point-set-topology

[unreal stratus]

yeah

And it's less fun that way anyway owo

[feral copper]

Yeah 😛

Although the Whitehead lemma does use the CW-topology

[solemn oar]

Yes, but working with a concrete complex is nice pedagogy.

[trail charm]

given a mapping $f : X \to Y$, why is the canonical map $r : M_f \to Y$ a strong deformation retraction (where $M_f$ denotes the mapping cylinder)?

[gentle osprey]

anamono for anamono

[unreal stratus]

(are these the pointed or unpointed constructions? not that it matters too much)

[trail charm]

not sure

if it helps, this is page 46 of bredon; the word "pointed" doesn't appear until page 128

[coarse night]

True by definition

[solemn oar]

If you look at the construction, there is a good candidate for the retracting homotopy. Can you see it?

[trail charm]

wdym construction? sorry

[unreal stratus]

like, the definition of Mf and what the map r looks like

[trail charm]

ah

bredon says r : Mf -> Y is a retraction of Mf onto Y induced by the projection X x [0,1] -> X x {0}, but i'm not sure what it means for a retraction to be induced by a projection here

[lunar yoke]

Best proof of this is that S^inf = |N(J)| is the geom realization of the nerve of the walking iso J, i.e. J has objects 0, 1 and one isomorphism 0 -> 1. Clearly J is equivalent to the pt, so this pulls through to see that S^inf is htpy equivalent to |Npt| = pt

[odd flame]

potentially silly quesetion

but what is the defining characteristic of a SES

like ok yes i see this definition

is the isomorphism a consequence or part of the definition though?

[hidden crag]

It’s applying the definition of an exact sequence and then using iso theorem

[odd flame]

homology is taking a while to get used to

[hidden crag]

Surjection means C is iso to B/Ker beta

Ker beta is im alpha

Im Alpha is iso to A because it’s an injection

The definition is only Ker = Im

Excuse the sloppy writing I’m on phone

[odd flame]

at the bar again timo

[long prairie]

answering maths questions at the bar

[hidden crag]

No I’m in a car

Not driving

[hidden crag]

Sometimes

[odd flame]

so the actual definition is the surj/inj stuff

one day im gonna take a shot before an exam

[hidden crag]

No that’s a consequence of there being 0s on the left and right

[odd flame]

it makes my driving better, should sharpen my mind for an exam too

[hidden crag]

Okay well I guess you could argue that they’re strictly talking about SHORT exact sequences

But they’re just a special case of exact sequences in general

[odd flame]

so it's the ker = im stuff

[hidden crag]

Yes

[hidden crag]

Prove this

[buoyant dew]

i come once again asking for help

can anyone explain why those two quotient spaces would be isomorphic

the statement clearly does not hold for arbitrary subspaces A,B. so what is the property of CW complexes that makes this work

[gentle osprey]

MyMathYourMath

[pseudo coral]

f it i might just present it and say this is left as an exercise to the interested reader lol

[buoyant dew]

hmm what did he mean by this

[unreal stratus]

Tbh this is a good question

Hm

So like there is certainly a continuous bijection from one to the other

[buoyant dew]

hold on

[unreal stratus]

Apparently you can do it just by looking at cell structures of each

[buoyant dew]

oh yeah i see the map X/A -> B/A n B

[unreal stratus]

I am too finite cw complex pilled for this and was gonna just say use compactness lmao

[buoyant dew]

ah cool, i see. do you have a reference to that

[unreal stratus]

I don't lol sorry but I mean you can create a cell structure on X/A by taking the cells away from A and having the basepoint as a 0 cell

[buoyant dew]

i shall ponder upon this a bit

thanks for the help @unreal stratus

[unreal stratus]

Hm ok I need to check lol

Nah I think it's fine

Take a homotopy A x I -> A from the identity to a constant map

And also the identity X -> X

These r compatible giving a map X x I -> X as desired

Then ig the point is at t =1 we have a map X -> X sending A to a point, giving a map X/A -> X which is as desired

Does that make sense

Yes sure

Sorry for not making that clear

Happened to look at this recently lol

Well there was a case where this lemma provided an easier proof if we work w CW complexes

Since then we get hep

[bitter smelt]

hatcher's formal discussion is partially in the appendix

[odd flame]

can i get a nudge on this pls

im guessing this is a purely algebraic question

Z -> Z + Z/2 im guessing is the map sending 1 -> (1,1)

what would Z + Z/2 -> Z/4 be

[nocturne basalt]

I mean you should be able to play around w what 1 and [1] are sent to

you only have 4 options for each

[odd flame]

but it is just algebraic right? like we dont need homology or anything?

[nocturne basalt]

yeah

[odd flame]

also just to be sure but this is the kinda thing where i should be fine just saying "clearly inj homomorphism" right

[nocturne basalt]

yes

[odd flame]

the only surj hom is the one sending (0,1) -> 3 and and (1,0) -> 1 right

sorry im def overthinking this

[nocturne basalt]

also oddly enough, this SES is something you can use to compute H_1(RP² with the boundary of a Mobius strip attached to RP¹) lol

well it is surjective

[odd flame]

yeah but it doesnt have ker = Z

am i thinking about this wrong? ker has to be Z for it to be exact right?

[nocturne basalt]

well you might not have the correct map from Z to Z oplus Z2

[odd flame]

O

but there cant be an injective map Z -> Z + Z/4 with image Z/2

[nocturne basalt]

this seems random but it might help

[silver umbra]

https://math.stackexchange.com/questions/1517052/a-homeomorphism-of-bn-fixing-the-boundary

[nocturne basalt]

at least if you're familiar with Mayer Vietoris

[silver umbra]

how is the mapping in the first answer here even a bijection? (let alone a homeomorphism)

[odd flame]

i am not, not in the scope of this hw

[odd flame]

this is true thought right? or is the tiredness getting to me

[silver umbra]

here is the map in question

how is the map phi given here even injective?

[nocturne basalt]

seems right

[odd flame]

ohhhh wait im just silly

am i

hold up

ok nvm

the image of the injective map has to be Z

but the kernel of the surjective map keeps being Z/2

but we need this equal no? what am i missing

nvm i got it

on a different note im having trouble understanding the degree of an automorphism of S^n

this is basically the definition i have

https://math.stackexchange.com/questions/2205452/local-degree-of-a-map-between-n-spheres this is the MSE post it's from but not the question im working on

that would be this

[willow viper]

hello quick question: suppose I have a topological space X and Y is a subset of X. I have to show that the collection of all open subsets of X that are contained in Y is a topology on Y.

For the intersection. Can I just say that $T_1 = {U\subset X | U\subset Y~open}$. And $\cap_{i=1}^n U \subset Y~open$ because the intersection of open sets is open again and $Y\subset X$ so $\cap_{i=1}^n U \in T_1$ and the same for union?

[gentle osprey]

b3s4d

[odd flame]

not sure what you mean by that T_1 but that seems mostly right

[willow viper]

I have defined T1 as the topology I have to prove to be a topology

[odd flame]

the subspace topology is an important and common example, munkres likely has some details about it

oh wait i think this is slightly different, ignore me

[willow viper]

Ok

[willow viper]

It just seems so weird because I didn't proof anything really??

[odd flame]

ok found this in hatcher but how is that a homotopy

plugging in 0 for t you get f(x)/f(x) = 1

and plugging in 1 you get -x/x = -1

[dry jolt]

Note that he takes the norm of the denominator

[odd flame]

ohhhh that's not absolute value

,av walter

[gentle osprey]

walter#1555's Avatar

Click here to view the image.

[dry jolt]

Yeah, you divide by the norm to ensure that the image of the homotopy is in S^n

The assumption that f has no fixed points ensures the denominator never vanishes

[odd flame]

so we just get f(x)/|f(x)| = f(x)

[dry jolt]

That's right

[odd flame]

cool beans

thank you waltuh

[dry jolt]

Happy to help

[odd flame]

still wondering abt this btw

like how useful is the degree of a sphere automorphism

i'll assume it's pretty important

can we use it to disprove stuff

https://tenor.com/view/waltuh-heart-locket-my-beloved-walter-white-gif-26530050

[dry jolt]

Oh degree is very useful

For instance, you can show that if n is even and f: S^n -> S^n, then either f has a fixed point or takes a point to its antipodal point

[odd flame]

i have that as a problem in a bit actually lol

[dry jolt]

Similarly, you can show that S^n has a non-zero vector field iff n is odd

Degree is also foundational in defining cellular homology which is invaluable for computations

[odd flame]

i see

is there any intuition for it? or is that an odd question to ask for thiss

is this defn of degree only applicable to automorphisms of Sn

this is more #groups-rings-fields

[quiet thorn]

hmm ok yeah i agree

[odd flame]

savin this for later dont mind me

[odd flame]

how does deg = -1 imply that there is antipodal identification

[odd flame]

<@&268886789983436800> idk if this warrants but self promo in multiple channelss

[ivory dragon]

it does

[gritty widget]

only the first one is bounded am i right? and the others are not bounded

[feral copper]

B and D are too

Wait

D is for sure

B is not indeed

[gritty widget]

oke thanks

[gritty widget]

is it correct ?

[feral copper]

Wtf why don't they allow you to use the pre-image of a closed set via a continuous map?!

That's torture x')

Didn't look the details of all of them but seems like you got the gist!

Only E is open, the others are closed

(unless empty in which case they're also open, I didn't check)

[gritty widget]

oke thank you again

[willow viper]

hello how can I start proving this?

[wispy veldt]

For the sufficient condition , pick a open set with respect with d and try to show its open with respect to d' ( how do you characterize open sets when you have a metric ? )

For the necessary condition , think open balls.

[willow viper]

And I need to prove the necessary condition?

[wispy veldt]

yes , its if and only if

[willow viper]

Yes, I've started normally with assuming that d and d' generate the same topology on M

[wispy veldt]

correct , now how can you define open sets with respect to a metric d ?

[willow viper]

so an open set is a collection of open balls for any x in M

[wispy veldt]

a union of open balls to be precise , so if $U$ is open then for any $x\in U $ you can find $r>0$ such that $B(x,r) \subset U$

[gentle osprey]

Susilian

[wispy veldt]

you already assumed they generate the same topology , what does that tell you about open balls ?

specifically $B_{r}^{(d)}(x)$

[gentle osprey]

Susilian

[wispy veldt]

here ofcourse i mean open in the topology generated by the metric d , and so the ball B(x,r) is defined using this metric

[willow viper]

yes so just x in M : d(x,y) < r

But I don't really understand the radii r1, r2 and d' and d

[wispy veldt]

ok lets start from the metric d

the open balls with respect to the metric d are open , do you know this?

[willow viper]

Yes

[wispy veldt]

sweet , now d and d' by our assumption , generate the same topology

[willow viper]

Ok so the metric d induces a topology

Like T(d) = { U \subset M | U is open wrt metric d }

[wispy veldt]

right , and the open ball $B_{r}^{d}(x)$ for any r you pick ( lets call it $r_{1}$ is open with respect to the metric d , and since d and d' generate the same topology then what can we say about $B_{r}^{d}(x)$ with respect to the metric d' ?

[gentle osprey]

Susilian

[wispy veldt]

do you know what happens when 2 metric generate the same topology?

[willow viper]

I'm thinking xD

[wispy veldt]

tyt , no pressure

[willow viper]

Ok

So Br^d(x) = Br^d'(x)

[wispy veldt]

not necessarily , but T(d)= T(d')

[wispy veldt]

according to your notation

[willow viper]

Oh wait

Ooook

Makes sense yea I missed that

[wispy veldt]

sweet

now $B_{r}^{(d)}(x) \in T(d)$

[gentle osprey]

Susilian

[wispy veldt]

can you pick it up from here?

[wispy veldt]

Remember to use this too ^

[gentle osprey]

Matplotlib

[feral copper]

Well I guess I could directly ask whether the pullback functor preserves Poincaré duals or something, but I'm no cat(egory theorist)

It seems to me that I'd need some sort of homology pullback and not pushforward...

[novel acorn]

there is

[feral copper]

Yeah I noticed https://mathoverflow.net/questions/115764/pullback-map-in-homology but it seems too untractable for me when it comes to actual computations...

[gentle osprey]

Matplotlib

[novel acorn]

I think it goes that $\Gamma_X = f_* \circ \Gamma_Y \circ f^*$

[gentle osprey]

Irony Incarnate

[feral copper]

Ah yes that makes sense

Wait no, this cannot be if f is not a homeo

[novel acorn]

it works for any continuous f

Like draw a diagram

and remeber that the f's go in different directions

so f* maps H^2(X) -> H^2(Y)
Then Gamma_Y maps H^2(Y) -> H_2(Y)
And then finally f_* maps H_2(Y) -> H_2(X)

[feral copper]

Maybe I'm missing something but what if either of f_* or f^* are not bijective? I mean, cannot it happen that the composite f_* G f^* is not an iso?

[feral copper]

Yeah well is that diagram commutative actually?

(it being commutative is equivalent to the formula)

I feel like there should be some splitting of the adequate homology groups by the kernel and cokernel of the corresponding f-maps

Whatever, I'll give it a thought in half an hour when I'm back 🙂 thanks for the discussion!

[novel acorn]

Oh sorry

I was wrong

the assumption is that the continuous maps preserve fundamental classes

[novel acorn]

You can probably show it by diagram chasing

[gritty widget]

Is there no algebraic structure on projective space?

[unreal stratus]

Wdym

RP^1 is a topological group for example

[gritty widget]

Oh interesting, I did not know. Is this the same for higher dimensions?

[unreal stratus]

Tbh this is an interesting question - I'm not sure when RP^n admits the structure of an H-space structure, let alone a topological group structure

[noble osprey]

Hi, i am trying to prove that the closure of W is a closed set. When we suppose the complement X\W is not open does that imply it is closed?

[viral atlas]

the closure of W is closed iff X\cl(W) is open

You may proceed by contradiction, assuming X\cl(W) is not open (which is what I think you were getting at)

[noble osprey]

Thank you.

[stark fog]

you can think of its definition as an intersection of all closed sets containing W

[feral copper]

Okay 🙂

Ty!

[noble osprey]

I think i am kind of confused why U is closed implies U is a closure because in my notes it says closure U is the set of points in X which are limits of sequence in U. So what does its got do do with it being a closed set? And when we mean a set of points in X is that a collection of sets with limits in X. I cant seem to get a picture in my head 😦

[bitter smelt]

What do you mean "why U is closed implies U is a closure"?

Are you suggesting that all closed sets can be realized as the closure of an open set? If so, this is not true. Take R with the standard euclidean topology. A point is closed. But a point here is not the closure of anything but itself.

Anyway the question amounts to definition unpacking. Write down the definition of closure (i.e., take your set W, include all limit points in X). Then write down one of the definitions of closed which might be most useful (contains all limit points in X)

A set of points in X is just a set of points which are in X. This set may or may not include its limit points.

[noble osprey]

Like the proposition states that in a metric space with U in X, U is closed iff U = closureU

[bitter smelt]

Right, so you mean to ask why does a set being closed imply that the closure of it is itself

[bitter smelt]

In which case, read this

[noble osprey]

yea

Okk. ty

[bitter smelt]

Feel free to ask follow up questions about the definitions

[noble osprey]

No probs ty for your help 🙂

[bitter smelt]

@gritty widget https://math.stackexchange.com/questions/3421788/topological-lie-group-structure-on-projective-spaces

(Elementary answer and hint, though you might want to verify for yourself the implicit claim)

[unreal stratus]

Oh lol

Didn't realise ye sure it lifts nice

Not totally sure how to show S^7 doesn't have a top group structure

Eh OK it'd just lead to an associative normed division algebra of dim 8 which is impossible

[coarse night]

😞

Wish I could understand the proof

It’s easy to show S2 is not a Lie group but how hard is it to show itself not a topological group?

[trail charm]

for defn 2, can "onto at all points" be replaced with "epimorphism at all points"?

i ask because defn 1 uses "monomorphism" instead of "injective"

[golden gust]

yes

[coarse night]

same thing

[unreal stratus]

Not even an H-space 😎

Read my mini thesis thing where I discuss H-space structures on spheres 🤓 jk

Okay this seems nice

https://math.berkeley.edu/~seewoo5/S2notLiegrp.pdf

for S^2

In fact it isn't gonna be too different to show you can't have H-space structures on S^{2n} for any n

[coarse night]

I'll have a read

[umbral panther]

If you want to show it’s not a Lie group, you should classify 2d Lie groups. There aren’t many

[coarse night]

But I haven't read H-spaces

[unreal stratus]

they cool

[feral copper]

Doesn't S^7 have an h-space structure induced by inclusion in the octonions?

[unreal stratus]

ye

[feral copper]

Maybe I didn't read properly what you said then 🙂

[unreal stratus]

S^0, S^1, S^3, S^7 are the only spheres w H-space structure

[coarse night]

bed time read for today

btw are a phd stud?

[feral copper]

Yeah the other ones are Lie groups 😄

[unreal stratus]

i am third year ug

[coarse night]

[unreal stratus]

Yeah idk how to show S^7 isn't a Lie group

how hard is that

[unreal stratus]

hbu

[feral copper]

PhD in what? 🙂

[unreal stratus]

like third year undergrad lol

[feral copper]

Not even a group (for the octonion structure), but idk tbh, I feel it's pretty hard

Ah sorry!

[coarse night]

1y phd

[feral copper]

😇

[unreal stratus]

ah noice coolie

[coarse night]

[unreal stratus]

Actually yeah I rckon it is just like

well i said it earlier lol

if S^7 were a topological group then we would have an 8-dimensional associative normed division algebra right

[coarse night]

everything is a Lie group

[unreal stratus]

but then if that is true like

wouldn't the S^2 group thing also be easy-ish

[feral copper]

Smth like this

[unreal stratus]

by showing there are no division algebra structures on R^3

[feral copper]

S^2 not parallelizable

[unreal stratus]

tho tbh idk how to show that is the case

[feral copper]

The vector space R^n possesses a bilinear product operation without zero divisors only f or n equal to 1, 2, 4 or 8.

[unreal stratus]

Indeed

[feral copper]

You gotta use this to at least restrict, and then you rule out S^7

Poor thing 😦

[unreal stratus]

Yeah I meant like given that theorem lol

[unreal stratus]

Basically I am writing smth up and prove that like only S^0,S^1,S^3,S^7 are H-spaces etc and deduce this

But never looked at S^7 aha

[feral copper]

I think this does it? https://math.stackexchange.com/a/12474/259363

Third cohomology should be nonzero

I had never seen this for non-abelian Lie groups

[unreal stratus]

Noice

Parallelisability of spheres is pretty fascinating

[feral copper]

Indeed

[unreal stratus]

It's cute how there's this story linking like Hopf invariant 1 and stuff with vector fields on spheres through that and others

Adams 😍

[feral copper]

This Cartan form would do a great exercise for an exam!

[unreal stratus]

What do you do Matplotlib?

[feral copper]

Tell me more about this sempai 😇

[unreal stratus]

What's this 👀

[feral copper]

In the MSE answer

[coarse night]

plt.plot

[feral copper]

Like for a 'living'?

[unreal stratus]

Oh I wasn't sure if you were like a PhD student or smth

or ug or like a big human who actually works xd

[feral copper]

2nd year phd yeah, in geometric topology I'd say 🙂

[unreal stratus]

Noicee

[feral copper]

What are you interested in? 🙂

[unreal stratus]

Oh well it's like

Cool that iirc Milnor and Kervaire proved which spheres are parallelisable (ofc 0,1,3,7), but then hopf invariant 1 shows that parallelisable => n = 0,1,3,7 <=> division algsbra structure on R^{n+1} etc and the vector fields on spheres problem gives you an explicit formula anyway

Like kinda cool you get the same results / different implications in different ways and it all matches up nicely

[feral copper]

Ah yeah it's nice indeed! Makes me think of the Milnor S^7-invariant; when he discovered all 28 possible distinct smooth structures on S^7, he showed that they were indeed distinct by computing an invariant, which happened to be Z/28-valued and surjective. Little did he know he had actually found them all!

[coarse night]

[unreal stratus]

oo where should i read about that lol

that sounds based

[unreal stratus]

Ah very cool :)

I'm interested in like topology + algebraic stuff ig, like leaning towards this alg toppy stuff

[coarse night]

It actually shows S² is not a Lie group not a topological group

There's also the Poincare Hopf index formula that also gives S² can't admit a non vanishing vector field and not hence not Lie

[feral copper]

Uhm, I remember reading this on a mathscinet review of Milnor's paper on the 28 S^7's

[umbral panther]

That doesn’t sound right to me. There’s an easy Z/7, and maybe he discovered that first, but by the time he got the Z/4, I thought he had a proof it was everything

[feral copper]

Seems like you're right, the mathscinet doesn't talk about that, I must've dreamt it x')

But still, I remember reading about a complete invariant in Z/28 somewhere... oh well

[umbral panther]

I don’t know any way to write down an invariant. The problem naturally splits into two pieces, the easy case detected by the cobordism invariant, and the hard part, bp, for bounds a parallelizable manifold. Now that I think about it, there might be an extension problem?

[feral copper]

Gotta be honest here, that's about as far as I know 🙂

[umbral panther]

Maybe you’re thinking of the fact that km computed the group of manifold homotopy equivalent to the sphere, but didn’t know that they were homeomorphic? until Smale proved the h cobordism theorem, at the same time

[feral copper]

Ah that's a possibility yeah, but tbh I didn't know Kervaire--Milnor did that x')

[coarse night]

How should I go about it?

[little hemlock]

Literally proved this today in class

[coral pivot]

the proof I know uses K theory lol, its deceptively difficult

[little hemlock]

we used stiefel whitney classes

[unreal stratus]

Oh i didn't realise you could do it with stiefel whitney

just do algebra 🤓

[willow viper]

But isn't U an element of T(d) and B an element of U?

[coarse kestrel]

Do double complexes arise from some natural topological construction?

[cedar pebble]

sure, consider the homology of a product XxY, this is computed as the homology of the total complex of a double complex coming from the homology of X and of Y

most spectral sequences come from double complexes

[coarse kestrel]

I see ok thanks

[coral pawn]

Can someone help me find a reference for something? Given an oriented real vector bundle V of rank r over an oriented manifold X of dim r and a section s of V with isolated zeroes, the Euler class of E can be written as $\sum_{z\in Z(s)}sgn_z(s)$.

[gentle osprey]

Finitely Many Bananas

[coral pawn]

where sgn_z(s) is positive if s preserves orientation near z and -1 if it reverses it

[coarse night]

You can look at Guillimen Polark(forgive spelling )in the oriented intersection chapter

It uses a little morse theory tho

[golden gust]

how is this proved?

[feral copper]

Ad was said by others far more clever than I am, Stiefel--Whitney or K-theory

[supple sable]

Is there a general approach to find the euler characteristic of a planar representation? I have one with the word abdacb^-1c^-1d^-1 and i struggle how to find the number of edges vertices and diagrams since some of them are identified with eachother

[coarse night]

it's 2cells - 1 cells + 0 cells

[supple sable]

Could you elaborate?

[coarse night]

what's your definition of Euler char?

[supple sable]

vertices - edges + faces

[coarse night]

that's what I said

[supple sable]

yes

but how did you get those numbers

"i struggle how to find the number of edges vertices and faces since some of them are identified with eachother"

Like in the diagram i see 8 edges, how did you know 6 of them are identified with the other two?

[coarse night]

only count the distinct ones

[supple sable]

ok so what i do for vertices i put a 1 at the end of edge a. This is then identified with the other front of edge a (beginning of c) so the other beginning of c must also be this vertex, then you get the end of edge d which is the beginning of edge a which is the beginning of edge d which is the end of edge b which is the end of edge c so i only get one vertex

since this process just gives me 1 vertex

?

4 letters means 4 edges?

and just a single face since its a polygon?

so euler is 1-4+1=-2

?

[solemn oar]

Yes, in the fundamental polygon of a surface all vertices are identified, and you have one edge for each letter.

[supple sable]

Okay so then this surface is non orientable since it has letter a twice and since Ng has euler characteristic 2-g it must be homeomorphic to N4 since any non orientable surface is at least homeomorphic to one Ng?

2-2g for orientable 2-g for nonorientable?

[coarse night]

yes

[solemn oar]

My mistake. But yeah, once you have the genus you can use the classification.

[honest bison]

What makes the difference y is odd

What if y is even

Thanks for your future help

[solemn oar]

You should try to solve it in both the even and odd cases. What did you try so far?

[unreal stratus]

I can sketch a proof if u want lol

[golden gust]

yes pls

[honest bison]

Im try to learn, ı am not good enough. I try but ı cant go so far

[solemn oar]

Do you know what it means for a subset to be dense in Q?

If not, start by looking up the definition.

Then see if you can apply it to your subset A here.

[gentle osprey]

potato

potato

potato

potato

potato

[unreal stratus]

*typo, Adams' not Adam's lol

but ye @golden gust if ur interested lol

[pearl holly]

Nice

[unreal stratus]

I think an alternative (topological) way would be that you can show S^{n-1} is paralellisable iff there is a normed division algebra structure on R^n

and then you can prove that only S^0, S^1, S^3, S^7 are parallelisable (though i don't know the direct way to do this, only that it is a consequence of Hopf invariant 1 and of Adams solution to the vector fields on spheres problem lol)

[golden gust]

very cool

[unreal stratus]

no u

[golden gust]

I assume something close to this is the subject of your essay

[unreal stratus]

Kinda cute how it's like lol, it all fits together nicely

[unreal stratus]

This is a chapter of my essay aha

[golden gust]

nice

[unreal stratus]

which i need to work on in ab it

[golden gust]

due in 12 days

[unreal stratus]

ho lee fuc

is urs going okay

[golden gust]

I think it's going well enough but like I wish I started writing earlier

bc there's always the sneaky little details

[unreal stratus]

Aw yeah sure

Me too lol

Idek how much the details matter cause I keep finding resources online that kinda skim over certain details lol

[golden gust]

I'm curious, are you able to like fully prove things

[unreal stratus]

• the word limit seems harsh on this

[golden gust]

or do you need to sketch / reference a lot

[unreal stratus]

this

I mean okay I basically either prove properly or refer

[golden gust]

yeah I'm glad it's not just me

[unreal stratus]

I am over the word limit and like half the stuff is blackboxed lol

ok not half but a decent bit

[unreal stratus]

Yeah

Idk what they expect though

[golden gust]

nice place to be in though

[unreal stratus]

one of my advisors seemed to want me to prove as much as possible lol

[golden gust]

I'm a bit under but approaching

[unreal stratus]

but this isn't like a research paper ig lol

[unreal stratus]

kinda lol, been trying to cut stuff out for ages xd

Pain

[golden gust]

I emailed a guy who's essay I found online and he said

As an undergraduate it is much more important to present that you understood the essence of a proof or concept rather than to just present all technical steps in detail. My biggest challenge was to stay within the word limit but to still include everything necessary to "tell the story".

so it seems universal

[unreal stratus]

Ah sure

[golden gust]

but I've been trying to at least give the key / moral ideas of a proof if it's not a good idea to copy it down

there's still a lot of blackbox though

[unreal stratus]

Yeah fair hm

No idea what the like marking is like lol

Feel like I'm going in a bit blind w that idk about you

[golden gust]

same

just closing my eyes and praying I'm doing the right thing

[unreal stratus]

Mood

[shadow charm]

Im gonna save this discussion for when I ask myself the same questions when writing my bachelor’s thesis in a year’s time

[long prairie]

same but in two years' time

[umbral panther]

That sounds like the definition of the Euler class. What’s your alternate definition?

[void gazelle]

Hi, guys, suppose we have a short exact sequence of chain complex, then why this is a commutative diagram?

[hidden crag]

Per definition

[unreal stratus]

ye

if you want like it is clear every square is commutative by definition of chain map

and then that shows the whole thing is

[void gazelle]

i still cannot see why each square is commutative

[unreal stratus]

I suggest you look at the definition of a chain map again

[void gazelle]

Oh, i see.....

Thanks!

[grim knot]

Let A be in X a retract and rho: X-> A it's retraction. Let now a be in A and x0 be in X. Why is this thing surjective? I wanted to show it directly. I mean we know that rho* is a grouphomomorphism, so it should be enough to show that im(rho*)=pi1(A,a)

[supple sable]

Let gamma be a loop in A with basepoint a, then it is also a loop in X

[nocturne basalt]

try using functoriality

[feral copper]

Hey! Given a non-singular algebraic curve in CP² (i.e. the zero-set of a non-singular homogeneous complex polynomial A(x,y,z)), I used that the function A defined on CP² has no critical values to say that we have a surface. What would be a quick one-line argument to say that it is closed orientable?

The thing is I'm used to real-valued functions, so I could've said it is the boundary of the pre-image of the open interval [0,+oo[ there, but not with C-valued functions x')

I could say it is a complex one-dimensional submanifold, so it is orientable this way, and for the closed part I could look at the quotient of the map on C^3\{0}, but I'm asking if there's a better argument

[golden gust]

it is closed because it is the preimage of 0 under A, but I can't think of a way in which you would show orientable without saying it is a complex manifold. what makes this argument unattractive?

[nocturne basalt]

this argument doesn't necessarily work, X can be simply connected while A is not

[novel acorn]

I don't quite understand you first question

But secondly i mean yea

The cell is a manifold so I don't see why the definition would be any different

A possible answer to your first question would be that an n-cell is homeomorphic to a closed n-disk
So the boundary of an n-disk is S^{n-1} and the interior is the open n-ball

[round dust]

Hey all, I’m having some trouble understanding how Hatcher writes down the third generator for the subgroup this covering space of S1 v S1 corresponds to

could anyone help me figure out how he produces that

[nocturne basalt]

those arent generators, they're relations

but basically you can follow the bottom b, the a, and then the bottom b in reverse and you get back to the basepoint

[round dust]

I see, that makes sense

How do I know that bab-1 is the right path to follow? I feel like I could have equally written down bab by going over the top one after

[golden gust]

it doesn't matter because b^2 is the identity so b^-1 = b

[round dust]

ohhh gotcha

thanks so much

cramming for my midterm

[golden gust]

good luck

[round dust]

ty!

[golden gust]

but you're asking for the actual method used to get those relations. he is computing the fundamental group of that graph, so he chooses first chooses a spanning tree (in this case, just the bottom b). then, he starts at the boldened vertex, travels some (possibly zero) amount along the tree, rides a single edge not in the tree, and travels back along the tree to the boldened point. doing that procedure by hand, you can check that the loops you get are exactly {a,b^2,bab^-1}

this is just as in the proof that graphs have free fundamental groups, if you've seen that

[round dust]

ahh

by spanning tree you mean a maximal one that isn't the whole graph?

[golden gust]

yes, a maximal subgraph with no closed loops

[round dust]

Yeah that makes perfect sense to me

tysm

[golden gust]

wonderful

[round dust]

will try to compute some myself

[round dust]

akhil mathew is rly putting me through it

[golden gust]

I did a very similar course last term, I'll drop the notes in case they're helpful

[round dust]

oh this is great

thanks for sharing

i’ll look at this after my midterm in an hour lol

[unreal stratus]

gl

the proof computing the fundamental group here is a big meh though tbf

well hm @golden gust i wonder what you think of it

like you know the stuff where he subdivides the graphs etc and plays around to make sure they're simplicial complexes or whatever

could he just have said like, lol cut up the edges without changing the topology such that WLOG we have a simplicial complex

I guess the only thing is like that wouldn't give you the same method for thinking about the fundamental group but you can just talk about how a spanning tree of the new thing relates to a spanning tree of the old one ig

[golden gust]

this?

[unreal stratus]

Actually tbf this is fine

I think I just remember it feeling overly tedious when he goes to define the bijection etc

Tbh the Chad method is just what May does I think lol

Like quotienting out the spanning tree being a homotopy equivalence

[golden gust]

I can't believe I never thought of that

I suppose this way is better for computing explicit generators right

but yeah that is clean

[unreal stratus]

It's sexy but needs more technology i think

And then u need smth like van kampen I think anywa7

[golden gust]

what why

[grizzled ibex]

i know you guys hate to answer that kind of thing but any good books for newbies on alg-top?

[umbral panther]

What’s your background?
Bott and Tu is the best. If you’ve studied differential topology, it’s the clear answer. If you haven’t … maybe you should

[bitter smelt]

Bott tu for alg top?

[umbral panther]

Yeah

[unreal stratus]

Bott Tu for alg top 👀

would be a bit of an idiosyncratic introduction

[nocturne basalt]

Can the universal coefficient theorem be restated in terms of some universal property that homology with Z satisfies?

[umbral panther]

Everything can be computed from Z because it’s the initial ring. Z-Chains give rise to other chains by simple tensoring. (Or hom for cochains)

Everything is nice in the world of chains / derived categories / oo-categories. That’s where you have universal properties. When you take homology you leave that world you get statements like universal coefficients

There is a general statement that if you have a chain complex over R and an R-algebra S, then there is a spectral sequence computing the homology after tensoring with S from the homology before. But it is more involved, with differentials and extensions. Whereas when R=Z, there are no differentials for degree reasons and it splits for, uh, reasons

[trail charm]

i'm trying to understand why two curves in $\mathbb R^3$ which intersect at a point are not transversal. i found a proof online:
Let $X$ and $Z$ be curves in $\mathbb R^3$. Suppose they intersect at a point $p$. Then
[ \dim(T_p(X) + T_p(Z) \le 2. ]

[gentle osprey]

anamono for anamono

[trail charm]

why does the last inequality hold?

another question, is the tangent space to a point on a curve in R3 a plane or a line?

[nocturne basalt]

what's it's dimension

[trail charm]

that's what im confused about

it would make sense if it was dim 2, but then why would the sum of them be at most dim 2?

[nocturne basalt]

are you familiar with the relation between the dimension of a smooth manifold and the dimension of its tangent space?

[trail charm]

they're equal, right?

[nocturne basalt]

yeah so whats the dimension of a curve

[trail charm]

one

oh

oops

[grim knot]

is this the right place to ask about functors?

[hidden crag]

Depends I guess

Just ask

[grim knot]

I need to show that the category U, which is given by Ob(U)= {C cat| Ob(C) is a set} and where the morphisms are the functors from a category to another is a category

now I wanted to show that if (C,D)!=(C', D'), then Mor(C,D) and Mor(C', D') are disjoint

but I don't know how to properly argue that

[rough cedar]

I think this should be in #category-theory

[grim knot]

oh, I'm sorry, didn't see that channel

[plain raven]

agreed that it's idiosyncratic but it's honestly fine

[quiet summit]

Hi i have a question regarding this problem. Aren't singleton sets countable and also open, so isn't it possible that int(A) = A?

pls help im confused

[hidden crag]

singletons are not open

assuming this is the standard topology on R

[quiet summit]

is standard topology the same as usual topology?

ok so is my proof correct
Let A be a countable subset of R, then A is either a singleton set or a union of singleton sets. Thus, the largest open set contained in A is null. Thus, int(A)=null

wait i think i need to show this. Let A be a countable subset of R, then A is either a singleton set or a union of singleton sets.

[nocturne basalt]

technically any subset is a union of singletons

I would try showing that if it had any point it would be uncountable

[quiet summit]

do you mean that if int(A) ≠ null then A is uncountable?

[unreal stratus]

Yes

[nocturne basalt]

yep

[naive hatch]

Hi - I’m working on an intro paper where I overview the fundamental group, covering spaces, and the simply-connected seifert van kampen theorem. At the end I’d like to include some examples that aren’t just pure math so people can see why we care about the fundamental group. Does anyone have any cool examples?

[golden gust]

in topological data analysis, one associates a topological space (or family of spaces) to some data set, and the topology of this space can tell you interesting things about the distribution of the data. I'm not an expert and I don't have an explicit example, but it seems that the fundamental group is one of the measurements people care about

[naive hatch]

This seems like a great example to include. I’ll look more into it!

[willow viper]

Hello I'm trying to prove the "<=" side here: Is this valid?? $\forall x \in U~\exists B_r^d \subset U\in \tau$ s.t. $B_{r_1}^{d'}(x)\subseteq B_r^{d}(x)\subseteq U \in \tau_{d}$, so $\tau_{d'} \subseteq \tau_{d}$. And then I'd do $\tau_d \subseteq \tau_{d'}$ and so d and d' generate the same topology?

[gentle osprey]

b3s4d

[willow viper]

What I'm also not sure of, is that if $U\in\tau_d$ and $B_r^d(x)\subset U$, does that mean that $B_r^d(x)\in\tau_d$? Not really right? But there must be some $B_r^d$ which are element in the topology right?

[gentle osprey]

b3s4d

[willow viper]

Did you mean "=>" like this: $\tau_d = {U\subseteq M:\mathrm{U~is~d-open}}$ and $\tau_d=\tau_{d'}$. Then $U, B_r^d(x)\in\tau_{d'}$. Since $\exists r_1 \le r$, then there exists a $B_{r_1}^{d'}(x) \subseteq B_r^d(x) \in \tau_{d'}$

[gentle osprey]

b3s4d

[nocturne basalt]

i think there might be some notation inconsistencies but seems like the right idea

like each open set in one is an open set in the other

and vice versa

[willow viper]

Yes that's kind of what I thought

But I'm not really sure if every open ball contains a smaller open ball in itself

And if both sides are correct anyways because I think I just assumed a lot of stuff

[nocturne basalt]

wdym, you are given that as a hypothesis right?

[willow viper]

For the <= direction you mean?

[nocturne basalt]

yeah

[willow viper]

yes

i think the first one should be correct yea

[willow viper]

although these questions I can't answer

[nocturne basalt]

well tau_d is the topology from the metric d right

[willow viper]

yes

[nocturne basalt]

so open balls are open

[willow viper]

right

Oh

So they must be contained in the topology

[nocturne basalt]

yeah

[willow viper]

But also in U which is also an element of the topology

are they contained twice?

[nocturne basalt]

wdym contained twice?

[willow viper]

or rather multiple times

[nocturne basalt]

open sets can be subsets of other open sets, that it not a problem

[willow viper]

Ok ok

thank u

On the => direction I'm not really sure tho

[nocturne basalt]

heres a hint for the first inclusion, B_r^d(x) is open in the topology from d and you know from the hypothesis that the topologies coincide

[willow viper]

so B_r^d is also open in In T(d')

And there also exists elements B_r^d' in T(d')

[nocturne basalt]

yep

the second inclusion is very similar

[willow viper]

Well

that's the hard part I suppose

Show that B_r1^d' is in B_r^d

[nocturne basalt]

you you just need an existence

so you can pick r_1

[willow viper]

right

[willow viper]

That's how I tried that tho

[willow viper]

So since there is always a smaller open ball in another open ball I can pick r_1 did you mean it like that?

[nocturne basalt]

no, you get existence just from being open in the metric topology

like a set is open iff any point in it has an open ball completely contained in the set

[willow viper]

hmmm

I don't really understand it

[nocturne basalt]

are you familiar with this result?

[willow viper]

Yes

In the metric space right

[nocturne basalt]

yeah so this should give you existence of some r_1

[willow viper]

Oh wait this makes sense

because the open ball is also just an open set xD

Then the proof is already done

No wait, there exists some r_1 yes but only wrt to the same metric

[nocturne basalt]

yeah its not immediate you need to use the fact that its open to then r_1 since it need not be a ball in the same metric

[steel glen]

working on hatcher 2.2.10
trying to find Hn(X), X = S^2 with the equator S^1 having its antipodal points identified with each other
is there any way to see how this works

X is obtained from the circle RP 1 by gluing on two disks, each of which wraps around the circle twice.

also, can this be constructed as a CW complex

i assume what was said above works as a CW complex if we just think of the RP1 loop as S^1, and glue the two disks going through twice

but idk if im being too loose by thinking of that RP1 bit as S1

[steel glen]

what do we get if we take the closed unit disk (in R^2) and identify antipodal points on the boundary? i gathered from stackexchange that it's RP^2 but i don't see how we arrive at that

[nocturne basalt]

yes this seems right

[nocturne basalt]

i feel like this just follows from the definition of RP^2 no?

[steel glen]

i've moved on from that but i guess so

just can't visualize it

[nocturne basalt]

yeah visualizing it is kinda hard but the correspondence between lines through the origin and antipodal points of the sphere should be immediate

[obtuse meteor]

Take RP^2 as lines through R^3. This becomes antipodal points on S^2. Delete the bottom hemisphere as it’s already accounted for. You’re left with a disk, and antipodal points on the boundary still need to be identified

Visualizing this identification is not really a thing you can do in 3D brain. But it’s kinda “twisty”

[steel glen]

still struggling but thank you all for your input

i had a question about covering spaces

what is the universal cover of S1 v S1

is it that one fractal that looks like a bunch of plus signs

[nocturne basalt]

yes

[gritty widget]

Most TDA doesn't really use the fundamental group, but instead considers homology groups.

[ancient hill]

suppose the interior is not empty. then A is not countable. fill in the blanks here

[gentle osprey]

ProphetX

[brisk epoch]

Can a square a smooth manifold?

[unreal stratus]

Depends what you mean

You can endow it with a smooth structure by importing charts etc from the disk, say

[brisk epoch]

I dont mean this to sound satirical but what could I mean?

[unreal stratus]

Well okay so

If you just try to give it charts in the "naive" way from its embedding in R^2 then you won't get anything smooth due to the corners

Both you can give it a smooth structure by, say, identifying it with the disk and then using charts for them

[brisk epoch]

I figured, just intuitively, if you can create a curve that goes through that corner point then surely it is a tangent vector and therefore a tangent space can exist everywhere on a square. But what would its curvature be?

[unreal stratus]

I'm not really sure what you mean with the first bit

[brisk epoch]

I'm using the definition of tangent space as the space of equivalence classes in which curves are equivalent with respect to having the same derivative under any chart.

I think my more pressing question follows this:

A circle of radius r could be uniquely defined with a manifold and a metric tensor. What is the minimal additional structure required to uniquely define a square?

[willow viper]

hello to prove this, is it fine if I just set r_1 and r_2 (using contained balls) to r_1 = r_2 = r/c?

like this I mean

[mortal briar]

Not sure what U is, but yeah basically a ball of radius r/c centered at x in the metric d is the same as a ball of radius r centered at x in the metric d'.

[willow viper]

I just defined U as an open set in the topology

[willow viper]

or like the topology consists of all open sets U

I suppose I could just write B_r^d \in T?

[nocturne basalt]

yeah i mean this + the previous problem you were working on gives what you want

[willow viper]

ty

But here I'm a bit confused. I know that d'(x,y) <= d(x,y) <= sqrt(n)*d'(x,y)

Does $d'(x,y) \le d(x,y)$ suffice that $B_{r_1}^{d'} \subseteq B_r^d$??

[gentle osprey]

b3s4d

[nocturne basalt]

not necessarily

[willow viper]

shiteee

but that's good because that's what I thought

[unreal stratus]

Other way round right

[willow viper]

wut

I'll need to find a concrete value for the existence of r_1 right

just like the previous where I chose r_1 = r/c

[nocturne basalt]

yeah but potato was saying that $d(x, y) \leq d'(x, y)$ so $B_r^d \subseteq B_r^{d'}$

[gentle osprey]

shd

[nocturne basalt]

note that the same radius is used

[willow viper]

But isn't d(x,y) bigger than d'(x,y)

[nocturne basalt]

no

[willow viper]

Now I'm reaaally confused

the euclidean metric is bigger than just taking the max

[nocturne basalt]

what are B_1^d(0) and B_1^{d')(0)?

[willow viper]

max{y_k} <= sqrt(y_1^2 + ... + y_n^2)?

[nocturne basalt]

try drawing it on a plane

[willow viper]

with 0 you mean set x to zero right?

[nocturne basalt]

yeah the ball around 0

[willow viper]

well that's x,y = 0 then so they're equal

[nocturne basalt]

well the euclidean ball is the standard circle with radius 1

but the infinity norm ball is not

[willow viper]

Wait I think I got it mixed up

I thought I could just use this

[solemn temple]

What is an example of a space that is T0 but not sober?

[plain raven]

Any infinite set with the cofinite topology.

The whole space is an irreducible closed set with no generic point.

[quiet summit]

Hello is my proof correct, or do i need to show that U is finite

let $f,g : X\ \rightarrow\ \ Y$ be continuous and Y be Hausdorff Now let $U\subseteq{Y}$ contains all points in Y s.t $f(x)=g(x)$ By theorem 1.5.21 U is closed and by theorem 1.6.7 the inverse image of U or the set $\left{x\in X\ :f(x)=g(x)\right} $ is also closed in X.
Th 1.5.21 finite subsets of a hausdorff space is closed
Th 1.6.7 if f is continuous then Inverse image of a closed set B in Y is closed in X

[gentle osprey]

Ji

[abstract saffron]

I'm not sure why it looks so complicated.

This amounts to $(f-g)(x) = 0$, so the set is $(f-g)^{-1}({0})$, which is closed since ${0}$ is, unless I'm missing something.

[gentle osprey]

Megumi_Tadokoro

[abstract saffron]

Oh, and ${0}$ is closed because $Y$ is Hausdorff. Unless I'm dumb, Hausdroff is equivalent to singletons are closed.

[gentle osprey]

Megumi_Tadokoro

[abstract saffron]

And I have no idea what Theorem 1.5.21 and 1.6.7 are

[quiet summit]

Th 1.5.21 finite subsets of a hausdorff space is closed
Th 1.6.7 if f is continuous then Inverse image of a closed set B in Y is closed in X

thiss

[abstract saffron]

And since $f-g$ is continuous, but that's trivial

[gentle osprey]

Megumi_Tadokoro

[abstract saffron]

I guess what you did is correct, though a bit phrasing will help greatly.

[quiet summit]

oh ok tnx very much

[hazy lotus]

Minus operation is not assumed here

This works for functions to (R,+,×) but not in general

[hazy lotus]

Instead prove the complement is open

Take a point x with f(x)=/=g(x) and then use haussdorffness of Y

Feel free to let me know if you need more help

[unreal stratus]

Also, an interesting what to do this is uh

Y is Hausdorff if and only if the diagonal - the set of points of the form (y,y) - is closed in Y x Y

I mean it amounts to pretty much the standard proof but this is a useful fact to know

[hazy lotus]

Cofinite topology on N has closed singletons but is not haussdorff. Closed singletons is equivalent to T1

[abstract saffron]

It's been a long time 😄 my memory is failing apparently

What's T1 again? Kolmogorov?

[hazy lotus]

No idea lol

The internet says Kolmogorov is T0

[abstract saffron]

Hausdorff is way higher.

I never remember these names

[abstract saffron]

Huh, strange... Where did I see that fact then...

[hearty jacinth]

Frechet

[abstract saffron]

Ahhhh

still remembers nothing

It's OK, I'll revise when it becomes necessary

[willow viper]

hello how can I start proving this

I know that the discrete metric generates the discrete topology on any set. I tried to show that the euclidean metric generates the discrete topology on Z but I don't know how

[coral pivot]

show the open sets are the same, you know what the open sets are in the discrete try showing those are also open in the subspace topology

(so in particular the singletons)

[willow viper]

the open sets as in P(Z)?

and the singletons as in 1-9 then??

[coral pivot]

right those are the open sets of discrete topology of Z, if that is what you are asking

I am not sure what you are asking

[willow viper]

I'm also not sure

[coral pivot]

ok so to be clear I guess, A topology on a set is a collection of "open sets", I am sure you are aware of this. Two topologies are the same if these collections are the same, so if you can show that the two topologies have the same open sets.

In our case, you want to show the subspace topology is the same as the discrete. In the discrete everything is open, in particular the singleton sets {n} are open. So if you show these are also open in the subspace topology, that would be a step towards solving this. You can recover the whole thing from this

[willow viper]

Ok I think I get it

So I have a $B_r^{d_E}(x) = {y\in\mathbb{Z}: \abs{x-y} < r}$ So there always exists a radius $r$ s.t. the ball consists of ${x}$ and is open then by definition?

[gentle osprey]

b3s4d

[willow viper]

well the ball would consist of a singleton {z in Z}

[coral pivot]

Yeah this is correct, so now that you know the singletons are open in the Euclidean topology on Z, can you see how this automatically implies that this is discrete

[willow viper]

yea because the discrete topology only consists of unions of the singletons

unions and intersections?

[nocturne basalt]

well everything is a union of singletons but as long as you have the fact that singletons are open then you have the discrete topology

[willow viper]

right thx

[willow viper]

hello I'm trying to prove for closeness under finite intersection and I'm using the fact that ${U_n}{n=1}^{N} \subseteq \tau_5$, so $ X\setminus (\bigcap{n=1}^N U_n) = \bigcup_{n=1}^N (X \setminus U_n)$. I know that X is infinite but what about U?

[gentle osprey]

b3s4d

[willow viper]

Is it that U is finite so X\U is infinite and thus element in the topology because the union of infinite sets is infinite?

or can U also be infinite? In which case I suppose that the maximum U can be is X which would X/X = emptyset then??

[unreal stratus]

U can be infinite too

I suggest you think of it in terms of like closed sets for ease

[willow viper]

What

So U is a closed set thus X is is an open set and element of the topology?

And I'm super stuck here I don't even know how to begin

[gaunt linden]

A reasonable approach might be to try to construct open-but-not-closed and open-but-not-continuous cases separately and then simply put them next to each other ...

(Though actually the first open-but-not-continuous case I can think of also turns out also to be not-closed, so perhaps just think of that first).

[unreal stratus]

No I mean it's easier to check whether it's a topology if you think in terms of closed sets

[willow viper]

That's the problem I don't know how

I tried doing stuff like X = {a,b,c}, Y = {a,b,c} with TX, TY being the discrete topologies for example which would make all the sets open

And then how can I prove continunity?

preimages being open?

[gaunt linden]

You probably need something richer than that, yes.
I can think of a modification of the "topologist's sine curve" function which satisfies the conditions (which needs only to be R -> R), but I'm not sure it's the simplest way to get through, so I won't dignify that by the name "hint".

[rancid umbra]

the finite union of infinite sets is infinite or all of X

[rancid umbra]

try
X = [0,1] x {0}
and
Y = S^1 = {(x,y) in R^2 : x^2 + y^2 = 1}

draw out Y and see if u can find a function that parametrizes Y

this doesn’t solve the entire problem, but if u can do this ur half way there. u just need an open map that’s not continuous now

[winged viper]

does anyone know an elementary way of proving that S^\infty x I is the direct limit of S^n x I

where S^\infty is defined as the direct limit of S^n

I think if we work in the category of compactly generated spaces or something like that, this just follows from the fact that - x I is left adjoint

but im looking for something more elementary, like proving that a subset of S^infty x I is open <==> its intersection with each S^n x I is open

[winged viper]

ehh i think the easiest way might just be doing a lot of point-set topology and using the tube lemma

[warm quiver]

Couldn't you also just show directly that S^\infty x I satisfies the corresponding universal property of the direct limit? Continuous maps out of S^\infty x I are precisely functions that are continuous on each S^n x I, so I imagine it shouldn't be that bad to show it's universal, unless I'm not seeing something.

[winged viper]

yeah but thats the same as showing that S^infty x I has the final topology

the way i proved that involves a bunch of point set

[winged viper]

like i think this is equivalent to showing that a map S^infty x I -> X is continuous <==> each map S^n x I -> X is continuous

this is how i proved the converse. im not sure if its the simplest way

[novel acorn]

maybe show the inclusion of S^n x I into S^m x I for n \leq m
And show it commutes

i.e. use the universal property

[winged viper]

right so i have this diagram

but its not obvious to me why it's universal

[novel acorn]

So like all these maps are inclusions if you take the right CW structure on S^infty
Now take maps S^n x I -> X for each n in this diagram
my idea would be to define the map S^infty x I -> X as being the corresponding map from S^n x I -> X on the n-cell x I of it's CW structure
And this map is actually well defined by the commutation assumption put on the maps S^n x I

the last point is like
The maps don't have like weird stuff going on between the cells

[winged viper]

but is it clear that the map S^infty x I -> X is continuous?

[novel acorn]

yea

Since all the maps that compose it are continuous

and the fact that there isn't funky stuff going on between the cells

means that nbhds map back well

actually

[winged viper]

im a bit confused what you mean here

[novel acorn]

I'll be back in a minute and I'll explain what I mean since I'm being a bit non rigorous

[winged viper]

sure thanks

[odd flame]

how is mayer vietoris "equivalent" to SVT

like what does the sequence we get from the thm tell us about X (X being the space we're breaking up into a union)

https://tenor.com/view/timi-gif-20112884

[bitter smelt]

What is SVT

[odd flame]

seifert van kampen

[bitter smelt]

Wtf is the T

[odd flame]

theorem

[bitter smelt]

Bro

[odd flame]

Bro

[bitter smelt]

Anyway, MV and SvK do pretty much the same thing

[winged viper]

if U \cap V is path connected (which is an assumption for van kampen), mayer vietoris tells you that H_1(U u V) = H_1(U) + H_1(V) / image of H_1(U u V), which is basically what van kampen tells you but more complicated because it's not abelian

[bitter smelt]

Like both decompose the space into two open subsets, both split the interesting group, etc

[winged viper]

the H_1 groups are the abelianized versions of the π_1 groups

which means you take a group G and you quotient by the "commutator subgroup," and turns it to an abelian group

when you abelianize the free product, you get the direct product

if you know category theory it's much clearer what the relationship between the two is

[odd flame]

perhaps i need to review the actual significance of homology/sequences

im still confus but that helped

i'll be back when i have a more concrete question

[winged viper]

just stare at the formulas and squint

[odd flame]

https://tenor.com/view/skeptical-futurama-fry-hmmm-i-got-my-eyes-on-you-gif-17101711

[winged viper]

like this

and the formula for van kampens

[odd flame]

what are k* and l* here

[winged viper]

The inclusion of A \cap B into A and the inclusion of A \cap B into B

K* - l* is the map that’s in the Mayer vietoris sequence

From H_1(A cap B) to the direct product of H_1(A) and H_1(B)

And you get the quotient by the first isomorphism theorem

[odd flame]

shouldnt that be more like

k* \oplus l* then

[winged viper]

Oh sorry

k is the inclusion of A into A u B

And l is the inclusion of B into A u B

And k* - l* is the map from the direct sum into H_1(A u B)

[odd flame]

wait wait

i thought it was the map into the direct sum

[winged viper]

no sorry i misspoke above

lemme just write it out

you have this sequence: $H_1(A \cap B) \to H_1(A) \oplus H_1(B) \to H_1(A \cup B) \to H_0(A \cap B) = 0$

[gentle osprey]

Frank

[winged viper]

The last thing is 0 because the intersection of A and B is path connected

exactness tells you that the map $k_* - \ell_* : H_1(A) \oplus H_1(B) \to H_1(A \cup B)$ is surjective

[gentle osprey]

Frank

[winged viper]

then apply the first isomorphism theorem

[odd flame]

are yoou sure it's \cup

oh that's just X

nvm

oh and there's the k* - l*

so if im asked to compute the MV sequence

is that asking me to literally define the homology groups at each part

[novel acorn]

I'll make this a big write out\
What I'm concretely talking abt is the fact that I'm defining the map $f:S^\infty \times I \to X$ as follows: take a point $x$ in $S^\infty \times I$, then if $x$ can be seen as an element of $S^n \times I \subset S^\infty \times I$ for some $n$, then $f(x) = f_n(x)$ where $f_n: S^n \times I \to X$. Now this creates some problems at first glance. Firstly look at what happens if we take $x \in S^1 \times I \subset S^2 \times I \subset S^\infty \times I$. Seemingly $x$ gets mapped to different points if we define $x$ as being an element of $S^1 \times I$ compared to it being included in $S^2 \times I$. However due to the fact that $f_1 = i_1 \circ f_2$ we can see that $x$ in fact gets mapped to the same point regardless of how we take it of being included in the CW structure.\
OK so the map is well defined, but now comes the problem of continuity. Take $U \in X$ to be open, and take $X_n$ to be the image of $f_n$. Now to show that $f^{-1}(U)$ is open, we need to show that $f_n^{-1}(U \cap X_n)$ is open (this follows from the fact that $S^\infty \times I$ has the colimit topology). To see this last part I thank this handy mathSE post that basically shows what we want: https://math.stackexchange.com/questions/724005/continuous-images-of-open-sets
Hence we have shown continuity!

[gentle osprey]

Irony Incarnate

[tough zephyr]

Is it just me or is this nonsensical?

[bitter smelt]

This is R^2 with two origins. What's the confusion?

[tough zephyr]

It's not even defining an equivalence relation on Y, it's defining an equivalence relation as a subset of X_1\cross X_2, that is ~ is the set of ordered pairs ((x,y),(x,y)), where x,y\neq 0

[bitter smelt]

huh?

[tough zephyr]

How is ~ an equivalence relation on Y?

[bitter smelt]

Elements in X_1 \times X_2 would look like ordered quadruples

or rather, ordered pairs of ordered pairs

((x_1,y_1),(x_2,y_2))

[tough zephyr]

Right, but I don't understand what it means by saying '~ is an equivalence relation on Y'

Usually when you say R is an equivalence relation on a set S you're saying R is a subset of S\cross S

It's a certain kind of binary relation

[bitter smelt]

Yes. They are telling you how to relate two points in Y. One point (x_1,y_1) is related to another point (x_2,y_2) iff x_1=x_2 and y_1=y_2 and neither are the origin

one point in Y looks like (x_1,y_1)

Is the "a point in X_1 is related to a point in X_2" whats throwing you off?

You can elevate it to Y in the way I wrote.

Since no points in X_1 (as a subset of Y) will be related to points in X_1 (as a subset of Y) except the reflexive relation

[tough zephyr]

As I understand it, an equivalence relation is a binary relation where the 'domain' and 'codomain' are equal sets. To say ~ is a binary relation on Y or X_1\times X_2 would mean that you're comparing 4-tuples with other 4-tuples

but you're not. You're comparing different kinds of objects than what the equivalence relation is 'on'

[bitter smelt]

First of all, Y is not X_1 \times X_2

just to be clear

[tough zephyr]

I think I found the source of my confusion

Sorry it's just that the definition of the disjoint union is a bit funny

[bitter smelt]

ok

No problem!

[bitter smelt]

RE this comment: You can think of it either way

[tough zephyr]

It's not exactly set theoretically the same as you're usually with an ordinary union of objects

I guess the right way to go about it is that the actual equivalence relation is supposed to be the 'completion' of what's mentioned in the problem

As in it's omitting the fact that it's also an equivalence relation on X_1 and X_2 when intersected with those sets

[bitter smelt]

Yeah they morally should have stated it on Y, but it doesn't super matter since nothing is happening on X_1 \subset Y as far as ~ is concerned

[tough zephyr]

Right, that's why there's two origins

[bitter smelt]

Yeah

[tough zephyr]

That's kind of annoying how they don't say that explicitly :/

Oh well now I can rest easily

[bitter smelt]

You'll get use to it, this sort of abuse of language is common

[tough zephyr]

Don't get me started on simplicial complexes

[bitter smelt]

You just have to see into the soul of the writer and read their mind sometimes

[odd flame]

going back to my old question

what does it mean to compute the MV sequence of a space X

does it mean the homology groups? the homomorphisms? or both

[nocturne basalt]

i think it generally refers to computing H_i(X) by using the MV sequence

[odd flame]

doing the first part of this

[nocturne basalt]

oh its just computing the groups in the sequence then

[languid patrol]

They just want you to compute the homology of one of the two spaces.

They should really have you compute both.

[odd flame]

i kinda have to do both to show them isomorphic anyways

[languid patrol]

Yes, it's very confusing the way it's worded.

[odd flame]

.

[nocturne basalt]

observe that the mv sequence for the space that you didn't do is the same 😂

[languid patrol]

Ah but I guess it's quite obvious looking at the Meyer Vietoris sequence even without really computing anything that the two will be the same

[odd flame]

ok ok lemme give it a try first

so if i get this correctly

we start with H_0 and build up the sequence right

[languid patrol]

Yes.

Or, what do you mean by that?

Also you already know what H_0 is so you can use reduced homology

[odd flame]

oh that was another thing i wanted to ask

when is that useful

i must confess that my understanding of homology is a bit lacking

but im working on it

[languid patrol]

Reduced homology is basically just saying "it's dumb that a point has homology, and that everything has at least some H_0, let's throw out one copy of Z/Q/whatever and only keep the important stuff"

You literally take the same chain complex but throw out a factor of Z

[odd flame]

ok i need to go back lmao

one sec

[languid patrol]

Anyway remember that the coboundary maps go from H_1 to H_0 and from H_2 to H_1

because you're doing homology

I believe in you.

[odd flame]

ok back to very basics

given a space X, what does it mean to compute its homology

[languid patrol]

oh dear

[odd flame]

i know i knowo

[languid patrol]

Well so given a space X if you reaaaaaallly get down to it (most people don't actually do this most of the time) you want to take a chain complex associated to X called C^*(X;Z)

Oh whoops thats a cochain complex

C_*(X;Z)

There we go.

[odd flame]

my prof taught stuff to us in terms of cubes n shit and it just didnt click for me

i know hatcher does it differently though

[languid patrol]

There are many ways to define this, perhaps the easiest is to take a triangulation of the space X by simplices, then the chains are just "formal Z-linear combinations of simplices"

And the boundary maps are the obvious boundary maps

There are technical subtleties about what kinds of triangulations should be allowed but they dont really matter

Anyway for a 2-dimensional X you get a complex

[nocturne basalt]

cubes for singular homology?

[odd flame]

it's in massey i believe

[languid patrol]

C_2(X;Z) \to C_1(X;Z) \to C_0(X;Z) \to 0 where the maps are assigning a triangle its boundary (which is a formal combination of triangles of dimension one less)

The H_2 is the kernel of the first map, the H_1 is the kernel of the 2nd modulo the image of the first, and the H_0 is C_0 modulo the image of C_1

Concretely tho in your problem you only need to know the answer for a 2-sphere, and some lines

And you also need to know that homology groups are a homotopy invariant

[odd flame]

i know that H_2(S^2) = Z and zero everywhere else, we did that in class

[nocturne basalt]

well H_0(S^2) = Z as well

[odd flame]

oop

[odd flame]

this is just a consequence of exactness right

[nocturne basalt]

wdym? the chain complex is not necessarily exact

[odd flame]

[languid patrol]

It's a consequence of the fact that the chain complex is a complex that the definition makes sense

E.g. part of being a complex is that the boundary of a boundary is zero.

But exactness only happens if all the homology vanishes, that's what homology measures.

This is a triangulation of a torus (because the sides a and b are identified) if you want to get a handle on what's happening this might be a nice thing to work out

You could also do S^2 explicitly as it's simpler.

Anyway intuitively H_1 is the group of "holes" in the space and H_2 is the group of 2-dimensional "holes" whereas H_0 is the group of components.

And so on for bigger homology groups

[odd flame]

ok this is the most tangible thing ive read so far lol

[languid patrol]

So for the torus you get H_1 = Z^2 (you should do this by hand!) reflecting that there are two independent holes on a torus.

Then H^2 = Z because the torus has no boundary

H_0 = Z because it has one component

All the higher stuff is trivial because it's two-dimensional

The other important thing to know is that it's a homotopy invariant, so often you can mush stuff around to make it simpler to compute.

[odd flame]

by this you mean it's like a surface right

like there is no border

like a disk would have a boundary

[languid patrol]

Yeah, I guess I really mean "compact oriented surface without boundary"

[languid patrol]

Yes exactly

Although thats a bad example because it has no homology lol

Except for H_0

Or sorry that's a good example

For that reason.

[odd flame]

im gonna guess that having no homology corresponds to the fact that it has trivial fundamental group too right

[languid patrol]

Yeah, even better: you can shrink the disk continuously to a point (it's contractible) so the homology has to be the same as the point.

[odd flame]

shd just lurking fr

[nocturne basalt]

you also have H_1 is the abelianization of pi_1 for path connected

[languid patrol]

Yes and in general there is an analogous statement for H_i as long as H_j vanishes for 0 < j < i

But this involves higher homotopy groups which you may not've learned yet.

[nocturne basalt]

thats hurewicz right?

[odd flame]

hurewicz is on another problem in my hw

im a little behind lol but this is helping a ton

so computing homology of a group is like asking how many n-dimensional holes does this space have

[nocturne basalt]

yeah thats one way to think of it

[odd flame]

and those holes can be represented by the nth homology groups

or well

they are the groups

[nocturne basalt]

err i think the correspondence is hard to quantify though idk that much

[languid patrol]

Yes exactly, although it's a bit complex how to interpret what it means when the groups are torsion, etc.

But if all the groups are free (like in your example) then that's basically what's happening.

The first version of homology people had was just the euler characteristic which is the alternating sum of the ranks of the homology groups, then people realized that there were actually numbers that made sense (the number of factors of Z in the H_i) then eventually realized that the best thing to do was to use the groups themselves for maximum information

For what you're doing H_i may as well just be a number, and that number is the "number of n-dimensional holes"

[odd flame]

but instead of a number it's a cyclic group

which is kinda the same i guess

[languid patrol]

Its the number of factors of the free cyclic group which occur

[nocturne basalt]

well generally if its finitely generated itll be Z^r + torsion parts

[languid patrol]

so if H_i = Z^d it's the d that's important to you

[silver umbra]

so i'm currently trying to prove that for a connected topological manifold, there exists a homeomorphism of it that sends one point of it to another

the first step is to pass to charts

namely, regular coordinate balls

and first show that there exists a homeomorphism of the closed ball sending one interior point p to say, another interior point q

that fixes the boundary

i'm struggling to prove this fact though

[languid patrol]

Since the manifold is only topological you can just move in a straight line.