point-set-topology

[rancid umbra]

yes

u wanna see the next part tho

[coarse night]

wanna see mine

[rancid umbra]

lol yea

anything but what im doing

[coarse night]

are you aware of the simplex algo?

[rancid umbra]

nope

[coarse night]

there was an assignment in part B that we had to solve

usig simplex

1 page forming the inequalities

[coarse night]

algo for solving linear optimization problems with linear inequality constraints

to solve it you make tables and look at the possible optimization opportunities

if you find one, you take it and repeat

which underneath is a gaussinan elemination type thing with some checks

so we had to compute a 4×11 matrix for each table

and we had to make 6 fucking tables

of 4×11

[rancid umbra]

dude

im laughing so hard lmao

why

[coarse night]

along with 4 extra cols total 15col

that's not the end

wait there's more

and it was a part of a 4 part problem where it was only the part 1

then you have to change some shit and do the whole fucking thing again

and the whole question carries 4 marks

means the process I just explained only gives you 1

out of 20

20 more to go

[rancid umbra]

i would simply refuse to do that assignment

[coarse night]

only solved p1

and then gave up

[rancid umbra]

thats the only right choice

[coarse night]

what an asshole of a prof

[rancid umbra]

fr

thats horrible

[coarse night]

and gave 23hrs only

[rancid umbra]

was he going to check them?

[coarse night]

will deduce marks for plagiarism for sure

bcz who tf can solve all of them,

I copied some

obviously

[rancid umbra]

can u not just apply the linear programming skills that you're learning and write a script that prints out the steps? or is it more complicated than that

[coarse night]

there are programs that do it

but he has specifically mentioned the method

not just simplex but dual simplex/2phase simplex

don't think there are specific implementation to that

also I'm too lazy to make my own for an exam

FUCK IT

[rancid umbra]

cry

[coarse night]

[rancid umbra]

next part of my problem is to prove that M is not a 2-manifold and idk how to do that

i give up

[empty grove]

What is M?

[rancid umbra]

from here

its a whole convoluted mess with the implicit function theorem and permuted graphs and shit and i dont care enough to try and get the points for this anymore lol. i cant see the motivation behind this example either

[empty grove]

[rancid umbra]

lol

just gonna drop this image as proof with the caption "its too pointy bro"

[empty grove]

Oh once you have an image you can solve it

The way to use too pointiness is to compute derivative as you approach point

[rancid umbra]

thats what one of the hints was saying but i do not really get it

[empty grove]

First try proving that L is not a manifold

Like L the letter

[rancid umbra]

whats a manifold

fuck loll

[sweet wing]

L is a manifold

[rancid umbra]

no ik

[sweet wing]

it is not a smooth manifold under like your usual embedding to R^2

[empty grove]

Ok I mean in with the standard smooth structure

[rancid umbra]

no but like

im just being sarcastic

[sweet wing]

standard smooth structure

[empty grove]

Of R²

[sweet wing]

implying a smooth structure is there alr

[empty grove]

[sweet wing]

yea you're looking at a immersed/embedded manifold

[empty grove]

manifold

bonk

[rancid umbra]

whats ur guy's preferred def of a manifold

[empty grove]

You should tell yours lmao

[rancid umbra]

its rather long

and unwieldy

[empty grove]

Whatever my prof gave

[sweet wing]

usually it's like

[empty grove]

set + atlas

[sweet wing]

hausdorff paracompact second countable and each point has a nbhd homeo to R^n

[empty grove]

That's just topological manifold though

[rancid umbra]

frantically opens wikipedia

[sweet wing]

yea

smooth manifold then transition maps need to be

smooth

[rancid umbra]

u guys are speaking another language lmao

[empty grove]

Speak your language then

🔫

[sweet wing]

or meme

manifold is some LRS whose functions at some point is isomorphic to that of R^n

insert continuous/differentiable/smooth/analytic functions R^n

[rancid umbra]

$M \subseteq\mathbb{R}^n$ is a $C^r$ $k$-manifold without boundary provided that for each point $p\in M$ there is an open subset $U$ of $\mathbb{R}^k$, a relatively open neighborhood $V\ni p$ of $M$, and a map $\alpha\in C^r(U,V)$ with continuous inverse and rank of $D\alpha(x)=k$ for each $x\in U$.

[gentle osprey]

c squared

[empty grove]

Nice

Then show L as subset of R² is not manifold

🙈

[rancid umbra]

uhhh

[rancid umbra]

let M = ([0,infty) x {0}) U ({0} x [0,infty)). then consider the point p = (0,0). let U be an open subset of R and V a relatively open neighborhood of p, and let f be a C^1 map from U to V with continuous inverse and rank Df(x) = 1 for each x in U.

and ur saying to look at partial derivatives of f

[empty grove]

Yes

M is union

Not product

Half x axis ∪ half y axis

Also L is a C^0 manifold

It's not a C^r manifold for r ≥ 1

[rancid umbra]

a C^0 1 manifold?

[empty grove]

I guess the definition you gave assumes that r ≥ 1

[rancid umbra]

yes

C^0 doesnt mean differentiable

[empty grove]

C^0 1-manifold would just be locally homeomorphic to R

Ye

L is homeomorphic to I

[rancid umbra]

how is it a manifold if i cant take the derivative

[sweet wing]

$(M,\mathcal O_M)$ is a [adjective] manifold when the stalks are isomorphic to the stalks of $(\mbb R^n,\mathcal O)$ for some $n$ ($\mathcal O$ here ring of [adjective] functions)

[gentle osprey]

ari 亲

[empty grove]

Topological manifold, not differential manifold

[sweet wing]

there are non smooth manifolds

consider

[empty grove]

Ari shut up

[rancid umbra]

oh

[empty grove]

[sweet wing]

graph of any continuous function

[rancid umbra]

wot in the stonks

[empty grove]

For topological manifolds you remove that last condition on derivatives

Then alpha just becomes a homeomorphism

[sweet wing]

yup

[rancid umbra]

okay, cool

[rancid umbra]

so uh

i feel like im not getting how to take the partial of f. like, Df(x) is a 2 by 1 column vector

[empty grove]

Yes, you compute the partials at p by computing partials around p and using continuity

Should be 1x2 vector I think

[rancid umbra]

no im going from R to R^2

thats two by one, right?

[empty grove]

But your definition has it the other way

Like both approaches will give the same result

But if you go the other way

You have an extra application of inverse function theorem or something

[rancid umbra]

i dont think so, right?

[empty grove]

To infer that partials even exist

Oh ok nvm

I misread

[empty grove]

To elaborate on this

Approach p along one of the axes

And you can actually compute one of the 2 partials on those points

[rancid umbra]

how, i feel like everything is backwards

like, im just in some open subset U of R

[empty grove]

Ok converge to the preimage of p from both sides

Lul

[rancid umbra]

am i supposed to be looking at f^{-1}

[empty grove]

Yes

[rancid umbra]

eff

why no makey sense

[empty grove]

F

So you have f^-1 p in U

[rancid umbra]

yes

and i wanted to do like, a line parametrized by t

[empty grove]

This may be assumed to be an interval without loss

[rancid umbra]

coming in from the left

[empty grove]

See how?

Yes

[rancid umbra]

so just like, g(t) = (t,0) as t --> 0

[empty grove]

Ye ye

We get there

[rancid umbra]

and you look at f^{-1} o g

[empty grove]

First gotta prove stuff about this map f

[empty grove]

Do you see this

[rancid umbra]

yes

[empty grove]

So the point divides the interval into 2 parts

Prove that one part maps to the x axis under f

And the other to the y axis

[rancid umbra]

f^{-1} : V \ {p} --> U\ {f^{-1}(p)} should be continuous

[empty grove]

Yes

And also surjective

[rancid umbra]

and injective

[empty grove]

So a connected component maps to a connected component

[rancid umbra]

oh fuck wait i didnt even realize

[empty grove]

And we have exactly 2 of those in each space

[rancid umbra]

wait but how is

how is the origin

even making sense here

[empty grove]

I don't get it

[rancid umbra]

wait nvm ignore me

[empty grove]

So without loss (see how) we may assume that left part of interval is mapping to y axis and right part to x axis

This isn't too important to do

But now you can talk about partials of f on the left part

Can you say anything about them

[rancid umbra]

they should be scalar multiples of each other because of the rank condition right?

but thats not whats important from what we just showed

[empty grove]

Yeah but you can say more

One of them is trivial to compute explicitly

[rancid umbra]

i wanna say one should be zero

[empty grove]

Yes

Which one

[rancid umbra]

so from the left part of U, the first partial is zero

[empty grove]

This isn't true actually

[empty grove]

Perfect

So the same on the other side

[rancid umbra]

wait how tho

this is just pure intution

[empty grove]

Lol

First coordinate function of f is constant on the left part

[rancid umbra]

ohhh

alr

this is gonna get wonky tho

[empty grove]

How wonky

[rancid umbra]

like super wonky like the derivative wont be continuous wonky

[empty grove]

Lol

[rancid umbra]

lol

[empty grove]

It can be continuous

But approaching from one side said first entry is 0

From other side second entry is 0

So rank 0

[rancid umbra]

um

so were using continuity of Df right

[empty grove]

Yes

[rancid umbra]

so Df(x) should have rank one in some neighborhood of p always

[empty grove]

Yes

At each point x in the chosen neighborhood

[rancid umbra]

shit. wait. so Df(x) = [Df_1(x) Df_2(x)]^T

[empty grove]

Yes

[rancid umbra]

and we just saw that Df_1(x) = 0 on left

[empty grove]

Because f is C¹

[rancid umbra]

Df_2(x) = 0 on right

[empty grove]

Yep

[rancid umbra]

so Df(x) = [0 0]^T

for all x

[empty grove]

Yes

No

Not for all x

For p

Or f inverse p

[rancid umbra]

where did we start zooming in on p

[empty grove]

We are approaching p from both sides

To find derivative at p

[rancid umbra]

and we can find Df(p) as the limit of partials

[empty grove]

Yes

[rancid umbra]

okay okay i think i got it

thank you so much man

[empty grove]

[drifting sundial]

just realised there's a small issue with my solution. the qn only assumes that Y is connected, but the result of 7-5 requires that Y be path-connected

[rancid umbra]

have not struggled like that in a minute . appreciate you walking me through that

[empty grove]

We want to show that given [f], equivalence class of a loop at x, [Ff] = [Gf]. You can do this by showing that Ff is path homotopic to g Ff g', where ' denotes the inverse path, and g is the loop at F(x) given by g(t) = H(x, t), H a homotopy from F to G, and use the fact that pi_1(Y) is abelian to get g Ff g' ≈ Ff

Might have to take H from G to F instead, I think this is right but you'll know once you work it out

[stone cipher]

https://math.stackexchange.com/questions/1354987/homotopic-maps-between-connected-spaces-inducing-the-same-homomorphism-between-t

[drifting sundial]

ah nice

[stone cipher]

Thx!

[stone cipher]

Indeed the lemma 7.45 kills the problem

[gritty widget]

is it hard to extend continous functions from subspaces to continous functions from the whole space in general

I can only see it in some pretty specific instances but can't really find an example where it's impossible

well I just thought of an example actually where it's impossible

the only continuous functions from R with standard topoogy to R with lower limit toplogy are constant

but the subspace (-infinity,-1]U[1,infinity) continuously maps to R with lower limit toplogy in a non constant way

and that couldn't possibly be extended

I'm also assuming is a closed subspace, cause I could do R{0} to R where negatives map to -1 and positives mao to 1

and since 0 is in the closure it's sort of getting an unfair deal

my real reason for wondering this is to work with continuous functions from compact hausdorff spaces to C

which seems like it should be super easy 😐😐

when the subset is a retract of the whole space I think I can do it by taking a deformation retract of the image of the boundary down to a point

[empty grove]

If you are in a normal space, then all functions from a closed subset to R (or a closed interval in R) extend to the whole space by Tietze extension theorem, and all compact hausdorff spaces are normal

[gritty widget]

and then making some dyadic filtration on the domain

[gritty widget]

yeah I was thinking normal for my dyadic filtration idea

[empty grove]

You should be able to infer the same for functions to C, by viewing C as RxR then breaking functions apart

[gritty widget]

yup

well I didn't know tietze extension theroem 😔

I would still like to do without that

I just need to somehow drop the requirement that the subspace is a retract

[empty grove]

The retraction thing isn't needed for Tietze, unless I am misinterpreting what you are saying

[gritty widget]

no, I'm just placing stronger conditions than the tietze theorem

I can extend maps as long as the domain is a retract of a normal space

like subspace

[empty grove]

right

by precomposing by retraction

[gritty widget]

yeah

😐

so basically when it's obvious I can do it 🤯

[empty grove]

so you want to weaken this hypothesis? Removing this would prove Tietze extension for compact Hausdorff, that can't be too much weaker than the whole theorem itself

[gritty widget]

well the retract thing still works for normal spaces

but my end goal is compact hausdorff spaces

[empty grove]

[gritty widget]

also I can assume it's connected but I don't see how that would help

oh right my dyadic filtration idea I forgot

basically my idea is

take the boundary of the subspace and watch where it maps in R

then do a little deformation retract in R down to a point

and put a nice little dyadic filtration of increasing neighborhoods around the subspace in the domain

like let's say the neighborhoods are U_d for dyadic rationals D

[empty grove]

dyadic filtration meaning increasing sequence of open interval with rational endpoints having powers of 2 in the denominators?

wait

ye idk what it is

[gritty widget]

so I can assign a real number to every point in X based on the inf d such that x in U_d, and I can map U_d\(U_(lower d) to the image of the deformation of the boundary at time d

I'm saying it in a really weird way

the thing is

let's say at time 1/2 or whatever, my boundary has been contracted down to [0,1] in R

and I have my thin "circle" of points that correspond to the real number 1/2

circle in the domain

I don't know how to distribute the points in the circle along the interval [0,1]

I can just say that I want them all to map there

so it doesn't help 😔

[empty grove]

you are taking pre image here?

I don't see why the image of the boundary is connected in R

[gritty widget]

it's not alwayd

wait it's gonna have an empty interior

wait no it's not

[empty grove]

That's not necessary

[gritty widget]

yeah I was just giving an example where it's [0,1]

either way you can deform it down to a point

and I'm using my dyadic filtration to get a continuous function from X to R such that on my subspace it's 0 and it gradually gets up to 1 as you move out

very similar to urysohn lemma

[empty grove]

If you already know urysohn why not just use tietze

[gritty widget]

cause idk tietze 😭

[empty grove]

tietze is easy from urysohn

[gritty widget]

it is?

let me thinky winky

[empty grove]

It is a bit tricky, but the proof is very short

[gritty widget]

ok so I have some wild function from subspace to R

and all urysohn gives me is a map where the subspace is 0

and some other closed subset that I don't care about maps to 1

[empty grove]

It is easier to do the [0,1] case

to prove tietze

then deduce R

[gritty widget]

yeah

ok

[empty grove]

Hint is, given a function f: X → [-1,1], find a function X → [-1/3,1/3] that is always within 2/3 of f

[gritty widget]

just compose it with the shrinky right?

[empty grove]

wait did I mess up

[gritty widget]

idk

I have to go to class 😔😔😔

[empty grove]

Wait

[gritty widget]

what

[empty grove]

oh ok

but I did mess up

[gritty widget]

you have 6 minutes

[empty grove]

A is closed subset of X

and we have f: A → [-1,1]

and we want to approximate it by a function X → [-1/3,1/3]

which is within 2/3 of f on A

[gritty widget]

ok

[empty grove]

that's a mouthful

[gritty widget]

just compose it with the shrink

[empty grove]

but that's not a function from X

it is still from A

[gritty widget]

yes

and I don't see where urysohn comes in

I just had a slight feeling that the techniques to prove urysohn might be useful again

[empty grove]

you separate the preimages of [1/3,1] and [-1,-1/3]

This gives a function on X which is 1/3 wherever f is > 1/3

is -1/3 when f < -1/3

and somewhere in between when f somewhere in between

So it gives a global approximation of f

now you subtract this from f, you get a function f_1 to [-2/3,2/3]

do the same with dividing this into 3 parts

you get a sequence of function g_0, g_1, ...

where g_i is approximation of f_i

and their sum uniformly converges to a function g because its bounded by geometric series pointwise

and the limit is continuous because uniform convergence

[gritty widget]

will read after class

[empty grove]

sure

[gritty widget]

class is out an hour early 😌

[gritty widget]

wait so are you saying to use ursyohn to separate those sets with values -1/3 and 1/3 as opposed to the usual 0 and 1

[empty grove]

Less than and greater than those, not just those values

[gritty widget]

well I mean you're taking the subset of A that maps to >1/3 and <-1/3 and using urysohn on those, where you have a function that is constant 1/3 on the first set and constant -1/3 on the second?

[empty grove]

Yes

And between the 2 everywhere else

[gritty widget]

yeah ok

I'm gonna draw this real quick

ok so

this seems a little strange

you have f:A->[0,1]

and you have your separation function U:X->[-1/3,1/3]

[empty grove]

[-1,1] in this case

[gritty widget]

oh yeah oops

so when you do f-u

that seems like a random ass function 😭

[empty grove]

Yes, but it is an approximation of f

Within 2/3 of f

And with range 1/3 of f

[gritty widget]

yes

[empty grove]

So when you repeat this process with f-u to get u_2

And then u_3 and so on

Each time range is multiplied by constant less than 1

Range of f-summation u_i

So f-summation u_i converges to 0 uniformly

So summation u_i converge to f uniformly

On A

[gritty widget]

ok wait

[empty grove]

And summation u_i is continuous because uniform limit of continuous functions

[gritty widget]

what do i do to get U2

[empty grove]

You have f-u1 with codomain [-2/3, 2/3]. Now apply urysohn to ≤ -2/9, ≥ 2/9

Splitting the range into 3 parts at each step

f - u1 - u2 then has range [-4/9, 4/9]

Each time, range becomes 2/3 of previous

[gritty widget]

I see it now

but idk how anyone on this planet thought of that

something just isn't clicking with me today

[empty grove]

Lol yeah it's creative

But it is pretty short of you write it down, assuming you don't mess up like me lul

[empty grove]

You just have to write this step for general n

[gritty widget]

it seems kind of genius

[empty grove]

And then everything follows from geometric series and uniform limit being continuous

Yeah

[gritty widget]

like you're just using the special urysohn functions to approximate f

that's nice

and each one carries along the rest of x with it

[empty grove]

Yep

[gritty widget]

that makes me feel warm and fuzzy

haha

[empty grove]

[gritty widget]

I was doing all of this for an algebra problem

[gritty widget]

ugh ew the only way I can see to extend this to R is really gross

basically I go back through all the details and show that if A originally mapped within (-1,1) then so will sum(U_n)

so basically the endpoints have no preimage

[empty grove]

There is a very elegant way

[gritty widget]

in that case I just use homeomorphisms from R to (-1,1)

[empty grove]

R is homeomorphic to (-1,1)

oh nice

[gritty widget]

butttt

[empty grove]

[gritty widget]

when you extend f

you get a function whose codomain is [-1,1]

and the boring part is showing it's range is actually within (-1,1)

[empty grove]

I think that wasn't too bad

[gritty widget]

it's not

[empty grove]

let me recall

[gritty widget]

in order for a point to map to 1, you need the error to always be maximum. in particular the error of f-U1 must be 2/3 at this point

and you can show that's impossible

wait nvm

I got that all wrong

cause the preimage will belong to X\A

so it's some point that belongs to the constant portion of every single urysohn map that you take

[empty grove]

You can do it by one more application of Urysohn I think

let me just see my notes

[gritty widget]

[empty grove]

you definitely don't need to go through the proof again

[gritty widget]

I have an idea

S is closed

the preimage of 1 is closed

[empty grove]

oh you do apply urysohn again

[gritty widget]

they are disjoint

[empty grove]

yes

preimage of {-1,1}

[gritty widget]

yeah sure

[empty grove]

and the closed subset on which the original function was defined

separate them by continuous function that maps preimage of boundary to 0 and that original closed subset to 1

[gritty widget]

then multiply?

[empty grove]

then multiply this separating function by the extension

much more annoying than I remembered

[gritty widget]

I was thinking map S to 0 and boundary to some mystery number and then add the functions

but I was like

oh no there's nothing you can add to bring -1 and 1 closer together

[empty grove]

lol

right

[gritty widget]

orrrr

3 way urysohn separation

[empty grove]

oof

[gritty widget]

0 on S, -1 on preimage of 1, 1 on preimage of -1

add it 😎

[empty grove]

lol that works I guess

nice

[gritty widget]

I'm not sure if it fucks the other points up though

like something that mapped to -0.8 preciously and then you subtract some more from it

[empty grove]

we don't care what it does to other points

[gritty widget]

why not

[empty grove]

images of points in S need to remain unchanged

others can do whatever

[gritty widget]

but isn't the goal to find a single extension that maps to (0,1)?

(-1,1)*

[empty grove]

yes

so we started with a map from S to (-1,1)

[gritty widget]

extended it to a map from X to [-1,1]

using your application of urysohn, we are all good

[empty grove]

ye but that 3 way separation should also work

[gritty widget]

using my application, the 3-way thing, some points might now be mapped to like -1.3

[empty grove]

oh

[gritty widget]

like we added the map

[empty grove]

fuck

true

rip

[gritty widget]

I guess yours wins

[empty grove]

oh no wait

simple solution

[gritty widget]

what

[empty grove]

things that now map to <-1

map them all to -1

and same on the other side

This is a continuous map on R

so this is fine

[gritty widget]

but then the codomain is [-1,1]

[empty grove]

oh I am very intelligent

[gritty widget]

yeahhhhhhh

[empty grove]

very sarcastic

[gritty widget]

hey you basically proved this all for me

[empty grove]

My professor proved all this for you

[gritty widget]

usually I actually think these proofs where you glue everything together and force it into the right place can be nice and fun and cute

but today with my sleep deprivation I just want everything to be perfect

[empty grove]

AG has taught me that gluing is pain

[gritty widget]

well we didn't actually do any gluing today

[empty grove]

ye lol

[gritty widget]

but like you know the wieirstrass approximation theorem (might be spelling the name wrong)

[empty grove]

eie

[gritty widget]

😳😳

I thought that one had a fun proof

like cutting and pasting trying to mold stuff in the function algebra to your desired function

[empty grove]

interesting

[gritty widget]

it's like arts and crafts 😊

wait which one is weierstrass approximation

[empty grove]

My probability course covered one probabilistic proof of it

[gritty widget]

maybe I'm mixing it up

[empty grove]

polynomials can approximate any continuous function on R

[gritty widget]

but is that the one about function alwgebras that separate points

[empty grove]

not on R

[gritty widget]

oh no I'm talking about the slightly more general one

[empty grove]

on closed interval

Stone Weierstrass?

[gritty widget]

ahhh yes

I think

[empty grove]

oh yes

I know what you are talking about with the gluing

[gritty widget]

yes

and you only need one polynomial approximation

[empty grove]

It is a neat proof

[gritty widget]

which is of |x|

[empty grove]

why polynomial

I remember that we had to prove stone weierstrass for abs value

[gritty widget]

you take polynomials in the actual functions to approximate their own absolute values

[empty grove]

I don't remember why polynomial

oh right

polynomials in the functions

ofc

[gritty widget]

I haven't used this theorem in a long time but stood out to me as being nice

i always tell myself that algebra is more interesting than analysis

but like

analysis still warms my heart sometimes

[gritty widget]

people might see me bouncing around channels asking questions about stuff that I am very inexperienced with, as I have already been doing. it's cause I'm trying to give myself a month long math buffet

[empty grove]

That is the way

[gritty widget]

I'm only on day 3 (since it's the 3rd of december) and it's already crazy fun

[long hornet]

Is the computation of pi_n(S^3) made easier by the fact that S^3 is a group, as in this case multiplication is the same as the one inherited from the group structures? I am pretty sure the answer is no, but I'd like to know why.

[shy moss]

why this made easier the calculation?

[shy moss]

if you have an H-group then the multiplication in the H-group is the same as the multiplication in homotopy groups

[shy moss]

this holds also for H-spaces and S^3 is an H-space

[plain raven]

have you studied the construction of the product of two schemes yet

[plain raven]

Topologists be like "This is topology, to me"

[pastel linden]

it's simplices all the way down

[coarse night]

why is this true?

is it like one of those space filling curves?

[hollow harbor]

wat

Squash S^2 down to (-1, 1) by projecting onto the z axis, and then wrap (-1, 1) around in a circle (with some overlap) to cover S^1

The map looks like f(x, y, z) = (cos(2 pi z), sin(2 pi z))

[coarse night]

oh,

[hollow harbor]

I agree that going the other way is hard

[coarse night]

it can be done in any dim right? like S^n+1 to S^n

[hollow harbor]

Maybe you can? Idk

Yeah

[finite heath]

u guys are op at this stuff

[long hornet]

Yes, my question was why this fact does not simplify the computation of the homotopy groups.

[fading vale]

"does not simplify" is kind of vague

but h-space structures here are nice

[long hornet]

I don't know whether it does or not

I mean I just guessed

[fading vale]

Like pi_1 of an H-space is always commutative for example

There are also larger technical reasons that this is nice

(an equivalent definition of an H-space X is that homotopy classes of maps [Y, X] for any Y have natural group structures)

(this equips pi_n(S^n) = [S^n, X] with multiple group structures on it)

(which has some technical importance)

But like there aren't some super strong general results on H-spaces that nuke pi_n(X) or anything and make it easy to compute

I can come up with other examples as to why the H-space structure is nice if u want

[long hornet]

Makes sense, the iterated loop spaces are also H-spaces

[fading vale]

Like if you've heard of the hopf fibration, there exist hopf fibrations S^n - S^m with fibers S^0, S^1, S^3, and S^7 and this is related to the fact that these are the only spheres with H-space structures

[long hornet]

But I was thinking maybe the group structure of S^3 in particular can be useful. For example if we could prove that all "spheroids" are homotopic to maps whose images lie in some conjugacy class, or whatever

[fading vale]

Tbh im not really sure what exactly you're looking for

[long hornet]

I only knew n = 3, m = 2 but that makes sense

[fading vale]

Like the fact that S^3 is an H-space is useful in a lot of ways but like... pi_n(S^3) isnt known for all n or anything like that

[long hornet]

Yeah, that's what I thought. What I was looking is why is that's the case?

I mean for example one might guess that we could represent homotopy classes by maps that are related to the group structure somehow

[fading vale]

So like

As i mentioned earlier X being an H-space furnishes [S^n, X] = pi_n(X) with a natural group structure for all n

and it coincides with the usual one on pi_n(X)

But that doesnt mean that this group structure is trivially known or easy to compute

[long hornet]

Yeah

[fading vale]

H-spaces will give you some useful extra structure on things like the hopf construction for making a fibration or (if you've seen it) the pontryagin product giving homology a ring structure

But in general theyre not going to just like, determine everything instantly

[long hornet]

I know it doesn't mean so (because the loop spaces L are H-spaces, and pi_n(X) = pi_1(L^(n - 1) X)).

[fading vale]

Its a condition furnishing extra structure on [Y, X] not a condition restricting narrowly what that structure is if that makes sense

we're not saying something like contractibility where we are demanding that elements of [Y, X] obey a very specific property

[long hornet]

Maybe I wasn't clear, but my point is not just that X is an H-space. It is an H-space like we understand.

For example I know that LS^2 is complicated H-space

But we completely understand S^3

[fading vale]

if X is an h-space then the induced structure is given by using the fact that like, we have u: X x X -> X which will induce maps [Y, X x X] -> [Y, X] by composing

and the former is equal to [Y, X] x [Y, X]

[long hornet]

I also know that we don't understand the homotopy groups of S^3 better, and thought that S^3 is a group was going to help. Since it doesn't, maybe I misunderstand where the difficulty comes from in homotopy groups.

[fading vale]

makes sense right

[long hornet]

Yeah

[fading vale]

But the thing is like, just because u is explicitly defined doesnt mean that u*(f, g) is going to be easy to compute for arbitrary f, g

if that makes sense

Like we still need to understand things about [Y, X] to say useful things here

[long hornet]

Of course, but that only proves that the computation shouldn't be trivial

[fading vale]

are you just like

[long hornet]

Why is this not possible?

[fading vale]

broadly asking why pi_n(X) isnt easy to compute in general

sorry if im not giving a satisfactory answer lol its a bit hard for me to say anything other than just "If it is we dont know how to do it and the machinery is extremely complicated"

[long hornet]

That makes sense to me, given how something like the Hopf fibration exists

I just realized that I don't know whether we can describe [X, S^1] or not!

[fading vale]

Oh ok i might be able to answer something in a satisfactory way then

So like

Basically you want to see a non trivial element of pi_4(S^3) for example

Like to see explicitly why these things can be ugly

even when they have H space structures

[long hornet]

Can we suspend the Hopf fibration?

[fading vale]

Yea exactly

In general the non trivial generator of pi_n+1(S^n) is given by suspension the hopf map

[long hornet]

Okay, I guess I can explain myself better now. It's not that the maps should be obvious (the fact that it's a group structure shouldn't help)

[fading vale]

And i guess more broadly the way homotopy behaves with respect to suspension means that non-trivialities in lower homotopy groups of spheres carry over under suspensions

[long hornet]

It's more like, having two spheriods a, b, we can represent them by canonical forms a', b' in a way that simplifies the question "Are a and b equivalent"?

[fading vale]

Ah yea

The answer is like in general no

F

[long hornet]

And what I expected was that "canonical" should be some property of the group

[fading vale]

the H space structure will help you find the product of two such spheroids

[long hornet]

For example, say that a'(x) is always in the center

[fading vale]

But the situations of like, [Y, X] is trivial and [Y, X] is uncountable

are gonna look massively different

Obviously lol

[fading vale]

well like, pi_1 of an H space is always abelian

and pi_n is in general abelian for n >1

So

[long hornet]

Wait

a' is a spheriod

I mean a map from S^n to S^3

[fading vale]

Sure

[long hornet]

Oh

Well, it doesn't work here because the center is trivial anyways

[fading vale]

Yea

[long hornet]

Imagine for example that there is a group such that all spheriods a are homotopic to spheriods with a'(x) have order 2

[fading vale]

Im not really sure what a'(x) means here, the notation seems to suggest you're looking at the image of a single element of S^n under a'

[long hornet]

Since the group structures coincide, that does give some value information

[long hornet]

Ohh, okay.
By spheriod b I mean a representative of an element in a (higher) homotopy group.
I write b' if [b] = [b']

[fading vale]

do you just mean like every element of [S^n, X] has order 2

[long hornet]

That's the conclusion, yes

But in my hypothetical example, we could prove it in a different way

Not by explicitly showing that the composition of the spheriod with itself is the identity, but rather because we can homotope the spheriod to another one in which all the values assumed are of order 2.

That's an extremely restrictive scenario, and I don't say that there is a useful realization of it. But my point is to use the group structure in such ways.

[fading vale]

There are definitely some situations where an H space structure can facilitate computations

Like maybe if you have a fibration between H spaces thats compatible with the H space structures and then the leray spectral sequence is a seuqence of Hopf algebras and that might help

Im sure those words mean nothing to u but im using it to illustrate the point that like while this is definitely not a never say never type situation

In general

just possessing an H space structure, even an explicitly known one, is not like a super strong nuclear bomb

[long hornet]

Well, the Hopf fibrations do arise from group structures

[fading vale]

U can say some things for sure in certain cases but its not like, that particularly strong an argument

Uh here fibration between H spaces is not necessarily the hopf fibration between spheres

or other spaces

[long hornet]

I know

I mean I get how it could facilitate computations

Although it seems reductive to say the Hopf fibration is a calculation haha

[fading vale]

well the fibration itself isnt but you can use it to compute a lot of homotopy groups of spheres because of the homotopy LES of a fibration

[long hornet]

I see! I should look up the definition of Hopf algebras (I know spectral sequences are daunting)

[fading vale]

I mega do not recommend you try to reverse engineer spectral sequences here bc theres just so much machinery youd need to learn first but like

Yea

Cool stuff

Hopf algebra is kind of but should not be inaccessible

[long hornet]

Too cool, I have been studying nothing else for a long time

[fading vale]

anyway idk if this explanation was very helpful? but i hope it was a little bit at least lol

[long hornet]

I did learn more

[long hornet]

The definition isn't, but it's essentially useless without examples

[fading vale]

something something group ring something something

[long hornet]

I saw the algebra/coalgebra thing

[gritty widget]

What are topics in classical stable homotopy theory versus modern

I know modern is spectra and browns represent ability

[fading vale]

Im not even sure if spectra properly counts as modern on its own

I feel like without it you have like freudenthals suspension theorem and related results

idek if u can prove freudenthals for arbitrary n-1 connected CW complexes without spectra

maybe im thinking of "stable homotopy theory" as too narrow a category though

[gritty widget]

What I mean is

It’s easy to see what the topics are in modern stable homotopy theory

But browns representatibilty is pretty crucial for any modern treatment

This came many years after fruendenthal suspension

So what were the topics in between

[drifting sundial]

just to confirm, is it true that every connected nbhd of S^1 (that isn't S^1) is evenly covered under the exponential map R -> S^1?

[empty grove]

yes

[drifting sundial]

thanks

[long hornet]

Consider the cell decomposition of S^n with two cells in each dimension, and let S^k denote the appropriate subcomplexes

Is S^n / S^k homotopy equivalent to something simple?

For example, according to Hatcher,we have S^2 / S^0 = S^1 v S^2
(note: R^n = 0)

I think it's S^(k + 1) v S^n.
If it holds for (k, n) and P denotes suspension, then
S^(n + 1) / S^(k + 1) = P(S^n / S^k) = P(S^(k + 1) v S^n) = S^(k + 2) v S^(n + 1),
and then one checks the "boundary cases" k = 0 and n = 0.

[gritty widget]

can every Rⁿ fibre bundle be turned into a vector bundle

I can see how to do it for a certain wide range of spaces (even wider if I can use choice) but it doesn't seem like a nice thing that you can just do

I can do it when each connected component of the base space is Lindelof

nevermind, I can't even do that. you need to make some annoying extension that I only know how to make when B is compact and hausdorff 😐😐😐😐

basically give a chart from U×Rⁿ you can put a vector space structure on each pi^{-1}(b) for b in U, but if you have another one for a neighborhood V, these vector spaces are not gonna match up on VcapU. I can change one chart to be compatible with the vector space structure of the other, but that only works on UcapV and can't see how to extend it to the entirety of either neighborhood

[long hornet]

Let X = (I x {0}) u (A x I), where A is the set of all rationals of the form 1/n, including zero. Apparently it is "not hard" to show that I x I does not retract to X. Any hints?

[rancid umbra]

what is I

[long hornet]

Yes

[rancid umbra]

okay, sorry for changing up my question on u. I is the unit interval

[long hornet]

lol

This is equivalent to (I, A) not having the homotopy extension property.

[empty grove]

look at where the points around (0,1) go

Nvm my argument has a flaw

[empty grove]

I mean you probably want to start here

I think I got it

Look at the image of the top edge I x {1} under the retraction, that will give the contradiction

@long hornet

[long hornet]

I can't see it

At first I thought its image is disconnected or something, but that's not obvious..?

[empty grove]

The image will contain all the tips of the lines, the (1/n,1) points

and by the intermediate value theorem, it will also contain the bottom edge

Take a neighbourhood of (0,1), there should be a neighbourhood of this point in the top line which maps into this neighbourhood

use the intermediate value theorem now to get a contradiction

[long hornet]

If there is such a neighborhood, I see why this will complete the proof. But I can't see why.

[empty grove]

Let U be an open neighbourhood of (1,0) in A. By continuity of the retraction, there must exist a neighbourhood V of (0,1) in I x {1} which maps into U under it. Then V contains some point of the form (1/n, 1) which maps to itself. Applying the intermediate value theorem to the x coordinate of the retraction, there must be a point in V that maps to (x,0) for some x not of the form 1/n, but U may be taken to be small enough that it doesn't intersect the x axis, so this is a contradiction

[long hornet]

Do you mean in X in the first line?

[empty grove]

oh my bad yes

[long hornet]

And (0, 1)?

[empty grove]

(0,1) is the point

in R^2

not the interval

[long hornet]

Yeah, but I thought we were working on the top edge

[empty grove]

(0,1) is on the top edge

it lies on the y axis

[long hornet]

Yes, but (1, 0) isn't, which is what is in the first line

[empty grove]

oof

yes

[long hornet]

Ohh, I get it now. Thanks!

An interval in the top edge around p = (0, 1) is connected, so it can only map to one strand

I mean it's mapped to only one strand

That's essentially the same idea (we require U to be small to force it to be disconnected)

[empty grove]

Well it could map to multiple strands, just not infinitely many

Oh ok

I see what you mean

[long hornet]

Actually the strand must be the left edge

So we have to get something like r(1/n, 1) = (0, x)

[desert bloom]

hiya

borrowing this from stack exchange:

I am struggling to find the equivalence between this and the axioms for topological spaces

the one that goes

A1: X and NULL are open

A2: intersection of two open subsets are open

A3: Arbitrary unions of open subsets are open

I think I proofed that N3 is equivalent to A2

and that N2 implies that X is open

but I am struggling to proof anything about the null set being open

or the arbitrary unions axiom

[empty grove]

You will first have to define what "open" means from the neighborhoods definition

[desert bloom]

will that be from N4?

[empty grove]

Yes with M = N

[desert bloom]

since there is a set M in N such that N is a neighborhood of all its point, then M must be an open set containing x

[empty grove]

This is not true

[desert bloom]

is it not?

huh

[empty grove]

M need not be open. Take the origin in R, M = [-1,1], N = R

[desert bloom]

oh yeah

[empty grove]

x = 0

[desert bloom]

of course

soz

[empty grove]

So define a set to be open if it is a neighborhood of every point it contains

[desert bloom]

oh, thats neat

I see

never considered that

how about NULL sets?

it's hard for me to proof anything about that because it feels like there is nothing about it in the axiom of neighborhoods

[empty grove]

That is vacuously a neighborhood of every point it contains

[desert bloom]

oh

[empty grove]

There are no counterexamples because there are no points

[desert bloom]

of course

thanks man

I'll try to do the arbitrary union one then

[light rivet]

So I have the following space: Consider $[0,1]^3$ the cube with the following equivalence classes $(x,1,z) \sim (0,x,z), (x,1,z) \sim (1,x,z), (0,x,y) \sim (x,y,1)$. So faces are identified with these diagonal faces across the cube. I want to show that the fundamental group of such a space is not isom to $\mathbb{Z}^3$, but the equivalence classes make looking at the loops rather difficult and annoying. Is there a easy loop here that I am not seeing?

[gentle osprey]

MasakaBakana

[light rivet]

Intuitively it seems that the fundamental group is Z X Z/2Z since the adjacent faces were glued while the top and bottom are also glued, does this seem right?

[bleak path]

Hi, have a simple question about proper maps, is the inclusion map a proper map?

Sorry clarification, the inclusion map of a LCH subspace of a LCH space

[rancid umbra]

LCH is locally compact hausdorff?

[bleak path]

Yes

[mint rose]

are you sure that's the correct equivalence relation?

because you're identifying (x, 1, z) both to (0, x, z) and (1, x, z)

[light rivet]

Yea, I realized that it was the adajance faces being identified, and the top and bottom faces. But is my intution that it looks like Z X Z/2Z ok?

[mint rose]

I'm not sure what the equivalence relation is exactly, for example you're not even considering the (x, 0, z) face

[light rivet]

Consider $[0,1]^3$ the cube with the following equivalence classes $(x,0,z) \sim (0,x,z), (x,1,z) \sim (1,x,z), (x,y,0) \sim (x,y,1)$. I made a type

[gentle osprey]

MasakaBakana

[light rivet]

This is much better, it should look like the 2 adj faces glued and the top and bottom glued

[mint rose]

okay

[light rivet]

So I thought it looks something like the real projective plane cross: Rp^2 X S^1 which is why I got that guess ofr the fundamental group

But im not sure if thats it

[mint rose]

If you look at a slice for some z

you get S^2

So it seems like this space is something like S^2 x S^1

[light rivet]

That is very bad for me... Because the questino I was asked was to show why that space is not homeomorhpic to the following:
$[0,1]^3$ with identification $(x,1,z) \sim (x,1,z), (0,y,z) \sim (1,y,z), (x,y,0) \sim (x,y,1)$ and that has fundamental group $Z^3$

[gentle osprey]

MasakaBakana

[empty grove]

I think I have a counterexample. Take N ∪ {∞} in the order topology (or if you are familiar with ordinals then ω+1). This is compact hausdorff. Take inclusion of N into this, and N is LCH, but not compact. Then the preimage of the whole set, which is compact, is a not compact set.

[light rivet]

So if that was the idenitication, they have the same fundamental group

[mint rose]

no

S^2 x S^1 has fundamental group Z

because the fundamental group of S^2 is 0

[light rivet]

Yes... mb

[empty grove]

If you also want the larger space to be non compact, then just embed this larger thing into the ordinal ω•2

[light rivet]

But is it true that the second thing I had is Z^3? It is the product S1 X S1 X S1 looking at the slice

[bleak path]

Hmmmm

I will think abou tthis

thanks

[mint rose]

Yes

that's just R^3/Z^3

so the fundamental group is Z^3

[light rivet]

Hmm, Im not seeing the S^2. I see the Z axis gets you the S_1, but noy why the xy axis looks like a S^2

nvm i see it

[finite heath]

Hey

Could someone draw out A, U, and V for me?

I can't seem to figure out how to draw the picture as I don't see how this is possible

[long coyote]

Show that the subspace $X$ of $R$, namely, $$X={0}\cup{1/n;n\in\mathbb{Z}}$$ is not a CW complex

[gentle osprey]

亜城木 夢叶

[long coyote]

i figured it out,

[finite heath]

Any feedback on my proof guys?

I learned this other characterization theorem of proving that a topological space is connected, but I used the definition and did a proof by contradiction because it seemed most simple

[empty grove]

Probably say why you care about the union of X \ P and X \ Q, since you are supposed to think about the union of P and Q. Also you should say P and Q are both non empty open, because otherwise the complement need not be finite (it would be the whole set)

[coarse night]

he did say non empty tho

[empty grove]

right, I was only looking at the next line

[finite heath]

yes sir

[gritty widget]

if $f,g:X\to Y$ are homotopic rel $x$ then I understand that they define the same homomorphism from $\pi_1(X,x)$ to $\pi_1(Y,y)$ where $y=f(x)$. but is there any meaningful way to say this when $f$ and $g$ are just homotopic?

[gentle osprey]

wren

[gritty widget]

the reason I'm saying this is that the homotopy classes of maps from the torus to itself are the 2×2 integer matrices, which are the same as endomorphisms of it's fundamental group. there is a nice way to see this if we pick a point on the torus and say we are only considering maps that are homotopic rel x but I can't see how to extend this reasoning

[fading vale]

What do you mean by "say this"?

[gritty widget]

asking this*

[gritty widget]

the reason I asked this^

[fading vale]

are you asking if theres a basepoint independent notion of induced homomorphisms on pi_1

[gritty widget]

yes I guess

[fading vale]

Since pi_1(T^2) is abelian the answer is yes

In general if i have a space X and a path from points x to y, call it g, then we can turn loops at x into loops at y

[gritty widget]

that's what I was thinking 😳 cause like the inner automorphisms are triviak

[fading vale]

yeah exactly

Paths between the basepoint induce non canonical isomorphisms on pi_1

through conjugation by paths

[obtuse meteor]

unique base change :)

[fading vale]

When pi_1 is abelian its canonical

[gritty widget]

ok here's the thing also

if we have maps f and g

and a path gamma:[0,1]->S¹×S¹

then like if we have a homotopy from f to g

at each time of the homotopy

the closed loops are gonna have different bases

so like to get paths from all f Base points to all g base points, do we just pick a path from f(gamma(0)) to g(gamma(0)) and then carry it through the homotopy?

[fading vale]

Sorry im not really sure what you're asking

f is a map S1 to X

[gritty widget]

ok if f and g are homotopic rel x then we have f(x)=g(x) and at all times of the homotopy we have the same thing

so now if they are just homotopic

[fading vale]

gamma(0) is in S1 x S1

f circ gamma isnt defined

[gritty widget]

we pick a path from f(x) to g(x) and we are all fine again

[gritty widget]

f is a map from the torus to itself

[fading vale]

ah

ok i see

[gritty widget]

omg oops I forgot to say f and g are maps from the torus to itself

it was in my brain but I forgot to say it 😭😭

[fading vale]

Oh I thought they were loops lol

[gritty widget]

yeah that made sense to assume 😔

[fading vale]

What is an f base point exactly

[gritty widget]

ok so in the domain I'm fixing a base point and considering the fundamental group based at that point

f sends closed loops at x to closed loops at y, g sends closed loops at x to closed loops at z

[fading vale]

Ah okay

[gritty widget]

so I pick a path from y to z

[fading vale]

Yes

[gritty widget]

wait a sec I just realized

isn't there a canonical path we can take that's given by the homotopy from f to g?

[fading vale]

yeah if we have H: T^2 x I -> S^1 with H(a, 0) = f(a) and H(a, 1) = g(a) then we can take H(x, t) with x fixed and t varying

well the path is not canonical unless the homotopy is somehow

[gritty widget]

oh you're right

[fading vale]

but the induced map on pi_1 will be because pi_1(T^2) is abelian

[gritty widget]

ok I see

so if f and g are homotopic then it will not necessarily be the case that closed loops are still closed at all times of the homotopy

right

they will be closed at time 1

oh no cause H(s,t) is a continuous map still for each t

I don't know what I'm confused about honestly

[fading vale]

sorry by closed loop do you mean like

we have a loop in pi_1(X)

we look at its image under f and g

and then take the homotopy between the images?

[gritty widget]

yeah I think

[fading vale]

so we have like ell: I -> X with ell(0) = ell(1) and f(ell(0)) = x, g(ell(0) = y, and a homotopy H: T^2 x I -> T^2 inducing, with H(a, 0) = f(a), H(a, 1) = g(a)

So that if we take H(a, 0) as a ranges across the image of ell we get f circ ell

and similarly g circ ell

we know that these are loops

[gritty widget]

yes

[fading vale]

and your question is, is H(a, t) a loop for any t, not just t = 0 or t = 1

Right?

[gritty widget]

yes

I think it should be from the fact that H(a,t) is a continous map in a

[fading vale]

Yea i think u pretty much got it lol

ell(0) = ell(1) so H(ell(0), t) = H(ell(1), t) for all t

not just t = 0 or t = 1

[gritty widget]

ok got it

[fading vale]

and since H is continuous its restriction to the image of ell defines a continuous map I -> T^2 -> T^2

i guess more precisely H(a, t) for fixed t as a ranges across the image of ell is the restriction of H to Im ell x {t}

[gritty widget]

ok now the harder question is whether two maps are homotopic given that they define the same homomorphism between fundamental groups (assuming the codomain is abelian)

[fading vale]

so we take ell: I -> T^2 and compose with the inclusion T^2 -> T^2 x I and then apply H

No not in general

just take any homomorphism between two spaces with trivial pi_1

but non trivial pi_n for some n > 1

[gritty widget]

oh yeah 😔

[gritty widget]

if the homomorphisms of every homotopy group are the same, are the maps homotopic?

[fading vale]

I believe you need them to be maps between CW complexes

[gritty widget]

moth is right but i just want to be extra careful and point out

that the precise statement is

there needs to be a map of spaces between the CW complexes

that induces the isomorphisms

i think moth implied this

[fading vale]

Yeah there are spaces that have isomorphic homotopy groups but which are not homotopy equivalent

[gritty widget]

but if you didn’t know this you might have missed it

[fading vale]

iirc S^2 x RP^3 and S^3 x RP^2

[gritty widget]

yeah that is the example that everyne goes to

because it even shows

it’s not because of

some strange pathological space

you couldn’t really ask for nicer spaces

[fading vale]

I have this general but incredibly imprecise intuition that homotopy groups are basically bad at distinguishing products

of spaces that is

[long hornet]

Speaking of which, out of idle curiosity, when do we have S^n = A x B for A, B nontrivial (i.e., have more than one point)?

One of them has to be a homotopy sphere, and the other a homotopy "point"

I think if we can transport the cellular structure to A and B, then maybe we can apply Whitehead's theorem somehow?

I saw a proof that this is impossible for n = 1, based on connectivity

[long hornet]

Apparently "homotopy sphere" means something different. I just meant a space with the same homotopy groups.

[minor hornet]

I'm thinking about how H < πX the subgroups of the fundamental group of a space (with nice properties) correspond to classes of covers. Is there a meaningful correspondence to subsets Y of X with πY = H?

[gritty widget]

I saw a cool proof to generate a topoligical space with a given finitely generated fundamental group. the same procedure can be applied to arbitrary groups, but I'm not sure if the fundamental group of the space will still be the desired group. does anyone know if this happens to be true?

[winged badger]

there is a way to generate a space with fundamental group G for any group G, even if it's not finitely generated.

[gritty widget]

theorem by eilenberg and maclane

[long hornet]

Hatcher proves it in page 52. He takes the wedge of circles (with as many circles as are there generators) and attaches the 2-cells in the right way

Given an abelian group A, can we find an H-space X with pi_1(X) = A?

[empty grove]

I like May's proof a lot more

[long hornet]

What's the general idea?

[empty grove]

He proves that a subgroup of a free group is free using Galois correspondence, then write the group G as F/H a quotient of free groups. Take a graph s with fundamental group F and the graph covering it corresponding to the subgroup H. Then the mapping cone of this has G as its fundamental group

It feels nicer because there's only one attaching happening, rather than attaching lots of 2 cells

[long hornet]

I didn't expect the mapping cone to show up - I guess I need to refresh my memory about covering spaces

[empty grove]

The mapping cone's fundamental group can be computed using svk on the cone part and the complement of the tip of the cone

[long hornet]

Makes sense

[empty grove]

It immediately comes out to be F/H

[long hornet]

It is nicer

Do we also get a CW complex? I think we do

Because we glue along a cellular map or something

[empty grove]

Yeah it should be CW, let me think

Isn't there a theorem that says that collapsing closed subcomplexes gives a CW complex

[long hornet]

Yes

[empty grove]

Ye that just inherits the structure

So first you make the mapping cylinder

Then collapse the top

And mapping cylinder is constructed by taking
A is domain
X is codomain
Then you taking A x I union X but you are gluing A x I along the edges

So for this I think you can argue cell by cell

For each 2-cell in A x I you are doing this gluing

[fading vale]

The wedges with circles approach generalizes for abelian groups with n > 1 i think

not sure about the other one

[empty grove]

Oh neat

[fading vale]

its also kinda nice cause its similar to the construction for moore spaces

[long hornet]

I think pusbouts exist in the cellular category, or something

[empty grove]

Seems legit, I'm willing to accept that as fact

[long hornet]

This is kinda dumb but if two spaces are homotopy equivalent, and one is H-space, so is the other, right?

I thought this was obvious (them being isomorphic in hTop), but the issue is that H-spaces are group objects in hTop*

So this is at least true when one is homogeneous

My motivation is as follows. If X is a K(A, 1), and we have an H-space G which is also that, then they are homotopy equivalent, and thus both are H-spaces.

So to prove that there is no H-space which is a K(A, 1), maybe we can just work with a specific construction, like Hatcher's or May's. But I don't have the sufficient machinery anyway, so this seems moot.

Another question is if we have two points x and x', do we have an isomorphism (X, x) ~ (X, x') in hTop*, the homotopy category of pointed spaces?

This is not true in Top* (of course, because not all spaces are homogeneous). The easy counterexample is X = (0, 1], but it doesn't work in hTop*.

[coarse night]

is the cont image of a path connected space path connected?

[long hornet]

Yes. Take a path joining two preimages and compose with your function.

[coarse night]

ok

was thinking something weird using sin(1/x), but looks like it's pc

[long hornet]

Well, the image of (0, 1] under this function is path connected

[coarse night]

yeah

[shy moss]

I think no

When X isn’t path connected (X,x) doesnt need to be homotopy equivalent to (X,x’)

[gritty widget]

yeah that's the construction but idek how to show it's fundamental group is the group

[long hornet]

Yeah, I just found that Q u S^1 is a counterexample. But what if we have path connectedness?

[shy moss]

if we have path connectedness and CW complex then (X,x) is homotopy equivalent to (X,x’)

[long hornet]

Oh, good point

[long hornet]

One proves that attaching a 2-cell has the desired effect (i.e., introducing the correct relation). Lee has a detailed proof.

[gritty widget]

how often do yall apply alg top to things?

and i call using theorems/techniques as applying

and i mean applying to other fields

[pearl holly]

this is probably a dumb question but if I have $f^(\alpha) \in H^n(X; G)$ then I can view this as a map $f^(\alpha): X \to K(G, n)$. What will the induced homomorphism of $f^*(\alpha)$ then look like?

[gentle osprey]

Tokidoki ✓

[pearl holly]

$(f^(\alpha))^ = f^* \alpha^*$ right?

[gentle osprey]

Tokidoki ✓

[pearl holly]

plz say this is right

[pearl holly]

I just want to make sure I'm not brain dead lmao plz

[onyx crow]

so I’m assuming you have another space $Y$ and $f^*$ is a cochain map

[gentle osprey]

BrokenPizzaDreams