point-set-topology

[zenith bay]

hahaha thanks

I'm sure that was like pulling teeth

[gritty widget]

you might also want to check that f_* is actually a homomorphism

[gritty widget]

all good

[zenith bay]

homomorphism ah yes that seems important

Yeah so I need to demonstrate $[f \circ (\gamma * \gamma')] = [f \circ \gamma]*_{\pi_1(Y,f(x))}[f \circ \gamma']$ right?

eesh notation ookay

[gritty widget]

woops pressed enter too early

f \circ (\gamma * \gamma') on the left, but otherwise yeah

[gentle osprey]

josh the 🐀

[zenith bay]

right okay

yeah that's useful

in that case I suppose it just boils down to whether function composition is distributive over loop concatenation

[gritty widget]

yup

[zenith bay]

$[f \circ \gamma]*_{\pi_1(Y,f(x))}[f \circ \gamma'] = [(f \circ \gamma) * (f \circ \gamma')]$ is definitional for the group operation

[gentle osprey]

josh the 🐀

[gritty widget]

👍

[zenith bay]

right yeah and I can see how the distributive property arises using the definitions

seems it'd be a chore to write out fully, however

[gritty widget]

pretty much how most of algebraic topology goes

[zenith bay]

lmao

good to know

I'm really just studying this for fun

It's like, the most distant topic from my actual research

[gritty widget]

then it'll be more fun when you discover a connection between the two

[zenith bay]

now THAT would be based

next up is Van Kampen for fundamental groupoids I suppose

Grothendieck would be proud

[gritty widget]

https://cdn.discordapp.com/emojis/889939602507460719.gif?size=96&quality=lossless

[gritty widget]

Lmao we’re doing the same topic rn

[mystic reef]

Can you maybe read my answer and tell me what you think?

I show this by proof by contradiction: I assume $(X,\mathcal{T})$ is disconnected. Then there exists nonempty sets $U_1,U_2\subset\mathbb{Z}$ such that $U_1\sqcup U_2=\mathbb{Z}$. Then it follows by the topology that $U_1=\mathbb{Z}\backslash U_2$ and $U_2=\mathbb{Z}\backslash U_1$. Thus $U_1,U_2$ should both be finite, but the union of two finite sets is again finite and since $\mathbb{Z}$ is not finite, I have reached a contradiction and therefore $(X,\mathcal{T})$ is connected.

[gentle osprey]

Ursus1234

[mystic reef]

Dont mind my english, its not my first language

[empty grove]

Ye works

Now try to eliminate the use of contradiction

[mystic reef]

what do you mean?

[empty grove]

I mean optional exercise lol

[mystic reef]

is it not a good way to proof something?

[empty grove]

You can do this in a more concise way without contradiction

It is

[mystic reef]

Yea maybe you are right

If I find time I will try lol

[empty grove]

One line proof: ||any 2 non empty open subsets intersect, so there can't be 2 disjoint non empty open sets||

[unreal stratus]

Or note equivalently you can consider closed sets

Or consider which sets r clopen actually

[plain raven]

https://github.com/agda/cubical/blob/master/Cubical/Homotopy/Group/Pi4S3/Summary.agda#L129

We're very happy to announce that we have finally managed to compute the Brunerie number using Cubical Agda... and the result is -2!
The computation was made possible by a new direct synthetic proof that pi_4(S^3) = Z/2Z by Axel Ljungström. This new proof involves a series of new Brunerie numbers (i.e. numbers n : Z such that pi_4(S^3) = Z/nZ) and we got the one called β' in the file above to reduce to -2 in just a few seconds. With some work we then managed to prove that pi_4(S^3) = Z / β' Z, leading to a proof by normalization of the number as conjectured in Brunerie's thesis.

Axel's new proof is very direct and completely avoids chapters 4-6 in Brunerie's thesis (so no cohomology theory!), but it relies on chapters 1-3 to define the number. It also does not rely on any special features of cubical type theory and should be possible to formalize also in systems based on Book HoTT. For a proof sketch as well as the formalization of the new proof in just ~700 lines (not counting what is needed from chapters 1-3) see:

https://github.com/agda/cubical/blob/master/Cubical/Homotopy/Group/Pi4S3/DirectProof.agda

So to summarize we now have both a new direct HoTT proof, not relying on cubical computations, as well as a cubical proof by computation.

Univalent regards,
Anders and Axel

I will try to explain what this says in my own words but i'm not an expert but i may fuck it up

It is possible to define the basic notions of homotopy theory purely internally within homotopy type theory because in this formal language the types share very similar formal properties with like, CW complexes or simplicial sets or other nice models for doing homotopy theory

This allows the homotopy type theory language to serve as a "direct internal language" for homotopy theory, i.e. a means to carry out homotopy theory axiomatically without worrying about the set theoretic difficulties of what a CW complex is, what a topological space is, what a Kan complex is, etc

instead you simply axiomatize the basic constructions

Most of this formalism is purely constructive/computational in the sense that the basic logical operations can be translated into computer operations; a proof can be "run" as a program

However Voevodsky proposed a powerful axiom called univalence which nobody knew how to give computational content to.

Research in cubical type theory has sought to give a programming language in which all the operations of homotopy type theory, including the univalence axiom, have computational content, and all proofs can be run/executed as programs.

The above link shows that Anders Mortberg and Axel Ljungstrom have, working in cubical type theory, written a proof that pi_4(S^3) is cyclic, i..e there exists n in Z such that pi_4(S^3) \cong Z/nZ. Their definitions of pi_4 and S^3 are both "synthetic", i.e. they don't rely on set theory but rather they take the notion of points and paths as fundamental. So S3 is something like

• two points,
• two paths between those points,
• two higher paths between those paths
• two higher paths between those paths

where the notion of point and path are undefined and axiomatic

Because their proof is constructive they are able to execute it and see what the n is.

The program returns -2, meaning that pi_4(S^3) \cong Z/(-2)Z, or Z/2Z

Very cool

[broken nacelle]

does the proof that the product of two compact spaces is compact use any theory besides the definition of compactness?

I'm trying to prove it by myself

[buoyant nova]

Does the proof you have use the fact that every sequence have a convergent subsequence iff the set is compact?

[broken nacelle]

I don't know

I'm trying to figure out the proof by myself first

it doesn't look too hard

[buoyant nova]

https://en.wikipedia.org/wiki/Tychonoff's_theorem

[buoyant nova]

Ok, well my hint to see the proof is to prove the equivalence I did

The thing is that in general it seems that equivalence doesn't hold (Instead of sequences it talks of nets)

[empty grove]

Ye sequences don't work pls don't use sequences

Nets are ugly

[buoyant nova]

But makes the equivalence hold 👀

[broken nacelle]

what're nets?

[empty grove]

It doesn't use any heavy set theoretic machinery (at least for the finite product case) but the proof I would say is still tricky

[empty grove]

Sequences are nets. You would assume that the subnet of a sequence is just a subsequence. You would be wrong.

[gritty widget]

no they aren't

[broken nacelle]

wtf are nets?

[empty grove]

Nets are a good idea but it also isn't much of a surprise that they work so well in characterizing things given that they mimic the order type of the neighbourhoods of a point

[gritty widget]

like sequences, but you index them using directed sets

you can speak about their convergence etc

[empty grove]

posts in analysis channels
Opinion discarded

[gritty widget]

I don't post in analysis channels

[broken nacelle]

hatcher has a proof of the result but they never introduced nets

[empty grove]

explain

[gritty widget]

I was probably helping someone

[empty grove]

That counts as posting

Just like you are posting in the anime channel right now

[gritty widget]

well, they post questions from topology there, so I have to sometimes

[broken nacelle]

the anime channel

[buoyant nova]

Maybe don't use it. I don't know the proof, what I said it was an idea, but may be wrong

[warped rover]

a generalization of sequences basically

[gritty widget]

I think it should be easier for product of two spaces

[warped rover]

yeah tychnoff is hard bc it's for infinite products

[buoyant nova]

Smells like AC

[warped rover]

yeah it is

idk if you need all that machinery for the finite case, but I can't be bothered to think up what the proof would require right now

[gritty widget]

I think something like Alexander's subbase theorem works

[buoyant nova]

My idea is to mimic the proof for the case when you can use sequences

But with nets

[broken nacelle]

[empty grove]

Finite case is tube lemma

Classical spiceless set theory

✓

[warped rover]

I have never heard the term tube lemma before in my life lmao

[broken nacelle]

I thought I was tripping

[warped rover]

point-set really has whacky lemma names

tube lemma, pasting lemma, etc

[empty grove]

They make sense though

You put the space in a tube

Wrap it up

Which is possible because it is compact

✓

[warped rover]

hmm it's too early in the morning to unwrap this in my head

[empty grove]

ok here is serious statement

If you have X x Y and Y is compact

And you have open set U in X x Y

And you know that a line L = {x} x Y is in U

Then there is some open set V in X

Such that the "tube" V x Y is contained in U

And x is contained in V

✓

Yes

Tokidrotocol

✓

[broken nacelle]

[gritty widget]

If you have a cover of X x Y by sets of the form U x Y and X x V, and neither the sets of the form pi(X x V) = V cover Y nor pi'(U x Y) = U cover X, then taking their union, clearly they don't cover X x Y. So we can assume without loss of generality that sets of the form pi(X x V) = V cover Y, and taking a finite subcover, we have a finite subcover of our space X x Y. By the Alexander's subbase theorem it follows that X x Y is compact

I don't think I am wrong anywhere here

Alexander's subbase theorem says that it's enough to check the compactness property for a subbasis of your space

[empty grove]

Nice

[broken nacelle]

hmm

[empty grove]

I guess that subsumes tube lemma

[broken nacelle]

imma bash my head onto this for a few more hours before I use those cheat codes hints

[gritty widget]

you shouldn't really try to deal with compactness of the product on your own imo

[broken nacelle]

coz I'm pretty sure hatcher didn't use any of that and I want to figure how they did it

[empty grove]

Here is another tube lemma based exercise though if you have a universal cover with compact base and finite fibers, then the total space is also compact

[gritty widget]

or at least, without any tools

[empty grove]

I would say have a go

[broken nacelle]

[empty grove]

As long as you are ready to face the consequences

[broken nacelle]

I'm the guy that picked up rudin to get the active role

I don't fear no consequenses

[empty grove]

[broken nacelle]

all right

thanks guys

imma go for a run now then come back

[empty grove]

Nice

[pearl holly]

I miss Mokiloki✓

[empty grove]

RIP 😔 🙏

[digital hedge]

I’m struggling to fill this out. I’ll have time to discuss this 4 hours from now.

[zenith bay]

I have been studying algebraic topology for two days so allow me to weigh in with my expert opinion

Specifically that $\pi_1(S^1) \simeq \mathbb Z$

[gentle osprey]

josh the 🐀

[zenith bay]

[gritty widget]

If D^n is n-dimensional disk, then D^n and n-dimensional simplex are homeomorphic and in fact contractible

[zenith bay]

I saw someone state elsewhere that spaces with isomorphic homology groups are homotopy equivalent - is this true?

[gritty widget]

hmm

clearly not

[pearl holly]

no I don't think so

[gritty widget]

the fundamental group of S^n is 0 for n > 1, this is a standard result for spheres

[zenith bay]

Two days I have been studying this! Two! I don't know what a homology group is yet!

[gritty widget]

homology is third day

homology

[pearl holly]

but if you work with CW complexes that are simply connected then a map between these that induces isomorphisms on homology is a homotopy equivalence

I think

[zenith bay]

When can you use homotopy/homology groups to infer homotopy equivalence?

[gritty widget]

when the spaces and isomorphisms are really nice

[zenith bay]

Hmm

Is it generally just super difficult to prove spaces are homotopy equivalent then?

Unless they are particularly well-behaved

[marsh forge]

Suppose X and Y are spaces and f:X--->Y is a continuous map inducing an isomorphism on homotopy groups. Then if X,Y are CW complexes, f is a homotopy equivalence. If X and Y are simply connected and CW, and f induces an isomorphism on integral singular homology, then f is a homotopy equivalence.

[marsh forge]

The spaces only have to be CW and the isomorphism doesn't have to be nice, just induced by a continuous map.

[gritty widget]

i see

[zenith bay]

Aah nice

Cool result

Thanks!

[marsh forge]

keep in mind that being CW is not a particularly difficult constraint, for two reasons

1: every decent space is already cw

2: every space is weak-equivalent to a CW complex

[zenith bay]

Alright cool

Can a CW space contain infinitely many 0-cells?

Hang on

I'm getting my terminology muddled

So a CW space is formed by gluing n-balls right

[marsh forge]

sure

[zenith bay]

Is there any finiteness condition on the number of n-balls for each n?

Alright

How about uncountably many?

[marsh forge]

Yeah I think you just need it to be set-indexed

[zenith bay]

Yikes

Fair enough

[marsh forge]

Yes

i double checked

its set indexed

[zenith bay]

So R^n can be seen as a CW space by placing 0-discs at each point in the space with integer coordinates, say, then gluing on the requisite n-discs to fill the space?

[marsh forge]

sure

[zenith bay]

Alright nifty

[long hornet]

A pretty cool fact is that you can't construct R^n using finitely many cells

[marsh forge]

Is this just because R^n is not compact

[zenith bay]

Right

[marsh forge]

A finite collection of cell attachments probably preserves compactness at each step

right

[zenith bay]

Mmh

That sounds sensible

[long hornet]

Wait, don't we need like k-cells for every k < n?

[long hornet]

Yes. It's nice how it generalizes "closed and bounded = compact."

[zenith bay]

Yeah I handwaved that away by "gluing on the requisite n-cells"

Probably should've said k-cells for K leq n

[marsh forge]

You only need to construct a cell structure on R^1

and then you can take the product structure for R^n

which gives you exactly the cell structure you're picturing

[zenith bay]

Yeah that makes sense

Cool

[long hornet]

Given a pointed map f : X --> Y, can we define a space T(f) such that T(f) is contractible iff f is an equivalence?

[zenith bay]

Are the homology groups obtained from the simplicial/cellular/singular homologies the same (up to isomorphism) for an arbitrary space? Or is this something else that is only true for CW spaces?

[marsh forge]

you can do it with two spaces, but not one

well

I guess you could just take the union

of the two spaces

Oh, we need X and Y to be CW

[long hornet]

What are they? Mapping cones?

[marsh forge]

when defined they all agree

singular is the only one defined on all spaces

[zenith bay]

Okay great

Right OK

[marsh forge]

(well, technically you can define CW for all spaces via replacement)

[marsh forge]

homotopy fiber and homotopy cofiber

[zenith bay]

Yeah I suppose the cellular homology is only defined in the category of CW spaces anyway

Ok

[marsh forge]

often times we only think of homology as being a functor out of the category of CW complexes

[long hornet]

Ohh, right..

[zenith bay]

Mmh my category theory is rusty. does this mean that different homology theories correspond to different functors?

I guess really I just have no intuition of what exactly that means

The only solution here may be for me just to learn more category theory lmao

[marsh forge]

yes

to the question

[zenith bay]

The functors are from the category of CW complexes into what? The category of homology groups?

[marsh forge]

the category of (graded) abelian groups

[zenith bay]

Right

Dare I ask what vaguely what cohomology is?

[marsh forge]

a homotopy invariant of spaces that acts nicely

[zenith bay]

The co prefix leads me to assume some functoral relationship with homology?

[snow cobalt]

Hello

I have trouble to understand something

[zenith bay]

Like, a contravariant functor to the homology functor?

Now I'm just guessing what cohomology is based on years old category theory lmao

[snow cobalt]

An open set U in a metric space (M,d) is a subset of M such that :
For all x in U, there exist epsilon > 0 such that B(x,epsilon) is include in U ?

[gritty widget]

yes

[marsh forge]

There are a few ways to think about the duality between them

In particularly nice cases you get an actual linear-dual

[snow cobalt]

Ok because my prof introduced a topology which is "the natural topology"

[zenith bay]

Oh neat

[marsh forge]

In general there is a short exact seqeunce / spectral sequence between the linear dual and the actual object

[snow cobalt]

And I was like ok so the open sets (the elements of the topology that are defined like that) so I wondered what is an open set in a more general way

@gritty widget

[marsh forge]

The term you're looking for is a general topology

on a set

you can read about it on wikipedia or any number of books

[snow cobalt]

I'm just confused as hell about the definitions

[gritty widget]

what do you mean?

[marsh forge]

Essentially it generalizes things like connectedness, closeness of points, and so on

[zenith bay]

Just wait till you see a non-Hausdorf space for the first time

That made me cry

[snow cobalt]

I want to sort things in my mind

I thought that

hum

I don't know how to explain

anyways xd

[marsh forge]

All spaces are hausdorf

[zenith bay]

Which definition in particular Christopher?

[gritty widget]

A topological space is a set together with a family satisfying certain axioms called the topology.
Any metric space usually has the topology that you defined above

[zenith bay]

Wait does the word space imply hausdorf

[snow cobalt]

"usually" so if it's not precised, it's implicit that we use this topology

?

[marsh forge]

In actuality no

But all spaces should be hausdorff

[snow cobalt]

I don't know I'm confused in general about topology now

[gritty widget]

Yes, we could of had defined the topology differently, but this is what we mean when we talk about the topology of a metric space

[snow cobalt]

I want to sort everything in my mind

[zenith bay]

Yeah topology is a weird one to get your head around at first

[snow cobalt]

The thing is

[zenith bay]

For me it was helpful to actually forget the natural metric topology to start with

[snow cobalt]

At the beginning of my year they introduced open sets in R^n so I was like ok that's cool

[zenith bay]

Though your mileage may vary

[snow cobalt]

Then they generalize everything and it's confusing me

[gritty widget]

it's probably the most natural one, because it appears as the topology for which the open balls of a metric space form a basis

[gritty widget]

open sets, we don't talk about open spaces because being open is a property relative to the space that we are contained in

[snow cobalt]

yes I meant sets, my bad

ok thx guys

[zenith bay]

Open sets generalise the notion of "nearness"

In a very loose sense

[gritty widget]

when I think about "nearness" I think about something like proximity spaces instead

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

[zenith bay]

Actually hangon

@snow cobalt https://mathoverflow.net/questions/19152/why-is-a-topology-made-up-of-open-sets this classic mathoverflow post might be helpful

[elder loom]

zariski topology

[zenith bay]

Trivial topology

[gritty widget]

algebraic geometry already has its own "hausdorff"

[marsh forge]

I am aware that my clearly false statement is false

But thank you

[plain raven]

all smooth functions are analytic

[snow cobalt]

Is it me or topology is just definitions that are stacked one after another

[gritty widget]

lmao

[broken nacelle]

you must be doing point set

[gritty widget]

you're not wrong about general topology, it really does just feel like a bunch of definitions over and over again

you're probably missing motivation for those

general topology was a mistake

[plain raven]

Terra

I have already had this fucking argument in another channel today

[plain raven]

You're on my shit list now.

[broken nacelle]

[gritty widget]

where

[gritty widget]

cry about it

I am crying

[marsh forge]

its ok tterra

i support u

[gritty widget]

general topology is just set theory buttered up with fancy words like "geometric" and "space" that evoke a vague sense that what you're doing is useful

[marsh forge]

damn coming for the throat

[gritty widget]

Depends on how much generality you're willing to allow

[broken nacelle]

I support tterra too

[gritty widget]

I wouldn't work with topological spaces on their own, but I think they're useful to built up upon

[marsh forge]

general topology was a necessary step toward the invention of stable homotopy theory

thats why its good

[gritty widget]

For example when defining quotients of a metrizable space, we need to consider some badly behaved topological spaces before we can get to upper-semicontinuous partitions

[marsh forge]

this is a sign that we were never meant to quotient metrizable spaces

[gritty widget]

A lot of constructions give upper-semicontinuous families of sets

Components of a compact metric space for example

Some attaching spaces etc.

They're common

[gritty widget]

Why is the argument made in the blue line the case. To give context , HATS means Hausdorff aleph-one size Toronto space. A toronto space X is a space in which its subspaces are homeomorphic to X if they have the same cardinality

[cedar pebble]

lmao Toronto space

[gritty widget]

Because if it's uncountable then it's of size aleph_1

So since it's a Toronto space, the subspace is homeomorphic to X

But X were assumed non-discrete

[plain raven]

Ok. So. @broken nacelle Take a skim.
The important stuff here is toward the very bottom of pg 1. Basically Brouwer was trying to solve Hilbert's fifth problem in low dimensions and he hit a brick wall because the foundations of topology weren't adequate to support his research, so he switched from his current research to point set topology in order to drive geometry as a whole forward. It's unlikely that we would have a rigorous proof of the Jordan curve theorem either, which has been the pitfall of many mathematicians.
The students of Lie were badly held back by their inability to comprehend the new paradigm. Lie's papers made some claims that were just literally wrong and other mathematicians wrote papers contradicting them, maybe because they thought in terms of the "generic case" and may have even expected counterexamples. Elie Cartan also tried to avoid topological methods but as a result left accidental gaps in his papers that require topological methods to fill.
The definitions and theorems were not given in a vacuum.
Weyl saw that you needed point set for Riemann surface theory, (and later rep theory of Lie groups) as you can see above, and he wrote "The Idea of a Riemann surface based on this" which was hugely influential. Hausdorff didn't postulate his famous separation axiom for masturbatory reasons, he postulated it because he was thinking about Weyl's work and the foundations of complex analysis, because Riemann surfaces obviously need to be fucking Hausdorff, and Veblen and Whitehead realized this was more generally essential for the definition of an n-dimensional manifold. Without Urysohn and Tychonoff, there is no Haar measure for LCH spaces, which von Neumann used to solve Hilbert's fifth problem for compact groups, and more generally no modern theory of harmonic analysis.

It was exactly Brouwer's research into questions in point set topology (the invariance of dimension theorem) which led him to the basic modern methods of algebraic topology.

The Hahn Banach theorem , the Banach fixed point theorem and several major fixed point theorems of functional analysis were direct results of research into point set topology and the efforts to extend Brouwer's fixed point theorem to the infinite dimensional setting.

[snow cobalt]

Let (E,tau) be a topological space and A C E.
I don't understand formally why the closed sets of the subspace topology tau_A are the closed sets of tau intersected with A

It seems logical but I want to get that from the definition

[plain raven]

what is the subspace topology in your def

[snow cobalt]

tau_A = {A intersected U | U in tau}

Let C be a subset of A
C is closed in tau_A <=> C^c is open in tau_A <=> there exists U in tau such that C^c = A inter U
<=> C = A^c union U^c

I get here and I'm blocked

I want C = A inter F where F is closed

[plain raven]

hm i can give you a hint but like

idk have you tried thinking about U^c ?

[snow cobalt]

it's closed

in tau

[gritty widget]

draw a picture

[broken nacelle]

oh wow

that's so thoughtful of you

[plain raven]

yeah history is interesting

mathemtics is way cooler when it's historically motivated anyway

this is from a big book on the history of topology full of various essays

[broken nacelle]

I should read more about math history

I'm definitely gonna do that in the summer

[plain raven]

Props to you, it's a good weapon to have in your arsenal. Knowing about the context in which a concept was invented is a powerful key

[broken nacelle]

[gritty widget]

@gritty widget so is the subspace of isolated points discrete?

[gritty widget]

Yes

[empty grove]

Cohomology operations of type (m, G, n, H) are defined as natural transformations of the form
H^m(- ; G) → H^n(- ; H)
But then the squaring map is given as an example of an operation of type (n, R, 2n, R) where R is a commutative ring. But squaring isn't even a group homomorphism though?

Are these just natural transformations as functors into Set?

[gritty widget]

@empty grove

[empty grove]

Ye squaring is the cup product squaring

But that shouldn't be a group homomorphism though?

Like that is just saying that (a+b)^2 = a^2 + b^2

So ab = -ba

But that isn't always the case for cup product

[pearl holly]

Ye they don’t need to be homomorphisms

They just need to natural

[empty grove]

oof

so weird

[pearl holly]

Ye I know lol

Actually I might be wrong lol

[empty grove]

[pearl holly]

Better wait for Max I think

I’ve only worked with them in Z/2 mostly

[empty grove]

lmao

Nice

[pearl holly]

ye okay I checked Hatcher, they don't need to be homomorphisms

[empty grove]

Wait so is H^n(X; G) = [X, K(G,n)] not a homomorphism?

Only bijection?

I thought [X, K(G,n)] also has a group structure

[pearl holly]

no that is a homomorphism right?

[empty grove]

ok wait

[pearl holly]

no wait lol

natural bijection I think

[empty grove]

Ye they shouldn't be if cohomology operations are not necessarily homomorphisms

Because that contradicts Yoneda

😵‍💫

[pearl holly]

ye

I shouldn't be answering question here lmao, I'm not qualified yet

[empty grove]

Lol I was gonna tag you if no one answered in a couple more hours

[bitter smelt]

What does it mean to be homotopic/isotopic relative to boundary

Homotopy/isotopy is identity on boundary?

[pearl holly]

I'm pretty sure it means that the homotopy should keep the boundary fixed but don't listen to me lol

[empty grove]

Toki be like correct answer but don't listen to me lol

Ye homotopic rel boundary means the boundary stays fixed throughout the homotopy

Same for isotopy

[pearl holly]

I'm doubting myself so hard when I answer question in this server lmao

[gritty widget]

you're probably more qualified on this topic than most people on the server

[empty grove]

If you answer with enough confidence, the poster will change their question so that your answer is right

[gritty widget]

be confident

[zenith bay]

Why the HECK is pi_3(S^2)

Nontrivial

[pearl holly]

cuz hopf map

[zenith bay]

[gritty widget]

Homotopy groups of spheres are weird

[zenith bay]

I though homotopy groups were for capturing holes and now my whole day is ruined

[pearl holly]

homology is better for that

[gaunt linden]

That basically counts the ways to map S^4 into S^2, up to homotopy, right? Intuitively one would think that has to be trivial due to the difference in dimension ...

[zenith bay]

Friendship ended with Homotopy. Now homology is my best friend.

[rugged rock]

Can anyone confirm that a retraction of a connected space is surjective?

[gritty widget]

Retraction is surjective

[zenith bay]

I should hope so

[pearl holly]

ye it has inclusion as right inverse by definition

[rugged rock]

Thanks guys, totally slipped out of my mind

[zenith bay]

So do homotopy groups behave more... intuitively on spaces that aren't spheres

I assume not

OK great

Designed for it?

Right. Is that covered in e.g. Hatcher?

I'm just self studying this stuff

Sounds about right lmao

I suppose so there's a clear thread to follow if the book is used for some a graduate course or w/e

[empty grove]

Consider π_2(T²). Clearly it's a hollow tube right? Surely that's a hole

[marsh forge]

Here’s a pitch for why they shouldn’t ever be well behaved in general

Whiteheads theorem tells us that for CW complexes, the homotopy groups know everything about the homotopy type

So in particular, the homotopy groups should be no less complicated than homotopy types

And figuring out whether two spaces are homotopy equivalent is usually very hard

Like homotopy groups essentially give you a complete description of the homotopy type, which is a very complicated amount of information

[empty grove]

Here's why you're wrong:

If something has 1 hole, it should be a sphere

See, homotopy types aren't that complicated

[zenith bay]

Riight okay

Yeah I can appreciate that line of reasoning

[marsh forge]

That would make my field much more boring

[gaunt linden]

It's probably just that physical experience with spheres up to S^2 gives shit intuition about how the higher ones behave?

[marsh forge]

Right so

For S^n—>S^m to be nonnullhomotopic

We need n>=m

The n=m case is easy

The issue is that the spheres we can picture are S1 and S2

And S2->S1 is always nullhomotopic

So you can’t really use any visual intuition

[zenith bay]

Weird

Spheres are weird

[empty grove]

Higher dimensions are actually wrong. 4 dimensional space was never supposed to be ℝ^4. Humans saw a pattern in ℝ^0, ℝ^1, ℝ^2, ℝ^3, and assumed that this must continue with n dimensional space being ℝ^n. This is a dangerous and baseless assumption, and needs to be rectified asap.

[marsh forge]

This is a good exercise if you’ve read like ch1 in hatcher

[gritty widget]

Are you memeing

Or actually making a point

Well. If it's a point then what should the continuation be then

[marsh forge]

R^n+1 should be R^n for n >2

[empty grove]

S^n should be defined to be K(Z, n), R^n should be defined to be S^n minus a point

[zenith bay]

cursed

[marsh forge]

What does K(Z,n) minus a point look like moldi

[empty grove]

It is a K(0, n)

[marsh forge]

I see

[empty grove]

Keep in mind the homotopy groups haven't been defined yet

[marsh forge]

The fact that every contractible space is eilenberg maclane has never occurred to me before

[empty grove]

The new insights novel approaches bring

[pearl holly]

Principia mathematica 2.0

[zenith bay]

Are all homotopy groups $\pi_n(S^2)$ known?

[gentle osprey]

josh the 🐀

[empty grove]

ye

[zenith bay]

well that's something at least

[marsh forge]

Wait

They are?

Are you sure

[zenith bay]

The strongest result I could find is https://arxiv.org/abs/1506.00952

Namely that for all n>1 the homotopy groups of S^2 are non-trivial

[empty grove]

oh F

[marsh forge]

Yeah

I was under the impression that we don’t know the homotopy groups of a single finite CW complex with nontrivial universal cover

[zenith bay]

Was this difficulty the motivation for homology?

Or am I missing the order of events here

[marsh forge]

Homology is older I believe

Well

[zenith bay]

Ah ok

[marsh forge]

Pi1 and homology for all n are roughly the same old

But pi_n was introduced later

[zenith bay]

Right ok

[marsh forge]

Fun fact

When pi_n was first defined people said it couldn’t possibly be the right definition

[zenith bay]

lmao why?

[marsh forge]

Because it was always abelian

For n>1

[zenith bay]

ah

[marsh forge]

Sorry I’m in an elevator lol

[zenith bay]

So presumable the homology groups of a space lose some information about homotopy type

[marsh forge]

Yes

[zenith bay]

I assume there is no equivalent to Whiteheads theorem for homology groups?

Right

[marsh forge]

There is if the space is simply connected and you take integral homology

I would think of the different homology theories as like

Different kaleidoscope lenses

Through which to look at spaces

They all have benefits and drawbacks

In some sense homotopy groups are a very similar lens, whose primary drawback is that they are impossible to compute hahah

[zenith bay]

lmao ok

[marsh forge]

This will be a sort of important aspect of the blue book group

[zenith bay]

Right

Well that's what I'm reading for lmao

Speedrunning algtop in 3 weeks

[marsh forge]

Lol

[zenith bay]

I'm just joking. I don't expect to be at the level to be able to meaningfully engage with the seminars in such a short amount of time. I'm aware my attendance would just cause issues

[empty grove]

How can attendance cause issues

[marsh forge]

You’re welcome to attend

I probably wouldn’t want you to give a talk but that’s not unique lol

[zenith bay]

I mean having read the syllabus it didn't appear to be that type of thing

[marsh forge]

I firmly believe that attending seminars you don’t have the prereqs for

Is a good thing

Even if you don’t understand what’s happening

Things tend to stick around in ur brain

Waiting to be awoken

[zenith bay]

Yeah I like to do that in general

I'd like to attend if it isn't a problem

[marsh forge]

You’re welcome to

[zenith bay]

Based thanks

Though I do still need to build some intuition as to why anyone bothers with homotopies when homology is so much easier

In particular what interesting information is lost in homology

[marsh forge]

Lots of spaces have the same homology but aren’t homotopy equivalent

[zenith bay]

Mm yeah

What theorems exist that can deduce homotopy equivalence from some homology

You mentioned one pertaining to simply connected spaces?

[marsh forge]

That is basically it

[zenith bay]

Ah lmao

[marsh forge]

The only other way to get homotopy from homology is essentially the Adam’s spectral sequence

The idea is to try to use homological algebra to go from homology data to homotopy data

[zenith bay]

Right, I see

So the goal of algebraic topology, broadly, is to classify spaces up to homotopy right?

[marsh forge]

I suppose so

I think it’s hard to really say there’s a goal

[zenith bay]

Yeah sure

[marsh forge]

I mean okay depending on who you ask

[plain raven]

that's more of one of the big research yardsticks that classifies our progress in the subject matter

[marsh forge]

A modern algebraic topologist might not care at all about spaces

[zenith bay]

Gotcha

[marsh forge]

Like stable homotopy theorists tend to sacrifice a good chunk of space-level data in return for working in a nicer category

And philosophically a lot of them find spectra to be more interesting than spaces

Or they study generalizations of spectra

That said, it’s always cool to solve space level issues with homotopy theory

And that’s the type of stuff that tends to make it to eg annals

[zenith bay]

Right

[marsh forge]

It’s similar to AG where ostensibly the field started by studying varieties but a lot of people are interested in schemes just for their own sake

[zenith bay]

Mhmm yeah that tracks

[marsh forge]

I personally think of AT being broadly about finding new ways to study complicated “geometric” objects via concrete algebra

[zenith bay]

What exactly is a spectra/spectrum? Some sequence of spaces chained by some type of interesting map?

[marsh forge]

The homotopy groups of spheres are a great yardstick / testing ground for these ideas

[marsh forge]

You can think of a spectrum as being a sort of space that has been suspended to the point that suspension stops changing the algebraic invariants

If you’re familiar with localization

The idea is to “localize” spaces by formally inverting the suspension functor

[zenith bay]

Yeah I'm not quite there yet

[zenith bay]

This makes a lot of sense though

[marsh forge]

It’s similar to algebra

Like I can take Z

And some prime p

And formally invert p

So that multiplying by p becomes a bijection

This of course loses some information from Z

But maybe it’s nice for other purposes

[zenith bay]

Huh, I'm not familiar with that

[marsh forge]

Oh! It’s like obtaining Q from Z

We just formally adjoin an element called 1/p

And then add all the like

n/p

For n in Z

And make sure the multiplication and addition still work

[zenith bay]

Analogous to a field extension?

[marsh forge]

Yeah similar

[gritty widget]

I hope I'm not interrupting, I only have one little question.
Why is the homotopy of chain complexes called that

[marsh forge]

It would be like

[zenith bay]

Though naturally not a field

[marsh forge]

Z[x]/px=1

[marsh forge]

You mean like chain homotopies?

[gritty widget]

yes, like when you write f = dh + hd

[marsh forge]

It ends up behaving similarly to normal homotopies

It’s an eq relation on morphisms that most relevant invariants can’t distinguish between

[zenith bay]

Adjoining 1/p to Z gives a Z-module right?

[marsh forge]

It turns out (ahistorically) that this is an example of a much broader phenomenon

[marsh forge]

Z algebra!

[zenith bay]

Aha right yes

[gritty widget]

oh, so it's more in the name than having actual relation to homotopy?

[pearl holly]

check https://ncatlab.org/nlab/show/chain+homotopy under 3. Properties

[marsh forge]

It’s supposed to be an analogy

[pearl holly]

that definition in nlab looks awfully similar to the standard homotopy definition

[marsh forge]

Is it the CxI one?

I’m on phone

[pearl holly]

what is Cxl?

[marsh forge]

Like you have some interval object

[pearl holly]

oh ye that one

[marsh forge]

Like a normal homotopy is XxI

X a space

[pearl holly]

yes right

[marsh forge]

Yeah this is like the model categorical generalization

Of a (left) homotopy

[empty grove]

I'm sure this definition came before model categories though

[marsh forge]

The spaces one and chain complex one might’ve

[gritty widget]

I see, there's more analogy than I thought then

[empty grove]

Ye

[marsh forge]

But I don’t know if the idea of an interval object is that much older

[empty grove]

Like it's just C tensored with simplicial chain complex of the unit interval

So interval object here is pretty simple

[marsh forge]

Anyway point is that the notion of homotopy ends up being vastly generalizable

And a lot of homotopy theory originates from like

Looking at chain complexes and the derived category

And looking at spaces and the homotopy category

And being like

Wow these are very similar

[plain raven]

one really short answer is that a homotopy between continuous maps induces a chain homotopy between the associated maps of chain complexes

(in singular homology)

[zenith bay]

plural homology drops when?

[marsh forge]

if you know about homology with coefficients, that counts as multiple homology theories

but there are far more

[heady grove]

Can anyone help me with this

Im having trouble with the Hausdorff one

[gritty widget]

what are your thoughts?

[heady grove]

I cant think of any counter examples

[gaunt linden]

Which non-Hausdorff spaces do you know?

[heady grove]

Only indiscrete one

[gritty widget]

discrete spaces are hausdorff

[heady grove]

yeah

mistake

so only the discrete one

[gritty widget]

i think you should research some examples of non-hausdorff spaces before continuing

maybe keep your problem in mind as you do so

[heady grove]

k

This will work right @gritty widget

[gritty widget]

you tell me

[heady grove]

It works just trying to understand why its not hausdorff

[gritty widget]

look at its open sets

[gaunt linden]

Note that "codiscrete" is another word for "indiscrete".

[heady grove]

lol

thanks

[gaunt linden]

Another famous (and not quite so destructive) example is the "line with two origins".

[heady grove]

hmm

ill have to research that one

[gaunt linden]

It suggests a bit more intuition than R/Q does.

[heady grove]

Any hints boys

[gaunt linden]

The space deformation retracts to a copy of S^1 perpendicular to the S^{n-2} you removed.

[marsh forge]

my hint is to draw it for S^2

[zenith bay]

Wait I am being dumb how is S^{n-2} being defined there

[gritty widget]

first two coordinates zero

[zenith bay]

Ooooooh right

[gritty widget]

x = (x_1, x_2, ..., x_{x+1})

[zenith bay]

Yeah cool thanks

What does that even mean when n=2?

Wait

Ignore that lmao

[marsh forge]

i have a very bad headache rn and dont want to think too hard, can anyone help me with a result

i believe that if you take the eilenberg steenrod axioms, and add one addition hypothesis, you get that all such theories are wedges of eilenberg mac lane spectra

but i cannot for the life of me remember what it is

[tight agate]

Homology theories “computable by chain complexes”

I don’t know the precise definition

But Peter May mentioned this

Probably something like if it factors through D(Ab)

I’ll try looking for a reference when I go back home

[marsh forge]

oh right thank you! this does sound familiar

[gritty widget]

how does being dense in a hausdorff space mean that f is determined unique

[marsh forge]

If X is hausdorf and S is a dense subset of X, then a continuous function is determined by its restriction to S.

[gritty widget]

$$f\mapsto f\restriction S$$ is an injection

[gentle osprey]

Blitz

Blitz

[gritty widget]

Being Hausdorff implies that it's closed (you can use an argument with the diagonal {(x, x) : x in X} being closed to prove that)

[empty grove]

What does this mean

"all the powers are tied together"

The formula (*)

[plain raven]

lmfao

get fucked

[empty grove]

[plain raven]

this is what you get for reading hatcher

no i'm just kidding

i'm just teasing you

all fun and games here

[empty grove]

Just yesterday we were discussing how Hatcher is completely rigorous

[patent bough]

if a topological space is not Hausdorff, does that mean that there can't exist a metric on R s.t. tau is the metric topology of the metric space (R,d)?

A metric topology is always Hausdorff, so I assume no?

[gritty widget]

Huh?

Is R real numbers

[patent bough]

yes

[gritty widget]

What's your question

I didn't get it

[patent bough]

if I have that for a topological space (X,t) this space is not Hausdorff, can there exist a metric space (R, d) s.t. t is the metric topology of (R,d)?

[gritty widget]

X = R?

[patent bough]

yes

[gritty widget]

Huh

[surreal lantern]

A metric space is always hausdorff

if that's your question

[patent bough]

X is not R originally

[gritty widget]

Topology induced by a metric is Hausdorff

In fact perfectly normal

[patent bough]

so there doesn't exist a metric ?

t is this btw

[surreal lantern]

I'm still not sure what your question is since you're somehow varying your base space?

[gritty widget]

If a space isn't Hausdorff then it's not metrizable

[patent bough]

X = (-1, 1)

(X, t) is not hausdorff

so its not metrizable?

[gritty widget]

Yes

[surreal lantern]

correct

[patent bough]

thank you

[heady grove]

Any ideas?

Would R^3 with two parallel lines removed work

I think so

I think it deformation retracts to r^2 with two points removed

which retracts to two circles of s_1

so pi_1=z^2

where z in the integers

I could be totally wrong here though

really

you sure

what is it then?

i know the theorem

ok

Guys i dont understand why its not Z^2

It is <a,b;> the free group on two generators

Isnt this isomorphic to Z^2?

ohh

i see

[gritty widget]

Z^2 is the free abelian group on two generators

[patent bough]

I don't really understand this question, not sure what result to use

here M is the manifold

[surreal lantern]

i think you'll get more answers to that in #diff-geo-diff-top

[west spindle]

idle curiosity of mine: let $\mathbf{H}{m,n}$ be the hypersurface in $\bR^{m+n}$ described by the equation $$x_1^2 + \dots + x_m^2 - x{m+1}^2 - \dots - x_{m+n}^2 = 1$$ (where $m, n \geq 1$). how does one find the number of connected components of $\mathbf{H}_{m,n}$?

[gentle osprey]

Ann

[west spindle]

i guess this is denoted by pi_0 so i'll just use that

obviously $\pi_0(\mathbf{H}{1,1})$ and $\pi_0(\bd{H}{1,2}) = 1$ and $\pi_0(\bd{H}{2,1}) = 1$ (this is the hyperboloid of one sheet) and also $\pi_0(\bd{H}{1,2}) = 2$ (hyperboloid of two sheets)

[gentle osprey]

Ann

[west spindle]

i think??

theres a 50% chance i screwed this up

$\pi_0(\bd{H}{1,n})$ should always be 2 and i \textbf{think} $\pi_0(\bd{H}{m,1})$ is always 1 but i am clueless as to what happens when $m, n \geq 2$

[gentle osprey]

Ann

[gritty widget]

@west spindle https://math.stackexchange.com/questions/2022156/how-many-sheets-can-a-hyperboloid-have-in-n-dimensions

[west spindle]

oh so it's always 1 unless m=1 in my notation

ok

ty

[gritty widget]

You're welcome

[mortal latch]

Hello folks. I'm trying to construct a topological space which is countable but not first-countable. I briefly looked online and glanced at a couple of non-trivial examples, but I really wanted to figure this out on my own and not just prove that a given space has this property. Can anyone give me a starting point for such a construction?

[plain raven]

By countable you just mean its underlying set is countable?

[empty grove]

I guess try to remove the topology from the problem. It suffices to study the neighborhoods of a single point. The conditions that the intersection of 2 neighborhoods is another neighborhood tells you that it might be worth trying to make all the neighborhoods small or all the neighborhoods large (intersection of large enough sets is large, intersection of small sets is small) so you probably want to involve finiteness/cofiniteness (but notice that there are only countably many finite/cofinite sets, so this won't work directly). Another thing you could think about is how uncountability of neighborhoods could arise. The easiest way to create an instance of uncountability is by making countably many choices, probably about what to include and what not to include in your neighborhoods. Combining all of this, ||you would want to make countably many choices about what cofinite subset to pick. Finite won't work because then you could pick small enough finite subsets for each choice and get a very small neighborhood basis||

@mortal latch

[mortal latch]

Thanks a lot! @empty grove I will think about this and get back to you with my progress.

[mortal latch]

Yes, that's what I meant.

[chrome gale]

Are the annular sets described by a larger ball minus a smaller ball, B(y,a,b) = {x | a < d(x,y) < b} where 0 < a < b, the basis for a weak* topology on R^n?

(This comes from trying to understand which parts of a basis of a weak* topology on the space of probability measures make it weak*)

[shut moat]

that just generates the standard topology on R^n

i don't fully understand what you're trying to do here. If you're looking at the weak * topology on the space of probability measures, you would want a base on that space, not on R^n

[chrome gale]

I am just trying to translate their definition on the space of probability measures to one on R^n.

[shut moat]

translate in what sense? It's a topology on the measures

[chrome gale]

Well they define it in terms of balls on measures in a metric they describe

I was wondering how it related to the typical metric on R^n. I guess my question is now, what's a basis for a weak* topology on R^n?

[high haven]

Is it possible to map a triangle injectively into the unit circle? I am trying to get intuition for homology and I have a feeling that you first need to deform the triangle into a line segment first and then map it into S^1 in order to get a continuous map

[gritty widget]

Filled in triangle?

[high haven]

Yup, I think it is also called standard 2-simplex

[gritty widget]

No, that's not possible

[golden gust]

a circle is disconnected by removing 2 points, and a 2-simplex isn't

[high haven]

Okay I understand now. Nice argument!

Is the reason why higher homology groups of the circle are trivial because that there is in some sense a lack of continuous maps into S1 in higher dimensions?

I know the Mayer-Vietoris proof but I would like a more down to earth approach as a sanity check

[golden gust]

I haven't studied homology so someone please correct me, but I don't think so? like, the second homology group of a torus isn't trivial but there isn't an embedding of the 2-sphere into the torus

oh wait you're not talking about n-spheres, just n-simplices

sorry I got confused bc of S1

I'll let someone else answer

[high haven]

I can't formalize it properly but since you can't wrap boundaries around the unit circle without sacrificing injectivity this should mean that you can't have too wild chains in higher dimensions

I feel like there is a simple geometric answer but I can't see it

[gritty widget]

How is the filter G concluded weaker than F^2?

[gritty widget]

Filter isomorphism = inclusion preserving bijection?

What's F?

@gritty widget

[gritty widget]

What's F though, in the second screenshot

@gritty widget

This is like a spiral of unwrapping the definitions

Yeah im just going to link the paper:
F is not an ultrafilter, F is homogenous, https://wrbrian.files.wordpress.com/2012/01/thetorontoproblem.pdf

its around page q12

page 12*

I might check it out later, this isn't something I can answer on the fly

So the filter portion of the paper just describes how the isolated points are determined I believe

Alright thanks

[shut moat]

Let $G_1$, $G_2$ be separable locally compact topological groups. Let $\pi: G_1 \to G_2$ be a continuous isomorphism. Using the Baire Category Theorem, I want to show that $\pi$ is necessarily a homeomorphism.

[gentle osprey]

Buncho Spheres

[shut moat]

It's been a while since i've used point set topology, so I'd appreciate a gentle nudge in the right direction

[tawdry valve]

this sounds like a variant of the open mapping theorem for Banach spaces. Maybe look at that proof and try to modify it? Not sure

[zenith bay]

Do all CW spaces admit a simplicial representation?

At least up to homotopy equivalence I suppose

[fading vale]

Yep, up to homotopy

[zenith bay]

based thanks moth

[fading vale]

Hatchet 2C.5

(I think that's the one)

[zenith bay]

I've had to give up on hatcher

[fading vale]

[zenith bay]

The guy's writing style winds me up big time

Am I reading a theorem? A proof? Remark? Some kinda stream-of-consciousness expo?

Absolutely maddening to me

[gritty widget]

hatcher is so fucking hard to read if you just want to find a result

fun pictures and discussions though

[zenith bay]

Yeah his illustrations are dope

Honestly it should've been published as a picture book imo

Maybe like in an adult coloring book format

[shut moat]

I’ll give that a try, thanks!

[plain raven]

@zenith bay We argue about hatcher like once every three days in this channel, so i probably shouldn't encourage this, but still, l m f a o

Gotem

[zenith bay]

All good, got it out of my system now. Giving Dieck a go now

[plain raven]

If you want the anti-Hatcher check out Spanier, but it's like Scylla and Charybdis. If Hatcher is long winded and rambly, Spanier is precise and formal to the point of it being overwhelming

i've heard tom Dieck and Spanier are similar

There's also not a lot of intuition from spanier but it's like the most encyclopedic reference i know of on singular homology/cohomology

[gritty widget]

Dieck has a very category theoretical approach

[zenith bay]

Hmmm I am interested in homology

Also category theory

I don't really know why

I think homotopy just scares me lmao

[gritty widget]

One thing Dieck covers that I heard books usually don't is cobordism

[plain raven]

Spanier's book is also categorical

It doesn't cover cobordism.

[zenith bay]

Was having a jolly grand time reading about fundamental groups and all that jazz then the higher homotopy groups of spheres came out of left field and now I have trauma

[plain raven]

Spanier's exposition of basic homotopy theory is actually very readable imo

relative to some of the stuff on homology and cohomology which is like "Here's this theorem stated in greater generality than you'll ever need. Also it's functorial/natural in all six of these variables."

[gritty widget]

I like how Dieck talks about the fundamental groupoid and states van Kampen as a property of functor preserving push outs, it feels more natural than its usual statement

[plain raven]

May also discusses the fundamental groupoid and gives a groupoid-based proof of van Kampen

Spanier discusses the fundamental groupoid but doesn't present van Kampen this way

[zenith bay]

Yeah I feel like some material out there really skirts around the groupoid thing

Which, like, seems important

[gritty widget]

I found some errors in Dieck though

Mostly in exercises, but one in a statement of a theorem. Only read till chapter 2 though

[plain raven]

Spanier develops like a kind of combinatorial approach to the fundamental group as well, like he defines a "fundamental groupoid" of a simplicial complex as being given by any formal chain of edges and proves a bunch of interesting result

i feel like this is helpful for giving a flavor of the combinatorial side of algebraic topology

[gritty widget]

Combinatorial topology, heh

[plain raven]

Also spanier is like... completely error free lol, the dude seems maddeningly careful about shit

i read that book for like 5 years and most like, he forgot to repeat hypothesis H in theorem 8 after stating hypothesis H in theorems 3-7

at that point it's clear from context what he's talking about lol

not five years but it feels like a lifetime

[zenith bay]

Sounds interesting. I'll give it a go also.

[gritty widget]

Sounds like a good book

[plain raven]

It's a compromise between being a textbook and a reference and it basically fails at this compromise imo.
The first three chapters are excellent, the exposition of homotopy theory in chapter 7 is really good. Chapters 4-6 on homology and cohomology need to be read with the understanding that like,

1. there's a lot of dense technical shit in there that is just not of general interest and you should skip over a lot of it
2. the theorems are stated in somewhat excessive generality and to really understand what they mean you need to specialize them.

Example, he defines a map between cohomology groups
H^n(X;G) \otimes H^k(X; G') -> H^n+k(X, G'')
for any three modules G, G', G'', together with any bilinear map G \otimes G' -> G''.
This is not important. You should just consider the case where G = G' = G'' = R, some ring, and then you have that there's a map
H^n(X;R) \otimes H^k(X; R) -> H^n+k(X, R)
which implies that
\bigoplus_n H^n(X ; R)
can be endowed with a multiplication making it into a graded ring.

[gritty widget]

What does it mean that open sets approximate each other?

That sounds like a really strong condition

In a metric space it can't really happen since we can take arbirary small open covers

Unless O1 is a singleton

As you noticed

Damn. You said the same thing twice, sorry 😓

[mystic reef]

I have following which I dont understand. $X=(0,1)$ and $\mathcal{T}={(0,1-\frac{1}{n}), |, n\in \mathbb{N},n\geq 2}\cup{\emptyset,X}$. Prove that $(X,\mathcal{T})$ is not $T_1$. this confuses me since I would by looking at it think that it was $T_1$. Since the compliment of any ${x}\in X$ would be open and thus ${x}$ would be closed. What am I missing?

[gentle osprey]

Ursus1234

[mystic reef]

why is this not the case?

[true robin]

note that every nonempty open set contains 1/3. In particular the complement of {1/3} is not open

[mystic reef]

but wouldnt the compliement to 1/3 be $(0,\frac13)\cup(\frac13,1-\frac{1}{n})$ is that not open or...

[true robin]

not in this topology

it is open in the usual topology but notice that this set is not in the topology you've been given

[mystic reef]

would this the be the compliment?

[true robin]

no, the first thing you wrote was the complement. In any case how can the complement depend on n?

[mystic reef]

yea, you are right

[true robin]

so do you get why the complement is not open now?\

[mystic reef]

no, I still understand it as a union of two open intervals. Cant quite see it

[true robin]

not all of the usual open intervals are open in this topology

only a few of them are open, the ones given in the definition of the topology

notice that (0,1/3) in particular is not in the topology

[mystic reef]

yea I see that

[true robin]

ok cool

[mystic reef]

just so I understand the intervals in this topology correctly it is (0,1/2),(0,2/3),(0,3/4) and so on right

[true robin]

yep

[mystic reef]

so since the complement to 1/3 or a part of the complement is not in the topology it cannot be open?

something along those lines?

[true robin]

yes the complement of 1/3 is not in the topology. (Don't bother with a part of the complement, that won't work as an argument)

[mystic reef]

ahh ok thank you

[gritty widget]

given this proof how is it obvious that the closure of V cannot intersect U

[mystic reef]

@true robin thank you for your patience 🙂

[true robin]

np

[gritty widget]

If closure of V intersects U, then U intersects V

Because then there is x in U such that every neighbourhood of x intersects V

In particular U and V intersect

[gaunt salmon]

If a space X has fundamental group G and H is a subgroup of G, does there exist a subspace Y of X that has fundamental group H? If not, does this hold for some set of groups? Like finite groups, or abelian groups? Sorry if this is a stupid question

[empty grove]

No subspace of S¹ has fundamental group 2ℤ (in a way that is preserved by the inclusion map)

I don't think the weaker versions you suggest would work either

But you can look at the galois correspondence for covering spaces

It says that if X is nice enough (connected, locally path connected, semilocally simply connected) then there's a unique connected covering space of X with fundamental group H for any subgroup H of G

So instead of subspace, you have a covering space

[gaunt salmon]

hmmm, thats interesting

thanks, i will look into that!

[gritty widget]

@gritty widget Thank you!

[snow cobalt]

Hello

I'm trying to prove that in a topological space (X,tau), V include in X
V is open <=> V is a neighborhood of all its points

=> is ok
for <=
I have :
V is a neighborhood of all its points <=> For all x in V, there exist U_x in tau s.t : x in U_x C V
And I have to prove that V is open i.e. V is in tau

[unreal stratus]

Just to check, by neighbourhood of x do we mean a set containing an open set containing x?

[snow cobalt]

yes

Can I say that

[unreal stratus]

Sure ^

Do you have any working towards the other direction?

[snow cobalt]

If V is empty it is open by definition
If V is not empty I create a family of U_x until I fill V ?

And by definition of tau, arbitrary Union of open sets are in tau

[unreal stratus]

I assume you mean if not empty? But yeah, there's no need to differentiate between cases anyway really

(also I'd say arbitrary not random)

[snow cobalt]

Yes xD

yep sure I had that word in french

My problem is

[unreal stratus]

But yes. Suppose V contains an open set U_x about each x in V - what is the union of the U_x?

[snow cobalt]

I don't know if it is a valuable argument because I can't disassociate my vision in metric spaces

[snow cobalt]

I differentiate cases so that I can take arbitrary elements x to create a family of U_x

[unreal stratus]

Well you haven't really given a proper argument yet ig

Eh you can still say 'for all x in V' even if there are no such x, but sure

[snow cobalt]

you're absolutely right

false => blabla so implication is true

so useless to treat cases

[unreal stratus]

Yeah, just try to formalise what your idea is, but it is the correct idea

The intuition from metric spaces does work here

[snow cobalt]

ok ok I see

thank you

[unreal stratus]

Np

[wary furnace]

is it true that the red set is disconnected? (black sets = the topology of X)

I'm probably blind, but to be disconnected there'd have to be 2 disjoint non-empty open subsets of {a,d,e} but... {a,d,e} has no open subsets to begin with

[rancid umbra]

@wary furnace open sets in {a,d,e} subspace topology are
{}
{a}
{d,e}
{a,d,e}

[wary furnace]

oh subspace topology

yeah now this makes sense

after all {a,d,e} had to have some topology, doesn't make much sense to talk about the connectedness of a set without any topology

[sharp frost]

Is it a thing that the 2nd homology group of a connected 2-dimensional space will always be Z?

[gritty widget]

H_2(mobius band) = 0

it is homotopy equivalent to S^1

[sharp frost]

This breaks my brain because I’m looking at the classification of 2D surfaces and they’re all meant to be equivalent to a sphere, a connected sum of tori, or a connected sum of projective planes and all of these have nontrivial H2 group

[gritty widget]

classification of closed 2d surfaces.

compact and without boundary

the mobius band has boundary

[sharp frost]

Oh that’s what it means

That makes more sense ty

[gritty widget]

Is the orthogonal group in n-dimensions a complete space?

O(n) ?

Isn't it compact even?

[hollow harbor]

Yes.

You can just see that it's a closed subset of R^(n^2) (it's defined by a level set of a continuous function A^t A - I = 0)

[gritty widget]

Thanks

[sharp frost]

how the heck is the hopf map defined? I can't seem to find an explicit form of it

something about regarding the unit quaternions as S^3 and then mapping each quaternion to some coset

[wide kayak]

There's an explicit one shown on the hopf fibration wiki page, under Direct Construction

where you think of a point of S^3 as a pair of complex numbers (z1,z2) with |z1| + |z2| = 1

Sick Dr. Octagon pfp btw

I was born on jupiter

[sharp frost]

oh tru how did I miss this

[sharp frost]

ayyyyyy it's very rare that someone recognises it

[winged viper]

hey, is anyone here familiar with hatcher's homology chapter?

im trying to figure out just how much of it i need to read to understand the proof of the jordan curve theorem (first prop in 2.B)

i have a final project where we basically just write a paper, and an option my prof said was okay is to basically learn homology and prove something related to complex analysis. and the jordan curve theorem would be great for that

it's just a bit hard for me to decipher the exact prereqs without reading everything, so any help would be greatly appreciated

it seems to me (although i truly have no idea) that the Mayer-Vietoris sequence is necessary, which seems to require machinery in the first section, namely prop 2.21

if i only need stuff up to prop 2.21 in the first section + mayer-vietoris sequence, then that would be fine, but if more stuff is required to understand then idk if id be able to finish the project (and keep it under the page limit)

[sharp frost]

yeah I just read through it and it looks like the proof uses simplices and complexes, Mayer-Vietoris sequence, knowledge of how to extract an isomorphism from an exact sequence, and general topological concepts like compactness

so reading up to prop 2.21 should be fine unless you are lacking some prerequisite knowledge

[winged viper]

Thanks so much! Does it require the important exact sequence (theorem 2.13)?

[sharp frost]

nah I didn't see them use it explicitly

I will double check (update: I'm pretty sure they only use the Mayer-Vietoris sequence)

[winged viper]

Thanks a lot dude

[gaunt salmon]

Can someone help me understand this theorem? If I have a 2-torus, my understanding is that the fundamental group will be the free abelian group on two generators.

Now lets say you have A, which is the red subspace in the image, which is isomorphic to the circle. This would have fundamental group Z, right?

Why not?

Like this, wouldn't that be a deformation retract or am i misunderstanding what a deformation retract is?

oh

no

I understand now

Sorry

Yeah, I am dumb. This wouldn't work. Sorry

Its just nonsense, the intermediate stages aren't even inside the actual topological space of the torus, they're just small toruses

thanks anyways

If the torus in the pictures included the inside, was a "solid torus" that included all the points inside the torus, this would work, right?

[plain raven]

FOLKS

Let X be a discrete topological space and beta(X) the set of ultrafilters on (the powerset of) X. beta(X) can be topologized in such a way that it becomes the Stone-Cech compactificaton of X.

beta(X) is compact, Hausdorff and totally disconnected.

i.e. it's a so-called "Stone space"

Equivalently, it's a profinite space.

What other properties does it have besides this?

How can it be topologically distinguished from an arbitrary Stone space? What properties does it possess that aren't shared by every Stone space?

[fair idol]

I was thinking about algebraic topology as a subject and it seems like it's only good for telling spaces apart. E.g. "well they have different invariants so the spaces are different". Is algebraic topology really only good for telling spaces apart? I suppose you could prove stuff like browers fixed point theorem via algebraic topology so maybe there are other interesting results that don't involve distinguishing spaces. Sorry if this is a bit open ended.

[golden gust]

why is this space (hawaiian earring) not homeomorphic to (N x S^1)/(N x {1}) ?

apparently it's the one-point compactification of countably many open intervals, but I don't see how that's not the same

[golden gust]

wait, is that quotient even compact?

I suppose that in the HE case, for all open sets there must be a uniform bound on the size of neighbourhoods in the preimage (just from looking at the picture)

whereas the image $\cup_n ( {n} \times U_n)$ is a valid open neighbourhood in the quotient space I mentioned, where $U_n$ is a ball around $1 \in S^1$ of radius $1/n$

[gentle osprey]

xdres

[wary furnace]

Hey guys is this a valid proof?
the question was "prove that the only topology that contains B as a subset is the discrete topology"

I'm still quite new to topology proofs so if there's something I could write in a better way lmk

[gritty widget]

use words to explain what you're doing

[wary furnace]

yeah good point lol

well the 2nd line in 1st column is a fact

[unreal stratus]

There's also no need for you to index singletons by something else btw

Like I'm not sure what your index set is either here?

[wary furnace]

well I just wanted a way to pick out an arbitrary collection of singletons from B

so I don't need indexing for that?