point-set-topology

[prisma garnet]

except at one point

I dunno ig I just feel like that's cheating

coz you can't exactly do the same for other spaces embeded in high dimensional euclidean spaces

[hidden crag]

This is not really true as stated imo

Just because removing one point makes it homeo R3 doesn’t mean it looks about the same „except at one point“

Maybe I’m too pedantic about the phrasing here

[high hill]

S^n is the one point compactification of R^n in general is it?

I dont see whats wrong with viewing it as that then

cheating

[prisma garnet]

I just want an intuition about what S^3 looks like

I was uncomfortable with that intuition to begin with

[high hill]

well theres lots of ways right

parallel slices of 4d space are 3d space

[river granite]

there's the hopf fibration, you could look into that

[high hill]

if you can view higher dimensional objects as some product of lower ones, thats always great

idk if u can for Sn

[prisma garnet]

sure but that doesn't retain the topological datum of S^3

[prisma garnet]

isn't that too advanced?

hmm, prolly not right

it's just higher homotopy theory stuff so chapter 4

[river granite]

not really, you can write it explicitly and for the "fibration" bit you can just rephrase that in more elementary terms

no homotopy memes required

the construction works only for some S^n though

[prisma garnet]

ooo

[river granite]

particularly n=3

[prisma garnet]

interesting

I bet there's an expository paper on that

[river granite]

would explain further but I just woke up

[river granite]

many probably

the wiki article is a decent intro even

[tiny ridge]

I think of S^3 just as everything around me aka R^3 and ignore the point at infinity until necessary

[prisma garnet]

that's what I said!

[tiny ridge]

Why is this unsatisfactory?

[prisma garnet]

I dunno I think I want a more geometric intuition

coz that completely ignores the geometry of S^3

I know that's what topologists do but like

¯_(ツ)_/¯

[tiny ridge]

It's unclear to me what you want

The best visualization for a high dimensional object depends on what your end goal is

Because you'll not be able to see all of it at once unlike orientable surfaces living in 3D

Typically we have 10 different ways to visualise something and switch POV depending on what we want to focus on

[prisma garnet]

that makes sense

frankly I don't know what I want

[tiny ridge]

Why do you want to visualise S^3

[prisma garnet]

I'm just very uncomfortable visualizing it

coz this still doesn't make sense to me intuitively ig

I do in paper but like

[tiny ridge]

Take R^3, this is where you live. Take a donut inside it.

The edible kind will suffice

Throw away the donut ie look at the complement of the donut

What does this space look like? Can you imagine a "pillar" which goes through the donut hole?

[prisma garnet]

by pillar do you mean like

$D^2 \cross \bR$?

[gentle osprey]

DarQ

[tiny ridge]

Yep

[prisma garnet]

oh wait the piller looks like a torus too no?

coz of the one point compactification of \bR^3

[tiny ridge]

Yep!

The pillar goes through the removed donut hole and the top and bottom melts with everything in the exterior

[prisma garnet]

oh wow hmm

that's trippy 😵‍💫

[tiny ridge]

The following might help: Take a ball inscribing the donut. The complement of the donut = (complement of the ball) U (pillar)

Pillar is inside the ball

See if you can see that

The following observation might also help: if you take a donut or a solid torus S^1 x D^2 and attach a pillar D^2 x I to the donut hole like a "plug", you get a ball.

[high hill]

wait wait wait back to this

if u assume this, how do you know ur on a ball, not a donut

ig thats all to do with paying attention what happens at infinity

[tough hamlet]

for this I'd say

think of it as the boundary of the 4-cube [0,1]^4=[0,1]^2×[0,1]^2

like that's essentially what he's doing

[tiny ridge]

Heres some pictures

The final picture here, at the bottom right, is actually a solid torus with a point removed

The removed point is at infinity

What you’re seeing is a stereoprojection of a solid torus in the sphere, and since it contains infinity, the stereoprojection makes the bulk of it bloat out and cover most of your visible space (exterior of the ball)

Just like cartographic projection of the globe makes Antarctica bloat up and seem larger than everything else

[tiny ridge]

The donut with a point deleted looks fundamentally different from R^3, if that is what youre asking. See above. The plug is an extra handle.

The handle is small compared to most of the bulk because stereoprojection does dumb things but its still there

[high hill]

If you ignore the point at infinity

for S3

youre in R3

[tiny ridge]

Yes

[prisma garnet]

oh wait omg

I got it

[high hill]

without visiting the point at infinity, you dont know if ur on a donut or not?

[tiny ridge]

You do

[high hill]

🤨

[tiny ridge]

You’ll see a donut with a point deleted

[high hill]

hmm

thonk

The point at infinity for R3

doesnt correspond to a point on a donut

[tiny ridge]

It would

[high hill]

😵‍💫 uhh

[tiny ridge]

Just see the above pictures

[high hill]

ok lets stick to S2 R2

Im saying if you decided to glue you paper into a torus

without reaching the edges of the paper, you surely wont know if ur on a torus?

or could u

[tiny ridge]

Deleting a point doesnt mean the entire perimeter of the square is inaccessible

[high hill]

But im not deleting a point from the torus

[tiny ridge]

It just means, say, the vertices are inaccessible

But youre deleting a point from S^2 to get to R^2

[prisma garnet]

it's like a magnetic field

[high hill]

im taking our perspective as we are imagining ourselves in R2

[tiny ridge]

Why would you delete more than a point to see the torus then

[high hill]

No no...

[tiny ridge]

In R^2 the discussion is moot because theres no analogue of the solid torus decomposition of S^3 in S^2

[tiny ridge]

Yup

[prisma garnet]

like, if you think of S^3 as [0, 1]^3 quotiented by its boundary

the piller should be attached to itself like this sortta

except that point that's deaad in the middle

[tiny ridge]

Yes lol

Spot on

[prisma garnet]

and that's attached to the boundary point

thankyuu

[tiny ridge]

The central circle of the other solid torus is actually the line going through the center of the circular magnet

In R^3 it is a line, but in S^3 it is a circle

[prisma garnet]

I see

I can finally move on lmfao

thx again this was very helpful

[tiny ridge]

This says btw that the central circles of the two solid torii link once like a Hopf link

Like the LHS

If you started thickening the two circles in the Hopf link on the left radially outwards, you'd get two solid torii inflatable balloons with ever increasing radii

In the limit they'll abut each other's boundaries

That's when you've written S^3 as union of two solid torii

[high hill]

Ok so like. My point was:

• To visualize S3, you can just imagine yourself to be on R3 until you need to worry about the point at infinity

Now until you worry about the point at infinity, I was arguing you surely can't tell the difference between this being S3 or S1xS1xS1

[tiny ridge]

You can

[high hill]

How.

[tiny ridge]

S^3 \ pt is distinct from S^1 x S^1 x S^1 \ pt. These spaces have different fundamental group

First is 0, second is Z^3

[high hill]

But I'm not comparing this at all to S1xS1xS1 - {0}

[tiny ridge]

So what are you comparing?

[high hill]

I'm saying R3 looks like S1xS1xS1 - S1vS1vS1 (if im not mistaken)

[tiny ridge]

False.

You need to throw away a lot more

[high hill]

ok idk what it is - but it certainly isnt 1 point.

[tiny ridge]

You have to throw away three 2-torii.

Sure, any two n-dimensional manifolds will look the same if you throw away sufficiently many things from both

[high hill]

But well my overall point was - until you hit the point at infinity, this visualisation is indistinguishable from a lot of 3manifolds

and so - does it really help?

[tiny ridge]

Well, yes. Because you remember you threw away a point.

And not anything larger

There is a unique 3-manifold obtained from R^3 by attaching a point at infinity. That is the whole point. No pun intended.

[high hill]

Id argue its the other way round. Youre viewing yourself In R3, and then deciding what happens at infinity. Add a point to get S3.
Do something else to get something else

Your perspective is in R3

(because thats how youre 'visualising' this space)

[abstract saffron]

Alenxandroff compactification or smth, right?

[tiny ridge]

My point is for this particular problem of DraQ, you don't need to care about what happens at the point at infinity

[high hill]

In which case, I feel like his problem doesnt even care this is S3!

It could be R3

some other 3-fold, etc.

[tiny ridge]

R^3 is union of a solid torus and a solid torus minus a point.

No, not some other 3-fold

Just R^3

It's a matter of what is important for the given problem. You focus your attention on the essential features of the space depending on what you want to prove about the space

There is a "functorial" way of translating things in R^3 to things in S^3 because the one-point compactification is so canonical. Not all things about a general 3-fold can be deduced from visualising a chart inside it

[broken nacelle]

I found this paper on the hopf fibration but I'm not sure if I'm its target audience

should I look for something that's presented in a more standard way?

particularly because I meet most of those prereqs to a reasonable degree

except perhaps vector calculus lol

[solar mason]

Can check ‘nonlinear systems’ by khalil

[broken nacelle]

nonlinear systems?

that seems far from topology hurb

[solar mason]

Oh I read Hopf bifurcation, not fibration. Don’t mind me

[red yoke]

https://youtube.com/shorts/rJZFQvuQ69c?feature=share

[opaque scroll]

Guess it depends on what you want out of learning about the Hopf fibration.

From an abstract algebra perspective the hopf fibration is a transitive group action of SU(2) on the sphere with isotropy group U(1). From a topological point of view it's a circle bundle on S^2 with total space S^3 , or more generally a circle bundle on CP^n with total space S^2n+1.

More intuitively the hopf fibration is a way to cut S^3 into disjoint unit circles, that link together like toruses.

It can't hurt to have many different perspectives.

[broken nacelle]

I don't really know which perspective I want to know, I've just heard the term thrown around a few times and I'm curious what it means

so prolly nothing to through

a good expository paper would do if anyone knows one

[quiet thorn]

maybe something like https://math.uchicago.edu/~may/REU2020/REUPapers/Lin,Chenjia.pdf ?

[broken nacelle]

ooooo

yea I'll check this out

[red yoke]

This one explains the visualization
https://youtu.be/-lZTnEgbVvI

[tiny ridge]

@broken nacelle Symbolically, the Hopf fibration is very easy to write down. Consider S^3 in C^2 as the unit sphere |z|^2 + |w|^2 = 1 and S^2 as the Riemann sphere (complex numbers together with infinity). The map S^3 -> S^2 is (z, w) -> z/w

Think about the fibers etc to understand the full picture

For instance what are the preimages of the meridians in S^2?

[silk raft]

hey, I'm reading Hatchers notes on K-theory and I'm struggling a bit with something regarding clutching functions. Apparently I'm not alone because I found my exact question on stackexchange without answer: https://math.stackexchange.com/questions/859302/filling-the-details-of-a-construction-via-clutching-function-of-a-vector-bundle

anyone here able to enlighten me?

[prisma garnet]

I should prolly ask this in #diff-geo-diff-top lol

[tiny ridge]

Not 100% certain and havent thought about this deeply but this looks like the following: Take E -> X and H^n -> S^2. Pullback H^n by X x S^2 -> S^2 and call that L, this is a line bundle over X x S^2. Then pullback E along L -> X x S^2 -> X. This defines a bundle over L, hence over X x S^2.

This bundle seems to be [E, z^n] over X x S^2.

[fading vale]

I also have not thought about this deeply but intuitively for n = 1 this clutching should look like fixing the 0 section still and then rotating the fiber as you wind around it which should introduce a twist corresponding to tensoring with H

And then taking higher n should rotate multiple times so twist multiple times

[tiny ridge]

Yup. Except H and E are over different bases.

Pullback to correct for that

[fading vale]

Right

[tiny ridge]

In general E o F for two bundles over the same base can be described as E pulled back over F, yes?

We get a tensor product of the transition matrices

[fading vale]

What's this circ notation

[tiny ridge]

Tensor, sorry

[fading vale]

Ah

Yes I think so

[tiny ridge]

Cool, everything checks out then.

[ebon galleon]

Just a sanity check: If $i : X \to Y$ is a relative cell complex (i.e., $i$ is a transfinite composite of pushouts of coproducts of disk inclusions, $i_n : S^n \to D^n$, but $X$ needs not be empty), then $Y-X$ is Hausdorff with the subspace topology, right? I know that the quotient $Y/X$ is a cell complex, so it should follow from this, since the subspace topologies of $Y-X$ and $Y/X - [X]$ should agree in this case?

[gentle osprey]

Ryx Egg (?!?)

[unreal stratus]

(S^{n-1} -> D^n i assume you mean)

But seems fine to me

[ebon galleon]

Yeah that's what I meant, my bad. Thanks!

[austere dirge]

does a metric space have a contraction mapping if and only if it is simply connected

[tough hamlet]

constant map to a given point

[austere dirge]

oh you're so right

lol i can't believe i didn't consider that

[unreal stratus]

Contraction mappings needn't be actual contractions of the space in the sense of algebraic topology

Not sure where you got simple connectedness from (could be simply connected but not contractible)

[coarse night]

what were you guys discussing?

[grave maple]

I was reading Beyond Topology and stopped at this proposition:

I don't see how that is immediate. Suppose J = {1} and I = {1, 2}. I don't see how f_1 being initial implies that f_2 is initial.

I suspect there's a typo: they meant J > I, not J < I.

[ebon galleon]

yeah that seems fishy as written, I would assume they switched it? Not hugely familiar with the exact details tho

[grave maple]

It's amazing how an obvious typo like this makes it through.

[austere dirge]

I think I was accidentally only considering homotopies as examples in my head

[obtuse meteor]

if you have a contracting flow then I would believe you're contractible

a contracting flow being a map phi_t such that phi_t(x), phi_t(y) are closer together than x,y. And then phi_t . phi_s = phi_{t+s}

this formalizes the picture I imagine you have in your head

[tiny ridge]

That sounds wrong. Take the Reeb foliation on an annulus

[obtuse meteor]

There might be some long line weirdness

ooo fun

a counterexample to this has to be fun tho :)

[tiny ridge]

Just check though, I said the first thing that came to mind

[obtuse meteor]

Yeah I mean. I also think a counterexample would have to be non-compact or else my intuition is utterly broken

[tiny ridge]

I feel you can have a limit cycle and get closer and closer without contracting to a point

[obtuse meteor]

if you're semi-locally contractible and compact then that's enough for me to have a proof

[tiny ridge]

Maybe?

[obtuse meteor]

limit cycle?

[tiny ridge]

Suppose all the forward limit sets of the flowlines is a single circle

[obtuse meteor]

hm no. Bc phi_1 has a unique fixed point under iteration and everything gets dragged to it

[tiny ridge]

Yes, you're right. You can end up at two diametrically opposite points in the limit cycle after sufficient iteration

[obtuse meteor]

this isn't as strict as I wanted to say. There's some 0 < q_t < 1 dependent on t so that the distance phi_t(x), phi_t(y) is at most q_t * d(x,y)

[obtuse meteor]

Two would mean non-unique though :((

[obtuse meteor]

Here's the proof in this case right. Let p be the unique fixed point of phi_1 (compactness + contraction mapping theorem). Then let U be a semi-locally contractible neighborhood of p. Given any x, define t_x so that phi_{t_x}(x) lies in U.

t_x should depend continuously on x. Then (s,x) -> phi_{s * t_x}(x) should homotope you into U, then contract from U

[umbral panther]

Usually you assume complete when talking about contraction (weaker than compact)

[obtuse meteor]

sure

I think we only need complete here

I thought maybe I would wanna max out t_x based on a compactness argument but that's actually unnecessary

(notably complete would still rule out a Reeb foliation on an open annulus)

[gritty widget]

Can a homotopy change the domain of a function? \phi here is undefined at t=1, because it maps (x,s) onto (x,0), no?

[unreal stratus]

The next line tells you what phi_1 is anyway

[gritty widget]

So homotopies are able to change domains?

[obtuse meteor]

no it's just the proof is written badly

it should be phi_t(x,s) = (x,(1-t)s) for 0 <= t < 1 and phi_t(infty) = infty and phi_1(x,s) = infty

[gritty widget]

oh ok, thank you sm

[unreal stratus]

I mean this wouldn't fit with the definition

[gritty widget]

Yeah that’s why I was confused

[frank bolt]

Is "knot theory" just the word we use to describe the topological study of simple closed curves?

I like to sit in on the seminars from my school's geometry and topology group, and it was odd to me how often knots came up.

A few times now I've seen speakers talk about manifolds with knots as their boundaries. It seems to me that knot theorists care about knots as knots, and everybody else uses them as closed curves.

[umbral panther]

Knot usually means a simple closed curve in 3d, up to an appropriate notion of moving it around. Sometimes people call any kind of embedding a knot, but codimension 2 is really special. For example, Zeeman’s unknotting theorem says that a sphere embedded in another sphere in codimension at least 3 is PL isotopic to the standard embedding. This is an exotic use of “knot” but it is proving that there’s nothing there. It’s fairly common to include embedding S^2 in S^4 as part of knot theory, though.

[fading vale]

one reason they come up very often is that 3-dimensional manifolds can be constructed from knots via a process called surgery: you take a knot in the three sphere (regarded as the 1 point compactification of R^3), you take a small tube running around that knot to obtain a solid torus in S^3, you cut the tube out, and then you glue the tube back in along some (non trivial, if you want something interesting) diffeomorphism of its boundary, which is the usual torus.

There is a theorem that says that by choosing various knots and various diffeomorphisms of the torus you can actually obtain every nice (read: closed orientable connected, if that means anything to you) 3 manifold this way

so you can often reformulate certain questions about 3 manifolds as questions about knots and torus automorphisms, which is sometimes easier since knot theory is generally fairly combinatorial and explicit

I hope that is adequate justification for why topologists in general care about the study of knots as simple closed curves in R^3 (or S^3)

[frank bolt]

Definitely, that's a cool theorem! There are a lot of low-dimensional topology people at my school, so it makes sense now why knots keep appearing.

[frank bolt]

Heh, that is exotic. Is it purely due to their codimension that these objects are treated as knots? Can they be studied similarly to knots?

[umbral panther]

If you define n-knots as S^n embedded in S^n+2, there is a map from n-knots to n+1 knots. So they are directly connected. There equivalence relation on 1-knots of becoming the same as 2-knots is popular. In particular, I think a knot that becomes trivial as a 2-knot is called slice

Codimension is essential. Codimension 2 means that the the complement of the submanifold has a different fundamental group than the total space. (Same for codimension 1, but simpler)

[frank bolt]

there is a map from n-knots to n+1 knots
What does this mean?

[umbral panther]

It’s a specific construction called spinning

[frank bolt]

a knot that becomes trivial as a 2-knot is called slice
Hadn't thought about it this way :o

[umbral panther]

I’m not sure I’ve got that right

[fading vale]

here is an example of a fun open problem related to this: an automorphism of the torus can be specified by taking a longitudinal curve about it and a meridian about it (one around each factor of S^1), and gluing the torus to itself in such a way that the longitudinal curve wraps around a times and the meridian wraps around b times. the ratio b/a, called the slope, is enough to specify the resulting 3 manifold you will get when you do surgery on a knot K. So a surgery can be denoted S_r(K), where K is a knot in S^3, and r is the slope.

cosmetic surgery is the following conjecture: if S_r1(K) is homeomorphic, as an oriented manifold, to S_r2(K), where K is a non trivial knot in S^3, then r1 = r2.

This is, if true, a pretty nice thing to know, because it means that if we produce 3-manifolds via surgery (which is a very concrete way to produce them) we can be sure that they are different if we are getting them via different surgeries on the same knot

the current status of this problem is that we have reduced to the case where r1 and r2 are 1/n and -1/n for natural n, and its done using some pretty fancy machinery in modern low dimensional topology (its called heegaard floer homology which may or may not be what some of the people at your school are doing)

actually the 1/2, -1/2 and 1/4, -1/4 cases were resolved recently using very fun fancy methods in mathematical physics-y stuff called instanton floer homology

[frank bolt]

surgery on a knot = dehn twisting it?

[fading vale]

Yep

[frank bolt]

Huh, usually I think of surgery as cutting and gluing.

[fading vale]

Well the idea is you are cutting out a tube about a knot, twisting it, and gluing it back in

[frank bolt]

Ah, right

[frank bolt]

They are doing heegard floer theory!

I hear it constantly. No idea what it means.

[fading vale]

Cool stuff

it is basically a combinatorial cut and pasty way of describing a certain invariant that is in general very hard to compute and confusing

(floer homology is the invariant)

(the combinatorics is in terms of a heegaard splitting of the 3 manifold)

(hence, heegaard floer)

[frank bolt]

I see

[onyx raft]

Is it true that if you post-compose by a nullhomotopic map that your composition is nullhomotopic?

Since suppose f:X->Y is nullhomotopic i.e there exists a nullhomotopy H:X\times I-> Y s.t H(x,0)=f(x) and H(x,1)=c. Suppose h:Z->X is continuous then h\times id: Z\times I-> X\times I is continuous as well. So we have a nullhomotopy H\circ h\times id of f\circ h as H\circ h\times id (x,0)=H(h(x),0)=f(h(x)) and H\circ h\times id (x,1)=H(h(x),1)=c

I was trying to construct a non-homotopic map from T^2 to RP^2 so I constructed a map such that sends a certain path in T^2 to the loop generating pi_1(RP^2) and this was my lemma needed to verify it was not nullhomotopic (as otherwise pi_1(RP^2)=e which is wrong)

I did it by T^2->S^1->S^2->RP^2 first map sending T^2 to its first circle in the product then S^1 to the equator then the last map is the covering map

And the path q:I->T^2 is just q(0)=( (1,0),b_0) to q(1)=( (-1,0),b_0)

[trim chasm]

as a Warwick student spending all their time in the Zeeman building, I am very pro Zeeman

[tiny ridge]

???? Codimension 3 sounds so wrong. There's a stable range when this happens, due to Haefliger

Certainly two embeddings of S^n in S^(2n+1) are isotopic. One can pick a homotopy, look at the movie S^n x I -> S^(2n+1) x I, make it self transverse and cancel the self-intersections by Whitney trick

This gives a concordance which implies isotopy by the h-cobordism theorem

n>=2

[umbral panther]

PL is different. Actually, smooth is different and PL and topology are the same
Check out the introduction by Zeeman surveying related results. By the time he published, Smale has proved the h-cobordism theorem which is category free and proves something very close to unknotting, which he mentions. Exercise: prove Zeeman’s theorem from this statement (necessarily using the PL hypothesis)
https://www.jstor.org/stable/1970538

Codimension 3 means that the fundamental group doesn’t change, which allows Smale to apply the h-cobordism theorem

[novel ember]

how should i prove that the inclusion map from π_1(S^1) -> π_1(X) where X is a quotient space of S^2 with the poles identified to a point has trivial kernel

i made the CW complex but i cant figure out how to show that the inclusion map has trivial kernel

[red yoke]

Trivial kernel?

[novel ember]

it should have a trivial kernel

[umbral panther]

Describe the universal cover

Alternately, show it’s a retract

[novel ember]

i havent learned covering spaces yet

hatcher does svk before covering spaces

[umbral panther]

Then show is a retract

[tiny ridge]

Very cool stuff

[red yoke]

Oh I thought you meant all pairs of poles identified

[novel ember]

yeah nevermind i showed that the homotopy class of loops of X is entirely determined by S^1

[unreal stratus]

By poles identified, do you mean just like north and south pole identified

and embedding it as the equator

or smth

[obtuse meteor]

this is quite a cool result from heegaard floer

[fading vale]

i forget who it is due to

i think hanselman?

the 1/2 -1/2 and 1/4 -1/4 case is due to miller i believe

[gray edge]

does anyone have any decent resources on the induced topology on graphs? I just need something that gives basic definitions that I can read through and also cite for my thesis

[hidden crag]

Hatcher has quite a bit of graph stuff

[stoic eagle]

What sort of definitions are you looking for

[gray edge]

Basically, take a graph and replace each edge with a copy of [0,1]. Embed it into R^2.

[gray edge]

I have a copy I’ll take a look thanks!

[stoic eagle]

Do you need the embedding?

[gray edge]

I’m doing stuff with paths and I need to be able to identify distinct points on the graph and its edges yeah

[stoic eagle]

Well R2 isn’t going to work for all graphs right

[gray edge]

The graphs are really Kripke frames so it’ll work for these

[stoic eagle]

Alright
Anyways I join the recommendation on Hatcher, although I’m still not sure what you’re looking for really

[hidden crag]

I mean hatcher cares about them from a homotopic point of view

You can use them to construct coverings of S^1 v S^1 or to solve problems about free groups

(Graphs are wedges of circle up to homotopy theory and they’re a nice representative)

But yeah he does have the basic topologization of them via homeos of [0,1]

[gray edge]

That’s really all I need, lol

[novel ember]

whats the cardinality of π_1(R^2-Q^2)

i know its at least ℵ_1

[ebon galleon]

I imagine you mean at least 2^\aleph_0 = |R|

Unless you know what \aleph_1 is

[novel ember]

i thought \aleph_1 was |P(N)|=|R|

[ebon galleon]

You're right, continuum hypothesis is obviously true

[coarse night]

R² \ Q² is a manifold so countable π_1

[tough hamlet]

what

[ebon galleon]

what

[novel ember]

what

[ebon galleon]

You might be thinking of R^2 \ Z^2 my friend

[coarse night]

Ok yes Q² not closed

[novel ember]

wait

π_1(R^2-Z^2) is COUNTABLE?

[ebon galleon]

[coarse night]

it is

[novel ember]

on god

[coarse night]

manifolds have countable π_1 (try to prove)

[novel ember]

i dont know what a manifold is

[ebon galleon]

countably many generators me thinks?

[novel ember]

when i do differential geometry i will come back

[tiny ridge]

Same as countable group

[ebon galleon]

Yeah that's what I had in mind

[coarse night]

try to prove @ebon galleon

[ebon galleon]

what, that countably many generators ==> countable group? Just a cardinality argument

[coarse night]

yes

[hidden crag]

This shouldn’t be too surprising when you draw a picture

[ebon galleon]

||coproduct of sequences of length n for all naturals n - countable union of countable sets||

[novel ember]

it should feel uncountable considering π_1(R^2-Q^2) is uncountable

[ebon galleon]

idk the whole manifold part or justification that these actually are generators tho Will get there eventually

[hidden crag]

Compare Q^2 and Z^2

[ebon galleon]

Even though Q^2 is countable, there's a fundamental different between them topologically, especially in R^2

[hidden crag]

If you remove Z^2 you’re just removing the vertices of each 1x1 square

[tiny ridge]

There is a homeomorphism of R^2 taking Z^2 to Z

And then to N

[hidden crag]

Removing Q^2 is fucked

Draw a picture cat bread

[tiny ridge]

And R^2 \ N is heq to countable wedge of circles

So same for R^2 \ Z^2

[hidden crag]

Nice argument Ibsen

[ebon galleon]

now prove this part

idk how bad it actually is, but seems a bit rough

[tiny ridge]

One at a time move the points around

Homeo(R^2) acts transitively on R^2

Moreover the homeomorphism can be chosen such that the support is small enough

Then compose countably many homeomorphisms whose supports are eventually disjoint and go off to infinity

Small work but good work

[ebon galleon]

okay yeah that makes sense. Only thing left to check is that you can take R^2 \ Z to be the colimit

[novel ember]

ok but why cant we sequence all the points in Z^2, choose an arbitrary point in R^2-Z^2, and then choose loops for each n such that the loop correlated with each n only loops around that point and nothing else

i used an argument similar to this for Q^2

why wouldnt it work for Z^2

[tiny ridge]

Those give countably many loops

[coarse night]

or you can think of like this, R²\Z² looks like bunch of squares attached by their sides, fox a point and take a path from the fixed point to every vertex and collapse

[tiny ridge]

I agree they generate the fundamental group

But that proves the group is countable

[novel ember]

oh wait

[coarse night]

you'll get countable wadge of S¹ (just a graph)

[novel ember]

i forgot concatenation

[tiny ridge]

Yup, infinite words not allowed

Easy to forget

[novel ember]

wait what

[tiny ridge]

Blah blah blah @coarse night

[coarse night]

follow up: find homotopy type of { (x, y, z): one of x,y,z is an integer}

[tiny ridge]

That's disconnected

[coarse night]

ah, don't take complement then

[tiny ridge]

💡

[coarse night]

looks like infinite apartment building

[tiny ridge]

Yup

[hidden crag]

Ooh ONE is an integer

[tiny ridge]

I like the infinite 3d jailcell which is given by taking a sphere at every integral lattice point of R^3 and for nearest neighbour edges joining them by a tube

[hidden crag]

I was wondering why that’s disconnected in the original question

[novel ember]

ooh

interesting question that i have 0 idea how to tackle because i cant use svk

[coarse night]

editted

[coarse night]

it's about collapsing some contractible set inside the set.

[novel ember]

oh wait

you changed the problem

[ebon galleon]

"I like the infinite 3D jail cell"

[hidden crag]

The original one would still have trivial pi_1 anyway

SVK wouldn’t have helped

[novel ember]

would 1.3 of hatcher have helped

covering spaces

[coarse night]

[novel ember]

yeah i havent read much of hatcher yet

only read 0 1.1 and 1.2

[hidden crag]

With the part is misread that is

Im being confusing

[ebon galleon]

I had a stroke reading this

[unreal stratus]

Bleak

[hidden crag]

Ignore it

[frank bolt]

I learned about Heegaard splittings of 3 manifolds, and as an example, my professor drew this to mean a genus 1 Heegaard splitting of S^3. I'm confused how H and H' glue together

H and H' need to have the same boundary to glue, but it seems to me that they don't

[red yoke]

S³ = Solid torus ∪ solid torus

[frank bolt]

Errr wait, yeah. They're both solid tori, so they both have T^2 as their boundaries. But I don't see how their boundaries glue still.

[red yoke]

https://youtube.com/shorts/rJZFQvuQ69c?feature=share

[frank bolt]

I don't see how this is relevant 😅

[red yoke]

This is a torus ye

The inside is a solid torus

The outside is a solid torus

And the animation shows how they are both solid tori by flipping the space

Or if you don't want to move things around

You can start by picturing a solid torus

Take another solid torus and slice it into discs along longitudinal lines

And glue the boundary of those discs onto the first solid torus

The 0th disc glues to the center of the torus

The π/2 -th disc is attached slightly above that

Eventually the discs explode towards ∞

And the π-th disc is an annulus

Then it zooms back from under

And completes the cycle

Or alternatively still

Take S³ = ∂(B⁴) = ∂(B²)×B² ∪ B²×∂(B²) = S¹×B² ∪ B²×S¹.

@frank bolt

Or you could also see this detailed discussion
#point-set-topology message

[novel ember]

ok it turns out

[novel ember]

i described the wedge of countably many circles

rather than the hawaiian earring

[unreal stratus]

Any tips/resources for actually computing homotopy groups of simplicial sets?

I'm specifically trying to compute stuff with tensor products of simplicial commutative algebras and am not sure where to get started w computations of stuff

[frank bolt]

Thanks! Had to run to class, will check later

[languid patrol]

It’s almost always impossible to do directly

It really depends on what your space is and how it’s presented

This is very much like asking “any tips for how to compute homotopy groups?”

The answer being: it’s impossible in general, for nice spaces it depends

[unreal stratus]

Yup sure

Thank

[frank bolt]

Heegaard splittings of S^3

[prisma garnet]

hatcher keeps talking about "the orientable surface of genus g"

does that imply orientable surfaces of genus g are all homeomorphirc?

[umbral panther]

Yes

[prisma garnet]

[dusky vault]

Hi I'm working on this point set topology question:

I've gotten this far into the proof, and I'm trying fo find a basis element from the standard topology on R^n that contains p, and is inside of B_(d')(x, epsilon)

[coarse night]

What about a “smaller ball”?

[dusky vault]

I thought a smaller ball centered at x might work, but potentially it could not fit p inside of it

I was thinking we might have to use a ball centered at p, but it looks a lot more complicated

[coarse night]

You have the right idea

Just draw picture and show you can do it

[dusky vault]

It's not enough to show that the basis element contains the other, it acutally has to contain p as well right?

That's the part that makes it a little more difficult

I was thinking we take the min of the distances to each side w.r.t p as the radius of our ball

[heady skiff]

what do they mean by "generating" the germ at the point a? as in its a representative of the equivalence class it's in?

[queen prism]

what subject is this again?

I didn’t learn germs in my functional class

[heady skiff]

this is for my algebraic topology class but as a prereq we’re doing some stuff on metric spaces and topological spaces

[gritty widget]

yes

[heady skiff]

And the reading he assigned us is from zorich analysis

[hoary breach]

does look like they're taking about equivalence class

[trail charm]

i'm really confused by this definition of homotopy by hatcher

i've seen the definitions where you have H : X \times I \to Y, or another definition with a family of maps h_t : X \to Y, but i've never seen it in the sense of a family of maps f_t: I \to X

i can't really connect this concept to those in my head

[unreal stratus]

It's a homotopy of paths

I.e. maps I -> X

[trail charm]

oh

wait yeah

yeah haha yeah you're right

[unreal stratus]

So it's a special case of the 2nd thing you said

[trail charm]

oops

thanks

[unreal stratus]

Np

[heady skiff]

i'm confused about how they're using the term "open set" is the second to last line - are they referring to the ball definition of an open set or the empty set and X

[paper wedge]

what

wdym

for the non-hausdorff example?

the open sets are the elements in your topology ( by definition )

[queen prism]

in the trivial topology the only open sets are the empty set and the whole space

[paper wedge]

there are only two open sets here

the whole space and the emtpy set

so the complement of a point is not an open set

[heady skiff]

ye

[paper wedge]

balls are the basis of the metric space topology

this is not the topology ur dealing with here

[heady skiff]

oh yea i forgot that open set refers to the subsets exhibited by the topology

i got confused between the two right

thanks

[paper wedge]

np :d

[heady skiff]

so just to clarify, if we have a metric space X with a defined metric then the open sets induced by this metric are exactly the open sets that define the topology on X?

[ebon galleon]

yes?

The topology generated by the metric is the topology generated y the metric, yeah

[paper wedge]

re read the definitions my friendo

[languid patrol]

yes there is a unique topology generated by the metric, that is that topology. However if a topological space happens to have a (possibly unrelated) metric on it it doesn't mean the topology and the metric topology are the same

[umbral panther]

Did you ever get an answer to this?
You proved that it was at least the cardinality of R, right? But you wanted an upper bound?
There are only R many continuous maps from the interval to R2, so that’s an upper bound

[novel ember]

my original proof was incorrect but i can fix it easily

turns out that i constructed the wedge sum of countably infinite circles

[umbral panther]

That only gives a countable lower bound

[tidal lynx]

If f: [a, b] -> R, g: [c, d] -> R are continuous functions and S = {(x, f(x) | x in [a, b]}, T = {(x, g(x)) | x in [c, d]}, we have that S and T are homeomorphic as subsets of R^2, right?

Would suck if that were false tbh

[ebon galleon]

Should be I think

[tidal lynx]

Ok yeah, S is homeomorphic to [a, b] by f (restrict the codomain to S), [a, b] is homeo to [c, d], and…

right

[ebon galleon]

Yeah that's what I was gonna suggest

since [a,b] is compact and R is hausorff

[a,b] --> S is injective, by above is closed, so a homeomorphism on it's image

and [a,b] and [c,d] are clearly homeo

[tidal lynx]

wait what’s the result you’re using

[ebon galleon]

If X is compact and Y is hausdorff, f:X --> Y is a closed map

[tidal lynx]

Oh shit

I think I just assumed that f had an inverse

Cuz it feels that way

Like say you have a string that starts out at [a, b] in R^2

then you perform some continuous operation to move it to S

Like all of topology says going from S back to [a, b] in a continuous manner is not guaranteed, but like, with our string, can’t we just reverse what we did on our way there

[ebon galleon]

intuitively, yeah. But have fun proving it with that 🙃

actually I think this is true regardless since it's just the projection onto the X component

[ebon galleon]

but this is an easy way to see it in this case

[tidal lynx]

Oh yeaaa

[ebon galleon]

yeah I forgot about that too lmao

[tidal lynx]

Wait

yeah lol

hahaha

wait on your solution

[ebon galleon]

https://math.stackexchange.com/a/3535566 exact justification for this btw

[tidal lynx]

don’t we need that f is an open map to show that f^-1 is continuous?

we got that f is a closed mapping, so what

[ebon galleon]

open + continuous bijection <==> homeomorphism <==> closed + continuous bijection

Remember: a function is also continuous iff the inverse image of a closed set is closed!

[tidal lynx]

ok the first <= and second => is easy

hmm I actually forgot the first =>

if f: X -> Y is an open, continuous bijection, and if V is an open set in Y…

f^-1(V) is

a set

that’s for sure

[ebon galleon]

f^{-1}V is open by continuity

[tidal lynx]

wait

true

Wait

I’m dumb

I’m supposed to look at

[ebon galleon]

If U is open in X, show that (f^{-1})^{-1} U (the inverse image of the inverse function) is open in Y

[tidal lynx]

(f^{-1})^{-1}(U), where U is open in X

and of course that set is open in Y, since it is just f(U) and f is an open mapping

ok

and the second <=

[tidal lynx]

oh this

ok

[ebon galleon]

yeah same idea for that one

[ebon galleon]

so because of this it's closed. It's clearly injective: (x,f(x)) = (y,f(y)) clearly implies x=y, so when you restrict the codomain to the image (which is S, the graph), it's a closed, continuous bijection. Hence a homeomorphism

[tidal lynx]

Yeah

Nice

And then if we consider maps with domain (a, b)

Of course the same result holds

But we no longer have this high power solution using the closed map lemma or wtv

[tidal lynx]

You can just provide the inverse (projection) as before, but I’m wondering if there’s a solution in similar taste as this one

[ebon galleon]

Nothing fancy comes to mind, no. The closed map lemma I had does not generalize to, say, locally compact Hausdorff spaces (even nice ones like Euclidean space!)

There's no general "open map" lemma along those lines (in more specific cases there are: for instance, the open mapping theorem in complex analysis, which states that any analytic function on an open domain is an open map)

[tidal lynx]

Right

[ebon galleon]

(functional analysis has one as well afaik)

[ebon galleon]

But yeah this is a better solution for general terms

[tidal lynx]

Wait you don’t really need the “universal mapping property of the product topology” right.

Like you can just check that those two morphism equalities are true

I guess maybe if you define the product topology using the universal property, then yea

[ebon galleon]

universal mapping property of product is just that maps into a product space correspond bijectively with maps into each component. Equivalently: A map (f,g):A --> XxY is continuous iff f:A --> X and g:A --> Y are

Just an easy way of checking continuity into product spaces

[tidal lynx]

Oh I know that the universal mapping property says that “for every top. space Z and family of morphisms into… there exists a unique morphism… such that … commutes” but I guess that is an easy consequence

[unreal stratus]

It is useful in practice to define maps so yeah you do sort of need it

[echo oyster]

Is (R - Z) connected ?

I can write it like Union of connected (p,q) where p,q are integers, but they have no common points

Is it sufficient to say it is not connected set

[hidden crag]

Yes

[echo oyster]

Ok, but what if we take product of (R-Z) x (R-Z)

It's like taking out vertical & horizontal lines from R2 right ? So it's also not connected

[opaque scroll]

Yes, in general if X is not connected, then XxY won't be either (unless Y is empty)

[echo oyster]

I see, thank you

[tribal palm]

people always be talking about point set topology, what other kinds of topology exists?

[unreal stratus]

Algebraic topology and geometric topology and various things

[hoary breach]

(see ⁠get-advanced-access to use this channel) point-set topology (topological spaces), algebraic topology (homotopy/homology/etc.), geometric topology, anime

[hoary breach]

anime

[tribal palm]

fascinating

thank

[trim chasm]

in my books, point-set topology is analysis

[tiny ridge]

Disagree with this take

[hidden crag]

Disagreeing as well

[languid patrol]

I'm going to agree with this take, I will propose a channel merge to the mods

[hidden crag]

Fuck off HAHA

[unreal stratus]

lol

maybe point set is to topology what analysis is to diff geo

No

[red yoke]

Point set is to topology what Lebesgue stuff is to analysis

[ripe lagoon]

So, i tried to venture into Topology today and got knocked out on second page itself.

So, they started with discussing Euler's formula for polyhedrons that is v+f-e = 2. I had known the formula earlier as well but i hadn't known that these polyhedrons are well defined and not any random shape.
So, They said that there are two conditions that a Polyhedron P would satisfy.
I got the first one but the second one stunted me.

It says - Any loop on P which is made up of straight lines segments(not necessarily edges) separates P into two pieces.

I can't wrap my head around this. They gave example of a polyhedron to show this and, yeah, that one didn't satisfy Euler's formula.

But, then, i think that a simple cube fails that second condition as well. If i take all the edges of one face, they form a loop but they don't separate P in two pieces.

What am i missing here? I appreciate any useful input whatsoever. Thank you.

[winged viper]

Why wouldn’t a cube be separated into two pieces? The loop you described should split the cube into: 1) top face and 2) everything minus the top face

[ripe lagoon]

Umm... I think that because of the example they had given to demonstrate the second rule.
Let me send the picture. Perhaps, i'm missing the distinction.

The figure on top right.

They said that loop shown doesn't divide figure into two pieces.

I apologise for the quality of picture. My hands shake a bit so i can't hold my phone static.

[winged viper]

I think #2 refers to just the surface of the polyhedron

Like it’s saying that a loop will split the surface of P into two components

So in figure 1.3, imagine the polyhedron is hollow

And cut along the loop

[ripe lagoon]

Yeah, they said that we'll rather be referring to the surface while using polyhedron, and not the solid.

[winged viper]

Okay cool

But do you see how the loop you described does actually separate the cube into two pieces

[ripe lagoon]

Umm... not really. I mean then how does loop separate in fig 1.3?

[winged viper]

It doesn’t separate in fig 1.3

It’s still one piece

[ripe lagoon]

A side question - by surface, do i imagine it like a plane having area, or like a loop with perimeter ?

[red yoke]

The surface of a cube are its 6 faces

Solid faces

[winged viper]

The surface is like a plane having area, hence the name “surface area”

[red yoke]

The square divides them into one face and five faces

[ripe lagoon]

Okay.

Wait... so, in example 1.3 - it's like a circular(just for eg) tube which is cut at one cross-section but it's still a whole thing. You can pick it up. Is it like that?

[winged viper]

Yep

[ripe lagoon]

I mean... pick it up by holding at any place and it'll get up as whole.

[winged viper]

When you cut it (and unfold it a bit) you get a cylinder essentially

[ripe lagoon]

While in cube, we have to pick both part separately?

[ripe lagoon]

Yeah. Yeah.

[winged viper]

Yes

[ripe lagoon]

Oh. Okay. I get it. Thank you very much.

[unreal stratus]

Hm, here's a weird question

View $S^n = (\mathbb R^n)^+$ (one point compactification); there's a natural action of $\Sigma_2$ on $\mathbb R^n \times \mathbb R^n$ which extends to one on $S^n \land S^n = (\mathbb R^n \times \mathbb R^n)^+ \cong S^{2n}$. How can we think about $S^{2n}/\Sigma_2$?

[gentle osprey]

potato

[unreal stratus]

For n=1, I'm pretty sure by cut and pasting that we just get D^2

I mean, we can model the 2-sphere as I^2 with boundary collapsed to a point, and then this action just folds along the diagonal

I guess in general you can do a similar thing and write like

$ S^{2n}/\Sigma_2 \cong { ((a_1,\dots,a_n), (b_1,\dots,b_n)) \in I^n \times I^n \mid \sum_i a_i \le \sum_i b_i}/{\sim}$ where $\sim$ crushes what's left of $\partial (I^2 \times I^2)$

[gentle osprey]

potato

[unreal stratus]

Why I'm interested is because it seems to follow from some result that (at least on homology) we should get S^{2n} for n even and smth contractible for n odd

[languid patrol]

One model should just send $(x_1, \dots, x_n, x_{n+1}, \dots, x_{2n}, x_{2n+1}) \to (x_{n+1}, \dots, x_{2n}, x_1, \dots, x_n, x_{2n + 1})$

[gentle osprey]

Topos_Theory_E-Girl

[unreal stratus]

Agreed

Though I'd write that with x0 being fixed aha

But yes

[languid patrol]

do you mean on rational homology?

[unreal stratus]

Lol you're good, the original source was indeed using rational homology

Do you think it's far worse in the non-rational case? I mean in the rational case they used the fact homotopy coinvariants and strict ones coincide and everything works out nicely

well like the action on the space gives you an action on cohomology which changes sign of a generator

[languid patrol]

I'm not sure, I was just guessing that this has to do with the resulting manifold being orientable or not: the point being just some dumb thing about the parity of the number of simple reflections that you need determining whether or not the switch is orientation preserving or reversing

[unreal stratus]

Okay sure

[languid patrol]

So when it's orientation reversing the quotient has top cohomology Z/2 and no other cohomology groups

which is why everything vanishes rationally

[unreal stratus]

Oh okay sure

lol

[languid patrol]

haven't calculated anything yet though so this is just my intuition

[unreal stratus]

Yeah so the full context was that if you have a (orientable) n-manifold $M$, let $W$ be the trivial n-bundle and let $\Sigma_2$ act on $W \oplus W (= M \times \mathbb R^{2n})$ in the obvious way, what does $(W \oplus W)^+_{\Sigma_2}$ look like

[gentle osprey]

potato

[unreal stratus]

And at least on rational cohomology this follows from applying Thom iso and the coinvariants playing nicely with rational cohomology

[languid patrol]

yep that makes sense!

[unreal stratus]

But my line of attack for thinking of it more generally was noting that $(W \oplus W)^+ \cong M^+ \land (S^n \land S^n)$ and I'm pretty sure it should work out that $(W \oplus W)^+{\Sigma_2} \cong M^+ \land (S^n \land S^n){\Sigma_2}$ lol

[gentle osprey]

potato

[unreal stratus]

Working with this gives the right thing for n=1 at least lol

😭

[languid patrol]

I guess if you were curious about computing this with integral coefficients you can just use meyer vietoris

where one open set is a neighborhood of the singular locus (which is I guess an n-sphere?) and the other open set can be computed with the usual spectral sequence

[unreal stratus]

Hm okay sure (thank)

[languid patrol]

what do the wedges mean in what you're writing?

[unreal stratus]

Smash

[languid patrol]

is that smash?

okay

Yes I think I agree that all of this comes down to just computing it over a point, since the whole space is just a fibration

over M

[unreal stratus]

Ye hm

Well

What led me to it was just the possibility that that could have some nice form, like a disk or a sphere depending on n but ye

I will think further lol

thank

[languid patrol]

sure!

[tiny ridge]

Any orientation reversing Z/2 action on the even dimensional sphere is homotopic to the standard Z/2 action

Quotient being RP

This is because an orientation reversing homeo of S^2n is homotopic to the antipodal map

Which is an easy exercise

[umbral panther]

Any orientation reversing map is homotopic to the standard orientation reversing map. But if it’s an order two map, it’s not homotopic through involutions. It might have fixed points and a path though involutions won’t change the fixed points

[tiny ridge]

Oh fair

If the action is free the quotient is still a homotopy RP but I suppose this isn't the case here

The action on S^2n is induced from the flip map on S^n x S^n after crushing S^n v S^n?

(S^n x S^n)/flip seems to me like a S^n-bundle over RP^n

[unreal stratus]

I'm thinking of it as permuting coordinates on R^n x R^n and extending by a point

[tiny ridge]

The diagonal is fixed and it's normal neighborhood is TS^n. S^n x S^n is the fiberwise one point compactification of TS^n

Quotient by Z/2 on the base

[unreal stratus]

Oh okay nice you're right with this yeah

[tiny ridge]

Oh no need to quotient on the base it just pointwise fixes the base

It acts by multiplication by -1 on the antidiagonal though

(x, -x) -> (-x, x)

[unreal stratus]

Oh lol homotopy RP would make sense

because that would account for dependence on n's parity & rational behaviour

[tiny ridge]

Trying hard to understand (S^n x S^n)/flip

Has to be a bundle over RP^n

[unreal stratus]

Sure hm

[tiny ridge]

Looks like a higher dimensional Klein bottle c'mon

[unreal stratus]

This is just funny cause I have some computation I have to do and the first step is computing the Fp homology of the space I mentioned here and this seems harder than I'd like

hm

Maybe it is a bit easier if we just care about Fp homology but I thought some geometric intuition would help but perhaps that's too naive lol

Feels like it should be relatively tangible

I mean it's just spheres and Z/2

[tiny ridge]

S^n x S^n fibers over the antidiagonal copy (-x, x) of S^n with fibers being S^n x {x} right

The flip map doesn't preserve the fibers. Need better fibers

I know (S^3 x S^3)/flip is SO(4)

This is an accident because S^3 is a group

[languid patrol]

I think the Meyer vietoris computation should work out very easily, it's S^n (fixed point locus) glued to RP^{n-1} (actually unorderd configurations of 2 points in R^n) along a little neighborhood also homotopy equivalent to RP^{n-1}

[unreal stratus]

Okay sure very nice

I will have a proper go at that then uwu

thank

[tiny ridge]

So basically it's a homology RP^n

Is it also a homotopy RP^n

That would be surprising

[unreal stratus]

Well for homology of this we can just do uh theorem on Fp homology of symmetric powers lol

but overkill

Not sure how to incorporate the smashing with M+ into the final computation tho (assuming homology of M+ is known lol)

hm

[languid patrol]

Isn't your space just the sum of two n-dimensional trivial bundles modulo a purely fiberwise action?

[unreal stratus]

True lol so a nice fibration

[languid patrol]

products are the nicest fibrations

[unreal stratus]

Actually no

Cause of the compactification right

[languid patrol]

Are you compactifying all of the fibers to the same point?

[unreal stratus]

M needn't be compact in my example, unfortunately

[languid patrol]

Oh wait

[unreal stratus]

Yeah I mean compactification as a space

Bleak

[languid patrol]

youre compactifying the entire total space with a point?

[unreal stratus]

Like thom space

Ye

[languid patrol]

I see

yeah that is worse

[unreal stratus]

So like we have a smash not a product

Ye

Hey ho

Life is pain

[prisma garnet]

is {0} u {1/n for all n} a CW-complex?

and is the suspension of any CW-complex a CW-complex?

I would try to prove that by myself but hatcher doesn't define CW-complexes properly

[ebon galleon]

0 dimensional CW complexes should be discrete afaik

[prisma garnet]

does discrete mean finite?

[ebon galleon]

like discrete topology: everything is open

so it's just like a collection of points

[prisma garnet]

ah

hmm

then no

wait {1/n for all n} is tho

[ebon galleon]

yeah

That's just a countable discrete space

[prisma garnet]

and the suspension of that is homotopy equivalent to the suspension of {0} u {1/n for all n}

[hidden crag]

Yes

[prisma garnet]

right?

[ebon galleon]

sure, but something that's h.e. to a CW complex needs not be a CW complex itself

In fact, broadly working with stuff that does have the homotopy type of a CW complex (read: is h.e. to a CW complex) is in some sense a convenient category of spaces for algebraic topology (whereas just working with CW complexes themselves is not)

[prisma garnet]

sounds hype

[hidden crag]

Or use weak equivalences

[ebon galleon]

( i do not know any details of what I just sent >.< )

[hidden crag]

[ebon galleon]

but Timo that's not cartesian closed

[hidden crag]

Surely this category doesn’t suck

[ebon galleon]

well

weak equivalence is a relative term I guess

If we are considering cgwh spaces to begin with it turns out nicely

[hidden crag]

[hidden crag]

I was talking in a topological sense

Not even sure if we’re making the same joke at this point

Lmao

[ebon galleon]

yeah ik the joke is that it's all of Top lmao

But we can fix that by restricting what we mean by "space"

[prisma garnet]

wait how do you actually prove this

I should fix my pointset

[limber wren]

(potentially dumb question) if you have continuous injections X -> Y and Y -> X are X and Y homeomorphic (kindof like Schroder-Bernstein but for continuous maps)

[tough hamlet]

no

take X to be R and Y to be (0,1)u(1,2)

[ebon galleon]

That's what I had in mind yeah

[limber wren]

oh yeah that's easy, thanks!

[tough hamlet]

np

[limber wren]

lol a random thought that popped into my head, next time I'll give it 2 seconds thought

[unreal stratus]

Also like, note this is still false even if we take them to be embeddings (lol same example works)

Lol said I wondered what happens with finite sets but then they're just bijections, wlog identity map after relabelling, and then it's easy anyway lol

[umbral panther]

X = {1/n} cup {0}
X is not homotopy equivalent to a CW complex. Its suspension is the Hawaiian earrings, which are not homotopy equivalent to a CW.

[prisma garnet]

no it's not

its reduced suspension is the hawaiian earrings

I know this coz I'm trying to prove pi_1(SX) is countable

[ebon galleon]

Huh, for some reason thought the Hawaiian earrings were a CW complex

[unreal stratus]

suspension usually means reduced

[prisma garnet]

bruh

[tidal lynx]

More on this https://mathoverflow.net/questions/1058/when-does-cantor-bernstein-hold

[umbral panther]

Ok, that has countable fundamental group, so it’s not obvious that it’s not homotopy equivalent to a CW complex, but it isn’t

[prisma garnet]

I see

why tho

[umbral panther]

If it were homotopy equivalent to a CW complex, it would have to be the wedge of infinitely many circles. So it would admit a map to that space. But it’s compact, so its image would have finitely generated fundamental group, so not be big enough

(You don’t need to know its fundamental group, only that it’s not finitely generated, so a CW model wouldn’t be compact)

[unreal stratus]

Actually lol that is interesting

i wonder about examples where X isn't a cw complex but the (unreduced or reduced) suspension is

both up to homotopy if you wish

I imagine there are boring extremes

[tiny ridge]

You don't need to take suspensions to argue {1/n} U 0 doesn't have the homotopy type of a CW complex

[unreal stratus]

I know lol

[tiny ridge]

There's a map from Z to this which is an iso in pi_n

Take the Whitehead continuum W in S^3. S^3/W feels like it's not a CW complex but if you cross this with R maybe it is?

Ipso facto good chance suspension is?

[umbral panther]

I think that if a space isn’t a CW complex, then its suspension isn’t, either. There are probably spaces that aren’t homotopy equivalent to CW complexes whose suspensions are contractible and thus homotopy equivalent to

[unreal stratus]

For the 2nd line that was my idea too yeah

[tiny ridge]

Yes ok R^3/W x R is homeomorphic to R^4

I bet S^3/W is not a CW complex

The quotient map is probably a weak homotopy equivalence I think.

I don't know if the suspension is well behaved

[umbral panther]

I was thinking something like taking an acyclic space whose suspension is contractible and forming something like the earrings. Then its suspension would be something like the comb, which is contractible

[tiny ridge]

Maybe. It seems possible

How do you prove R^3/W x R is homeo to R^4? Probably show somehow that W x R in R^3 x R is a shrinkable set

If thats true it seems plausible SW in S(S^3) = S^4 is also shrinkable

Bing shrinking says cellular sets are shrinkable

I don't see how to write W x R as intersection of cells

Probably have to look at Freedman's notes

There might be problems at the two suspension points

Maybe the spirals become like logarithmic spirals there, winding infinitely often around the two poles

But there's no diffeomorphism of the sphere which straightens a foliation by logarithmic spiral loxodromes to the longitudinal foliation, is there?

Oh but we only care about homeomorphism. So maybe?

The flow of the discontinuous vector field (-y/sqrt(x^2+y^2), x/sqrt(x^2+y^2)) on R^2 is a self homeomorphism fixing the origin after all

That straightens an infinitely winding spiral arc from the origin to the x axis

Conjecture: Unreduced suspension of S^3/W is homeomorphic to S^4

[unreal stratus]

Conjecture: topology is hard

Lol I am feel dumbs

Oh yeah I had a question lol

I keep seeing this situation where you have a finite group G act on a space and the homotopy coinvariants = strict coinvariants rationally

And I assume same for invariants / fixed points lol

is there any like theorem/fact this is a consequence of hm

I mean I assume somewhere it's cause of BG having torsion homology for n >= 1

from what i've heard

[tiny ridge]

There's a fibration X -> (X x EG)/G -> BG

If G is a finite group, H*(BG, Q) is 0

Because of what you said I suppose

[umbral panther]

Fixed points are pretty different. Take a free action on the sphere. The fixed points are empty, but if the dimension is odd, the action is probably trivial on homology

[tiny ridge]

That's strange. The E^2 page is just the row H*(X, Q)?

What am I doing wrong? I seem to get that rational cohomology completely ignores the fact that you're taking finite quotients

Is this correct

[unreal stratus]

Hm how does this prove what I wanted exactly?

[tiny ridge]

I'm just running the spectral sequence on the above

[unreal stratus]

Like sure this lets us compute X's homology with serre spectral sequence or smth but uh

how about X/G

[umbral panther]

The spectral sequence for this? You need to use twisted coefficients. You need homology of BG, ie, coinvsriants

[tiny ridge]

Ah yes thanks

There we go

Homology of BG with rational coefficients is trivial but

So I'm still confused

It seems to say H*(X_G, Q) = H*(X, Q) which is most certainly wrong

[unreal stratus]

well i know you you can just show H*(X/G; Q) = H*(X;Q)^G

from somewhere lol

iirc

[tiny ridge]

Yeah I'm missing the G invariants above

Why

[unreal stratus]

is it invariants yeah i think

yeah

[tiny ridge]

But what's happening in the specseq

I guess H^0(BG; H^n(X; Q)) is the only E^2 row

Am I mistaken to make this equate to H^n(X; Q)

Yes, because of the twist certainly?

This should be H^n(X; Q)^G

0th group cohomology of a G-module M is M^G

This is the mistake I was making

[tiny ridge]

So that gives you the proof

[unreal stratus]

Poggers

nice thank

[tiny ridge]

Proof for X_G though

Homotopy quotient

Here's a way to show now that the cohomology groups of X/G is the same as X_G. There's a map (X x EG)/G -> X/G. Totally not a fibration

But fibers are rationally acyclic

[unreal stratus]

Nice

hm

[tiny ridge]

If you have a map between CW complexes with rationally acyclic fibers it's an iso in rational homology

[umbral panther]

This works, but I think it’s overkill and better to by hand

[tiny ridge]

Follows from derived pushing forward the constant sheaf and using the Grothendieck SS

[umbral panther]

Or you could call it the leray ss

[tiny ridge]

Yeah there is probably a transfer argument I'm not seeing

Anyway this gives cohomology of X_G = cohomology of X/G = G invariants of cohomology of X (rationally) in one shot

Also the map X_G -> X/G is an iso in rational cohomology so it's a rational homotopy equivalence I suppose

Maybe some fundamental group assumption

[unreal stratus]

https://mathoverflow.net/questions/120351/equivariant-cohomology-for-actions-with-finite-stabilizers found this, nice

recommends leray too

[tiny ridge]

It's the same proof as the one I wrote just more careful

[umbral panther]

Compose projection with transfer and you get multiplication by |G|. Therefore the cohomology of the quotient is a summand of the invariant cohomology. More specifically, the composition in the other direction is the sum of the action of each individual group element, this the invariants

These formal properties of the transfer are pretty much exactly the statement we’re trying to prove

[tiny ridge]

What is the transfer map if G is not a covering action?

[umbral panther]

Yeah, the whole point is that it exists

[tiny ridge]

Why

[umbral panther]

I mean, the difficult part is that it exists

[tiny ridge]

Well what's the proof

Or where can I see it

[umbral panther]

Choose a triangulation that is equivariant. Construct it in each simples

[tiny ridge]

I'm not convinced that works without more details. Certainly works for covering actions

Some triangles might have nontrivial stabilizers

Need to multiply by stuff

[umbral panther]

Yeah, you have to subdivide

[unreal stratus]

You guys inspire me lol

[broken nacelle]

[limber wren]

Nice, thanks for the reference friend!

[tidal lynx]

And you could also ask the similar question “For what structures are bijective homomorphisms automatically isomorphisms?” Again topological spaces famously don’t satisfy this, but most of our favorite algebraic structures do.

[narrow cairn]

im stuck on a problem: show that given a surjective map q from X to Y, if Y satisfies the universal property of the quotient topology, then Y has the quotient topology

all ive shown so far is that q must be continuous

[ebon galleon]

Let Y' be the set Y with the quotient topology under q. You know that Y' also has that universal property, and you'd like to show that Y = Y'

In other words, you'd like to find a homeomorphism between them (hopefully using the the fact that they both have the universal property!)

[narrow cairn]

hmm, i tried using q^-1 of q in the universal property but the inverse order is wrong, for a surjection its the other way around

[ebon galleon]

But notice: You have the map q: X --> Y. This map satisfies the condition of the universal property (check if you're unsure!) on Y', so this should imply that there is a map f : Y' --> Y that fits in

(i.e., so f q = q, by the universal property)

[narrow cairn]

oh shit i see so the id map is continuous and so is its inverse

im so dumb lmao

thanks

[ebon galleon]

Well

Why is f the identity map?

A priori, there can be a map f with f q = q, but f is not the identity. What (purely set-theoretic) property implies f = identity?

[narrow cairn]

doesnt matter, the identity works so since the identity is a homeo they are eq

ofc just reverse Y and Y' here to prove its a homeo

[ebon galleon]

Oh true lol, uniqueness gives that huh

I was thinking surjectivity but yeah (since surjective <==> right-cancellable)

[ebon galleon]

And yeah, just swap the roles of Y and Y' to get the other direction

[tidal lynx]

do people actually care about the lower/upper limit topology of R, the K-topology of R, and the finite complement topology? Or are they just examples to get topology students’ hands dirty

[ebon galleon]

Stuff like that could be useful counterexamples to keep in mind

But I don't think people particularly care much about them

[tidal lynx]

Like, I remember in one of the first sections of Munkres he introduces them and shows that one is stricter finer than another, two of them are incomparable, etc.

But on the other hand… who asked?

[ebon galleon]

I mean they're good for understanding which properties are different

For instance, R with cocountable topology has the property that limits have unique sequences but it is not Hausdorff. I believe a (countably?) infinite set with cofinite is T_1 (frechet) but not sober (conversely, it's set of points (Locales shit, dont worry about it) is sober but not T_1)

But for, say, algebraic topology purposes? Who cares lmao

[narrow cairn]

what spaces does algtop use

besides like R and stuff

[ebon galleon]

Uhh I have not properly taken an alg top course but they are largely interested in stuff called "CW complexes" (more generally, relative CW complexes, perhaps). Essentially, spaces which can be formed by "gluing" together copies of (closed) disks together, of arbitrary (and in increasing order) dimension. So for instance, picture you have a closed unit disk (in R^2), and you "glue" together the boundary, the circle, to a single point. If you picture it curving into 3D so you can do that, you'll get a sphere.

And so for a variety of reasons, CW complexes have turned out to be important in algtop (some complicated ofc, but the general idea is that most spaces of interest can be constructed as (or are homotopy equivalent to) CW complexes)

Also, unit interval and unit cubes (again, of arbitrary dimension). Reason being: Homotopy 🙃

now someone will come in and tell me that I am wrong

[umbral panther]

People care about the cofinite topology as a special case of the Zariski topology in algebraic geometry, but not really in topology

[ebon galleon]

fair, Zariski topology is not generally Hausdorff (or T_1) even, is it

(also of interest: stone spaces, but this of more interest to logicians I believe)

[quasi forum]

I have a quick question. If X is a finite space with some topology on it, would it be vacuously limit point compact since there are no infinite sets in X to even look at?

[trim chasm]

topologists really like R^n and things modelled on it

[coarse night]

yes

finite spaces are anyway compact

hence compactness is regarded as "finiteness" in Top

[gritty widget]

I understand the open-set definition of a topological space (as in, what each of the requirements mean), however, I can't really get any intuition for why it is the way it is. My doubts are:

1. How does this have anything to do with the closeness of two points ? I assume that two points A and B are close only if they appear "together" in all elements of T (as in, there is no element of T which has A but not B nor vice versa) - however, this still doesn't really help with the intuition.

2. Why do we require any arbitrary union of the elements of T to be in T, but not do the same with intersections ?

3. Why do the null set and the set on which the topology is made, have to be part of T ?

4. Why are the elements of T called open sets ? I assume they are some kind of generalized open intervals on R, but why ? And why do they have to be open, and not closed ?

Any intuition for why this definition is the way it is will help - even if it is not as abstract, and just a "special case" of the definition. At least knowing what this definition generalizes still helps.

[tough hamlet]

Do you know about metric spaces?

[gritty widget]

Yes, I know about metric spaces.

[tough hamlet]

and you know how the metric gives us a topology?

[gritty widget]

Unfortunately, no.

[languid patrol]

It’s a weaker notion of closeness, topology is more qualitative

It’s more like “nearness”

Intuitively two points are closer if there are “more neighborhoods” containing the two of them

[gritty widget]

I searched how metric spaces induce a topology (via open-balls), and I understand that particular aspect now.

[red yoke]

In a lot of contexts you don't really talk about nearness of just two points, you talk about some property of the space as a whole

And open sets generalize the notion of "localness"

[gritty widget]

How so ? Why not closed sets too ?

[gritty widget]

I see, so we can say that, in certain topologies, A is closer to B than it is to C, if A and B appear more often "together".

[red yoke]

In a lot of topologies it doesn't really make sense to talk about two points being "closer" or "further" by looking at their topology

E.g. all topologies of metric spaces

Cuz we've abstracted beyond distance to "localness"

[red yoke]

As an example recall that continuity for metric spaces is

For any x in domain and open ball Bε centered at f(x), there exists an open ball Bδ centered at x such that f(Bδ) ⊂ Bε.
For topological spaces this generalizes to
For any x in the domain and open subset N containing f(x), there exists an open subset M containing x such that f(M) ⊂ N.
Or more succinctly,
The preimage of every open subset is open.

[gritty widget]

I see. I guess I will have to think about it, but it does make a little more sense now.

Okay, now I realise that I have never asked myself why these "open-balls" in metric spaces have to be open in the first place - why does it matter whether the distance between the center and any arbitrary point is smaller than epsilon (the ball is open), or smaller than or equal to epsilon (the ball is "closed") ?

[red yoke]

To generalize the real line

The reason an open set is a useful notion is that it doesn't just represent "localness" at one point - it represents "localness" for every point in the open set

Consider the open interval (0, 1). For any real 0 < x < 1, there is some ε > 0 such that (x-ε, x+ε) ⊂ (0, 1)

So (0, 1) can be interpreted as "some region around x" for any x ∈ (0, 1)

Ig this also answers your 3rd question

[red yoke]

Requiring <ε instead of ≤ε aligns with this notion, so we define these to be the "open" balls

[red yoke]

An example is defining differentiable functions Ω ⊂ R^m → R^n. Since an open Ω represents a region around x for any x ∈ Ω, we can easily talk about the derivative at x.

[hollow sail]

If we have a bounded sequence that diverges, in finite dimensions... Does it have necessarily at least 2 limit points?

[red yoke]

In R^n it can have 1 limit point

[hollow sail]

Can you give me an example?

bounded

[red yoke]

a_n = 0

[hollow sail]

Ah yeah i forget to mention that if it doesnt converge

fixed it

[hollow sail]

Now?

[red yoke]

Then yes in R^n

[hollow sail]

Amazing thank you

yeah its easy to prove

[red yoke]

Also #real-complex-analysis exists

[hollow sail]

Yeah sorry

[gritty widget]

Oooooooh, I think the open set idea makes sense now.

All I have to figure out now is why do we require to have the intersection (and also union) of any two members of a topology, inside that topology.

Is it so that we get "all the possible open sets" we can form starting from some "base" open sets ?