point-set-topology

[foggy heath]

opps, I was getting the wrong answer. it's not 0, because there there is no alternating sum involved. Sorry, I was to say a Kahler mfd always has odd-dimensional Betti numbers even.

[cursive flume]

how does, therefore F closure subset F hold?

if x in F =>x in overline {F}, that means F subseteq F overline

not the other way around

x is some element in F per se

then they show x in closure F

how does this imply closure F subset F?

[empty grove]

x is an element of closure F in the beginning

[cursive flume]

why?

[empty grove]

By 10.4 this means that there is a sequence x_n in F that converges to x

[empty grove]

That's the assumption, when they are proving the converse

[empty grove]

Then by the hypothesis, x in F

[cursive flume]

No, the converse should be: whenever (x_n) is contained in F and x_n->X, then x in F => F is closed

[empty grove]

Yeah, so they start by taking an element of closure F and then showing that it is in F

[cursive flume]

how is that the proof

we have to start from whenever (x_n) is contained in F and x_n->x

and show F is clsoed

[empty grove]

We have to start with that is a weird to say lol

Like you want to show that F is closed

[cursive flume]

that's how a proof works no?

a=>b. Suppose a is true

prove b

[empty grove]

ie that closure F is contained in F

If we want to show this

Then we take an element of closure F and show that it is in F

[empty grove]

And the a is used along the way

[cursive flume]

Let x in the closure of F, then by 10.4 there is a sequence in F converging to x, so by the hypothesis x in F

[empty grove]

Yes

[cursive flume]

this is the proof I made up

but

this is not the one they have

[empty grove]

This is it though

[cursive flume]

so strictly speaking this is false?

[empty grove]

No?

Lol I don't get it

[cursive flume]

'it follows from theorem 10.4 that x in F overline, hence .

[empty grove]

Like they have the same proof, just phrased differently using the fact that 10.4 is also an iff

[empty grove]

But this phrasing is better

[cursive flume]

how does their hypothesis give me F bar subset F

I agree with all there is written,it's true

[empty grove]

Because of 10.4, starting with "Let x ∈ closure of F" is the same as "Let x be a point and x_n a sequence in F converging to x"

[cursive flume]

except F overline subset F

[empty grove]

You started with the former, they started with the latter

[cursive flume]

correct. and by hypothesis, this means x in F

[empty grove]

Ye

[cursive flume]

you can't start with the latter though

that's a consequence

[empty grove]

It's the same thing as the former

Like 10.4 is an iff

[cursive flume]

ohhh

lol

so they jumped abig step

they used a<=>B

[empty grove]

Yeah kinda

[cursive flume]

and immediately said x in closure is same as ...

not just =>

i mean fair,it is true

[empty grove]

Yep

[cursive flume]

but who would think of ths

[empty grove]

[cursive flume]

isn't this way clearer

[empty grove]

A lot of analysis proofs just open like that lmao

[empty grove]

Ye

[empty grove]

They just start with a statement that is the converted to sequences version of the one they are supposed to start with

[cursive flume]

yeah,now I see

it is true nevertheless,so it's good proof

I just didn't get why they do that

[empty grove]

[cursive flume]

topology go brr

[cursive flume]

here When they say "by letting"... why is that true?

I don't think that's true in general

[unreal stratus]

There should exist a nested neighbourhood base in this situation right?

[cursive flume]

ahh

the one given by V_n:=cap_{i=1}^{n} U_i

[cursive flume]

is it sure that the author is minimal here?

[cursive flume]

In this proof seems i didn't use first countability of Y.

I don't think it's needed

X being first countable suffices for a,b,c to hold,right?

[gritty widget]

This definitely doesn't hold in general

[cursive flume]

what do you mean?

I am convinced this doesn't hold if X,Y are not first countable

that's clear

but where do we need Y be first countable?

[cursive flume]

as far as my proof goes and is correct, only first countability of X is used.

[gritty widget]

Yeah you don't need it

[cursive flume]

so the book is not minimal

[gritty widget]

I read it as just first-countability

[grave maple]

If you replaces sequences with nets/filters, you don't need first-countability.

[gritty widget]

What are cool topologies which induce the "pointwise convergence" of sequences

On the product. Besides the product topology

Is there a description of all the topologies which induce pointwise convergence for sequences?

[grave maple]

I'm not sure I understand. You can always define a topology where some sequences don't converge.

[gritty widget]

On the product

[coral pivot]

Blue book LaTeX group

[gritty widget]

if you have a space

and you take its 1-point compactification

then you remove a point

will this be homeomorphic to the origional space

damn no you can do silly things

[gritty widget]

not at all

yeah

thanks

Best way to see it is that [0, 1) has one-point compactification [0, 1]

but removing any point from the interior leads to a disconnected space

thats a nice example

[stable kite]

As Blitz showed, this cannot be the case - However, if $X$ is a compact Hausdorff Space, $x \in X$ arbitrary, then $X \setminus {x}$ is a locally compact Hausdorff Space whose one-point-compactification is homeomorphic to $X$. :)

[gentle osprey]

AzarTheKas

[gritty widget]

thank you

the following should be true

but im not sure where to look

or what the precise statement is

if we have some colimit of homology groups

such that the all the maps in the colimit

can be given as

iterated multiplcation by some element

then

the colimit should be isomorphic to inverting this element

[cursive flume]

Bourbaki moment

He has nice remark that X,empty need not be assumed to be in the topology in this sense

[empty grove]

I see nothing wrong with that

[coarse kestrel]

It literally says that the empty set and the whole set are in the topology

[grave maple]

Maybe ProphetX is saying that the definition does not explicity state that X and the empty set are open sets because they are implied by the definition?

[cursive flume]

Yes that was my point,i.e. is minimal

[feral copper]

Hi! Is it obvious to you what the fundamental group of the complement of RP² in S⁴ is?

Is it Z/2?

[gaunt linden]

Is there even a canonical way to consider RP^2 a subset of S^4? At least a priori, different embeddings don't need to have homotopy equivalent complements.

[feral copper]

That's also what I've been wondering...

My RP² at least is not necessarily standard x')

Maybe the Euler number classifies the embedding? (But I know nothing when it comes to euler numbers...)

[gentle osprey]

engineersanonymous

[bitter smelt]

well its the entire space

by definition of topological space, the entire space and the empty set are always open

so, trivially, yes

[gentle osprey]

engineersanonymous

[bitter smelt]

afaik open/closedness of a map does not depend on continuity. Might depend on author?

open map is a function which sends open sets to open sets

this does not require continuity

sure

try to come up with a counterexample

what do you think?

you'll need to be a little more precise in your language

"not closed" in what topology?
"discontinuous" on what domain?

[gentle osprey]

engineersanonymous

engineersanonymous

[bitter smelt]

what are the topologies here

generic?

then try Y with indiscrete toppy

and X = R

[golden gust]

I think Y has to be Hausdorff for the graph to be a closed subset of X x Y

[bitter smelt]

and then the answer is spoiled

[golden gust]

oh

[golden gust]

seems like they already knew that

but still, my bad

[gritty widget]

how do we know $f[$\omega$+1]$ is compact?

[gentle osprey]

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

[foggy heath]

I don't know what is \omega but I guess the fact may be related?

the continuous image of a compact set is compact.

[gritty widget]

omega is the ordinal of countable sets

[gritty widget]

general topology

you don't really need to know any deep algebra at the introductory level. familiarity with groups and what not would help, but you can also pick this up as you go

[stark creek]

Okay, Thank you.

[hot mirage]

Can someone tell me what this map

Means

The id,\delta

[gritty widget]

x -> (id(x), \delta_i(x))

[gritty widget]

In general it means the morphism induced by those two from the universal property of products

[gritty widget]

omega+1 is compact

If you cover it by sets of the basis, one of them needs to be of the form (n, omega] for some n

But [0, n] is finite so just pick an open set for each m in [0, n]

This gives a finite subcover

[gritty widget]

Yeah. They say it's embedding so it's precisely a statement that omega+1 is compact

https://math.stackexchange.com/a/3009993/476484

In this answer it's shown that [0, a] is compact for any ordinal a

That is, a+1 is compact

[surreal lantern]

what maps represents F->E here? Is it the inclusion of F into E after identifying it with the pre-image of pi?

[empty grove]

Yes, but it wouldn't hurt to just think of that as notation

It means that E → B is a map with mapping fiber F and the arrow F → E denotes the mapping fiber projection, but in fiber bundles it just becomes the inclusion of F as the fiber of the basepoint

[surreal lantern]

I see, thank you

[surreal lantern]

Is it particularly easy to see that the sequence of pi_k's is exact between F,E and B? Because im kinda not seeing it but they don't give a proof of the exactness in that pdf

[empty grove]

No it isn't trivial

The idea is that up to homotopy, the fiber of the fiber projection F → E is the loop space on B, which I will denote by LB, and you get a fiber projection LB → F. Continuing similarly, repeatedly taking fibers, you get the sequence of spaces
... → L^2 B → LF → LE → LB → F → E → B
Let [X, Y] denote the pointed set of based homotopy classes of based maps from X to Y. For any fiber sequence F → E → B, one nice property is that when you apply the functor [X, -]: Based-Top → Based-Set to this for any space X, you get an exact sequence [X, F] → [X, E] → [X, B]. Since the above is a fiber sequence up to homotopy, applying this functor gives you a long exact sequence. Then take X = S^0, and use the fact that [S^0, L^n Y] = pi_n(Y).

Any book that covers higher homotopy groups should cover this

Exactness of a sequence of pointed sets makes sense because you can talk about kernels and images, but this also turns out to be a sequence of abelian groups if you cut out the last few terms. The exactness of [X, -] applied to a single fiber sequence isn't hard to prove, so the most non trivial part here is probably proving that repeatedly taking fibers of maps gives you that sequence up to homotopy

[surreal lantern]

Hmm okay that's a lot of words i don't know but i'll try to make sense of it somehow, thank you

[west spindle]

take a solid that consists of a spherical shell into which 12 holes have been punched and consider the surface that is its boundary

what is the genus of this thing?

it's 11, right?

namely that if you punch one hole into a spherical shell like this its surface becomes a sphere

and each additional punched hole increases the genus by 1

or am i off my rocker here

[golden gust]

I think that's right

[surreal lantern]

what if i only focus on this part of the sequence, is there some basic reasoning why this would be exact or is that difficult as well

[golden gust]

this is definitely true

[surreal lantern]

because the pdf basically says that part is trivial and then they define a connecting homomorphism for the pi_k+1 to pi_k parts of the sequence (without proofing the exactness of it )

[empty grove]

Yeah this should be easy. Suppose you have a map S^k → F. Composing this with the inclusion then fiber bundle, you get the constant map to the basepoint of B, so the image at E is contained in the kernel at E. Now suppose S^k → E is such that composing with E → B makes it 0. Then there is a homotopy in B from the constant map at the basepoint to this map S^k → B. Lift this homotopy to E, and you get a homotopy from a map in the pre image of the basepoint to the original map, so the map S^k → E is homotopic to a map S^k→F included into E, so kernel is in the image

[surreal lantern]

Alright that really isn't too hard then, thanks again

[dim kite]

I have a question, does a ring exist such that its additive group is isomorphic to its multiplicative group?

[pastel linden]

Well, then youd have a group iso sending 0 to 1

[stable kite]

The zero ring {0} trivially fulfils this, but I'm strongly assuming you're searching for something actually meaningful.

[dim kite]

Yes

[dim kite]

Yes

[gritty widget]

topology question, it isn't

[dim kite]

you forgive me?

[gritty widget]

I don't forgive, I only judge

[dim kite]

[dry jolt]

R works, the isomorphism is given by the exponential map

[dim kite]

🤝 thanks king

[gritty widget]

R doesn't work, it doesn't have an element of order 2

[dry jolt]

walter

[gritty widget]

you're thinking of isomorphism from R_+ with multiplication to R with addition

[dry jolt]

you are correct, I have dementia

[dim kite]

XD

[dry jolt]

methinks it is not possible; let R+ denote the additive group and R* the multiplicative group and suppose there exists an isomorphism f: R* -> R+.
Then 0 = f(1) = f((-1)^2) = f(-1) + f(-1) so R has characteristic at most 2. For any nonzero element r, f(r^2)= f(r) + f(r) = 0, hence r^2 = 1

methinks this implies R is finite ring?

[dim kite]

Wow

[dry jolt]

no that is not true because Z^N has infinitely many solutions to x^2 = 1

this is a tough cookie to crack

[dim kite]

groups and rings 🤢

[dry jolt]

wait i'm trolling, in char 2 every element satisfies r^2 = r so r = 1

right?

no i'm double trolling

[dim kite]

[dry jolt]

i forgor

[gritty widget]

well, there is few points here I want to make

1. what does characteristic mean for arbitrary ring
2. either -1 = 1 and indeed R has a characteristic 2, or simply f(-1) is some non-trivial element of order 2 in the additive group of R

[dry jolt]

the characteristic of a ring is the smallest multiple of 1 that equals 0

more formally, there's a unique ring homomorphism from Z to R, the characteristic is the non-negative generator of the kernel of this map

[gritty widget]

sure. I use this concept in 2) so I'm not sure why I was asking.

[dry jolt]

mmm yes methinks you are onto something

i have class but i will think about this more

[gritty widget]

One thing we can certainly deduce is that R has to be infinite

[dry jolt]

yeah, i was trying to get a contradiction by showing that R must be finite

[dim kite]

oh i find it, in the exponential function in complexes

thanks for the help guys

see you other day

[gritty widget]

the multiplicative and additive groups of C aren't isomorphic

for the same reason as real numbers

as was said already, a ring of this kind would have to have a non-trivial element x such that 2x = 0

[dim kite]

oh

good point

[gritty widget]

so if R has no zero divisors, then its necessarily of characteristic 2

[dry jolt]

Methinks R x Z/2Z

Struggling to construct explicit iso

[gritty widget]

bravo, that works

[dry jolt]

YES

[gritty widget]

The multiplicative group of R x Z/2Z is isomorphic to the multiplicative group of R

and that one is isomorphic to R x Z/2Z - simply send r to (log(|r|), sign(r))

[dim kite]

eureka

[west brook]

Why is this also an anime channel?

[stark creek]

Don't ask questions, just accept the truth.

[gritty widget]

why cant any neighborhood of xn contain cofinitely many point sof Vn?

[ripe pecan]

Very true.

[gritty widget]

W contains cofinitely many points of V_n means that their intersection is infinite

By assumption it was finite

It's not that any neighbourhood can't contain cofinitely many points of V_n

It's that it must contain cofinitely many such points

And that's in contradiction with our assumption

[gritty widget]

Because that's the joke. No point in questoning it.

@gritty widget because finite sets are closed?

[gritty widget]

What is because finite sets are closed?

Most likely, no

Oh nvm i got it. Thanks for the explanation. I confused Vn as being finite.

[weary shoal]

I'm not sure how to approach this question.

The definition of a regular map throws me off here too.

[gritty widget]

what about the definition of a regular map here throws you off?

do you know a definition of a regular map?

to approach the question, it might help to recall that "closure" equals "smallest containing closed set"

[weary shoal]

A function f : X -> Y is regular if there exist m regular functions f_1, ... f_m on X such that f(x) = (f_1(x), ..., f_2(x)) for all x in X

[flint sinew]

are there any good places to get beginner-level topology assignments?

[frigid patrol]

Elementary Topology by Viro is a good problem textbook

[weary shoal]

I’m still not sure how to solve this

[gritty widget]

what have you tried?

[hollow yew]

Hello, does anyone know a resource with intuitive descriptions of nearby and vanishing cycles?

[weary shoal]

$V_X(I)={x\in X: \forall g\in I, g(x)=0}$ then $f(V_X(I))={f(x): \forall g\in I, g(x)=0}$. Now, $I^c=(f^*)^{-1}(I)={g:g\circ f\in I}$, so that $V_Y(I^c)={y\in Y:g(y)=0, \text{for all } g \text{ such that }g\circ f\in I}$, and you want to show that the closure of the first set is the second set.

But like ain't sure

[gentle osprey]

Mat_

[gritty widget]

@gritty widget For case 1 I am still confused, how do we know every neighborhood of x contains cofinitely many points of V'?

[cursive flume]

how can remark 5.22 make sense?

in order for the collection B to be a filter basis,we must ensure that the collection F is a filter already.

[cursive flume]

is prop 7.47 what they actually mean?

[viral atlas]

Looks like it

[cursive flume]

would be nice

cause I made 7.47 up,it's written by me so i'm unsure if it's OK

[gritty widget]

They're saying that if B doesn't satisfy those conditions then it's not a basis of some filter

[flint cove]

Please someone tell me whether I'm stupid
But taking the specialization preorder is clearly a functor Top\to PreOrd, right?
Because if x in cl(y) then f(x) in cl f(y) because lim x = y when viewed as the constant sequence

[gritty widget]

Yes

[gritty widget]

Under the homeomorphism, x gets sent to omega

But any neighbourhood of omega is contains (n, omega] for some n

So its complement is contained in [0, n], so it's finite

[cursive flume]

If it satisfies those cond it is basis for a filter

If it is basis for a filter then it satisfies this cond-why?

[cursive flume]

Ah this is the contrapositive of my qn?

[clear storm]

Is $X\vee X \to X\times I/ ({x_0}\times I)$ where $(X,x_0)$ is a pointed space a cofibration?

Like it is a cofibration if there is a retract in <- direction

Like intuitively speaking not because if you take $S^1\vee S^1$ and $X\times I$ and collapse $x_0 \times I$ to a point

[gentle osprey]

Sei

Sei

[lunar yoke]

so im pretty sure if A -> X is a cofibration, then A_+ -> X_+ is also one.

now note that $X \vee X = X \wedge S^0_+$. Since $S^0 \to I$ is a cofibration, by the above $S^0_+ \to I_+$ is one, and then $X \vee X = X \wedge S^0_+ \to X \wedge I_+$ is one

[gentle osprey]

Phil

[gritty widget]

does this argument work

[cursive flume]

how did you conclude sum and product functions are continuous?

[gritty widget]

it's mentioned in the text ill screenshot and send so you can verify if its assumable

[cursive flume]

I agree with projection maps

[cursive flume]

then your solution is correct.

but now I am confused a bit not gonna lie

I don't see why (x,y)->x+y is continuous

your notes claim it is easy to see using preimage of open interval is open

but i don't see it

[gritty widget]

neither do i

hatcher is not for the faint

[cursive flume]

could you send your notes?

maybe there is some part which clarifies it more

[gritty widget]

http://pi.math.cornell.edu/~hatcher/Top/TopNotes.pdf

page 13 and 14

i think its right at the end of page 13 that Hatcher states why (if and only if)

[cursive flume]

wait

so I see everything,except why (x,y)->x+y is continuous

which part is your confusion?

[cursive flume]

from here

[gritty widget]

my argument for $x+y$ being continuous is that if $f$ and $g$ are continuous on $\mathbb{R}^n$ then their limits exist on $\mathbb{R}^n$. Since $f$ and $g$ are continuous with limits at every $x,y$ in $\mathbb{R}^n$ then $f(x,y)+g(x,y)$ exists for every $(x,y)$ but $f(x,y)+g(x,y)=(f+g)(x,y)$ and since $(f+g)(x,y)=x+y$ then $(x,y)\mapsto x+y$ where the map is called $f+g$ is continuous by equality

[gentle osprey]

engineersanonymous

[gritty widget]

Hatcher did define a metric and say what a limit is in $\mathbb{R}^n$ tho...

[gentle osprey]

engineersanonymous

[gritty widget]

heres some stuff we can use

[cursive flume]

@gritty widget

maybe this?

[gritty widget]

that is...

beautiful

[cursive flume]

@gritty widget how would ya prove it

[gritty widget]

epsilon delta

[cursive flume]

we are not allowed

[gritty widget]

this isnt hatchers analysis text its his point set topology chapter

gotta use open sets

then hide the epsilon delta proof behind the whole pre-image of opens nonsense

[cursive flume]

wdym hide

[gritty widget]

i don't feel like elaborating

[cursive flume]

in formulas mb like this

[chilly crypt]

Im not able to see how an open cover of a compact metric space has a lesbegue number... it feels like the radius of the open balls near the boundary of 2 intersecting open sets will tend to zero as we go closer to the boundary

[gritty widget]

could you draw a picture to explain what you mean?

the way i see it, the space being compact should prevent "going to zero" problems, but i'm not quite sure of what you mean

[chilly crypt]

Ah it got resolved, shouldve tried drawing it with my hand instead of my mind lol

[swift fjord]

Your proof os circular then. You want to prove the sum and product are continuous, but you used a claim that uses the fact that the sum and product are continuous

Like your.original sc is circular

[dusk tree]

What exactly does this "stable situation" mean?
The context is the orthogonal group, the infinite orthogonal group and their respective path spaces from the identity to minus the identity.

[empty grove]

Stable situation means that a lot of homotopy group related things are not very well behaved with respect to suspensions, but they become well behaved if you only look at a certain range of higher homotopy groups. So there is a way to get this nice behaviour in all order by switching from spaces to spectra

I think that's what they are doing at the end? But then they also take a direct limit so I am not sure

My impression is that in general stable usually means well behaved with respect to suspension

[dusk tree]

What sort of properties, for example?

[empty grove]

Intuitively if you take suspension, n dimensional holes should become n+1 dimensional holes. This is what you see with homology, but is not what happens with homotopy groups

But it does happen in a range of dimensions, see Freudenthal suspension theorem

There is an excision theorem in homology which allows you to cut out pieces of your space. A homotopy excision theorem also holds but only in a limited range

Yeah that seems to be the case, so I guess we are dealing with the non stable behaviour of cohomology with respect to loop?

[dusk tree]

For the orthogonal group, one can prove that the space of complex structures is homeomorphic to the space of minimal geodesics from the identity to minus the identity.

One then iterates this procedure to produce the \Omega_k s.

The whole thing is about the Bott periodicity of the infinite orthogonal group.

[gentle osprey]

SK2099

[dusk tree]

I have no idea what role this stability condition plays in the whole thing.

[cursive flume]

It is different though

He wants to prove sum on R^n and prod on R^n is continuous using sum and prod on R are

Right?

[swift fjord]

The question was to prove that the sum and product on R^2 are continuous, and they used this result which shows the sum of functions on R^n is continuous, but this result uses the fact that the sum and product on R^2 are continuous

[cursive flume]

Oh hm. For n=2 it is circular indeed

Yeah you're right

I'm unsure how to do the proof then

[empty grove]

You prove that the preimage of an open interval is open in ℝ²

You can sketch the preimage, it's quite simple

[cursive flume]

Ya but how does that not go back to calculus eps delta?

[empty grove]

Ultimately you have to do something like that

It's not quite that

[cursive flume]

It goes back to metric balls anyway as open sets in standard topo are given as such

[empty grove]

Actually it is

Yeah like it has to

How would you formally prove a result about a concretely define operation

[cursive flume]

So @gritty widget was right

[empty grove]

Maybe you can find some big brain alternative formal definition of addition

[cursive flume]

But for any X=R^n, n>2 this proof isnt circular,right?

[empty grove]

Ye

[cursive flume]

Is this legit proof for R^2?

I mean I don't see it using eps delta where is it hidden?

[gritty widget]

thanks for the ping, it was really important that i see this

[hot night]

Is there a difference between the Hausdorff dimension of the Koch snowflake and the filled Koch snowflake?

[gritty widget]

The filled Koch snowflake, obviously, has dimension 2

[hot night]

Why is that obvious

[gritty widget]

Because it has positive 2-dimensional Hausdorff measure

it has non-empty interior

Anyway, I've just noticed, this has nothing to do with topology

[hot night]

Then real analysis?

[gritty widget]

It's not real analysis per se I'd say. I think #advanced-analysis would work

[hot night]

Do I see you there?

[gritty widget]

You can continue here, doesn't really matter

[hot night]

I switched haha

[cedar pebble]

we're looking at how the homotopy groups of the orthogonal group stabilize (i.e. for fixed k, π_k(O(n)) stabilizes as n≥2k)

so for n small relative to k, π_k(O(n)) is fairly complicated, but for large n the behavior stabilizes and you get Bott periodicity

π_k(O) is just π_k(O(n)) for n in the stable range (i.e. for n≥2k)

[dusk tree]

So, what exactly does "stabilize" mean in this context?

What gets stabilized as you take the direct limit?

[cedar pebble]

the homotopy groups

the inclusion O(n)->O(n+1) induces a morphism π_k(O(n))->π_k(O(n+1))

that this stabilizes just means this map is eventually an isomorphism for n large enough

[dusk tree]

Ooooh

So, for small n, it's only a homomorphism?

[cedar pebble]

sure

[gritty widget]

Let X be a space and A a subspace

Is something like the following true

The pair (X,A) is homotopy equivalent to (X/A,A/A) if and only if the inclusion of A into X is a cofibration

[lunar yoke]

the quotient map is a quasi-isomorphism if the inclusion is a cofibration, and a homotopy equivalence if A is also contractible

i'd have to think about the converse

The inclusion of S^0 into D^1 is a cofibration, but (D^1,S^0) is clearly not homotopy equivalent to (S^1,*), so you really need A to be contractible or something similar if you want a homotopy equivalence

[gritty widget]

thank you

[lunar yoke]

lmao i just noticed but the statement (X,A) homotopy equivalent to (X/A,A/A) entails in particular A homotopy equivalent to A/A, i.e. A conctractible

so its actually an iff in the sense that given a cofibration A -> X, we have (X,A) ~ (X/A,A/A) iff A contractible

[cursive flume]

why does halfing the kernel give a double cover?

[sacred laurel]

That like to consider S1 as [0,2Pi] and consider S1 is covered twice by [0,4Pi]

you consider the set with two elements to give representative of one

[cursive flume]

oh wait

is this like Z/4Z and Z/2Z?

like one has twice as many elements as other?

and we know that Cx SL_n(C)/ker p is a cover of GL_n(c)

hence the 2 times p i double cover?

[sacred laurel]

not really

[sacred laurel]

Yes, the thing is just to say that looking at [0,4Pi) gives each time two distinct represenative of the same angle

Representation of GLn(C) is very important to study Holomorphic differential Equations

and Differential Galois Theory

it is nice you are looking at it

[cursive flume]

Do you have any intuition why the map p is surjective?

[sacred laurel]

consider any 1/n-th power of the determinant of a Gln(C) matrix

then consider "det(a)^{-1/n}" * A

put log womewhere

and you are done

(thus you may have to be careful about log determination etc.)

[cursive flume]

wait i'm confused

so we want to show that any matrix B in GL(n,C) can be written as e^{z}A for some A in SL(n,C)

right?

[cursive flume]

how is this related to this?

[sacred laurel]

For such B

"det(B)^{-1/n}" * B lies in SLn(C)

B = "det(B)^{1/n}" * "det(B)^{-1/n}" * B

e^z = "det(B)^{1/n}"

solve it

but this is kinda trivial since exponential is onto on C*

[cursive flume]

this would mean that det(B)^{-1/n} *B has determinant one

how do you conclude that?

all we know det B is non zero

[sacred laurel]

B is in GLn(C) by assumption

[cursive flume]

yes,that means det B non,zero

[sacred laurel]

hence by definition, invertible

[cursive flume]

right

[sacred laurel]

yes

[cursive flume]

ohhh wait I think I see

lemme wriet

[sacred laurel]

You should check out your DMs

[cursive flume]

so det(B) ^{-1/n}*B . Take det of this to obtain : (det(B)^{-1/n})^{n} det(B)=det(B)^{-1} det B=1.

is this the trick?

[sacred laurel]

y

[cursive flume]

ok all of it makes sense

thanks

[sacred laurel]

np :)

[gritty widget]

Let E be a spectrum

I want to look at E-cohomologies where i invert some elements

so I want to consider E^*(X)[1/f]

can i realise by doing some bousefield localization on the spectrum E

[hot night]

I read that the idea singular homology is purely non-geometrically and just a tool for and in contrast to simplicial homology. Is this correct?

[gritty widget]

i would say it the other way around and call simplical homology a tool for (computing) singular homology

[uncut surge]

Saying it's "non-geometric" cannot be accurate, since it manages to capture geometric behaviour (read: holes) of spaces, but it's not quite as easy to visualize arguably

and ye I agree with TTerra

Simplicial homology is a very direct & comparatively easy to interpret thing because the things you're studying are literally built by simplices, whereas singular homology can be thrown at literally everything that has a topology - they both describe geometry, and for simplicial complexes they describe the same thing, but singular homology can do more

[hot night]

Are they equivalent?

Meaning isomorphic

[hidden crag]

https://math.stackexchange.com/questions/229036/difference-between-simplicial-and-singular-homology

maybe this will help?

[hot night]

I have already read it.

[gritty widget]

did it answer your question?

[hot night]

Not really. I also once read that there actually exists spaces for which the homologies are not the same.

[gritty widget]

they are equivalent for spaces which can be triangulated (given the structure of a simplicial complex). for a counterexample, you would therefore want to look at spaces for which this condition does not hold

but one of these homologies isn't even defined for spaces for which there exists no triangulation, so it's not clear what the remark means

does this answer your question of whether they are equivalent?

[hot night]

Yes thank you.

[gritty widget]

can anyone explain this?

[unreal stratus]

What do you want explained - why we choose that as the definition of convex? (though half the definition is cut off)

[gritty widget]

Let me explain, those are definitions and naming conventions of those objects

Other than that there's nothing to explain here. The definitions aren't some kind of hard constructions

Explaining a definition makes sense in some circumstances, but not really here

[gritty widget]

In that capacity, what would a 3-cell be?

[gritty widget]

[a, b] x [c, d] x [e, f] where a < b, c < d, e < f

that's a parallepiped with all sides parallel to the axes

But how come do we denote the Euclidean space as convex iff the aforementioned condition is satisfied(for the latter part of the definition)

Can you give me a proper statement in English?

the Euclidean space is R^k

denote R^k as convex iff the aforementiond condition is satisfied - what does that mean

uhhh

sorry, what i mean is the statement ... We call a set $E \subset \mbb{R}^k$ convex iff $\lambda \mathbf{x} + (1-\lambda)\mathbf{y} \in E$.

[gentle osprey]

Renegade

[gritty widget]

for all $\lambda\in (0, 1)$ and $x, y\in E$

[gentle osprey]

Blitz

[gritty widget]

here is the full definition

I guess that's the question; why is this the case?

what is? That's the definition. Are you asking for motivation for the definition?

I suppose

I mean it has to be derived somehow

affine functions preserve convex combinations I guess

the notion of convex set is one of the most important in analysis

like someone said before, geometrically it just means that if we have x and y in E, then the whole segment from x to y is also contained in E

okay, so the definition of a k-cell makes sense but geometrically i'm having somewhat of a hard time picturing it

you can also give relation with a convex function and its epigraph being a convex set

https://en.wikipedia.org/wiki/Epigraph_(mathematics)

[gritty widget]

Just, every coordinate is from some closed interval

fair enough. so a $2$-cell implies $a_i < b_i$ for $1 \leq i \leq 2$, right?

so we have, for the Euclidean space $\mbb{R}^2$, and ordered pair $(x_1, x_2)$ which satisfies $a_1 \leq x_1 \leq b_1$ and corresponding $a_2 \leq x_2 b_2$?

[gentle osprey]

Renegade

[gritty widget]

A $2$-cell is a set $${(x_1, x_2)\in \mathbb{R}^2 : a_i\leq x_i\leq b_i, i = 1, 2}$$ for some $a_i, b_i\in\mathbb{R}$ with $a_i < b_i$ for $i = 1, 2$

ah fair

[gentle osprey]

Blitz

[gritty widget]

I

I'll try to find some geometric representation of this

thank you

in my experience, a 2-cell is anything homeomorphic to that

so what's topologically a disk

And $n$-cell is just $${x\in \mathbb{R}^n : |x|\leq 1 }$$

[gentle osprey]

Blitz

[gritty widget]

when doing topology this definition can be more pleasant, because the boundary of an n-cell can be defined more clearly

[gritty widget]

to be honest, one can put any norm in here

[gritty widget]

but this implies that the bounds are -1 and 1

wouldn't the definition of a k-cell be a more generalized version of this

[gritty widget]

what do you mean

[gritty widget]

what is the distinction between a k-cell and n-cell then

i guess

[gritty widget]

the number

of coordinates

[gritty widget]

why are we incorporating the norm here?

[stable kite]

What are the compact subspaces of the Long Line?

[hollow yew]

Here's an attempt to reduce to the Euclidean case. Maybe you could point out errors.

1. A compact subset must be contained in a long-interval, since otherwise it tends to ±∞ (then we can produce an open cover without finite subcovers). Hence, a compact subset must be contained in a closed long-interval.
2. Closed long-intervals are monotonically homeomorphic to closed Euclidean intervals.
3. Hence, compactness of a subset of a closed long-interval is reflected by a monotone homeomorphism.
4. The compact subspaces of a closed Euclidean interval are just the closed subsets, because compact subsets of Hausdorff spaces are closed.

In summary, the compact subspaces of the long line are monotone homeomorphic images of the compact subsets of Euclidean space. In other words, they are just "stretched images" of the Euclidean setting.

[hot night]

Why is the Lebesgue covering dimension a topological invariant?

[golden gust]

it's defined in terms of open sets

[hot night]

Can you please elaborate. It's not immediately obvious for me. Would it possible to write a short chain of logical arguments?

[gritty widget]

a homeomorphism lets you go back and forth between the open sets of two spaces

as far as you should be concerned, homeomorphic spaces are the same with just the points and open sets relabeled

[hot night]

Yup, I can see that.

[gritty widget]

nevermind, i don't know where this was going. i was just going to say "and because it's defined in terms of the open sets..."

this was not bound to help

why not just write down the definition of lebesgue covering dimension and use your homeomorphism to translate it between the two spaces?

it might not be immediately obvious, but have you tried to prove it at all?

[empty grove]

Suppose X and Y are homeomorphic. Take an open cover of Y. Pull this back to an open cover of X. Take refinement so that the order of the refinement is at most dimension+1. Then take the covering of Y induced by this refinement

[hot night]

What do you mean by "Pull this back to an open cover of X"?

[empty grove]

Take image/preimage of all the open sets under the homeomorphism

[golden gust]

suppose {V_i} is an open cover of Y. then we can write it as {f(U_i)} for some open sets U_i of X. the pulled back open cover of X is {U_i}

[empty grove]

Here U_i = f^-1(V_i)

Specifically

You don't want to take just any preimage

[golden gust]

(recall that since f is a homeo, U is open in X iff f(U) is open in Y)

[empty grove]

oh yeah I guess there is only one full preimage for a homeomorphism

[hot night]

"then we can write it [V_i] as {f(U_i)} for some open sets U_i of X" Why is that? Why does homeomorphism between X and Y imply homeomorphism between its subsets of covering?

[gritty widget]

take preimages of open sets

[golden gust]

a homeomorphism induces a bijection of the topologies

[hot night]

Okay I get now that we can say X homeo Y implies U_i homeo V_j=f(U_i), where f is a homeomorphism and U_i,V_j are subsets of the open coverings of X and Y respectively. But how can I deduce that the dimension is equal?

[gritty widget]

But how can I deduce that the dimension is equal?
by reading the rest of moldilocks' message

[hot night]

I'm not quite sure what "covering of Y induced by this refinement" means exactly?

[gritty widget]

take image

[hot night]

Got it.

Thanks.

[hot night]

Does this idea work to show that the Koch curve is homeomorphic to [0,1]? Consider the second iteration E2, it's homeomorphic to E1 - by the map which at every spike/triangle maps it to its base giving E1. Since a homeomorphism of a homeomorphism is another homeomorphism I could show that E2 = E0 = [0,1]. Now I could do it for every natural n to show that E_n = [0,1]?

Equal sign "=" means here homeomorphism.

[golden gust]

yeah this shows that each E_n is homeomorphic to [0,1], but I think you would have to show that this is preserved under the limit

[hot night]

Wouldn’t induction suffice?

[pastel linden]

the koch snowflake is E_infinity

induction applies to every finite index

[unreal stratus]

I claim infinity is finite. Because 1 is finite and if n is finite so is n+1

[gritty widget]

[hot night]

Idk just looking for hopeless ideas

could i finish the proof in any way

Maybe it‘s better to go backwords, so proving first that E0=E1, then E1=2 and so on

[unreal stratus]

The issue is more that proving all En are homeomorphic to the unit interval doesn't show Einfty is

[golden gust]

honestly idk how to do this formally, but maybe something like "the endpoints of each E_n are connected by injective paths f_n : [0,1] --> E_n. the limit of the sequence (f_n) is an injective* path f : [0,1] --> F. then f is a continuous bijection from a compact set to a hausdorff set, so is a homeomorphism".

• the only problem is in showing f is actually injective

[hot night]

Would it suffice to prove that homeomorphism stays invariant under infinite compositions?

[gritty widget]

infinite compositions 😵‍💫

[hot night]

lol

[golden gust]

you're going to run into limits whichever way you spin it

[gritty widget]

can't you just parametrize the curve and argue with that?

[golden gust]

maybe I'm being too anal, I thought you couldn't straight up assume an injective parametrisation of F

because that's the result immediately

[gritty widget]

why would you willingly work hard?

[golden gust]

oh nice, that works

[honest narwhal]

#advanced-analysis is the channel for that

[golden gust]

suppose $B$ is closed. then $\overline{B} = B$ so (3) gives us $\overline{f^{-1}(B)} \subseteq f^{-1}(B)$. but we already have the reverse inclusion so in fact these two sets are equal, i.e. $f^{-1}(B)$ is closed

[gentle osprey]

Average J∘du=du∘j enjoyer

[golden gust]

by the way, I think the existence of a parametrisation is guaranteed by this lemma I found (without needing clever object-dependent constructions)

[gentle osprey]

Nobody

[proud herald]

I don't think Zariski topology on $\mathbb{A}^1_\mathbb{K}$ is Hausdorff.
Then why is my teacher always using that regular functions $f,g\colon V\to\mathbb{A}^1_\mathbb{K}$ that coincide in an open dense set $U\subseteq V$ also coincide on the whole $V$?

[gentle osprey]

SkyTwX

[gritty widget]

Z(f - g) is closed and contains U, so it contains its closure, V.

[fading vale]

Probably worth stating is that there is an analogous property to hausdorffness for varieties/schemes called separatedness which implies that morphisms into it which agree on a dense subset are equal, just like hausdorffness does for continuous maps into a space

[gritty widget]

separatedness is basically making this true by definition

funny how math work

[mint rose]

Ah! I've seen this used in our course but I couldn't understand why!!

The prof said that continous maps that agree on an open subset (Zariski topology) are the same (maps from alg. varieties to a field)

[fading vale]

Intuitively: hausdorff says that the diagonal map X -> X x X is a closed embedding. Separated says that the diagonal map X -> X x_{Spec Z} X is a closed immersion

[mint rose]

I still don't have the vocab to understand scheme lingo tho

[fading vale]

If you're working in variety land then everything is separated

[hot night]

Can I directly prove that the Koch curve has (manifold) dimension = is locally euclidean to R? (or do I run into the same problems as earlier?) Take the atlas to be $\lim_{n\to\infty}\bigcup_{i=1}^{4^n}S_i$, where $S_i$ are straight lines with length $1/3^n$, where $S_i\cap S_j=\text{singleton}$. Take for every point $x\in S_i$ in the Koch curve its neighbourhood to be $S_i$. Also, every $S_i$ is homeomorphic to $\mathbb R$. Doesn't this prove local euclideaness?

[gentle osprey]

epic_morphism

[fading vale]

so you dont have to worry about it

[mint rose]

Okay

Well in the end we just consider polynomial fractions so I didn't really look further

I was just confused by the fact the argument is (continuous + coincide on dense subset) => the same

[fading vale]

would you like a proof of it? its not so bad

[mint rose]

sure

[gritty widget]

does the reduced homology of a pair make sense

[mint rose]

In which way?

reduced relative homology?

[gritty widget]

yes reduced relative homology

[mint rose]

Yeah it does

I was just reading about it in Hatcher

http://pi.math.cornell.edu/~hatcher/AT/AT.pdf

page 118

it's the same as normal relative homology though

unless the subspace is empty

[gritty widget]

ah so

\tilda{H}_n(X,A)=H_n(X,A)

for A non emepty

or maybe do we need A not a point

[mint rose]

Yeah, although I might have to revisit the argument, since I didn't completely understand his justification 😅

Gimme a minute, I'll think about it

now that I think about it

he never actually defined reduced relative homology

that's why I'm so confused about it

[gentle osprey]

Moth

[mint rose]

he ends the sequence with 0 -> Z -> Z -> 0 -> 0 which I don't completely understand as that would imply reduced relative homology is the same by definition....

maybe I'm missing something

[mint rose]

Thanks, pretty succint

So normally we use that Y is hausdorff to show that the diagonal is closed

[fading vale]

Yes

Thats why its analogous pretty much

[mint rose]

but now it's just obvious that it's just V(Y_1 - Y_2)

Thanks

[fading vale]

it would be when A is nonempty

For a pair (X, A)

[mint rose]

you understand the argument?

why does he augment by that short exact sequence

and not something else

because for me that implies directly that for him the definition of $\tilde{H}_k(X, A)$ are the homology groups of $\dots \to C_1(X, A) \to C_0(X, A) \to 0$, but that's just normal relative homology....

[gentle osprey]

Digiteraat

[mint rose]

Unless I missed it he never actually defines it

[fading vale]

The augmentation means that we're setting C_-1(A) to be Z, C_-1(X) to be Z, and C_-1(X, A) to be 0, right?

[mint rose]

yes

why that choice?

[fading vale]

Well its sort of the logical analogy with the usual definition of relative homology

where we set C_-1(X) to be Z

and then take the augmentation map from C_0(X) to Z

[mint rose]

so you're saying it's just because $C_{-1}(X)/C_{-1}(A) = \Z/\Z = 0$?

[gentle osprey]

Digiteraat

[mint rose]

because if that's the case I don't understand why he brings in the long exact sequence

[fading vale]

the LES is how hes defining (reduced) relative homology

[mint rose]

for me the definition is this

then he proceeds to prove they fit into the l.e.s.

in any case it's presented in a strange way, unless I'm missing something

I guess what he's saying is just that we can replace normal by reduced homology in the L.E.S (for the terms not the relative homology) and it still holds...

[fading vale]

uh basically the point is that the reduced relative homology H^tilde(X, A) should be characterized by the LES relating it to H^tilde(A) and H^tilde(X)

[mint rose]

Yeah I guess that makes sense

[gentle osprey]

Moth

[fading vale]

there should be an i under that last H but you get what i mean

[mint rose]

Yeah no worries

[fading vale]

basically the only thing that changes is that when i = 0

[mint rose]

I have an exam in algebraic topology on monday xD

and algebraic curves on tuesday

great times

funnily enough both topics (and even questions I was asking myself) appeared in the span of a few minutes on this channel

[gentle osprey]

Moth

[fading vale]

the difference being there is one less factor of Z

[mint rose]

ahh yeah that makes sense

[fading vale]

as youd expect because then you're just looking at ordinary reduced homology

[fading vale]

Good luck! im sure you'll do well

[mint rose]

Thanks!

[sturdy notch]

pretty sure thats because we’re all following the exact same course at the exact same uni

[mint rose]

lol

t'es qui?

[sturdy notch]

gars de 2eme

[mint rose]

david

moi c'est matthew mdr

[sturdy notch]

j avais devine 🙂

[mint rose]

hmmm

[sturdy notch]

c est bizarre que le prof n ait pas remarque que son argument de = sur un dense n est pas suffisant, je lui croyais sur parole jusqu a aujourdhui

[mint rose]

je lui au posé la question pendant la pause et il était genre tkt c'est un argument de mettop

[sturdy notch]

mdrr je vais demander/dire ca sur piazza demain, je crois ce serait utile pour les prochaines annees

[mint rose]

c'est une bonne idée

[sturdy notch]

bonne chance pour algtop 😉

[gritty widget]

[mint rose]

Merci!

[proud herald]

C'est pas la seul cacahuète dans le cours pfff

[mint rose]

j'avoue

[gentle osprey]

Digiteraat

[gritty widget]

is there some nice condition where i can say that

(X/A,*) is quasi isomorphic to (X,A)

i know that if A into X is a cofibration then (X,A) is quasi isomorphic to X

ah nevermind

[lunar yoke]

This is false btw, but i suspect lime_soup knows this and just stopped typing midway

just as a heads up for people reading this

the correct statement is that if A into X is a cofibration you get that (X,A) is quasi-isomorphic to (X/A,*)

[bright acorn]

If $p : \tilde{X} \rightarrow X$ is a covering map and the singular homology groups (with coefficients in $\mathbb{Z}$) of $\tilde{X}$ are all finitely generated, do we have that the homology groups of $X$ are finitely generated as well?

[gentle osprey]

MisterSystem

[bright acorn]

We can suppose p is an n-sheeted covering for some integer n if it is necessary

[coral pivot]

I dont have a counterexample for n-sheeted cover but

take X= K(G,1) for G infinitely generated

then universal cover is contractible, but H^1 is abelianization of G which should also be infitely generated

[mint rose]

anyone understands why d_1 = 0? Page 141 in Hatcher http://pi.math.cornell.edu/~hatcher/AT/AT.pdf

[cedar pebble]

d_1 sends a 1-cell [a,b] to the 0-cell [b]-[a] but if there is only one 0-cell we always have [a]=[b]

[mint rose]

Hmmm I guess that makes sense - but how do I know the map sends [a, b] to [a]-[b]? is that always the case for 1-cells?

because that's the definition for delta-complexes no?

[cedar pebble]

yes

sorry I should say [b]-[a]

the differential in the cellular chain complex sends a simplex [a_1,...,a_n] to the alternating sum of [a_1,...,a_n] with a_i omitted

so lika d_2 would send [a,b,c] to [b,c]-[a,c]+[a,b]

so in particular it's always going to send a 1-cell to its boundary, the difference of its two 0-cells

[mint rose]

but that's simplicial homology

Here we're talking about cellular homology

so for CW-complexes

[cursive flume]

so I've read up a bunch on characteristic classes. The basic idea is that we have a principal G-bundle, which in particular has a fixed structure group G. We consider the Lie algebra $\mathfrak{g}$ of this structure group. We then have specific polynomials on this Lie algebra, namely, the ones with are Ad-invariant. Then, basically, the curvature form of the principal G-bundle is a form in particular, hence it makes sense to ask in which DeRham cohomology class it lies. The rough idea of the Chern-Weil theory is to write down a concrete group homomorphism Between the space of invariant polynomials on the Lie algebra of the structure group to the DeRham cohomology group. Then characteristic classes basically will be some specific invariant polynomials. A good example is the Chern Class, which is defined as $$c(F):=\det \left( I + \frac{i F}{2 \pi} \right),$$
where $F$ is the curvature 2-form. This makes sense to define it as such, because the curvature 2-form is a Lie algebra valued 2-form. So far, this is all COHOMOLOGY WITH REAL coefficients, which is the big problem. The rough idea how to move to cohomology with integer coefficients is that we define the chomology group with integer coefficients using singular cohomology. We have the canonical inclusion Z->R, which induces by functoriality of singular cohomology a natural map $\psi$ between singular cohomology in integer coefficients $\psi:H(\mathbb{Z}) \to H (\mathbb{R})$. So far so good, but now comes a strong point: we know that singular cohomology group and deRham cohomology group with real coefficients are isomorphic by the famous DeRham theorem, let's denote this isomorphism with $\phi$. We therefore, have the chain:
$$\psi:H(\mathbb{Z}) \to H(\mathbb{R}) \cong H_{dR}(\mathbb{R}).$$
Finally, we come to define integral DeRham classes as images of singular cohomology classes with integer coefficients under the homomorphism $\phi \circ \psi$.

[gentle osprey]

ProphetX

[cursive flume]

can someone give feedback, whether this makes sense, or I am missing some point?

[cursive flume]

to be more explicit, the characteristic classes will be images of the. invariant polynomials under the Chern Weil homomorphism

[cedar pebble]

this more or less makes sense yes

though do keep in mind that Chern-Weil theory does not remember the integral structure

it only recovers the characteristic classes in real cohomology, not integral cohomology

[cursive flume]

right

where can I learn more about integral cohomology?

like classifying spaces

[cedar pebble]

basics of integral cohomology will be in any basic algebraic topology book like Hatcher
for classifying spaces it depends on what you need. I'm trying to remember what is the good reference for the cohomology of some common classifying spaces of Lie groups, this appears in a lot of textbooks

I will also say that for Chern classes, the nlab page is surprisingly well written and goes through all the stuff you need about cohomology of classifying spaces for unitary groups https://ncatlab.org/nlab/show/Chern+class

[pale frost]

The differential they gave is for cellular, for simplicial its practically the same but you alternating sum remove the nth vertex of given simplex

[hot night]

What is the very first step to define and compute homolog? I see that homology (in general) is defined by the quotient of boundary and cycles and done. However I have also heard of the Eilenberg-Steenrod axioms? Or what‘s the idea or relation between them? Can someone just clear this up for me - I‘m very new to alg-top. Thanks.

[cosmic socket]

People came up with a bunch of different homology theories that were similar and Eilenberg Steenrod kind of unifies them and allows you to prove results given that your homology theory satisfies those axioms.

[gritty widget]

-usually singular homology comes first, with simplicial homology introduced as a simpler (and equivalent in many cases, we had this conversation) theory for computation
-the "general homology" you see is something purely algebraic, working for any chain complex of abelian groups or whatever. the homology groups here measure how badly exactness of your chain complex fails
-the eilenberg steenrod axioms are a collection of properties a lot of homology theories on topological spaces satisfy, and you can usually compute the (singular, etc.) homologies of spaces using just these axioms (spheres by an easy induction, for example).

[hot night]

Is a homology theory for example singular or simplicial homology meaning in what setting you work in or analyse the topological space?

So basically the idea is - if I shows smth is true in the ES axioms, then the result follows for most homology settings?

[pastel linden]

yes, and there are different homology theories for various settings that usually agree on "nice" classes of spaces

So for example if a space is triangulizable (e.g. is homeomorphic to a simplicial complex), the singular and simplicial homology groups are the same

same kind of idea with CW complexes with cellular homology

and with cohomology you can have for example de rham cohomology on smooth manifolds which is very computable but only applies to the smooth setting

[hot night]

Just for clarification: We can only compute simplicial homology for triangulizable spaces, right?

[pastel linden]

Yes, simplicial homology is defined in terms of the simplicial structure

It's a nontrivial theorem to show that the homology is independent of the choice of triangulization (or you show it's equivalent to singular homology, which is independent)

[stable kite]

What are the compact subsets of the Long Line?

[hot night]

Interesting. Does it have a name?

[hot night]

Also, I‘ve seen that the homology of a torus is computed by gluing two triangles. How does this work?

[gaunt linden]

You can glue a torus together from one rectangle by identifying opposite sides. Now suppose you've initially made that rectangle by gluing two triangles along one side ...

[odd flame]

https://www.youtube.com/watch?v=zbSKhv5xmNY getting taught topology by a 17 year old makes me just a little sad

[stable kite]

Don't worry about it - different people are different levels of fast and proficient with different topics. Learn at your own pace, and don't compare yourself too much ^^

[odd flame]

i know lol just a bit of a sigh

should i think about a topology on a set X as coming from a basis, or about a topology on a set X that we can find a basis for

basically what "comes first", the basis or the topology

or is it depending on what context it's being discussed in

[stable kite]

A basis generates the topology on your space, it is completely determined by the basis elements, so conveniently you often define topologies as being defined by certain basis elements.

But a topological space still is the pair of a set X and a collection of open sets, we don't need to mention a basis to define a space at all

[odd flame]

but that basis still has to exist

and it could be that topology we've defined

[stable kite]

I mean yeah, there is some basis for your topology in all cases, even if you might not know it

[formal tide]

I've been told there are two closed paths with same basepoint in the sphere, and two homotopies between them, such that the homotopies are not homotopic

can anyone please elaborate on that?

[empty grove]

Sounds like a rephrasing of the fact that the 2nd homotopy group of the sphere is non trivial

[formal tide]

indeeed

I've not seem the proof of that, but in any case I wanted some geometric intuition of that

it seems to me that a homotopy between two paths would be a sheet covering the two paths

and it would seem to me that I can elongate any two sheets so that they match

[empty grove]

Not if the sheets are like upper hemisphere and lower hemisphere

The boundary of the sheets should stay fixed

Otherwise the first statement is false

[formal tide]

okay that solves it

;p thanks!

[golden gust]

often there's no canonical basis for the topology. so like for the plane R^2 with the standard topology, you can take your basis to be

• all open balls (this arises from the standard metric on R^2)
• all open 'rectangles' (this comes from viewing R^2 as RxR with the product topology)
• all open balls with rational centres and radii (this one is useful for second-countability arguments)

[formal tide]

hello, I've been trying to solve this, but I think we can only conclude H1(U u V) is either 0 or Z

this is the diagram I have in mind

[fading vale]

I'm assuming you can use mayer vietoris?

[formal tide]

yup though it's a more basic version I think, it only talks about these arrows

[fading vale]

Okay

Im going to assume you cant use the reduced version of this sequence then

[formal tide]

yeah, not sure what's that so I don't think I can

[gentle osprey]

Moth

[formal tide]

hmmm how does this imply the result?

[fading vale]

working out the point set details took me embarrassingly long here lmfao

but the point is that if we write U and V as the union of their connected components C_1, ..., C_n and D_1, ..., D_m then since U cap V is nonempty, C_i cap D_j is nonempty for some i, j. then any connected component of the union must contain one of the C or one of the D because these are connected sets

Hence k <= n + m

However because C_i cap D_j is non-empty, the connected component containing one must contain the other, and hence there are strictly fewer connected components of U cup V than n + m, that is, k < n + m

so the surjective map Z^n+m -> Z^k has non-trivial free abelian kernel equal to the image of Z in H_0(U) oplus H_0(V) which implies that the kernel of Z -> Z^{n+m} is trivial bc we know that the sum of the image and the kernel of that map is Z and since Z is rank 1 and so is the image the kernel is rank 0

then its just applying exactness

the kernel of Z -> Z^{n+m} is 0, so the image of delta: H_1(U cup V) -> Z is 0

and since delta is injective, H_1 is trivial

@formal tide

Literally spent like 15 minutes thinking about the connected components argument

ur brain on algebra

[formal tide]

lol thanks

my brain on algebra works fine tbh, it's when mixed with topology that it lags/bugs :p

[fading vale]

the easy approach to this problem is to apply the reduced mayer vietoris sequence which is the exact same thing constructed the exact same way but you replace H_0 with H_0 tilde

then reduced H0 of the intersection would have just been 0 and it would have been solved immediately

[formal tide]

hmm I'll check that out then, thanks again

[formal tide]

silly semi-pedantic question, the 0-skeleton of a CW complex must be a "discrete set"; does that mean finite with the discrete topology or have any cardinality with the discrete topology?

[empty grove]

Any cardinality

[formal tide]

ty

[odd flame]

just wanna double check my understanding:
a base of a topology is just some collection of sets such that taking infinite unions and finite intersections of them we can generate a topology

this is munkres' definition of a basis, and im pretty sure im right about what i said b4 but i dont see how it equates to this

in particular i dont really understand what we gain from the second condition or why it's necessary

[odd flame]

also this doesnt necessarily mean closure of the basis right? reading about R^2 as an example and taking the union of open balls doesnt yield and open ball

sorry if that's a lot but pls ping if anyone can help otherwise i wont see till like tm

[golden gust]

actually for a basis you don't need to take intersections. arbitrary unions are enough

[odd flame]

ok i can see that

but im still having a hard time seeing the point of that second condition

[odd flame]

actually how do you get the empty set from a collection of subsets if not by taking an intersection? or is it usually just already included

[golden gust]

empty union

[odd flame]

ok take a random set X = {a,b,c,d,e}

a collection of subsets B might be {{a,b,c},{c,d,e}}

[golden gust]

is this a top space? if so what's the topology

[odd flame]

do you need a topology in order to think about a basis?

i was under the impression that you dont necessarily need a topology to think about a basis of a topology on a set, rather a basis must generate some topology

i was gonna say why can't B be a basis

[golden gust]

a basis determines a topology

[odd flame]

like obviously it violates the condition that c in the intersection isnt part of its own basis element

but how does that prevent it from being a basis

idk if im making sense lol feel free to tell me if im not

[golden gust]

if those are basis elements then they are open, so their intersection must be open

[odd flame]

ahhhhh

[golden gust]

but you can't write their intersection as a union of basis elements which is the problem

[odd flame]

but a basis doesnt have to be a topology itself right?

[golden gust]

yeah exactly, if your basis is a bunch of balls then of course the union of two balls is not always a ball

it just generates a topology

[odd flame]

https://tenor.com/view/balls-lol-wtf-gif-4254879

basis for R^3

[odd flame]

why do we actively exclude writing elements of a topology as intersections of basis elementss

[golden gust]

it's just a definition that turns out to be useful. sometimes we want to be able to say "U is open, so write it as the union of some basis elements" and argue using that

idk if I can comment on the "why" of a definition beyond that

personally I like to think of basis elements as "obvious neighbourhoods". so like in a product space X x Y, the obvious/"least complicated" open neighbourhoods are open rectangles, i.e. sets of the form U x V for opens U of X and V of Y

and the collection of all such rectangles forms a basis for the product topology

[unreal stratus]

There is also the idea of a subbasis which does p much what you want and is more convenient in some circumstances, where you take arbitrary unions of finite intersections of those sets

But if we use bases we only need to worry about unions which I think is simpler to check and more tangible

e.g. the idea of open intervals being a basis for the topology on R is much simpler (to me) than thinking about sets you can form from unions of finite intersections of open rays lol

[swift fjord]

A subbasis is also just a fancy nake for a cover

Name

[grave maple]

Yeah, I guess it depends on the space that you're studying. Maybe the open sets are "obvious" and you don't need a basis. Or maybe the space was defined using a subbasis from the get go.

[unreal stratus]

Ye

[odd flame]

this is what i was asking about!! how does it avoid intersections

like do we define a basis like this because it's more useful only talk about unions?

actually on thing i should make sure im getting right first - a collection of subsets is said to be a basis if it has certain properties, which in turn tells us that it generates a topology on X by taking arbitrary unions (and only unions - this is the part im thinking about) of elements of the basis

[unreal stratus]

I'm not sure what sort of answer to give in terms of 'because' but in any case the notion of a basis is useful because we only need to talk about unions of elements, I'd say

But also very useful in terms of showing sets are open - just take any element x of the subset A and show x is contained in a basic open set contained in A

[unreal stratus]

Yeah there are sort of two ways in which we talk about bases i guess

This is when we talk about being a basis for a topology and then yup we can generate a topology from it

But you can also be given a topology and find a basis for that topology, e.g. the collection of open intervals is a basis for the standard topology on R

[unreal stratus]

Oh well the way we define a basis means the topology it generates will actually be closed under finite intersections - Suppose an element x is in the intersection of two open sets V and W. Then in particular x is in a basic open subset B contained in V and a basic open C contained in W. We need x to be in a basic open subset (since V cap W is open) but the way we define the basis stipulates that this'll be the case! (that is, that we can find a basic open containing x contained in B intersect C in fact)

[odd flame]

wait arent B and C the same there

the condition for being a basis is that if an element x is in the intersection of two basis elements (subsets of X obv) then x must be contained in a separate basis element that is a subset of that intersection

[unreal stratus]

That's what I go on to say

[odd flame]

https://tenor.com/view/face-palm-oh-no-oh-god-damnit-damn-gif-10780870

[unreal stratus]

I'm just trying to word it so it seems natural why we have that condition in the definition of a basis, lmk if I'm unclear / if that's not smth you worried about

[odd flame]

you're wording what i was thinking about better than i was in my own head i think im getting it

[unreal stratus]

Cool nice, I think bases are easier to understand once you've used them more anyway?

[odd flame]

yeah i starte ch2 of munkres today lol

[unreal stratus]

Noice

Oh for example when proving continuity of functions X -> Y it suffices to check preimages of basic open subsets of Y

[odd flame]

spent like the entire day on the section on basis

[unreal stratus]

Ah dw

[odd flame]

so essentially in generating a topology from a basis, the condition about subsets of intersections guarantees that the topology we end up with is in fact a topology and is closed under finite intersections

[empty grove]

One way to phrase the basis condition is to say that the intersection of 2 basis elements is a union of basis elements. So if we define open sets to be unions of basis elements, the condition says exactly that the intersection of 2 basis elements is open. Then intersection of 2 open sets is open because intersection distributes over union.

I find it more useful to state it this way than saying "for every point in the intersection, you can find a basis element containing it contained in the intersection". One is easier to remember, the other is easier to check

This is kinda what potato said too but in a different way

[odd flame]

moldi

im assuming the latter is easier to check right

[empty grove]

Yeah, because it's exactly what you get after you expand the definition of a set being a union of a collection

Try to expand that definition and you'll arrive at the definition a basis

[odd flame]

actually im tryna see how to get from a basis to a topology

munkres seems a bit obtuse about it

should i worry too much about the proof he gives

[empty grove]

The open sets in the generated topology are defined to be exactly the unions of basis elements

[odd flame]

homie munkres spends a page and a half doing that rigorously it seems

[empty grove]

[empty grove]

If you can check that the definition of a basis is the exact condition which makes this topology closed under finite intersections, you can skip whatever munkres says ||he probably says the same thing||

[odd flame]

it seems like it just harder to parse for the sake of rigor

[empty grove]

It seems that he defined the generated topology earlier

Since in the first image he's already assuming that you know what T is

[odd flame]

oh yeah

[empty grove]

oh god

[odd flame]

though i also found that a bit funky cuz he doesnt really mention unions

ok so it's not just me right, it is terse af

[empty grove]

That definition is exactly what it means for the open sets to be unions of basis elements

It's not terse, he should at least mention that this is an unravelled definition

[odd flame]

i just wanted to say terse

[empty grove]

Terse would be to just give the shorter definition

[empty grove]

Understandable

[odd flame]

namely

[empty grove]

Open sets are unions of basis elements

[odd flame]

yeah i can see that in my head

[empty grove]

You should check that these 2 really are the same

At least once properly

[odd flame]

this and his definition of generation?

[empty grove]

Yes

[odd flame]

tomorrow problem

[empty grove]

,ti nitezba

[gentle osprey]

The current time for nitezba is 10:19 PM (EDT) on Wed, 06/07/2022.
Moldilocks is 9 hours and 30 minutes ahead, at 07:49 AM (IST) on Thu, 07/07/2022.

[odd flame]

im gonna assume it's useful to know how to go from a basis to a topology and vice versa right

basis to topology just being arbitrary unions

but i looked ahead and it seems necessary to be "rigorous" about proving that something is indeed a basis so i gotta do that too

[empty grove]

Ye but before being rigorous you should understand why the rigorous definition lines up with intuition

[odd flame]

epic that's what i tried to do today

[empty grove]

There's no vice versa here

[odd flame]

hoping the next three sections will go by faster since theyre examples

[empty grove]

Examples

[odd flame]

from a topology to a basiss generating it?

[empty grove]

Yeah I mean the only natural assignment of basis is the full topology

[odd flame]

ngl examples be lowkey tedious but imma force myself to speedrun tomorrow

[empty grove]

And that's a completely useless basis

[odd flame]

then why give it a lemme mr munchers

[empty grove]

That's not a natural assignment

This is a lemma for checking whether a collection C is a basis or not

Not for producing a basis from the topology

[odd flame]

https://media.discordapp.net/attachments/958525246992961566/959867718583062608/it_is_pointless.gif

i'll bother u tm moldi thank you for the help

and potato

[empty grove]

[unreal stratus]

np

[long hornet]

I find this proof in Hatcher confusing. He's proving Poincaré duality for noncompact manifolds (p. 248).

I thought we are already done by the (3) because we always have countable charts. Why is the "completely general noncompact manifold" different?

[empty grove]

His definition of manifold might not include second countability

[long hornet]

Right! Thanks. I checked and it turns out you're right.

[elder edge]

Anyone has solutions to morris book topology without tears?

[hot night]

Are there any fundamental homology theories I should know about besides simplicial, singular and cellular homology?

[pastel linden]

those are the big ones to start with

[fading vale]

Depends heavily on what you're interested in

[pastel linden]

does anyone have a good resource to get started on classifying spaces?

I was reading this article and it sounds incredibly neat

https://math.ucr.edu/home/baez/calgary/BG.html

[fading vale]

The nlab page

Unironically

[pastel linden]

[fading vale]

It depends on what you want to do with them I guess but the homotopy theory side of things ive seen from there was done pretty well

[pastel linden]

I'll take a gander

[fading vale]

If u want to do bordism the thom spectra page is also pretty good

[coral pivot]

https://math.mit.edu/~mbehrens/18.906spring10/prin.pdf

its where i learnt things like BG from^

[odd flame]

it's probably late where you are but can you elaborate a bit on what you meant by this pls

[odd flame]

also how would this be done "properly"? thinking about open balls on R^2 it's kinda obvious

suppose you have a subset U of a topological space X, and going with munkres' defn, it is open if every element in U is contained in a basis element B that is fully contained in U

then simply by definition U is literally a union of basis elements

or is a basis element itself

[empty grove]

To say that X is a union of sets from the collection C is the same as saying that for every x in X, there is U in C such that x ∈ U ⊂ X

No topology here, just set theory

@odd flame

[odd flame]

ok that's what i thought

[odd flame]

i think i was overcomplicating this

he literally is just checking that some collection of open sets is a basis by making sure that every element of the topology on X (ie every open set) is either an element of the collection, or the union of elements of the collection

cant believe i spent an obnoxious amount of time on his ch1 on set theory only to get tripped up by set theory when i do the actual topology

[empty grove]

That's not the entire proof though

[odd flame]

just to make sure asking something like "find a basis for this specific topology" is unreasonable in the sense of "generating" the basis right - the only thing you can do is take a collection and check if it's a basis

[empty grove]

Well it kinda is nvm all that's left is to say that nothing else is a union of those elements but that's obvious

[empty grove]

You can still ask to find a basis

[odd flame]

oh i didnt even read the proof i was just reasoning through this

[empty grove]

But then there is always a stupid answer of whole topology

[gritty widget]

In practice we don't give topology explicitly, but provide a basis/subbasis that generates it

It's sometimes interesting to find some basis which satisfies desirable properties

For example, a countable one

Examples of this are everywhere. Product topology being the most obvious example

example

Similarly topology of any metric space or similar constructions are given by an (obvious?) standard basis

[odd flame]

gotcha

separate question: im reading about comparing bases of topologies, is the topology of open balls on R^2 comparable to the topology of all rectangular regions of R^2?