point-set-topology

[pearl holly]

ye

[onyx crow]

what is $K(G, n)$ ?

[gentle osprey]

BrokenPizzaDreams

[pearl holly]

wait f^* is the induced map

[onyx crow]

oh sorry that’s what I meant

[pearl holly]

K(G, n) is a CW complex such that pi_n(K(G, n) = G and all other homotopy groups of this space are trivial

[onyx crow]

Oh you’re using the isomorphism $H^n(X;G) \cong [X, K(G,n)]$

[gentle osprey]

BrokenPizzaDreams

[pearl holly]

ye that's right

[gritty widget]

induced by what?

im a little confused

isnt f*(alpha) a function acting on a n-chain alpha

f* is the element if H^n right?

[pearl holly]

f^*: H^n(Y) --> H^n(X)

[gritty widget]

yuh

oh

lol

[pearl holly]

and \alpha is an element in H^n(Y). So f^*(\alpha) lives in H^n(X) so I can view it as a map X--> K(G, n)

[gritty widget]

yea mb

topology of type K(G,n) is a CW complex with n-homotopy group being only nontrivial one

or wait

K(G,n) is equivalence classes of CW complexes

[pearl holly]

wait what?

[gritty widget]

right?

[pearl holly]

K(G, n) is uniquely determined up to homotopy right?

[gritty widget]

yeah

[pearl holly]

oh that's what you mean

[gritty widget]

A type K(G,n) CW complex is determined up to homotopy equivalence

so my conception is that you can find homotopy equivalence between two CW complexes and they are both of type K(G,n)

btw why are you learning alg top

i have slowly lost some motivation for learning it

feels very handwavy at times

[pearl holly]

idk lol

[pearl holly]

but ye it do be getting messy sometimes

and handwavy

but I like the fact that a lot of theorems can be understood intuitively by just drawing a picture

[gritty widget]

i sort of like that

[pearl holly]

it's like rubbery geometry with algebra

[gritty widget]

but at a certain point it feels fake

like any person can only draw so many things

and interpret the visuals

tbh it just feels like algebra

atleast for me

[pearl holly]

ye lmao

like weird algebra

[gritty widget]

ye without fail

if you were to say 4 essential theorems what would they be

im saying excision, svk, UCT for cohomology and idk last one tbh

[pearl holly]

1 donut = 1 COFFIS CUP

[gritty widget]

and homotopy invariant theorems

[gritty widget]

tbh thats my main takeway

i have no intuitive understanding of the big theorems

[pearl holly]

okay I need to go, see you! 👋

[gritty widget]

😎

[plain raven]

abstract simplicial complexes and simplicial homology

[gritty widget]

ty

[obtuse meteor]

I think algtop classes should introduce more classical problems in algtop to see why things are the way they are (tm)

I've been reading about the hopf invariant 1 problem and I really feel like it should be something one is exposed to as like. A thing to care about (tm)

in an algtop class

[plain raven]

I think that alg top classes should introduce some basic concepts of smooth manifolds and sketch de rham cohomology, if not precisely

but like

hatcher's introduction to singular cohomology is decently geometric and nice but it's so limited that it doesn't really suggest at all how much richer things get when you start adding in differential forms

singular cohomology becomes so much easier to visualize when you think about cochains as formal integral operators.

You could honestly introduce singular cohomology by asking students "How would you prove Stokes' theorem?"

and then like, talk about it a little bit

and then be like "Well, an arbitrary manifold could be complicated to prove it for, so why don't we try triangulating the manifold (or reducing it to simple primitive shapes), proving it for simplices and then reducing it to simplices using the triangulation"

and in the course of answering this question and proving stokes theorem you'd get a chance to naturally introduce with lots of motivation some constructs from singular cohomology

[empty grove]

Talk series when

[plain raven]

i have already promised to do a talk on spectral sequences so this would come after

[true robin]

I’ve never seem this notation before, but if I were to guess, it would be path concatenation

yes it is the same, concatenation works well with homotopy

[cold vine]

Can someone clear things up for me regarding HEP for CW-complexes? I'm working in Spanier and relative complexes but this shouldn't really matter.

So following Spa I have shown that X formed from A by attaching cells, gives a strong deformation retraction $X\times I \to X \times 0 \cup A \times I$. Now by corollary $A\subset X$ is a cofibration, but I have argued differently that since we have the def retraction we have a retraction - and now since we have a retraction we can compose it with the identity and gain an extension of the homotopy (this is in Hatcher). I think these two things should be synonymous by definition since HEP = inclusion is a cofibration. Now since CW-complex is obtained by attaching cells (I think I need to show that each k-skeleton is closed in the next one) we have that each $X^k \text{ or } (X,A)^k$ has the HEP and since $X=\cup X^k$ so does the whole complex. But in Spanier it is argued that since each skeleton inclusion is a cofibration and that we have topology coherent with the skeleta now we can use the induction. Is in my case when doing with the HEP it relevant that I need to examine the topology here? I don't see why it would be?

[gentle osprey]

PAHUS

[coarse night]

what is the Stone-Cech compactification of (0,1]

sin(1/x)

[empty grove]

nope

I doubt there is a nice description

There usually isn't

[coarse night]

but there is one right?

[onyx crow]

afaik all top. spaces have Stone–Čech compactification if that’s what you’re asking

But apparently they are not easy to describe

[coarse night]

so for C[-1, 1] we can take n's to be all the even integers, we'll get only even functions.

does it contradict Mintz's theo?

[long hornet]

I read it requires choice to prove its existence

[coarse night]

can't run from AoC, lol

[empty grove]

This is irrelevant

choice is a standard axiom

Yes it exists but the constructions I know are all very bad

You work with the universal property

[coarse night]

[long hornet]

Well, if the existence of something depends on choice it seems natural to assume that you can't find a nice description of it

I think this applies particularly to topology because you have to conceive a space as a "whole," in some sense. One can "begin" to think of a well-ordering of the reals, for example, more clearly

[empty grove]

Oh sorry I misinterpreted what you were trying to say

That is fair

[coarse night]

[long hornet]

It's more suited for the analysis channel. Also, what are "these" functions?

[coarse night]

polynomials of degree n_i

[arctic plover]

by Homeomorphism?!

[wanton marsh]

?? no ?

homeomorphisms are bijective

by a quotient map

[cold vine]

I have shown that X obtained from A by adjoining/attaching n-cells has HEP and thus for a CW complex each k-skeleton has HEP. How do I argue from this that the whole CW-complex has the HEP property. It is by some induction argument but what is required to make it work?

[lunar yoke]

I think it just works as you'd expect

starting with some homotopy H_{-1} on A x I, you can inductively extend it to H_n on X_n x I

and since X is the colimit of all the X_n, if i recall correctly you also get that X x I has the colimit topology of all the X_n x I since I is locally compact, hence the functor Y -> Y x I is left adjoint to Y -> Y^I and thus preserves all colimits

so you can define a map from X x I by just sending (x,t) to H_n(x,t) if x is in X_n

and its continuous since all the restrictions to X_n x I are continuous

[cold vine]

mmh yeah I think that makes sense. Is colimit topology synonymous with coherent/weak topology? Thanks!

[lunar yoke]

yeah by colimit topology i just mean final topology with respect to the inclusions. I think its also called the weak topology sometimes

[cold vine]

Yeah that makes sense! Thanks a lot I wasn't quite sure that works but I think you verbalized it so now it makes sense!

[lunar yoke]

although i dont like that name since for me the weak topology is the one from functional analysis

[cold vine]

yup true

[fickle marsh]

So I have a thing that I'm thinking that I know can't be true:

A perfectly normal space is a normal space where each closed set is the union of a countable number of open sets

The problem I'm working through is showing that all metraizable spaces are perfect normal

Now I know that not all metraizable spaces have a countable basis (2nd countable), but it seems like perfect normal implies a countable basis, since being normal implies being hausdorff, and hausdorff spaces have that X and the empty set are the only sets that are both open and closed

Ah okay so I figured it out as I was typing but I will continue anyway, my problem was that X being closed means that it can be written as a countable number of basis elements

but that does not mean you can write the topology on X as a countable number of basis elements always

[lunar yoke]

No you're thinking of something else. Hausdorff tells you that any two distinct elements can be separated by disjoint open neighborhoods

[fickle marsh]

Being normal implies being hausdorff

[lunar yoke]

X and the empty set being the only open and closed sets is called connected

and generally has nothing to do with being hausdorff

[fickle marsh]

I think that there's a theorem that shows that hausdorff spaces have that property

[lunar yoke]

no

take two disjoint open balls in R^n for example

clearly hausdorff, but not connected

i.e. the set that contains precisely one of the balls is both open and closed

[fickle marsh]

I see yeah I got that confused

[plain raven]

i haven't really read much of goerss jardine

which seems likek it's hurting me more and more as like 90% of the time i'm thinking about simplicial homotopy shit

i figured out that there are two reasonable definitions of the suspension of a simplicial set and I wanted to know if they were homotopy equivaelnt. so i started writing a mathoverflow post. but then I opened G-J and sure enough, by golly, they're like "There are two reasonable definitions of the suspension of a simplicial set and they're homotopy equivalent"

By Jove!

[empty grove]

You are the alien discovering the same math as humans 👽

[hot pivot]

If only I saw this 30 second explanation of a boundary earlier, I'd understand so much of the motivations behind topology (mostly in R) in an instant. 3b1b is an excellent educator.
https://youtu.be/-RdOwhmqP5s?t=1104

[pearl holly]

is there a nice way to illustrate/draw [0, 1] x S^2?

[lunar yoke]

Doesnt D^3 - some smaller open ball give you something homeomorphic?

[pearl holly]

oof that might be true I have no idea

[lunar yoke]

like a "thick" S^2

[pearl holly]

ye okay I think I see it

[lunar yoke]

yeah i think for details you can use spherical coordinates

[pearl holly]

I think I will skip the details tho lmao

[lunar yoke]

relatable

[pearl holly]

Okay but thank you so much!

[long hornet]

Related: S^2 x (0, 1) is R^3 - {0}

So I want to know what's the one-point compactification of R^3 - {0}

It has one because it's lch

[lunar yoke]

so i think the one point compactificatio of R^2-{0} would be something like this horn torus

and then you try some higher dimensional analogue?

[lunar yoke]

this is based purely on intuition and pictures in my head though

[long hornet]

Aha

I can barely think of this one

But I remembered the OPC(X x Y) = OPC(X) ^ OPC(Y) thing

[lunar yoke]

whats OPC?

[long hornet]

Here ^ is the smash product

One point compactification.

[lunar yoke]

well i dont know about smash products yet :/

[long hornet]

Their definition is straightforward but I find it hard to work with them

[lunar yoke]

yeah ok ive seen the definition, but havent really done anything with them yet

started learning algtop 2 months ago

[long hornet]

Cool!

Well, it seems that the smash product here has a copy of S^3

Now S^2 x [0, 1] and the smash product both contain S^2 x (0, 1), so both are compactifications

But the OPC is the minimal compactification

So S^2 x [0, 1] contains a copy of S^3...?

If so, we can conlude that it can't be embedded in R^3

[lunar yoke]

that sounds plausible

[long hornet]

It seems false. S^1 x [0, 1] doesn't contain a 2-sphere

[lunar yoke]

i meant that it cant be embedded into 3-space, not sure about the sphere

but is it even the case that given two compactifications one must be included in the other?

seems wrong

[long hornet]

No, but I thought the OPC is minimal. I think it is, but maybe in a different sense.

At any rate, S^2 x [0, 1] is obtained from R^3 - {0} by "adding" two 2-spheres somehow

[lunar yoke]

yeah

i think its easier to think of R^3-{0} \cong S^2 x (0,1) as the "open thick S^2" and then S^2 x [0,1] as its closure, the "thick S^2"

and the latter you get by adding a little S^2 on the inside, and one big one on the outside, as boundary of the open thick S^2

[fading vale]

You can just use the formula you found above: OPC(S^2 x (0, 1)) = OPC(S^2) smash OPC((0, 1)) = (S^2 cup infty) smash S^1, where the union is disjoint union of topological spaces because S^2 is compact. our base point should be infty so our end result is like

$(S^2 \cup \infty \times S^1)/(S^2 \cup \infty \vee S^1) = (S^2 \times S^1 \cup \infty \times S^1)/(S^2 \cup S^1)$

[gentle osprey]

Kosher Nostra Chaya Moth

[fading vale]

Which will basically look like a higher dimensional analog of the horned torus like phil said, bc we're basically taking our extra disjoint copy of S^1 coming from our extra point and one copy of S^2 in our S^2 x S^1 and squishing them together to a point

So obviously we can basically ignore the extra disjoint copy and its just squishing one slice of S^2 in S^2 x S^1

So this comes from the 2 dim case but replacing S^1 with S^2

[plain raven]

both of these suspensions have a right adjoint, which we can call a 'loop space'. is there any good abstract nonsense argument that allows you to prove that if F, G : SSet -> SSet are naturally homotopy equivalent then their right adjoints RF, RG are also homotopy equivalent?

today i figured out an interesting monoidal product on simplicial Abelian groups

[gentle osprey]

Nobody

[plain raven]

thank you. I am having trouble understanding what you're saying here tho

the adjunction between left and right homotopies exists regardless of whether Y is Kan right

Oh. Ok I think i get what you mean

I don't like this way of phrasing it tho

To me the way I read this is that in general, we have a left homotopy from $f$ to $g$ iff we have a right homotopy from $g$ to $f$, and if $Y$ is Kan then the homotopy relation is symmetric

[gentle osprey]

Kanga Gang Made Man

Nobody

[long hornet]

That makes sense. So essentially S^2 x I = D^4 - open subball?

[long hornet]

Amazingly, this makes sense
I really wanted to show that there was S^3 embedded, but we found that this is not true.

Let m(M) be the least k such that M embeds in R^k as a closed submanifold. I wanted to show that m(S^2 x [0, 1]) = 4. In general, m(M x N) <= m(M) + m(N), of course.
When do we get equality when M = S^n? My motivation is that S^n "pushes embedding to it limits" so when we take a product we generally have to just start a new and just use a new Euclidean space for N.
Also, m(R^n - {0}) = n + 1, not n, which miraculously saves the conjecture (because otherwise we would use the S^n x R = R^(n + 1) - {0} homeomorphism).

Oh, actually m(S^n x S^k) <= n + k + 1, so equality doesn't always hold.

[fading vale]

yea the torus embeds in R^3 but S^1 embeds in R^2

[finite heath]

how is {x} x Y topologically equivalent (homeomorphic) to Y?

🐻‍❄️

[empty grove]

The homeomorphism is deleting the first coordinate

It's like in R², any {x} × R subset is just a vertical line, and any R × {x} subset is a horizontal line

And then it should be intuitively clear why it's true

[finite heath]

as in

f: {x} x R --> X is homeomorphic?

f: R x {x} --> X is homeomorphic

ugh

my intution for homeomorphisms SO bad.

[empty grove]

Not to X, to R

If you sketch this subset of R²

You'll see that it is just a line

And a line is like

Real line 🤡

So your set is set of all pairs (x,a) where a ∈ ℝ

And the homeomorphism is (x,a) ↦ a

And its inverse is a ↦ (x,a)

Like the x axis in R² looks exactly like R does

Same for y axis in this case because we are taking product of 2 copies of R

And any other horizontal line looks like the x axis (homeomorphism is translation)

Vertical lines look like y axis

Now replace R × R with X × Y

[pearl holly]

So there's something I don't understand. Hatcher starts with an element $\nabla^(\lambda(\alpha)) \in H^2(RP^\infty \times S^1) \cong Hom(H_2(RP^\infty \times S^1), \mathbb{Z}_2)$ where everything is with $\mathbb{Z}_2$ coefficients. Then he writes "A basis for $Hom(H_2(RP^\infty \times S^1), \mathbb{Z}_2)$ is represented by $RP^2 \times {x_0}$ and $RP^1 \times S^1$". Now he shows that a cocycle representing $\nabla^(\lambda(\alpha))$ takes value 0 on $RP^2 \times {x_0}$ and 1 on $RP^1 \times S^1$ and then apparently we get that $\nabla^*(\lambda(\alpha)) = \omega_1 \otimes \alpha$ where $\omega_1$ is a generator of $H^1(RP^\infty)$. I don't understand the last equality. Where does the tensor come in?

[gentle osprey]

Tokidoki ✓

[pearl holly]

\alpha is a generator of H^1(S^1) btw

[shy moss]

in what page of Hatcher is that?

[pearl holly]

506

lmao I skipped like half the book tho

why were you muted yesterday moldi?

[empty grove]

Literally 1984

[pearl holly]

[empty grove]

My grad top course is following Hatcher

Can never escape

[pearl holly]

If I would guess the last equality comes from the fact that $H^2(RP^\infty \times S^1) \cong \mathbb{Z}_2 \oplus \mathbb{Z}_2$ where the first $\mathbb{Z}_2$ comes from $H^1(RP^\infty) \otimes H^1(S^1)$, but this "corresponds to" $RP^1 \times S^1$ and $\nabla^(\lambda(\alpha))$ is 1 on this space and the other $\mathbb{Z}_2$ comes from $H^2(RP^\infty) \otimes H^0(S^1)$ but this "corresponds to" $RP^2 \times {x_0}$ and $\nabla^(\lambda(\alpha))$ is 0 on this. This means that $\nabla^*(\lambda(\alpha))$ is a generator of $H^1(RP^\infty) \otimes H^1(S^1)$ and so it can be written like that. But I don't know if this is right

[gentle osprey]

Tokidoki ✓

[pearl holly]

the "corresponds to" seems weird

I guess those spaces just correspond to the dual basis of Hom(H_2(RP x S^1)) but this still feels really handwavy

[stable yarrow]

Mine did too

I honestly didnt mind hatcher but I am very CS and algorithms brained

Fulton is a good book for some places that hatcher drops his spaghetti

But it has a lot more universal property stuff

[onyx crow]

Jk lemme check Hatcher

[pearl holly]

I don’t think the symbols matter tho, nambla(lambda) should just be considered as an element

But Max helped btw

[onyx crow]

oh it was too advanced for me anyway. I haven’t gotten to cohomology operations

[pearl holly]

Ye the details require some previous stuff that Hatcher talked about

[fading vale]

The tensor just comes from the kunneth formula w coeffs in a field

[pearl holly]

Ye I know, my question was more of like why the nabla thing was exactly equal to that thing

[fading vale]

Its 0 on RP^2 hence a generator for its first cohomology in Z2 coeffs

[pearl holly]

Oh ye that’s right

Wait I’m on a train hol up

[clear storm]

a quick question for you topology folks, namely, is pushout "preserved "under taking unions? Namely under what condition is taking union of pushout preserve the commutativity of the diagram? For example, I can imagine easily that under taking disjoint union, the pushout is preserved. The pushout of the disjoint union is just the disjoint union of the pushouts.

But I am wondering if this also holds in the case where the union is not necessarily disjoint where you have some compatibility condition on the intersections.

[pearl holly]

Okay I’m back

[pearl holly]

But isn’t that just this?

Or am I missing something?

[fading vale]

I think you mean that its a generator of H^1(RP^infty)

[pearl holly]

But it can be written as “generator” tensor “generator for H1(S1)”

[fading vale]

Uhh not necessarily, it might be tensored with the 0 element in H^1(S^1) for example

but it can be written as generator tensor some element of H^1(S^1)

[pearl holly]

But alpha is a generator tho

[fading vale]

does hatcher say that

Im confused as to what hes claiming

[pearl holly]

Ye, I forgot to say that in the post

[fading vale]

Ok

Uhhhh

Oh im silly

its Z2 coefficients

so yes

[pearl holly]

Okay nice

[fading vale]

Obviously it cant be tensored with the 0 element then it would be 0

[pearl holly]

Okay but thank you so much!

Also when replacement emote for :catlove:

[fading vale]

[coarse gazelle]

Im very beginner of learning topology. can anyone help me with my questions please 🥺 I have some more

[grave maple]

What have you tried?

[coarse gazelle]

Im tryin right now

dont know where to start

[grave maple]

You're asked to show that something has a property. What is the property here?

[coarse gazelle]

I guess I can create topology by using B1

[coarse gazelle]

does anyone have a proof for this? I couldnt find in books

[winter falcon]

Besides some typos if that proof of the propositionn justified?

[winter falcon]

I feel like in proving 1) => 2) you assume the first and prove the second but the proof seems to take some stuff from second condition to prove the implication?

[little hemlock]

I think they switched the proofs of 1) => 2) and 2) => 1)

err, not "i think." they definitely did

[winter falcon]

lol otherwise the proof is good?

[winter falcon]

For me the proof doesn't look solid though, the flow of logic is kinda mixed

[stone cipher]

Since the unit ball $B$ of $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}^n$, then can I say that $B\cap \mathbb{H}^n$ is homeomorphic to $\mathbb{H}^n$, by restricting the previous homeomorphism to the half unit ball?

[gentle osprey]

RaD0N

[stone cipher]

$\mathbb{H}^n$ denotes the half plane $\left{(x_1,...,x_n)\in\mathbb{R}^n: x_n\ge 0\right}$.

[gentle osprey]

RaD0N

[empty grove]

Not necessarily, the homeomorphism you are restricting may not have these exact things in correspondence

[stone cipher]

consider $B\to \mathbb{R}^n$ as $x\mapsto \frac{x}{1-|x|}$ and its inverse is $y\mapsto \frac{y}{1+|y|}$.

[gentle osprey]

RaD0N

[empty grove]

Yes

[stone cipher]

I guess it's not changing the direction.. so it is an homeomorphism, indeed.

[empty grove]

Yeah

[stone cipher]

🙂

So, in a $n$-manifold with boundary, $M$, a point $x\in \partial M$ admits an open neighbourhood that is homeomorphic to $\mathbb{H}^n$ and maps $x$ to $\partial \mathbb{H}^n$?

[gentle osprey]

RaD0N

[empty grove]

Ye

[stone cipher]

thx. I'm studying these things on Introduction to Topological Manifolds of Lee's book. Honestly, I don't like his approach...

[pearl holly]

yo if I have a map f: X --> Y which is uniquely determined by the restriction f|_S and I have a triangle

X
| f
Y---> Z
that I want to show commutes, why is it sufficient to prove that the triangle

S
| f_S
Y ---> Z
commutes instead?

the "diagonal map" in the second triangle is restricted to S

[empty grove]

You need maps from X to Z to be uniquely determined by their restrictions to S

Not maps to Y

[pearl holly]

oh wait

but why is that?

oh ye right

lmao

[empty grove]

Finally a tinky question I could understand

[pearl holly]

I might have a similar question hol up

ye okay never mind, I just mixed up my maps lmao

but thank you so much!

[empty grove]

✓

[finite heath]

^ this guy is built different

[plain raven]

These days I spend a lot of time messing around with simplicial formulas and not really getting anywhere

very demoralizing

Often I feel that I don't have enough intuition for a construction in homological algebra. There are two ways I can deal with this: Go learn more homotopy theory so I have more geometric intuition for the construction, or do a bunch of computations to see if I can rework it into something more manageable.

I am kind of obsessed with the Alexander-Whitney map (and I guess its inverse the Eilenberg-Zilber map) because it appears everywhere in homological algebra and homotopy theory, and every once in a while I come back to it to see if I can get better intuition for it.

[dusk heron]

If we have a mixed-degree cohomology class $u\in H^*(X)$ and a homology class $v\in H_k(X)$ of degree $k$, does the pairing $\langle u,v\rangle$ mean by convention that we throw away all other degree parts (than $k$) in $u$ before we do the pairing?

[gentle osprey]

gustavn64

[honest terrace]

Guess a metric

[long hornet]

Can you use metrization theorems?

For example this space is compact Hausdorff and locally metrizable (I think the latter is easy to see, but I don't know how off the top of my head)

[empty grove]

I don't see local metrizability, the basic open neighbourhoods can't a priori be metrizable because they are homeomorphic to the whole space

Urysohn doesn't seem to apply because does not seem second countable

Nagata Smirnov I don't think applies because not separable? not sure

😵‍💫

[empty grove]

do you have one in mind

[long hornet]

Ohh, I thought it was the opposite

Maybe just use the sup thing

[honest terrace]

Wait I'm dumb, no I don't have any, I brainlagged

It isn't metrizable

[empty grove]

😵‍💫

ye doesn't seem metrizable lol

[honest terrace]

As a more than countable product of spaces of metric spaces with more than 2 points

it isn't first countable

[empty grove]

oh right

neat

It is only given to be uncountable

[honest terrace]

1. X is only assumed to be uncountable

2. How does that not contradict what I said above ?

[empty grove]

ZFC is inconsistent

[long hornet]

Oh, right

[empty grove]

always bet on triangle inequality

[honest terrace]

I believe your answer is correct and moldi is right

everyone is right

[long hornet]

That doesn't give the product topology

[honest terrace]

ZFC is inconsistent

It's not about countable or not

it's about a property of neighborhoods at a given point of a metric space

[empty grove]

metrizable implies first countable

😌

[honest terrace]

Any point in a metric space has a countable nghbrhood basis

(check wiki for exact def if you don't know it yet)

[long hornet]

Shika's argument shows that X^Y is not metrizable if X has more than one point and Y is uncountable. If Y is countable and X is metrizable, then this is metrizable iirc.

[honest terrace]

Yes it is

define d((x_i), (y_i)) = sum_i min(1, d(x_i, y_i))/2^n

And that works

[long hornet]

Are direct (or maybe I should say inverse?) limits of metrizable spaces metrizable?
I'm always confused by the limit/colimit thing. Are products limits or colimits?

[honest terrace]

products are limits, sums (or coproducts or whatever) are colimits

Are direct (or maybe I should say inverse?) limits of metrizable spaces metrizable?
(And no idea about that)

[long hornet]

So you get maps from the limit to its constituents

Okay thanks

So limis are quotients of the product, that's true in modules

I think essentially because of the first isomorphism theorem, which doesn't hold in Top

[honest terrace]

It doesn't even make sense, does it ? 🤔

[long hornet]

I mean if f : X --> Y is surjective, then the quotient X / ~ isn't necessarily homeomorphic to Y. We only get a continuous bijection.
Here ~ is the relation that "makes f injective"

I didn't mean the quotients themselves, but rather some construction of limits: We take the product of all M_i and then quotient by a certain relation and that will be the limit.
Or maybe I got it wrong and we just take a subobject of the product?

Great, this almost settles it then: If this holds in Top, then since metrizability is preserved by countable products and subobjects, countable limits exist.

[lean marten]

Don’t you need equalisers or something?

[empty grove]

Equalizers of Top are subobjects

Well they are subobjects in general

Not just in Top

[lean marten]

Ah ok

[empty grove]

I guess in this case we also want the equaliser maps to be embeddings, rather than just injections

So I should say in Top lol

Because metrizable is not preserved for arbitrary injections

Even though those will be monic so subobjects in the category sense

[pearl holly]

it's not generally true that H^n(X; G) = H^n(X^n; G), right?

X^n being the n-💀 of X

so in what cases does that hold?

[slate hound]

I'm studying topology with Lee's Introduction to Topological Manifolds and I'm currently doing this exercise.

I get that E is a covering space of E/~ and that if this is a universal cover it's not hard to conclude that the fundamental group of E/~ is Z/Z2, but I'm having troble proving that E is simply connected. Anyone have any pointers for where to go from here if I'm even going about this the right way?

[empty grove]

You at least need to go to X^n+1 because attaching disks can make your paths contractible. I think Hatcher had a theorem that the nth homology group only depends on the n-skull and maybe then use univ coefficient or just try to do it directly. Maybe there were some extra conditions I don't rember 💀

[empty grove]

This should be R⁶ - (R³ x {0}) via some linear homeomorphism. Then projection onto the last 3 coordinates would be a deformation retraction onto R³ - 0

The linear homeomorphism should come from proving that the diagonal of R³ x R³ is a 3 dimensional vector subspace

[slate hound]

I got as far as it being R6 minus some copy of R3, but missed the part about the deformation retraction, that fills in the gap for me

thanks!

[uncut surge]

Something like (x,y) maps to (x, x - y) should identify it with R3 times R3 minus 0 i think?

but i'm gaming right now this is just a quick thought maybe it's wrong

[pearl holly]

oh sorry I didn't see that you responded

ye I see

so my question comes from this: Hatcher has shown that a certain map $f: H^n(S^n) \to H^n(S^n)$ is the identity and now he says that the fundamental class of $H^n(K_n)$ where $K_n$ is a $K(Z_2, n)$ space with $n$-skeleton $S^n$ and $n-1$ skeleton a point is fixed by this map too. I don't see how tho

[gentle osprey]

Kanga Gang Swagger Tokidoki ✓

[pearl holly]

so maybe H^n(K_n) is a subset of H^n(S^n) since the n-skeleton of K_n is S^n or something idk

he says "since S^n is the n-skeleton of K_n, this implies that f is the identity on the fundamental class"

and the fundamental class is an element of H^n(K_n) such that viewing it as a map K_n --> K_n is homotopic to the identity

[gaunt linden]

In general R^n (for n ge 2) with countably many points removed is path connected: Between any two points you can find continuum many disjoint paths in R^n.

[gaunt linden]

It's probably not considered deep enough to mention.

[plain raven]

blessings

I'm glad i'm not the only one lmfao.

Today I decided I don't know as much about fiber bundle theory as I should

especially in terms of how the classifying space BG of a group has historically been understood

and so I spent some time reading Husemoller's Fibre Bundles.

Good book. Elementary enough that I can make steady progress without getting too bogged down in technicalities

[long hornet]

If A and B are disjoint, closed and discrete subsets of a manifold M whose dimension is at least 2, and have the same cardinality. Does there exist a self-homeomorphism that sends A to B? I know this is true when A and B are finite (i.e., the action is n-transitive for all n).

Hey, my name is blue!

[rancid umbra]

hi blue

[honest terrace]

Is this false when the dim is 1 ? 🤔

[long hornet]

Yes. For example you can't send 3-tuples to others, or something similar, in the real line

[honest terrace]

Oh huh yeah

for order reasons

yea ok I see

[long hornet]

I think one could inductively build such a map

[honest terrace]

Do you have a reference for how it's done in the case where A and B are finite ?

[long hornet]

It was an exercise in Lee's top manifolds book. The rough idea was to prove it for disks first.

[honest terrace]

Alright ty, I'll go check the book

[gritty widget]

does anyone have an example of two sequences of spaces

...-> X_2 -> X_1 and ...-> Y_2 -> Y_1

such that X_n and Y_n are homotopy equivalent

but the limits of these sequences are not

[stable yarrow]

good example

[gaunt linden]

It sounds like you want the self-homeomorphism to extend a previously given bijection A to B? In that case, here is a counterexample: Take a "long ribbon" -- that is, the product of an open long ray with the open unit interval. The long ribbon is glued together from omega_1 copies of the square [0,1)×(0,1) except that the zeroth square doesn't include a left edge. We can coordinatize the long ribbon as omega_1×[0,1)×(0,1), but remember that even though each point has three coordinates, the whole ribbon is a two-dimensional manifold. Remove from the ribbon all points of the form (alpha,t,½) for a nonzero alpha, and also all points of the form (alpha,0,¼) and (alpha,0,¾). That still leaves a path-connected manifold. Now let A = omega_1×{¼,¾}×{¼} and B = omega_1×{½}×{¼,¾}. These sets are definitely discrete, and we have already removed the limit points that would prevent them from being closed. Is there a self-homeomorphism that maps (alpha,x,¼) to (alpha,½,x) for all alpha in omega_1 and x in {¼,¾}? No, there cannot be: for each of the uncountably many alphas, the image of {alpha}×[¼,¾]×{¼} is a path from (alpha,½,¼) to (alpha,½,¾), and all of those disjoint paths (which have disjoint open neighborhoods) must pass through the zeroth square because that's the only place you find any points with last coordinate ½.

[long hornet]

[long hornet]

I don't quite understand how the coordinates work, but I only have a vague idea about omega_1. I know that it's the minimal uncountable well-ordered set, or something.

Is (alpha, x, y) the point (x, y) in the alphath square?

[gaunt linden]

Yes.

[long hornet]

And does omega_1 have maximal element?

[gaunt linden]

No.

You might need to read up on the long line before the example really makes sense.

[long hornet]

I read about it before, but not "rigorously"

Okay, so we delete three lines, two of them are "full"

My bad, we don't do that really

[gaunt linden]

(It feels plausible that the property might hold for countable A and B, so a counterexample would need to involve a manifold that can contain an uncountable discrete set in the first place -- those do require somewhat weird constructions such as the long line).

The first deletion just slices the ribbon in half for most of its length. I could also have glued two ribbons together at their finite ends instead, but the coordinates wouldn't work out as nicely.

The second deletion removes all points that are at risk of being a limit point of the A or B I define later. It's basically just a way to get around your specification that A and B must be closed.

[long hornet]

Yeah, I actually didn't add countability because I was assuming manifolds are second countable.

[gaunt linden]

Ah, then my idea doesn't work at all.

[long hornet]

So your map between A and B requires crossing the middle line.

[gaunt linden]

It moves half of the A points across the middle line, but the other half stays almost what they are. So simply mirroring the entire manifold won't work.

[long hornet]

My confusion now is that I can't even imagine a path from the top half to the bottom. I thought this required going along the (missing by definition) leftmost edge.

[gaunt linden]

The slice-in-half cut omits the entire leftmost square.

[long hornet]

Ohhh

I have drawn a typical square. It makes more sense now.

[gaunt linden]

Yeah, I was trying to convey geometric intuition through a text-only medium. Sorry.

[long hornet]

It's a really nice example, though. Thanks!

[gaunt linden]

By the way, the requirement that A and B are disjoint doesn't really add anything to the problem. If they are not disjoint from the beginning, we can just precompose everything with a homeomorphism that moves each A point to an allowed location within its local neighborhood. Since A is discrete anyway, this can surely be done for all the A points at once.

[long hornet]

Makes sense.

Discreteness is relevant, though, right?

Well, I guess so

For example if we take A = Q x Q or something

[gaunt linden]

Definitely relevant, because homeomorphisms preserve limits, so the required connection between As and Bs cannot be arbitrary othewise.

One could require the fixed bijection between A and B to be a homeomophism between the subspace topologies -- I cannot immediatly see whether that would make the problem harder or not.

[long hornet]

At first I thought of two convergent sequences, and we add suitable conditions.

[long hornet]

I think the sequence case is a special case of this

[gaunt linden]

Yes, I agree.

[long hornet]

It seems like a long shot, but is the function from Tietze's lemma controllable?

Oh, actually it works for real-valued functions to begin with

[gaunt linden]

Sorry, I don't know that.

[long hornet]

I suppose I can work in the disk

With the added condition that the boundary should be fixed by the homeomorphism

It seems possible to stick together infinitely many piecewise linear maps

[regal zealot]

Hello all. I was hoping to get some help with this problem. Let T denote the group of maps of the plane $\mathbb{R}^{2}$ generated by the maps f, s.t $f((x,y))=(x,-y)$. I want to show that T acts on $\mathbb{R}^{2}$ as a group of homeomorphisms. \par

I know that I have to show that for all $t_1,t_2 \in T$ that $t_1 t_2 (x)=t_1(t_2(x))$, e(x)=x where e is the identity element of T, and that f is a homeomorphism. This feels pretty simple, but the computations are weirding me out a little bit. For the first part, would it be $f(x_1,y_1)f(x_2,y_3)((x_3,y_3))=f(x_1,y_1)(f(x_2,y_2)(x_3,y_3))$? Or am I misunderstanding something.

[gentle osprey]

CornOnTheCob

[empty grove]

Isn't f just a single map

You check that f is a homeomorphism

So it is an element of the group of homeomorphisms from R² to R²

And you are just taking the subgroup generated by it

Anyway if it weren't a homeomorphism then you couldn't talk about a group generated by it

[regal zealot]

Okay. Then I am way over complicating the matter. Essentially all I need to do is show that the map f is a homeomorphism.

[empty grove]

Yeah

I mean they say group generated by f

That immediately implies everything is a homeomorphism

Assuming group under composition

So it's a weird question to ask

[regal zealot]

Gotcha. Things are starting to make sense. I will agree it is an odd question.

[gritty widget]

you can assume A is not compact and try to get a continuous function with no maximum

sequential compactness may be handy

[finite heath]

@gritty widget question sir!

https://cdn.discordapp.com/attachments/885780514626220043/920123824580726874/unknown.png

So would (-1, 0] in X* be mapped to like {0}, (-1, 0), or (-1, 0] in X?

as in

p-1((-1, 0])

i know quotient map is surjective

so are all the sets in X* open and closed?

@pearl holly

please Todd

help a brotha out!

💯

[finite heath]

no!

[long hornet]

Call the equivalence classes V_n (this contains n). Note that the union of V_n for all n >= n0 is open, and the union for all n <= n0 is closed, and these unions are saturated, so their images are open/closed (resp.), as well.

[wraith horizon]

Are rational numbers in R open, closed or neither. They'd be neither right?

not sure if this is supposed to go in #multivariable-calculus 🤔 but this channel is labeled topology

[ivory dragon]

That is technically a topological question so it's fine I guess

Anyway, yes, the set of rational numbers is neither open not closed in R

This follows from density

[lament steppe]

New to Topology and this question is stumping me a bit.
My first thought is that if 2 sets are open and disjoint, how could the union of the 2 not have a gap in them?

For instance, if a < b < c < d, and the set we want to be equal to the union of disjoint sets is (a, d) then the best I can think of it doing (a,b) U (c,d).
Even if b = c, that would mean that b would not be included in the union: (a,b) U (b, d)

Am I suppose to be taking the complement of these disjoint sets?

[empty grove]

(a,d) itself is already written as a union of disjoint open intervals

[lament steppe]

Huh, Im not sure I understand what you mean. Can you give a concrete example?

My first thought is that maybe (a,d) = ( a, b) U [b,d). This would avoid the "missing gap at b", but the second interval isnt open (since it is closed at b...)
Sorry, if my question seems trivial, but I have a flimsy grasp of this stuff at the moment 🙃

[limpid leaf]

{(a,d)} is the disjoint union

vacuously

i think proving this for any open in R is kinda annoying/hard? but maybe i’m misremembering. but for just intervals this is all you need

[lament steppe]

So you are saying if that you have 1 open set, it is not "overlapping" another set and therefore is disjoint by definition?

[limpid leaf]

well no surely (a,d) intersects uncountably many other open sets, but thats sort of irrelevant

all that matters is that you can write (a,d) as a disjoint union

and the way you do that is literally just “(a,d)”

like a set containing a single element is pairwise disjoint

vacuously

so {(a,d)} is pairwise disjoint vacuously

and the union of {(a,d)} is (a,d)

[lament steppe]

right, cause if you are only considering the 1 set in your collection, then taking a union of just that 1 element gives the open set we want, and since its not "overlapping" another set it is by definition disjoint.. (I keep putting overalpping in qutoes because im not sure if there is a better word to describe this.)

does that sound correct?

[limpid leaf]

yes that’s correct

the right word is intersecting but it doesnt matter

anyways you haven’t actually finished the question

this is sort of the easy part

cause these are the basic opens in usual topology

but you want to do this for arbitrary open sets in R

[lament steppe]

gotcha ok, ill spend some more time with it. thanks for some clarification

[limpid leaf]

np

[long hornet]

I think it was, because it seems equivalent to saying intervals are connected...? It's proved in Apostol's analysis book. @lament steppe

[limpid leaf]

yeah but if you get this as an exercise you surely dont know what connected means i think

maybe you do and idk what a normal topology course does

[long hornet]

Yeah, I just meant it's roughly the same difficulty

[limpid leaf]

oh yes

is it obvious that they are equivalent

it feels like it but i cant think

hm

yea I think it is

[long hornet]

Maybe if we know that R^1 has the open intervals as a basis, and we know that only countably are needed, the proof would be easier.

[limpid leaf]

yeah that is the assumption here i think

[empty grove]

That is the proof yeah, you start with second countability then combine intervals that intersect

[limpid leaf]

oh that’s nice

topolojee

[long hornet]

And that's the first exercise

[limpid leaf]

[empty grove]

This was a homework problem for me in real analysis sem 1 lmao

[limpid leaf]

you were talking about countability axioms in analysis?

or you did this the other way

lol

[empty grove]

Without using those words

[limpid leaf]

oh yeah i guess its just like use basis given by Q and combine lol

[empty grove]

We had the hint that we have to prove it without the word disjoint first

Then make it disjoint

Yes exactly

[long hornet]

And this is just false in R^2

[empty grove]

Yeah no disjointness there

[gritty widget]

By “induction” you can get a version of Mayer vitoeries for finite unions

But there is no hope for countable unions is there

Because manifolds are a thing

?

[gritty widget]

does the smash product work well with cohomology

if i know the cohomology of X

do i know what the cohomology of X smash sphere is

[tight agate]

yes

it just shifts the degree

[X smash S^1, E_n] = [X, maps(S^1, E_n)] = [X, E_{n-1} ]

where E_n represents the n^th cohomology

[gritty widget]

so theni guess

[X smash S^q, E_n]=[X,E_{n-1}]

[tight agate]

n-q, yes

[gritty widget]

opps thougth i wrote tht

that

[tight agate]

some people take this to be one of the axioms of a cohomology theory

[gritty widget]

and we also have

[X wedge Y, E_n]=[X,E_n] times [Y,E_n]?

[tight agate]

no

oh

wedge

yes

[gritty widget]

okay thanks

so

if we we smash with a wedge of spheres

its just

the product of a bunch of shifts

[tight agate]

yes

[gritty widget]

thank you

[tight agate]

it should be

as smash is like the tensor product

and wedge is like direct sum

[gritty widget]

oooh

[tight agate]

and tensor products commute with direct sums

[gritty widget]

wait

[gritty widget]

so here i should men direct sum

but it doesn;t matter because because

oh no

[tight agate]

it's product

not direct sum

and it does matter

in any category Maps(X coprod Y, Z) = Maps(X,Z) prod Maps(Y,Z)

that's the definition of coproduct

[gritty widget]

and the wedge is the coproduct?

for based top spaces

[tight agate]

yes

[long hornet]

How obvious is that? I mean should double, triple, etc. intersections appear individually? Because I think simply taking the intersection of a the subspaces loses quite some information.

[pearl holly]

I know that the isomorphism $H^n(X; G) \cong H^n(X^{n+1}; G)$ is given by the induced homomorphism of the inclusion. But if I view this as $[X, K(G, n)] \cong [X^{n+1}, K(G, n)]$, will be the isomorphism simply be restricting maps?

[gentle osprey]

Kanga Gang Swagger Tokidoki ✓

[pearl holly]

ye it will lol

[winter falcon]

I have done the first exercise which is proving that $\mathcal{B}_x is a base of neighborhoods of x, now Im stuck on question 2, how do you show that the function $x_0 = 0 is a limit point of set A

[light rivet]

So I have a sort of intuitive question. We know the fundamental group of R^3 is trivial. Intuitively it seems that the fundamental group of R^3 minus a line in R^3 is Z, based on how many times you loop around this missing line. Then if I erase another line parallel to the first, what should the fundamental group be. I was thinking it was either the free group on 2 generators or ZxZ, but I cant decide between the two

[long hornet]

@light rivet

[light rivet]

What's a homotopy type?

[long hornet]

Using Moldilocks' idea, we get R^2 minus two points

[light rivet]

well R^2 minus two pts has fundamental group F_2

[long hornet]

It's easier to explain what I meant: I wanted an easier description of R^3 - line, up to homotopy.
Formally, it's just an equivalence class of the homotopy relation.

[long hornet]

Exactly

[light rivet]

Hmm, I don't think this situation changes even if the lines intersect right?

It seems like its still F_2, but that seems weird for some reason

[cedar pebble]

it's F_2

R^3 minus two lines deform retracts onto R^2 minus two points

if the lines intersect this changes somewhat

[light rivet]

So the -2 parallel lines I can kinda see just intuitively why the situtation is similar to R^2 minus two points. Does intersecting lines have an analogous picture?

[cedar pebble]

in this case it's enough to think about e.g. R^3 minus two axes, and the deform retract this onto its intersection with the unit sphere

in this case you'll get a sphere minus four points, which has fundamental group F_3

[light rivet]

Why does this idea not work for two parallel lines? If we still consider a large enough sphere which intersects both lines, it would still be a sphere minus 4 points

[cedar pebble]

I mean you just don't get a deform retraction onto a unit sphere here

the property you need is that these lines going through the origin only vary in radius, and have a constant angle

so then you are retracting by collapsing the radius onto the unit sphere

but if you have two non-intersecting lines in R^3 you can't arrange them like this

[coarse gazelle]

how can I show the continuity on this function? topological

[cedar pebble]

wrong channel, try #real-complex-analysis

it's also very clearly not continuous

[coarse gazelle]

can you please tell me why?

[light rivet]

The rationals are dense, so you're jumping points in the picture

[cedar pebble]

it's nowhere continuous, since the irrationals are dense in the reals

[coarse gazelle]

thank you!

[limpid leaf]

,tex I cannot figure which of these are homeomorphic to whcih other ones. I think they are all not homeomorphic, but not really sure
\begin{itemize}
\item $X = (0,1) \times (0,1)$,
\item $X' = S^2$,
\item $X'' = \bR$ with upper limit topology
\end{itemize}

[gentle osprey]

jesse

[limpid leaf]

Cause I think R with upper limit is totally disconnected(?), so can't be homeo to X, X'. And S^2 is compact, so can't be homeo to X, so none are homeomorphic?

is totally disconnected even an invariant?

yeah it msut be

[gaunt linden]

Yes, "totally disconnected" can be defined only in terms of the topology, so it's homeomorphism invariant.

[limpid leaf]

okay nice so none are homeo to each other

very cool.

[gritty widget]

How to model categories relate to infinity categories

If by infinity category i mean quasi-category

so a simplicial set with inner horn fillers

i thought model categories only helped you create a homotopy category

and the homotopy category of a quasi category forgots alot of information

[limpid leaf]

I have this thm that if i have a continuous function from a metric space (X,d) to (Y,d'), then if X is path conn then so is Y.

can I just take like the topologist sin curve as a subspace of R^2, and then any continuous f: R^2 -> R^2 can be restricted to the TSC, which isn't path connected, so is false?

[tight agate]

if you have a simplicial model category, then the homotopy coherent nerve gives you an infinity category

in some sense the infinity category sorta sits in between the model cat and its homotopy category

the infinity category remembers some of the information that the homotopy category forgets

[gritty widget]

ooh

i thought it infinity categoires had more information than a model category

can every infinity category be realised as the homotopy coherent nerve of a simplical category

[tight agate]

no

the ones that can be realized that way are called presentable infinity categories

I think

https://mathoverflow.net/questions/78400/do-we-still-need-model-categories

there's some discussion about the relation between model cats and infinity cats in that post

[gritty widget]

thanks

i saw this but was confused about where to start

[gritty widget]

this helps alot

[tight agate]

the infinity category you get out of the model category remembers all the "homotopical information" that you can extract from the model category

the nlab page on the homotopy coherent nerve should also have something on it

[gritty widget]

thanks alot

i appricate it

[tight agate]

no worries

[hollow harbor]

[tight agate]

an easy to digest example is the derived infinity category

if you're familiar with classical derived categories

it's also interesting to see how a dg-enhanced derived cat can do some of the stuff that the derived infinity cat does

I'm not sure if this is right, but you can get away with just using dg-enhancements to do everything if you only care about derived cats

of course, the general infinity cat takes care of derived cats and stuff like spectra simultaneously

which is nice

[empty grove]

You need it to be surjective

[limpid leaf]

sorry this is a "prove or disprove" question, so I figure the theorem is false

i guess my counter ex doesn't work?

[empty grove]

oh

I'm not sure what you are doing in that example

[limpid leaf]

[empty grove]

But you can always take a constant map

[gaunt linden]

That goes in the wrong direction. Your continuous fct needs to be defined on all of X.

[empty grove]

To any non empty space

[limpid leaf]

ah yeah you're right

[gaunt linden]

It also needs to be surjective for the conclusion to hold.

[limpid leaf]

what do you mean Moldi

[empty grove]

Take Y not path connected

[limpid leaf]

Like just let Y be literally any non-path connected space, and thne theres a constant map?

[empty grove]

Take X path connected

Constant map to anything

Is a counterexample

[limpid leaf]

oh yeah that works

lol

okay much nicer

[empty grove]

Constant maps are always continuous

[limpid leaf]

is the inclusion into a subspace always continuous?

sorry, can I always write down a continuous inclusion

this is unrelated

[empty grove]

By definition of the subspace topology yes

[limpid leaf]

right yeah

okay so I could modify my sine curve example to just be an inclusion

since TSC is a subspace of R^2?

[empty grove]

That also goes in the wrong direction

[limpid leaf]

lmfao fuck

[empty grove]

You want domain path con

[limpid leaf]

wait what

R^2 -> TSC?

domain is path con

[empty grove]

What is the map

[limpid leaf]

inclusion

[empty grove]

The inclusion goes the other way

[limpid leaf]

LMFAO

fuck

soryr its eraly

[empty grove]

😵‍💫

[limpid leaf]

can i not fix this example somehow

[empty grove]

Why do you wanna work with such a complicated example

[limpid leaf]

idk

[empty grove]

[limpid leaf]

always nice to have

examples

i guess

[empty grove]

The most general thing you can say is

If Y is indeed path connected

You can get a counterexample

By just adding path components to Y

Like taking disjoint union with other space or whatever

Like the whole point is that f could just be mapping into a single component

While Y could have many

All we know is that im f is path con

[limpid leaf]

ok

[empty grove]

Lmao if you really wanna fix your stupid example

Take f: (0, infty) → TSC

x ↦ sin(1/x)

Codomain being closure of TSC in R² ofc

[limpid leaf]

heck yeah

😎

[empty grove]

[gentle osprey]

Cleetus-Wazoo

[empty grove]

f^2 is identity, so you know T precisely as {1, f}

It is just folding R^2 along the x axis

quotient is homeomorphic to a half plane

which is simply connected

[torpid fox]

Okay. So if I am following correctly since the quotient is homeomorphic to a half plane and T is {1,f}, then the fundamental group of the quotient would be T correct? Or am I missing something?

[gaunt linden]

Once you know what the quotient is (homeomorphic to), T is not relevant anymore.

[wanton marsh]

R² -> R²/T is not a covering map

[torpid fox]

So T is not relevant anymore since the quotient is homeomorphic to the half plane and the half plane is simply connected.

[gaunt linden]

Yes? What is the fundamental group of a simply connected space?

[torpid fox]

The fundamental group of a simply connected space is the trivial one element group for some point, in this case, it would be for a point in the in the half plane.

[gaunt linden]

I don't know about "for some point". In general the definition of the fundamental group depends on choosing a root point, but when the space is path-connected (which this one certainly is), that choice doesn't actually matter (up to isomorphism).

[torpid fox]

I think am starting to understand. I'll have to mull over it a little bit to let it sink in. Thank you for your help. Topology isn't my strong point, so my apologies if I came across as a bit daft.

[gritty widget]

Quick question: If X is a topological space, and U is a dense and proper subset of X, then U has to be open, right? Since the closure of U is X. If U was closed, then it would coincide with its closure so U = X

[rancid umbra]

Q as a subset of R

[gritty widget]

Ah so it can be neither closed nor open

[rancid umbra]

yup. same with R/{0} as a subset of R. R is closed and open in R

that’s open

[gritty widget]

Thanks

[rancid umbra]

i think tho, the only closed dense subset of X is itself

[gaunt linden]

Yeah -- its complement would be open and not intersect the ostensibly dense subset. That's only allowed if it's empty.

[rancid umbra]

alternatively, if the set is closed, then it’s equal to its closure, which is X since it’s dense

[plain raven]

i wanted to read a proof that if EG -> BG is a principal G-bundle, then this bundle is the universal principal G-bundle iff EG is contractible

so I found a reference for this, there's a version of this theorem due to Dold and I got his paper and tried to read it last night

it's all like

barycentric subdivision and coordinatewise homotopies

this took me so many rereads to understand

need to find a good more modern treatment of this EG and BG stuff from a homotopical perspective

[golden gust]

what on god's green earth is this

does it mean something different from \geq

[gaunt linden]

No, it's just a typographical variant that some people like better.

[Or at least, anyone who wants to use it for a different relation had better point very loudly to their definition earlier in each document they're writing].

$\geq$ $\geqq$ $\geqslant$ btw

[gentle osprey]

Troposphere

[shy moss]

I think this is in the Mitchell’s notes

[plain raven]

alight i'll take a look at them

[shy moss]

https://sites.math.washington.edu/~mitchell/Notes/prin.pdf

[plain raven]

👍

[limber ravine]

I say it isn't because $\emptyset \notin \Omega$

[gentle osprey]

mns

[limber ravine]

What do you think?

Is $\emptyset \in \Omega$?

[gentle osprey]

mns

[hollow harbor]

Thats the empty union

I would count that

[limber ravine]

what do you mean with empty union?

[hollow harbor]

Union over 0 things

Its dumb but

Thats what they had better mean

If they want this to be a topology

[limber ravine]

well but omega is the set of unions of all intervals, 0 things isn't an interval (?)

[hollow harbor]

0 intervals is a collection of intervals that you could union together

Its stupid

They should have just said the empty set was included

[fading vale]

you can just take the interval (a, a)

[hollow harbor]

But

[limber ravine]

can that be proved? like in a stupid way

[hollow harbor]

Fuck moth is literally a genius

[limber ravine]

oh I see

(a,a)

I was thinking about (b,a) with b > a

[fading vale]

but like ryc said its kinda common to just

not specify the empty set

but assume its included

[hollow harbor]

It's just a notational thing, like people would say the sum of anything over the empty set is 0.

[fading vale]

sometimes the whole space too

it depends on context i guess

[limber ravine]

gotcha

thanks guys

[hollow harbor]

Anyway moths reasoning is better

[trail tiger]

What are some interesting things we can say about a space all of whose homology groups are trivial? "Basically Euclidean" I'm guessing is one of them by Poincare's lemma.

[gritty widget]

What is “basically Euclidean”supposed to mean?

[trail tiger]

What is “basically Euclidean”supposed to mean?

no idea, hence the quote and question. it obviously goes the other way, i was wondering how strong the opposite direction was

The keyword to Google for this is "acyclic space".

thanks!

[violet sonnet]

hiii

can someone show me why this is true for a graph like the following:

i just don't see it

i can see homotopy equivalence but not homeomorphism

[pearl holly]

I literally don't understand anything here

Hatcher is trying to show that Sq^1 is the Z/2 Bockstein and I just don't know what's going on in the proof. K_(n+1) is K(Z_2, n+1) with (n+1) skeleton S^(n+1) and n-skeleton a point. Why is the Bockstein on the generator of H^1(RP^2) the generator squared? Why is the n-fold suspension of RP^2 the (n+2)-skeleton of K_(n+1)?

ah yes, the cursive shit

[plain raven]

toki you are really churning through this book

have you been doing like nothing but Hatcher cohomology for weeks

[pearl holly]

lmao I skipped like half the book

I just read what's fun

I jump around

[plain raven]

ok

Question: Let $G$ be a topological group, and let $p : E\to B$ be a principal $G$-bundle. Whatever niceness assumptions you want, i.e. $B$ is paracompact, $p$ is locally trivial. Forget $p, B$ (one can recover them from the quotient of the group action anyway)

[gentle osprey]

Kanga Gang Made Man

[plain raven]

Can $X$ be expressed as a retract (in the category of right $G$-spaces) of the total space of a trivial $G$-bundle $G\times A\to A$

[gentle osprey]

Kanga Gang Made Man

[trail tiger]

Stop reading Hatcher

start reading tom Dieck

[plain raven]

I don't think i'm asking this well, and I might have the wrong question.

Let $\mathcal{C}$ be the category whose objects are fiber bundles with structure group $G$, where $G$ is some topological group. Can every object in $\mathcal{C}$ be expressed as a retract of a trivial bundle?

[gentle osprey]

Kanga Gang Made Man

[empty grove]

That is false The graph you gave is a counterexample because it has a point where 3 lines meet, which can't happen in wedge of circles in which always an even number of lines meet at any point (and there can't be a homeomorphism because the number of lines meeting at a point can be expressed as the point having a connected neighborhood such that when you remove the point, the neighborhood has that number of connected components)

[plain raven]

ok i'll think about it

[empty grove]

Thank you clerk for thinking about my answer

[gaunt linden]

I think that should be simply connected neighborhood for the number of lines.

[empty grove]

True, alternatively any neighbourhood of that point contains a connected neighborhood with that property

[clear storm]

If you have a relative CW-complex say $(X,A)$ with $A\subset X$, I believe you can have cell attachments that are "redundant" in a sense that the attached cells are already in $A$. Am I correct? I don't believe the lack this redundancy to be required for something to be a

[gentle osprey]

Sei

[clear storm]

cw-complex

[shy moss]

yes, there are cells which are already in A

[plain raven]

i thought about it and i get what you're saying

[long hornet]

Is that the rank of the local homology at x?

[empty grove]

might be but I don't have good intuition for local homology 😵‍💫

[limber ravine]

does it proves (a)?

(the new set that I want to prove is Omega')

[empty grove]

You can't take union from 1 to infinity

It can be a larger union

[limber ravine]

it needs to be finite?

the union I mean

[empty grove]

no it can also be an uncountable union

[limber ravine]

no I mean

[empty grove]

1 to infinity is just a countable union

argument seems fine

just remove 1 to infinity

and make it an arbitrary indexing set

[limber ravine]

like from 1 to n?

[empty grove]

no

[limber ravine]

$\bigcup A_i$

[empty grove]

That is still a countable union

[gentle osprey]

mns

[empty grove]

yes

[limber ravine]

just this?

[empty grove]

well

you can put some indexing

like $\bigcup_{i\in I} A_i$

[gentle osprey]

Moldilocks ✓

[empty grove]

where I is some set

not necessarily N

[limber ravine]

interesting

So in the beginning could I say: Let $A_i \in \Omega'$ with $i \in I$ for any set $I$. Then ...

[gentle osprey]

mns

[empty grove]

Yes

[limber ravine]

hum ok, thank you very much!

just to confirm because the book doesn't have solutions. this is indeed a topological structure, right?

[empty grove]

Yes

oh no

ok they add in empty set separately

yes

[limber ravine]

nice!

[supple sable]

My next topology course recommends two books, i would like to pick one of them. They are

Hatcher, Algebraic Topology

and

Lee, Topological manifolds

The course will cover homotopies homotopy equivalences and the fundamental groups. The van kampen theorem and homology.

This is in no particular order, just what i got from the course descriptions

[empty grove]

Hatcher is good assuming that you have no idea what a category is

[fading vale]

Lee is not an AT book to my understanding so i imagine it wont cover most of the topics you listed

or if it does theyll be pretty spread out

[supple sable]

So Hatcher it is

[limber ravine]

I must show why $[a,b]$ is closed

[gentle osprey]

mns

[limber ravine]

in R tho

it is if $R \setminus [a,b] \in \Omega$

[gentle osprey]

mns

[limber ravine]

Omega is the union of all intervals (a,b) with a,b in R

Notice $R \setminus [a,b] = (-\infty,a) \cup (b, \infty)$. Since both intervals are in Omega, then we deduce by the properties of the topologic structure that $R \setminus [a,b] \in \Omega$. Thus $[a,b]$ is closed

[gentle osprey]

mns

[limber ravine]

Just want to make sure about the infty being there

[empty grove]

You can't directly say that (b, ∞) is in Omega

∞ is not in ℝ

[limber ravine]

could I make it as a union of infinite open intervals?

[empty grove]

Yes

[limber ravine]

for instance (b,bi) with bi < bi+1

then their union would be (b, infty) ?

[empty grove]

This doesn't guarantee what you need

[limber ravine]

b < bi < bi+1

[empty grove]

Because you could have bi = 1-1/n

b = 0

Then the union only goes up to 1

Just take all the (b, b') with b' > b

[limber ravine]

but then bi wouldn't be lesser then bi+1 ?

gotcha

[empty grove]

It would be

1-1/n is strictly increasing

[limber ravine]

as n increases??

[empty grove]

Yes

[limber ravine]

wtf

oh yeah

lol

[empty grove]

Because 1/n is strictly decreasing

lol

[limber ravine]

I see xd

[empty grove]

So yeah this makes it an uncountable union

Which is allowed by the axioms of a topology

[limber ravine]

so we have $(-\infty,a) \cup (b,\infty) = \bigcup(b',b) \cup \bigcup (a,a')$ with $b' > b$ and $a < a'$. By axioms of a topology we have \bigcup(b',b) \cup \bigcup (a,a') \in \Omega$. Thus $R \setminus [a,b] \in \Omega$ and $[a,b]$ is closed

[gentle osprey]

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

[empty grove]

(b, b') and (a', a)

And a' < a

[limber ravine]

I was thinking that b' and a' was the ones that increase

In fact it should be b' < b and a < a'

[empty grove]

b' should be to the right of b

[limber ravine]

why?

[empty grove]

We are taking the interval to the right of b

To infinity

If you take b' < b

Then you are including b in this for example

And b shouldn't be in this union

Because that is in [a,b]

[limber ravine]

so b' and a' are the fixed numbers?

[empty grove]

a and b are fixed

[limber ravine]

oh fk

[empty grove]

a' < a < b < b'

[limber ravine]

yeah wtf

[empty grove]

[limber ravine]

I was doing [b,a] in my head

[empty grove]

oof

[limber ravine]

that's why the b' < b and a < a'

lol

[empty grove]

Lol that makes sense

[limber ravine]

xd

thanks

so we have $(-\infty,a) \cap (b,\infty) = \bigcup(a',a) \cap \bigcup (b,b')$ with $b' > b$ and $a > a'$. By axioms of a topology we have \bigcup(a',a) \cap \bigcup (b,b') \in \Omega$. Thus $R \setminus [a,b] \in \Omega$ and $[a,b]$ is closed

[gentle osprey]

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

[empty grove]

Yes

[stone cipher]

,tex Suppose that $q:E\to X$ is a covering map. If $X$ is an $n$-manifold, then $E$ is too.

[gentle osprey]

RaD0N

[stone cipher]

I have some trouble proving that one :/

I need just to prove that E is 2nd countable.

[pearl holly]

I know that $\pi_1^s, \pi_3^s, \pi_7^s$ are nontrivial, but is there an easy way to calculate what they actually are?

[gentle osprey]

Kanga Gang Swagger Tokidoki ✓

[pearl holly]

I mean like, what are some examples of pi_1^s and how would we calculate them?

actually nvm

[shy moss]

I think you need more information

[pearl holly]

ye I just wanted a example of a space where the the stable homotopy ghroups where nontrivial

like just any random space, expect the sphere's because that's already known

[shy moss]

\pi^s_1(RP^{\infty})=Z-2

[pearl holly]

oh ye right

that's a good example, thank you so much!

but what about like pi_3^s or pi_7^s lmfao?

I'm too noob at homotopy groups, I know almost nothing about them

[shy moss]

pi_3^s is Z_8 and pi_7^s is Z_16 \oplus Z_2

it was calculated using adams spectral secuence

https://www.jstor.org/stable/1993859

[pearl holly]

oh daim so it's invariant of the space?

oh I guess you mean RP^\infty

[shy moss]

yes

[pearl holly]

oh wow, this is very cool

thank you so much!

[long hornet]

I think this is not trivial to prove

It's a theorem that pi_1(X) is countable, so the covering has countably many sheets (because pi_1 acts transitively on the fibers)

[empty grove]

As written there is not enough information because E need not be a connected cover

So you could have uncountable topological sum of X covering itself which is not second countable