this set is a subset of int cl S, but int cl S is not always this set, for example, take S = Q, int cl S = R but there are no intervals in S

[ebon galleon]

That's just int S I think

Unless you mean x \in R on the left

[cursive tendon]

oh sorry

you are right, edited

[opaque scroll]

It's pretty close to being the closure of the interior

[cursive tendon]

https://media.discordapp.net/attachments/844681108473118750/1136087835955449988/336361466376617986.png

[ebon galleon]

It should be this:
$$
{x \in \mathbb{R} \mid x \in \text{int} (S \cup {x})}
$$

[gentle osprey]

Ryxiann

[opaque scroll]

I believe it should be the interior of the closure of the interior

[cursive tendon]

yeah for example when S is (0,42) U (42, 100), it's exactly int cl S

[ebon galleon]

since the condition could be refined to \exists (a,b) with x\in (a,b) and (a,b) \subseteq (S \cup {x})

[opaque scroll]

Not quite I guess, if S is something horrible like R\{±1/n, 0}, then 0 is in the interior of the closure of the interior, but doesn't satisfy your definition.

[ebon galleon]

should at least be a subset of int cl int S

[cursive tendon]

$$ \bigcup_{x\in\mathbb{R}} \operatorname{int} (S \cup {x})$$

[gentle osprey]

Mattuwu

[ebon galleon]

I think that works?

[cursive tendon]

how about
$$ \operatorname{int} S \cup \operatorname{iso} S^c $$

[gentle osprey]

Mattuwu

[cursive tendon]

interior of S and isolated points of S complement

[runic sun]

That should work also

[cursive tendon]

that's an interesting construction. Does it have a name? It seems like a nice nontrovial example of a measure zero set

[steel glen]

that set has infinite measure
{+-1/n, 0} is countable

[ebon galleon]

I imagine he meant {±1/n, 0} lol

[tidal lynx]

If S is an infinite subset of [0, 1] then does there exist a convergent sequence consisting of distinct elements of S?

this has a nice proof

and I just came up with the problem too!

though does anyone have an alternate proof that doesn’t reference directly the ||nested-interval property of the reals|| (though obviously you have to appeal to completeness somehow)

[viral atlas]

Pick an increasing sequence in S, use Bolzano-Weierstrass to obtain a desired convergent sequence from it

[ebon galleon]

It's infinite --> there's a sequence of distinct points. Therefore, there's a convergent subsequence by compactness

[tidal lynx]

Nice, that’s even cleaner

[ebon galleon]

Yeah it's compactness you need, not (just) completeness

[ebon galleon]

By the way, this is true in any infinite subset of a compact metric space, if you know much about those. The same argument generalizes. In fact, it's even equivalent to compactness!

[steel glen]

do you even need compactness?

it's sufficient but it's definitely not necessary

[ebon galleon]

Yes, if you want the limit point to be in the set

For instance, {1/n} in (0,1)

[steel glen]

ok i guess we implicitly want the limit point in S then

that's on me lol, that's obvious now that i say it

[ebon galleon]

If you just want a limit point, then boundedness is sufficient in any locally compact metric space (I think?)

[steel glen]

this is what i had in mind

[tidal lynx]

Wait I’m being dumb why does (1) imply (2); it’s not like the set of elements in a sequence of reals is necessarily closed? :
(1) a subset of the reals is closed and bounded subset iff it is sequentially compact
(2) every bounded sequence of reals has a convergent subsequence

these are the two main (equivalent) formulations of bolzano weierstrass

[steel glen]

the union of the elements in the sequence is not necessarily closed, but just take the closure
then you have a closed, bounded set which by (1) is sequentially compact, so the sequence must converge

[tidal lynx]

oh

that feels like cheating

but it works!

[ebon galleon]

Lmao

[steel glen]

just sprinkle in the proof that clS for bounded S is bounded, then it shouldn't be as cheat-y

[tidal lynx]

No I meant how we fattened up our sequence so that we could apply Bolzano-Weierstrass