point-set-topology

[plain raven]

i just don't know the rigorous definition or the categorical universal property

[marsh forge]

Alternatively a colimit over a diagram of cofibrants

[fading vale]

Im not really sure what I want to do other than like

I think Im going to read this book on characteristic classes

because I know I want to learn that

[marsh forge]

It’s the same but you put an infinity symbol places

[fading vale]

I feel like I have like 0 understanding of where to go from here broadly in topology or like what homotopy theory I should be learning

[marsh forge]

Yes

It’s tricky

I mean genuinely if you want to do any serious algebraic topology

You gotta learn infty cats

[fading vale]

Thats kind of why ive been doing a lot of AG for the past few months

[marsh forge]

But you might wanna start w model cats

[fading vale]

Yeah like

I was just about to ask is that the "first step" i should be taking?

_>

[marsh forge]

Maybe hovey

I hear that book is good

I mean you really need like

A bare minimum of actual model cat knowledge imo

The literature is basically entirely rewritten for infty stuff

W the exception of old papers

But you can black box hard model cat stuff

[fading vale]

https://people.math.rochester.edu/faculty/doug/otherpapers/hovey-model-cats.pdf this?

[marsh forge]

Yes

Actually

You know

I think riehl has a book that like

Tries to do model cats into basic infty cats

I forget the name

[plain raven]

She has a book "Categorical Homotopy Theory"

[marsh forge]

But that could also be good

[plain raven]

is that what you're thinking of

[marsh forge]

That sounds right

Still am drives

Red lights

[fading vale]

Maybe thatll be good

is this the stuff I should be starting on?

Its kinda like you know because theres all this homotopy theory stuff and model cats and crap and then also I think Im supposed to be learning about spectral sequences but the literature is not very accessible at least to me

[marsh forge]

If you wanna read something else

Try orange book

Like green book but less bad and more orange

[fading vale]

the nilpotence thingy?

[marsh forge]

Yes

[plain raven]

i'm still intending to make these videos lol

but i should stop talking about it and actually do it

[marsh forge]

So maybe a pitch for model cat thinking

Is

You have some construction that is not well defined

Like on some eq relation

You have options

You can try to quotient the target

To make thing agree

Or you can pick which represents count in the domain

That’s what model cats do

[fading vale]

[marsh forge]

One sec

[fading vale]

Max do you know the book "introduction to homotopy theory" by paul selick

its really short but the latter half of it seems like it might be helpful

[marsh forge]

I do not

Send toc

[fading vale]

This is part II

theres a part I but its just stuff i already know

like intor AT

[marsh forge]

It’s old school

But all good

[fading vale]

It seems really fast so i dont know if thats good pedagogically but I feel like maybe itll let me sort of get a grip on like. the breadth of these things and orient myself a little bit

Even if i dont learn everything super well

I think I might try it out

[plain raven]

chapter 12

triples and cotriples are synonyms for monads and comonads

[fading vale]

[fading vale]

Maybe after this itll be a little less daunting trying to figure out where to go next

[marsh forge]

Very weird that this doesn’t cover spectra

Am I missing it

[plain raven]

13.3

[marsh forge]

Oh so late

Interesting

[fading vale]

Well to be fair the book seems mega short so

It can only be so late

[marsh forge]

Oh shit

I didn’t read the page count lol

[fading vale]

Yeah

anyway like

I dont know

It doesnt seem like a huge time sink per chapter so i think ill give it a go sometime soon

[marsh forge]

I also think at some point in AT u just gotta read papers

Like textbooks are in some ways not a great way to learn math imo

You learn faster in some ways through papers

[fading vale]

Yeah thats fair

Its just hard because like

I have no idea what im doing

But maybe I will have slightly more of an idea what i should be doing

[marsh forge]

I think you just have to like

Pick ansubfirld of at

Subfield

And then decide to just learn that story

And maybe you blackbox some stuff

But that’s whatever

[fading vale]

Mhm

also wait

[marsh forge]

I’m waiting

[fading vale]

_>

U know what u were waiting for

[marsh forge]

okay so here is a more thought out description of model categories. Basically, you have a category with some notion of weak equivalence. Of course, after forcing weak equivalences to be isomorphisms, a lot of stuff breaks, not the least of which are (co)limits. Since we want colimits in the homotopy category to be computable in the starting category, we are forced to say that only certain diagrams "count" and as long as every diagram is equivalent to one that counts, we are good

this formalism works and is pretty useful, but it is like, kind of awkward and annoying

to make sure that you replace your diagrams properly and stuff

and like, thinks like equivalences of homotopy categories are kind of annoying to reason about

Then, on the flip side you have the infinity categorical perspective, and im just gonna paste a message from AT server you should read

[fading vale]

[marsh forge]

The first thing that I think is important to point out is that (oo,1)-category theory is not really higher category theory. (oo,1)-categories behave much more like 1-categories than 2-categories, so maybe a reasonable way to approach them is more as "weak 1-categories" - in fact this is where the quasicategory model somehow comes from, and also how people actually use them. In that sense, their definition and use is intriniscally tied to homotopy theory, and areas that can use tools from homotopy theory (mainly in geometry, algebra and topology).

They're a way to make a lot of intuitions precise, and to say a lot of things that couldn't be said before. For instance, homotopy co/limits : I don't know about you, but it was very disheartening to me at first to realize that they didn't have a universal property if you define them using model categories. Of course people in the field knew what it "morally was", but precise statements were kind of hard to come by. oo-categories provide a way to do that, as well as provide a way to ensure that your constructions are homotopically meaningful (in this setting, if you can define them, they have to be !). Here's an example : multiplicative properties of algebraic K-theory. The history of those is kind of a mess, and there seemed to have been quite the number of mistaken attempts. Compare to the oo-categorical approach, where the multiplicative structure just comes from the universal property (see Gepner-Groth-Nikolaus for instance)
[12:24 PM] Maxime: An example of an intuition that they allow to translate faithfully is the idea that a morphism in the arrow category is a pair of morphisms together with a homotopy between the two composites - this allows to define functorial cones for instance, as opposed to triangulated category. In the homotopy category of the arrow category of a model category, a morphism between two given arrows is
[12:24 PM] Maxime: "a co/fibrant replacement of the corresponding things and a strictly commutative square between those" (up to a certain equivalence relation). That's really far from the intuition and the way you want to use it.
In that same kind of example, look at how complicated the definition of a symmetric monoidal triangulated category is, and compare to the conceptual simplicity of a stably symmetric monoidal oo-category. Granted, there's a lot of technicity, but conceptually it is much simpler, and 1000 times closer to the intuition/the things you do with it.

[12:25 PM] Maxime: Another thing that made the previpus frameworks unsatisfactory is the lack of "fornality" (not in the sense of rigour) : "category theory is about making what is formal formally formal", and many "formal-looking" arguments just weren't formal in homotopy theory, but became so with oo-categories.

@plain raven might also be interested

[fading vale]

This seems really cool

[marsh forge]

Main takeaway and my favorite part of this is that infinity categories are like, tailor made for homotopy theory

[fading vale]

When riehl does this is she basically introduction infty cat theory

[marsh forge]

this is a double edged sword

some things are like impossible to construct in an infinity category

but thats because they are homotopically poorly behaved

on the flip side, anything you can build or anything you do build is automatically homotopically well behaved

since i only do homotopy stuff, this is kind of just free lunch

[plain raven]

It was an interesting read, Max. Thank you.

[marsh forge]

yeah

[plain raven]

I'm not doing mathematics right now but I'd like to get back into it.

[marsh forge]

wow i havent gotten to talk about exciting stuff on discord for awhile lol

that was fun

[fading vale]

[plain raven]

Maxime seems like a nice guy. I have talked to him a few times on twitter

[fading vale]

thank you!

[marsh forge]

topologists seem like

very nice in general

everyone ive met

save myself

i think topologists tend to be chill bc we just have our little spaces and we play with them and we don't have as much glory or wide reaching applications as other fields but we are having fun and that is what matters

[fading vale]

Kind of need that in my life

[marsh forge]

its a good vibe

[warped rover]

Is there a good way of defining the long line without getting into the messy ordering stuff?

[gritty widget]

what is the long line

lines at the dmv!

[coarse night]

long lines are funny

[coral pawn]

Is b gonna be a 3 sheeted or 4 sheeted cover?

I couldn't come up with a 3 sheeted cover for b, but my 4 sheeted cover seems to have loops that are too long

The subgroup I got was $<a^2,b^2,a^{-1}ba^{-1}, baba, a^{-1}bab^{-1}>$

[gentle osprey]

Finitely Many Bananas

[coral pawn]

Which is bad because it contains b

[heady grove]

Can anyone help with the last one

Thats what i was thinking

you can always scale an element in Y to an element in X and vice versa

so they are homeomorphic sets

thats the tricky part

y = tan(x*pi/2 )

i think this works for (i)

(x_1,........x_n) \mapsto (tan(x_1pi/2),........,tan(x_npi/2))

could this work?

its the only thing i can think of

true

but which number?

oh

does this seem right

hold on this has length one

Nah im not sure to be honest

agreed

ill come back to this later

[unreal stratus]

I think scaling does work nicely if you wish

Try to find a pretty much rational map for (i) (there is a specific simplest one up to scaling) and then you can alter that

[frosty ermine]

Could anyone help me solve this?

dist(x, C1) = || x - P(x)|| where P is the projection to the relevant space

[lunar yoke]

How about taking the lines with slope 1/2M and -1/2M in R^2, intersecting only at the origin. Then let x = (2M,0). Distance to intersection is 2M, whereas distance to the lines is at most 1, as they have the points (2M,1) and (2M,-1) lying on them

[frosty ermine]

Cool, thanks!

[gritty widget]

How do I get

Klein bottle from annulus

by identifying the antipodal points

Of both outer and inner circle

[lunar yoke]

thats just the usual construction of the klein bottle i think?

annulus is basically a cylinder

[gritty widget]

hmmm

[lunar yoke]

just a bit flattened down

[gritty widget]

Oh shit

Yes

[lunar yoke]

but homeomorphic

[gritty widget]

yeah

Wew

[gritty widget]

I know that for any $\varepsilon>0$ and $1>t>0$ there is homeomorphism $f:[-1, 1]^n\times [-1, 1]\to [-1, 1]^n\times [-1, 1]$ such that $f \equiv \text{Id}$ on $[-1, 1]^n\times [-1, t]$ and $\text{diam}(f([-1, 1]^n\times {1}) < \varepsilon$ for $n = 1$

[gentle osprey]

Blitz

[gritty widget]

Can I use induction to prove this for all n?

Maybe I can use something like. Let $h$ be a map for $n = 1$ and $h_k(x_1, ..., x_n, x_{n+1})i = h(x_k, x{n+1})_i$ for $i\in {k, n+1}$ and $x_i$ otherwise

[gentle osprey]

Blitz

[gritty widget]

And then set $f = h_1\circ ...\circ h_n$

[gentle osprey]

Blitz

[gritty widget]

First property is satisfied

[lunar yoke]

what do you need this for?

i feel like visually the claim is obvious

but might be tedious to write smth down

[gritty widget]

To prove the Hilbert cube is homeomorphic with its own cone

[lunar yoke]

[gritty widget]

That's what they wrote in my book. But I want a correct proof, with no hand waving

[lunar yoke]

ok

[gritty widget]

They mentioned induction should work somehow

Actually the map for $n=1$ is such that $h([-1, 1]\times {1})\subseteq {1}\times [-1, 1]$

[gentle osprey]

Blitz

[lunar yoke]

yeah

you shrink one side of the cube

and interpolate linearly between that shrunken side and the part of the cube where the last coordinate is <= t

at least thats what i'd do

[gritty widget]

I've used barycentric coordinates

So it's identity where I want it to be

And I can map sides to other sides

Homeomorphically

This works for n = 1. Would be too difficult for n > 1

[lunar yoke]

cant you just say $f(x,s) = (x,s)$ for $s \leq t$ and $f(x,s) = \left(x\left(\varepsilon\frac{s-t}{1-t} + 1-\frac{s-t}{1-t}\right),\ s\right)$ for $s >= t$

[gentle osprey]

Phil

[lunar yoke]

nvm i think i misunderstood the problem

the map f in you statement needs to be surjective onto the cube again, i completely overlooked that

[lunar yoke]

what does it look like for n=1?

[gritty widget]

Well, I didn't write it explicitly, just used that I can map polygons onto other polygons homeomorphically while keeping some sides intact + subdivision. Using barycentric coordinates

[lunar yoke]

but where do you map [-1,1]^n x 1 then

[gritty widget]

I can draw a picture tomorrow

Blue region is being mapped onto the green region

@lunar yoke

[lunar yoke]

wow

that seems pretty arbitrary

i thought of something like this

the first step is the function i gave above

[gritty widget]

I've used this homeomorphism in different proof

Of homogenity of the Hilbert cube

[lunar yoke]

i mean sure if it works then nice

[gritty widget]

So maybe a little arbitrary

[marsh forge]

Is there a specific reason for this approach?

This seems like it would probably easier to prove for the normal infinite cube

[gritty widget]

I'm not sure.

I assume so if they wrote it like this

[gritty widget]

This almost works

Take $h_k(x_1, ..., x_{n+1}) = (x_1, ..., h(x_k, x_{k+1}), ..., x_{n+1})$

[gentle osprey]

Blitz

[gritty widget]

Then $f = h_1\circ ...\circ h_n$ should work

[gentle osprey]

Blitz

[gritty widget]

No wait, it doesn't work

[hazy lotus]

I'm having trouble proving that given a metric space (X,d) the vector space of continuous functions from X->R is finite dimensional if and only if X is finite.

The backwards direction was simple because the topology on X is just the discrete topology and every set is clopen and every function is continuous.

This means a spanning set is the set of functions that send one value to 1 and everything else to 0I'm not sure at all how to do the forward direction.

A hint was to use Urysohn's lemma

[lunar yoke]

the same argument as for finite X should apply to show that if X is infinite it has unbounded dimension

i.e. pick n points, then you find n functions such that the i-th one is 1 on the i-th point and 0 on the others

hence these n functions are linearly independent

[hazy lotus]

Ok so I can construct these functions by Urysohn

[lunar yoke]

if X is finite you can pick n arbitrarily large

[hazy lotus]

I'm not immediately seeing that they are linearly independent

[lunar yoke]

right for this you can use urysohn

[hazy lotus]

oh wait

nvm

OK I see thanks!

[lunar yoke]

no, sadly not

take for example R^2 -> R, (x,y) -> xy/(x^2+y^2) for (x,y) != 0 and sending (0,0) to 0

this is continuous everywhere except the origin

but there its continuous if you fix either variable

[coral pawn]

This cover is normal right?

Since rotation by 0, 2pi/3, and 4pi/3 are deck transformations which make the thing act transitively

So the conjugacy class should just be the set <a^3,b^3,ab,ba>

Also, since this is a 3-sheeted cover, 0, 2pi/3, and 4pi/3 must be all deck transformations so this group should be isomorphic to Z/3Z

Can someone verify this?

This seems too easy

[tough imp]

Does this look like a reasonable amount of material for a 2 semester AT course?

[plain raven]

Idk honestly, hard to tell how superficial the coverage will be

[tough imp]

Gotcha

Overall the courses at Columbia have scared me because they cover far more topics than the equivalent at UW

Here’s their geometry class for example, at UW we spend the whole year covering like topological manifolds and the basics of topology, followed by just Lee’s Intro to Smooth Manifolds

So I don’t think we really hit any of the latter 3 groups, so I’m a bit worried that the pace might be much faster than what I’m used to

Granted they don’t have to cover the basic point-set stuff UW spends the first quarter doing, but even then the list of topics covered is not comparable

[arctic relic]

Both look pretty standard for a 2 semester course tbh

Or I guess standard for a Columbia tiered university

[marsh forge]

yes

[marsh forge]

this looks like a great class, honestly and about the speed mine went at

although different and arguably better material

Keep in mind maybe that adding smooth structures and stuff makes everything slower and harder

[swift fjord]

1st semester looks a bit much. Depends on how deep you go into every topic of course

[marsh forge]

I assume a lot of the material won't be in depth

[tough imp]

Oh okay, cool

👍

[lunar yoke]

If X is locally compact, then the functor - x X preserves quotient maps

this is usually shown in pset-style fashion like this

does anyone know whether its possible to prove this using left-adjointness of - x X and some fancy category theory?

im pretty sure it works for easy cases like quotient maps where we contract some subspace cause thats a pushout anyway

[marsh forge]

hm

now im confused

bc quotient spaces are colimits and left adjoints preserve colimits

so i dont even know why the locally compact assumption is necessary

[lunar yoke]

so that - x X is a left adjoint at all

if its not locally compact that doesnt work in Top i think

you would have to go to compactly generated spaces or smth

[marsh forge]

ohhhh

[lunar yoke]

the adjoint is the mapping space (-)^X

[marsh forge]

lmfao

i was thinking about the wrong assumption

i see

then yeah this is immediate as long as xX is a left adjoint

(i always forget that Top is not actually a good category)

[lunar yoke]

now im confused

how do you represent a general quotient space as colimit

coequalizer?

[marsh forge]

yes

[lunar yoke]

oh nice

how did i not think of this before

[marsh forge]

hahaha

[lunar yoke]

anyway thanks

[marsh forge]

it was one of those things where i didnt have a construction in mind but like

quotient topology 100% has colimit vibes

[lunar yoke]

yeah it feels like it for sure

[shadow charm]

« Best thought of »

[empty grove]

Accurate 😼

[gritty widget]

what does "closure of a linear set is meager iff it's nowhere dense" mean

what does "linear set" mean

I think we're in R^n here

[empty grove]

Probably vector subspace

[gritty widget]

that does not correlate well with the context

eh, no need to overthink it

I am wondering how to prove this theorem by Nadler

Let $F_n$ be a sequence of contractions of a locally compact complete metric space, convering to a contracton $F$.
Let $x_n$ be fixed points of $F_n$ and $x_0$ be a fixed point of $F$. How to prove that $x_n\to x_0$

[gentle osprey]

Blitz

[gritty widget]

I've showed that there is an increasing sequence $N_k$ of natural numbers such that $F^{N_k}(x_{n_k})\to x_0$

[gentle osprey]

Blitz

[gritty widget]

note: this doesn't hold without local compactness, so it plays an essential role here

[gritty widget]

the convergence is pointwise btw

If $L_n$ is Lipschitz constant of $F_n$, then this boils down to showing that $\frac{1}{1-L_n}d(F_nx_0, x_0)\to 0$

[gentle osprey]

Blitz

[gritty widget]

at least I think so

[gritty widget]

$d(x_n, x_0)\leq d(x_n, F_n x_0)+d(F_nx_0, x_0)$

[gentle osprey]

Blitz

[fading vale]

@marsh forge dude why are the details of anything in homotopy theory so insane

that selick book i found is only so short because 80% of the proofs are "see this other book" and then when i go look them up theyre horrible

[pearl holly]

what are you doing moth?

[fading vale]

i am learning classical higher homotopy theory i guses

Not even maybe idk

I am doing fibration and cofibration stuff

[pearl holly]

I see I see 🙈

[marsh forge]

hahaha whats an example

[fading vale]

Everything

[marsh forge]

nice

[marsh forge]

yeah i feel like the modern treatment of this stuff is hard to find

homotopy groups are just an artifact of the natural t-structure on spaces

[fading vale]

Homological algebra...

[marsh forge]

yeah all this technical stuff is just gross lol

[fading vale]

Yeah

Its bad

also i genuinely think more statements are unproven than proven in this section

[marsh forge]

this is a good place for you to learn to not focus so much on details

focus on vibes

[fading vale]

sometimes they dont even tell u where to find the proof theyre just like

They just state it

[marsh forge]

yes

ur fault for using a book this old tho

[fading vale]

This book is from like 98!

[marsh forge]

thats older than me

too old

[fading vale]

max admits to being old and decrepit

[marsh forge]

i mean honestly ive never read anything like this material

seems dry and boring

[fading vale]

I guess i will just have to be less picky about details

[marsh forge]

idek what numerable category means

[fading vale]

Its just semilocally simply connected

[marsh forge]

[fading vale]

[marsh forge]

this might actually be like

too old

if its referencing concise

[fading vale]

Thats not concise

[marsh forge]

bc concise's treatment of this stuff itself is bad

[fading vale]

Its like uhhh

[marsh forge]

what is may2

[fading vale]

"classifying spaces and fibrations"

[marsh forge]

gross

[fading vale]

Peter

The stuff after this chapter seems less

Like this

I mean i dont even hate this really

[marsh forge]

is there a reason u just ignore my suggestions for reading

and read this awful stuff instead

[fading vale]

What suggestions

[marsh forge]

orang book

blue book

papers

[fading vale]

I tried blue book but it was

[marsh forge]

higher topos theory

did u start with section 3

[fading vale]

Yea

[marsh forge]

what tripped u up

[fading vale]

i didnt go very far

[lunar yoke]

what is orange book

[fading vale]

Idk just like the general typesetting and vibe and not feeling like i knew enough about fibrations/cofibrations/etc

[marsh forge]

something periodicity something nilpotence by ranvenel

[empty grove]

Why they called color book

[fading vale]

Ill probably pick it up again i guess

[marsh forge]

well the green book isnt even green so who knows

[fading vale]

I only decided to try out selick because its short as fuck

[pearl holly]

ye same

but my cat theory was shit as well lmao

[marsh forge]

the fibrations/cofibrations stuff isnt that big of a deal

you can think of them as just nice inclusions and "covering" maps, respectively

[lunar yoke]

bruh its officially called orange book i thought it was some kind of meme

[marsh forge]

yes

not to be confused with higher topos theory

which is also orange

[fading vale]

I will read riehl probably soon

[empty grove]

They have a copy of HTT in my college library

[marsh forge]

honestly moth like

[empty grove]

Saketh borrowed it as a meme in our first year

[marsh forge]

it might be more productive to do like

Baby riehl, half of big riehl, bluebook + orang book, finish big riehl

then do real homtopy stuff

[empty grove]

And we went around the hostel showing it to everyone and going "hey I have a doubt on page 563 here" or "I read this book but I found it to be quite lacking in content, do you have better recommendations"

[marsh forge]

at least riehl would teach you model category stuff in the correct generality

[empty grove]

[pearl holly]

there's big riehl?

[lunar yoke]

which of these is big riehl

[empty grove]

Categorical homotopy
Homotopical category

[marsh forge]

oh wait

[empty grove]

[marsh forge]

actually @fading vale

https://arxiv.org/pdf/1904.00886.pdf

this is a great short read probably

I think learning about cofibrations from old topology books

is bad

and also sometimes like

incorrect

[fading vale]

Yeah so like i basically want to do this but my thought process on this is that I think id feel more comfortable if i had a better sense of the stuff that this came out of + had seen a taste of these things and selick is so short that I feel like i can read it on and off while i do other things before i jump into these like

much larger undertakings

[marsh forge]

if u say so

i mean genuinely like the right intuition

is that cofibrations are cellular maps of CW complexes

[fading vale]

[marsh forge]

and that fibrations are what they need to be for this to be a model category

and once you replace everything in a diagram by (co)fibrations you get homotopy colimits

e.g. if you want a homotopically meaningful answer you should always be using CW complexes

i don't think it has to be much deeper than that

like I think there is a big difference between learning the "classical" theory and just learning a theory developed when we didnt have the proper machinery

some stuff is just outdated and backwards

[fading vale]

Wheres the boundary between the "not having proper machinery" and classical stuff

Ah fuck it if this material isnt gonna be helpful i may as well just jump into riehl why not

@marsh forge Whats baby riehl, the book you recommended me earlier?

[marsh forge]

riehl categories in context -> the arxiv link above

would be my recommendation

[fading vale]

the category book

[marsh forge]

idk how comfy you are with baby riehl

theres also hovey's model categories

this is probably more than any modern researcher needs to know about model categories tho

[gritty widget]

I have a course from topology which touches on simplexes etc.
And I asked my lecturer if we'll be having cells
He said it's not a lot different, so we won't be having those. Is that true

[marsh forge]

no

[fading vale]

[marsh forge]

its not at all true lol

[fading vale]

Yeah uh. not sure about that one lmfao

[marsh forge]

i mean in some sense moth like

the right way to learn topology

is to find a paper you find interesting

and try to read it

and when you get stuck google stuff

and then go back to the paper

etc

[fading vale]

not sure about that one re the cellular thing lmfao

[marsh forge]

this will sort of automatically filter out what is/isnt important

yeah i got u

[lunar yoke]

Schwede recommended me this as intro to model categories https://math.jhu.edu/~eriehl/616/DwyerSpalinski.pdf

[fading vale]

for baby riehl i think that i will probably generally understand shit up through chapter 4 but ch5 and 6 seem new

[lunar yoke]

havent looked at it yet tho

[gritty widget]

that's sad. Hopefully I'll have cells in another course

[marsh forge]

this seems reasonable

i might consider blackboxing some of that stuff

like all of ch7 lol

everything else looks good tho

i guess ch7 is short

proving model category axioms is though

[marsh forge]

6 is important if you care about how certain constructions are made (e.g., smash products of spectra)

5 is less important immediately

you'll want to know that stuff eventually

[lunar yoke]

Clerk had lots of rants about monads somewhere in #category-theory which made the stuff make more sense for me

[lunar yoke]

kan extensions

[marsh forge]

Kan extensions really aren't that scary

they are just like

[lunar yoke]

yeah i believe you

[marsh forge]

I have a functor C->D

and a functor C->C'

[lunar yoke]

just havent had the time to properly look at them yet

[marsh forge]

what is the closest thing i can get to a functor C'->D making the diagram commute

or what is the best thing i can get

[fading vale]

is big riehl the book you told me about before

the stuff about model categories and quasi-categories

categorical homotopy theory

[marsh forge]

i think you can start with the arxiv link above

[fading vale]

i see

it looks pretty much like a condensed version of the book

[marsh forge]

idk if the book i had in mind actually introduces model cats or takes them as given

[fading vale]

I think it takes them as given so

I should probably read the paper

[empty grove]

I borrowed this from library

I read like 2 pages but I liked the author's philosophy in the preface 😌

[hazy lotus]

Given a metric space, I'm not sure how to show that if the closure of a ball B, say cl B, is contained the closure of a set A, then cl B intersects A.

[empty grove]

Suppose cl B doesn't intersect A. Show that the center of the ball is not in cl A

[hazy lotus]

right... that makes sense.

thanks!

[gritty widget]

since cl(B) is contained in A, the only possibility of cl(B) not intersecting A would be that cl(B) is empty, so B is empty

but a ball contains its center by definition

[hazy lotus]

At the start we only has cl B contained in cl A

We’re trying to prove cl B intersects A

It doesn’t have to be true that cl B is contained in A

For example A=the rationals and (X,d) is R with the standard metric

[gritty widget]

I misread

[marsh forge]

i am begging you to change your username muda

[coarse night]

jojo fan

[swift fjord]

it's useless

[empty grove]

SHINSHINSHINSHINSHINSHONSHISNSYINSGISNSHINSHISNHOSMUSINSHIN

[hollow harbor]

This is actually so based

[mellow prairie]

why is this not a triangularisation of the dunce hat? apparently it can’t be done with less than 8 vertices but i don’t see why this doesn’t do it

[gritty widget]

is this true

[X,Y disjoint Z] =[X,Y] \oplus [X,Z]

assuming X, Y, Z connected

[marsh forge]

Assuming connected yes

it should suffice to only assume X is connected I think

[gritty widget]

yeah i think you are right about only needing X connected

[fading vale]

The issue is that you have things like these two red lines which are technically being identified

You want to add one more vertex around the center to prevent this identification from happening

[mellow prairie]

yeah thanks now it makes sense. i was trying to make sure no vertices are the same but forgot to look inside lol

[rugged rock]

Anyone know a good resource for someone who is just starting topology and is very comfortable with algebra? I am sort of struggling being able to effectively write proofs

I am currently doing homeomorphisms and continuous functions. Chapters 18 and 19 of Munkres.

For example I dont really understand the whole open ball arguments

[marsh forge]

Munkres would probably be the recommendation, have you done chapters 1-17 of munkres?

[rugged rock]

Yeah i did 12, 13, 15, 16

I had a brief look over the set theory one, but perhaps I should go back and restudy it

[marsh forge]

You should probably do the rest if things like open balls are tripping you up

[rugged rock]

While that is true, I did not have much trouble with chapters 12,13, 15 and 16, but now I got into continuous functions and im a bit more clueless

Besides Munkres, which is the book I'm currently using, are there any other resources that may be helpful?

[marsh forge]

I mean... continuous functions are ch2 of munkres

just bc you didnt struggle with 12 or 13 doesnt mean you dont need prereqs for 18 etc

But hatchers notes are a good secondary, short resource

[jolly garden]

What should one have in mind when thinking about the quotient space given by (x,y)~(x,y+1)~(x+1,y) on IR², namely IR²/~. I know that it is homeomorphic to a Torus ...

I get that each of these points are equivalent in the quotient space, but how does this form a Torus?

[fading vale]

You can imagine the quotient map here as taking each square and mapping it onto the central square containing 0 so that we'll be left with a single square containing the origin

so think of this square as being given by like, (t, s) for 0 <= t, s <= 1

The problem with this is that the edges of our square might be given by, for example, (t, 0) and (t, 1)

But if you look at the quotient map in question we actually have to identify these together as well

since (t, 0) ~ (t, 0 + 1) = (t, 1)

similarly (0, s) ~ (1, s)

so the actual resulting quotient is the square here (filled in) with sides identified like so

This is equivalent to the torus

by these identifications

[jolly garden]

Totally makes sense now. Thank you.

Kinda reminds me of: https://en.wikipedia.org/wiki/Elliptic_function#Period_lattice_and_fundamental_domain

[wintry heart]

why is $\bR$ complete if the Cauchy sequence $x_n = \sum_{i=1}^n \frac{1}{i}$ does not converge in $\bR$?

[gentle osprey]

meg sam

[gritty widget]

harmonic series ?

[wintry heart]

yeah

[gritty widget]

it's not a cauchy sequence

[wintry heart]

wait... why?

[gritty widget]

think of the cauchy sequences as all the sequences that converge

with a few exceptions when the value to what it converges doesn't belong to your space

[wintry heart]

isnt the definition of cauchy sequence just $\forall \epsilon \exists n (\forall m>n d(x_m,x_m+1)<\epsilon)$?

sorry, too lazy to do latex

[gritty widget]

like the cauchy sequence $\left( 1 +\frac1n\right)^n \in \mathbb{Q}$

[gentle osprey]

Entelechy

[gritty widget]

yea basically

the distance between terms whose indices are greater than n

[wintry heart]

but the harmonic series does have that property

[gritty widget]

it does not

[wintry heart]

huh?

[wintry heart]

...........oh, it's not this

i just forgot that the 2 terms dont have to be, like, adjacent

[gentle osprey]

meg sam

[gritty widget]

ahah based

[plain raven]

In general topology was invented to formalize arguments about continuity which appear in analysis and (differential) geometry. So for intuition you should read real analysis or functional analysis, and perhaps an introductory book on manifolds (Introduction to Topological Manifolds by Lee)

A good real analysis book will be self contained in terms of topology so you will be able to read it without problems. The book by Lee assumes background knowledge of topology, I believe, but you will be able to get geometric intuition even if the formal arguments are not clear.

A good complementary book on topology is the one by Klaus Janich, not that there's anything wrong with the Munkres book.

[gritty widget]

Topology books are usually not very exhaustive, I think

Because I read through Dugundji, and there's still so many things to learn in Engelking Sieklucki for example

I think I might be just imagining things

Never mind

[coral pivot]

these are linked, you use the period lattice Lambda to get an isomorphism between C/Lambda and an elliptical curve, so elliptical curves are just torii

[still vessel]

Hello, physics student here on his first steps with topology, so i am having a slight beginner problem, i dont know how to prove something is an open set. Normally what i would expect is to have some condition that need to be met or to prove some equalities and such, but the definition of open set in the book we use is pretty much non existent (Morreti book) so i dont know how to go about proving something is an open set

and a closed set is define by having a complement which is open :/

[plain raven]

Do you understand the definition of a topological space?

It will help if you post a specific example, it is easier to argue with specific examples and then you can extrapolate from there.

[still vessel]

i believe i understand the definition of topological space at least at face value

the example is to prove that an interior of a subset A of a topological space is open set

my only idea is that i need to somehow prove that interior of a subset is also a subset of X, which would make it an element of a topology an would have to be by definition an open set?

is that a correct thinking process?

[plain raven]

Can you give me the definition of interior

[still vessel]

where interior point of S means there exists an open neighbourhood in S which contains x

wait is union of open sets an open set? It sounds like it should be

[viral atlas]

Yes, this is one of the conditions on a collection of subsets to be a topology on a given set.

[still vessel]

Ah, so can i say that an interior is basically a union of open neighbourhoods and as such is open itself?

[viral atlas]

The standard definition of a topological space, from Wikipedia.

[viral atlas]

Precisely!

[still vessel]

Nice, ok thanks for the patience and letting me come to it myself xD

[viral atlas]

#2 here tells you that the union of any collection of open sets is an open set in a topology.

[still vessel]

yeah, did not really think of it like that

[gritty widget]

I just read proof of the Sierpiński theorem about characterization of rational numbers using some homogenity properties of the Cantor set

damn that was nice

just everything falling out of the earlier results

[fading vale]

These are very related things! Elliptic curves over C correspond to embeddings of the torus into C^2

[golden gust]

you got any resources on this?

[fading vale]

There are lots of ways to do it

It kind of depends on whether u want to take a complex analytic or algebraic geometry approach

(theyre equivalent but the machinery is pretty different lol)

The intuitive explanation is that elliptic curves are smooth projection compact connected curves (so dimension 1) of genus 1 hence diffeomorphic to a torus

But not biholomorphic

So the complex structure can look very different

[gritty widget]

was mentioned here

[cedar pebble]

intuitive

[fading vale]

[gritty widget]

I have a space $X$ with countable basis of clopen sets $\bigcup_{m=1}^\infty \mathcal{B}_m$ where $\mathcal{B}m$ is a countable partition of $X$ by sets of diameter $\leq 1/m$, and $\mathcal{B}{m+1}$ refines $\mathcal{B}m$.
How do I prove that if $X$ is $\sigma$-compact, then it can be partitioned into compact sets $X = \bigcup{n=1}^\infty A_n$ where diameters of $A_n$ go to $0$?

[gentle osprey]

Blitz

[fading vale]

Is there some result that says that if a hurewicz fibration is a homotopy equivalence it has the form Y^I -> Y or something?

Im trying to understand why cofibrations, hurewicz fibrations, and htpy equivalences form a weak factorization system for Top

I get 2 of the axioms but the actual lifting is confusing to me

To use the HEP for cofibrations you need this diagram to have the form A -> Y^I -> Y through the top and right and A -> X -> Y through the left and bottom

[fading vale]

Ok thinking about this for a little bit didnt work so its time to call upon @marsh forge

[marsh forge]

There are explicit models you can use to do this easier, I don’t think it is obvious from the axioms

The idea should be to use mapping cones for cofibrations and a sort of fattening for fibrations

I believe the details are in concise

[fading vale]

Okay

[sturdy terrace]

do the homotopy classes of maps from T² to S² have a nice enumeration and/or group structure?

[marsh forge]

I do not think so.

to both questions

well i don't think there is a natural group structure, obviously any set has a group structure

[sturdy terrace]

it feels to me like any map from S¹ x S¹ essentially splits via the projection to S¹ ∧ S¹

so you'd have some mostly-natural identification of [T², S²] and [S², S²] = pi_2(S²)

[marsh forge]

a based map $X\times Y \to Z$ descends to a map $X\wedge Y\to Z$ only if it maps $X\vee Y$ to a point.

if the original map comes from a product of based maps $X\to Z$ and $Y\to Z$ then this is satisfied, but not necessarily otherwise

[gentle osprey]

ShamsOnlyTwitchSub(Prime)

ShamsOnlyTwitchSub(Prime)

[marsh forge]

Even working up to homotopy, I don't think there is actually a comparison that works here

(I could be wrong though I am quite tired)

[lunar yoke]

So I tried recalling the proof of the Whitehead theorem and noticed that somehow i was convinced one can assume the map f : X -> Y between the (lets say connected) CW complexes to be cellular by cellular approximation. Most proofs i could find also do this, but without elaboration.
But why is the assumption that f induces isomorphisms on all homotopy groups preserved by such an unbased homotopy?

If the homotopy could be chosen to be based then it would be obvious, but im not sure if that can always be done.

For instance, what if the map we start with has image disjoint from the 0-skeleton of Y?

[marsh forge]

Okay I keep trying to write things and keep confusing myself lol

[lunar yoke]

relatable

i literally only find proofs that either completely avoid this or dont elaborate

[marsh forge]

Okay I think the lemma needed is the following, which I was drastically over-complicating. Given any unbased homotopy between $f,g:X\to Y$, where $X$ has a basepoint $x$, we get an induced base-change $\pi_n(Y,fx)\to \pi_n(X,gx)$ given by restricting the homotopy from $f$ to $g$ to the restrictions to $x$. Then the following should commute

[\begin{tikzcd}[ampersand replacement=&]
{\pi_n(X,x)} & {\pi_n(X,fx)} \
& {\pi_n(X,gx)}
\arrow["\cong", shift left=1, from=1-2, to=2-2]
\arrow["f", from=1-1, to=1-2]
\arrow["g"', from=1-1, to=2-2]
\end{tikzcd}]

[gentle osprey]

ShamsOnlyTwitchSub(Prime)

[marsh forge]

in particular if one of f or g is an iso so should the other be

sorry

the RHS should be Y

not X

[lunar yoke]

oh i know that lemma

[marsh forge]

oh its true?

sick

i was just guessing

[lunar yoke]

sick guessing skills

[marsh forge]

I mean i am sure i heard it before

i just don't think about this stuff often

does that make sense phil?

as to how it answers the original Q

[lunar yoke]

yes, thank you

i just noticed that i almost drew this exact diagram in my own attempt

but i had a homotopy in the middle

and then it became messy

i should maybe try to do the more simpler thing first in the future

[marsh forge]

lol fwiw i did not think of this approach for like 15 minutes bc I kept trying to do something fancier

[lunar yoke]

well i guess the fancy proof of whitehead is that yoneda style one

and this is more a hands-on approach

[lunar yoke]

this is in Dieck p.127 btw

[lunar yoke]

well if y is in B_r(x) then d(x,y) < r and you have a ball around that y in your union

[rancid umbra]

this proof is pretty straightforward. another route your could take is to note that d(x,B) is a continuous function from X to R

[long grail]

What's the significance of the hairy ball theorem

[marsh forge]

its neat

also you can model a lot of things as continuous vector fields on a sphere

[gritty widget]

the classifying space of line bundles is RPinfinity with the canonical line bundle

but how do i find what the classifying map for the mobius bundle is

i assume it should be

include S1 into RPinfty as RP1

[shadow charm]

"toplogically, a dog is a sphere" has to be the most based thing i ever heard

[gaunt linden]

We all know that it is cows that are spherical.

[shadow charm]

exactly

[gritty widget]

im a bit confused with the topology on R^infinity

why can we say that the map (x1,x2,...) goes to (0,x1,x2,...) is continous

[gaunt linden]

There are several different topologies to choose between. The common ones do all make the shift map continuous, but the precise arguments will depend on which topology you want it to work for.

[gritty widget]

i'm using the topology via the direct limit of R^ns

[gaunt linden]

That's an unusual choice. According to https://math.stackexchange.com/questions/3025251/what-is-the-topology-on-the-direct-limit-of-bbb-ri-rightarrow-bbb-rj, the resulting topology is not metrizable, so you might need a somewhat ad-hoc argument there.
Perhaps you can show that the shift map is a homeomorphism onto its image, though?

[gilded crane]

let's say i have two topological spaces X,Y and two open subsets A in X, B in Y, why is A x B an open subset of X x Y ?

(in product topology)

[hollow harbor]

The product topology on X x Y is generated by preimages of open sets in X or Y under projections. What is the intersection of (preimage of A under projection onto X) with (preimage of B under projection onto Y)? Write out what points are in this set.

Since finite intersections of open sets are open, this is open.

[gilded crane]

hmmm, doesn't this intersection only contain one point?

[midnight echo]

does the intersection contain a subset of the product topology?

[gritty widget]

is the pullback of a trivial bundle trivial?

[midnight echo]

if a set contains an open ball does it imply that the set is open? or can closed sets also contain open balls? Thanks

[gritty widget]

1. no, (0, 1) is contained in [0, 1]
2. wdym by "also"

[midnight echo]

ahh yes of course

[gritty widget]

Sets don't divide into open and closed

[gritty widget]

A set contains an open ball iff it has non-empty interior

[midnight echo]

cause the open ball has radius > 0?

so it must contain something?

[gritty widget]

Well, yes

The open ball is non-empty, can have one point

But no less

[fading vale]

Closed under coequalizers here means that if f,g: X -> Y are cofibrations for example then so is the coequalizer q: Y -> Q? Not that q circ f = q cirg g

That basically makes sense I guess, I can try to come up with a specific counterexample maybe

Maybe theres something for abelian groups by looking at cokernels

[limber ravine]

Why isn't the indiscrete space connected?

For if X = {a}, the indiscrete topology would be {0, {a}}

why can't we make a partition like {{0}, {{a}}. Then the intersection would be 0 and their union the whole space

One question: is {0} open in the topology above? I see it isn't an element of such set but I am not getting it

Moreover, {0} and {{a}} are both nonempty subspaces of {0, {a}}

[fading vale]

Indiscrete spaces are connected

discrete spaces with more than one point are not

[limber ravine]

I mean is*

sorry

[fading vale]

is 0 supposed to be the empty set?

[limber ravine]

Ye

[fading vale]

Cause to be disconnected it has to be the union of disjoint nonempty open sets

[marsh forge]

(every space would be disconnected if empty was allowed)

[limber ravine]

but {0} is not empty?

[marsh forge]

{0} is not an open set

its not even a subset

i mean assuming your space is not some cursed bs w the empty set as a point

[limber ravine]

so the subset would be from X or from the topology?

[marsh forge]

open sets are subsets of X

to show a space is disconnected you must produce two nonempty open sets with trivial intersection whose union is X

[gentle osprey]

UltraLilith

UltraLilith

[fading vale]

The latter isn't

Otherwise that would imply that the empty set itself is an element of your space, like max mentioned

[limber ravine]

I see, the confusion was that I've been taking subsets from the topology

That's why I was refering {0}

[fading vale]

Yeah

You want elements of the topology

Not subsets of it

[limber ravine]

I see, then that's the reason for the discrete topology on {a,b} being disconnected

because {{a,b}, 0} , {{a}, {b}} satisfy the requirements

[fading vale]

Yes if I put the discrete topology on X, then any subset of X is open e.g is in the topology, viewed as a set of subsets

So i just need to write X as the union of any nonempty disjoint subsets

And if X has more than 2 points this is obviously always possible

[limber ravine]

Question: Can the partition be of more than 2 sets?

[fading vale]

Functionally speaking yes because the union of open sets is open

[limber ravine]

so for {a,b} a possible partition would be {0, {a,b}, {{a},{b}}?

[fading vale]

Thats not a partition because these are not disjoint

and they're not all non-empty

[limber ravine]

so {0, {a,b}} and {{a},{b}}

[marsh forge]

no

[fading vale]

[marsh forge]

you cannot include the empty set in a partition

{a},{b} is the only valid partition of the discrete space {a,b}

[limber ravine]

lol

[marsh forge]

you should really think about this geometrically

[fading vale]

Yeah I was about to say that if you're amenable to it I think you should draw a picture

[marsh forge]

this should be thought of as finding a separation of your space into two "connected" components

[fading vale]

The idea is that you are literally dividing up your space

[marsh forge]

it makes no sense to consider the empty set a connected component

[marsh forge]

you should also draw a picture if you aren't amenable to it

[fading vale]

[limber ravine]

uh, ok so for X = {a} we can't like divide X in two (or more) subsets such that their intersection is trivial and the union is X

but for Y = {a,b} we can make {a},{b}

[marsh forge]

a discrete topological space is connected if and only if it only has one point

[limber ravine]

For Z = {a,b,c} we make {a,b},{c}

for instance

[marsh forge]

yes

[limber ravine]

Interesting

[limber ravine]

Consider Q with the relative topology induced from R, letting $A = Q \cap (-\infty, \sqrt{2})$ and $B = Q \cap (\sqrt{2}, \infty)$ we see that Q is disconnected.

[gentle osprey]

mns

[gritty widget]

You can even show something stronger

Q is zero-dimensional

So it has a basis of clopen sets

[pearl holly]

they are so good at explaining https://www.youtube.com/channel/UCvyNnSLtoHOzeEwnOZWGXWQ/videos

anyone knows any other youtube channel about AT?

[viral atlas]

Wildberger

[marsh forge]

echt

[gritty widget]

Is the following equivalent to requiring a Mayer-Vietoris axioms

If $X$ is obtained by $B \cup_f Y$, where $A$ is a subcomplex of $Y$ and $f:A \rightarrow B$ is cellular

Then $\ldots h^q(A) \rightarrow h^{q+1}(X) \rightarrow h^{q+1}(B)\times h^{q+1}(Y)\rightarrow h^{q+1}(A) \rightarrow \ldots$ is a long exact sequence

[gentle osprey]

lime_soup

[fading vale]

Is that supposed to be a direct sum? anyway you can definitely recover normal mayer vietoris. for X = A cup B write X as the pushout of the inclusions A cap B -> A and A cap B -> B and apply this to recover the usual mayer vietoris

For going the other way around uhh

Gimme a sec

ok back

[marsh forge]

This should be a statement about preserving homotopy pushouts, which should be true

it might actually just be stronger than MV

I havent had coffee yet tho so im not sure

but homology should preserve arbitrary homotopy colimits

[fading vale]

Yeah this is how I wanted to think about it

Is there some result like

pushouts of CW complexes are homotopy pushouts

[marsh forge]

yeah

homotopy pushouts are computed with cofibrations between cofibrants objects

cellular and cw suffices

[fading vale]

oh right

[feral copper]

Hey folks! Does anyone know about the relative Thom conjecture (https://mathoverflow.net/q/75381/137265)? The question and answers are a 11 years old, maybe some things have come up in the meantime... Thanks!

[upper basalt]

like a continuous function maps a connected set to connected set , what can we say about complex analytic functions , and what are some topological properties of complex analytic function, are they behave same as continuous function?

[gritty widget]

Imo people around discord generally wouldn't know things like these, you'd get more information by asking directly on math overflow

[gritty widget]

Every complex analytic function is continuous of course, but they don't quite behave like continuous functions, they're much more rigid

[marsh forge]

You can also ask on the AT server if you are willing to use your real name

(I don't know anything about this conjecture)

[gritty widget]

I wanna be on AT server

[marsh forge]

There are invites floating around

[midnight echo]

if $E \subset X$, then is $CL(E) \subseteq X$?

[gentle osprey]

ArtyLeAardvark

[midnight echo]

where X is a topological space

[marsh forge]

If X is the entire space then the closure of E in X is necessarily in X

Just by definition

[midnight echo]

yes but what I meant was does the closure of the set then mean that CL(E) could be a subset or the set X itself?

whereas before E was a proper subset of X

does taking the closure then remove the fact that it's a proper subset?

[marsh forge]

It could be all of X yes

[midnight echo]

ahh ok

thanks

[marsh forge]

For example

If you consider Q inside of R

The closure of Q is all of R

Such a subset is said to be dense

[midnight echo]

ohhh that's what dense means

[marsh forge]

Yes

There are other equivalent definitions

[midnight echo]

ok I was kinda struggling to fully grasp what that meant

thanks alot

[marsh forge]

Np

[light rivet]

So I know S^n minus a pt is homeomorphic to R^n via sterographic projection, and S^n minus its north and south poles is homemorphic to R^n minus a point. I don't see why that is homemorphic to S^{n-1}

[lunar yoke]

Its not homeomorphic, but homotopy equivalent

im pretty sure this follows from invariance of domain

basically S^2 without north and south pole is still mostly 2 -dimensional, but S^1 is completely 1-dimensional

coincidentally, i also have a question regarding something similar

Let I = [0,1] and say I have a cont. map f : I^n -> I^n so that f restricted to the interior is an open embedding

is it true that f(boundary I^n) is disjoint from f(interior I^n)?

im going throught a very lengthy proof of hurewicz right now and this is claimed as true by "pointset topology"

i feel like this should also somehow follows with invariance of domain or something similar but im not sure

[light rivet]

yes, but S^n minus 2 point is not homemorphic to S^{n-1}, only homotopy equiv

[lunar yoke]

*two points

without a point its just D^n and contractible

[light rivet]

yep, mb

[feral copper]

I don't really care about my real name, I use it on the MathOverflow already! What's this AT server? 🙂

[lunar yoke]

i figured it out, its true, and really is just basic topology, not even invariance of domain

[lunar yoke]

https://discord.gg/U3fjYwyk

[feral copper]

Thanks!

[light rivet]

I have the following question. A continuous map f: S^n \to S^n induces a homomorphism of their nth homologies, so a homomorphism Z \to Z. I don't see why this is clear. I know that if we have chain complexes C,D, then a chain map f certainly induces a homomorphism of homologies, but why is that also the case here. A chain map is more than just continuous.

[lunar yoke]

So given a topological space X you can form the singular chain complex C(X) which has C(X)_n = free abelian group on the continuous maps from the n-simplex into X for n >= 0 and C(X)_n = 0 for n <0

using the inclusions of faces of simplices you can build natural boundary maps between these

and the homology of this complex is by definition exactly singular homology of your space

taking the singular chain complex is functorial, so sends continuous maps to chain maps

and homology as well is functorial as well, as you already noted

so continuous maps of spaces induce maps between the associated homology groups

i mean there are lots of details missing and maybe you should just consult a book

[light rivet]

Ok, I think I understand that now. I have the following other question. If I have S^n, and I take a hemisphere of it, why is that hemisphere homeomorphic to a n-simplex? This is clear to be in S^1, where the heimisphere is a closed line is the 1-simplex

[fading vale]

A hemisphere of an n-sphere is homeomorphic to D^n which is homeomorphic to an n-simplex

The latter homeomorphism is explicitly given by some kind of like barycentric subdivision projection thing but it also just follows from the n-simplex being convex

I think technically the barycentric projection thing is how the homeomorphism between convex sets and balls in R^n is constructed but like

Oh I guess ill ping since its been like 4 hours sorry if you already figured this out lol @light rivet

[light rivet]

I got it. ty

[fleet arch]

I have a question about borsuk ulam. There's this proof of it in this one 3b1b video and it feels deceptively simple compared to every other proof I've seen. I was wondering if someone could point out where the rigor breaks down

https://youtu.be/yuVqxCSsE7c

[marsh forge]

timestamp?

Seems fine to me @fleet arch

[fleet arch]

huh, so does this not generalize to S^n by doing basically the same thing?

oh wait, I guess I cant be certain about the loop cutting the origin anymore huh?

[gritty widget]

Hi, if I have a sequence of function $f_n:X \to Y$ that converges uniformly to $f: X \to Y$ ($X$ and $Y$ are metric spaces) how would I go about showing that if all $f_n$ are continuous, then $f$ is continuous too ?

[gentle osprey]

Entelechy

[bright rampart]

hey so I am a bit confused with question 3 in section 1.10 Well Ordering section of Munkres.

This is my depiction of what is happening

I know the ordering is not the same but i am having a hard time picturing the ordering relationships I suppose

oh wait i think i know where i may have messed up with the second one is it…

[lunar yoke]

I would say this is more of a real analysis question, but note that |f(x) - f(y)| <= |f(x) - f_n(x)| + |f_n(x) - f_n(y)| + |f_n(y) - f(y)| holds for all x,y,n and see if you can figure the rest out

[gritty widget]

Oh I see, two triangle inequalities

Thanks, for some reason I had trouble understanding where this inequality came from

[bright rampart]

I am not sure if my visualization here is correct

[bright rampart]

what exactly is it about the first picture that stands out?

the problem when we increase by n, is when we move up in y, I think we still need to follow x < y for each row

so how would you illustrate {1,2}xZ^+

hmmm

we have a jenga tower now more or less, does that count? haha

wait a minute

i think i figured something out

i was missing some arrows?

lmao

we have more stable tower now

😂

wait do i get rid of a vertical set of arrows?

oh wait

so we are taking {1,2} individually paired with Z^+

Which was something I have a hard time following a bit because wouldn’t our 2 always be greater when we climb many y times

also might ask about problem 10 shortly

[gritty widget]

This is topology?

[bright rampart]

its around the end of the first chapter in munkres

its just abstract deconstruction leading to point-set stuff

[bright rampart]

I mean its already posted here and its in a topology book in the second to last section of the deconstruction chapter leading to the topology part of the book... I don't see the point in reposting it

also I finished every problem in the section otherwise so I am not struggling elsewhere I don't think

[marsh forge]

Can you post the question

[bright rampart]

sure h/o

[marsh forge]

h/o?

[bright rampart]

hang on

[marsh forge]

Ah I see

Interesting

What do you think?

[gaunt linden]

This is getting near. Now you only need an arrow that declares the bottom right element to come after the entire left stack.

[bright rampart]

ok im going to take another look. I finished the other problems I cared about in the same section

[gritty widget]

Just because it's in a topology book doesn't make it topology

😅

[rancid umbra]

is there anything wrong with this argument?
Let X be a non-empty set and let B(X,R) be the set of bounded real valued functions on X equipped with the uniform norm.
Fix f,g in B(X,R) and consider the function F : [0,1] --> B(X,R) by F(t) = f + (g - f)t.
||F(t) - F(s)|| = ||g - f|| |t - s|, so F is lipschitz and F(0) = f, F(1) = g, thus B(X,R) is path connected.
furthermore, B(X,R) is a vector space over R

this seems like such a strong result. not sure if im missing anything subtle

[gritty widget]

Doesn't seem very strong

[rancid umbra]

i guess it seems strong to me because i assumed nothing about X other than it was non-empty. even if it was empty, this claim still follows. why does it not seem strong to you?

[gritty widget]

It just doesn't. The proof is easy and very general

And yes, it's absolutely correct

It works for X = empty too

Why it works is because R is a vector space, pretty much

Like, replace B(X, R) with B(X, Y) where Y is a normed vector space over R

[rancid umbra]

what norm are you placing on B(X,Y)?

[gritty widget]

||| f ||| = sup of ||f(x)|| for x in X

It doesn't matter much that f is continuous in the proof

The only place where that matters is in the definition of F

To make sure it's into B(X, Y)

[rancid umbra]

f isn’t assumed to be continuous tho

[gritty widget]

Topological structure of X is basically absent

Sorry, I read this as "B(X, Y) are bounded continuous functions"

But if you take X to have discrete topology then it's the same thing

[rancid umbra]

right. maybe surprising is a better adjective. i thought that the structure of X would influence B(X,Y), but it doesn’t

it’s really just the structure of Y that’s important

[gritty widget]

Because Y is a pretty strong space to have to begin with

Usually when you have C(X, Y) the emphasis is put on X being compact instead

[rancid umbra]

can Y be anything weaker than a real normed vector space?

[gritty widget]

If you take Y = l^p for 0 < p < 1 then I think it works too

So if you have topology defined in terms of some generalized norm that's not necessarily homogenous, but it satisfies triangle inequality and stuff, it should still work

Then your function F should be Hölder continuous instead of just Lipschitz

[rancid umbra]

but Y still needs to be a vector space over R, correct?

[gritty widget]

Not necessarily, you can still get some path-connectedness properties, even if it's not a vector space over R, I think

[bright rampart]

Ok I think I understand that problem I was struggling with.

Now I’m trying to visualize better more abstract orderings. For instance

{1,2} x (Z^+)^2… and so on

[lunar yoke]

as stated above this isnt really the channel for order theory. That said, if you're interested in well-orders, you could take a look at ordinal numbers. They give a nice way to describe all well-orderings

of course except for this based take

[bright rampart]

Thanks. Also I don’t think it’s necessary right now for me to think too hard about it

[sterile owl]

Does anybody know a good video explaining 'the circle not being contractible' proof?

[gritty widget]

I doubt there's just one universal proof of that

the one I know proves that the map $z\mapsto z$ of $\mathbb{S}^1$ is essential iirc. This is by proving it doesn't have a continuous logarithm

[gentle osprey]

Blitz

[sterile owl]

Hmm, alright. The one I'm looking at uses the spiral staircase metaphor but I'm unsure of the algebraic proof behind it.

[gritty widget]

it should be similar

having a continuous logarithm means something like, it factors through the map $x\mapsto e^{ix}$

[gentle osprey]

Blitz

[sterile owl]

Oh yes, that's exactly what I'm doing

[gritty widget]

this only uses some connectedness, nothing big

[sterile owl]

I have some holes in my knowledge from jumping around in the book, that may be where my issues come from lol

[gritty widget]

you suppose that there is $\varphi:\mathbb{S}^1\to \mathbb{R}$ such that $z = e^{i\varphi(z)}$

[gentle osprey]

Blitz

[gritty widget]

then take $z = \exp(it)$

[gentle osprey]

Blitz

[bright rampart]

alright boys, last section of the deconstruction chapter leading to the meat of Munkres. Wish me luck. I'm about to learn topology for real soon lol

[rancid umbra]

remember kids, topologies are just glorified semi-lattices

[gritty widget]

If we are working with an generalized cohomology theory

is the cohomology of the empty set always going to be

0

[high hill]

If we define the measurable sets of a Measure Space to be open, when does this not give us a topology? (an example would be appreciated)

(trying to link Measure Space and Topological Space)

[lunar yoke]

doesnt this follow from H^n(0) = H^n(0,0) and the pair sequence for (0,0)?

[gritty widget]

ah clever thank you

[high hill]

Feels surprising given the parallels in how a measurable vs continuous function is defined

[marsh forge]

Its somehow more natural to work with pointed spaces

Where the empty set really isnt a pointed space

In which case we are really looking at reduced theories

and the (co)homology of the point will indeed be trivial

This will correspond to the unreduced (co)homology of the empty set, I guess

[high hill]

It's been a long while since I've done measure

ah ok rereading the defns makes more sense

Sigma

• X
• Complement
• Countable union
  Topology
• 0, X
• Finite intersection
• Arbritrary union

===
Discrete, Indiscrete certainly always work

Hmm the topology has to be symmetrical (open and closed coincide), not sure of a name for that

It sounds like this topology has be a disjoint union of a bunch of indiscrete spaces

idk I feel like that's everything? If we have a clopen set, the space is disconnected and we can split into a disjoint union?
So I feel all the Topologies which are also Sigma algebras have to be arbritrary disjoint union topology of indiscrete topologies 🤔

[old girder]

Does anyone know of a proof for sequentially characterising interior points

[gritty widget]

Is the goal to find out which topologies can also be sigma algebras?

Because the criteria you gave seems strong

But I guess it makes sense. At the end of the day you want only complement to be satisfied because unions come easy

you want every open to be clopen too I think

Yeah I think the conditions you gave might be too strong

only because of the finite intersection condition

[high hill]

🤔

Can you give an explicit example of something I didn't include that could be Sigma

[gaunt linden]

{X,Ø} is a topology on X that's also a sigma-algebra, but that makes a connected space.

Hmm, that might not be an answer to what you actually asked.

Ah, shutting up, then.

Turn it around: Which sigma-algebras are topologies?

[gaunt linden]

Actually this does sound right to me.

[gritty widget]

This only works for finite disjoint unions

[high hill]

My topology isn't so strong. Can we not turn the discrete topology on R into an uncountable disjoint union of indiscrete topologies?

(for example)

[gritty widget]

Obviously yes

[high hill]

Given a topology, as long as it satisfies the complement axiom, it is also a sigma algebra

As for 'simplifying' this, I could only think of splitting the topology up (using the idea of disconnected -> disjoint union topology)