point-set-topology

[novel acorn]

something wrong with my beain

[ebon galleon]

it happens

[novel acorn]

anyhway illum for topological groups if your subgroup is open then the quotient G/H is discrete iirc

but I want to hear you explain to me why Z_p ain't discrete :3

[rough cedar]

which is true

[ebon galleon]

no

[rough cedar]

wait nvm Zp isn't finite lol

[ebon galleon]

Z_p is not discrete

[novel acorn]

Z_p is infinite and compact

[ebon galleon]

It's totally disconnected, however

[rough cedar]

ah fuck

[novel acorn]

so {0} is only closed in Z_p :P

[ebon galleon]

And by translation, all other points are only closed as well, not open

[rough cedar]

I need to show that given a subgroup of Zp with finite index it's closed

I was tryna use contains nbhd of 0 => closed

[languid patrol]

that seems like a good proof

[novel acorn]

don't you mean open?

[languid patrol]

it will be both

and showing one is equivalent to the other for finite index subgroups

[novel acorn]

hmmmm
Because I only know the result that shows that a subgroup of a compact topological group G is open iff G/H is finite
Ohhhh but every open subgroup is also closed

okay now it makes sense

[rough cedar]

what nbhd do we use

[abstract saffron]

a concise course in algebraic topology

[languid patrol]

This doesn't seem to be true

I think for instance you have \prod_{i = 1}^{\infty} \mathbb{Z}/2 which is compact with the product topology and has infinitely many index 2 subgroups which are not closed, and thus not open.

[umbral panther]

A closed subgroup of a compact group is open iff G/H is finite. But there are profinite groups with finite index subgroups that are not closed. Not profinite groups that contain a finitely generated dense subgroup (aka reasonable profinite groups). But take the infinite product of Z/2. This is a vector space and the axiom of choice says that it has unreasonable finite index subgroups

[languid patrol]

it sounds better when you say it

[rough cedar]

[novel acorn]

Hmm
I mean the reference I used was Ramakrishnan Valenza prop 1.4 (iv)

[haughty yew]

where do you get that

[languid patrol]

probably it proves the statement that bw mentioned, the assumption of "closed" is crucial

[novel acorn]

it doesn't assume the subgroup is closed

[languid patrol]

then it is wrong

[novel acorn]

well I wouldn't be surprised it has a bunch of errors

[languid patrol]

or perhaps there is some kind of Noetherianness assumption implicit in the theorem

[abstract saffron]

it's a joke 😄 a concise course in algebraic topology is actually a book, by JP May. But it's actually pretty hard and not concise at all. Maybe complete tho.

[languid patrol]

yes it is a false statement then

[haughty yew]

so you do need that first?

[abstract saffron]

I wouldn't recommend Hatcher, I don't like it, but a lot of ppl do. Just go with it, there's no prereq maybe except Algebra and Topology (duh)

[abstract saffron]

No, not at all. Def not.

[ebon galleon]

Hatcher chapter 0

[haughty yew]

what would you recommend then?

[languid patrol]

What do subgroups of Z_p look like?

[novel acorn]

the only problem is that I can't seem the find where the error in their proof is

[abstract saffron]

Actually A concise course might not be a bad book. I haven't gone through it, but I haven't heard any bad rap. I like Munkres, but that's because I only care about homology.

[hidden crag]

Concise course is not a bad book but bad for a first pass

[abstract saffron]

My prof in Diff Geo recommended Bredon

It's good, but might be a bit heavy. You need to be really solid in Topo and Alg.

[abstract saffron]

So is Hatcher, tbh 😄

[languid patrol]

post proof

[pseudo ocean]

Wait

Is {0} a manifold?

A 0-manifold right?

[languid patrol]

the unique simply connected 0-manifold 0.0

[abstract saffron]

If you want a taste of it first, I recommend Nakahara

Chapter 2, 3, 4

[pseudo ocean]

Why is it called "unique simply connected"?

[pseudo ocean]

Cuz it's just 0?

[ebon galleon]

A 0-manifold should just be a discrete space, right?

So if you have 2 or more points, it's not simply connected. And there's only one topology on a single point

[pseudo ocean]

Yeah that's what im wondering

Why is it classified as a simply connected space

Also speaking of simply connectedness

[ebon galleon]

The singleton is simply connected (in fact, all homotopy groups are trivial)

[pseudo ocean]

Is it purely a global characteristic of a top space or can it be defined locally?

[hidden crag]

Because it’s both simply connected and unique 👍

[unreal stratus]

Global

[pseudo ocean]

Ah makes sense

[unreal stratus]

Think of a circle for example

[tough hamlet]

finite doesn't imply discrete without something else (like Hausdorff)

[pseudo ocean]

Not simply connected, yes?

[unreal stratus]

You can cover it by two contractible opens

But it ain't simply connected

[ebon galleon]

T_1 is sufficient even

[unreal stratus]

Like you need "the whole thing" to tell

[pseudo ocean]

"Opens"? Open sets?

[tough hamlet]

the quotient will be Hausdorff if H is closed for example but not in general

[unreal stratus]

Well this is basically tautological

Actually nah

Just in my mind a space X is T1 iff topology contains cofinite one

Which is true

And then with finite sets that has a further corollary lol

But ye

Topology is cute

[ebon galleon]

Well, it's mostly tautological, just observing that everything T_1 and above coincide. Unsure if sober implies discrete for finite tho

I don't believe so

No it doesn't, Sierpinski is sober

[unreal stratus]

Was gonna say lol Sierpinski may help

My beloved

[languid patrol]

🤯

[ebon galleon]

I know, fascinating

[feral copper]

But: is the singleton contractible?

[tough hamlet]

I would guess it just wants discrete+finite anyway

maybe that's just what people meant by finite back then

[languid patrol]

Too advanced for this server

[unreal stratus]

Is the empty set contractible

Find out more at 8 !

[queen prism]

the answer may surprise you

[languid patrol]

it must have non trivial \pi_0 because it's disconnected... right?

[unreal stratus]

Rethink what it means to be trivial

[pseudo ocean]

like you? ayup!

[unreal stratus]

[ebon galleon]

I mean they're not even weakly equivalent so...

[rough cedar]

I have no idea, I only know that the only closed groups are 0 and p^mZ_p, of which the p^mZ_p are also open

[languid patrol]

Okay what do the finite quotients of Z_p look like?

[rough cedar]

I wanna show that they're all cyclic

[languid patrol]

So what happens to the image of 1?

[obtuse meteor]

locally simply connected is an important condition

[grim knot]

guys pray for me for tomorrows topology exam, I hope to survive it

[rough cedar]

under quotening?

[obtuse meteor]

(altho really the "correct" condition is semi-locally simply-connected, which makes me wanna throw up)

[languid patrol]

Almost as important as semi locally simply connected!

A famous Farb quote is “every space has a universal cover” I have heard

[obtuse meteor]

lel

if it doesn't you don't have the right space :)

[rough cedar]

image of 1 becomes a generator no?

[languid patrol]

So first of all why is every finite quotient of Z_p a p group?

Everything follows from this

[rough cedar]

does it have something to do with when we have like e.g. (0, 0, 1, ....) in the subgroup then we can get any valid (0, 0, x, ...)?

[languid patrol]

It’s simpler

Don’t need to be so ad hoc

[rough cedar]

no idea, any hint?

[languid patrol]

So what finite abelian groups are there?

[rough cedar]

structure theorem?

[languid patrol]

And what is the p-adic valuation of n when n is coprime to p?

[rough cedar]

0

uh

[languid patrol]

Okay so what does that say about n in Z_p?

[rough cedar]

it's a unit

[languid patrol]

Okay so what about in any quotient? Can it not be a unit?

[rough cedar]

its inverse in the quotient will just be the image of its inverse, no?

phi(a)phi(a') = phi(aa') = phi(1) = 1

[languid patrol]

Yes so can the quotient have any factors of the form Z/n?

[rough cedar]

wait, how does unit even make sense here? are we viewing it naturally as a ring now or?

[languid patrol]

As a Z module

[rough cedar]

right

[languid patrol]

Aka an abelian group

[rough cedar]

wait I still don't quite understand what we mean by unit here lmao

there's a scalar a st ax = 1?

[languid patrol]

Ahhhhhhhhhhh

So just that n acts invertibly on Z_p

[rough cedar]

what does invertibly mean

[languid patrol]

It is an invertible map of abelian groups

[rough cedar]

I'm not sure how to make sense of that. if we were looking at it as a ring I would've said that 0 can't be a unit

[honest narwhal]

If n has an inverse m

Oh wait

[languid patrol]

It is the padic integers

[honest narwhal]

You actually don't use cursed notation

Thank god

Actually more general

Let's say you have a ring R

And u is a unit in R

Can you show that multiplication by u is invertible?

[rough cedar]

yeah the inverse is just multiplication by its inverse

[honest narwhal]

That's what acting invertibly means. You act by "multiplication by u"

[rough cedar]

oh

I always get confused when people don't specify the action

[honest narwhal]

Often when they don't it's because there's only one "natural" action in sight

Often your first guess will be correct, and this is a skill you hone

[languid patrol]

So Dami the confusion is coming from the fact that we’re trying to consider whether or not there are finite index subgroups of Z_p which are not closed

As an intermediary step we were showing that there are no prime to p finite quotients of Z_p as an abelian group

[honest narwhal]

I guess if you know the subgroup is also open

[languid patrol]

So multiplication by n as a Z module has an inverse but it is not in Z

[honest narwhal]

Then it's just union of cosets

[languid patrol]

Yes but we don’t know that

[honest narwhal]

So presumably we're considering a "not necessarily open subgroup"?

[rough cedar]

group action as a module?

[grim knot]

what are some common examples of first countable (but not second) and second countable spaces?

[languid patrol]

Don’t think of a group action, just think of the map n as being some abstract map

But every map of abelian groups is equivariant for multiplication by n

[languid patrol]

First countable but not second is a discrete uncountable space

Second countable is a one point space, or any manifold, or R^n

[honest narwhal]

Metric spaces are auto first countable, and second countability is equivalent to separability fwiw

If that helps you root the ideas a bit in your head

[rough cedar]

but how do we consider that as a Z module

[honest narwhal]

Every abelian group is a Z-module

[rough cedar]

yes but isn't it an abstract map and not an abelian group

[honest narwhal]

The point of that was

We're not using yet the group structure on Z (I belive, I'm jsut reading scattered bits)

So the map "multiplication by n" makes sense for every abelian group

[rough cedar]

you lost me here #point-set-topology message

with the unit

[honest narwhal]

So one way to see it is this

n in Z_p

It's a unit right?

Now if H is a subgroup of Z_p, I form the quotient Z_p/H

[languid patrol]

Then nH = H because H is a subgroup, so n is injective on a set of representatives for Z_p/H so multiplication by n is bijective on the quotient

This means that by fto fgab the quotient is a p group since n is arbitrary and prime to p

But that means that there is some p^n which annihilates the quotient which means that p^n Z_p is contained in H

So that means H is open

This is what I was going for @rough cedar

[steel glen]

show S^p * S^q = S^{p+q+1}
i was going to do this by showing the suspension of S^n is S^{n+1} and just using associativity, like a successor function kind of thing
is there a better way to do it?

X * Y is the join of X and Y

[languid patrol]

That’s the best way to do it

[unreal stratus]

That's the easiest way imo yes

[languid patrol]

Just show associativity and win

[steel glen]

ok thanks

[honest narwhal]

Oh yeah I stepped out for food but TTEG stepped in

[unreal stratus]

i think it's slightly more convenient to reindex and write it as S^{n-1} * S^{m-1} = S^{n+m-1} but yeah

Idk

[unreal stratus]

Any nice references for the Thom isomorphism theorem in the case of not-necessarily-oriented bundles?

In particular, I was looking at the isomoprhism $\tilde H_(W^+; \mathbb Q) \simeq \Sigma^d \tilde H_(M^+; \mathbb Q^{w_1})$ where $\mathbb Q^{w_1}$ is the orientation local system, $(-)^+$ denotes one point compactification and $W$ is a $d$-dimensional vector bundle over $M$

[gentle osprey]

potato

[unreal stratus]

In the orientable case this just collapses down to normal stuff

[umbral panther]

My answer to all questions is Bott and Tu

[unreal stratus]

So true

does bott-tu cover sheaves

I really should've just read this a year ago lol

[umbral panther]

I don’t think it covers general sheaves, but it does cover local systems

[unreal stratus]

Ah lol it just states general thom iso at the end of chapter 7

Still, good idea lol

I think Spanier has some proof of smth similar actually, so that answers it ig

[umbral panther]

There exist orientable vector bundles X, Y such that W+X = det(W)+Y. The orientable Thom iso shows that the Thom spaces of W and det W have isomorphic homology, reducing the question to line bundles

[unreal stratus]

Very nice

[nimble portal]

[unreal stratus]

Btw, how does one get that first statement?

[umbral panther]

This is called K-theory

Easier to prove the stronger statement that for any V, W there exists X trivial, dim about that of M and Y uncontrolled so that V+X=W+Y. Just take a random map from V+X to W and it will be surjective. Let Y be the kernel. In this example, you have to prove it’s orientable

[unreal stratus]

How do you prove it using K-theory, interesting

[unreal stratus]

and thank you sure hm

[umbral panther]

I meant this is the first lemma of K-theory. You can define K-theory as formal differences of vector bundles, but each one has a representative as a a vector bundle minus a trivial bundle

[unreal stratus]

Yes agreed

but the orientability doesn't seem to come in there

I assume it's smth like w1 of E and det(E) agree?

[umbral panther]

Yes, orientation is determinant. Determinant turns sum to tensor product

[unreal stratus]

Sure dope

I am too used to the complex case where I dind't have to worry about orientability lol

[haughty yew]

https://cds.cern.ch/record/640926/files/0750306068_TOC.pdf this book?

[abstract saffron]

Yeah, this one

[pseudo ocean]

How? Almost all manifolds are locally simply connected afaik

That isnt particularly helpful in categorizing them

[obtuse meteor]

Yes and sort of

[pseudo ocean]

For example the torus and sphere are locally simply connected but they have differing homology groups

[obtuse meteor]

The universal cover of a space is very useful for classifying it

[pseudo ocean]

How does that connect with simple connectedness?

[obtuse meteor]

And universal covers are constructed using the fact that you’re semi locally simply connected

[pseudo ocean]

Agh—

[grim knot]

What is the Fundamentalgruppe of the empty set?

[tough hamlet]

doesn't have one

[grim knot]

Okay so just to make sure, when I use Seifert va kampen I need to make sure that the intersection of the two sets is not empty, right?

[tough hamlet]

yes and path connected

[grim knot]

I just noticed that we always take our basepoint in A n B, which implies that it can't be empty

[tough hamlet]

yeah

[hidden crag]

On the contrary

The universal cover is very helpful in classifying coverings of a space and (semi) local simple connectedness is one condition for it to exist

So a lot of spaces having that property is better

[fading vale]

It's not that property P distinguishes spaces it's that you need property P to define objects to distinguish spaces in the first place

[cosmic comet]

Please help with 11.17, I suppose that 11.16 should be used, but don't know how.

[pseudo ocean]

I think a sketch would start with constructing a homeomorphism from R to I

Then have another homeomorphism that goes from the empty vertices out to the infinite legs of the cross

Then combine em into a map that is a homeomorphism of the entire space

Buuut the universal covers thing implies you start out from universal covers over the cross and square then work the homeomorphism between those universal covers

@languid patrol help

[languid patrol]

What is the question?

[pseudo ocean]

Question right above me

[pseudo ocean]

This^

[tall mason]

Hi, I'm trying to solve a hatcher question and would love a hint because I'm a bit stuck
the question is 1.3.9 - Show that if a path-connected, locally path-connected space X has pi_1(X) finite, then every map X to S^1 is nullhomotopic.

now, we let f: X -> S^1 and p: R to S^1 is the cover
now since X is path connected, locally path connected, the lifting criterion states that a lift f~: X to R exists iff f_*(pi_1(X))) is a subset of p_*(pi_1(R))
since R is a universal cover of S^1, p_* is the zero map and so this theorem would imply that a lift exists iff f_* is the zero map

as that means f o g is homotopic to the constant loop in S^1 for all loops g in X, then f is nullhomotopic

however, I'm not sure how to prove that this lift exists/the image of f_* is contained in the image of p_*
I haven't used any properties of S^1, or more importantly the fact that pi_1(X) is finite anywhere - does anyone have a hint as to where I should use these two pieces of information

[languid patrol]

The point is: if there exists a nontrivial loop which maps to a nontrivial loop in S^1 under f_* then you are done by what you’ve already written

Otherwise the image of f_* is trivial so you can lift the map f to R, so it’s null homotopic

[tall mason]

im so sorry - i dont really understand this
if f_* maps a nontrivial loop to a nontrivial loop, no lift exists - but where would I go next? I dont see how what I wrote deals with this case

[languid patrol]

Yes but that’s impossible, do you see why f_* has to have trivial image?

You have to use the assumption about X

[tall mason]

im afraid not haha

right sure

ok i will think about this 🤔

[languid patrol]

So f_* is a homomorphism of groups

Which groups?

[tall mason]

pi_1(X) and the integers
so we're mapping a finite group into the integers

hmm

[languid patrol]

How many finite subgroups of the integers are there?

[tall mason]

oh

hahah

none right?

they're all nZ

wait imsilly

[languid patrol]

Only the trivial subgroup

Since every element of Z has infinite order there is no map from a finite group to Z

[tall mason]

ohh of course

wow

thats so cool

tysm

[pseudo ocean]

How are yall so good at topology i suck ass at it 😭

[ebon galleon]

Practice

[prisma garnet]

dude wtf is hatcher saying here?

[hidden crag]

More context

[prisma garnet]

one sec lel

[unreal stratus]

Van kampen

I think I've seen people ask about this specific step online

So hopefully you will find more complete stuff

[prisma garnet]

this is in the proof of SvK, F is a homotopy, A_alpha are path connected open sets of our whole space, X (basically the sets in the assumptions of SvK) and R_ij are defined like this

I think that's all the context needed?

I don't want to post the whole proof lol

[prisma garnet]

first of all, why is a neighborhood of R_ij mapped to A_ij?

I \times I is a neighborhood of R_ij, no?

[unreal stratus]

A small enough neighbourhood

[prisma garnet]

oh it must be this image

[steel glen]

is IxI not the whole domain, why would that be a neighborhood of R_ij

[prisma garnet]

this is stupid lol

[prisma garnet]

it is wdym

F is a homotopy between two paths

[steel glen]

oh sorry, i mixed up what you were saying

[prisma garnet]

I see

[obtuse meteor]

This is (essentially) a consequence of R_{ij} being compact and A_{ij} being open

[obtuse meteor]

I believe this should be "is mapped entirely into a single A_alpha." Aka only one A_alpha satisfies F(R_{ij}) is a subset of A_alpha. Otherwise this seems bunk

[prisma garnet]

what if F is a homotopy between two loops in A_alpha n A_beta?

[obtuse meteor]

Ah sure that’s fine sorry

I misread the original phrasing in a different way :p

[prisma garnet]

I'm starting to understand why people hate hatcher

yo @eager herald

can you help me out rq?

[prisma garnet]

the sentence just proceeding the highlighted one is straight up just nonsense btw

there's no mention of \Phi in Lemma 1.15

[eager herald]

dude im ngl i forgot what he does here

i wroked it out before

and i fucking forgot

i hate my life

and hatcher

but i remember that being complete bullshit

[prisma garnet]

lmfaooo

[prisma garnet]

I stg there's a lot of "complete bullshit" in this one proof

[eager herald]

fr...

[broken nacelle]

ig I'll ask a proper question after my workout

[queen prism]

but what about asking a proper question while working out

[broken nacelle]

On my phone?

[red yoke]

Lemma 1.15:

Every loop is homotopic to a product of loops each of which is contained in a single Aα.
Elements in the Grp coproduct of the Aα's are finite strings of elements in the Aα's, and saying a loop is path-homotopic to another is the same as saying that they are equal in the fundamental group.

I think Hatcher likes to assume people have good algebraic / geometric intuition

I find having read Aluffi quite helpful for reading Hatcher

[prisma garnet]

ikr!

it's like the perfect algebra reference for hatcher. especially the homological algebra subsection from the chapter 3 read right before homology in hatcher is just perfect

[prisma garnet]

altho I'm not sure which question this remark is answering

[red yoke]

Explaining the first sentence

[prisma garnet]

what I'm confused by is where in lemma 1.15 is \Phi mentioned

[red yoke]

It's not, it's a reformulated statement

[prisma garnet]

ohh

[red yoke]

The highlighted sentence basically means there are multiple ways to do this

E.g. if you take the union of two intersecting squares and draw a loop in the region of intersection, you can consider it a loop solely in the first square or as a loop solely in the second square

[umbral panther]

Spaces don’t have fundamental groups. Based spaces have fundamental groups. Every (path) component has a different group. The empty space has no component, so no group. Even on a single component, the fundamental group depends on the base point and paths between them. They’re isomorphic, but not canonically isomorphic. Different paths give conjugate isomorphisms. The right object is the fundamental groupoid

[pseudo ocean]

Yeah do math while on the treadmill 😎

[pseudo ocean]

Cant I just pick a point in a space and use that as my base point?

[ebon galleon]

That's what is meant by based spaces: a specified point

There is no point you can specify in the empty space

[umbral panther]

If the space is not connected, the point matters a lot. If the space is connected, you can, and people often do say take any point, it doesn’t matter. But it does matter and you can make mistakes this way. The different fundamental groups are not the same but isomorphic. Not canonically isomorphic, but isomorphic up to conjugation. If the group is commutative, there is no conjugation, but if it isn’t, this is real indeterminacy. Sometimes the isomorphism matters.

Sometimes the base point matters. Consider a fiber bundle, like the Hopf fibration. Think about the base, the fiber, and the total space. Think about the fundamental group of the fiber. In some sense all the fibers are the same, so that all have the same fundamental group, but what does same mean? You can’t choose a base point for every fiber simultaneously

[next crystal]

isnt it that base point doesnt matter if the space is path connected? but if its just connected it could matter

[umbral panther]

By connected I mean path connected

[pseudo ocean]

Speaking of non-path connected spaces, how do we deal with em?

[nimble portal]

This was such a good explanation

Thank you

[next crystal]

we dont

[cerulean oriole]

If X, Y are topological spaces with X discrete, is the compact-open topology on Y^X (the space of *all * functions from X to Y) the same as the product topology?

[quick bough]

consider the path components

idk hahah

[cerulean oriole]

And if X is indiscrete, is it the same as the box topology?

These are more of “checking my work” questions, really.

[quick bough]

that is true, yeah

[pseudo ocean]

Suppose i want to deal with em.

[lime sable]

you deal with each path component separately

and if your space is weird then you don't do algebraic topology with it

[languid patrol]

What do we do with R^2 - Q^2? This is path connected and I want to know it’s fundamental group

[lime sable]

then it's weird

i don't like weird things

😦

everybody is always talking about "nice" topological spaces, what about weird ones

wait i forgot that the word "pathological" exists

too many syllables

[tough hamlet]

if I had to guess I'd say free goup on continuum many generators lol

could be much worse tho

[languid patrol]

That would work for R^2 - Z^2! Every loop in this one contains infinitely many holes though

[prisma garnet]

I can't believe this is path connected hurb

[tough hamlet]

move along axis directions

[prisma garnet]

yea yea

I figured out why, I still can't believe it

[tough hamlet]

in fact R^2-(R-Q)^2 is also path connected

[pseudo ocean]

Gimme a weird topological space we dont do alg top with

[prisma garnet]

the real line with the topology generated by half open intervals

... I hope

[ebon galleon]

Pick your favorite infinite cardinal \aleph, and give it the co-\kappa topology for \kappa < \aleph, e.g. R with cofinite or cocountable

[high hill]

free group with Q^2 generators right

[high hill]

wait nahhhh what

[prisma garnet]

wait why

I would've thought only countably many generators would do

[high hill]

hecks the answer

[high hill]

oh I see the problem, you are only allowed to consider finite combinations of generators. I 'want' to consider infinite combinations (only). and a few more restrictions

There's a bijection between loops and open sets in R^2 I uh. think.

aaa not quite

I'm not even sure if you're allowed a fractal of finite length . Don't see why not.

[prisma garnet]

I can see how this deformation retraction work but it just feel too...

unrigorous

[unreal stratus]

welcome to hatcher

[prisma garnet]

is there a way to make it more rigorous?

and should I be looking for a rigorous explanation for why the deformation retraction exists in the first place?

[quiet thorn]

Well if you really want to do it rigorously, you write out the map but often that can be a pain

[prisma garnet]

yea no thx lol

[high hill]

rigor 😒

[prisma garnet]

shuri gettin' americanized

[languid patrol]

The usual yoga is that in your first course someone makes you write them out really concretely then you earn the right to just hand wave later

[quiet thorn]

Usually the problem isn't too bad though

[prisma garnet]

holy shit one of the examples in SvK is longer than its proof

[high hill]

is this true

loops can be fractals?

in R^2 say

[languid patrol]

The loops can be as bad as you want really

[high hill]

fairly certain u can construct map explicitly

[languid patrol]

Probably can be space filling curves

[high hill]

what, but isnt infinite length an issue

[languid patrol]

Not really

[high hill]

oh

[languid patrol]

Pi_1 is just about continuous maps

There are no rules

[high hill]

nvm yh i didnt realise. . .

Then regarding RR - QQ...

[abstract saffron]

koch's snowflake go vroom

[high hill]

well I feel certain that you cant loop around a hole

without also looping around holes in its neighbourhood

I was trying to think if that was a possibility

[abstract saffron]

You can, but thinking about it will break your mind

[high hill]

hmm so bijection with open sets idea doesnt rlly work

[high hill]

wat

[abstract saffron]

One nice thing about Q is we can construct it. We know exactly where the points are

[prisma garnet]

[abstract saffron]

So for example, you take Stern-Brocot tree, but consider only the first n levels, and consider all points with coordinates in this set

As it's finite, you can define a loop avoiding all those points

Now there is some more work to do, but if my intuition is correct, you can take the limit of this process

[high hill]

an idea like this?

[abstract saffron]

Similar to how you define Hilbert's curve

[high hill]

i shall read . . .

[abstract saffron]

the thing is Stern-Brocot tree is semi-fractal 😄 (it's not a fractal, but the process generating it is)

So there should be something similar to generate a family of curves avoiding them

Hence taking the limit of this family should be possible

actually this can be a good topo exercise 😄

not too hard, but definitely blow students' minds

Now the question is, must such a curve be of infinite length?

[high hill]

i now have no clue how we might compute this...

it isnt open is it

[tough hamlet]

no but it's not closed either

[abstract saffron]

I have a feeling that this group is simple

Then we classification-of-simple-group the shit out of it

Oh but that classfication is for finite simple groups :sadge:

[languid patrol]

What classification of simple groups are you imagining?

[high hill]

i see a mo thread and i cant make a head or tail out of it https://mathoverflow.net/questions/67549/fundamental-group-of-r2-q2

[languid patrol]

Because it’s uncountable

[prisma garnet]

what's uncountable?

the number of holes?

[obtuse meteor]

I have no intuition of this sort

[high hill]

The problem is you can have an infinite number of holes in a loop

that is not possible for RR - ZZ

so freegroup doesnt work

[tough hamlet]

bro recieved a message from above

[prisma garnet]

yea, I was talking about R^2-Z^2

[high hill]

freegroup - you can only have finite strings.

[prisma garnet]

why is the free group on uncountably many generators when there are only countably many holes

[obtuse meteor]

Basically they’re skirting the question for a good reason

[high hill]

wait this should be doable with countable many right

[obtuse meteor]

It’s free on countably many yes

This was just misspoke in the conversation

[obtuse meteor]

The point is this space is so bad bc it doesn’t have a nice local property which guarantees a universal cover. Universal covers are part of why fundamental groups are good

So instead you might want to consider a structure which controls the possible covering spaces (similarly to a universal cover)

The constructed pro-group in the post does this

[prisma garnet]

oh aight

[high hill]

ic

why maf hard

[prisma garnet]

no u

[high hill]

back to being a finitist...

[prisma garnet]

so uh

a combinatorialist? 🤢

[quiet thorn]

Classification of finite simple groups

[prisma garnet]

oh yea finite group theory is a thing

[quiet thorn]

"Finite group theory is a dead field, there's no one working in it that isn't over 60" - paraphrased from one of my profs

[prisma garnet]

LOLLL

wew doesn't exist then ig

[obtuse meteor]

But you should carefully consider why there are uncountably many for R^2 \ Q^2

And what’s markedly different for this

[prisma garnet]

oh that exercise is right on the section I'm reading now lel

[obtuse meteor]

For a devious hint: ||As an example if we take the standard embedding of R^2 into S^2 and then look at S^2\ Z^2 then that has uncountable many||

Also try to prove R^2 \ Z^2 only needs countably many generators. It’s hard

[umbral panther]

||I think it might be more useful to say that the Hawaiian earrings embed in that space ||

[obtuse meteor]

Sure

[obtuse meteor]

Giving away the game / why I like this hint for bw ||the point is that in R^2 \ Z^2 if you try to make a combination of infinitely many generators they can’t form an actual loop, because they go off to infty. If you Compactify first then you’re in business||

[broken nacelle]

Wait really? I thought since every loop is contained in a bounded ball around the origin we can ignore the holes outside the ball hence the loop is an element of a free group with finitely many generators

Which is subset the free group with a generator for each hole

Doesn't that work?

[ebon galleon]

the bounded ball depends on the loop though. There's not single ball that all of the loops would be in, since they can have arbitrarily large (yet still individually bounded) image

[languid patrol]

It’s not free on countably many generators, the group is uncountable

[obtuse meteor]

Oh no but… hm no I think this argument works

[broken nacelle]

The idea is that the fundamental group of any ball minus the finitely many holes would be subset the group generated by loops going around each hole once

[obtuse meteor]

Take any loop gamma : [0,1] -> R^2 \ Z^2 with base point (pi,pi). It’s image is compact. Hence bounded. Hence can be broken as a finite product of loops in the bounded part of R^2 \ A (A a bounded subset of Z^2)

[broken nacelle]

Even R^2-Z^2?

[obtuse meteor]

So every gamma splits as a finite product of the generators

[broken nacelle]

Yea exactly

[languid patrol]

No that one Faye is right about, and you can prove it by writing it as a limit of compact balls with punctures

[obtuse meteor]

This was what I was replying to TTEG :p

[languid patrol]

Ooooh then Faye is right about both 🙂

[obtuse meteor]

For Q^2 you can’t do this trick

[tough hamlet]

do we actually know what the group is for Q^2

[obtuse meteor]

I’ve never seen this example before it’s fun

[obtuse meteor]

Press x to doubt

[languid patrol]

It depends on what you mean by “know”

[obtuse meteor]

I mean that kinda depends what you mean by “know a group” probably

Hahaha

Sniped

[languid patrol]

You can describe it in terms of more down to earth objects

But it’s not going to have some concise answer

[tough hamlet]

any group that warrants this response I don't wanna know

[ebon galleon]

I may have misinterpreted what you meant, lol, sorry

[obtuse meteor]

It’s okay bc all spaces have universal covers :)

So this never happens :)

:)))

[broken nacelle]

Analyst moment

[tiny ridge]

Let X(m, n) denote the fundamental group of the complement of (1/m Z) x (1/n Z) in R^2. Note pi_1 X(m, n) is a free group on the set (1/m Z) x (1/n Z). If m1 divides m2, and n1 divides n2 then there is an inclusion map X(m1, n1) -> X(m2, n2) which induces a map on the fundamental group (throw in "roots" of the generators). The fundamental group of the complement of Q^2 maps into the projective limit of this system. This should be an embedding of groups, and the image should entail infinite words in which the same letter does not appear infinitely often

Similar to the fundamental group of the Hawaiian earring as a subgroup of the projective limit of F_n's

Much worse actually. It's possible to have an infinite word with no letter appearing infinitely often which is not contained in the image (think of a sequence of loops which limit to a loop)

It's probably easy to write down a condition and thus also the group explicitly. It won't be useful in any sense.

[languid patrol]

I read the stack exchange post too Ibsen!

[tiny ridge]

Which post?

[umbral panther]

Isn’t there a theorem that every subset of the plane has free fundamental group?

[broken nacelle]

[tiny ridge]

Not true

Hawaiian earring

I remember something of this sort though, by Cannon

Probably some more hypothesis is required

[languid patrol]

It is locally free probably (any finitely generated subgroup is free), perhaps that is what you’re thinking?

[tiny ridge]

That sounds much more believable

The first shape group is an inverse limit of free groups.

At least for one dimensional continua

And I think \pi_1 objects into \pi_1^Shape

Injects lol

[red yoke]

I think it works if every point on the plane has an open neighborhood whose intersection with the subspace is path-connected and simply-connected

[languid patrol]

This is what I was thinking as well

[tiny ridge]

Jeremy Brazas would know on the spot

[red yoke]

Cuz then its homotopy equivalent to a graph

[tiny ridge]

Right, Arki. Seems like then you can just take a cover by simply connected fellows and make a nerve

Simply connected open subsets of R^2 are contractible so this nerve is homotopy equivalent to the full space by the nerve theorem

[red yoke]

[tiny ridge]

Wait but is it true that simply connected not necessarily open subsets of R^2 are also contractible

I think this is one of those basic questions I never learnt how to do

Oh I guess this follows from some sort of Hurewicz?

H_n of nice enough planar subsets is 0, probably by Alexander duality

[red yoke]

No

[tiny ridge]

Example?

[red yoke]

Like a comb but the lines bunch up on one side

[tiny ridge]

Ah ok I meant all higher homotopy groups are 0

That's enough for nerve theorem to go through weakly

[languid patrol]

Seems quite hard to prove that theyre weakly contractible even

[tiny ridge]

True for locally contractible compacts by Alexander duality though yes?

• Hurewicz

[languid patrol]

Yes for sure

[tiny ridge]

Yeah no clue how to do it generally

[languid patrol]

But the general question was about simply connected planar sets

[umbral panther]

This is special to 2d. The version of the Hawaiian earrings with 2d spheres as a subset of R^3 has homology in high degrees, unlike its shape

Whereas subsets of R^2 have vanishing higher homotopy groups, just like their shape, the shape being the approximation by reasonable spaces

[languid patrol]

http://at.yorku.ca/i/d/e/b/15.pdf

I assume this is not the simplest proof but it looks like the most amusing

[tiny ridge]

The last page is like Solomon's Key

Sigil into the other side

[umbral panther]

Rumor has it that the simplest proof uses the monotone light factorization

[coarse night]

[nimble portal]

How do you know if a space is based?

[tough hamlet]

a based space is just a pair (X,x) where X is a space and x is a point of X

the fundamental group of (X,x) is defined using loops in X based at x

so based isn't an adjective that applies to spaces so much as a based space is a space with a specified base point

[unreal stratus]

Let $X$ be a $G$-space (in this case the action is free too). It's true that (on singular chains) $C_(X_G) \cong C_(X)_G$, where on each side I take coinvariants, right?

[gentle osprey]

potato

[unreal stratus]

It looks like this should be obviously the case but I'm not very used to equivariant stuff lol so I may be wrong

[gritty widget]

would u guys know by any chance if there is course material i could find from a university on algebraic topology?

[unreal stratus]

But because stuff is defined in terms of maps into X I assume G is compatible with taking stuff

[unreal stratus]

Depends on what you want but:
https://people.maths.ox.ac.uk/ritter/algebraic-topology.html Oxford 2020 lecture notes + problems [(co)homology mostly]
https://www.dpmms.cam.ac.uk/study/II/AlgebraicTopology/ - Cambridge problems (mostly fundamental group)
https://www.dpmms.cam.ac.uk/study/III/Algebraic Topology/index.html Cambridge (more (co)homology

[nimble portal]

Ah ok LOL

[nimble portal]

this is awesome tysm!!

when you say (co)homology you mean homology + cohomology right?

[unreal stratus]

I do indeed

[tiny ridge]

What is coinvariants? Is X_G the Borel construction?

In any case C((X x_G EG) is chain equivalent to (C(X) o_ZG C(EG))_G if that is what you have in mind

[unreal stratus]

I just mean the normal like X/(g.x ~ x)

But in fact yeah X is just a smash product and the symmtric group acts by permutation so it's closer to what you mean

Specifically I wondered if, say, $[C_(X) \otimes C_(X)]{\Sigma_2} \simeq C*( X \times_{\Sigma_2} X)$ and stuff like that, using Eilenberg-Zilber and that fact that (afaik) homotopy coinvariants and normal coinvariants should coincide due to freeness of the action lol

[gentle osprey]

potato

[unreal stratus]

And the key step was what I mentioned above

i.e. $C_(X \times_{\Sigma_2} X})$ vs $C_(X \times X)_{\Sigma_2}$

[gentle osprey]

potato
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[tiny ridge]

Ah gotcha. Yeah if G acts freely on X then C(X/G) is isomorphic to C(X)/I_G C(X) where I_G is the augmentation ideal of ZG

Chain equivalent*

[gritty widget]

thanks i really appreciate it

[unreal stratus]

np

[gritty widget]

are there similar notes for different courses?

[unreal stratus]

Hm

where is this from btw?

Interesting

[tiny ridge]

There is certainly a chain map C(X) -> C(X/G). This is surjective because given a simplex Delta^n -> X/G you can pullback the G bundle X -> X/G over the simplex, trivialize, and take a section to get a simplex Delta^n -> X lying over the original. The kernel consists of...

A linear combination of simplices map to 0 if they all cancel out after projection. Which means you can collect the terms so that each collection maps to the same simplex, and their coefficients cancel out

Which means that the linear combination was in I_G C(X)

It's not even upto chain homotopy, the complexes are literally isomorphic

[unreal stratus]

Pog

Interesting

[tiny ridge]

Weirdly tautological

[unreal stratus]

Though now I'm wondering if modding out by augmentation is related to taking coinvariants

[tiny ridge]

I think that's essentially what co invariant means right

If M is a ZG module, M_G is just M/I_G M

[unreal stratus]

Yes

Nice!

So that does answer my question then thank you

Though I've a feeling it might be more direct to say like

Hm nvm

[tiny ridge]

As a corollary of this you have a very cool description of Borel equivariant homology. Just take C(X), tensor with some acyclic G-complex over ZG, then take co invariants

Cohomology of this chain complex is Borel equivariant cohomology

[unreal stratus]

Nice, so you can do the stuff on the algebraic side which feels a lil cleaner

[tiny ridge]

This is very useful if X is a manifold

[unreal stratus]

Here X is a manifold :)

[tiny ridge]

Oh great. There is an extremely concrete acyclic G-complex for a Lie group G, just C[g^*], with zero differentials. g being the Lie algebra

You could tensor this with the de Rham complex of X

Unwrapping the tensor product to a double complex, you then have \Lambda^i T^*X o Sym^j g^*

But this is cohomology so you have to take G invariants

Borel cohomology is modelled by (\Lambda^i T*X o Sym^j g^*)^G where G acts diagonally - on the first factor by extending the smooth action on X tensorially to forms and on the second factor by ad

So you get equivariant de Rham theory

Which is quite useful in many setups.

Has anyone ever written down Witten's SUSY Morse theory in the equivariant setup?

[unreal stratus]

Interesting thank you hm

[bitter smelt]

What does it mean for a space E to carry a stable hamiltonian structure induced by a fibration E -> S^1

How's it induced

(fibers are a surface)

(E is 3 dim)

[tiny ridge]

Probably the foliation associated to the stable Ham structure is the fibers?

[broken nacelle]

what's the fundamental group of the fundamental group of the fundamental group?

is a functor a topological space?

[unreal stratus]

Depends

Simplicial sets here we come

[broken nacelle]

wait what

[ebon galleon]

Simplicial sets my beloved

This is a modern view on homotopy theory. They're two different model category that are Quillen equivalent (i.e. they give the same homotopy theory). So for many homotopical purposes, one might as well just look at simplicial sets instead of spaces since they're """the same"""

[umbral panther]

The fundamental groupoid is idempotent

[brittle rapids]

:

[pseudo ocean]

Im a bit lost when it comes to Homology

How do you actually get those abelian groups in the chain sequence?

Is there a procedure for it?

[unreal stratus]

Do you mean in the exact sequence?

Or in the actual chain complex

[pseudo ocean]

Yeah i believe so

I mean arent chain and cochain (which generate the homology and cohomology) exact sequences?

[unreal stratus]

No

Do you know what it means for a complex to be exact

[pseudo ocean]

OH WAIT

ugh

Clumsy ass notation from wikipedia

[unreal stratus]

Oh lol

[pseudo ocean]

Either way

I still dont know a solid way of computing those abelian groups

[unreal stratus]

Anyway basically the chains and cochains are a nightmare and massive and you aren't really gonna do much with them directly

But with the actual (co)homology groups

[lime sable]

you usually do not directly work with singular homology

[unreal stratus]

You'll use theorems like Mayer Vietoris, excision, etc

[pseudo ocean]

Then...what

[unreal stratus]

You use formal properties

[pseudo ocean]

To actually find them?

[unreal stratus]

Like the ones I gave above

And use easy examples and work up from there

[lime sable]

you do something like simplicial or cellular homology if you want to do all the computations directly, and using the tools potato mentioned

[pseudo ocean]

Hn

[bitter smelt]

huh?

The sHs is what I'm trying to get at, not the fibration

[coarse night]

exercise

[tiny ridge]

What's your definition of sH?

[coarse night]

fundamental group of fundamental group of π_1

[tiny ridge]

It involves a pair consisting of a 1-form and a 2-form. The foliation given by the fibers should be given by the kernel of the 1-form

More precisely, any fibration M^3 -> S^1 can be equipped with a fiberwise area 2-form \omega. Let \theta be the volume form on S^1 pulled back to M.

These define a stable Hamiltonian structure

[silver umbra]

currently struggling to prove that the mobius band with its boundary circle collapsed is homeomorphic to RP^2

i guess for the purposes of the problem it suffices to show homotopy equivalence

nvm, got it

[coarse night]

draw the diagram

[unreal stratus]

Yeah it's pretty cute lol

[raw cargo]

This is not exactly about about topology, but something related to a topic that was mentioned in this channel.

#point-set-topology message

So, in the Wikipedia page, it says that "There is no known simple formula to compute T(n) for arbitrary n"

Does this mean someone has come up with a formula which is too long or an algorithm of time complexity too large?

It really is confusing because if there is such a formula, I can't find it.

[opaque scroll]

I mean it's easy to write down an algorithm like take all possible subsets of P(X) and check if it's a topology.

I guess this will have complexity something like 2^2^2^n

[raw cargo]

But like, where does someone draw the line. Is an exponential solution simple enough?

[opaque scroll]

I mean often the benchmark for "efficient algorithm" is polynomial time. But what exactly is meant by "simple formula" isn't always so precise.

[raw cargo]

Yeah, like the references don't seem to provide any such formula or algorithm as far as I've seen.

[opaque scroll]

Right, but the statement was that no such formula is known, so then it would be hard to provide I guess

[raw cargo]

I didn't word that correctly. What I meant to say was that there is no example to draw that line.

But yeah

I will try to read the references in case I find something useful anyways

[unreal stratus]

I mean it'd be nice to just have some simple formula like we do for other combinatorial things

but ye simple is vague i guess

This makes me want to write a program to do smth lol

[tribal palm]

eek the topology course i've signed up for had a 100% failing rate last year

[abstract saffron]

if there's an algorithm, there should be a formula. You can engineer it into a formula. Doesn't mean it'd be insightful, or simple, or even practical.

It's like, we do have formulae to compute n-th prime and pi(n). Doesn't mean they're practical or insightful.

[broken nacelle]

jesus christ

GL!

[unreal stratus]

At least you can't get a below average grade

[raw cargo]

I suppose so. It just feels weird to say that a formula has some time complexity.

[onyx raft]

When you glue two finite CW complexes along some boundary, you have a CW complex still right?

e.g a standard thing to do is you take RP^2 and you glue it along a circle (e.g the one lifting to the equator in S^2) to some other circle like the boundary circle of a mobius strip and then you are asked to probe at it with algebraic tools

Since you have two spaces you'll want to use van-kampen/mayer-vietoris type results

[unreal stratus]

what do you mean by "some boundary"

[onyx raft]

Hmm that's a good question

[unreal stratus]

You could glue them in some perverse way, like glue [0,1] and [0,1] in a horrible way to get smth non-Hausdorff

[onyx raft]

Ahh and of course that's not CW

So nice gluings that are checked by hand is probably what I need to do

[unreal stratus]

I guess in the example you give though you're gluing along (part of) the 1 skeletons which is much nicer but yeah

[onyx raft]

Yeah, I don't think I've really seen anything worse during these styles of problems

I just want the existence of the N_epsilon's that hatcher proves exists in the appendix for CW complexes

[umbral panther]

Here’s a mistake you can make with the fundamental group if you don’t pay attention to base points. Homotopic maps can induce different homomorphisms

You can have two maps f,g: X -> Y with f(x)=g(x) so they both map the fundamental group based at x to the fundamental group based at f(x) so you can compare them and see that they are different homomorphism. But there can be a homotopy between f and g. There can’t be a homotopy that holds the base point constant, because then they would induce the same map on fundamental group

[winged viper]

Does anyone know if there is a way to do this problem without using the lefschetz fixed point theorem?

I’ve been trying to prove that f has no fixed points => f is not null homotopic (since the assumption in the problem implies f is null homotopic by lifting to the universal cover)

[obtuse meteor]

I think you can do this by carefully choosing your universal cover

Cover by the hyperbolic plane in the Poincaré disk model. Any map of the disk to itself has a fixed point if you include the boundary. f being nullhomotopic should tell you something abt that

(This is my suspicion in terms of how I’d do this without lefschetz)

[winged viper]

Hmm sorry I’m not too familiar with hyperbolic geometry but the Poincaré disk doesn’t include the boundary right? Are you saying that f being null homotopic may allow extension to the boundary

[obtuse meteor]

Possibly yeah

This feels like the right technique simply bc brouwer’s fixed pt theorem is the underpowered version of Lefschetz

[winged viper]

Right yeah

I’ll think about it some more thanks

[umbral panther]

If the map on fundamental group is trivial, the image in the universal cover is bounded. The universal cover consists of a bunch of copies of a fundamental domain, indexed by the fundamental group. Each copy does the same thing shifted by the image of the fundamental group. Since the image is trivial, they do exactly the same thing

Boundedness should let you pass from the open disk to closed disk and apply Brouwer, but I’m not sure of the details

If the map on the fundamental group is the identity, then the map extends to the boundary disk at infinity. That leads to coarse geometry, like Mostow rigidity. But first Milnor
https://projecteuclid.org/journals/journal-of-differential-geometry/volume-2/issue-1/A-note-on-curvature-and-fundamental-group/10.4310/jdg/1214501132.pdf

[tiny ridge]

T^2 is the torus, how on earth is it covered by the Poincare disk

It's a flat manifold

[umbral panther]

Boundedness means that you can restrict to a big closed disk containing the image and apply Brouwer to that

[red yoke]

Alternatively, f having no fixed points means f - id maps to a punctured torus which has free fundamental group, so WLOG some generator of Z² gets sent to 0. But f also sends it to 0.

[unreal stratus]

Wdym by f - id

Like modelling it by R^2/Z^2 ?

[red yoke]

Ye

[winged viper]

So in the end you use the fact that if c and c’ are null homotopic loops on the torus, then c + c’ is null homotopic?

[red yoke]

Ye

[winged viper]

Oh cool

That’s a nice solution

[obtuse meteor]

Oops

I mean it’s still universally covered but

This is kinda silly :)

(The open disk and R^2 are the same thing oops)

[obtuse meteor]

Ah this is quite good

I still feel as if there has to be a good way to extend to something where you can apply Brouwer

Yeah here’s a formal proof along the lines I originally was thinking

Take f : T^2 -> T^2 which is 0 on fundamental group. Take a lift f’ : T^2 -> R^2

T^2 is compact so this lies in some convex compact subset A of R^2

Now f’ . projection : R^2 -> R^2 has to map A into A

So it has a fixed point in A

Both of these arguments work for any genus surface which is nice :)

[high hill]

curious if this series is written by someone well known or anon https://unapologetic.wordpress.com/2007/03/01/free-products-of-groups/

I feel like I've seen gowers write similar series, but iirc had his name in the url

[umbral panther]

The author’s name is John Armstrong

[high hill]

ty, silly me. i blame phone for not finding it

[urban zinc]

Why are proper maps (preimages of compact sets are compact) important? Like what's some interesting property about them (besides just the definition, which I guess is a natural one)?

[gritty widget]

some categorical equivalences only work when you take proper maps as morphisms

[tough hamlet]

some technical results about quotients by group actions

[urban zinc]

Can you give an example

[urban zinc]

What kind of result?

[tough hamlet]

like

If a Lie group G acts freely and properly on a smooth manifold M

the quotient of M by G is regular

i.e. the set of orbits has a (necessarily unique by other results) smooth structure making the quotient map a submersion

[gritty widget]

There is an equivalence of categories between Compact Hausdorff spaces and Unital commutative C*-algebras. You can drop the Hausdorff and unital assumptions if you take only proper maps as morphisms, and unital star-homomorphisms

[urban zinc]

Hm

Interesting

[urban zinc]

is there any connection between proper maps and properly discontinuous actions

[tough hamlet]

uh yeah I think that's just what happens when G is discrete

I think it's equivalent

oh I should say

an action being proper means the map GxM->MxM given by (g,m)->(gm,m) is a proper map

anyway yeah I think that properly discontinuous actions are what the above becomes when G is discrete

[tough hamlet]

do you know anything about closures of connected sets

try to see if you can say something about them then

np

[unreal stratus]

Poggers!

You can also prove it "by hand" by considering an arbitrary continuous function Y -> {0,1}, say

It kind of nicely shows where you use all the hypotheses

[unreal stratus]

(Have you seen the characterisation of continuity I'm alluding to?)

But yeah so I mean once thing is - forget about the density and just say f(X) = Y. How dodes the argument in terms of a map into {0,1} go?

And you'll see the argument works just as well

[frail venture]

anyone here read any of SAG chapter 1 (the stuff on solution sheaves) and is able to help me with something that is probably very simple?

[abstract saffron]

Someone actually read the Old Testament?

I bow to you with my utmost respect

Wait, I was thinking about SGA. What's SAG?

[frail venture]

its lurie's spectral algebraic geometry. very hard for me, but has the benefit of not being in french

[unreal stratus]

https://en.wikipedia.org/wiki/Saag

[abstract saffron]

Good choice. Avoid French if you can.

[high hill]

If a space is compact path connected, does there exist a surjective path?

If not, say with stronger conditions? (idk what)

[tough hamlet]

no:
take indiscrete topology on a set bigger than R

[high hill]

ok, lets stick to manifolds

but then again, that kinda fails what I wanted
I wanted to justify the existence of a space filling curve for say the shrinking wedge of circles

Other than explicitly constructing it

I felt like since its embedded in R^2, bounded, path-connected, it should be easy to do

[tough hamlet]

for a Hausdorff one do a bigger than R wedge of copies of [0,1] at 0

[high hill]

[tough hamlet]

oh w8 compact

[high hill]

ok, I mean the countable shrinking wedge of circles embedded in R defo has one where you basically draw the shape out

[tough hamlet]

ah I have a nice (compact) Hausdorff one

[high hill]

ig thats even almost bijective, i think

[tough hamlet]

take the unit ball of the dual of a huge Banach space

in the weak-* topology

[high hill]

it seems like explicit construction of the space filling curve is needed to show its existence then...

[ebon galleon]

Explicit construction

[brittle rapids]

we have SGA, SAG, now it is time for GAS

[tough hamlet]

for manifolds at least it's a yes

[high hill]

but manifolds is too restrictive kek

[tough hamlet]

(just cover by finitely many interiors of compact subsets of charts and do space filling curve stuff on each basically)

[high hill]

ig the situation im after could be characterized by subspace topo of manifold maybe

compact pathconnected subspace

[brittle rapids]

i think you can do the space-filling curve outside, and just connect the points where the curve leaves the subspace with a path

[high hill]

interesting

not fully convinced by that rn ngl

[tough hamlet]

yeah that sounds sussy

[high hill]

the subspace is path connected

so ig...

[tough hamlet]

actually no I think it works

[high hill]

cool im convinced

[tough hamlet]

the components of the set on which it's outside the subset are open intervals

hmmmm

might mess up continuity at the other points tho

like if you make too many of these excursions in a neighbourhood of a point

you could lose continuity

[high hill]

im thinking now what if the path outside the subspace

[brittle rapids]

i think it's fine

[high hill]

could be very short

[tough hamlet]

yaah

[high hill]

infinitesimally so

[tough hamlet]

could be very close in R^n

but very far in the subset

[high hill]

its probably not possible with appropriate restrictions

doesnt compactness of the subspace stop it? nvm

[brittle rapids]

what's an example where this happens

[high hill]

topologists sine curve kinda thing

[high hill]

as you approach, this becomes more and more true

with compactness, this property has to be in the limit

[high hill]

ig take a fractal is best way to construct such an example

The other day, it was discussed there were space filling curves that evaded a single point. Id imagine these would fit the bill

[tough hamlet]

ok I have an example where we lose continuity of a curve through the subset

[brittle rapids]

i want it

[high hill]

manifolds are hard...

[tough hamlet]

this blue curve

[brittle rapids]

no, they're bendy

[tough hamlet]

where the subset is the topologist's sine circle

when approaching from the right

when you get forced to leave the subspace

you have to go the long way round every time

which violates continuity at that point

(this point)

[brittle rapids]

ya you're right

lmao tfw you used this proof on an assignment before

[tough hamlet]

anyway

[high hill]

im failing to follow

can u give me an open set

[brittle rapids]

wait now i'm confusedc

[tough hamlet]

if the subspace is locally path connected you might be in business

[high hill]

that doesnt have an open preimage

[brittle rapids]

oh i get it

if you push the blue open intervals onto the sine curve, the total length can't be normalized

wait no

i don't get it

[tough hamlet]

the blue curve is though R^2

like it's just

(t,c) for some constant c

the point is that

[brittle rapids]

it definitely fails ya

[tough hamlet]

whenever it leaves the curve

if we try to fill in the blank by staying in the curve

we can only go clockwise

[brittle rapids]

why

[tough hamlet]

which leads to a minimum distance being travelled from the red point, arbitarily close

[tough hamlet]

uh basically

if we don't then we're gonna fail continuity at the red point right

because we have to go up with the sine wave

[brittle rapids]

just to get my bearings, this would work if the points of intersection were locally finite right

[tough hamlet]

yeah

though locally finite=finite becuase interval compact

[brittle rapids]

oh right

[tough hamlet]

ig you don't even need to think of clockwise vs counterclockwise now I think about it

regardless, moving between intersection points involves passing though a crest/trough of the sine wave

[high hill]

Yes.

[tough hamlet]

which is gonna ruin continuity of the y coordinate at the red point

[high hill]

btw i forgot to say, isnt this thing not path connected

[brittle rapids]

you can go the long way

[high hill]

ah.

[tough hamlet]

the topologist's sine curve is connected but not path connected

[high hill]

ah...

[tough hamlet]

the topologist's sine circle is path connected but not locally path connected

[brittle rapids]

is that really the name

[high hill]

ahh....

[tough hamlet]

yes

[brittle rapids]

now is this fixable with first-countable, locally path connected, and path connected (closed subspace of a space that admits a space-filling curve)

[high hill]

shurisleep, but this has been enlightening thanks

[brittle rapids]

good night

[tough hamlet]

[high hill]

im once again reminded topology doesnt get u away from analysis

[tough hamlet]

this space is a counterexample to the surjective curve thing as well I think

[brittle rapids]

which surjective curve thing

compact?

heine-borel i guess

[brittle rapids]

also i think so

[tough hamlet]

yeah I think that's enough

[brittle rapids]

what do you do with a point of intersection in the closures of the open intervals, but not any one of their endpoints

oh i guess you don't need to worry about those

[brittle rapids]

i guess if a space admits a space-filling curve then it is first-countable, and any subspace is first-countable already

[prisma garnet]

yo can someone explain how hatcher got this equality?

this looks llike the product rule on derivatives

tbh I'm not even sure if I should go through this example

it's like twice as long as the proof of SvK and it concerns knots which I'm not sure I'm interested in

[scarlet turtle]

idk what D is, but it's probably this: https://math.stackexchange.com/questions/3162026/boundary-of-a-times-b

[tiny ridge]

Yes, boundary of a product of manifolds-with-boundary satisfies the Leibniz rule

Exercise

You should go through these examples. Integral to reading Hatcher is figuring out what he’s talking about

Otherwise you’ll just learn a whole bunch of technical terms which you have no real feel for

Draw some pictures

[prisma garnet]

I am I'm just not sure I should go through all of em

[prisma garnet]

it's the disk

the points inside S^n

• S^n ofc

[scarlet turtle]

ah D is closed? in that case then it's precisely that formula in that MSE post

[tiny ridge]

The observation d(M x N) = dM x N U M x dN is pre-algebraic topology. A lot of what you should know from point set topology before learning AT are scattered in the examples of Hatcher

If you dont go through these examples youll not know what you dont know

IMO, in some sense the book is optimally written, wherein you can pick up a lot of prerequisites just by reading it patiently.

[prisma garnet]

I see, yea, my pointset is definitely weak

[tiny ridge]

Most point set topology courses spend a lot of time on things which will never be useful rather than emphasizing things like this which are

No one really cares about T3.5 spaces

Or whatever

[prisma garnet]

[tiny ridge]

Anyhow, draw a simple minded picture. Let M be D^2 and N be D^1

Verify the above formula

convince yourself that this is actually a representative picture in all cases

[prisma garnet]

yea, I'm doing that rn

thx

[tiny ridge]

Its interesting to ask if this really is a coincidence or does this have anything to do with the Leibniz rule for derivatives, by the way

I will let you ponder that

[prisma garnet]

I've heard of something like that

De rham's cohomology or somesuch

[scarlet turtle]

i've only done very basic point set topology, so i have no clue about this. but is it something to do with derivatives being like the "boundary" of function?

[prisma garnet]

I dunno if you could derive the connection with less machinery

[opaque scroll]

#Stokestheorem

[tiny ridge]

There is a simple explanation, you don’t need to know anything

Think like a high schooler

[tiny ridge]

Why are derivatives like boundary of a function?

[scarlet turtle]

it's kind of like

uh

the "rate of change" would happen at the boundary of a volume

?

[tiny ridge]

Very close, illustrate by an example?

[scarlet turtle]

like a ball would have the boundary of a hollow sphere

and if i infinitesimally increased/decreased its volume the amount of change is just at the surface

[prisma garnet]

<@&268886789983436800>

spam

[ivory dragon]

just ping mods and dont react

[prisma garnet]

why did I scroll up

[tiny ridge]

Exactly

Thats it!

Now explain d(M x N) = dM x N U M x dN using this

If you draw a square its automatic given what you said

[scarlet turtle]

💀

[prisma garnet]

it's \partial

[scarlet turtle]

oh yeah the rate of change of the square's area is just it's perimeter

[prisma garnet]

not \del

[gentle osprey]

jagr2808

[scarlet turtle]

and the RHS traces out the perimeter

[tiny ridge]

Indeed

[scarlet turtle]

:o

[tiny ridge]

Infinitisimally increasing the width and the height is the same as doing one of them holding the other constant, and adding the two up

[tiny ridge]

This simple minded thing is the real content of the Stokes’ theorem above by the way

Its just a complicated way of writing it by people who want to feel good about writing complicated formulas

[opaque scroll]

Tbf it also looks extremely elegant written like a formula like this.

[tough hamlet]

it's too late jagr

you're already cancelled™

[opaque scroll]

You caught me

[tiny ridge]

It’s ok to do something if you feel good doing it. I prefer simplicity.

[opaque scroll]

You can have both

[tiny ridge]

I’m a high schooler from 20th century Moscow

[tiny ridge]

Elegance clouds simplicity, historically proven fact.

Symbolic elegance, more precisely

[opaque scroll]

I don't know, there's a lot of power in good notation

But it should not replace the intuitive understanding ofc

[tiny ridge]

Probably. My experience with people who are good at symbolic computation is that they have a simpler way of understanding the notation, at which point they only carry that understanding around and throw the notation in the dustbin

I’m not good at symbolic computation so I cannot say

[prisma garnet]

how am I supposed to visualize S^3?

I get that topologically it's looks about the same as \bR^3