point-set-topology

[gritty widget]

this means that $\pi_q(\Sigma^n X)=0$ for $q<n$

[gentle osprey]

lime_soup

[gritty widget]

i need to know that $\operatorname{colim}n(\pi{n+k}(\Sigma^{n}X)=0$

[gentle osprey]

lime_soup

[gritty widget]

ah okay but $\pi_{n+k}(\Sigma^n X)=0$ for $k<0$

[gentle osprey]

lime_soup

[gritty widget]

does this realyl work for any space X or do we need some coniditon like finite CW complex

[cosmic prism]

any topology experts here right now i need help because i got very lost in pushouts

[lunar yoke]

can you specify your question?

[cosmic prism]

Let $\alpha:A\rightarrow X,;\beta:A\rightarrow Y,;\gamma:B\rightarrow Y,;\delta:B\rightarrow Z$ be continous maps of topological spaces. Prove that $(X+{A}Y)+{B}Z$ is homeomorphic to $X+{A}(Y+{B}Z)$

[gentle osprey]

Zorty

[lunar yoke]

did you cover the universal properties of pushouts

it would be a lot easier to prove via them

[cosmic prism]

thats just this right (didnt want to write it out lol)

[lunar yoke]

yes

and the fact that the pushout is determined uniquely up to unique isomorphism through this property

[cosmic prism]

we barely looked at that a few weeks ago when we briefly covered category theory

[lunar yoke]

so using this, if you show that the 1st one satisfies the universal property of the second one, you get that they are isomorphic

and with pushouts statements like this follow from / are known as "pushout pasting"

so your diagram looks like this and you can put both of the things in the corner down right

[cosmic prism]

yeah i already drew that but i still dont get it lol

[lunar yoke]

and you can prove all of this by just using the universal properties and so on, without actually looking at the topology

[gritty widget]

Quiver

[lunar yoke]

i guess it really depends on your tastes and how familiar you are with category theory whether you find this easier than working with quotient spaces of quotient spaces

[cosmic prism]

yeah i have absolutely no clue about category theory

[gritty widget]

Category theory is just playtime with arrows

[cosmic prism]

but universal property of pushouts doesnt mention anything about isomorphisms and homeomorphisms in my script so im confused

[gritty widget]

Well, just try to get what a pushout is, you don't need much category theory knowledge

[gritty widget]

Ah. You can prove the isomorphism though

[lunar yoke]

well you can show that if 2 objects satisfy the same universal property you have a unique isomorphism between them

thats not special to pushouts, but generally anything similar

for example tensorproducts if you know them

but ok yeah if you dont have much time then its probably easier to actually work with the topological spaces

since you're not familiar with category theory

[cosmic prism]

so i have to prove that one of the pushouts fufills the universal property of the other one and conclude that theyre isomorphic from that?

[lunar yoke]

that would work yes

one thing you can show that is helpful for these kind of exercises is the pushout pasting lemma i mentioned above:
In the pictured diagram, if you know that the top square is a pushout, then the bottoms square is a pushout if and only if the whole rectangle (i.e. the diagram made from only the corners) is a pushout

[cosmic prism]

yeah i think they expect me to do it this way

still have no clue 😵‍💫

[crimson path]

If S is a non-empty compact subset of a metric space and x is some point in the complement of S,
it is possible to show that min{d(p,x) | p in S} exists by invoking continuity of the map f(p) = d(p,x).
Since f is continuous on a compact set, it attains a minimum. Done.

Does anyone know a way of showing this without reference to continuity of f?
I attempted to create an open cover of S by taking a union of balls at each q in S
of radius d(q,p), but could not make much progress beyond obtaining a finite
subcover.

[little hemlock]

Try this cover: ${U_\epsilon}{\epsilon > 0}$ where $$U\epsilon = { p \in S : d(p,x) < \epsilon }.$$ You'll have to justify openness of each $U_\epsilon$ and at this point you are just kind of redoing the proof that $p \mapsto d(p,x)$ is continuous and extreme value theorem.

[gentle osprey]

kxrider

[little hemlock]

wait well openness is obvious

also, this is just the same cover you came up with ig? Wasn't reading that closely

[crimson path]

@little hemlock: Hm maybe, not sure. Yours is the set of all points in S within epsilon of x, so I suppose we next want to make epsilon smaller and see how small we can make it before we leave S. Mine is somewhat extraneous and cycles through all elements in S and their distance to x.

@little hemlock: But I did try another approach, though I could not escape referencing continuity. Since x is outside S and S is closed, there is some delta>0 such that B(x,delta) is in the complement of S. Then delta is a lower bound of {d(p,x) | p in S} and thus the infimum exists. We can make a sequence that converges to the infimum, and that will provide a sequence in S. S is compact so it has a convergent subsequence and continuity now will show that the infimimum is the minimum. But I could not do it without continuity of f. I guess it is necessary

[little hemlock]

yea i mean, if you want something which requires a weaker condition than continuity, mine should suffice. You only need that preimages of open intervals of the form (0, eps) are open

but also any proof which basically reproves continuity of d(p,x) and extreme value theorem in the process is not "referencing" continuity of d(p,x). At least in the sense that continuity is not a premise.

@crimson path

[crimson path]

@little hemlock: ok thank you very much for your input!

[little hemlock]

npnp

@crimson path hm im sorry, i just thought more about this and i realize it doesn't work

sorry if you were racking your brain over what i wrote lol

yea this is like a heine borel thing. it would be much much easier to assume continuity. Otherwise I think the program would look like: prove every open cover of f(S) has a finite subcover, and then apply heine borel to see f(S) is closed and bounded

[lament needle]

I was watching a 3blue one brown video and it gave a challenge to show that there is no continous mapping from the edge of a mobius band to a circle. So I tried to think about this in terms of the fundamental polygon of the strip. In this qoutient space the top left and bottom right corner of the square are identified and the bottom left and top right corner are identified. So In my head im thinking of this as gluing two line segments together but that would just give us a circle. I would think this provides us with a continous map since these corners are identified?

[lament needle]

After thinking more about the question in the video I dont think the question asked about the boundary of the mobius band. But actually that gluing the boundary of the mobius band to a circle required the mobius band to intersect itself

[arctic relic]

If someone asks to find homology of a map f and f:x->y, is that the same as the homology of f(X)

Or I guess it might just be the degree of f?

[marsh forge]

They probably want you to compute f*

the induced morphism on homology

[gritty widget]

hatcher page 453

I'm not sure how to see this

[gritty widget]

I know that the cofiber of $X \vee Y \rightarrow X\times Y$ is $X\wedge Y$

[gentle osprey]

lime_soup

[gritty widget]

But i see no reason why the stable homotopy groups of X smash Y should vanish

If X and Y are now spectra, does $X\vee Y \rightarrow X\times Y$ always induce isomorphism in homotopy

[gentle osprey]

lime_soup

[gritty widget]

or do we need that X and Y are connective/

[marsh forge]

Stably yes

[gritty widget]

can you help me see this?

i can't see how to show that the part from the smash does not contribute

[marsh forge]

Proof might depend on your Prefered model of spectra—it’s immediate if you already know that spectra is additive

Fwiw the smash product of spectra is not the cofiber of this map

[gritty widget]

at the moment im just working with sequential spectra

[marsh forge]

The highlighted claim should essentially be like

By definition

Ie you have you cells you define for a product

And the low dim cells are exactly the ones in the inclusion

It should follow from a connectivity argument on the suspended spaces

[gritty widget]

hmm

[marsh forge]

I can try to write it out I just feel strangely sleepy atm lol

But basically just argue that the cells in dimension n in the product are exactly the cells in dimension n of one of the two spaces

[gritty widget]

no no its okay

thank you very much though

[marsh forge]

The reason being that there’s a big range w no cells

[gritty widget]

i think i see

and then

if i want to do this for general spectra

not just

suspension spectra

[marsh forge]

I would start by proving the htpy category is additive

There’s probably a direct way

But honestly it’s easier to do the additive thing

[gritty widget]

yeah

i would like to do it the direct way though just because

when i try to do it the direct way

its seem that we need connective spectra

which i think should be wrong

so there is something I am messing up that i want to fix

[marsh forge]

You don’t need connective

That doesn’t mean your argument doesn’t need connective tho

Oh, here is a direct argument I think

There’s a sort of obvious way to describe a wedge as a homotopy pushout (include points)

Then pi_n will preserve this homotopy colimit

This also induces natural maps into the product and a universal map into the product

It’s also clear that pi_n(XxY) is the direct sun

Sun

Sum*

So you should get exactly that the homotopy groups agree and the universal map is an iso

[gritty widget]

okay

i get that $\pi_(X\vee Y) = \pi_(X)\oplus \pi_*(Y)$

[gentle osprey]

lime_soup

[gritty widget]

via this arguement

and and i get that $\pi_*(X\times Y)=\pi(X)\oplus \pi_(Y)$

[gentle osprey]

lime_soup

[gritty widget]

so indeed $\pi_(X\times Y)\cong\pi_(X\vee Y)$

[gentle osprey]

lime_soup

[gritty widget]

this is very foolish of me

im not sure why we can say that this isomorphism is induced by the map X wedge Y to X \times Y

[marsh forge]

I think you argue this is the same thing as the isomorphism of direct product and sum in Ab

[gritty widget]

oh so

X wedge Y to X \times Y

is a map between

coproduct and product

hmm no nevermind

becaue the product in spectra should be smash?

[marsh forge]

No

Product in spectra is \times

Smash product is not a categorical product

It’s a monoidal structure

[gritty widget]

oh okay

so because X wedge Y to X \times Y is the map between the coproduct and product

[marsh forge]

Yeah

[gritty widget]

pi_* should preserve things

[marsh forge]

Basically the functor preserves this whole diagram

[gritty widget]

thank you

[marsh forge]

np

[fair idol]

Does the covering space correspondence say that, if the spaces are "nicely connected", there is a bijection between n-sheeted covering spaces of a space X and index n subgroups of X's fundamental group? I'm having trouble interpreting the theorem because it's very wordy

[coral pivot]

Yeah, I can explain in more detail when I get home

But for now search like

Galois theory of covering spaces

[gritty widget]

if a space has sufficiently many adjectives then algebra = topology

[fair idol]

Thank you so much. I'll keep googling!

I have a feeling that if X is sufficiently nice and $\Tilde{X}$ is an n-sheet covering space of X with it's (Galois)corresponding subgroup Y of $\pi_1(X)$ then there is an isomorphism $\pi_1(\Tilde{X})\cong Y$. I think this is what the correspondence is saying

[gentle osprey]

limbostar

[marsh forge]

Yeah essentially the theorem tells us that subgroups of pi 1 should be covers and that if $X_H$ is the cover corresponding to $H$ we should have an exact sequence $1\to H\to \pi_1 X \to \pi_1 X_H \to 1$

[gentle osprey]

i_hate_printers

[marsh forge]

I’m ignoring some conditions for all this to be true

Sorry that last group should be the deck group

And the first group should be pi_1 of cover

Whoops

Technically the deck group is the normalizer kill H but if the cover is normal what I said is true

I always mix this up

[fair idol]

Thank you so much! I really appreciate that explanation 😃

[mint sage]

HELP. I don't even know how to study the topological properties of this quotient space...

[wide kayak]

hmmm

good question, I don't know off the top of my head but I know this is important

you can represent two distinct points of X/G by [x] and [y], where there is no g in G for which g*x = y

there's no such g because otherwise we'd have x ~ y, i.e. [x] = [y]

now you want to find two disjoint open sets of X/G containing [x] and [y]

since X is hausdorff there are disjoint open sets U containing x and V containing y

now compactness of G will have to be used somehow

[mint sage]

I've thought of all that

but no next😮‍💨

[gritty widget]

you want to get some open sets in G, somehow, so you could try using continuity of G -> X, g -> gx for each x

that's the first thing that comes to mind for me

it might also help to understand what the open sets in X/G are in terms of the action of G on X. i'd suggest writing them down, it'll help you figure out what you need to do

since you said you don't know how to study the topological properties of this quotient

i'll start: "by the definition of the quotient topology, W is open in X/G if and only if..."

[mint sage]

Let me try

[mint sage]

but that's what I don't know☹️

[gritty widget]

then open your textbook and get reading

you can't do a problem on the quotient topology if you don't know what it is

[gritty widget]

is it possible to write suspension in temrs of attaching a cell

[lunar yoke]

I dont think so. Recall that the suspension of S^1 is S^2. To get that by attaching a single cell we would have to attach a 2-dimensional cell, and we would get a pushout
S1 ---> S1
| |
D^2 -> S^2
where the left vertical map is the inclusion of the boundary and the right vertical map is the inclusion of the equator. But then you get H_n(D^2,S1) = H_n(S^2,S^1).
This cant be, since Z = H_2(S^2) = H_2(D^2,S1), but H_2(S^2,S^1) = H_2(S^2 v S^2) = Z + Z

there is probably a way easier solution to this

[gritty widget]

is there a slick way to shwo that the n-th suspension of a space is n-connected

by slick i mean slicker than

for any space X \Sigma X is path connected

then use the induction

[marsh forge]

Cellular approximation w the correct cell structure should work I think

[gritty widget]

can you give a slight more i’ve a hint by what you mean by this

[marsh forge]

I’m not confident it works but I think you should be able to put a cell structure on a suspension that has no cells in the desired range

And then by cellular approximation all maps out of the sphere are nullhomotopic

I’m about to go surfing and haven’t had any coffee so this is the best I can do rn hahah

[gritty widget]

oh la la

enjoy

thanks anyway

[marsh forge]

I want to argue that If X is a CW complex then Sigma X has a cw structure in which the cells are just one dimensional higher cells from X

Like same cells but one dim higher

This will force one extra level of connectivity

I’m not actually sure this works tho lol

Surf time

[lunar yoke]

in retrospect it should work, since an n-connected cw complex is htpy equivalent to one with trivial n-skeleton

[gritty widget]

hihi

Can someone help me find a finite topological space with the fundamental group Z2?

aka with two non-homotopic paths

Maybe we could say it is disconnected and it is the union of two sets that are each simply connected? Maybe that would work?

[runic sun]

If it's not path connected its fundamental group might be ill-defined, in case both these components would be simply connected it would be trivial at every point.

[lunar yoke]

Maybe this helps https://en.wikipedia.org/wiki/Pseudocircle

[wide kayak]

I don't think you mean finite. Maybe compact?

Oh what

Ive never seen that pseudocircle thing. Freaky!

[gritty widget]

I did mean finite, sadly. 😢

[gritty widget]

Alright, I'll look at it now!

[wide kayak]

Yeah I see. Trippy, I've never thought about homotopy in a finite space

[gritty widget]

Interesting. Do all pseudocircles have pi(X) = Z2?

ohhh

[fair idol]

If van kampen applies to $X=A\cup B$, can I be real sloppy and take the denominator (some kernel of a map) to be $\pi_1(A\cap B)$

[gritty widget]

There are two paths, one going left and one going right?

[lunar yoke]

no? the name comes from the space being weakly homotopy equivalent to the normal circle, as the article says

so it has pi_1 Z

not Z2

[gentle osprey]

limbostar

[lunar yoke]

im not sure if you can use this to construct the thing you want

[gritty widget]

Okay. So I can construct a set with pi_1 = Z2 from this?

[gritty widget]

Ah, okay.

[lunar yoke]

although the article mentions there is a functorial way to give a finite topological space that is weakly homotopy equivalent to the realization of any finit simplicial complex

and RP^2 should be a finite simplicial complex

(and RP^2 has fundamental group Z2)

[gritty widget]

Oh, I see!

So there's a way to find a finite space weak homotopic equivalent to RP2, you say?

Can I use this same map?

[lunar yoke]

the article cites McCord, Michael C. (1966). "Singular homology groups and homotopy groups of finite topological spaces". Duke Mathematical Journal. 33: 465–474. doi:10.1215/S0012-7094-66-03352-7.

[gritty widget]

I see...

It's more complex than I thought!

[lunar yoke]

if this is some sort of homework question then there is definitely some easier way

but I also have never thought about things like this

[gritty widget]

It's an extra credit problem.

So it's much more complex than homework, I'd guess.

[lunar yoke]

well it should still be solvable on your own

probably

[gritty widget]

lol yeah

hopefuilly

This one looks easier:

I think it's just a set theory problem, basically.

[lunar yoke]

yeah thats pretty straightforward i would say

[gritty widget]

Do you have any advice for making it formal?

Or any ideas?

[lunar yoke]

oh wait i misread

i thought it said to show that if V and W and V n W are simpy connected then V u W is

[gritty widget]

Is a simply connected space minus another simply connected space always simply connected itself?

[gritty widget]

That was HW a while ago, I think.

[runic sun]

No, consider a disc inside a larger disc

[gritty widget]

Or D^2\{x}

[runic sun]

Homotopy equivalent

[gritty widget]

We have simply connected minus simply connected has pi_1 = Z.

These problems are unexpectedly difficult, it seems.

I mean, I have some lower hanging fruit.

Does any one of these three seem easier?

[runic sun]

These problems look pretty fun, do you mind telling where they're from?

[gritty widget]

From my professor, lol.

I can just send them all to you if you want.

[runic sun]

Oh, fair enough

[runic sun]

I'd appreciate that if you don't mind

[gritty widget]

Oh, apparently he doesn't accept problems after yesterday.

Well, I might as well try to solve them anyway.

[lunar yoke]

what did you cover in the course? Just to get an idea of what tools you are supposed to use

[gritty widget]

General topology up to basic algebraic topology.

[lunar yoke]

covering spaces?

homology and cohomology?

[gritty widget]

So topologies, connectedness + compactness + etc.

[lunar yoke]

kk

[gritty widget]

Then

Path-connectedness, homotopy, covering sapces, fundamental group, etc.

Well, rip.

Now I have to actually study for my exam instead of just solving a bunch of problems like I did for my first mid-term. 😢

A problem is 20/0, so 75/100 + 2 * 20/0 = 105% on an exam I barely passed, buahaha.

Oh, well. I really like what we've learned recently anyway.

[lunar yoke]

this looks like the lebesgue lemma could be of use

ok yeah i think that works

starting with a loop in V, you take the homotopy in V u W, subdivide I^2 with Lebesgue, and then substitute the interior of the squares that are mapped to W by extending their boundary that is mapped to V n W, which overall results in a homotopy with image in V

[gritty widget]

I'm not familiar with the lemma.

[lunar yoke]

its this https://en.wikipedia.org/wiki/Lebesgue's_number_lemma

[gritty widget]

But it's okay. It's too late to get credit for the problems anyway, so I won't die trying to solve it.

[gritty widget]

Thanks! Looks interesting.

[lunar yoke]

you can try proving it yourself, its just pointset topology and not too hard

moreoever, its very useful

[gritty widget]

It looks like it's about metric spaces.

Not all t-spaces?

[lunar yoke]

yeah we apply it on I^2

the open cover we use consists of H^{-1}(V) and H^{-1}(W) where H : I^2 -> V u W is the homotopy

basically, it tells you that you can subdivide the square I^2 into m^2 little squares of sidelength 1/m for some m, so that each little square is mapped either completely to V or completely to W by H

[gritty widget]

Could you help me solve a HW problem or two I'm stuck on?

[lunar yoke]

not now, its 2:30 am here ^^

[gritty widget]

Gn!!!!

[lunar yoke]

but you can always just post them here and see if ppl will respond

[gritty widget]

Thank you for all of your help today!!

[gritty widget]

Okay, will do.

Where p: X-tilde --> X is a covering map and X-tilde is path-connected:

Suppose pi_1{X} is Z (integers) and p-1(x0) is finite. Find the fundamental group of X-tilde.

If X is simply connected, then p is a homeomorphism.

[flint cove]

Does anyone know of a proof of Markovs Topologization theorem (a group is topologizable iff no finite set of inequalities has precisely one solution)? I read the proof in Zelenyuk's book but I have absolutely no Idea what is going on

(ie, I have no Idea why this constructed topology works precisely as we need)

*countable group

[gritty widget]

A characterization of topologizable countable groups? That's still really nice

[flint cove]

I don't even know which paper originally contained the proof, that would help as well

(A quick search on scholar yielded surprisingly little. Perhaps I just don't know the right keywords.)

[gritty widget]

Sometimes they don't have the articles

[flint cove]

Ah, hold on.

[5] Markov, A.: On unconditionally closed sets. Mat. Sb. 18, 3–28 (1946). (Russian)

Found in

Topologizable structures and Zariski topology by Joanna Dutka and Aleksander Ivanov
Gonna read this next. This should provide some context.

Oof, okay, this doesn't look much more digestible tho lmao

[summer terrace]

Hey, is a topology basically a 'generalization' or 'idealization' of continuity and convergence?

A book called Topology through Inquiry motivates the definition through this explanation.

[gritty widget]

I mean yeah, it's in the names, almost all functions we care about in topology are continuous or relating to continuity in some way

Convergence comes out more in study of metric spaces but it's there too, sure

[summer terrace]

Hmm, got it.

[summer terrace]

The finite complement topology on the reals is just {R}, right?

[hollow harbor]

Ummm

That is not a topology

Topologies include the empty set

[summer terrace]

Oh, my bad.

[hollow harbor]

In any case, the finite complement topology includes the empty set as well as any set whose complement is finite

So R\{t_1, ..., t_n} for any numbers t_1,...t_n in R will be an open set in the topology.

You can see that this is a topology since the intersection of finitely many of these sets will just exclude all the numbers from each one. There are finitely many of those.

[summer terrace]

But R - {t_i} isn't finite, or is it?

[hollow harbor]

Another way to think about it is that the closed sets are exactly the finite sets (as well as all of R)

[hollow harbor]

Right. It has finite complement.

[summer terrace]

I think I get it.

[hollow harbor]

For me the easiest way to think about this is in terms of the closed sets instead of the open ones

[summer terrace]

Not quite there yet, guess I'll read more

[gritty widget]

I wanted to say that just now

Got sniped

[gentle osprey]

Wew "DarQ" Tbh

[broken nacelle]

nvm

it's coz f has to be well defined

is the converse not always true?

nvm

it's answered in my textbook

[gritty widget]

I have a problem.

I need to classify all homeomorphisms from the Sierpinski triangle to itself.

My idea is that an continuous map from the ST to itself is clearly a homeomorphism, and thus is open and continuous, and the only open map from ST to itself would be one which maps the triangles starting at generation X to another generation of the traingles...

But I'm not really sure where to go from there.

Well, Id is clearly one.

[gritty widget]

I think if we look at the first iteration of the Sierpiński triangle, we have 3 local cut points and they need to be mapped to each other

And the 3 triangles similarly need to be mapped to each other

If we look at the second iteration, I think the same pattern occurs

[gritty widget]

Ohh, I see! So three total? Or 6?

I think the only homeomorphisms of Sierpiński triangle are the rotations, but I'm not sure

Three

[gritty widget]

Makes total sense!!

Thanks for the idea!! That seems very correct.

Because of all the cut points of the smaller triangles.

They must all be mapped to cut points.

[gritty widget]

Okay, can you also help me find covering maps f: X -> Y and g: Y -> Z s.t. g o f is not a covering map?

I need to go to work tomorrow

Sorry

npnp

tysm!

Yes, thank you, I was thinking of S_3

My bad for saying the wrong thing

Ohh, so we do get S(3)?

All permutations on three elements?

Thanks, Nobody!

I'm going to tell my professor that "Nobody" helped me solve this one. 😂

I had help from "Nobody."

lol jk

[coral pivot]

I mean is this all of them? Can u prove it

[gritty widget]

Well... It has to map cutting points to cutting points.

They're not actually cut points, just local ones

Also, for the record, he's very abstract and presenting the problem will probably end up like:

Me: "Cutting points map to cutting points, which are preserved by rotations and-"
Prof: "Yep, you've got it."

[coral pivot]

Right but there’s infinitely many such things

[gritty widget]

Local cut points? Oh, per triangle?

Hmm

[coral pivot]

So you gotta rule those out ig

[gritty widget]

A cut point breaks connectedness

A local cut point breaks connectedness locally

Got it.

[shadow rampart]

Is the injection of $\mathbb{Z}^2$ into $\mathbb{R}^2$ a topological embedding? $\mathbb{Z}^2$ has the usual Euclidean metric.

[gentle osprey]

dumbo

[shadow rampart]

It's distance-preserving which seems like it should be enough

[unreal stratus]

Yes, more generally we see Z^2 has the subspace topology from R^2 and subspace inclusions are top embeddings in general

[arctic relic]

Do covering maps lift the identity loop in X to the identity loop in X tilde

I said yes by functoriality of p* but I’m trying to find counter examples and it’s sort of a black swan situation

For every subgroup G of H, pi_1(X), st it can be realized at a covering space of X has the trivial loop in its group presentation so e has to lift to e

[icy schooner]

Let $\gamma(\theta) = \begin{pmatrix} cos(\theta) & sin(\theta) & 0 \ -sin(\theta) & cos(\theta) & 0 \ 0 & 0 & 1 \end{pmatrix}, \theta \in [0, 2\pi]$ a loop in $SO(3)$

[gentle osprey]

Iteribus

[icy schooner]

how do you get a homotopy from $\gamma^2$ to the trivial loop

[gentle osprey]

Iteribus

[icy schooner]

I don't think I can scale the angle

especially continuously cuz then it wouldn't be a loop at all at some point

[gritty widget]

Hello

I am looking for anyone who is studying or is interested in knot theory

[still vessel]

Can some one check if my proof is valid?

Thing i want to prove is following:

Let $f: X \rightarrow Y$ be a continuous map then if X is connected so is f(X).

The definitions i am working with are the following:
Contiuous map: $f: X \rightarrow Y$ is said to be contiuous if

$(\forall O_c \in X) (\exists O_{f(c)} \in Y) : (x \in O_c) \land (y \in O_{f(c)})$

where $c \in X$ and $f(c) \in Y$ and $O_c$ is an open ngbh around c.
Connected space: X is said to be connected if

$\nexists U,V \in X$ such that $(U \cup V = X) \land (U \cap V = \emptyset)$

The proof:
Assume Y is not connected, then there exist $U,V \in Y$ s.t. $(U \cup V = Y) \land (U \cap V = \emptyset)$. Let $x,c \in X$ and $y \in U$ and $f(c) \in V$. Since f is continuous, $x \in O_c$ and $y \in O_{f(c)}$. If $y \in O_{f(c)}$ then $U \cap V \neq \emptyset$. Hence Y must be connected.

[gentle osprey]

FireLi0n

[unreal stratus]

Perhaps I'm misundrstanding something but your definitions don't seem to be correct? For example your definition of continuous doesn't make any sense to me - it's certainly not the standard one

Similarly the definition of connectedness is incorrect

[still vessel]

isnt this definition of continuous maps standard definition with open neighbourhoods? Also the connectedness is the one we use, i just failed to write that U and V are open sets

This is from a book

[unreal stratus]

No, if you want it to be pointwise you've switched the roles of Y and X for starters (and not actually mentioned what x and y are either) but the typical definition would be that the preimage of open sets under f is open

[unreal stratus]

And sure, that's the typical definition, but also note when you write it in symbols you need to say U, V are subsets of X rather than elements thereof

[still vessel]

ah, okay that makes sense

[gritty widget]

DOES ANYONE HERE STUDY KNOT THEORY

I have a few non-hw questions

[broken nacelle]

how do we know that path components are open subsets?

[unreal stratus]

If x is in a path component P and a neighbourhood of x is path connected then that neighbourhood will be contained in P

this is by the definition of path components

[broken nacelle]

I see

[bronze lake]

I did an REU in knot theory, I may be able to answer

[gritty widget]

What's REU?

Anyways

What's a square knot and is there a trick to tying it faster/more accurately?

https://www.youtube.com/watch?v=Gu8e0nArRuY

I'm trying to understand the math behind this

Tying knots isn't knot theory afaik

[grave maple]

It probably helps though.

[golden gust]

probably knot

[bronze lake]

Square knot? I only know $3_1#3_1^*$

[gentle osprey]

cgodfrey

[gritty widget]

are you saying there aren't tricks I could use to make it better?

[sharp frost]

no he's saying that the study of physically tying knots in string is not really what mathematicians refer to as "knot theory"

[gritty widget]

ah

what does knot theory study?

[wide kayak]

physical string knots that are infinitely thin and have the ends tied together

[sharp frost]

knot theory does coincide with the real practice of tying knots somewhat, one main difference is that in order for something to be counted as a "knot" in mathematics it needs to be joined up at both ends

mathematicians basically look at two of these knots and go "hmm what topological invariants can I use to show that these are not the same knot"

or "how many different knots is it possible to tie if you work in 3-dimensional space"

[gritty widget]

wait so

if I have a loop

are u saying Ican tie all those without interrupting the loop?

[sharp frost]

no you can't untangle mathematical knots

[wide kayak]

not allowed to have the string pass through itself

[sharp frost]

if you were able to untangle a knot into a loop then the knot would be considered to be identical to a loop

[gritty widget]

oh

[gritty widget]

hmm ok

[sharp frost]

yeah like in that picture the top left knot is considered to be identical to the loop in the bottom left

on an unrelated note does anyone know whether there's a way of embedding the direct sum $A\times B$ into the free product $A*B$ as a subgroup

[sharp frost]

assuming A and B are finitely generated

[wide kayak]

this old book I have has some cool pictures

they're links - multiple knots tangled up

[gritty widget]

ok so to summarize

knot theory doesn't actually teach u how to tie knots or how strong the knots are?

[sharp frost]

thats right aha

[wide kayak]

definitely not how strong the knots are

[gritty widget]

would u happen to know which field does this?

[sharp frost]

sailing

[gritty widget]

hahaha im from a related field

[wide kayak]

boy and girl scouts

[gritty widget]

the thing about marine knots are they're good at tying the stuff but they dno why they should tie a certain knot

it's all just

do this do that

so I figured I should dive further in

[sharp frost]

I think determining the strength of a knot can be done using some things in physics

because physics studies tension and friction, which are what make knots work

or engineering probably

[gritty widget]

how about researching more efficient ways to tie/untie knots?

[sharp frost]

I think that mathematicians do do that but it's probably overkill to use that sort of theory on knots in rope (I think at least, knot theory isn't my field)

there are people trying to use knot theory to work out how to unravel DNA

[gritty widget]

well, it's not so much on sailing knots but on surgical sutures

@sharp frostWould it be possible to do a quick 15 mins or 30 mins informational interview on what you do and what knot theory is about in vc?

whenever you're free ofc

[sharp frost]

I don't know much about knot theory I'm afraid, you can get a lot more from a youtube video than you can from me https://www.youtube.com/watch?v=aqyyhhnGraw

[gritty widget]

okay will check it out

thanks anyways :)

[gritty widget]

Can something be said about inductive dimension of subsets

Say, closed subsets

For covering dimension it can

Which is the same in what I'm doing

[gritty widget]

why is this true?
X a CW complex then

$[\Sigma^n X, \operatorname{hocolim}k \Omega^k Y{n+k}] \cong \operatorname{colim}[\Sigma^n X,\Omega^k Y_{k+n} ]$

[gentle osprey]

lime_soup

[gritty widget]

is it true in general homotopy classes of maps into a homotopy colimit is the same as the colimit of the homotopy classes

seemingly we need X compact

(manifesting max to appear)

[cedar pebble]

@gritty widget are you sure you want colimits here, rather than limits?

[gritty widget]

i’m (was) pretty sure?

[cedar pebble]

hmmm

[gritty widget]

this is this confusing thing with

inverse limit and colimit

[cedar pebble]

like just in standard category theory we have [X,lim Y]=lim[X,Y]

if you have an inverse limit (<-) then this is a limit

[gritty widget]

okay yes then this is not an inverse limit

this really is a colimit

yeah because

[cedar pebble]

okay that's weird

so like this statement would be automatically true for (homotopy) limits rather than colimits

[gritty widget]

you want it to be whatever the groups stabilise to for large n

[cedar pebble]

right yeah

ahh yeah okay

yeah you need X to be compact I think

[gritty widget]

oh so

we do something like

[cedar pebble]

or a compact object I guess

[gritty widget]

write X as a colimit

over subspaces

sorry no this is nonsense

[cedar pebble]

yeah so you need X to be a finite discrete space I think

[gritty widget]

i have no idea how we use compactness

actually

can i ask you about compact objects

[cedar pebble]

you're not using compactness, you're using that X is a compact object; these are different notions

sure

[gritty widget]

i hear this thrown around

but don’t know what this means

[cedar pebble]

so if C is a locally small category with filtered colimits, an object X in C is compact if Hom_C(X,-):C->Set preserves filtered colimits

[gritty widget]

in top spaces the finite cw complexes are the compact objects?

[cedar pebble]

that is, Hom_C(X,colim Y)=colim Hom_C(X,Y)

which is the sort of thing you're trying to do

[gritty widget]

ooh okay

[cedar pebble]

In Top, the compact objects are finite discrete spaces

[gritty widget]

hmm okay

maybe

cw complexes are

generated under colimits by the compact objects

[cedar pebble]

right

[gritty widget]

maybe one last question because this is getting away from alg top

but is there some relation between

compact objects

and a size arguement

in a large category

that the compact objects are small?

[gritty widget]

Hew, how can I find the fundamental group of RP2\{0}?

Any tips?

what is 0 in RP2

[gritty widget]

Any point, sorry.

I'm guessing it's Z, that's what it looks like in my head...

Or maybe it's Z x Z, actually?

That would make intuitive sense. Or Z x Z2.

using the description of RP^2 as the disk with antipodal boundary points identified: remove a point, retract outwards to the boundary, and you're left with something homeomorphic to S^1

[gritty widget]

Okay, so my initial idea was correct!

Many thanks!

[marsh forge]

Small is sometimes used as a synonym but has nothing to do w sets

[gritty widget]

Are there theorems of the form dim(XxY) >= dim(X)+dim(Y)-r

[compact bluff]

lebesgue covering dimension?

if so, it never holds (unless r=0) for compact spaces, nor when both X and Y are metric

otherwise there are examples where X is metric (and even separable) and Y isn't where the dimension of the product is strictly larger than the sum of the dimensions

[gritty widget]

@compact bluff compact metric spaces

Yeah, covering dimension

I've made an error in my calculations because I thought the equality holds, but maybe it can be saved a little

[compact bluff]

equality doesn't hold (in general), but you get dim(X x Y) <= dimX + dimY for compacta

(or even if at least one of X or Y is compact)

[gritty widget]

How bad can it go?

[compact bluff]

I wouldn't be surprised if there exist Y of arbitrarily high dimension but you get dim(X x Y) = dim(X) + 1

[gritty widget]

What about dim(X^n) ?

[compact bluff]

you might be able to get better estimates on that (though maybe still not perfect)

though apparently this paper gives a necessary and sufficient condition for when dim(X x Y) = dim(X)+dim(Y) https://www.jstage.jst.go.jp/article/pjab1945/36/7/36_7_400/_pdf

Looks like if the Q-(cech) homological dimension equals the dimension of X

might be easier to compute this condition using Lemma 1 in that paper

[gritty widget]

Hmm...

I think there might be some results in one of the newer versions of books about dim theory

[gritty widget]

Yeah, apparently dim(X x Y) >= dim(X)+1 can't be improved

For compact spaces

[dusky pawn]

https://github.com/RandomAnass/TDA-DL

An application of topological data analysis

[still vessel]

So i dont know if i understand the set of open balls correctly. So the open ball is a set of points which are located a certain distance from a chosen origin of that open ball ( if i understood that correctly). So if i have set of all open balls of a metric space, isnt that just a union of all points as you can just have different open balls with different origins and different distances? You dont even have to care about the distance in the definition of the ball, just the origin, if you have enough balls you will cover all the points by just those points being the origin, no?

[lunar yoke]

I think you may be confusing the set of all open balls with the union over that set

the set of open balls contains open balls, not points

but you're right that if you take the union over all open balls, you get the whole space back

i.e. every point of your space is in some open ball

[still vessel]

hmm ok, so how exactly do those open balls form a basis of space then? My understanding of a basis comes from linear algebra where you have vectors from whose linear combination you can produce any other vector in the space. How does this concept extend here?

[lunar yoke]

no the basis from linear algebra is only the same in idea

the definition is completely different

A collection of open sets of a topological space forms a basis for that topological space, if every open set can be written as the union over open sets in your basis

so in the case of metric spaces, the collection of all open balls form a basis, because you can write any open set as a union of open balls

that may not be obvious immediately, but its not hard to prove

[still vessel]

ah okay, that makes more sense. And since the open ball can be as small as possible in "distance" you can form any open set you want with them

[lunar yoke]

yeah thats the idea

[still vessel]

now it makes more sense in my head

thanks

just a technical question, are the words "collection" and "set" synonyms or is there some technical difference?

[lunar yoke]

i used them synonymously above

i think most people do

[hollow harbor]

People often say "collection" when they're working with a set of sets or a set of functions or something else that can get confusing if you don't use separate words.

[gritty widget]

Is it?

[lunar yoke]

i meant in the sense that they determine the object you are working with

or do you mean that there is more analogy there?

[gritty widget]

The second is how I understood it, and were questioning

[cosmic prism]

Let $\alpha:A\rightarrow$ and $\beta:A\rightarrow Y$ be continous maps of topological spaces, where $\alpha$ is surjective and $Y$ connected. Show that the pushout $X+_{A}Y$ of $\alpha$ and $\beta$ is connected.

[gentle osprey]

Zorty

[cosmic prism]

My understanding of connectedness using the more set theory based definitions and visual meanings of pushouts is quite okay, but it's just questions such as these where a very "formal" approach is needed where i really struggle... Can someone help me?

I suppose it would be enough to prove that $Y\rightarrow X+_{A} Y$ is surjective which makes sense since every point in $X$ is identified with some element in $Y$ since $\alpha$ is surjective but im really not sure

[gentle osprey]

Zorty

[lunar yoke]

since you are working with pushouts, there is a characterisation of being connected that fits quite nicely

a space X is connected iff every continuous map f : X -> {0,1} is constant, i.e. not surjective. Here we equip {0,1} with the discrete topology

if you know that X is connected iff the empty set and X are the only clopen subsets, then this criterion follows basically immediately

and so now we can work with maps out of the space. Recalling the universal property of pushouts, a map X +_A Y -> {0,1} induces maps X -> {0,1} and Y -> {0,1} that commute with maps from the pushout

since Y is conected, the map Y -> {0,1} is not surjective, and since alpha is surjective, the map Y -> {0,1} already completely determines the map X +_A Y -> {0,1}, from which you can conclude that that map is also not surjective

hence X +_A Y is connected

[lunar yoke]

this works too though

i only saw it now xd

epimorphisms are stable under pushouts

[cosmic prism]

How would alpha being surjective imply that the map from the pushout to {0,1} is determined by Y -> {0,1} alone? i dont quite get that part

[lunar yoke]

ok so if you have x in X, then you can write it as alpha(a) = x for some a in A.

[cosmic prism]

yes that makes sense

[lunar yoke]

now use that the square commutes

its a bit hard to write down without names of the maps xd

sending x from X to {0,1} is the same as sending it first to X +_A Y and then to {0,1}

but since alpha(a) = x, this is the same as going A -> X -> X+_A Y -> {0,1}

[cosmic prism]

right

[lunar yoke]

which agrees with A -> Y -> X+_A Y -> {0,1} since the diagram commutes

but then this is A -> Y -> {0,1}

so the map X -> {0,1} is already determined by Y -> {0,1}

and the universal property of the pushout tells you that maps out of X +_A Y are uniquely determined by the corresponding pair of maps out of X and Y

if you look at the definition topologically, when you map some element in X+_A Y to another space, you choose a representative in X or Y, and then use the maps from there

[cosmic prism]

ok i think i got it lol

[cosmic prism]

doesnt this mean that X is connected as well?

[lunar yoke]

no

[cosmic prism]

oh no nvm yeah its just a specific map to {0,1} which is constant of course it doesnt hold for all

[lunar yoke]

if you want to show X is connected with this method, you start with a map X -> {0,1}

but that doesnt give you a map Y -> {0,1} or X +_A Y -> {0,1} for that matter

[cosmic prism]

Also: do you by any chance have an example in mind for surjective, continous functions $f: X\rightarrow S$ and $g: Y\rightarrow S$, where X and Y are connected such that the Pullback from f and g is not connected? I've been looking for an example in R but I couldnt find any unconnected curves, where both sides were surjective functions

[gentle osprey]

Zorty

[lunar yoke]

yeah so consider S = S^1 and X = Y = [0,1] with f(t) = exp(2pi i t) and g(t) = f(t+1/2)

Then the pullback is X x_S Y = {(x,y) in X x Y | fx = gy} = {(x,x+1/2) | x in [0,1/2]} u {(x,x-1/2) | x in [1/2,1]}

so you get the blue region, which is clearly not connected

[cosmic prism]

ohh yes that makes so much sense thankssss

[kind depot]

any idea how to prove this. im not sure what is meant by induced in this context tbh

[gritty widget]

there exists a map $\tilde{f}\colon\bR P^{n-1} \to \bR P^{m-1}$ such that $\tilde{f}([x]) = [f(x)]$, brackets denoting equivalence classes in projective space

[gentle osprey]

TTerra

[kind depot]

mmm where our equivalence is $f(x) \approx \lambda^k f(x)$

[gentle osprey]

Optimism

[kind depot]

? seems like the proof is rather trivial

[gritty widget]

it should be straightforward

it might be a good idea to write down the details

[kind depot]

so like is it simply saying let $x \equiv \lambda x$ then noting $f([x]) $ where we take a representative implies taht $f(x) = f(\lambda x) = \lambda^k f(x)$ which is identical to defining a function $\tilde{f}$ over projective spaces?

[gentle osprey]

Optimism

[kind depot]

assuming qoutient topology

????

[still vessel]

Am i right in understanding that every point in a metric space is a limit point of some sequence in that space?

[coral pivot]

yeah but thats not saying much right, if a is the point

take the sequence a,a,a...

[still vessel]

haha, i didnt think about that one xD

what if we limit it to non trivial sequences such as that?

[coral pivot]

you can find sequence such that a repeats infinitely often

your metric space in general could have like an, isolated point

that might be the type of image you are trying to get at

[gritty widget]

Yep. Like in the one-point metric space

Then it's necessarily trivial

[still vessel]

hmm, what i am trying to do is to prove the following:

Let $S \subset X$. Prove that $z \in \overline{S}$ iff there is a sequence of points $(x_n)$ in S converging to z.

[gentle osprey]

FireLi0n

[still vessel]

so i am trying to understand how to approach this

isnt this basically saying that every point has a sequence that approaches it

[gritty widget]

It's not true in a topological space

[still vessel]

its a metric space, sty

[gritty widget]

Are you studying metric spaces?

Okay

[still vessel]

yes

[coral pivot]

well how are you defining closure/closed sets

[still vessel]

Closure is a union of Interior and Boundary of a set

at least that is the only one i know of

[coral pivot]

how did you guys define boundary lol

[still vessel]

give me a sec to find it

$\partial A = { a \in X | \forall O_a : O_a \cap A = \emptyset \land O_a \cap (X-A) \neq \emptyset }$

[gentle osprey]

FireLi0n

[still vessel]

where $O_a$ is an open neighborhood of a

[gentle osprey]

FireLi0n

[coral pivot]

I think you wanted both to be not empty

but yeah makes sense

if you combine this and the interior, notice that a closed set C has the property that every point has each neighborhood intersect C

so lets work with this definition of closed

[coral pivot]

so what progress did you make on this

[still vessel]

i proved the one way, starting from a sequence and proving that it is in closure

[coral pivot]

right

[still vessel]

the other way i dont know how to begin with

[coral pivot]

so what do you need to show for the other way

[still vessel]

in my mind i have to show that there is a sequence that converges to that point

[coral pivot]

sorry this is badly written lol. what i wanted to say was, each point that has all neighborhoods intersecting C is in C

[still vessel]

i dont know how to show that there is a such sequence

[coral pivot]

ok so thats not it

[still vessel]

yeah, i understood what you meant

[coral pivot]

the idea is that, you are closed if you contain each limit point

so rather than each point having a converging sequence

you want that for each x\in X such that a sequence in Abar converges to it, x\in Abar

does that make sense

[still vessel]

okay, so i have to prove that a closed subspace contains all its limit points?

i mean, that is kinda trivial, no?

[coral pivot]

if you mean prove this for general closed sets from your definition, sure. good practice to convert between defs

in particular check that this holds for Abar to show Abar=closed and hence the closure right

(you can directly see this from the neighborhoods defintion btw)

[still vessel]

Abar is the closure, i dont have to show that its closed

what i meant is that we already showed it earlier, so i can take it as true

[coral pivot]

sorry when i am saying Abar i mean the set of z such that a sequence in A converges to it

not the closure yet

you want to show its the closure

[still vessel]

but isnt this the other way around?

i am starting from the assumption that z is element of closure, not that it is a limit point

dont i have to prove that z is the limit point?

[coral pivot]

Oh sorry thats what you are doing, i was just gonna say that, you already showed the set has to be in the closure right

so now if its closed you get the other inclusion

cause remember by definition, closure is contained in all closed sets containing A right

[still vessel]

i must admit i am confused at this point

[coral pivot]

yeah sorry let me write this all out

So let $S$ be the set in question, let $$S' = {x\in X: \text{ a sequence in S converges to x}} $$ and let $\bar{S}$ be its closure, i.e the smallest closed set containing $S$. Now you have already noted that $S'\subset \bar{S}$ as it must contain all the limit points of $S$ right. Now if you show that $S'$ is closed, then $S'\supset \bar{S}$, as all closed sets containing $S$ will contain $\bar{S}$, this is what it means to be the closure

[gentle osprey]

JohnDS

[coral pivot]

so really all that remains is to show that this set is closed

[still vessel]

okay i think i understand now, so the point was to prove $S' \subset \overline{S}$ and $S' \supset \overline{S}$

[gentle osprey]

FireLi0n

[still vessel]

i was missing that from my logic

[coral pivot]

yeah probably should have written this out earlier, but yeah

[still vessel]

okay, thanks for the help

[still vessel]

Just to follow up on this as i thought more about it. How can i be sure that S' contains all of S? That doesnt sound right to me.

[coral pivot]

every point in S is a limit of a sequence in S right

namely the constant sequence

[still vessel]

Ah, forgot about that crap sequence again

So every point is S is a limit of a sequence in S but not every sequence in S has a limit point in S

got it

[coral pivot]

mhm

[unreal stratus]

Lol that notation S' seems potentially confusing though

since S' would normally be the set of limit points instead of the set of all limits of sequences

[coral pivot]

I mean the convention I follow these are the same things

[still vessel]

whats the difference?

[ashen talon]

What's the difference between a Weyl chamber and an alcove?

[unreal stratus]

Typically in a top space X, a limit point of a set S is a point p of X such that every open neighbourhood of p intersects S in a point besides p. In metric spaces, this is equivalent to there being a sequence in S \ p converging to p

(Every source I've used including Munkres, Rudin and Lee use this)

The difference is that e.g. elements of S needn't be limit points (for example, every point of the cantor set is a limit point thereof, whilst no point of Z is a limit point (both in the subspace topology from standard top on R))

[marsh forge]

What convention is this?

[coral pivot]

The convention where limit points just mean there is a sequence converging to it. I don’t remember from where I got this, but never caused me an issue

[unreal stratus]

what about limit points in a general top space?

[coral pivot]

Same thing, all nbds intersect the set, I just don’t exclude the actual point

[marsh forge]

This is definitely wrong

Like

Against any course in topology or definition I’ve ever seen

[coral pivot]

What’s the advantage of ur definition over mine

[marsh forge]

Well for one, every point in a set is the limit of its constant sequence

So your definition is useless lol

Like

In the set {2} \cup [0,1]

2 should not be a limit point

Plus all of the like

Limit point point set yoga is just wrong if you don’t use the standard defn

And a bunch of subsequent defns

[coral pivot]

Like?

[marsh forge]

I mean I can go through munkres later and tell you lol but you’re just being obstinate

Like Topologist’s have all agreed on one defn of limit point

And it’s the one used in all the papers and it is not equivalent to yours

Which I’ve never seen anywhere

I mean even in like

Brin and Stuck

A ton of stuff goes wrong

Using your defn

[coral pivot]

Ok, I mean it doesn’t change any of my explanation to his problem cause I explicitly defined what I wanted S’ to be

[marsh forge]

Sure

It’s just not the limit point set

[gritty widget]

Given some co/homology $H_n(X)$ what would it mean to have the co/homology $H_n(X;\bZ)$ I don't really understand the notation if anyone could help!

[gentle osprey]

TheEternalKittenLord

[gritty widget]

Like I know what X,M would mean (group cohomology)

but Im not too sure about ;

co/homology as in cohomology or homology btw so ignore the fact that cohomology is referenced with H_n aswell

[marsh forge]

I’m confused

What’s your definition of H_n(X)

Normally that notation is not enough, we need a coefficient group like Z

Often times we just ignore this when the choice is obvious but it’s odd that you’d be comfortable with H_n(X) rather than H_n(X;G)

G just tells us the coefficients we are using

[marsh forge]

If X is a space it doesn’t really have a natural notion of group cohomology?

[compact bluff]

cohomology is usually given by H^n(X) as opposed to H_n(X)... the definition usually depends on which (co)homology theory you're working with. if you look around, people are usually using (integral) singular homology

I've personally found Cech cohomology (for compact spaces) to be useful, because I work with nasty spaces. You can also take advantage of the continuity of the Cech functor in this case.

[gritty widget]

Continua?

[compact bluff]

yea

[gritty widget]

I've started learning about them recently

[compact bluff]

they can get pretty wild

like when you first see the pseudoarc, you think its crazy (idk if you've ever read the paper "Drawing the pseudo-arc" by Lewis) there comes a time in your life when you realize that the pseudoarc is actually very well-behaved...

[gritty widget]

Like the circle of pseudo-arcs?

[compact bluff]

yea, also the generalizations to solenoids of pseudo-arcs

pseudo-solenoids, higher dimensional hereditarily indecomposable continua

ultra-arcs and ultracoproducts of continua

there's a whole world out there

[compact bluff]

no ive been studying model theory of continua for the past little while

Not really, I've been more into co-e.c. continua/infinitely generic/model-completeness type things

sure thing

[pastel thistle]

Any help with the counterexample!! Im going crazy

:(

[gritty widget]

Can anyone help me find two covering maps f and g s.t. g composed with f is not a covering map?

[gritty widget]

you got an example in #diff-geo-diff-top

[gritty widget]

there's a result that, under some hypotheses on the spaces, the composition of covering maps is a covering map. something like local path connectedness. try working with some spaces where those fail

[gritty widget]

What's a good example of such a space?

since i don't remember the exact hypotheses, i can't tell you. you can find it online probably

[gritty widget]

Okay, thanks!

[wide kayak]

one hypothesis that works (for the composition to be a covering map) is if both covering maps have finite sheets

you can find a counterexample involving an infinite sheeted cover

[gritty widget]

Oh, I did find one!

But thanks!

Actually, i need help with the Sierpinski triangle problem again.

So, I need to prove more rigorously that only three points are cut points, basically.

[wide kayak]

hmm

[gritty widget]

The problem basically asks for what all the homeomorphisms from the Sierpinski triangle to itself are, and the answer is S(3).

But I need to prove it formally.

[wide kayak]

that makes some sense, given the symmetry group of an equilateral triangle is S3

but I haven't done that problem before and don't know how

it feels intuitively clear that all six of the usual symmetries of an equilateral triangle also are symmetries of the sierpinski triangle

since it's made of all these essentially independent triangles

maybe you can argue that any further symmetry of sierpinsky would give an additional symmetry of the equilateral triangle, which is impossible

[gritty widget]

Hmm...

[gritty widget]

It's homeomorphisms.

Not symmetries.

Hence why connectedness plays a role.

homeomorphisms are bendy symmetries and the sierpinski triangle isn't very bendy

[wide kayak]

a self-homeomorphism is a symmetry

in a way

bendy symmetry, I like that

is it actually possible to have a self-homeomorphism that's not a rigid body motion?

(for a subspace of R^n)

probably yes but I can't think of one

maybe like... take (0,1)... and start sliding all the points within itself to the right

in a non uniform way

yeah that should work

[marsh forge]

There are tons of self homeomorphisms of a disk

That just like

Mess w stuff

But yes yours works too

[wide kayak]

does it work on a closed disk?

I guess yes

but I don't think a closed interval has any except reflection

[marsh forge]

Closed interval has a lot too

Almost all homeomorphisms are not rigid body

Rigid body is super strong

[wide kayak]

interesting

I figured there would be trouble with the endpoints

like, in [0,1], if you're gonna move 0, you'll have to replace it with something, and then it's not clear how you can keep the mapping one-to-one (if it's not a reflection)

on the other hand if you keep 0 where it is

you'll have to move nearby points away from it

and that could cause a discontinuity

it could be that you're only moving nearby points closer to it

but then I thought you'd have trouble over at 1

[marsh forge]

Here’s an exercise

The orientation preserving self homeomorphisms of [0,1] are exactly the ways to draw paths from the bottom left to the top right of a square

Such that the path satisfies the vertical and horizontal line tests

[wide kayak]

ohh

that's really nice

so the closer a point is to 0, the less it moves away, which saves continuity

[marsh forge]

Sure

[winged viper]

this may be a really stupid question--I'm trying to understand the topology generated by a family seminorms on a vector space, and I'm a bit confused. If I'm considering the topology generated by only a single seminorm, then wouldnt the initial topology on this seminorm only contain open sets that are unions of circles? (because every preimage of a seminorm is a union of circles)

[winged viper]

my bad, circles centered at the origin*

i guess, rather union of annuli about the origin

i imagine its supposed to be everything generated by the open balls, but i'm not sure what im missing

[winged viper]

^ realized im really stupid and that the definition of an initial topology on a TVS requires the addition/scalar mult functions to be continuous

[bright acorn]

I am currently reading the book "Complex Topological K-theory" by Efton Park. This is an introductory text and mostly deals with K-theory of compact hausdorff topological spaces. And from what I've noticed, a major theme in this book is that when we are doing K theory of compact hausdorff spaces, we can identify the monoid of isomorphism classes of complex vector bundles under whitney sum with a certain monoid of idempotent matrices with coefficients in C(X), so that we have at least a certain explicit description of the K-groups of X in terms of certain algebras of matrices. (I am attaching some screenshots from the book where the author discusses this)

So, I'd like to ask if this relationship between between the K groups of a topological space with certain algebra of matrices still holds up in more general settings. Can we still somehow describe the K groups of a more general topological space in terms of certain (possibly more complicated) operator algebras?

[compact bluff]

you can relax compactness to locally compact Hausdorff spaces, by noting that for LCH X, the inclusion map i : {p} -> X union {p}, the inclusion of p in the one-point compactification of X, induces a map on the K-groups, K^0(X cup {p}) -> K^0({p}) = Z. Then you define K^0(X) to be the kernel of this map. then K-theory will still give a cohomology theory, but now on LCH spaces

or by more generalized, maybe you mean operator K-theory? In this case, you can define K-theory for C*-algebras and get K-theory for these "noncommutative" spaces, in which case the operator K-theory of C_0(X) for an LCH space X will agree with the topological K-theory of X

[bright acorn]

Oh, I really meant for more general topological spaces, I didn't even know "operator K theory" was really a thing lol.

Yeah, the locally compact hausdorff case is as interesting as the compact hausdorff case, but this book seems to treat mainly K-theory for compact hausdorff spaces.

Which is fine I guess for an introductory book.

[compact bluff]

yea I'm not sure you can generalize further to get interesting/useful things... I mean if you do, you're probably either going to break the fact it is a cohomology theory, or you're going to break bott periodicity, which are really the only way to compute K-theory

or if there is, I haven't seen it.

though i'm not super well-versed in this stuff

[bright acorn]

https://mathoverflow.net/questions/156372/a-generalized-k-theory-via-generalized-idempotents

This is the best I could find for the time being.

[compact bluff]

You might be able to find some answers in this book: https://www.math.uvic.ca/~hemerson/NCG_2.0.pdf

[bright acorn]

Thanks a lot, I will take a look into it when I get the time

👀

At the time I am mostly interested in studying K-theory in order to understand better some stuff in differential topology related to cobordism and surgery theory.

But nothing too fancy

Yeah, I will probably have to study foliation theory at some point, since I probably will focus on trying to do geometric topology in the future.

But tbh, idk a lot of references for foliation theory.

Do you know any introductory references to the subject? that would be helpful too.

I will probably not read it soon, but I will keep these ideas in mind until I encounter orbifolds and foliations in the future.

Alright, thanks Ultra. I will try to search for it.

Ah, also Ultra. Do you know any references for studying elliptic operators?

With like, more of a focus on diff top and K-theory, I know K-theory is useful for studying elliptic operators and these also appear a lot in diff top.

So I was thinking studying these would be the next thing to do.

Alright, thanks again Ultra.

[honest narwhal]

Wait we doing k theory now

[broken nacelle]

how can we be sure of that line?

(I hope this is the right channel for such questions, it seems far too basic for what's usually discussed here lol)

[gritty widget]

L's definition as a supremum

if you could not find such c, then L would not be the least upper bound

[golden gust]

note that's not always true (e.g. 2 is the supremum of [0,1] union {2}) but it's true here because the set for which L is a supremum is an interval, I think

[deep hearth]

Given a compact space X, can we always embed X into a product space [0, 1]^A for some set A?

[wide kayak]

I don't think non-hausdorff can embed into hausdorff

[river granite]

yeah

the existence of such an embedding is equivalent to X being completely regular and Hausdorff

[gentle osprey]

Hausdorff

[river granite]

@vast estuary sure, take arctan() and compose with a linear function that sends (-pi/2,pi/2) to (a,b)

[gritty widget]

Equivalent to being completely regular

So any compact Hausdorff space embedds into [0, 1]^A for some A

Or more generally any LCH space

[deep hearth]

Got it! Thanks

[gritty widget]

why do we [X, hocolim Y_n]= colimit [X, Y_n]

if this was not a homotopy colimit

and if X was compact object in whatever category

we would just say

for compact objects

the corepresentable functor preserves filtered colimits

[gritty widget]

but i’m not sure what to say in this model category case

[still vessel]

Is this statement true for any topological space?

$$A \subset A'$$, where A is a subset of a topo. space and A' is the set of all limits points of A.

[gentle osprey]

FireLi0n

[tawny hazel]

Let A be the closed interval [0, 1]. Every point in (0, 1) is an interior point of A.
The points {0, 1} are, again, boundary points of A. A point x ∈ R is an exterior
point of A if x < 0 or x > 1.

Consider x ∈ (0, 1). Let r = min{x, 1 − x} > 0. If y ∈ B(x, r) =(x − r, x + r), then x − r <y < x + r. But r ≤ x, so
y > x − r ≥ x − x = 0,
and r ≤ 1 − x so
y < x + r < x + 1 − x = 1.

Therefore y ∈ (0, 1) so that B(x, r) ⊆ (0, 1). Hence x is an interior point of (0, 1).
Now suppose that x = 0, and fix r > 0. Then x ∈ B(x, r), and x /∈ (0, 1). Let
y = min{1/2 , r/2}. Since r > 0, it follows that y > 0. But y ≤ 1/2 <1 so we conclude that y ∈ (0, 1). Furthermore, |x − y| = y ≤r/2 < r, so y ∈ B(x, r). Therefore, for any r > 0, B(x, r) contains a point in (0, 1), and a point not in (0, 1). Hence x = 0 is a boundary point of (0, 1).

This is a question in my notes. The given set is A and this example proves its interior, exterior and boundary points. I understand the logic behind this, but could someone please help me understand how r is chose for proving x is an interior point
and how y is chosen for proving when x is a boundary point

[gritty widget]

No

It's not true for R

[still vessel]

why not?

[gritty widget]

Take any finite subset of R for example

Then A' is empty

If you want this property to hold, you need to take $A\cup A' =\overline{A}$

[gentle osprey]

Blitz

[still vessel]

but cant you just make a trivial sequence for every point making it a limit point? I feel like i am confusing mulitple things here

[gritty widget]

Which is the closure of A

This is generally different from derived set A' of A, that's the problem

[gritty widget]

No way

[still vessel]

this is actually what i am trying to understand, as its a bit confusing with the theorem that point is an element of the closue iff it has a sequence converging to it

[gritty widget]

Limit points are different from points of closure

[still vessel]

isnt a limit point a point which has a sequence that converges to it?

[gritty widget]

No

[still vessel]

what? i am utterly confused now

[gritty widget]

If you're not studying limit points then idk why you need to do this rn

[still vessel]

becouse i need it to prove a theorem

[gritty widget]

Limit point of a set A is a point x such that for any neighbourhood U of x there is a point y which is in U, A and is distinct from x

Being distinct from x is what makes A' and closure of A different

[still vessel]

yes, i know that, i just thought its a property not a defintion

[gritty widget]

It's a definition

For sequences we also have limit points, but they don't relate in any direct way to limit points of sets

For sequences those are just "all possible limits"

More precisely, all possible limits of subsequences

This can be generalized to filters and/or nets

Without trouble

[still vessel]

so the proof that i am following and trying to understand refrences a theorem which states that a limit point of a set is also a limit of a convergent sequence

isnt this the same as what i said?

This might be only for a metric space i guess

[gritty widget]

No

For metric spaces this would be $$A'\subseteq \overline{A}$$

[gentle osprey]

Blitz

[still vessel]

So the theorem states:
Let $S \subseteq X$, where X is a metric space. Then $z \in \overline{S}$ if and only if there is a sequence of points $(x_n)$ in S converging to z.

[gentle osprey]

FireLi0n

[still vessel]

doesnt this imply both $A' \subseteq \overline{A}$ and $\overline{A} \subseteq A'$ ?

[gentle osprey]

FireLi0n

[flint cove]

Is the property „a_n converges to a“ preserved under the sup of topologies?

I tried thinking about it in terms of the closure operator which we can construct explicitly for the sup, but my example doesn't quite seem to work (T_i = cofinite topology adjoining k\N as a closed set, and then hoping that the sup of the Ti is discrete)

Perhaps someone knows how the neighborhood base of a must look like in the sup of the topologies

[gritty widget]

The basis for topology here is $\bigcap_{i=1}^n U_i$ where $U_i$ are open in their respective topologies

[gentle osprey]

Blitz

Blitz

[gritty widget]

This proves the supremum preserves this property

[gritty widget]

This is possible because the intersection is finite

[gritty widget]

Assuming the previous basic open set contains a

[flint cove]

Ohh, yeah, that's pretty easy once I flip my mind back to think about the open sets

Thanks @gritty widget

This easily generalizes to filters then, since if all U_i contain some filter element F_i in F, so does their finite intersection

With the special case being the tails filter here

[gritty widget]

👍

[bright acorn]

Could you guys give me some motivations for introducing the notion of homotopy equivalence between morphisms of chain complexes of modules? I mean, I guess one of the reasons is that when two morphisms of complex are homotopy equivalent then they induce the same maps on homology. But like, how is this notion of homotopy equivalence related to the notion of homotopy equivalence of maps in the usual sense of topology?

[lunar yoke]

well for one thing any homotopy in spaces induces a chain homotopy between the singular chain complexes

you use this to show that singular (co)homology is homotopy invariant

If you do the intermediate steps Top -> sSet -> sAb -> Ch, then they all preserve their respective notion of homotopy, where this is only nontrivial for the last map

if you believe that simplicial things are topological, then Dold Kan, which says that the category of (bounded) chain complexes is equivalent to the category of simplicial abelian groups, also gives motivation to have some notion of homotopy

[bright acorn]

Wait, what is the category of simplicial abelian groups?

[lunar yoke]

Fun(Delta^op, Ab)

equivalently, the category of abelian group objects in sSet = Fun(Delta^op, Set)

[bright acorn]

Ah. So like, is this a general construction? If we take any category C, then it is "natural" to consider the category of contravariant functors from the usual simplicial category into C (I.e Fun(Delta^op, C))?

[lunar yoke]

yes, usually we call them presheaf categories

they pop up everywhere

of course not necessarily with domain Delta^op

but any small category

[bright acorn]

Ah, ok. I know about sheaves of abelian groups in the context of topology.

But idk about sheaves in category theory.

[lunar yoke]

E.g. Fun(C^op, Set) is the category of presheaves on C

[lunar yoke]

well to look at sheaves instead of presheaves you need some sort of notion of what it means that a collection of objects covers some other objects

i think you do that with sites

though they can tell you more about that in #algebraic-geometry

[bright acorn]

Alright, then. Thanks for the insights 👍

[marsh forge]

It is a generalization of it

Homotopy theory extends far beyond spaces

And really shows up anywhere you have some notion of things being “weakly equivalent”

So the idea is that like, chain homotopy to homology is more or less the right analogy to homotopy and pi_*

[fair idol]

Can somebody remind me how to introduce a cell structure on a genius two surface? I think there is a standard 4n-gon structure but can't find the details

[coral pivot]

and you keep generalizing for higher genus like this

[hollow harbor]

it's always these adjacent abab strings right?

i don't really have intuition for why that works

somehow each is like

a petal of the surface

if you think of the surface organized like a fidget spinner or something

yeah i think i see now

[fair idol]

The way I think about it is when you identify a_1 you turn one copy of b_1 into a loop, same with a_2 and b_2. I think these are where the two holes come from

Given a topological space X and a subgroup $H<\pi_1(X)$ how do I find the corresponding $X_H$ covering space?

I have a concrete example where $X=RP^2\times RP^2$ and $\pi_1(X) = Z_2\times Z_2$. Its subgroups are the identity ${0}$, itself $\pi_1(X)$, and $<(1,0)>$,$<(0,1)>$,$<(1,1)>$.

I have a hunch that the identity of X is the covering space corresponding to X as a subgroup of itself. Further the universal covering space is the one corresponding to the trivial subgroup. There must be some lattice structure here. Any hunches on how to calculate the corresponding connected covering spaces $X_{<(1,0)>},X_{<(0,1)>}$, and $X_{<(1,1)>}$?

I read something about orbits but really couldn't comprehend it.

[gentle osprey]

limbostar

[coral pivot]

Right so the galois corrospondence acts like this:

for a cover Y->X, for a subcover Y', you send it to aut(Y|Y') and for a subgroup H of aut(Y|X), you send it to Y/H

Y/H is the quotient space of the action of H on Y

i.e space of orbits

and you give it the quotient topology

(its like, a "fixed subcover of H" if you will, similar to something you might know from field theory)

[warped rover]

Is there no automorphism of a stone-cech compactification that fixes the original space?

oh yeah there is not due to density

[warped rover]

I thought those were automorphisms over the subspace?

Any automorphism that fixes the original space extends uniquely to that spaces closure

oh yeah

mb

I'll self-sully

Yeah that's a good point though, I should be careful about that

[gritty widget]

It's a neat result though. I'm assuming by "continuum hypothesis implies there is some" you mean there is a topological space which has multiple such automorphisms

[feral copper]

Hi! If I have a Kirby diagram of a 4-manifold X, it induces a surgery diagram on Y=(bdy)X.
But... what is this diagram? I mean, I understand that the 2-handles give a framed link, but what happens with the 1-handle unlink? Does it simply become a 0-framed unlink?

[feral copper]

The computations that I did work fine if I 0-frame that unlink, but idk in general...

[gritty widget]

Does $\mathbb{R}^n$ embedd into $SO(n+1)$? Or $SO(m)$ for some $m$.

[gentle osprey]

Blitz