point-set-topology

[brave geode]

i'm not sure what you mean

these are homotopies

[red yoke]

Every subject has details

[knotty vine]

Maybe the simplex category and then use Dold-Kan

At least for complexes that vanish in negative degrees (connective? I forget)

[brave geode]

i think so

i have no idea how dold-kan works lmao

[knotty vine]

Its not that hard if youve seen simplicial sets before

To go from chain complexes to simplicial abelian groups you do basically the same thing as with singular simplicial sets. The other way is a little more difficult, but this "reduced Moore complex" is quasi-isomorphic to the usual alternating chain map complex

[plain raven]

you want to work in the setting of enriched category theory probably and consider additive functors.

[knotty vine]

Is there a connection here?

• A monoid is a lax monoidal functor from the terminal monoidal category 1 but also a strong monoidal functor from the augmented simplex category \Delta_a. That is, 1 is the "laxification" of \Delta_a.
• A chain complex is a functor from Delta to Ab but also an additive functor from the "walking chain complex"

[plain raven]

Can you explain what you mean by "a functor from Delta to Ab"

Do you mean up to equivalence of categories

[knotty vine]

Oh that shouldve been Delta^op

[plain raven]

Yeah, I mean, that wasn't my concern lol

[knotty vine]

Yeah, with Dold Kan, but thats not a problem I dont think?

[tidal cedar]

Like a simplicial abelian group

[plain raven]

Because the question is conceptual, it makes a difference because imo the answer is best understood in terms of Dold-Kan; and so if you make certain identifications already based on Dold-Kan it can obscure it.
Here is an interesting reformulation of Dold-Kan that you may find helpful in understanding this -
Dold-Kan can be understood as stating that, working in the bicategory of Ab-enriched categories and Ab-valued profunctors, there is an equivalence between Z[\Delta] and the (opposite) of the walking chain complex. As an immediate corollary, there is an equivalence of categories between their associated presheaf categories, i.e., SAb \cong Ch(Ab)

So I think one way of rephrasing the question is:
How does the relationship between Z[\Delta] and the walking chain complex compare to the relationship between Delta and the one-object 2-category (or bicategory)

Here by Z[\Delta] I mean the free Ab-enriched category on Delta, given by taking free Abelian groups of homsets and extending composition to be bilinear in the obvious way.

[gentle girder]

https://tenor.com/view/mind-blow-galaxy-explode-boom-fireworks-gif-5139389

[plain raven]

(And by the way, the Moore normalization definition can easily be adapted to work on presheaves on \Delta_a, there is an equivalence of categories between augmented simplicial Abelian groups and chain complexes as well, so let us just forget about Delta and focus on Delta_a)

This is not the time to get into this but the standard formulation of Dold-Kan is morally wrong and it should be presented as an equivalence between augmented simplicial Abelian groups and chain complexes

Btw I see what you're getting at by "laxification" but it seems more easy to formalize this by saying that to say that Delta is constructed from 1 by "strictification" in some sense than to say that 1 is constructed from Delta by laxification. It is like looping/delooping, it is clear what is meant by the loop space, some homotopy theory is required to establish that delooping is a coherent concept, or that the delooping is well defined.

[knotty vine]

Im always willing to go on a detour: do we get a different chain complex if we take the left vs. the right Kan extension along the inclusion Delta -> Delta_a (i.e. either add a single (-1)-point or add a (-1)-point for every connected component)?

[knotty vine]

Right, but I just like the word laxification

But certainly not every monoidal category has a laxification, so strictification is better ye

Maybe "strongification"

[plain raven]

Yes, the difference is a shift in degree, a suspension or desuspension in chain complexes. In my subjective opinion, there is a suspension "buried" or "hidden" in the usual definition of Dold-Kan which is removed by modifying the definition to use the augmented simplex category

The usual textbook definition of Dold-Kan is not a strong monoidal functor. The definition I would propose to replace it with is a strong monoidal functor. The usual textbook definition of Dold-Kan asserts the weaker statement that it is "monoidal up to homotopy equivalence" - I claim that this is a red herring and this is really a statement about the relationship between the Cartesian product of spaces and the join of spaces - the smash product of spaces agrees with the join of spaces up to a suspension, up to homotopy.

I can be more formal about this if necessary (and desirable) lol

[knotty vine]

If I understand correctly, the textbook Dold-Kan would be a Quillen equivalence (on the usual model structures)? While your version already works for the 1-categories?

And that the only reason the textbook version doesnt work for 1-categories is not an "interesting" fact but only the relationship you mentioned?

[plain raven]

Yes, I'm talking about 1-category theory here, I personally find a lot of mathematical value and content at the 1-categorical level, for example there are more or less computationally practical resolutions, even if they are all the same up to weak equivalence. Or for example, you might want to talk about a "cofibrancy structure" on an object, working constructively, rather than just saying an object is cofibrant as in the model category language.

[knotty vine]

Ofc the lower dimensional a fact works the easier it is

[plain raven]

Dold-Kan in the standard presentation is already a true equivalence

but it is not an equivalence of monoidal categories

[knotty vine]

Right thats what I meant ye

Funny that model categories are themselves already a 1-category approximation of infty-cats

[umbral panther]

Dold-Kan is an equivariant of 1-categories
But there is a very serious issue about the monoidal structures. The categories of algebras are Quillen equivalent, but the categories of symmetric algebras are not Quillen equivalent, so they are substantially different

[plain raven]

Interesting, ok. I see

[umbral panther]

In fact, commutative dgas (over a field of positive characteristic) have no known model structure and the forgetful functor to chain complexes does not admit a left adjoint in the infinite sense

[knotty vine]

yikes good thing I dont need them

[plain raven]

For concreteness here's the definition I would propose.

1. We know that presheaf categories are a "free cocompletion"
2. We know that the augmented simplex category is the free monoidal category with distinguished monoid
   It is possible to combine 1 and 2 into a single universal property of the category of (augmented) simplicial sets : every cocomplete monoidal category C with a distinguished monoid M, whose tensor product is cocontinuous in both arguments, determines in a canonical way a functor SSet -> C

One then checks that this works in the Ab-enriched setting, that is, simplicial Abelian groups are the free Ab-enriched, "cocomplete-monoidal" category (where here by "cocomplete-monoidal" I mean to imply that the cocompleteness structure and monoidal structure are appropriately compatible, i..e, the tensor product is cocontinuous in both arguments)

Cocomplete should be understood in the sense of enriched category theory here, using weighted colimits

The category of chain complexes has the usual tensor product, and we take as distinguished monoid the complex
0 -> Z -> Z -> 0
which is concentrated in degrees 0 and 1.
Intuitively here the degrees are counting the number of points in the simplex for that dimension, so 0 = the empty simplex, 1 = the simplex at a point.
So this complex represents the space with one point (using the reduced homology convention, I guess, where each space has a single canonical instance of the empty singular simplex)

Geometrically, the tensor product represents the join of simplicial complexes, you can check that the chain complex functor from simplicial complexes to Ab sends the join to the tensor product up to isomorphism

There is then a unique functor SAb -> Ch(Ab) which is cocontinuous, sends the one point space to the one-point space, and sends Day convolution of presheaves to the tensor product of chain complexes

and this is a monoidal equivalence.

[knotty vine]

Very cool!

I remember that the usual tensor product of chain complexes is itself already a Day convolution https://mathoverflow.net/questions/359306/is-the-tensor-product-of-chain-complexes-a-day-convolution

[umbral panther]

Just to be clear, cdga sits between two good categories, simplicial commutative rings and E_infinity rings. In characteristic zero, they’re all the same. If you’re in positive characteristic and you find a natural cdga, it actually has the extra structure of a simplicial commutative ring. If you want a cdga, you should settle for an E_infinity ring

[knotty vine]

I am guessing this story is basically the same one, translated via Dold-Kan

[plain raven]

Yeah. I would say it's not just a matter of "after translating via Dold-Kan, you can see that these stories actually agree", the answerer is literally saying "Well, if you translate via Dold-Kan, then the answer is yes."

What I said earlier about Z[\Delta] being equivalent in the bicategory of profunctors to the walking chain complex, you cannot transfer monoidal structures along profunctors because they're not functors.
I regard this answer as saying
"the walking chain complex is not a monoidal category, but Z[\Delta] is, and it's equivalent to the walking chain complex in the bicategory of profunctors, so here's a definition which is precisely engineered to let us transfer monoidal structure along profunctors, at which point the answer becomes yes, because you can transfer the monoidal structure from Z[\Delta.]"

[unreal stratus]

Society if it were lol

[plain raven]

Yeah but the point of my post is that it can be and it should be!
I don't mean to be dismissive but Moore just fucked it up and was off by a dimension shift of one like. Sometimes definitions are wrong

Like more seriously the existing alignment is based on the "geometric" intuition that degree 0 of chain complexes should correspond to 0-simplices, degree 1 should correspond to line segments, degree 2 should correspond to points and so on.
If you just change your intuition so that the object in degree n corresponds to the convex hull of n points it is fine

it just means we have to reindex our homology theories by one which people working with homology theories based on monads already do

ok i'm going to get some work done lmao

[unreal stratus]

Nice

Me too, might do some enriched cats

[white oxide]

What about working class cats tho 🥺

[hot locust]

Okay I heard that you can define compactness in R^n in terms of sequences, I.e every sequence in the A (compact) has a converging sub-sequence with a limit in A.

Isn't this just rephrasing heine borel

If you use this to prove heine-borel isn't this a circular argument

This is the post that prompted this question.

https://math.stackexchange.com/questions/4529598/proof-on-the-heine-borel-theorem

[tender halo]

It is not the same thing, it just so happens so in metric spaces sequential compactness (which is the property you stated) is equivalent to compactness

[hot locust]

No but if you are specifically talking about R^n then?

[tender halo]

You do need to prove either sequential compactenss or compatness somehow though

Sequential compactness is a pretty easy consequence of Bolzano-Weierstrass

[hot locust]

That they are equivalent here?

Or we can take it for granted here since it's a metric space

If we are using the result

[tender halo]

You do need to prove the equivalency too of course

If you want to use compactness that is, most theorems use compactness rather than sequential compactness

[hot locust]

Yea fair

[hot locust]

So say we prove the equivalence, then this argument is fine?

[tender halo]

yeah

[hot locust]

Cool, thanks.

[bright acorn]

what's a codisc bundle?

[umbral panther]

It must mean the unit disk bundle in the vector bundle that is the cotangent bundle. That was my guess just from the word codisk, but the fact that is a symplectic manifold clinches it. Why codisk, rather than disk? Maybe for some sufficiently exotic Finsler structure, you need to use the fact that it is dual to a metric space, rather than is a metric space in its own right?

[hexed steppe]

i think codisk is used instead of disk because the symplectic structure is quite important there

idk about the finsler setting

but my impression is that symplecticness of the space of oriented lines is central to making geometric optics work

[umbral panther]

There phrasing implies that there is a general notion of a codisk bundle, which is false

It would made more sense to say the disk of the cometric

[hexed steppe]

not sure i see the difference

how is "codisk bundle" not the same thing as "bundle of disks of the cometric"

[quartz horizon]

This is a bit of a silly question, but

Take a closed disk, x^2 + y^2 <= 1

Remove an open rectangle from it so that it looks like this

How would you show, explicitly, that this is homeomorphic to the original disk?

[cedar pebble]

how explicitly are we talking about lol

it's kind of painful to write down an explicit map but it's obvious that you can do this

[quartz horizon]

Writing down an explicit map :)

Like, I think it’s common to just say “it’s obvious” and move on, but it’d be nice to actually prove

[cedar pebble]

sure okay you can write down a piecewise map that stretches the middle part back into the disk you want

[quartz horizon]

What do you mean?

[cedar pebble]

[hollow stump]

When you are talking about R^n a compact set can just be defined by being closed and bounded, but when we talk about more general topologies it makes more sense to say that a set is compact if there always exists a finite subcover

So Heine borel becomes the definition and not a theorem

Lol

But yeah, a closed and bounded set is not necessarily compact in a non-complete metric space for instance

[quartz horizon]

Wait

[quartz horizon]

Compact subspaces of Hausdorff spaces are always closed

And the metric is continuous, so a compact subspace will be bounded too

In any metric space

[red yoke]

Jordan curve theorem

[quartz horizon]

I feel like this is too much machinery!

Like, don’t you think this would be a reasonable problem on a first problem sheet?

[hollow stump]

Hmmm... I don't really know what is a Hausdorff space? Lol

[quartz horizon]

In particular, a compact subset of a metric space is always closed and bounded

[hollow stump]

But I'm just saying that bcs the Heine borel proof (the general idea behind it) ceases to work when you eliminate completeness

[quartz horizon]

The point of heine-borel is to prove the reverse implication for R^n with the Euclidean metric

[steel glen]

no, it's rather cumbersome

with little to be gained from it

[quartz horizon]

That’s surprising to me

I feel like being good at topology means that you should be able to work with shapes like this

[steel glen]

it would be more illustrative to ask them to draw the homeomorphism

[cedar pebble]

not in the way you're insisting on no

[red yoke]

Animate Bad Apple by finding a parametrization for the boundary (3 points)

[cedar pebble]

insisting on writing down explicit maps like this kind of a waste of time

[quartz horizon]

Still, people are suggesting quite complicated pieces of machinery to solve it..

[steel glen]

my class introduced deformation retractions on the first day

[cedar pebble]

I mean yes because it's something you should be able to solve with a general method instead of writing down explicit maps in every example

[quartz horizon]

That’s for homotopy equivalence, not homeomorphism, right?

[hollow stump]

No yeah

You are totally right

Let me change the order real quick

[cedar pebble]

like if you understand topology you should be able to quickly convince yourself of how you might write this down, and then not continue to waste time actually writing this down

[steel glen]

yes, but that would be the intuition for how to produce the homeomorphism, since it will be the same map in this case

[quartz horizon]

I’m not actually sure how you get from the homotopy to the homeomorphism

[cedar pebble]

because the homotopy is a homeomorphism in this case

[quartz horizon]

Hmm

[red yoke]

Is the homotopy always a homeomorphism though Missed context

[quartz horizon]

I don’t think that types well though

[cedar pebble]

no nothing about this is unique

[quartz horizon]

A homeomorphism doesn’t use [0, 1]

It only uses the two topological spaces X and Y

[hollow stump]

Now it's correct

[quartz horizon]

There’s no “time” parameter involved

[hollow stump]

@quartz horizon thanks for giving me the heads up

[cedar pebble]

you can think of this as writing down a 1-parameter family of homeomorphisms

[quartz horizon]

No problem! These things can be tricky

[cedar pebble]

where at one endpoint you have the identity and at the other endpoint you have the map you're asking for

[hollow stump]

I just wasn't thinking that much when I wrote it lol

[quartz horizon]

But this is from the space to itself, right?

Like, the homotopy is between two maps X -> X

I don’t see how it gives a map X -> Y

At least, that’s how a homotopy equivalence works

[cedar pebble]

anyways if you were forced to write down an explicit map it's not that hard

write it as some composition of easier to understand/write down maps

[quartz horizon]

That’s… kinda what I’m asking lol

[cedar pebble]

well you should think about it if you're so curious

you won't gain anything by having someone else explain this to you

just as you won't really gain much by thinking about this problem

although you should probably think about it if it's not so obvious

[red yoke]

Note that S^1 x [0, 1] quotiented by S^1 x {1} is a disk, so if you have an appropriate homotopy S^1 x [0, 1] -> X such that S^1 x {1} gets mapped to one point, then you induce a bijective map from the disk to X, and you can apply "compact -> Hausdorff"

[red yoke]

But imho the point of elementary alg top is to make intuitive things easy to prove

So for something intuitive like JCT one would just apply it without hesitation

[hollow stump]

looks at the test
"yeah, I know how to solve this"
leaves

[cedar pebble]

this but unironically

[steel glen]

apart from being a bit against the spirit of why we do topology, writing an explicit map for this is also not a useful skill for later topology ventures
in general you care about showing/using existence of homeomorphisms or homotopy equivalences, so the experience of painstakingly writing this map out wont be very useful

[cedar pebble]

I've done topology and read topology papers for close to a decade now and I can count the number of times writing down an explicit map like this was actually useful on one hand

[hollow stump]

I remember watching a 3b1b video and thinking that space filling curves were necessarily bijective and continuous so I just spent a week trying to prove that such a curve was possible for R^2 💀

[steel glen]

oh man we had a small discussion about this and also continuous bijections between R and L2[0,1] in this server not too long ago

[hollow stump]

Lol

[cedar quest]

Any idea what this notation means?

Specifically the ['s

[unreal stratus]

This seems very non standard to me at least

I've never seen someone write like this, no sure what the [ means or why they are using curly brackets to indicate intervals

[languid patrol]

You don’t do low dim top though

[unreal stratus]

I thought low dim top people were even less likely to write smth out lol

[cedar quest]

I am thinking those are just typos

[quartz horizon]

This is quite interesting to me

I guess I’m a physicist, so I care about concreteness

So I feel like if I couldn’t write down a map, or an algorithm for producing such a map, I’m not sure if I understand the concepts well enough..

[tribal palm]

well they’re not saying they can’t write down such a map, just that they don’t need to

though, there certainly is very often in mathematics, theorems establishing the existence of maps without any means to actually construct them

Riemann mapping theorem comes to mind, but there are dozens

[quartz horizon]

Mhm I’m aware of them

But I don’t think you need such heavy machinery for this

[tribal palm]

i admit i did not read more than a couple messages up and don’t actually know the context

[quartz horizon]

Oh so

[quartz horizon]

The problem is simply to show this shape is homeomorphic to a closed disk

[tribal palm]

just wanted to stress the difference between knowing a map exists, and knowing you could explicitly give it

[quartz horizon]

Mhm

[tribal palm]

there is a tradition for constructivism at my uni

admittedly very weak, but certainly present

[grim knot]

Guys I am confused about something (nothing new haha), if I have a path connected space and I know that for example (1,0) is a strong deformation retract of X with given homotopy H. I thus know that I have a path gamma between every point y in X and (1,0). How can I now connect the given homotopy with gamma such that I get that y is also a deformation retract?

I get the concept behind it but I don't know how to formly define the homotopy

[gentle girder]

if i’m getting your question right, for $0\leq t \leq\frac{1}{2}$ then define $H_2(x,t)=H(x,2t)$, then for $\frac{1}{2}\leq t \leq 1$ we let $H_2(x,t) = \gamma(2(t-\frac{1}{2}))$

[gentle osprey]

kahler smaynifold

[tidal lynx]

if f:X -> Y and g:Y-> Z are maps such that f and “g compose f” are continuous then is g continuous as well?

[quartz horizon]

Not necessarily!

[unreal stratus]

No, let f be a constant map

and g be anything you like

[tidal lynx]

Sad

[unreal stratus]

True for homeomorphisms at least lol

[tidal lynx]

Yeah cuz g = (g o f) o f^{-1}

[quartz horizon]

Uh

[tidal lynx]

So here then how do they establish the forward direction

[quartz horizon]

No yes this is right

[quartz horizon]

This is cause of the universal property of the quotient

[tidal lynx]

why is that property true

It seems to be a statement of the form I described above (which is false in general), but ofc there’s more structure here

[quartz horizon]

Essentially by how the quotient topology is defined

It’s defined such that this statement is true

[formal tide]

Hatcher describes CW complexes as (pic). Two questions about this, can we have any number of n-cells attachments phi? E.g. an uncountable number of attachments. Second, we only admit a countable number Xn? Maybe we could start attaching cells again to X^omega and continue the process all again so that we have X^alpha for each ordinal. (it's not immediately clear to me that attaching a cell to a X^omega space is in X^omega, this latter question is silly if this is the case)

[ebon galleon]

2 out of 3 my beloved

[unreal stratus]

2 out of 6, a little more devious

[red yoke]

We can have arbitrarily many cells of each n yes

It's only for finite n though

[gentle girder]

countable n

[quartz horizon]

Ah, this brings back memories

I remember wondering how you’d prove that the order in which you attach cells doesn’t matter

[red yoke]

Countably many finite n

[quartz horizon]

And this is what got me into category theory

[unreal stratus]

wdym lol

Oh oka what arki said lol

[gentle girder]

sorry yeah whenever i talk about algebraic topology i am super unrigorous and imprecise (i learned from hatcher)

[unreal stratus]

But then it is sort of redundant lol

[red yoke]

Hatcher is mostly rigorous

[unreal stratus]

Well I meant the set theory of it was weird lol

But fair

[formal tide]

ohhh seems I could prove this to solve my second question

[gentle girder]

yeah i am just mad at hatcher

[ebon galleon]

As you should be

[quartz horizon]

It turns out there’s an abstract nonsense proof

[red yoke]

But do transfinite CW complexes exist

[quartz horizon]

Namely that “colimits commute”

[brave geode]

please no

[formal tide]

oh right

lol

[red yoke]

Transfinite ∞-groupoids

[formal tide]

yeah it's that easy

[quartz horizon]

Mhm!

But trying to do this hands-on

In terms of constructing an explicit homeo

Is horrible

You have to consider equivalence relations on top of equivalence relations on…

[formal tide]

tbf the proof of colimits commute give an explicit morphism

so just compute that (;p

[quartz horizon]

It does!

[gentle girder]

i bet countably many X^i are just equivalent to a “true” CW complex no matter what order type

[ebon galleon]

Cell complexes my beloved

[quartz horizon]

In fact, at any stage, you can partition the cells you attach any way you like, and add them in any order

[unreal stratus]

Eh you can add uncountably many cells of each dimension

[ebon galleon]

Transfinite composites of pushouts of coproducts of sphere inclusions

[unreal stratus]

Should be able to add as many as you want

[gentle girder]

oh wait what would like e^omega be

[unreal stratus]

Wdym

[gentle girder]

nvm

[gritty widget]

hmm okay

[gentle girder]

i learned nothing in topology apparently

[gritty widget]

Guys, how difficult is measure theory?

[unreal stratus]

Wrong channel but uh

[gentle girder]

idk but topology is pretty cool

[unreal stratus]

Depends how advanced you do it

[gritty widget]

Oh ok

[gritty widget]

Oh mb no worries

[formal tide]

like 3

[gritty widget]

3?

[unreal stratus]

Lol

[formal tide]

+- 1

[gritty widget]

Hm okay

[gentle girder]

for a second i forgot that you had to attach e^n when making X^n, so i thought something like “uh X^omega would just be something like attaching a 2-cell somewhere” idk (don’t think about it it doesn’t make sense)

[unreal stratus]

Lol ok

[gentle girder]

sorry

[unreal stratus]

Dw

OwO

[ebon galleon]

UwU

[red yoke]

Is Ryx be UwUer now

[gentle girder]

uwu

[unreal stratus]

UwU

[red yoke]

Wait is UwUer an adjective or a noun

[unreal stratus]

Neither

It is an exclamation

Oh

[ebon galleon]

Noun

[unreal stratus]

Changed the message to make me seem wrong

Lol

Then noun sure

[red yoke]

Edit sniped

Me, subtly changing the question on the finals when your head is buried in the paper

[brave geode]

it suffices to show that we can deform a piece x^2 + y^2 \leq 1, y \geq mx + b, y \geq c, where b, c are small enough, into the shape x^2 + y^2 \leq 1, y \geq c. then perform horizontal stretching; i.e. if y = mx + b intersects the semicircle at p, then for everything below p draw a horizontal line, and stretch them uniformly to the right

[grim knot]

oh that was really easy, I oversaw it, thank you very much for answering!

[brave geode]

@quartz horizon

i think this generally will treat convex constructions

[quartz horizon]

Uh, I can’t really visualise what you’re saying..

[brave geode]

if C is a closed set in R^n, and we have continuous functions f, g: C -> R with f = g on the boundary, then given a continuous function h that is below f, g we have a homeomorphism between the volumes {(v, y) \in R^n+1 : v \in R^n, h(v) \leq y \leq f(v)}, {(v, y) \in R^n+1 : v \in R^n, h(v) \leq y \leq g(v)} that is realized by a homotopy relative to the "boundary"

$V_f := {(v, y) \in \bb{R}^{n+1} : v \in \bb{R}^n, h(v) \leq y \leq f(v)}$, $V_g := {(v, y) \in \bb{R}^{n+1} : v \in \bb{R}^n, h(v) \leq y \leq g(v)}$

[brave geode]

where the boundary is the graph of h, and the "cylinder" in ∂C × R; i.e. points (v, y) such that v ∈ ∂C, h(v) ≤ y ≤ f(v) = g(v)

[rigid path]

Hi, I have a question. Let $f\colon X\to R$ a real valued continuous function and suppose that $f$ cannot be continously extended over $\beta X$ (Cech compactification). Can I conclude that $f$ is not bounded? If yes how?

[gentle osprey]

Milo

[brave geode]

is X nice

[rigid path]

Wym

Oh sorry I forgot to say

[ebon galleon]

If f was bounded, you could consider it as a function X --> [-M, M] for some M>0, for which there would be an extension to βX by its universal property.

[rigid path]

X is not compact obv

[ebon galleon]

Shouldn't need any assumptions on X for this.

[brave geode]

oh yeah

[rigid path]

I didn't get it, what do you mean with "universal property"?

[ebon galleon]

That maps X --> C for a compact Hausdorff space C extend to maps βX --> C

[rigid path]

Ohh okay thank you a lot

But I need X Tychonoff to use that property right? (Which anyway is my case I forgot to mention sorry)

[ebon galleon]

Nope, you can define Stone-Cech compactification of non-Tychonoff spaces as well, which still have this property (not sure of an exact construction, but they exist)

[rigid path]

Okok, didn't know about that. Thank you again

[tender halo]

its author dependent whether they call Stone-Cech-like compactifications of non-Tychonoff spaces a Stone-Cech-compactification

we want the space to be Tychonoff so that the inclusion morphism is an embedding

[formal tide]

I just tried to use this to prove that the order of attachment does not matter but you could attach an n-cell to an m-CW complex at a point that doesn't existed at the (n-1)-skeleton stage; so commutativity of colimits doesn't help

[quartz horizon]

Yeah, so it does matter what stage you’re at

[formal tide]

particular example: take a point, attach a loop at it, then attach another loop at some point different than the initial one

we get a figure 8, which is a CW complex

[quartz horizon]

Mhm

[tender halo]

for T1 spaces you can take Wallman extension, but it wont be nice, in fact it wont even be hausdorff if the starting space is not normal

[formal tide]

I'm not sure the problem can be solved using pure category theory

[ebon galleon]

I am of the opinion that special objects like this should be defined via their properties, from which it becomes a theorem that they exist. So here, to me the defining property of the Stone-Cech compactification of any space is this extension property, and one can prove (using, say, the Special Adjoint Functor Theorem) that this exists for any space.
And then, using this universal property, one can still show that the inclusion morphism is infact injecting and an embedding.

[ebon galleon]

So I was told once that all cell complexes (cells attached without regard to the dimension at a specifica attaching step) can be realized as CW complexes, but I'm not 100% confident that's actually true lol, and idk if there's a proof written anywhere (looking on MSE or MO might be good).
As you've noted, it's not as simple as colimits commuting with colimits, so it's not purely categorical.

[formal tide]

The result is false! https://mathoverflow.net/questions/23415/cw-structure-on-spaces-obtained-by-attaching-cells-wildly

[ebon galleon]

(at the very least, it is true that any cellular complex is homotopy equivalent to a cw complex, by Whitehead's theorem)

[formal tide]

(up to homeo, true up to homotopy)

[ebon galleon]

Interesting, I'll keep this in mind, thanks. I guess I should take handwavy comments from profs with a grain of salt lol

[grim knot]

I am trying to solve this exercise, but I do not understand what they request me to do(?)

oh wait the mean the sphere with S3 right?

Why did I think about singular 3-simplex

[unreal stratus]

yh

[red yoke]

You don't even need infinitely many cells, just glue a disk to another disk via Hawaiian earring

Idk how to prove it but it doesn't look like a CW complex

[formal tide]

I'm not sure what do you mean, attach D2 into D2 by S1->Hawaiian? Not sure if this latter map exists

[red yoke]

Well we got rid of the condition that it glues to the 1-skeleton

If gluing to the right skeleton is required it should be a CW complex since order of quotienting shouldn't matter?

[red yoke]

I think you could probably do "assume CW structure exists → earring is in X¹ → X¹ is not locally contractible"

[red yoke]

(Maybe that was what your Prof was alluding to?)

[empty grove]

Whitehead gives you this up to weak equivalence, but in fact this is true up to homotopy equivalence

Because of cellular approximation, the attaching maps might as well always go to the lower skeleton

And a homotopy between attaching maps gives a homotopy equivalence between the complexes in question

Needs a bit more work if there are infinitely many cells, then you gotta play with colimits a bit

[ebon galleon]

Well you have that there's a weak homotopy equivalence from a CW complex to your cell complex, which by Whitehead's theorem (the general version about cofibrant objects in Top_Quillen) we know this is a homotopy equivalence

That's what I had in mind, at least

[empty grove]

Ah sure

[tight ocean]

what is the homology group $H_0(\phi)$ for empty set? Also, what about reduced homology group?

[gentle osprey]

contrapositive

[empty grove]

Its unreduced homology is trivial in all degrees (the chain complex is 0). Its reduced homology, if you do bother to define it, is ℤ in degree -1, 0 elsewhere.

Usually reduced homology is only defined for pointed spaces (its the homology relative to the basepoint) so you wouldn't care about the empty space when you use reduced homology. It's the only space which has negative homologies.

[tight ocean]

okay got it

[formal tide]

It's not hard to see that (coprod Ai) x B = coprod (Ai x B), this is not true in any category, but can I still get a "high level" proof of this for Top?

[ebon galleon]

In the category of all spaces I'm not sure there's a general way to see this. It really relies on how the product & coproduct topologies are formed.

Cartesian closed subcategories ofc admit the usual "left adjoints preserve colimits proof"

[hexed briar]

Can non alternating knots have 3/4 crossings?

I can’t figure out how to draw one

[hexed briar]

Figured out. 4

Can’t do 3

[red yoke]

There's only one knot with 3 crossings

What did you find for 4

[solemn oar]

To be pedantic, there are two.

[tribal palm]

the trefoil knot and the trefoil knot?

[solemn oar]

Trefoil knot is chiral, so yeah. 😌

[tribal palm]

lovely

[solemn oar]

Oh I just read the original question. The answer is no. Any prime non-alternating knot must have 8 or more crossings.

[grim knot]

If I want to analyze $S^2$, where the north and south poles are identified, why does there exist a neighborhood U in S2/~, such that ${N} \sqcup {S}$ is a strong deformation retract of U?

[gentle osprey]

damn_guuurl

[unreal stratus]

That seems quite false

[grim knot]

I rewrote it

maybe that makes a bit more sense

[unreal stratus]

Oh lol that's very different

[grim knot]

sorry, it didn't change the TeXit

[unreal stratus]

Try doing it for S^2 i.e. find open set deformation retracting onto {N} u {S}

this should be clear with a picture

then the same proof will work for S^2/~

[grim knot]

Okay that makes sense, so it should be enough to take once the upper hemisphere and the hemisphere down without the equator, then it would automatically retract right?

[unreal stratus]

Yes sure

in my mind I just imagine a lil ball around each point lol

[grim knot]

That makes sense, thank you, I have another question though. I saw in Hatcher that the above quotient space is homotopy equivalent to S^2 v S^1 and I completely don't see how

[red yoke]

They are both a quotient of S² ∨ arc

[unreal stratus]

Yeah one way I would visualise it is like

you consider S^2 an attach an arc from N to S

That arc is a contractible subcomplex, so quotienting out by it is a homotopy equivalence and gives you S^2/~

But then you can also move that arc so that both ends at the same position

and then you get S^2 v S^1

[red yoke]

If f, g: A ⊂ X → Y are homotopic then X ∪f Y, X ∪g Y should be homotopy equivalent (?)

[unreal stratus]

In the case that the relevant things are cofibrations yes

So for CW complexes yes that works well

[grim knot]

that is much easier than I thought

[grim knot]

that's crazy haha, thank you very much

[unreal stratus]

Well it's sort of tricky on the point set level i guess

[red yoke]

Ig you just need to show cofibrationness

[grim knot]

I am not that advanced yet

[unreal stratus]

Well that should be the easy bit since we have cw complexes etc

[red yoke]

You mean proving this?

[unreal stratus]

Oh no I meant proving they're cofibs

[red yoke]

I meant this

[unreal stratus]

Actually this is reminding me how poitnset toplogy with unpointed stuff can be a pain lol

[red yoke]

How does having a point help

[empty grove]

Unpointset topology

[red yoke]

Hatcher has some weird variant of cofibration

Called a "good pair"

I think he defines cofibration (actually "HEP pair") in ch.0 and proceeds to forget about it

[empty grove]

Those are not quite equivalent

Oh ig you didn't claim that they are

[grim knot]

This is exactly what I wanted to show

So you read my mind

[plain raven]

We've had this conversation about the relationship between good pairs and cofibrations before. Somebody on Stackexchange posted a nice answer explaining some of the distinctions between them.

I can link it (both the conversation in this channel and the stackexchange links) if anybody wants to see the gory details
just would take me like 180 seconds
and i'm lazy 😄

[unreal stratus]

Yeah this has just made me think about cofibrations etc again lol

[red yoke]

Iirc for closed A ⊂ X, cofibration is equivalent to:

• Open neighbourhood N of A
• Homotopy N × I → X with A fixed from id into A
• A and X\N precisely separated
  While "good pair" is:
• Open neighbourhood N of A
• Homotopy N × I → N with A fixed from id into A

[unreal stratus]

i guess it depends on what you mean by cofibration

:^)

Actually lol this has made me wonder like is there a model category of spaces where Hurewicz cofibrations are the cofibrations

For Strøm you restrict to closed ones

I guess it's just that it seems like the cofibrations in model structures aren't quite in line with what people mean by cofibration etc in Hatcher-level texts

But I suppose the former is an abstraction of the latter so like fair

[unreal stratus]

Eh okay if you use Strøm for CGWH then Hurewicz cofibrations are closed already

But still, not quite Top

[ebon galleon]

The naive guess is that you might try to restrict your fibrations then to be "strong" fibrations: maps with the right lifting property against all acyclic cofibrations. Unsure if the proofs dualize well enough for this to actually work.

But also CGWH or a variation thereof is the correct category of spaces lol

[red yoke]

Why is it nicer than say, CW complexes

[umbral panther]

It has an internal hom. The space of maps from one cw complex to another is not a cw complex, although it is homotopy equivalent to one

[red yoke]

Is is naturally homotopy equivalent to one?

[umbral panther]

Every space has a natural weak equivalence from a cw complex, namely the realization of the simplicial set of singular simplicies

[ebon galleon]

Yeah you could also do spaces which are homotopic to CW complexes, iirc Milnor has a paper showing this is convenient

[grim knot]

what is the best CW complex one could assign on this space?

[red yoke]

Split into two 2-cells probably

Using a line tangent to both disks

[grim knot]

I don't see why that is needed

As a first question, how many 0 cells would you take?

[red yoke]

4

Oh actually one could do three

[grim knot]

so one for each disc

then you attach 1 1-cell to every point right? plus the two tangen lines you were talking about

so you get five 1-cells (?)

I still kind of not see why the two tangent lines are needed

[grim knot]

Is this what that thing woulf look like?

jeey

[red yoke]

?

[grim knot]

This is the space I cited right?

[red yoke]

Yea

[knotty vine]

Why not a single point with a loop a and a 2-cell with boundary aaa

[knotty vine]

I dont really understand what this means

[latent flare]

hey guys, so sorry to interrupt, but i'm taking algebraic topology for the first time this semester and i have literally no idea what's going on. my teacher isn't great and this pset is due pretty soon, so i'd appreciate if anyone could help 😭 if not, that is okay as well

[red yoke]

Ah rip I misread

[knotty vine]

Just ask!

[latent flare]

ok awesome, didn't want to interrupt but thank you so much

i feel like i just need hints to push me in the right direction for all of the problems

i have a vague idea for some of them but i'm really not sure what i'm doing lmaoo

[red yoke]

1. Take two arbitrary paths and induce f, g: π1 → π1'. When is (f^-1)g the identity?

2. For the backwards direction, there is a trick to turn any homotopy S¹ × I → X into a homotopy relative to any finite number of points

3. You just check where each path maps to ig

4. Find a map Klein bottle → S¹ that induces a surjection on π1

[latent flare]

hey i'll take anything at this point im not complaining 😭

[red yoke]

5. Be inspired by (4)

[latent flare]

lolllllll

thank you so so much

this was very helpful

🙏

[grim knot]

which equivalence relation is needed to identify the k-tori with the genus g?

[red yoke]

Euler characteristic determines compact orientable surface up to homeomorphism

[grim knot]

didn't do euler characteristic yet

I see the homeomerphism actually, but I wanted the equivalence relation to compute the homology groups

[red yoke]

Wait what's your definition of genus

[grim knot]

😄

we literally only did it in one exercise, but I know that by identifying the edges one gets a torus

[red yoke]

Wait what do you mean identify k-tori with genus

[grim knot]

I wanted to find an equivalence relation such that k-tori is isomorphic to $\Sigma_k / \sim$

[red yoke]

You mean given any k-fold torus, determine g?

[gentle osprey]

damn_guuurl

[grim knot]

I wanted to find an homotopic space to the connected sum of a k tori, and I thought about a quotient space

[tribal palm]

does this result have a name?

[red yoke]

Classification of compact surfaces

[tribal palm]

thank you

[red yoke]

The second section of Munkres point set has it

[red yoke]

Rip I'm still not sure what you mean

A connected sum of k tori is Sigma_k

[grim knot]

hahaha it's okay, I ask in reverse than, maybe that could give me an hint. If you were asked to compute the homology groups of the conencted sum of k-tori, what would you do? I can use singular and cellular homology

[red yoke]

This gives a CW complex

[grim knot]

okay wait a sec, do we have equality?

[red yoke]

Yes

[grim knot]

well that is much better

I thought it was kind of a quotient space

[grim knot]

it is this construction right?

[red yoke]

Yes

Try to identify it with quotient of a polygon

[solemn oar]

This problem doesn't ask about the fundamental polygon, but about an inductive proof.

[livid escarp]

You need to find an appropriate decomposition for MV such that $U^\circ \cup V^\circ = \Sigma_g$ and you could do it by letting $U = \Sigma_{g-1}$ and $V =\Sigma_1$ with some intersecion at the handle.

[gentle osprey]

tracy

[obsidian vigil]

basic question, I often see U\inTau and then other times I see U\in(X,Tau)
Are these the same thing with the first being a bit of sloppy notation?

[knotty vine]

Theyre the same, but the second is much more sloppy

[unreal stratus]

I've never seen the second

It seems to be almost intentionally misleading

Since you could be in X or in tau

And (X, tau) is a set in its own right

Just don't write the second one

It's also more standard to say U is open in X unless you are explicitly working with different topologies on the same set or smth

[plush folio]

x is in cl(A) iff there exists a sequence a_n such that a_n is in A and a_n converges to x. This is true in metric spaces, but is it true in a topological space in general?

[quartz horizon]

Not in general, no

[plush folio]

Not even in Hausdorff spaces?

[red yoke]

It's true if the space is first-countable

[safe torrent]

Maybe a crank question but is it possible to use homology/cohomology tools to work with non-topological objects? Just by having some abelian group and some function that acts like the boundary function giving you dd=0 as usual

[red yoke]

The closed long line is compact Hausdorff but there's no sequence converging to ω1

Except those that are eventually ω1

[safe torrent]

I know cohomology is a contravariant functor from Top to Ab iirc so I guess it's just pure crankery

[red yoke]

In general you use nets instead of sequences

There are many cohomologies

Not just on Top and not just Ab-valued

[quartz horizon]

There are Hausdorff spaces where this is false, I think

[safe torrent]

So really you can use it as a purely algebraic method to do other stuff outside topology?

[quartz horizon]

Mhm, you can

[safe torrent]

https://tenor.com/view/big-brain-markiplier-gif-14835823

[red yoke]

https://en.m.wikipedia.org/wiki/List_of_cohomology_theories

[safe torrent]

I've seen it before but the terminology looks foreign to me

At least the non-ordinary ones

But thanks though 🌸

[red yoke]

Perhaps easier ones might be de Rham or Dolbeault cohomology

Which come from differential geometry

[safe torrent]

Isn't de Rham kind of topological?

[red yoke]

Yes

[uncut surge]

Hochschild cohomology is really easy to understand for a newcomer!

At least, the definition, and the low-level meaning of it. You really only gotta understand algebras and modules

[safe torrent]

I'm reading a paper regarding Cech cohomology but I'm having a hard time understanding it

[uncut surge]

HOO BOY i love cech cohomology

[safe torrent]

Computational Cech cohomology

[uncut surge]

do you wanna share the paper?

[safe torrent]

I can send the paper

[uncut surge]

Now I'm also not a pro on cech cohomology (if there is someone on earth that is?)

[safe torrent]

I'm stuck on page 25 where the author essentially starts defining the matrices

I don't understand how they got them really

[uncut surge]

Gotta say after reading the abstract, I'm not sure what the value of that paper is b/c all of what he's saying sounds like it is well-known since the 1950s at least

but let's go

ooh big matrix

[safe torrent]

Yeah if you go further there is craziness going on

[uncut surge]

Let's see, which example are you stuck at?

[safe torrent]

Page 25

S1

delta 0

[uncut surge]

Is it clear to you that the matrix is just the matrix representation of the linear map \delta^0 that he defines at the end of page 24?

[safe torrent]

yeah it is

f_j-f_i

For U_ij

[uncut surge]

Hm yeah but it takes a while to understand how you translate that linear map into the bases that he chooses

and why the bases even make sense

[safe torrent]

It's kind of analogous to simplicial/cell cohomology

[uncut surge]

Once you get the hang of Cech cohomology, the ideas are really clear, but it's really annoying to write down

[safe torrent]

if you have a-b then for a you denote 1 and -1 otherwise. That's the way you go about it at least

[uncut surge]

Okay aha yeah I see what oyu mean, the bases simply come from the fact that we have the constraint i < j

[safe torrent]

What each column represent?

f_i or f_ij?

[uncut surge]

Yeah, sounds like you get the idea!

What are you struggling with then? Making it rigorous?

Because that is then exactly how you define the matrix elements

[safe torrent]

No I just don't understand how they got the first column for example

[uncut surge]

I'll abuse notation a bit, but: e.g. $\delta^0((1,0,0,0)) = (-1,-1,0,0)$, because $f_1$ contributes with $-1$ to the $f_{12}$ component and with -1 to the $f_{13}$ component

[gentle osprey]

Lartomato

[uncut surge]

And the vector $(1,0,0,0)$ in the domain of $\delta^0$ represents the $f_1$ element

[gentle osprey]

Lartomato

[grim knot]

does somebody know a good "way" or algorithm on how to find the functions in meyer vietoris sequences?

Like here, how do you know how the generators behave(?) Take for example phi

[uncut surge]

b/c on $U_{12}$ you have $\delta^0(f_1,0,0,0) = -f_1$ and on $U_{13}$ you have $\delta^0(f_1,0,0,0) = -f_1$ and on all other $U_{ij}$, we have $\delta^0(f_1,0,0,0) = 0$ because $U_1$ does not contribute

[gentle osprey]

Lartomato

[red yoke]

You could draw the generators

[grim knot]

fair, but how do you recognize afterwards?

you just see the way it changes in the other space?

[red yoke]

It should follow from the definition of the maps

[grim knot]

I mean phi in this case would be the direct sum of the inclusion

so we take the generator and send it in the other space to see what it does right?

[red yoke]

Yea

[grim knot]

this thing

[uncut surge]

It's at least true that there's no super easy way, you have to understand how the generators of the cohomology classes act when included into the image space

you really have to do some topological work, there's no algebraic trick that's guaranteed to work

[red yoke]

The inclusion into Möbius strip is ×2 and the inclusion into torus maps a generator to a generator

Which you can see from drawing the maps

[safe torrent]

Ohhh I finally get it! Thank you very much 🌸

[uncut surge]

Happy to hear 🙂 If you've already made it through the rest of the theory, I think you've understood the most important parts of Cech cohomology!

[safe torrent]

Yep I really like Cech cohomology

[uncut surge]

The calculations are annoying to do explicitly but it's good to understand how it can work

"
The formal proof for Theorem 5.7 is currently beyond the author’s understanding" oh my god what a cute baby

i wish him well

[hollow geyser]

Is this correct/satisfactory?

[unreal stratus]

I mean I would just use the fact that (f^-1)^-1 = f more directly ig

I don't think the way you've phrased it is ideal though, like

[hollow geyser]

My intuition was that it was true, but I was uncertain I had actually bothered to prove it before

[unreal stratus]

You've said like, homeomorphism => bijective => has an inverse

and then said the inverse is continuous

But the definition of homeomorphism just says "there is a continuous inverse"

So you should just appeal to that

(especially since inverses of continuous bijections needn't be continuous, and the way you've written it makes it sound like that is the case (to me))

[hollow geyser]

I see

okay

[unreal stratus]

But yeah, this is basically why homeomorphism is one of the "right" notions to study

Since open sets correspond to one another

in this way

so stuff you can say about the topology of one carries over

[unreal stratus]

You can just say "f has a continuous inverse f^-1"

[hollow geyser]

This was my thought, I just wanted to be sure. I actually need this property in another proof, but my topo book never explicitly mentioned this property of homeomorphisms (maybe it assumed it was obvious?)

I was stuck on some stupid problem for a few days and I just now realized homeomorphism can probably save me.

[safe torrent]

By the way if you see the same matrix again. Don't you think the last column is kind of wrong? Following the logic the last column should be (0,0,0,1) because f_4 contributes 1 to f_34

[hot locust]

Okay so to show that [-inf, inf] with order topology is compact, is it sufficient to show that all open covers contain a set of the form [-inf, a) and (b, inf] and since the give space inherits hausdorfness due to the order topology the rest can be written as [a, b] so it has a finite subcover. It seems like I have something here but honestly idk how to go about it or what I have is even right, for starters, sure [a, b] is compact, sure. But it's not going to be contained in any of the open covers. Can I please get a hint?

[uncut surge]

Yeah true I think. wonder if that changes the rank of the matrix?

[safe torrent]

I'll try computing it and see what I get

Nope it's still the same rank

[uncut surge]

Ah no the matrix still has rank 3. It's all good

[safe torrent]

I guess it's just a typo then

[uncut surge]

Lmao yeah I think the guy didn't notice he got it wrong because both matrices have the correct rank

[safe torrent]

[uncut surge]

You don't even need to do anything smart, if you add all the columns of the correct matrix, they add up to zero

So automatically non-full rank

I wonder if that's a general feature of these matrices

Oh lmao of course! Because on every component U_{ij}, you have one +1 for inserting f_i, and one -1 for inserting f_j; since the columns of the matrix correspond to these f_i and f_j, every +1 is counterbalanced by a -1

Nice. Not useful but nice

[hot locust]

Is this even compact?

Is [-inf, inf] with order topology compact? No right?

[alpine nest]

it is

[uncut surge]

Yeah I guess you just need to think about what open sets in the set can even contain +inf or -inf

[hot locust]

Wait it is?

[uncut surge]

If you can prove that a set [a,b] is compact for real numbers a and b, I'd guess the same proof should work here as well?

[alpine nest]

You can even exploit that fact

[alpine nest]

This is mostly right

Your open cover of [-inf,inf] can be adapted to give an open cover of [a,b]

Which has a finite subcover

Take that, throw in those two "halflines" and there's your finite subcover of [-inf,inf]

[uncut surge]

ah sorry your commentary is way more useful than mine, i failed at reading comprehension lmao

[hot locust]

So should I try showing that every open cover would contain a set of the form [-inf, a) that's the right approach?

[alpine nest]

Yeah

Well, a set of that form or a set that contains a set of that form

[hot locust]

Won't that just be sth like [-inf, a) union (a, b) or sth like that?

Or is there some other possibility here?

[alpine nest]

Well, the basis for the order topology are open intervals, so any open set containing -inf has to contain some open interval around -inf

I.e. some set of the form [-inf,a)

Actually those are called rays, but the point still stands

[hot locust]

Oh yea gotcha, got it, bingo, hallelujah. Thanks.

It just hit me.

Thanks

[hot locust]

Okay small question here, is it fine if I just use basis elements

[hot locust]

Or is it a good idea to just use arbitrary open sets?

Would choosing a basis here cause some loss in generality?

[vast tulip]

hey guys. taking a course in algebraic topology for the first time here. I'm dont intuitively understand the concepts of "quotient topology" and "product topology" yet. Its not well explain in the course textbook either. could someone suggest a good source video/book/lecture notes etc to understand this topic

[cedar pebble]

This is covered in basically any point set topology book

[tribal palm]

i also found the quotient topology rather hard to wrap my head around, but these digital lectures were very helpful with lots of examples giving intuition for the topic: https://www.youtube.com/watch?v=P7mTDR8FRMc&list=PLd8NbPjkXPliJunBhtDNMuFsnZPeHpm-0&index=15

[vast tulip]

thanks!

[alpine nest]

Well, the standard definition of compactness says that any open cover has a finte subcover. Whether or not that's equivalent to "any open cover using basis sets has a finite subcover" is a nice exercise

[sick wharf]

Is there such a thing as a combined covering-packing number? I mean a subset (x_i) in e.g. a hilbert space that is covering with B(x_i, epsilon_c) and Packing with B(x_i, epsilon_p)
epsilon_c > epsilon_p ?

Whereby epsilon_c assumes an upper bound, and epsilon_p assumes a lower bound. (Or maybe a bounded ratio epsilon_c/epsilon_p < ∞ or something.)

l2 space has packing number omega and covering number omega, but can we cover it without resorting to e.g. all finite support rational functions?

Maybe a weaker statement would be: There is a covering B(x_i, epsilon_c), so that {x_i} is not dense (?)

Thoughts?

[ebon galleon]

For an extra challenge, prove that it suffices to check this on a subbasis, or find an example of a subbasis for a topology where every open cover in the subbasis has a finite subcover, but the topology generated is not compact 🙃

[prime elbow]

Let (X,d) be metric space and I want to show that the open ball is open set.

let for any x_0(belong to X) and r>0 , B= B(x_0, r)={ y | d(x_0, y) < r).

Now i take for any y belong to B, there exists an r_0 > 0 such that such that for any element of B(y,r_0) also belong to B
So I want to show that let any x belong to B(y, r_0) , x will be an element of B .

I am stuck at these inequalities
i have d(x,y)< r_0 and d(y,x_0) < r and want to show that d(x,x_0) < r holds

Any hint ?

So what I take r_0 such that it will be holds

[merry geode]

You need to utilize the forall qualifier

[prime elbow]

In what definition, for all y belong B there exists r_0> 0 such that B(y,r_0) is a subset of B?

[merry geode]

Surely that is not the only one

[prime elbow]

Can you please explain more ?

[merry geode]

Ah wait, is that the only qualifiet that is involved

Yeah, my bad

Then you need to use the existential(exists qualifier) properly

[prime elbow]

What is this?

[merry geode]

Basically, you need to show the existence of a ball at x contained in the bigger ball, right

[prime elbow]

May be

[merry geode]

"May"?

[prime elbow]

At x means x be center, right

[merry geode]

Yep

[prime elbow]

Existence of a ball at y is contained in the bigger ball which at x_0 ?

[merry geode]

Ah, right. It is y

Yea

[prime elbow]

So how we proceed?

[merry geode]

For existence, you can take the radius quite arbitrarily.

[prime elbow]

But I cannot find the way to show that inequality holds

[prime elbow]

Got it thank you

[merry geode]

Good job!

[coral pawn]

Can someone explain the connection between the different model structures on the category of simplicial sets? There's one where the weak equivalences are maps that induce isomorphisms on all homotopy groups, one where X --> Y is a weak equivalence if [X, K] --> [Y, K] is a bijection for all Kan complexes K, and one where we require the bijection to hold for all weak Kan complexes. Are all of these somehow equivalent? Are the infinity categories associated to the different model structures equivalent?

[radiant walrus]

It's not written here, but X must also be a vector space, yes?

[paper wedge]

why is that

@radiant walrus

[radiant walrus]

To have W being a subspace

[paper wedge]

I think you got confused here

[radiant walrus]

My idea was trying to create an isometry with some vector space W, and then expand from there to create X~? I don't know

[paper wedge]

if a proper subset of X becomes a metric space when equipped with the same metric then we call this proper subset a subspace of X

[radiant walrus]

So X doesn't have to be a vector space?

[paper wedge]

No, this is just completing a metric space.

Maybe you got confused by the word "subspace"

[radiant walrus]

Yeah because like

[paper wedge]

here is a definition

[radiant walrus]

Oh

Wait...

I seriously thought that a subspace was a subset with addition and scalar multiplication

[paper wedge]

Yeah, I figured.

[radiant walrus]

Damn

The book didn't really explain it so I thought it used the same definition

Why is it called the same thing? 😔

[paper wedge]

I think it would be vague to assume X is a vector space given this context.

What field is this vector space over? What is the addition operation? etc

[radiant walrus]

Idk I thought it'd be any

But um

[paper wedge]

I guess it's an honest mistake :D.

[radiant walrus]

I was thinking of, like, the way to tackle this problem. I was thinking about creating a function that takes points from X to some set X~ and then make W be the image of the function?

That's my first attempt of tackling the problem at least

[paper wedge]

Try to think about what this "some set" should look like first.

[radiant walrus]

Well, we're using completeness here so it should probably involve Cauchy sequences of X

[paper wedge]

Yes

[radiant walrus]

I was thinking like

This is really difficult..

[paper wedge]

I think you can do it.

[radiant walrus]

So let's see

I gotta create a function that takes inputs in X and outputs to X~

[paper wedge]

Think out loud, and for every approach work out it's details.

[radiant walrus]

Then we have the image of the function also be a metric space, so that'll be the W

And then we have to prove that X~ is complete

And then show that it is unique

[paper wedge]

In my opinion, knowing what X~ should look like should be your first concern

Once you know what X~ should look like, this function you are talking about will come naturally

[radiant walrus]

Okay then let's see how X~ should look like

I have to artificially create a complete metric space, given only a random metric space

[paper wedge]

Yes, you have to someway brute force this.

[radiant walrus]

So the Cauchy sequences of X~ have to converge...

[paper wedge]

Let's think about that for a minute

The cool idea is that YOU get to define what it means for the cauchy sequence to converge

convergence is defined in terms of the distance which you get to define

with distance i mean metric.

[radiant walrus]

Yeah, I just have to define a distance

Let's see

[paper wedge]

a distance in which any cauchy sequence of your new metric space should converge

[radiant walrus]

I have an idea

So W has to be dense in X~, yeah?

Since I'm gonna try to make W be the image of the isometry, I could define the distance function d~ to be about the distance function d and then relate to it with arbitrary points of X~ by using the triangle inequality
d~(w1,w2) <= d~(w1, x~1) + d~(x~1, x~2) + d~(x~2, w2)
Something like that

I think I'll use ' instead of ~ to make it easier to type

[paper wedge]

So your elements X~ are the same?

[radiant walrus]

What?

[paper wedge]

What isometry are you talking about?

[radiant walrus]

I'll think about the isometry in a bit

But it's just a placeholder for now

[paper wedge]

The isometry condition is a more general condition of the case that X itself should be a subspace of it's completion

or nvm

So what does your X~ look like tho?

[knotty vine]

A Quillen equivalence is the right notion of equivalence for model categories: it induces an equivalence on the associated infinity categories. The first two model structures are Quillen equivalence, see https://ncatlab.org/nlab/show/model+structure+on+simplicial+sets#characterisations_of_weak_homotopy_equivalences . The second model structure is the one for quasicategories (infinity categories defined as weak Kan complexes) and it is certainly not equivalent to the first, but it is related of course, see https://ncatlab.org/nlab/show/model+structure+on+simplicial+sets#comparison

[radiant walrus]

I'm thinking of, like, the set of all Cauchy sequences of X that converge

Wait, that's just the set of all convergent sequences

[paper wedge]

You also might have metric spaces that do not have ANY convergent sequences at all and still cauchy ones.

[radiant walrus]

Take x in X. x = x1 = x2 = x3 = ...

[paper wedge]

Yeah my bad

But you might have spaces that are so incomplete , like for example {1/n} with the usual R metric

this has cauchy sequences but no convergent sequences ( other than the ones you mentioned )

so by your construction the completion would be the empty set ( or just the set itself )

[radiant walrus]

Ahhhhh

[paper wedge]

You are very close.

[radiant walrus]

How about just the set of all Cauchy sequences?

[paper wedge]

Okay. What metric would you define on this set?

[radiant walrus]

If (xn) and (yn) are in X~, I could maybe set the distance to be the standard distance definition of sets of points?

[paper wedge]

Lets call this X~ , Y.

[radiant walrus]

That is, the inf of {d(x,y); x in (xn) and y in (yn)}

[paper wedge]

so an element , x , of Y would be a cauchy sequence of elements of X , x = {x1,x2,....}

[radiant walrus]

Oki

Yes

[paper wedge]

d ,being the metric on X, is defined over XxX.

[radiant walrus]

I guess I have to use d when making d', huh

[paper wedge]

So you are suggesting to define d_bar ( the new metric ) as the infimum of d(x_n,y_n) as n varies?

or maybe the limit?

[radiant walrus]

Um, we can try both

[paper wedge]

as in d(x,y) = lim d(x_n,y_n)?

[paper wedge]

Must this even always exist?

[radiant walrus]

Uhh

I don't think so

Because Cauchy Sequences might not always converge

Wait

Hold on

[paper wedge]

hint: d:XxX ---> R

[radiant walrus]

I don't know how that's supposed to help.. I mean, R is complete, I guess

[paper wedge]

Yes

[radiant walrus]

Do I uh

Is Y supposed to be a subset of R, or?

[paper wedge]

No. That wouldn't make sense

To recap , we have a very good candidate choice for the set of our completion

and we are trying to define a metric on this set such that all cauchy sequences of this set converge.

[radiant walrus]

The set of all Cauchy sequences

[paper wedge]

so we know that this new metric , call it d' must be a function from YxY --> R such that some conditions we will verify later

that means

if x and y are elements of Y then d'(x,y) must be some element of R

we are trying to see if d'(x,y) = lim d(x_n,y_n) makes sense

as in does this lim always exist

given that x_n and y_n are cauchy

so now

the question is: if x_n and y_n are cauchy sequences that may not converge , does the lim_n d(x_n,y_n) exist?

We know R is complete , so we must show that { d(x_n,y_n) | n is in N } is bounded

correct? @radiant walrus

If we do so, then by completeness we have our limit.

[radiant walrus]

Oh my god

Wait

Lemme check my notebook

I think I

[paper wedge]

given x_n and y_n are cauchy, ofc.

[radiant walrus]

Okay

You will not believe this

So, last time I studied this was like 2 months ago or so

And I was still doing homework

I accidentally skipped it now that I think about it. I should probably go back to it

But

The question I stopped on

Which I only wrote the statement, not the proof

If (xn) and (yn) are Cauchy sequences in a metric space (X, d), show that (an), where an = d(xn, yn), converges. Give illustrative examples.

I'll try to work on this

[paper wedge]

Yes, if you prove this then you have proven that your metric indeed always exists.

Well, we can't call it a metric untill we have verified all 3 conditions, so that's later

candidate metric, i should say.

[radiant walrus]

Given e > 0, we have that there are N' and N" such that n,m > N' => d(xn, xm) < e/4, and n,m > N" => d(yn, ym) < e/4. Let N = max{N', N"}. We then have that d(xn,yn) <= d(xn,xm) + d(xm,yn) <= d(xn,xm) + d(xm,ym) + d(ym,yn) => an - am <= e/2 < e. Using the same logic, we have that am - an < e, so |an - am| < e, for all n,m > N, therefore (an) is a Cauchy sequence and, since (an) is a sequence in the real numbers and that R is complete, (an) converges.

Does that work?

[paper wedge]

d(xn,xm) + d(xm,ym) + d(ym,yn) => an - am <= e/2

why is that

why does d(xm,ym) not blow

[radiant walrus]

I may have skipped steps there

So an = d(xn, yn), correct?

So am = d(xm, ym)

We have
d(xn,yn) <= d(xn,xm) + d(xm, ym) + d(ym, yn)
an <= e/4 + am + e/4
an - am <= e/2
an - am < e

[paper wedge]

Okay

this is correct

So now what does this imply

[radiant walrus]

It implies that the distance function makes sense

[paper wedge]

Okay

[radiant walrus]

It always outputs a non-negative finite real value

[paper wedge]

So we have four things to verify

1. positivity

2. d'(x,y) = 0 <--> x=y

3. triangle inequality

4. completeness

for this to be a complete metric space

[radiant walrus]

Positivity comes free

[paper wedge]

correct

triangle inequality as well, no?

[radiant walrus]

Well yeah
d'( (xn), (yn) ) = lim d(xn, yn) <= lim (d(xn,zn) + d(zn, yn)) = lim d(xn, zn) + lim d(zn, yn) = d'( (xn), (zn) ) + d'( (zn), (yn) )

[paper wedge]

okay

now lets show completness

this is going to get slightly hairy

[radiant walrus]

Wait

[paper wedge]

lets think about 2) first

is 2) even true?

[radiant walrus]

It's not

[paper wedge]

correct

how do we fix that

[radiant walrus]

The problem comes from this

lim d(xn, yn) = 0 does not imply xn = yn for all n

[paper wedge]

correct

so how do we fix that

hint: brute-force it

or not brute-force it

[radiant walrus]

Am I doing okay so far >.<

[paper wedge]

You are doing great

[radiant walrus]

Um

So I was thinking like

We just go with it, like

If the Cauchy sequences (xn) and (yn) are such that lim d(xn, yn) = 0, then we just say that they're the same sequence in Y. They're the same point

[paper wedge]

Exactly.

[radiant walrus]

We just get rid of unnecessary points that were in Y anyway

[paper wedge]

We prove that x ~ y iff d(x_n,y_n) = 0 is an equivalence relation

partition our Y into the equivalence classes of our relation

[radiant walrus]

Sorry you lost me there

[paper wedge]

and then define our metric over the equivalence classes

[radiant walrus]

I don't know what an equivalence relation or equivalence class is

[paper wedge]

Okay.

[paper wedge]

Thats very smart considering you do not know equivalence relations , imo.

I have to salute you

okay so

[radiant walrus]

In my mind I'm like

"They're going to the same place anyway"

[paper wedge]

Lets go back to basic set theory

do you know what a relation is?

[radiant walrus]

I might have forgotten

But I did learn it at some point

[paper wedge]

a relation of on a set is just any subset of the cartesian product of the set with itself

[radiant walrus]

Rightt, that

[paper wedge]

so for example if our set is Z then a relation would be the subset {(1,2),(2,3)}

another notation for (1,2) is 1R2

or 1 ~ 2

okay?

[radiant walrus]

Mhm I remember that from classes

Not the 1 ~ 2 but yeah

[paper wedge]

okay

an equivalence relation is a special kind of relation

an equivalence relation is a relation such that it is reflexive , transitive and symmetric

a relation is reflexive iff a ~ a for all a in your set.

a relation is symmetric iff a ~ b implies b~a for all a,b in your set

a relation is transitive iff (a~b and b~c) implies a ~c for all a,b,c in your set.

if you have all three conditions then you call your relation an equivalence relation

cool?

[radiant walrus]

lim d(xn, xn) = 0, so (xn) ~ (xn). That works
lim d(xn, yn) = 0, so lim d(yn, xn) = 0, which means (yn) ~ (xn). That works
lim d(xn, yn) = 0 and lim d(yn, zn) = 0, so lim d(xn, zn) <= lim (d(xn, yn) + d(yn, zn)) = 0 + 0 = 0, so lim d(xn, zn) = 0, so (xn) ~ (zn). That works
Okay

[paper wedge]

The main point of defining equivalence relations in such a way is that in this way a~b is the same as saying a is equivalent to b

now

let X be a set and ~ be an equivalence relation on it

let a be an element

define [a] to be the set { b | a ~ b }

this is called the equivalence class of a

now the main point is this, you can partition X into the set of all equivalence classes

[radiant walrus]

Ahhh

So an equivalent class of x are all the points y in X that are equivalent to x

[paper wedge]

yes

and now as sets X can be thought of as just the set of all equivalence classes

did you ever study group theory?

[radiant walrus]

Nope

[paper wedge]

do you know the clock works

how*

[radiant walrus]

I think so?

[paper wedge]

yeah lets consider Z

we define an equivalence relation on Z by a~b iff 12 divides (a-b)

[radiant walrus]

Woah, you're very active now

[paper wedge]

yeah

now

what do the equivalence classes look like?

what does 1 get identified too?

u look at ur clock and it says 13

13:00

[radiant walrus]

Yeah!

13

Because

12 divides 13 - 1

[paper wedge]

yeah

[radiant walrus]

And 25

[paper wedge]

so 1 gets identified with 1+12,1+12+12, and so on

similar with negative integers

[radiant walrus]

I see I see

[paper wedge]

so you have partitioned Z into these equivalence classes

you have somehow "glued" these points together

so this is what you did above.

you glued the sequences that have distance 0

[radiant walrus]

You turned Z into just {1, 2, 3, ... , 12}

[paper wedge]

not quite

into {[1],[2],...,[12]}

equivalence classes.

[radiant walrus]

I guess you need that []

[paper wedge]

ofc

as we defined , [1] is the set of all y such that 1~y

[radiant walrus]

I feel like it's a bit of an abuse of notation but I get it

[paper wedge]

so [1] = {1,13,25,...}

[paper wedge]

it is

this is properly written this way:

if X is our set and ~ is our relation

[radiant walrus]

It'd be more like [1] U [2] U [3] U ... U [12]

[paper wedge]

we define the quotient as X/~

as the set of all equivalence classes

so our set would be called Z/~ which is now {[1],[2],...,[12]}

so back to our original problem

[radiant walrus]

Ahh oki

[paper wedge]

you proved that x ~ y iff d(x_n,y_n) = 0 is an equivalence relation

so lets consider the new set now Y/~

once we are over Y/~

we have "glued" the sequences that have distance 0 now

and now they are effectively the same point

so d(x,y) = 0 <--> x=y works

and now we have a metric space.

[radiant walrus]

Can't we take X~ to be Y/~

[paper wedge]

this is exactly what we are doing

[radiant walrus]

Oki :3

[paper wedge]

regular Y wouldnt work , as there are diff sequences that have distance 0

but now that we have literally glued them together

that is

we have literally put out the word that "sqeuences that have distance 0 are now the same"

then , well, they are the same! :D.

[radiant walrus]

Reality can be whatever I want

[paper wedge]

give or take yeah

equivalence relations are everywhere

i just learnt about some kind of equivalence relation yesterday

they are everywhere

so now

our completion is Y/~

[radiant walrus]

We gotta prove that Y/~ is complete now

[paper wedge]

yeah

and we are almost done

[radiant walrus]

I got this

[paper wedge]

yeah

[radiant walrus]

Lemme see

[paper wedge]

it might be a bit hairy

try to draw stuff

introduce new notation

etc

good luck

[radiant walrus]

Yeah because like

I'm having a Couchy sequence of Couchy sequences

[paper wedge]

not just this

yo uare having a sequence of equivalence classes of cauchy sequences

[radiant walrus]

Almost the same thing

[paper wedge]

yeah.

[radiant walrus]

I'm sorry, I'm really tired right now

I'll go sleep and try to think about this later

[grim knot]

Does somebody have an hint on how to compute the homology group of the torus without a discrete set of points?

[umbral panther]

The first point is the hardest

[grim knot]

I mean by removing one point I just get S1 v S1 right?

mmh but what happens if I remove more?

[red yoke]

you could probably adapt the method you used for the first point

[grim knot]

there is like a line in the middle right?