point-set-topology

[brittle rapids]

you're saying something but it's not quite phrased correctly

[white oxide]

Yeah, you are missing the mark a bit there

You phrased it as "there are numbers inside [0,0.99]"

[gentle girder]

hi kerr :3

[white oxide]

Hi :3

[quartz crater]

Okay so to answer your question the union of [0,r] through r\in (0,1) contains 0.99 because there exists an x s.t x \in [0 ,0.99] ?

[placid jackal]

if anyone could help explain how the resulting space is S^2 thatd be super appreciated--just a bit confused because it doesnt look like S^2 in my mind

[gentle girder]

draw the 1-skeleton on S^2 and see that the "outside" corresponds to the last attaching map

[unreal stratus]

Lol funny exercise

[gentle girder]

i guess my worry would have been that you get something that's like R^2 but note that D^2 is compact

so the last attaching map does get you the last bit of S^2 instead of a plane

(the attachimg map is literally "walk around each petal, go up the stem and then back down the stem")

[placid jackal]

ah i think i see now

that last bit helped a ton

thank you so much! I really appreciate it

[gentle girder]

i feel like talking in this channel helped me so much with algtop (and made me fall in love with it more, which was like less the case last semester)

[tribal palm]

in the plane, closed simple loops have an inside and an outside, while on a sphere this is not the case (consider the loop around the equator), does this property for a space have a name?

[white oxide]

I am not sure I follow, the loop around the equator still dissect the sphere into two components?

[quartz crater]

maybe this is better:

[knotty vine]

But on the plane, one is bounded and the other is not

[tribal palm]

right

so does metrizatibality and compactness necessarily play into this?

oh, the Jordan-Brouwer separation theorem seems to be the the topological generalization of the R^2 case i had in mind

or, i’m not quite sure

what is a topological sphere in R^(n+1) ?

a subspace hom to some S^k ?

oh nvm it says, lol

actually. can we have a injective, continuous image of S^n with the subspace top in R^(n+1) not hom to S^n ?

[tribal palm]

either way this is wrong since we’d clearly need k=n

[hidden crag]

No

It’s an injective map from a compact space into a hausdorff space

[tribal palm]

how is that enough to conclude the map is a imbedding?

[knotty vine]

It implies that the map is closed, and a closed injective continuous map is an embedding

[unreal stratus]

yeah this is a really useful ting to remember

[white oxide]

mad ting that theorem

[tribal palm]

i… did not know this

thanks

[tribal palm]

what about generalizations into other spaces than euclidean space?

i am probably not even at a point to grok the above theorem but am still very curious about what theorems are established

[white oxide]

The closest thing that comes to mind is the excision theorem for good pairs, although I am not sure if you can choose the pairs appropiately?

spiritually you wanna show that H_0(X\A,A\A) is non-trivial where A is your embedding of the sphere and X your ambient space.

[tribal palm]

spiritually

[white oxide]

I guess I should say "in spirit" but eh

[tribal palm]

i am unfortunately not really familiar with even basic algebraic topology (besides the intuitive “the fundamental group contains some information about the holes in a space”)

nor do i know what a homology is

[white oxide]

Well, the jordan curve theorem for the plane is proved using the fundamental group usually

But more generally homology groups capture the information present in brouwer separation theorem

[tribal palm]

i have only seen a sketch for the proof, using that the parity of the number of times any line intersects the curve is invariant to rotation around points (under quite strong assumptions about both those points and the curve)

[white oxide]

hm

Actually, if you had homology you ought to be able to nuke the problem

[tribal palm]

swag

whatever nuking the problem means

[ebon galleon]

i love nuking problems

[tribal palm]

can you explain what it entails

[tiny obsidian]

Using really strong results to solve problems that could be done with simpler methods

[tribal palm]

waw, sounds like fun

[wispy veldt]

very fun

[ebon galleon]

[unborn dagger]

I'm studying basic point-set topology right now, and I have a question about definition of homeomorphisms (a bijective function such that both it and its inverse are continuous). For topological spaces in general, is it true that the inverse function of a given bijective continuous function is always continuous? (if you think I should have asked this question in a different channel please inform me)

[ebon galleon]

No.

[unborn dagger]

Yeah I guess not too lol. Trying to prove this but I can't prove that a bijection continuous function is always an open map so I guess that's where the problem comes from?

[unreal stratus]

Lmao

[cedar pebble]

take the identity function from R with the discrete topology (where every subset is open) to R with the Euclidean topology

[unreal stratus]

identy functions are often good for these things

[ebon galleon]

nG snipe

wg

stealing my clout

[unreal stratus]

Here's a more natural example (imo): Consider the function [0,2π) -> S^1 sending φ |-> (cos φ, sin φ). This is obviously continuous and bijective.

But S^1 and [0,2π) are not homeomorphic (and specifically, the image of [0,1) is not open, for example)

[unborn dagger]

thanks a lot! will think on your examples XD

[brittle rapids]

there is a characterisation of when the identity function (X, τ) -> (Χ, τ') is continuous

which will clarify the intuition

[unreal stratus]

although saing "identity map is not continous" hurts me inside

i'm glad nG said function

[brittle rapids]

hot potato

[ebon galleon]

indeed

[unreal stratus]

lol

[hexed briar]

Does there exist a knot with unknotting number n for any natural number n?

I'd assume so because isnt the unknotting number just roughly half the knotting number?

[heady skiff]

can somebody tell me if this is continuous in the box topology or not? I'm pretty sure it is, and here was my reasoning: I wrote $h$ as the composition $g \circ f$ where $f: (x_1, x_2, \dots) \mapsto (a_1x_1, a_2x_2, \dots)$ and $g: (x_1, x_2, \dots) \mapsto (x_1 + b_1, x_2 + b_2, \dots)$. so i took an arbitrary basis element of $\mathbb{R}^\omega$ under the box topology, and looked at the preimages, and found that it happened to be an open set of $\mathbb{R}^\omega$

[gentle osprey]

okeyokay

[unreal stratus]

Yeah I agree

Like the preimage of an open subset of R under a linear map is open in R

[heady skiff]

ye

[unreal stratus]

then because we have the box topology that means preimage of a basic open is a basic open

And then I guess it should still be a homeo right

because ||you can just write down the inverse in the same form||

[heady skiff]

yeah i think i already did that part

ty

[unreal stratus]

np!

nicely done

[heady skiff]

tyty :)

wouldn't the coarsest topology on $A$ just be the topology generated by the subbasis $\mathcal{S} = \bigcup_{\beta \in J} \mathcal{S}_{\beta}$

[gentle osprey]

okeyokay

[tribal palm]

yes

[tribal palm]

my prof told to just ignore everything about the box topology saying it’s completely irrelevant and i’ll take it, mwahaha

[heady skiff]

lol yeah i'm tempted to skip this section

but i promised myself i would do a decent amount of problems from each section

[tribal palm]

but yeah this is the actually important topology

[heady skiff]

is this a generalization of the product topology to arbitrary sets which need not be a product or smt

[tribal palm]

waht

[heady skiff]

cuz in the product topology you take as a subbasis all preimages of open sets in each X_\alpha under the projection map

[ebon galleon]

product topology = coarsest topology such that all projections are continuous

[heady skiff]

but there you're considering the topology on an actual product

anyways i'm prolly yappin

[ebon galleon]

This is the definition of "coarsest topology containing a collection of sets"

[white oxide]

havent heard of the box topology in years, jesus

[ebon galleon]

Or in this case I guess, coarsest topology such that a specified set of functions are continuous. This gets into uhh some categorical stuff I guess (it relates to how you construct limits in Top)

[tribal palm]

i mean if you’re asking if the product top on one space is the same as the topology on the space then the answer is yes, but not in any interesting way

[white oxide]

It being the coarsest is more of a consequence of the universal property than any construction

[tribal palm]

just in the same way that you can identify points with 1-tuples of points

[white oxide]

I am bit confused, you described essentially the same subbasis twice?

[heady skiff]

nvm ignore it

[gentle girder]

lol box topology mentioned

[red yoke]

The point of box topology is to show you why product topology is better

[balmy field]

So true

[somber falcon]

This construction is actually used. Box isn't important, but weak top is.

[white oxide]

May we hear from it again in like 3 years

[somber falcon]

This and the order topology on the reals, another munkres example

[red yoke]

I thought order topology be pog

[somber falcon]

I never saw it again...but I study manifolds

[alpine nest]

isn't order topology on the reals the same as the Euclidean topology?

I mean, I agree that the Euclidean topology is pog, of course

whatever pog means, I'm in my 40s and no longer down with da youth

[alpine nest]

Pandora's Box topology

[rigid path]

Anyone here familiar with realcompact spaces?

[naive trench]

Real compact?

[rigid path]

yes

[gleaming warren]

I saw them in a book, for what do you need them?

[rigid path]

I need to study them, so I was wondering if anyone has some links or stuff to understand them deeper

[gleaming warren]

is there a deeper reason why you study them or just to know them?

[rigid path]

It's for my thesis

[gleaming warren]

What's the topic?

I encountered them too for my thesis, since they somehow help to understand prime ideals

in C(X)

[rigid path]

It's not well defined yet, but I'll surely need them

[gleaming warren]

Is it more topology or more analysis?

[rigid path]

The first one

[gleaming warren]

okay I saw them more from a analysis point of view

[somber falcon]

Yeah that was the punchline to the munkres thing

I had to look up pog

My first thought was those coin thingies

[buoyant dew]

what's he trying to say here?

[umbral panther]

Most long exact sequences are of the form
A_n -> B_n -> C_n -> A_n-1 -> B_n-1 -> C_n-1
Where the index changes every 3 terms

[buoyant dew]

ah right, so it's an intuitive remark - he's not saying anything profound

[tidal cedar]

yeah

[buoyant dew]

ok thank you both

[tidal cedar]

I guess you can see this as coming from like

[buoyant dew]

yeah yeah snake lemma and such

[tidal cedar]

the SES of chain complexes inducing the LES of homology

SES being 3 nonzero terms, yadda yadda

it's interesting they say "the" homology and homotopy LES

[umbral panther]

The LES associated to a SES of chain complexes; the LES associated to a fibration of spaces

[buoyant dew]

bit unrelated but i just want to appreciate how good of a book tom Dieck is

[white oxide]

oh no, rat escaped the containment zone (discussy) 🙀

[tidal cedar]

yeah

[tidal cedar]

YES

I HAVE NEVER SEEN YOU BEFORE BUT FROM YOUR NAME AND THIS MESSAGE I ALREADY LIKE YOU

[buoyant dew]

<3

we are friends now

[fading vale]

:copium:

[quiet thorn]

Munkres and hatcher have the same number of letters

[alpine nest]

Coincidence?

[unreal stratus]

Well it is kinda profound lol

[tribal palm]

that "munkres" and "hatcher" have the same number of letters ?

[grand monolith]

Hi I have some questions about basic topology

I understand that there are open sets and closed sets, and that it's possible for a set to be open and closed at the same time (or neither)

Sets are open in a top space if finite intersections are open and unions are open

Sets are closed in a top space if finite unions are closed and intersections are closed

The complement of an open set is a closed set

And vice versa

So I was reading this topological proof that there are infinite primes

And it feels weirdly too abstracted, like it feels more like notational magic that might obscure some faulty logic or something

So they defined the space as such

$N_{a,b} = {a+nb : n \in \mathbb{Z} }$

[gentle osprey]

CosmoVibe

[grand monolith]

Where b > 0

And a and b are both integers

All of these sets are both open and closed

[white oxide]

You can't seriously think that's it's just a coincidence

[tribal palm]

impossible

[grand monolith]

And so to prove that there are infinite primes, all they have to do is show that, assuming finite primes, the finite union of the N sets for $b \in \mathbb{P}$ is closed, but the complement is finite and open, a contradiction

[gentle osprey]

CosmoVibe

[grand monolith]

How do we justify these topological axioms

[grand monolith]

Is there a reason we can assume things like this

[grand monolith]

And this

At the same time

Is this something in topology 101

Is it just a simple set theory proof, and these sets happen to agree?

[devout karma]

Assuming X and Y are metric spaces and X is compact, for two continuous functions f, and g between them is it necessarily true that d_y(f(x),g(x)) for x in X is bounded

it would be really cool and awesome if d_y(f(x),g(x)) was in C(X) but idk how to show that off the top of my head

[naive trench]

Since f and g are continuous then f(X), g(X) are compact. Now compact implies boundness but distance betweem images mmmm

[merry geode]

Distance between compact sets is bounded, I believe

[devout karma]

if d_y(f(x),g(x)) itself was continuous from X to R then you could use a cover of preimages on X and show it’s bounded

[naive trench]

Ahg wait Idk how is boundness on a metric space

[merry geode]

Boundedness, as being inside a big enough ball?

[merry geode]

Ah wait, this argument works as well

[devout karma]

Idk how to show it lmao

[merry geode]

Thank Hausdorff >.>

[white oxide]

I mean, its still a cts function from a compact set aint it

[devout karma]

How the fuck do I prove d_y(f(x),g(x)) is in C(X)

[white oxide]

oh wait you havent shown the metric to be cts yet?

[merry geode]

X -> X x X -> Y x Y -> Y

[devout karma]

i literally just forgot because last time I did the proof was a year ago lma

[white oxide]

well, i guess take a refresher there

you know that (f,g) is cts so thats the missing piece

[devout karma]

i haven’t proved the metric itself from Y x Y is continuous in a long ass time

[merry geode]

Ah, fwiw me too. I just think every map in topology is continuous (unless it was constructed otherwise)

[white oxide]

well, as a hint: try to imitate the epsilon-delta way

[devout karma]

my goal was to show that the space of continuous functions from X (compact) to Y, both metric spaces, is a metric space if it’s true

[white oxide]

as a different question or still for the original one?

[stuck geyser]

The original one

[white oxide]

wtf

[stuck geyser]

Like a year ago I wanted to see if Arzela Ascoli could be generalized in a specific way and I had some stuff down but I did a funny and skipped over a step that I forgot how to do

[white oxide]

who is the real one 🔫

[stuck geyser]

It won't let me fuckin login to my main on my phone

i'm pissy but too lazy to fix it

[white oxide]

I mean thats way too overkill for showing that it is bounded

Nor am I sure how they would be even related

[naive trench]

Right I was trying to see it witv the distance and not open sets

[stuck geyser]

$d_{C(X,Y)}(f,g) = \mathrm{sup}_{X}\left(d_Y(f(x), g(x))\right)$

[gentle osprey]

Mizalign

[white oxide]

d_Y(f(x),g(x)) would be a cts function form X into R, so a cts function from a compact set into R. So it achieves a maximum and a minimum

[naive trench]

I think its just the difference in norm of f and g respect Y

With the supremum over X ofc

[stuck geyser]

The image of any compact set under any cts function should be compact. For a cover, the cover set preimages cover X, so chose a finite subcover of preimages over X. Take the image. Cash money

nah wait

[white oxide]

nah dont wait

the finite subcollection must cover the image

[stuck geyser]

I thought the image of a preimage is a subset of the original

not necessarily the original

a refinement, mayhaps

[white oxide]

okay so you have found U_1 , ... , U_n s.t. their pre-images cover X

[merry geode]

You can use characterization of compact sets on a metric space.

[stuck geyser]

nvm fair

[merry geode]

Or are you trying to show that the image is compact?

[white oxide]

then U_1, ... , U_n cover f(X). For if x isnt covered by any of them, then it is disjoint from the U_i and hence its pre-image is disjoint from them too

[stuck geyser]

fair

Forgot they were preimages of something lol

but yeah

I wanted to see if you could generalize Arzela Ascoli to C(X,Y) if it's between metric spaces with the domain compact

To characterize compactness on C(X,Y)

I at least wanted to see if being compact in C(X,Y) (as a metric space) implies equicontinuity and pointwise boundedness (OR the pointwise IMAGE of the valuation is compact)

Since the valuation of X x C(X,Y) to Y I'd assume is continuous lol

lol wait X x F would be compact in the product topology cuz Tychenoff lmao

so the image would have to be bounded because it'd be compact

[merry geode]

Yea

[stuck geyser]

Equicontinuity is a hmm moment

[grand monolith]

Ok I think I get it

Between these three axioms:

• unions and finite intersections of open sets are open
• intersections and finite unions of closed sets are open
• the complement of an open set is a closed set and vice versa

If you assume two of them you can prove the third

$-(A \cup B) = -A \cap -B$

[gentle osprey]

CosmoVibe

[grand monolith]

So that's why it works

[white oxide]

You have to assume the third

Otherwise you could choose a smaller collection of closed sets, for example take the standard topology on R and for the "closed sets" take trivial topology on R. Then it satisfies 1 and 2 but 3 fails

[grand monolith]

Ahh

Oh I know what I'm missing now

You need an open set membership definition

Right?

Wait that has nothing to do with this

Ohhhh I see

The open set membership is qualified if it induces these three properties

Is that it?

So if I pick a definition for open sets that doesn't induce these properties, then it's a bad definition

Brain working?

[white oxide]

You either need to define closed sets as the complements of open sets, or equivalently

[grand monolith]

Got it

Thanks

Uh one more question actually

I noticed that this particular topology proof doesn't make use of metric spaces even though I know they are foundational to topology

Is this because it is abstracted away but necessary from first principles?

Is there a way to demonstrate that a top space that has a notion of open and closed sets must have a valid distance function?

Or am I misunderstanding something

[ebon galleon]

Metric spaces have a distance function, which gives us a way to talk about its open sets; that is, a metric induces a topology on you set.
They're important in topology because metric spaces are typically quite nice and a lot of things are easier in metric spaces than in general topological spaces. But there are many topologies which don't come from metrics (arguably, most don't)

[grand monolith]

So it is true then that all metric spaces are topological spaces or conversely?

[pallid delta]

any metric induces a canonical topology, but not all topological spaces are metrizable

[grand monolith]

Thanks!

[valid escarp]

i mean i get this but, cant i just find an arbitrary metric on the topology and call it induced? what does induced mean?

[pallid delta]

A metric induces a topology whose basis is the set of open balls

[valid escarp]

that it has come in a "natural" way?

[knotty vine]

Yeah, the continuous functions between metric spaces are precisely the same as the continuous functions between the induced topological spaces

[gritty widget]

In this proof,$ L_1(\mathbb{R})$ refers to f being inside the family of $L_1$ norms, right?

[gentle osprey]

radar ashe

[gritty widget]

This proof just confuses me

I understand what it is saying, but why does this only work when the equality holds in the L2 norm

so we have f which is an L1 norm, we take the fourier transform of f

and then we can say f(x)=the summation of a bunch of discrete values

but it's an infinite amount of discrete values

[feral copper]

That's not #point-set-topology, even though it speaks of 'norms'. It's (functional) analysis

Also what it is saying is: $$\left|f-\sum_{n\in\ZZ,;|n|\leq N}[...]\right|_{L^2}\tendsto{N\to+\oo}0$$

[gentle osprey]

Matplotlib

[feral copper]

So: you're assuming that f is L^1 (integrable), to be able to look at its Fourier transform with the actual integral. Then, if it happens that F(f) is compactly-supported, then the series of functions where sinc appears converges in the L²-norm to f

[buoyant dew]

homological algebra question, not sure if this is the right channel

is this meant in the intuitive sense or is there an interesting categorical naturality going on

so I suppose we can write the exact sequence as a functor

[unreal stratus]

If we have sequences A -> B -> C and A' -> B' -> C' and a map from one of those to the other, then it should induce a map between the exact sequences in the "obvious" way

i.e. be a functor from sequences of the appropriate shape to sequences of the other shape

[buoyant dew]

potato how come you always save the day

[unreal stratus]

Because I don't 😭

Alhough, to be fair, I don't see which this should be the case lol

[buoyant dew]

i guess it works out

[unreal stratus]

No yeah it does

Cool

Like, say we have a square uh

$[\begin{tikzcd}
A & B \
{A'} & {B'}
\arrow["f", from=1-1, to=1-2]
\arrow["\alpha"', from=1-1, to=2-1]
\arrow["{f'}"', from=2-1, to=2-2]
\arrow["\beta", from=1-2, to=2-2]
\end{tikzcd}]$

[gentle osprey]

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

[unreal stratus]

Then like

if f(x) = 0, then

f'(alpha(x)) = beta(f(x)) = 0

so alpha(x) is in ker(f')

and the map alpha restricts to a map ker(f) -> ker(f')

Then similarly for all the other things

[buoyant dew]

So the functor is just R-Mod(A,B) x R-mod(B,C) -> Exact right?

[unreal stratus]

Uh wdym Top

lol

[buoyant dew]

shit

[unreal stratus]

But uh

No so like

If you want to make it more detailed

You consider the category of sequences A -> B -> C

with morphisms being the ladder diagrams i mentioned

Then you consider the category of long exact sequences with morphisms again being ladders

This construction is a functor from the first category to the second

[buoyant dew]

ok ok i understand

[unreal stratus]

But writing it out like this is often more painful than illuminating so often people just say natural like this

[buoyant dew]

see the last time i was taking AT I was reading Hatcher

[unreal stratus]

Sure

Yeah that doesn't have much hom alg in it i suppose

[buoyant dew]

yeah

so I have to spell everything out rn

thanks for the help though. how long have you been in topology?

ah you're an undergrad? sorry i assumed you were a grad student

[hidden crag]

who would've thought

[quiet thorn]

No, you just simply abhor hatcher for the sake of abhoring

[buoyant dew]

no no that book is actually really bad

ive actually read half of it

[white oxide]

What do you like then?

[buoyant dew]

tom Dieck has been pretty good so far

very good even

[buoyant dew]

I see this but I don't quite understand what the claim of naturality is supposed to mean here

[unreal stratus]

The correct approach to AT is to start with luries higher topos theory

[buoyant dew]

painful

surely "natural" doesn't just mean the above construction is functorial, right?

[hidden crag]

That is what it means

[buoyant dew]

when I see 'natural' i assume the maps are actually components of a natural transformation/isomorphism

verily, i am confused

[white oxide]

If you want to do algebraic topology you need some experience with algebra jestie

I recommend going through luries higher algebra first

[unreal stratus]

It often means functorial

[white oxide]

I think of it as being the same as canonical, choice-free

[buoyant dew]

damn

[white oxide]

Although it usually does end up being a map that's part of a larger functor/natural transformation

[buoyant dew]

thanks ppl

[ebon galleon]

So true! I don't understand why this isn't done more often

[hidden crag]

of course ryx is summoned by that message

[quiet thorn]

Is that actually what ryx did

[fading vale]

Hatcher haters... Sad

Visual is good

[grand monolith]

Ok maybe a weird question

What makes a math problem "topology"?

So I saw a proof of the infinitude of primes using topology

It uses open and closed sets to show a proof by contradiction

[naive trench]

Ohh thats a cool problem

[grand monolith]

I also understand that top spaces are not always metricizable and that metric spaces often induce top spaces

But like

What then is the common thread between that proof and the coffee mug bagel animation

[white oxide]

I mean, yeah. They often do

[naive trench]

Not often, a metric space induces a topology always

[grand monolith]

Why are these both "topology"

[grand monolith]

Oh

[white oxide]

You might even say the set of exceptions has measure zero, as required by the first measure axiom

[naive trench]

The open balls in the metric are the open sets in the topology

[naive trench]

Thats a homeomorphism

So you can "deform" your space into a another

[ebon galleon]

No lol, but I did learn model categories before algtop

[stuck lagoon]

Can anyone tell me where to read about what is known about discrete metric spaces? By this I do not mean metric spaces defined via the discrete metric (although this is one example). I mean metric spaces where every point is disconnected (meaning, for every point x, there is an open ball of size d > 0 such that x is the only point in the ball).

The topology induced by any discrete metric space in the above sense is always the same, the discrete topology. However, it seems to me that there is a lot more potential structure in the metric space that is lost by passing to topology.

I have been having trouble looking for resources on this, because it is one of the rare cases where the metrical objects at hand truly do not have good topological analogues.

[grand monolith]

Ok but what makes it topology then, the notion of open and closed sets?
Also I'm guessing that homeomorphisms require some kind of notion of continuity so it's probably very difficult to find any relevant application of homeomorphisms in a domain Z^n for any n, right?

Surely homeomorphisms are just specific tools for analyzing particular topologies then?

[naive trench]

Homeo are to other importart stuff. Is like a isomorphism but we ask to the map be continuous and their inverse too. There are topological properties that are invariant under homeo so homeo are quite important

[buoyant dew]

this does not answer your question but if I am not mistaken, there is a very natural Borel measure on this topological space which allows you to assign 'density' to residue classes which is pretty heavily used in analytic number theory

[umbral panther]

Why do you care?
Why does the metric matter?

One popular topic is the large scale geometry of groups. Take a finitely generated group and fix a generating set and define the distance to the origin to be the word length in the given generating set. This depends on the choice of generating set, but only up to “coarse quasi-isometry”

[buoyant dew]

you're looking for totally disconnected spaces i believe

[umbral panther]

Totally disconnected spaces includes the cantor set, which the given definition excludes

[naive trench]

Topology is the generalization of continuity and other stuff from R^n or C^n to other spaces with a topology or metric defined. Imagine that you are working with a map between spaces that are not R^n or C^n, how would define continuity? That way topology comes and give us a lot of properties as: completeness, compactness, continuity, connectness, density, etc. Topology is a powerful tool that is used like... almost everywhere

[ebon galleon]

Disconnected was the wrong word. I think "isolated" probably fits better

[grand monolith]

Not too familiar with some of these terms, is this basically saying that for any two sets of integers, we can somehow define a distance function between them the loosely describes their "density"?

If so that partially answers my question

[buoyant dew]

no, that's not it.

[stuck lagoon]

Why do you care?

One example of this is the metric $\max(K(x | y), K(y | x))$, the max kolmogorov complexity "between" two strings. This clearly has a very rich structure that isn't captured by the discrete topology.

[gentle osprey]

wikiemol

[buoyant dew]

don't worry about what I just said above, i just wanted to point out that even the 'weird' topological space (natural numbers with residue classes as basis elements) has a really good reason to be called a topological space

[buoyant dew]

which definition?

[grand monolith]

Ok, so would a better description be that topology abstracts notions of continuity and other properties like how abstract algebra abstracts notions of algebra through their properties?

[ebon galleon]

The given definition is that it induces the discrete topology.

[grand monolith]

And it does this through defining metric spaces?

[buoyant dew]

ah yes

[buoyant dew]

i am blind

[umbral panther]

I’m suspicious of taking something asymmetric and forcing it to be a metric by taking the max of the two directions. It doesn’t want to be a metric

[ebon galleon]

:( it can be a metric if it wants

[naive trench]

Generalizes some properties that we already know in R^n and C^n yeah

And i dont think makes sense that abstract algebra abstracs notions of algebra

[stuck lagoon]

This is exactly what we do with the metric |x - y|, over the real numbers. I see no problem with this.

[grand monolith]

Btw thanks for being so patient with me

[buoyant dew]

no you ask a good question

[naive trench]

Before define topology one see first metric spaces. To understand the notion of open set with open balls, the same with closed ones and so on

[grand monolith]

Ok

So now one more thing feels weird

[naive trench]

Metric and topology spaces are both important ofc

[grand monolith]

If metric spaces are so foundational to topology, then how come the top spaces that don't have metric spaces are also top spaces?

[ebon galleon]

Yeah I think this is a reasonable way to make something symmetric

[buoyant dew]

the topological space is a generalization of the metric space

[grand monolith]

Ohhhhhhh

[naive trench]

Topology spaces have their own definition and a metric space induce a topology space as a consecuence of the defintion

[grand monolith]

Ahhhhhh got it

Okay thanks so much

[naive trench]

I wouldnt say its a generalization

[grand monolith]

Still have some haziness obv since these explanations are very loose and non technical but I appreciate the responses and I think I have what I need to work with for now

[white oxide]

I would say so

[buoyant dew]

what book are you following?

[white oxide]

Just a nuclear generalization

[grand monolith]

Currently none

Any recommendations?

[naive trench]

Munkres

[buoyant dew]

ah. then the standard book is Munkres

really it's very hard to grasp these concepts without a proper introduction

[white oxide]

Man I wish there was a standard book for a first course of AT

[buoyant dew]

tom DIeck

tamm otomd ieck

i guess it's not the standard but as far as i can tell thats the book preferred by actual AT people

the only thing the book lacks is a proper introduction to category theory

[white oxide]

Lmao

Love the shade

[buoyant dew]

wym

what does shade mean I only know scottish slang

[stuck lagoon]

I am not sure this is what I am looking for, again, this seems like a purely topological concept, but every discrete metric space induces the same topology, the discrete topology, so it seems that topology is not (entirely) the right tool, even though intuition from topology probably applies

[buoyant dew]

yeah no i misread your initial message

[stuck lagoon]

Its really weird that I am coming up blank on searches for this

I would expect there to be more about this online

[buoyant dew]

i wonder if we can embed each such metric space (of cardinality <= continuum) into R^n for some n

https://math.stackexchange.com/questions/4201166/not-every-finite-metric-space-embeds-in-an-mathbbrk aha!

doesn't even work in the finite case. interesting

[white oxide]

Is there? You can recover the metric from the discrete topology tho?

[stuck lagoon]

That is very weird. Although, the solution there could be just, make it an assumption! haha "Embeddable" metrics may be mostly what I am interested in actually, but its hard to say

[stuck lagoon]

And no, they can all be induced by the discrete metric (the metric which is 0 when x = y and 1 otherwise)

[white oxide]

the "is there" refers to the additional information

[ebon galleon]

Consider a finite graph. To each pair of points, you can assign a distance by the minimum length of a path

This induces the discrete topology, but the distance itself is meaningful and cannot be recovered from the topology

So yes, there can be non-topological data given by the distance function itself

[buoyant dew]

well the graph need not satisfy triangle inequality

but yeha

yeah** damn

[white oxide]

howdy

[ebon galleon]

Why would it not?

[buoyant dew]

crap

[pallid delta]

but the metric does

[buoyant dew]

yeah

yeah yeah lol

i thought you were giving weights to edges

[pallid delta]

even then

[ebon galleon]

All graphs are transitive so true

[buoyant dew]

lol

[ebon galleon]

And yeah that should still work with weights

[stuck lagoon]

If there are weights it does not necessarily,
consider a triangle with the hypotenuse weighted larger than the sum of the other two sides

[pallid delta]

still a metric

[white oxide]

well, they said minimum length

[stuck lagoon]

Oh haha true, yeah

[white oxide]

so its fulfill the triangle inequality by def there

[ebon galleon]

Oh uhh I guess you might need connected graph actually

[pallid delta]

just make it infinite

[buoyant dew]

yh I was thinking of an entirely different, stupid construction

[ebon galleon]

Sure, if you allow metrics to take infinite distance that also works

[pallid delta]

or greater than any achieved distance, and argue that makes no difference

[white oxide]

Reals? extended

[buoyant dew]

of course in the infinite, weighted case the topology might just turn out to be non-discrete right?

[pallid delta]

yeah

[knotty vine]

If continuous maps are the "right" morphisms to consider for the category of topological spaces, but metric spaces hold more information than just their topology, what is the "right" kind of morphism for metric spaces?

[pallid delta]

but since the question came from CS, graphs may be assumed to be finite

[ebon galleon]

Nonexpansive maps

[buoyant dew]

non expensive maps

[knotty vine]

Why not isometries?

[white oxide]

turn the real numbers into a sufficently cursed graph

[ebon galleon]

I.e., d(f(x),f(y)) ≤ d(x,y)

[buoyant dew]

my prof has a paper like that

[knotty vine]

Seems kinda lax! I smell double categories!

[buoyant dew]

those preserve the entirety of the metric structure. Those should ideally be the 'isomorphisms' of the category

[pallid delta]

looks like 1-Lipshitz, but without a norm

[ebon galleon]

Isométries would force everything to be injective also, so it's a bit too strong

[white oxide]

well, you still get them from inverse non-expansions

[knotty vine]

Aha that makes sense

[ebon galleon]

That's exactly what it is, you can talk about Lipschitz between metric spaces

[ebon galleon]

Double categorical foundations of analysis sotrue

[stuck lagoon]

I have no idea what we are talking about now lol

[knotty vine]

nevermind, Im just -brained

[white oxide]

not long and you will write about continental philosophy modelled by higher topoi

[buoyant dew]

the marxist taco

[pallid delta]

they moved on to another question

[knotty vine]

Eat your heart out Wittgenstein!

Always funny seeing the giant Hegel pages on nlab

[white oxide]

some hegelian dialectics fans said its summary of the philosophy is bad, so schreiber probs has a target audience of like 2-3 people for it

[knotty vine]

Sorry, what was the original question?

[stuck lagoon]

Sounds like a question for stackexchange then, only have to figure out how to phrase it without getting downvoted to hell

[ebon galleon]

Yeah I'm not sure of any great resources for what you're looking for, I was just trying to motivate why you'd care about the distance function and accidentally derailed it

[knotty vine]

@buoyant dew Stallman pfp and Hatcher-hater name... my arch nemesis

[buoyant dew]

@stuck lagoon have you considered looking up finite metric spaces? there seems to be some work on the combinatorial structure

[ebon galleon]

Are you a Hatcher apologist

[stuck lagoon]

Interestingly, I was hoping the metric I mentioned as an example (conditional complexity) would satisfy something akin to this property by the data processing inequality, but sadly only applies to mutual information, which I can't make into a useful metric

[knotty vine]

I liked Hatcher, then again it wasnt my first AT book (which was Lee)

[buoyant dew]

man that book, albeit indirectly, gave me depression. like i legitimately began therapy

[stuck lagoon]

I will look into that! Could be a good starting point. My main motivating example is definitely countably infinite spaces (spaces of strings, computable functions, etc), but perhaps I will find useful things that generalize to the infinite case from studying the finite case

[white oxide]

The deformation retract visit me when I sleep sometimes

[knotty vine]

I admit that "Delta-complexes" were a bad move

[buoyant dew]

mind if I ask where you got the idea from? are you working on a problem or just casually interested?

[buoyant dew]

reading hatcher is like reading persian poetry by someone who only knows english

[knotty vine]

Have you looked at the pretty pictures though?

[knotty vine]

Sounds interesting!

[buoyant dew]

what field are you in btw? topology or something else? @knotty vine

[knotty vine]

I mostly do category theory now, but with applications in CS, type theory, and homotopy theory

[buoyant dew]

aha interesting

[stuck lagoon]

Mostly just casually interested. I feel like there is something to be gained from understanding computable functions in terms of information theoretical metrics.

I think one motivating problem was in AI. Generally speaking, often the way AI is trained is via a minimization of a loss function, and this generally corresponds to the topology induced by continuous functions with an Lp norm. This is how, e.g. the universal approximation theorem was proved for single layer neural networks.

However, I realized that we do not actually know for sure if minimizing the loss function actually corresponds to minimizing information distance. So I was attempting to prove that, then immediately realized that this can't be the case, since information distance is discrete. Which is a weird paradox that I wanted to find the solution to.

[buoyant dew]

ah i see

[white oxide]

what kind of "tools" does that usually involve?

[buoyant dew]

very interesting

[knotty vine]

For homotopy and type theory stuff its the usual higher categories/Lurie business. For CS it's monads, (co)algebras, distributive laws

[pallid delta]

Just asking since you seem to be familiar, but what kind of math appears when studying theoretical AI and ML stuff ?
Asking as a 3rd year math/CS double major (who understands the things you just mentioned)

[ebon galleon]

You're so fucking based holy shit

[knotty vine]

[stuck lagoon]

I am not as familiar as I might seem lol. I work at a company with a (very fledgling) AI research department, that I am very tangentially involved in. But it does seem that most of the work in AI right now is being done by engineers, not mathematicians (so lots of empirical/statistical results, things like "we tried this! It worked! Wow!")

But the few serious "mathematical" papers I have seen on the subject use mostly computability/information theory (thats probably obvious) and also functional analysis.

[knotty vine]

There's a bunch of interesting serious-math papers on automatic differentiation

I only know of them because categories were mentioned in them...

[stuck lagoon]

Oh I hadn't heard of automatic differentiation before.

[knotty vine]

The idea is to be able to automatically differentiate any given function in some programming language so that when you write function computing some neural network in that language you automatically get backpropagation. There's programming language semantics involved and, as I mentioned, category theory. I believe some people have even completely abstracted away from differentiation.

[pallid delta]

How is category theory useful for this (or CS in general) ?
As someone who basically knows nothing about it

[stuck lagoon]

Pretty interesting! Yeah looking on the wiki article I can see how that should be possible in theory and potentially lead to both better accuracy and better performance

[pallid delta]

cause to me it sounds like this very abstract, pure math thing, and then I see that it's used in lambda calculus, and apparently even in ML

[stuck lagoon]

I am curious about this too. I know type theory is very much involved in programming language design (to the point where in some broad sense they may even be considered equivalent from a mathematical perspective at least). And type theory has a very categorical "flavor" to it but I have never been quite able to formalize the relationship between them, other than "types are objects with morphisms computable functions between them"

"(and functors and monads are here too)"

[pallid delta]

I just learned of type theory's existence

[stuck lagoon]

I am one of those "type theory should be taught instead of set theory" people so your statement physically hurt me

[knotty vine]

Dont let Exomnium hear that! I purposefully wont @ him!

For a logic-category theory connection, very vaguely: categories with certain properties provide semantics for certain kinds of logics and conversely, certain logics provide an "internal logic" of certain categories so that we can reason in that category synthetically (without having to refer to objects and morphisms explicitly)

There are many connections between category theory and computer science, but one I'm interested in is using monads to model computational effects. Mathematical functions dont have any of those usually: theyre total, deterministic, and dont have side effects, but functions in a programming language (often functional programming languages) may not be any of those. It turns out that you can model those kinds of functions as Kleisly morphisms for specific monads. Partiality is modelled by the (-)+1 (aka Maybe) monad, nondeterminism by the powerset monad, side effects by the Reader/Writer/State monads.

[tribal palm]

type theory is objectively superior to set theory, also much less confusing than cat theory

[knotty vine]

Often functions dont have a single kind of computational effect so we would like to be able to combine monads together to get for example probabilistic nondeterministic functions. But it turns out there's no good way of combining the Distribution monad (for probability) and powerset monads!

[knotty vine]

Lets not get too carried away or I will @ him!

[tribal palm]

[pallid delta]

Guess I'll just start taking that program semantics class and see if I like it
Same for information theory

[stuck lagoon]

Lol I personally don't have that strong a view, but I do think its a shame that type theory isn't taught to mathematicians/programmers typically. Especially for CS people I think type theory has a lot of practical applications to the "working" programmer. Learning it just made me a better programmer.

[tribal palm]

this is applied in, like, haskell already right

[pallid delta]

tbf I just finished my first semester as a CS major (3 days from now technically)

[knotty vine]

Yes!

[tribal palm]

i never quite understood the IO type

[knotty vine]

You just live with it

[stuck lagoon]

Yeah monads are awesome, I have never really understood them categorically though, at least not completely. I understand intuitively that there is a category of "non-determnistic" (in the sense of multiple outputs) morphisms but thats about where my understanding ends from a category theory perspective.

[knotty vine]

Do you know what a monoid is?

[stuck lagoon]

Don't you dare

[knotty vine]

[pallid delta]

Even I knew that 2 years ago

[knotty vine]

Its not that bad once you get it

[stuck lagoon]

Lol I do know what a monoid is. And I have had "the category of endofunctors" explained to me before and I was like "wow that makes a lot of sense" and then promptly forgot it

[knotty vine]

Well you only have to remember the catchphrase to impress (annoy) people!

Here's another connection: data types (specifically inductive types) can be modelled as initial F-algebra for some endofunctor F. An F-algebra is a pair of an object X and a morphism f : F(X) -> X. Morphisms between F-algebra are maps g : X -> Y so that the square that involving all four obvious morphisms commutes. This forms a category F-Alg. If it has an initial object, this is the initial F-algebra. For example, the natural numbers are the initial F-algebra of the functor (-)+1 : Set -> Set

[stuck lagoon]

And the way I've been explaining monads to programmers these days is "Its a type that represents multiple (perhaps even uncountably infinite) outputs"

[tribal palm]

[knotty vine]

Or no output at all, or a probability distribution of some outputs, or just a single output but it also modifies some state, etc...

[tribal palm]

but in reality it’s all just some mutable state

[stuck lagoon]

Well, in the case of modifying state, I think its best to think of a monad as "returning all possible worlds", which is why it remains "pure" for the sake of the functional programmer's (my) ego

[stuck lagoon]

And is this basically a generalization of a K-algebra

an F-algebra I mean

[knotty vine]

Thats where the names come from

[unreal stratus]

For a field F

[tribal palm]

Körper

or something

[knotty vine]

Body

[stuck lagoon]

Ah I see, so in the case of a K-algebra the functor is _ x _ (cartesian product with itself), and the category is field extensions, or vector spaces.

[stuck lagoon]

Wait a second, another example! Kleene's recursion theorem?

functor being any total computable function, initial object being the fixed point

wait, actually that doesn't make sense I don't think haha

[knotty vine]

You mean the Kleene fixed point thm?

[stuck lagoon]

Yeah

[knotty vine]

It is related. We can construct the initial algebra with the same process as is used in Kleenes thm

Similarly, the proof that the underlying morphism of an initial F-algebra is an isomorphism is completely the same one a the proof that an initial pre-fixed point (I think it's called) is an actual fixed-point

[knotty vine]

This one's called Adameks fixed point thm apparently

[stuck lagoon]

Interesting, I have been trying to wrap my head around Kleene's theorem proof for a while now. I gave up on the Y combinator one, but the one he originally gave (iterating) made more sense. Still not complete sense, but more sense non-theless.

[knotty vine]

Now heres a nice thing we get from category theory: all this stuff can be dualized for free. It turns out terminal F-coalgebra are models of coinductive types (not surprisingly) such as infinite streams

This is related to Haskell again since lists in Haskell may be infinite as well

So this Haskell List type is an algebra an a coalgebra at the same time in some compatible way (but I dont remember the details)

[tribal palm]

haskell just super lazy

[knotty vine]

eepy haskell

[tribal palm]

someone should make that a technical term

[merry geode]

Why is this haskell thing here

[knotty vine]

Got carried away

[merry geode]

Ah sorry I wasn't saying that to judge

I am interested in how topology discussion came down to mention of FP

[knotty vine]

I think we went from metric spaces to information theory to "serious" math in AI research to automatic differentiation to applications of category theory in computer science

[merry geode]

Woah

[knotty vine]

All roads lead to category theory

[merry geode]

Indeed, but I mean current era AI-adjacent CS has quite a distance from CT

[knotty vine]

Turns out not!

Well, not further than most other fields

[merry geode]

How?

[knotty vine]

https://arxiv.org/abs/2106.07032

[merry geode]

I mean, this paper is exploring using category theory in ML, right. Not necessarily connected to ML in practice.

[knotty vine]

As was said earlier, it seems like ML research is currently mostly engineering (whatever works is good, dont need to understand why). But ML is very young, and just like with the engineers that built cathedrals in the middle ages, eventually foundational research like this will become important (if the singularity doesnt happen before then).

[buoyant dew]

hey

so the part in the blue is just the picture on Hatcher right? and would the red be

nevermind i never fully understood the explanation in hatcher

lol

can anyone explain these to me? a brief intuitive explanation is ok

[knotty vine]

Blue means: a homotopy for topological spaces induces a chain homotopy on the singular chain complexes

[merry geode]

Hmm, idk about that. Programming has been always "whatever that barely works is the best"

[buoyant dew]

ah great

[knotty vine]

Red means: if you have a good notion of tensor product for chain complexes and you have that the unit inverval looks like I_* := 0 <- Z*Z <- Z <- 0, then you can make chain homotopies just like you can do with regular homotopies: a homotopy between f,g : X -> Y is a morphism H : X*I -> Y such that H(-,0) = f and H(-,1) = g

[buoyant dew]

that's a mouthful

[knotty vine]

Replace I with I_* and the product with the tensor product and you get chain homotopy

[buoyant dew]

this I have no idea about

thanks a lot

[knotty vine]

Do you see how the interval chain complex idea works?

[buoyant dew]

i don't think i do

[knotty vine]

The interval has two points, so the chain complex Z*Z at order 0. It has a single line segment, so we have Z at order 0. It has nothing else, so 0's everywhere else. The boundary map from Z -> Z*Z is k |-> (-k,k)

[buoyant dew]

hang on

[knotty vine]

There's no homology here yet

[buoyant dew]

yeah lol. and it's contractible so those cant be the homology groups

yeah yeah I see

[knotty vine]

You know about simplicial sets?

[buoyant dew]

yup

[knotty vine]

actually nevermind that doesnt help lol

[buoyant dew]

wait wait

nae bother. he explains it couple chapters later

much appreciated

[knotty vine]

For the tensor product: we want a product of chain complexes which corresponds to the usual geometric product

[buoyant dew]

yeah he explains the rest of the construction as well

[knotty vine]

This can't just be the degree-wise product because we ought to take the product of say, a line (1d) and a triangle (2d) to obtain a prism (3d)

[buoyant dew]

ah ok

we want the category to be monoidal right? together with the tensor product of chain complexes

[knotty vine]

So instead we define $(X_\bullet \otimes Y_\bullet)n \coloneqq \bigoplus{i + j = n} X_i \otimes Y_j$

[buoyant dew]

ok i see

[gentle osprey]

semer

[knotty vine]

This in fact makes the category of chain complexes a monoidal closed category

Now remember that a usual topological homotopy is just a continuous map X*I -> Y

[buoyant dew]

yeah

[knotty vine]

Now replace X and Y with chain complexes, the product with the tensor product, and the unit interval with the unit interval chain complex we just defined

If you work this out you get a chain homotopy!

[buoyant dew]

this is cool, thanks for the help

[knotty vine]

yw!

[buoyant dew]

i need to step up my category game though

idk what a closed category is

it probably doesn't help that one of the 2 profs at my institution who knows about CT actually hates it

[knotty vine]

Doesnt really matter for this, just that theres a notion of internal hom

[buoyant dew]

yeah yeah thats what the wiki says

again much appreciated

[buoyant dew]

actually nevermind lol

[buoyant dew]

so I have a bunch of questions here

We're considering $(C_\bullet)$ with the finite support R-module structure right? That is, $C_\bullet \cong \bigoplus C_n$. Then, a homomorphism $C_\bullet \to R$ corresponds to choosing a homomorphism from each $C_n$ because we have $$\left [ \bigoplus C_n, R\right ] \cong \prod \left [ C_n, R\right ]$$ in modules.

[gentle osprey]

I Abhor Hatcher

[buoyant dew]

did I get this right? Also what is the most general categorical setting $\left [ \bigoplus C_n, R\right ] \cong \prod \left [ C_n, R\right ]$ holds?

[gentle osprey]

I Abhor Hatcher

[buoyant dew]

And each chain complex is actually a cochain complex under this definition when we negate the indices right? That is, if $C_\bullet=(C_n)$ is a chain complex then $C^\bullet = (C_{-n})$ is a cochain complex, or am I missing something?

[gentle osprey]

I Abhor Hatcher

[buoyant dew]

and finally if someone could explain the footnote I'd be very thankful

[unreal stratus]

Ye

[buoyant dew]

hey potato

[unreal stratus]

This actually holds in any category (provided the relevant things exist lol)

Basically the "Hom functor preserves limits"

[buoyant dew]

ah alright

[unreal stratus]

It's just that here the first limit takes place in the opposite category, i.e. it is a colimit in the original category so you get the direct sum

pretty sure this is a typo on nlab btw right like this should be LLP not RLP

[buoyant dew]

thanks a lot

and i guess the footnote is just a terminological remark

[unreal stratus]

Yeah

[ebon galleon]

Correct

Or uhh you're correct that it should be LLP

[ebon galleon]

If [ , ] is Set-hom, then yeah always. If [ , ] here means an internal hom in Mod-R then that's a property of multivariable adjunctions.

[unreal stratus]

Phew lol thanks

[unreal stratus]

Sure yeah i meant straight up hom impoverished over sets

[ebon galleon]

"impoverished"

[buoyant dew]

heya

the way he puts it, there seems to be an immediate reason for checking this but I don't see it

this is just after constructing homology of pairs and triples

[hexed steppe]

can you post more context

is there a theorem or something

[buoyant dew]

well no I don't think so

hang on

this is the theorem right before it. and after it it's just examples

i guess i just don't see the utility of showing that the boundary operator is natural transformation between the functors $H_k(X,A)$ and $H_{k-1}(A,B)$

[gentle osprey]

I Abhor Hatcher

[buoyant dew]

hmm

[white oxide]

Knowing that maps induce maps of long exact sequences seems pretty useful to know

[buoyant dew]

yeah i have no objections to that

[hexed steppe]

yeah idk why it is phrased as “it remains to verify”

[buoyant dew]

but the naturality of the boundary operator is not really necessary to show that right?

that basically comes for free (snake lemma)

ah

ah

[white oxide]

Fair, I don't see why either

[buoyant dew]

sorry i keep misreading stuff

[white oxide]

Snake lemma? For that you need to know the diagram commuted already

[buoyant dew]

im not sure, i think that follows from the fact that the following is a SES

[buoyant dew]

the first sequence here I mean

[hexed steppe]

whats (11.3.2)

[buoyant dew]

good question

so this says whenever you have a SES of chain complexes you can write the long exact sequence. snake lemma right?

#calculus is the right channel i believe

[white oxide]

As I understood it, f_* is induced from maps from one short exact sequence to another

[unreal stratus]

Wait Kerr can I check smth with ou

[buoyant dew]

yup

[unreal stratus]

Cause I don't know if I'm smoking

[fleet wolf]

Sorry, I clicked the wrong channel. My bad.

[white oxide]

?

[unreal stratus]

Someone asked like "what is the double cover of the torus"

but isn't it just... the torus

Maybe I'm being silly

[hexed steppe]

maybe they mean the orientation double cover

[unreal stratus]

But then it's orientable so it's just two copies of T^2

[hexed steppe]

yeah

[unreal stratus]

the double cover i have in mind is like

S^1 x S^1 (x,y) |-> (x,y^2) i guess right

like the fibre of any poitn (x,y) is (x, square roots of y)

[buoyant dew]

sorry I just understood what you're saying

[unreal stratus]

i guess the cheeky way is that like

T^2 = R^2/Z^2 so the cover corresponding to a subgroup H of Z^2 is R^2/H

[unreal stratus]

Naturality means you can do things like 5 lemma for example

[unreal stratus]

oh okay nvm sure

Lol

[buoyant dew]

yeah ok this makes sense. the naturality of ∂ basically tells us that the given LES construction is a functor from Top(3) to the category of chain complexes, right?

[unreal stratus]

Yeah the process of showing it's natural is probably not helpful, but the fact it is natural is :^)

[buoyant dew]

yeah I see it now

thanks

[buoyant dew]

i guess it tells us a bit more than that

[white oxide]

Sounds all right, doesn't the cheeky way correspond to a general way of constructing covers from an universal covering space?

[unreal stratus]

Yes exactly

I more just mean that it feels like overkill

Though here it's fine ig since T^2 is often presented as R^2 / Z^2

lol

T^2 = S^1 x S^1 = BZ x BZ = BZ^2 :trollface:

[white oxide]

My thought was just that checking the naturality was the last missing piece of checking if you have this

You get induced maps between their LES

[buoyant dew]

fantastic

[white oxide]

Idk what Top(3) is

Been a while since I watched watchmojo

[unreal stratus]

it isn't standard notation i.a.h.

[buoyant dew]

we don't need to check whether these commute with f* because they are just induced maps that operate on representatives, right?

so the only thing left to do is to check whether ∂ commutes with f*

[buoyant dew]

hahahaha

[white oxide]

You probably shown something like that before?

[buoyant dew]

i mean i guess it's clear

im just too sleep deprived and want to verify each step

[red yoke]

On a noncompact manifold, singular homology is dual to compactly supported cohomology and locally finite homology is dual to singular cohomology; On a compact manifold, the general form of Poincare duality is relative Cech cohomology (K, L) = relative singular homology (M\L, M\K) for compact K, L. Is there a (reference for a) similar generalization for noncompact manifolds?

[buoyant dew]

thank you @white oxide

[unreal stratus]

What

[white oxide]

oh nvm keep going

[unreal stratus]

LOL

[white oxide]

the pensivebreed incident will remain secret

[unreal stratus]

Lol

[hidden crag]

lool

[stuck geyser]

Hello chat.

[white oxide]

Hi chatette

[stuck geyser]

If $F \in C(X,Y)$, $X$ and $Y$ both metric spaces, and $F$ is compact. If each function in $F$ is uniformly continuous in $X$, then is $F$ necessarily equicontinuous

[gentle osprey]

Mizalign

[unreal stratus]

i guess F \subset but ye

[stuck geyser]

subset

Trying to go about how to prove it

[hidden crag]

might be better off in #real-complex-analysis

[stuck geyser]

I suppose?

[buoyant dew]

Where do I start reading about the De Rham cohomology and differentials? Like what's the 'modern beginner' book?

[unreal stratus]

Bott Tu is great

Like diff forms in alg top

[buoyant dew]

thanks!

[buoyant dew]

god i wish there were homology type theory

[knotty vine]

homology is just a special case of homotopy

[white oxide]

What's the definition of homotopy here that makes it a special case?

[buoyant dew]

well it's obvious, logy is a special case of topy

i am so braindead

final question today

is the mayer-vietoris thm just a pushout preserving functor like the van kampen theorem? I don't immediately see why but I was told it was the categorical analogue of van kampen thm

[lime sable]

https://ncatlab.org/nlab/show/homology two on this page:

• homology groups of a nonnegatively graded chain complex are the homotopy groups of the corresponding simplicial abelian group under the dold-kan correspondence. (when the chain complex is unbounded, it instead corresponds to "stably simplicial abelian groups")
• homology groups of a space X with coefficients in a spectrum A are homotopy groups of the smash product of the suspension spectrum of X with A. if you start with a coefficient object like an abelian group, you can take its eilenberg-maclane spectrum to get a coefficient spectrum

[buoyant dew]

no no that makes no sense, nevermind

[plain raven]

"just"

[plain raven]

not directly related but mayer vietoris is actually equivalent to the excision axiom i think. or at least directly related

Both of these are kind of topological reflections of Noether's second isomorphism theorem

the diamond isomorphism theorem.

[jade pagoda]

Does anyone mind checking this proof? The question is: let X be a set, F a collection of real valued functions on X, and T the weak topology generated by F; then T is Hausdorff iff for all pairs of distinct points x, y, there is some f in F with f(x) ≠ f(y).

If my proof is correct, only the requirement that T be T0 is required for the forward direction, but this would mean that T being T0 implies that it is Hausdorff, which shouldnt be true. I'm a bit confused, as I am blind to my own mistake.

[golden gust]

is there a criterion for when (the total space of) a fibre bundle can be viewed as a bundle over the fibre instead of the base? does the bundle need to be trivial?

[red yoke]

If your bundle F → E → B has a cover of compatible local trivializations p^-1(U) → U×F → F that glue to a map E → F, then you get a homeomorphism E → B × F

Which I assume is what "bundle over the fiber" means

[wispy veldt]

Seems fine , also in a topological vector space T0 does imply T2 (and T3.5 too)

[unreal stratus]

In what way do you mean?

As in, what forces it to be compatible with your original bundle

[red yoke]

Bump

[jade pagoda]

Oh seriously? Neat. Thanks!

[grim knot]

hey guys, can somebody tell me, why it follows that [c]=0? I don't see it. I'm looking at the zeroth homology groups and in this case epsilon is an augmentation

[knotty vine]

The homology group is $H_0 = \ker \epsilon / \im \del$. So since $c = \del \sum_x n_x \lambda_x \in \im \del$, we have $[c] = 0$.

[gentle osprey]

semer

[grim knot]

I am having a brainless moment, how are quotient groups defined again?

these are meant to be factor groups right?

[knotty vine]

The quotient A/B means "everything in B becomes 0"

[grim knot]

well thank you very much, that makes obviously sense now !!

[azure thunder]

Is there anything within the equation that would make you think it would yield a value of 1.0 easily 🤷‍♀️

[lucid geyser]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[civic verge]

I would like to know if I made a mistake in any of these.

[fickle elm]

X is a a set and {X} is another set with one element the set X? Then {X} should have nothing to do with the topology on X.

[red yoke]

I like how it asks you to induction

[fickle elm]

And the notation is a bit weird. E.g. $a\subset \tau$ is not a good expression and there is no meaning in judging the statement since it does not make sense.

[gentle osprey]

Dong_Valentino

[knotty vine]

Formally it makes sense

it could even be true, depending on what a is

[proud pawn]

discrete topology on X is just the power set of X so basically it's just asking what objects here are subsets (or collections of subsets) of X

Also has anyone heard about the terms such as metacompactness, orthocompactness, mesocompactness etc.?

I've just found out about them and they look interesting but I'm not sure how practical they really are

[civic verge]

Are you saying that b is false?

[knotty vine]

ye

[civic verge]

I have 7 false, and in reality there are 6, what would be the other error?

[knotty vine]

K

[civic verge]

The empty set is a subset of t

mmmm

why

[knotty vine]

No, the empty set is an element of t. So {0} is a subset of t

[unreal stratus]

And also a subset of t tbf lol but that is irrelevant ig

[knotty vine]

Also N should be false

[civic verge]

why

cannot be a proper subset of t

[knotty vine]

a is not an element of t (but {a} is), so {a} is not a subset of t

[civic verge]

sorry it's the translator
but it would be because a cannot be an OWN subset of t

[knotty vine]

I dont know what you mean, can you rephrase that?

[knotty vine]

Thats P which you correctly said was false

Ah, there is no problem with "strict inclusion" here

[knotty vine]

But that's unrelated to N

[civic verge]

but it tells me that an element a is a subset of t and it wouldn't be

So if
{a}⊆t false

This is what I am understanding

[knotty vine]

The following are true: $a \in X$ and ${a} \subseteq X$ and ${a} \in \tau$ and ${{a}} \subseteq \tau$.\
The following are false: $a \in \tau$ and ${a} \subseteq \tau$.\
The following is very false: $a \subseteq \tau$.\
The same things hold for $\subsetneq$

[gentle osprey]

semer

[civic verge]

oouh

I understand

[civic verge]

look

[knotty vine]

Looks good. What about M and O?

[civic verge]

M is true

and O is false

[knotty vine]

Why?

[civic verge]

Because the set X is a subset of t
then the set with the element X would not be a subset of t

[civic verge]

The set X belongs to t by definition, so X is also a subset of t.

[livid escarp]

I'm trying to calculate homologies of complex projective spaces, but the lecturer kinda skipped over the construction of its CW structure, so I have a question:

The Wikipedia says that we can regard CP^n as a quotient of S^{2n+1} under the action of U(1) so the space looks like a 2n-dimensional sphere. Why?

My understanding is: CP^n is defined as 1-dim subspaces of C^{n+1}. Any such subspace can be identified with a point in C^{n+1} and as such it is described by 2n+2 real numbers. We norm the point (?)by projecting it to a unit sphere losing 1 dimension, so the result looks like a 2n+1-dimensional sphere. And then since we have to identify the points under complex multiplication we have to mod out another dimension because complex multiplication is something that looks like S^1 and S^{2n+1}/S^1 = S^2n.

[languid patrol]

it doesn't look like a 2n-dimensional sphere that is a lie

in one case this happens CP^1 is S^2

CP^n has cohomology in every even degree less than 2n so you know it can't be a sphere

for n > 1

to compute the homology the nicest way is to use the following cell decomposition: CP^n is just C^n \cup CP^{n-1}, where CP^{n-1} is the set of complex "vanishing points at infinity"

this can be seen by letting $C^n$ be lines of the form $f(t) = (a_1t, \dots, t)$ with $a_i \in \mathbb{C}$ for $i \leq n$, and letting $CP^{n-1}$ be lines which lie in $\mathbb{C}^{n}$

[gentle osprey]

kålrot

[languid patrol]

then you can use meyer vietoris/excision

and induction

[livid escarp]

Why does my textbook say that as a topological space its equivalent to a quotient space of S^2n+1 under the S^1 action?

[languid patrol]

that is true

[livid escarp]

the caveman logic I had is that sphere mod sphere is sphere

[languid patrol]

but the quotient of S^{2n+1}/S^1 is not S^{2n}

[livid escarp]

but where does the S^[2n+1] come from? This is the part that is just stated but never explained

[languid patrol]

the point is that up to multiplication by a positive real number every complex vector in n+1-dimensional space lies on S^{2n+1}

[languid patrol]

this is just the unit sphere in n+1 dimensional complex space

[livid escarp]

how do I see that?

[languid patrol]

$C^{n} \cong R^{2n}$

[gentle osprey]

kålrot

[livid escarp]

but n+1 = 2n+2?

[languid patrol]

yes but the sphere in R^{m} is S^{m-1}

[livid escarp]

holy... this does make sense now

[languid patrol]

happy to help

[livid escarp]

yeah its of codimension 1

makes sense

I was thinking really hard why its 2n+1 and all that time I just had to think of S^2 and R^3 haha

[languid patrol]

So they are just taking C^{n+1}, quotienting by the positive reals first, then quotienting by S^1

and using that $C^* \cong R_{> 0} \times S^1$

[gentle osprey]

kålrot

[livid escarp]

by quotienting by the positive reals you mean the projection onto the S^2n+1?

[languid patrol]

that is one way of interpreting it

[livid escarp]

im just thinking of the natural way to identify RP^2 which is to normalize the vector

= take one canonical representative from each 1-dim subspace

[languid patrol]

the map $f$ sending $v = (z_1, \dots, z_{n+1}) \to v/|v|$ has the property that $f(v) = f(w)$ if and only if $v = rw$ for $r$ a positive real

[gentle osprey]

kålrot

[languid patrol]

so it is quotient map for the group action coming from scaling by positive reals

this shows that $C^{n+1}/R_{>0} \cong S^{2n+1}$

[gentle osprey]

kålrot

[livid escarp]

yess because if v = rw for a positive real then by interpreting r as a complex number theyre the same in the projective space too

which does make total sense

[languid patrol]

yep

[livid escarp]

on an unrelated note: are there any norms with values in complex numbers?

(the one that people work with and make sense)

[languid patrol]

not really

there's nothing wrong with the notion it's just not as useful

the point of a norm is to be able to order elements by size

so you want the target to be an ordered field

[livid escarp]

oh yeah

I did that in my model theory class

real closed fields I think?

in a sense the biggest field that can be ordered

[languid patrol]

real closed fields? I don't know what those are

but yeah anyway for most norms it's fine to use \mathbb{R}, although more complicated ordered abelian groups can show up

[livid escarp]

its the field F such that if you extend it by \sqrt(-1) it becomes algebraically closed

so in a sense its one element away from becoming algebraically closed

and it has the property, to quote "There is an ordering on F that does not extend to an ordering on any proper algebraic extension of F."

[languid patrol]

interesting, are there fields like that that you can write down which are not just R?

[livid escarp]

do you have any way to visualize how the cell structure on CP^n work?

[languid patrol]

ah okay yeah

infintiely many

[livid escarp]

I get that you have a cell in each even dimension, but it's not entirely clear to me how do boundary maps work

[languid patrol]

well since they are only in even dimensions the boundary maps are not very interesting

since there are no odd dimensional cells

[livid escarp]

Okay, I realized that I do not need to understand it to compute cohomologies because theyre all mapped to 0 in an odd dimension.

In fact, the cohomology is somewhat trivial

[languid patrol]

yep

[livid escarp]

one more question: the computations of cohomologies with coefficients are somewhat tedious => I have a space X => find CW structure => cellular chain complex => its dual cochain complex => compute cohomologies => compute cohomology for general coefficients.
I understand that there are lots of computational tools to do that but the question is do you at some point learn it all by heart? Right now just to compute Ext for something simple I have to think of so many different things at once it's somewhat unsatisfactory

[languid patrol]

i mean with totally random coefficients i dont know what you are supposed to do

but for finitely generated Z-modules you will just memorize all of the possibilities and all of the Exts and Tors

or at least Ext^1 and Tor_1

anyway if the complex has zero differentials with Z coefficients it has 0 differentials with any coefficients

so for this particular calculation it's easy

[livid escarp]

yeah I understand, it's more of a soft question because it's end of the semester in Europe and I'm wondering if people who did that a lot just learn it by heart and I just have to get used to it.

although one of my more advanced friends mentioned that he liked the algeometric approach with G valued locally constant functions more

but I kinda spaced out as soon as I heard it

[languid patrol]

I mean Tor_1(Z, Z/n) = Tor_1(Z/n, Z) = 0 and Tor_1(Z/n, M) = M[n]

[livid escarp]

at the first glance it does seem like "here, for each topological space X there are 10 different objects and to compute something you have to work with all of them"

[livid escarp]

I just use Tor1(R/I,R/J)=(I∩J)/IJ

[languid patrol]

yep sure

that is a consequence

[livid escarp]

would it be a mistake to think of Tor as of something that makes my SES exact on the left after I apply \otimes N to some SES?

and Ext on the right after Hom(-,N)?

[languid patrol]

yeah idk there are many approaches to defining cohomology of a space, they all have certain advantages and disadvantages

but in terms of computability it's hard to beat just finding a cell decomposition if one is available

[languid patrol]

yes that is a good way of thinking about it

[quartz crater]

For an euclidian metric space, would tweaking the definition for an interior and boundary point have any consequences if instead of an open ball it instead was defined using a closed ball?

At first glance it seems wrong but now I’m starting to doubt myself

[unreal stratus]

I'm not sure what definition you are using

But often things don't matter since every open ball is contained in a closed ball and vice versa

Also just a quick question: thinking of $\mathbb{CP}^{n-1} = \mathbb{R}^{2n}/\mathbb C^{\times}$ and $\mathbb{RP}^{2n-1} = \mathbb{R}^{2n}/\mathbb R^{\times}$, we get a fibre bundle $S^1 \to \mathbb{RP}^{2n-1} \to \mathbb{CP}^{n-1}$ right?

[gentle osprey]

potato

[unreal stratus]

Or am I missing something

[quartz crater]

For interior point for example:

Let M be a subset of R^n. A point a in R^n is called an interior point in M if a has neighborhood that lies completely in M.

And in my scenario neighborhood is a closed ball

[unreal stratus]

Ah, sure. Then yes, it doesn't matter. Normally I'd say that a has an open ball about it lying in M, but it doesn't matter for the reason I just gave :)

Like given a point p and an open or closed ball about p, you can always find a smaller ball about p which is open or closed

[quartz crater]

Ah I see

That makes sense

Is there a reason we use open balls instead of closed?

For the definitions

[unreal stratus]

Hm well often one assumes neighbourhoods to be "open" and the simplest example of an open set is an open ball (a closed ball isn't)

Also, you often want a neighbourhood of your point to "Look like R^n", and open balls are the same in many ways (homeomorphic ) to R^n

[quartz crater]

I see—oh interesting

[granite slate]

im 1 week into alg. top and wanna prove something with homology. I want to prove that there is no continuous function f from S^3 to S^2 satisfying a certain property. So far i've got that the singular 3-simplex composed with f will give an induced homomorphism between the third degree homology groups, which is isomorphic to Z for S^3 but trivial for S^2, so the homomorphism will be trivial

im not sure how to go from here to get a contradiction though

[hidden crag]

it would be helpful if you told us what property you mean

[uneven bronze]

I'm reading the following exercise in Bridges Foundations of Real and Abstract Analysis. I simply can not make sense of this exercise, and was wondering if anyone knows what the gist of it is about, and if possible, if there are any similar questions on the Internet about this. I simply do not know what to search for.

[gentle osprey]

Philip

[uneven bronze]

It just seems very unintuitive to me. If K intersects S at a single point, how can there be a ball, concentric with K, and radius greater than K, but that is disjoint from S?

the n in R^n and K_n are unrelated by the way.

[granite slate]

If S^2 is an embedding then f(x) = -x for every x in S^2

[fathom steeple]

how do they know that f(sup)\in A?

[unreal stratus]

What is A

Pretty sure whatever it is should be closed which is why

[alpine nest]

Something does seem wrong here, any ball concentric with K and with radius greater than K will be its superset, so at the very least it will contain that one point x

[safe torrent]

Any ideas how to setup coboundaries for Cech cohomology as matrix representations?

For simplicity let's say you have a chain complex
{0} -> Z^10 -> Z^10 -> {0}
The maps being d_0 to d_2 respectively. You can pretty much see that im(d_0) = 0 and ker(d_2) = Z^10 but how about the other cases?

[somber falcon]

Like all things in math, it is always better to break down your space in terms of some type of structure into simple parts. Maybe your space is some kind of bundle, or maybe it is a product of simpler spaces. You can take advantage of that, and then you memorize the homology for some of the simple component spaces, and you can quickly compute quite a lot.

The overwhelming majority of spaces we use are products/quotients/bundles over etc a handful of simple spaces.

Mayer-vietoris is a very handy one

Kunneth formula

Once you learn poincare duality you realize the options for what can happen are pretty limited for manifolds

You end up with a lot of different partial info that you can often quickly piece together

Very rarely do you actually write out simplices and all that, typically only for the sake of learning ime

Connected sum is another way to break things down that comes with a good cohomology decomposition

Imo one should focus more on the eilenberg-steenrod axioms rather than the direct definitions. Some books even start with those.

For tor and ext I think this has been pointed out but you are almost always ending up with homology that is finitely generated abelian groups, and these are classified and split into free factors and torsion factors, so all you need to know is ext and tor for those, then you know "everything".

It's not as much stuff as it seems like...but yes I was overwhelmed at first. You don't really know what is important and what can kind of be forgotten when you first learn.

[tidal cedar]

My first AT class after self studying from Hatcher defined the bare minimum for singular cohomology, then introduced ES axioms, proved them for singular, and went on from there

I think it was really good pedagogy

[solemn oar]

My first at course we spent most of our homology time on singular and treated it pretty geometrically, and just mentioned the axioms and uniqueness theorem at the end. Then the following course picked up by treating cohomology theories using omega-spectra.

[tidal cedar]

wow that's a pretty wild jump

[signal fox]

Hello i had to show that the set of continuous function M from [0,1] -> R with the sup metric is not compact.

I wanted to know if my reasonning is correct

so i defined a function h from M -> R by h(f) = max (f) on [0,1]

I showed that h is continuous, and h (M) = R

Thus if M was compact, R would be compact, which is not the case and thus M is not compact

[livid escarp]

thank you for answering! yeah we did start with axiomatic homology theories, then did simplicial homology, then singular, then cellular, then euler characteristic, then lefschetz numbers, then simplicial/singular/cellular cohomology, then cohomology rings then tor then ext then universal coefficient theorem so im just trying to coherently piece everything together without getting lost 😵‍💫

[signal fox]

Thanks in advance for any feedback

[grim knot]

hey guys, can somebody tell why both functions need to be continuous, shouldn't it be an or since we have a disjoint union? I mean we are either in one set or another