point-set-topology

[heady skiff]

especially in topology it seems

are there any subjects of math where pictures don't help

something like foundations/metamathematics maybe

[queen prism]

refer to https://www.youtube.com/watch?v=VYQVlVoWoPY

[limber wren]

mmm I dunno, I think understanding things visually is so natural to people that it's never a bad idea to TRY to be able to draw a picture

lol it's hard sometimes and definitely more so for weird advanced shit, but personally I always try to draw a picture to get an intuition

[heady skiff]

yeah that makes sense

also what's the best way to see that U is open in S^n? i started involving stuff with metrics but that always seems like a pain, so i guess i should consider S^n as a subspace of R^{n + 1}

[limber wren]

Yeah, it's the intersection of S^n with an open set

the open set being { x | x.p > 0 }

since dot product is continuous

[heady skiff]

ohhh

wait so

it's the image of an open set essentially right

is that how you can see the set { x | x.p > 0 } being open

[limber wren]

inverse-image, yup

[heady skiff]

oh right

ye

[limber wren]

inverse image of (0, infinity)

[heady skiff]

coo coo

[limber wren]

yeah that open set in particular is an open half plane, so it's like an open hemisphere of S^n I think

[meager frigate]

Could I get some help on proof of a topological space?

[limber wren]

you can normally just ask the question directly, people are always happy to help

[meager frigate]

Yeah I tried help but I’m realizing it’s more for hs stuff

And people aren’t so eager to help

Ty tho

I think I got the first part but I’m not sure about the

Finite Intersections part

[limber wren]

what is C_X(U)?

[meager frigate]

Complement of U

[limber wren]

ahh

[meager frigate]

Where X is the universal set I guess

Weird notation 👍

[heady skiff]

are you proud

[limber wren]

1. is X\X either X or countable?
2. X \ emptyset either X or countable?
3. If X \ U_i is either countable or all of X for each i, what can you say about X \ ( union of all U_i)? Use demorgans law for this
4. If X \ U_i is either countable or all of X for each i (only finitely many i) what can you say about X \ (intersection of the U_i). Probably use demorgans law again for this

[meager frigate]

1. Yes
2. Yes
   3/4. I’ll try that

I forgot there was demorgans law for those unions as well

[limber wren]

yup

[meager frigate]

It’s called “demorgans laws +” in my class lmao

Ty!

[limber wren]

lol why ++?

[meager frigate]

Just one

My bad

[limber wren]

because it's objectively worse than demorgans law?

[meager frigate]

But bc the first was taken

[limber wren]

(that was a c++ joke)

(I appologize)

[meager frigate]

…

Idk C++ but I think I’m gonna learn C before it somehow

Maybe I’ll get the joke then

[limber wren]

lol the best c++ programmers program basically c with a small simple subset of c++ functionality

imo of course

[heady skiff]

isn't this just saying that p is a quotient map? since surjective continuous map + open map => quotient map

wait a minute

wait

isn't this proposition just a specific case of this theorem

[limber wren]

It looks like the the universal property of a quotient space

similarly

[meager frigate]

@limber wren

Does this look right?

[limber wren]

demorgans law flips unions and intersections

so C_X(intersection of U_i) becomes union of C_X(U_i)

sorry fixed that

[meager frigate]

Shoot yeah forgot to flip it but wouldn’t it be the same logic

For showing unions

Like that could be my arbitrary union proof

[limber wren]

no, for that one you have C_X(intersection of U_i) = union of C_X(U_i)

so you need to show the unions of the C_X(U_i) is either all of X, or countable

you only have finitely many

so consider the case where none of those C_X(U_i) is all of X, or at least one of them is all of X

[meager frigate]

I’m gonna do the intersection before I confuse myself even more gimme a sec to fix it

[limber wren]

lol np

[meager frigate]

Fixed?

I forgot to put they are in T_c in the last line but

Is the main logic of the inclusions close?

[meager frigate]

@limber wren sorry for the ping

[limber wren]

np

it's correct but I would explain a bit more why if C_x(U_i) is countable or X then the union of all the C_X(U_i) is either countable or all of X

lol like your teacher will probably take a mark off and write "why?"

[meager frigate]

Is it because the largest subset is X and so taking the union of X with any smaller subset would only give X

And since they’re all countable the union of finite sets is countable as well

[limber wren]

I would think of 2 cases:

1. all of the C_x(U_i) are countable
2. not all of the C_x(U_i) are countable

in each case, you need to conclude that the union is still either countable or X

that's all you're missing from your proof, actually proving it, you're kindof just stating the result without explaining why

[meager frigate]

Don’t they have to be countable tho if the complements of the subsets are in T_c?

Like if they are subsets they should already be countable

[limber wren]

no remember U_i is in the topology if C_x(U_i) is EITHER countable or all of X

so some of them could be countable, others could be all of X

[meager frigate]

Yeah so wouldn’t the 2 cases be about those 2 things?

So if they aren’t countable im guessing I can’t just say “if it’s not countable then it’s X”

[limber wren]

you can say that for each C_x(U_i) yeah, for each one if it's not countable, then it's X

but you need to be able to say that about the union of them too

just look at case 1

[meager frigate]

Yah

[limber wren]

you have the union of each C_x(U_i), and they're all countable

so what can you say about the union?

[meager frigate]

Countable as well

[limber wren]

why?

(but yes lol)

[meager frigate]

Because the number of elements

Is less than Infinity so unioning them should be too?…

I’m not too familiar with the diff between countable and finite

Ik finite is finer

[limber wren]

okay, countable basically means you can write the elements in a list, like {x_1, x_2, x_3, .... }

so suppose I have a finite number of lists

how can you write a list that contains all of the elements of each list?

(ie write the union as a list)

[meager frigate]

Just taking the finite union from list one to list n

[limber wren]

sort of, like if I had lists { x1, x2, x3, .... } and {y1, y2, y3, .... } I could write {x1, y1, x2, y2, x3, y3, .... }

and if you had N lists, you could do the same thing

[meager frigate]

Wait is that not the union then?

[limber wren]

exactly

[meager frigate]

The finite union of these lists

[limber wren]

the finite union of countable sets is also countable

so if each C_x(U_i) is countable, their union is also countable

[meager frigate]

I thought I said that tho

[limber wren]

you did, you just didn't explain why, that's it

but yeah, finite unions of countable sets are countable

[meager frigate]

Makes sense

And case 2?

[limber wren]

so if each C_x(U_i) is countable then the intersection of them is countable

so in case 2, you have (at least) one of the C_x(U_i) = X

so what is their union?

[meager frigate]

X

[limber wren]

yep, why?

[meager frigate]

Because it’s the largest subset

Of X

And so you’re just adding repeat elements to a list?…

[limber wren]

a better explanation would be that each C_x(U_i) is in the union of all of them

so if one of them is X, then X is contained in the union

but X is the full set! so then the union has to be X as well

[meager frigate]

Yeah since X is the largest subset of X then union of them would be X

[limber wren]

yeah

so in case 1. the union is countable

[meager frigate]

That’s what I’m trying to say yeah okay and so then that shows the intersection is in T

[limber wren]

and in case 2. the union is X

so by definition, the union is in the topology

[meager frigate]

The intersection right?

[limber wren]

sorry yeah, the intersection

[meager frigate]

That makes sm sense now yeah ty

I’ll rewrite the proof do you mind if I ping you later when I do finish?

[limber wren]

sure

[meager frigate]

Ty 🙏

[unreal stratus]

Silly question perhaps, but does anyone know why the cup product (and for that matter, the cap product) is so named?

I had a feeling it could be something to do with how the relative cup product involves unions and perhaps it's mimicking that somehow, but i'm aware the cup product originally had a more geometric meaning in terms of linking numbers anyway

[coral pawn]

What is the precise definition of the fiber in the context of simplicial sets? Let X be a simplicial set and x be a 0-simplex of X. Given f: Y --> X, what is f^-1(x)?

What are the n-simplices of f^-1(x)?

The main problem here is defining x as a simplicial set

Of course, we could just say x is just the smallest simplicial set in X generated by x

But that not very explicit

It can't just be {x} in the 0th slot and empty everywhere else

because there are not maps to the empty set from the singleton

The mth slot must contain the image of {x} under (the image) of any map [m] --> [0]

I think this is the correct definition

But idk if it's the standard

[unreal stratus]

Yeah I'd define it as the pullback of {x} <- Y -> X but then you need to define what we mean by {x} yes

[coral pawn]

Flipped arrows?

[unreal stratus]

which i would define as you said, the smallest simplicial set containing x in weight 0

[coral pawn]

okay cool

[unreal stratus]

Oh yeah sorry lol am used to maps going X -> Y lol

[coral pawn]

And that would agree with the explicit definition i gave as well right?

[unreal stratus]

I think so yes, though of course there's only one map [m] -> [0] :)

But yeah basically x and its image under degeneracies

(which also makes it clear that when e.g. you realise it you just get a point)

[coral pawn]

Ah so it would just be a singelton at each level

[coral pawn]

idk why I didn't realize this lol

[umbral panther]

I don’t know, but I’d guess that the cap product came first and the cup product was an auxiliary operation with a similar symbol. The goal was intersection of cycles, represented by the cap symbol, but that didn’t work out, so they retreated to the cap product and then the cup product

[unreal stratus]

Sure that makes sense

[hot locust]

What might be a good example of a topological game which uses the notion of compact spaces from point-set topology?

is this the right place to post this question?

[muted arrow]

Topological game huh...

Someone ping me if this is actually a thing 👍

[distant lichen]

There are notions of games used for definability & distinguishability in logic

[quiet thorn]

Look up Banach Mazur game

But I'm not familiar enough with topological games to answer the question above that one, but I think this is a fine place to ask

[signal stratus]

Hi how do I prove this

I tried to prove $(\bar{A})^{o}=A^{o}$ first but apparently it’s wrong

[gentle osprey]

www

[merry geode]

Maybe you can prove for closed set E,

bar{circ{E}} = E.

[signal stratus]

Ohhhhh makes sense thx!

[hot locust]

Look up telgarsky's article on 50th anniversary of Banach Mazur game

[feral copper]

https://eudml.org/doc/214826

(that's Telgársky's paper)

[feral copper]

So I know there's a relative version of Mayer--Vietoris, if $(X,Y)=(A\cup B,C\cup D)$ with $C\subset A$ and $D\subset B$: $$\cdots\to H_n(A\cap B,C\cap D)\to H_n(A,C)\oplus H_n(B,D)\to H_n(X,Y)\to\cdots$$
Is there a cohomological version of this?

[gentle osprey]

Matplotlib

[feral copper]

So the answer must be yes: I should be able to dualize the chain complex by taking Hom(-,Z) and then looking at cohomology.
So is the LES the same as the cohomological absolute MV sequence?

[feral copper]

(basically, reversing all arrows in this and taking cohomology groups instead)

[opaque scroll]

You can show that for an open set U

\bar{U}^o contains U, so the LHS contains the RHS

And for a closed set E, you have that \bar{E^o} is contained in E. Therefore also \bar{E^o}^o is contained in E^o, so the RHS contains the LHS.

And it's not true that \bar{E^o} = E, so you shouldn't try to prove that.

[merry geode]

Idk what I was thinking here, I forgot ^\circ on each side

I was not intending to drop that ^{o}

Anyway they might have figured out >.>

[signal stratus]

Thank you!

[pseudo ocean]

is there a comprehensive list of homology groups of topological spaces?

like how theres a comprehensive list of solutions to the EFE's

[fickle elm]

You can see the answer of this question.

https://mathoverflow.net/questions/188025/are-there-some-tables-or-handbooks-of-homology-and-homotopy-groups-of-every-mani

[mighty loom]

hey could sb help verify my understanding for these before proceeding to justification thank you
I got that
a) open and closed bc its natural number set in R
b) open but not closed we get (-inf, 0) U (0,+inf) are union of open set -> open
c) empty set so open and closed

[surreal bane]

Hi @mighty loom

[mighty loom]

hi

[surreal bane]

Nice to meet you.

Are you solving maths problems?

[ebon galleon]

a) why would this set be open?
b) open but not closed yes
c) this is not empty.

[mighty loom]

thank you

for c) i think i mistaken Q for integer

[ebon galleon]

Ah lol

Yeah so since nonempty that'll change whether it's open and/or closed too ofc

[mighty loom]

i guess its closed but not open for c

if it said to justify it will my explaination of set notation be good enough

[ebon galleon]

Not open I agree with. But is it closed? If it were closed, you would be able to take some neighborhood of - (1 - ε, 1 + ε) - that's disjoint from it (since complement would be open and 1 is in complement)

[mighty loom]

sorry i meant open but not closed, but since you imply that its not open i dont understand because for every x in C you can find its neighborhood that is a subset of C

i think i get why its not open is it because its a bunch of isolated point

[opaque scroll]

How would you find a neighborhood of 1/2 for example?

[ebon galleon]

"isolated" has a technical meaning which doesn't fit here, but you might have the right idea(?) So if you take a rational number a/b between 0 and 1, is a neighborhood of a/b going to only have rational points in it?

(paying attention to what sets things are in here is important)

[mighty loom]

yes this was my understanding my attempt to explain that there is some real number, so from these i could conclude that neighborhood would not be subset of C therefore is not open

thank you

[trail charm]

in the bottom lines, why do we have $H_1(X) \to 0$? in the Mayer-Vietoris Sequence, we should have
[ H_1(X) \to H_0(S^1) \to H_0(S^1) \oplus H_0(S^1) \to H_0(X) \to 0 ]
which simplifies to
[ H_1(X) \to \bZ \oplus \bZ \to \bZ \to H_0(X) \to 0 ]

[gentle osprey]

ana(functor)mono(morphism)

[fickle elm]

I think in the last part you wrote Z and Z+Z should swap place.

And X is connected, so H_0(X)=Z. Since H_0 is just the connected components, the last part is already exact, H_1(X)--> Z should just be the zero map.

[trail charm]

good point, makes sense, thanks

[fickle elm]

Or you could use the reduced homology version of MV sequence,which ends H1 maps to 0.

For unreduced version you need to argue like the inclusion has 0 as kernel so the image from H1 is also 0.

[rustic lava]

I'm still not very good at subspace topologies, what is this subspace topology?

Is it like {(0,1) U (6,7) U (100,105)}?

[thorny agate]

What's the definition of the subspace topology?

Say S is a subset of a topological space T

What does it mean for a set A to be open in S?

[rustic lava]

{S n U : U is in T}

[thorny agate]

Cool

[trail charm]

Z \to Z (+) Z given by 1 to (2, -2) is injective thus has kernel 0

oh wait oops that's to show H2(K) is 0

[thorny agate]

Ok so intuitively what do you think the answer is? Try to visualize what open sets are in R and how intersecting them with A in your problem changes them

Draw pictures

I think it may be easier to think about clopen sets

[rustic lava]

Intuitively just like {A,empty set,(0,1],[6,7],(100,105),(0,1]U[6,7],[6,7]U(100,105)}

But I'm not convinced that's right

[thorny agate]

Are you sure those are all the open sets? Is something like (4, 5) not open in A?

[thorny agate]

I gotta go but my gut says this is the right approach

What are the only clopen sets in a connected space?

[rustic lava]

How?

[trail charm]

well i guess the image from H1 being injective isnt hard to see

[rustic lava]

(4,5) isn't A n Anything

[trail charm]

since H1 is Z (+) Z_2

hmm wait im thinking about kernel

[thorny agate]

Wait no I misread, not (4, 5)

But like (6.1, 6.9) idk

[trail charm]

passing to reduced is obvious since you just add the Z summand to go back to unreduced

[thorny agate]

(101, 104)

[rustic lava]

Ah right

I'll have a think about the clopen sets then

[fickle elm]

My first thought is that in H0 every copy of Z corresponds to a connected component. And the inclusion of the intersection induces the diagonal inside Z+Z.

I think the reduced version is easier because everything you have is connected.

[trail charm]

yeah

[fickle elm]

Saves a lot of trouble manipulating in H0.

Well not a lot of.

But doing fewer is always better

[thorny agate]

Actually no, even easier, is A connected in R? Can you use this?

[trail charm]

ive already proved H0 = ~H0 (+) Z and Hn = ~Hn, so yeah it does save a lot of trouble

alright thanks

[rustic lava]

There was something about "a subset of R is connected iff it's an interval"

But I'm not sure we've done the proof of it yet

[thorny agate]

Plz draw it out

It should be ready really obvious

[rustic lava]

I'm bad at picturing things lmao

I don't know the drawing of this

Partially related, what is the complement of (0,1] wrt the subspace topology here?

[ebon galleon]

Don't worry about the topology. Draw this out as a collection of intervals in R

[rustic lava]

Like on a number line?

[ebon galleon]

Yes.

[thorny agate]

Hence why I said draw not visualize

[rustic lava]

Yeah I just never know what to actually draw though

[queen prism]

draw the set on the number line and then erase the part of the line outside the set

[rustic lava]

Is this just {[6,7] U (100,105) n S : S is in T}?

I feel like that can't be how this works

[tribal palm]

it’s so weird to me how a space can be path-connected but not locally path-connected

[opaque scroll]

Idk, isn't that just like saying even though there's a wall between us I can still just walk around the wall.

[tribal palm]

yes but the weird thing is exactly how that wall is somehow “part” of the space

[opaque scroll]

Yeah I guess either make the wall part of the space like the topologist sine curve, or just have so many walls that you can't get away from them, like RxQ

[red yoke]

Cuz path-connectedness is inherently a global feature

You can't tell whether two points are connected if you zoom in too much

[tribal palm]

yeah the sine curve is my example and it’s clear enough from the picture, but it’s still weird to me

[trail charm]

damn RxQ is a good example

[opaque scroll]

But I feel the weirdness there is being connected but not path connected. The fact that you can connect some things with paths when they're not locally path connected is less surprising I feel

[tribal palm]

true but i just have no understanding of that so i can’t even make sense of the weirdnes

[opaque scroll]

Hmm, it's like the topologist sine curve and the compactified long line sort are in this same situation. Where there should be a path from one end to another, it just that it would take so long that you have to give up before you reach the end.

Except I guess the problem with the topologist sine curve is not that it's too long, it's just that you get dizzy

[tribal palm]

[hot locust]

How do I prove this?

[novel acorn]

Wdym prove it

What do you wanna prove?

[feral copper]

Your picture is cropped

It's incomplete, there's no statement to be proved

[hot locust]

Okay yea bad wording, my bad. I wanted to ask why this makes sense. Like, how can the only open sets containing {xi} be {xi}UN and {x1, x2}UN? How do I show that explicitly?

Or is it sufficient to just write it the way it's written or should I put more details here?

[novel acorn]

that's just a definition

[hot locust]

It feels obvious but I feel like I should write more

[hot locust]

Ahh

Okay

[hot locust]

Okay, more like, what kind of topology does this definition give on {x1, x2}UN?

[novel acorn]

uh

it doesn't really have a name

[hot locust]

Ah okay

But it does right?

[novel acorn]

ye

[hot locust]

Alright, thanks 👍

[desert lodge]

I'm trying to add to my list of examples of open sets in R

I've added (0,1), (0,1) U (2,6), R, phi

is there a type which I missed or less obvious?

[gritty widget]

R - {finite set} is a good one

why are you compiling a list of open subsets of R?

[desert lodge]

i compile examples whenever I come accross defintions

just some examples and non examples

[desert lodge]

This is great

[umbral panther]

What’s phi?

[desert lodge]

null set

[umbral panther]

Oh

[desert lodge]

i think it's vacuously open

[umbral panther]

R minus Z
R minus {1/n} minus {0}
R minus the cantor set

[desert lodge]

R- {1/n} won't work?

[gritty widget]

pick an open set around 0

[limber wren]

Try to think of an open set U in R which is

1. made up of the union of an infinite number of disjoint open intervals
2. is contained in [0, 1]

[desert lodge]

let me think

[limber wren]

thinking is good

I'd be impressed if you could do it without thinking

[queen prism]

proof by looking up the answer

[desert lodge]

sorry, maybe a little hint ?

[limber wren]

sure

you know that 1/2 + 1/4 + 1/8 + 1/16 + .... = 1?

[desert lodge]

yes, the series 1/2^n

[limber wren]

yeah

so you have an infinite sum which adds up to 1

I'm asking for an infinite number of open intervals whose union is inside of [0, 1]

and the length of [0, 1] is 1

lol thats my hint

[desert lodge]

same reasoning, intervals having length 1/2, 1/4/ 1/8.. ? as they are disjoint the length of union is sum ?

i need to be more precise

[limber wren]

yeah, lol to both

like you're right and have the right idea, and could be more precise

but thats good, start imprecise but with the right idea, and then try to make it precise

that's literally how to figure out proofs!

[desert lodge]

An= (1/2^n, 1/2^(n+1)) ?

idk if that works

[ebon galleon]

Swap them around

(1/2^{n+1}, 1/2^n)

[since otherwise they would be empty since 1/2^n > 1/2^{n+1}]

[desert lodge]

makes sense

(1/2^{n+1}, 1/2^n)

@limber wren

[limber wren]

seems right yeah, start with (0, 1/2), add (1/2, 1/4), then (1/4, 1/8), etc

oh actually

a little problem

you want to do (1 - (1/2)^n, 1 - (1/2)^{n+1}), starting at n = 0

so you'd have (0, 1/2)

then (1/2, 3/4)

[ebon galleon]

huh

it's symmetric

[limber wren]

ahh right

[ebon galleon]

first should be (1/2, 1)

[limber wren]

oooh yup, totally

[ebon galleon]

which is why it looks wrong

[limber wren]

okay then yeah

lol I think left to right, mb

[desert lodge]

@limber wren

sorry i was away

[limber wren]

np

[desert lodge]

so (1/2, 1/4)U (1/4, 1/8) U (1/8, 1/16)U...U works?

[limber wren]

well writing (1/2, 1/4) is a little weird, you normally don't write intervals like (2 , 1)

lol I do know what you mean though

[desert lodge]

other way around

[limber wren]

yeah, I got you

[desert lodge]

this is empty

[merry geode]

It could mean a fractional ideals generated by 2 elts

Sorry, took time to get what you mean..

[limber wren]

yeah, if a > b then (a, b) is empty

since (a, b) = { x | a < x < b }

and there clearly are no such x, since a > b

so it's a well defined interval, it's just empty

[limber wren]

it is

but yeah, (1/2, 1) U (1/4, 1/2) U (1/8, 1/4) U .....

[desert lodge]

sort of montone decreasing intervals?

idk if thats the correct usage lol

[limber wren]

normally that would mean each interval is contained in the previous one

so like (0, 1), (0, 1/2), (0, 1/4), (0, 1/8), etc

[desert lodge]

i see

[desert lodge]

here all are disjoint

[limber wren]

yep

so you can have an infinite number of disjoint balls, and when you combine them their total size adds up to 1

next question: prove that you can have an infinite number of disjoint open balls, whose total size is as small as you want

[desert lodge]

here An= (1/2^(n+1), 1/2^n) works right?

where should i index from?

[limber wren]

that would be (0, (1/2)^n)

[desert lodge]

not that one

[limber wren]

always starting at 0, but the endpoint goes 1/2, 1/4, 1/8, etc

[desert lodge]

wrong pin

[desert lodge]

this one

[limber wren]

ahh yeah, that would be ((1/2)^{n + 1}, (1/2)^n), starting at n = 0

[desert lodge]

[0,infy) is the index set?

[limber wren]

no, {0, 1, 2, 3, .... } is the index set

[desert lodge]

yeah, indexing set are countably infinite or finite

[limber wren]

[0, infinity) are all numbers greater than 0

I mean not necessarily, you can have an uncountable index set

[desert lodge]

i'm actually just getting started to all of real analysis, sorry in advance

[limber wren]

that's actually leads to a harder question

lol not at all, it's all good

so in the example I gave, you had a countable index set right?

[desert lodge]

yes

[limber wren]

an infinite number of disjoint open balls, indexed by {0, 1, 2, 3, ... }

is that possible if you index by an uncountable set?

like can you have uncountably many disjoint open intervals?

(this is not easy lol, but it's good to think about)

[desert lodge]

i dont know if i can trust my intuitions on this.

[limber wren]

lol whats your intuition?

[desert lodge]

i'd say disjoint open intervals are countably infinite

[limber wren]

yeah

okay, you know Q is countable?

[desert lodge]

yes,

[limber wren]

cool, use that to prove (by contradiction) that it's impossible to have uncountably many disjoint open intervals

lol I know I'm throwing questions at you, don't worry if they're difficult

[desert lodge]

all good

i appreciate

[silver umbra]

hey all, so im currently trying to wrap my head around regular finite covers of trivial surface bundles, but im not sure what the fiber is supposed to look like

so if we have two surfaces, B and F

then B x F -> B is the trivial fiber bundle with F the fiber

but if we were to take some regular finite cover E -> B x F

then composing this covering map with the projection B x F -> B

gives us a fiber bundle E -> B

but what does the fiber of this look like?

[tawdry widget]

inverse image of {b} times F I suppose.

[merry geode]

How do you get E as a base space?

[pseudo ocean]

ok im stuck

how do i continue computing the homology for this? in fact, is there a general procedure for computing the homology group of an orientef graph?

[silver umbra]

but what does this look like?

[tawdry widget]

Idk, I just thought about definition . p: E-> B times F, q: B times F->B, (qp)^-1({b})=p^-1q^-1({b})=p^-1({b} times F)

[silver umbra]

but beyond that is there anything else about it that we can say?

[tawdry widget]

Seems like a universal covering of (B times F,(b0,0)) is just B’ times F where (B’,b0’) is a universal covering of (B,b0). So E=B’ times F, fiber of E looks like multiple layers {b’} times F: image of b’ is b, so F^|inverse image of {b}|=F^|π(B,b0)|

(any path (f,g) : I -> B times F, from (b0,0) to (b,x) is homotopic to (f, line segment from 0 to x), which can be viewed as (f,x) from B’ times F)

[pseudo ocean]

guys does anyone know the homology groups of a complex containing n-2simplexes?

so far all I can do is break em up into some triangles and compute the individual homologies

[desert lodge]

U= (1/(n+1), 1/n) also works right?

i think he is away, can someone check if free?

[pseudo ocean]

can anyone help me out pls?

fuck it I'll do it myself

[jaunty summit]

Is every second countable topological group homeomorphic to a dense subgroup of a Polish group?

[opaque scroll]

Can't you do something like choose a system of neighborhoods of the identity that are closed under taking inverse. Then use that to define a metric and take completion?

[jaunty summit]

Maybe? The issue is that the metric might not be well behaved.

Like Q is homeomorphic to Q ∩ (0, 1)

The completion of Q is R, which is a topological group.

[opaque scroll]

I don't understand the objection

[jaunty summit]

We can define a group operation on Q ∩ (0, 1) by just stealing it from Q (via homeomorphism), but the completion of Q ∩ (0, 1) is [0, 1], which is not a topological group.

The limit of a bunch of homeomorphisms is not necessarily a homeomorphism, so there's no guarantee that the completion actually has a well-defined group operation.

[opaque scroll]

The completion of a topological group only depends on it's topology

[jaunty summit]

Yeah.

[opaque scroll]

And gives you a topological group

[jaunty summit]

So I'm asking if its always possible to metrize a second countable group in a way such that its completion is also a topological group.

It will have a Polish completion (clearly), but it might not be a group.

[opaque scroll]

The completion of a topological group is always a topological group

[limber wren]

yeah I would think you would extend the group operations to the completion using continuity of the operations

[opaque scroll]

You have to assume the group is Hausdorff if you want it to be metrizable though

[limber wren]

at least for locally compact ones

[jaunty summit]

Supposedly first countability and metrizability are equivalent for topological groups, so this isn't a problem (Birkhoff–Kakutani theorem).

[west brook]

How is “alg top” pronounced?

With ɡ or with dʒ?

[opaque scroll]

Then they must include Hausdorff in their definition, because you can make any group into a topological group by giving it the trivial topology

[jaunty summit]

L

[queen prism]

alj top

[opaque scroll]

I usually go with g, but mostly because there's no dz in my language 🙂

[gritty widget]

do you say "al gay braic topology?" no, you say "al jeh braic topology"

dʒ

[limber wren]

al juh bray ick taw paw luh gee

[gritty widget]

topologay

[jaunty summit]

who pronounces it with "g"

ive never heard anyone say it that way in my life

[jaunty summit]

this is wrong tho

[west brook]

I’ve never heard anyone say it at all in my life

[opaque scroll]

Why do you say that?

[jaunty summit]

(0, 1) is a topological group (if you steal its operation from R), but its completion is not.

[0, 1] can't be a topological group because every homeomorphism from [0, 1] to itself fixes a point.

What I'm asking about is the existence of a "nice" metric that allows you to extend group operations to the completition.

[opaque scroll]

I see, you're worried that completing with respect to the metric will be different than completing the group

[jaunty summit]

yeah

[opaque scroll]

So yes, you have to choose the metric based on the system of neighborhoods

So you choose a system of neighbors U1 > U2 > ... then say that the distance between x and the identity is 1/n if x is not contained in Un

(but is contained in Un-1)

Then it should be clear that being a Cauchy sequence in the group is equivalent to being a Cauchy sequence with respect to the metric

[jaunty summit]

Ah, I see.

[rustic lava]

I now realise just how easy this was and am disappointed it took me this long 😆

[thorny agate]

Happens to me all the time lol

[heady skiff]

here if K is a homotopy between k and k' and H is a homotopy between h and h', G(x, t) = K(x, t) o H(x, t) wouldn't make sense right cuz for a fixed x you can't compose em

[tawdry widget]

kh homotopic kh’ homotopic k’h’ (transitivity)

[limber wren]

no but you can compose K(x, t) and (H(x, t), t)

[heady skiff]

yea that works

[limber wren]

that kindof thing

[heady skiff]

is your canonical topological space a cylinder

[ebon galleon]

Homotopies have a cylinder for a domain.

[heady skiff]

lol

would the constant map f_0: X --> [0, 1] given by f(x) = 0 for all x in X be continuous? since the preimage of any open set in [0, 1] (regarded as a subspace of the real line) would either be all of X or the empty set right

an open set either contains 0 or it doesn't

[ebon galleon]

Constant maps are always continuous yes.

[limber wren]

no, it's a blob

a potato

but my canonical picture of homotopies is a cylinder between 2 potatos

with a squiggly tube running down the middle

[narrow cairn]

objectively incorrect

i mean thats correct but your conclusion is wrong

is "algtop" not "algetop"

like you say "lol" not "læl" even though (at least when i say it) the o in out is pronounced as æ

[heady skiff]

so i understand that the fundamental group is the set of path homotopy classes of loops. but i'm trying to see visually when two loops based at a point x_0 are not homotopic, what would be an example? because intuitively it seems like we can always deform one into the other

(ie i'm trying to see why there exist distinct classes)

[narrow cairn]

so

take S^1

famously not contractible

take the base point (1, 0)

[heady skiff]

what does contractible mean

[narrow cairn]

take l_n to be the loop going around the circle n times clockwise for an integer n, where l_0 is the constant map at (1, 0) and n negative represents going counterclockwise

then l_n is not homotopic to l_m for all n ≠ m

[narrow cairn]

homotopy equivalent to a one-point space

or equivalently, space where the identity map is nullhomotopic

[heady skiff]

i see... i'll think about this

thank you

[narrow cairn]

just keep in mind the maps are given by f(x) = exp(2iπnx)

i think its nontrivial to prove these arent homotopic but its intuitive

[gritty widget]

if it makes you feel better, i meant to reply to their other message

[narrow cairn]

it doesnt

you are still very wrong

if anyone walks up to me and says "im studying aljtop" i will punch them

[umbral panther]

Do you punch everyone who pronounced geese like goose?

[heady skiff]

oh shit my b i've been saying aljtop in my head all along

😹

[ebon galleon]

But alj top is correct

[rustic lava]

What actually is a maximal connected set?

There's literally nothing else in my notes that mentions it before this point

[queen prism]

being maximal is measured with respect to some order relation

if x is maximal with respect to the order <= it means that there is nothing bigger than x
in other words if x <= y then y = x

when you talk about sets, this order is the subset relation

[narrow cairn]

why is everyone here wrong

[queen prism]

so a connected set is maximal if it is not contained in any larger connected set (other than itself)

[rustic lava]

Ah right, thanks

[queen prism]

take for example (0, 1) u (2, 3)
(0, 1/2) would be connected but it is not maximal because (0, 1/2) < (0, 1)
on the other hand (0, 1) is maximal

[narrow cairn]

this reminds me of the situation of the main character, winston smith, from the famous novel 1984 by george orwell. winston finds himself in a world where everyone else has been fed propaganda from a government called "the party", so that they all uniformally believe something incorrect. winston finds himself at odds with their beliefs, and he questions if he truly remembers the past. while winston disagreed with everyone around him, his account of history was actually the real one; he was correct. this is directly analogous to the situation im in, since im right and everyone else is wrong, because theyve been fed lies by the party. the party in this case is the moderators of this server, who hold total power over this server's citizens

[rustic lava]

Ah, makes sense thanks :)

[queen prism]

me when i try to explain why tau > pi

actually more accurate to that example is why x! > gamma(x)

[heady skiff]

what do they mean by y_n projects under pi to a loop based at 1? that the composition pi o y_n is a loop based at 1?

[narrow cairn]

they mean pi o γ_n i believe

[heady skiff]

ye

makes sense

wait

this is fucking sick

why does bro say anticlockwise tho

[opaque scroll]

Im with you. I dont even think Im able to pronounce aljtop without putting a vowel between j and t

[narrow cairn]

finally a sane person

[heady skiff]

new jagr pfp

[gritty widget]

there is no difference between "aljtop" and "aljetop" when spoken, to me. they're the same

and i didn't even say anything about this either

i was going after the "hard vs. soft g" part

[gritty widget]

sketch the curve (cos t, sin t)

[heady skiff]

nah yea i know

i'm just used to counterclockwise

[opaque scroll]

So you're telling me you're actually pronouncing it aljetop? Like with the same rythm as algebra but with top as the last syllable? That sounds completely bizzare to me?

[queen prism]

algtop is harder to say

[gritty widget]

"aljtop" and "aljetop" wasn't even what i was talking about. this is completely orthogonal to what i was getting at originally

[opaque scroll]

I thought you where arguing for the use of j instead of g in the pronunciation? Is that not what we're talking about?

[gritty widget]

.

[opaque scroll]

j is the symbol we're using for soft g though right?

Or are we just taking past each other?

[gritty widget]

probably talking past eachother

i don't see how "hard vs. soft g," which is what i was originally talking about, has anything to do with the pronunciation of "aljtop" and "aljetop"

[opaque scroll]

Okay, then let me switch up the notation. G = hard g, dz = soft g.

Then my stance is
aldztop = unpronounceable
aldzetop = bizarre
alGtop = the way to go

[gaunt linden]

"Bizarre" describes this whole conversation.

[heady skiff]

nah it's funny

[umbral panther]

alGtop is unpronounceable. Try alGdop

[narrow cairn]

whats your opinion then

[west brook]

I can easily pronounce all of those

[novel ember]

alGtop

[hidden crag]

interesting language differences

alg top is perfectly pronouncable for me because the g is pronounced "harder" in german

[valid escarp]

in algtop the g is followed by an o so its a hard G

so alGtop

[queen prism]

🇩🇪

[hidden crag]

in the original word algebraic topology it'd be a soft g

[valid escarp]

then say the original word

if not then its alGtop

[hidden crag]

i'm all for that since the g is hard in my language anyway (in the original term as well)

but the point the others are making is obviously the one i said

i can't believe i'm arguing about this

[gritty widget]

it's too easy to get baited by this server

[hidden crag]

maybe we call it alk top now

[gritty widget]

just call it top

[gaunt linden]

B-but ... point set?

[knotty vine]

It's Alχtop, like LaTeΧ. Cccchhhhhh!

[unreal stratus]

I just say com top for combinatorial topology

[distant hollow]

Homo toe pee.

[tidal lynx]

I say Al Top

Al as in Alan

Alan Dershowitz

[narrow cairn]

okay but the g in homology is a hard g we can all agree there

(im joking but also /'hoʊməloʊgi/ is funny to me)

[opaque cloud]

homologif

[red sage]

can you give me examples of a limit point of a set?

I have got 3 is a limit point of [1,3] aswell as (1,3)

can you provide me examples other than such?

R is a limit point of Q right?

[fickle elm]

R and Q, if you mean the set of real numbers and the set of rational numbers, are both sets. Limit point is an element. I guess I know what you mean. For any fixed real number r, any open neighborhood (open intervals) of r contains at least a rational number, so r is indeed a limit point of Q.

[pseudo ocean]

guys can any collection of 2-simplices connected be reduced to a wedge sum of 1-spheres?

[red sage]

Yes

I’m looking for more examples

[fickle elm]

well, a simple example that comes to mind. Consider the set {1,1/2,1/3,...,1/n,...} with subspace topology from R. 0 is a limit point.

[red sage]

That works, thank you

Can a finite set posses limit points?

My gut says no.

[fickle elm]

I feel like it is possible if it had some particular topology

The open set is very large

[red sage]

I’m in R btw

[fickle elm]

If you are in R with the usual topology

The limit point of a finite set should only be the elements from the set itself.

[red sage]

A={1,2,3}

[fickle elm]

It is easy to prove the open set can always "shrink" somehow

[red sage]

1 isn’t a limit point right?

as the neighbourhood (0.9,1.1) doesn’t intersect A other than 1

[fickle elm]

1 itself also counts.

(0.9,1.1) intersects A

It is non empty

That is what is important for a limit point

[red sage]

Can I show the definition i’m having?

[fickle elm]

Every nbhd of the element has a non empty intersection of A

That's the def I am thinking

[red sage]

This is what I have

It says intersect the set A in some point other than x.

[fickle elm]

Ah I see

I may make a mistake

You are right

[red sage]

All good, i just had the textbook with me

So with this in hand, i feel like finite sets can’t have limit points

is that true?

[fickle elm]

Yes in that case you are right

Because every 2 point in R can be separated by open sets

2 different point

[red sage]

If I need to prove, what choice of epsilon should I take arbitrary

[fickle elm]

You have distance in R

[red sage]

yes

[fickle elm]

Take half of the smallest distance since points are finite

[red sage]

Given {a,b,c,..n}

how do I know which has least distance

Suppose minimum exists

[fickle elm]

You do not need know which, you just need to know it exist. And you can get it by finite steps.

All possible distance between 2 points in this case is finite

[red sage]

Makes sense

[fickle elm]

Unless you need to write an algorithm or smth

But since it is finite, you can always claim you do it one by one

[red sage]

Makes sense

Just one more question

I’m studying all these part of real analysis course

this is from chapter named Basic topology of R

but i didn’t got the motivation of what is topology yet

What does it mean to have a topology ?

[fickle elm]

Well, for a set, say R, the topology is a collection of subsets. We call them open sets and force them to satisfy some condition.

You can think of open set is somehow choose what kind of points is "near" to a given point.

I am not an expert in analysis. For me, I think the comcept of topology helps with how to glue geometric structure together

You could try asking this in #real-complex-analysis. They may know more what topology helps in analysis.

[red sage]

Thank youu

[tulip bluff]

so im trying to show that R³ and the upper half space with z ≥ 0 are not homeomorphic. can i do this with removing points? if they were then removing the origin from R³ would give me a space homotopy equivalen to S² but removing the origin from a point from the upper half space would give a space homotopy equivalent to S¹ i think and the first space is contractible while the latter is not

[high hill]

sounds good - the point removal idea

well

im not sure what u mean by homotopy equiv to s1 for the upper half, that doesnt sound right to me
And i also dk what u mean by S2 being contractible to a point

[knotty vine]

That doesn't work because the removal of a point is not a topological property: image a unit disk centered at 0 and a disk at 5, removal of the point 0 will give two different spaces.

[high hill]

it is though

depending on how you phrase it

For one space its possible to remove a point such that (some topological property preserved by homeomorphism)
For another space its impossible ...

Hence they are not homeomorphic. Is a valid argument

Youre observing something about the subspace topologies. And subspace topologies are preserved via homeomorphism.

[knotty vine]

Youre totally right!

[high hill]

So i guess yes, their argument could do with some work

u don't wanna just consider removing the origin in R3

[tulip bluff]

the upper half space is homotopy equivalent to the upper hemisphere of S². If you remove a point from the upper hemisphere i think you end up with a space homotopy equivalent to S¹

[high hill]

Ok so like see what I said about the kind of argument you wanna make

You mentioned specifically about removing the origin from the upper half plane and no thats not -> S1

[high hill]

It depends on which point you remove.

In fact yeah i still dont get your argument completely

you're removing a point after you make the homotopy equivalence which I think will rarely make sense

you can R^3 -> open disc -> remove centre -> S1 if you like

its uh... meaningless without being specific about what this means topologically (i don't think it tells u much)

[knotty vine]

So, what is the quickest argument? Maybe: suppose there is a homeomorphism f : H3 -> R3. Then we also have a homeomorphism of H3 with the origin removed and R3 \ f(0). But the first space is contractible while the second is not (it's homotopic to S2)

Then you don't even need any algebraic topology

[uneven bronze]

I have a very basic question. I have two analytic functions defined on $\mathbb R$ and suppose they agree on $[0,\infty)$. I want to verify that I can use the identity theorem, which states that $[0,\infty)$ needs to contain a limit point in $\mathbb R$. Grateful if someone could confirm and give an example of a limit point. Moreover, what would be an example of a subset of $\mathbb R$ where I could not use the identity theorem?

[gentle osprey]

sunside

[uneven bronze]

https://en.wikipedia.org/wiki/Identity_theorem#

It seems to me like every point in $[0,\infty)$ is a limit point in $\mathbb R$. So can I use the theorem for every subset of $\mathbb R$?

[gentle osprey]

sunside

[opaque scroll]

If the subset doesn't have any limit points then you can't use the theorem. So any finite set for example, or the integers.

Indeed sin(2pi x) and f(x) := 0 are both analytic and agree on the integers

[queen prism]

i wanna say
"... which states that [0, inf) needs to contain a limit point" sounds like a conclusion, when it's really a condition in the hypothesis

[tulip bluff]

why is H3 with the origin removed contractible?

[opaque scroll]

It's star shaped, for example

[merry geode]

Yea, that’s kind of obvious

You can even construct explicit deformation retraction

[tulip bluff]

isn't it homotopic to the upper hemisphere?

[merry geode]

And that upper hemisphere is contractible as well.

Upper hemisphere is homeomorphic with a disc, and a disc is contractible.

[narrow cairn]

is this from abbott?

[narrow cairn]

ooh wait does this work for Rn?

[sudden flower]

Is there an easy way to see that two simply-connected covering spaces \tilde{X} and \tilde{Y} of connected and locally path-connected spaces X and Y for which X is homotopy equivalent to Y are also homotopy equivalent?

[uneven bronze]

Basic question. I'm reading Rudin's PMA. His definition of separated sets goes like this: "Two subsets $A$ and $B$ of a metric space $X$ are said to be separated if both $A \cap \bar{B}$ and $\bar{A} \cap B$ are empty." Then later on in a proof in chapter 8, he claims that two open, disjoint sets $U$ and $V$ are separated. How does this follow from the definition?

[gentle osprey]

sunside

[unreal stratus]

So you want to show that closure of V doesn't intersect U

This is equivalent to showing that closure(V) is contained in X \ U

Can you see why that is the case

[uneven bronze]

yes, hmm, ok, show closure(V) is contained in X \ U

hmm, from our assumptions, what do we know about the closure(V)?

[unreal stratus]

Well, can you name a closed set containing V which is contained in X \ U

[hot locust]

Okay quick confusion, since every topological space is open in itself then doesn't every space have a finite cover, i.e, itself?

Like, I know it's wrong somewhere but I am not absolutely sure why. 'cause recently I got a question where a set could be a cover of itself since it was the only open set covering it, and it was a valid argument there.

But it's not generally, imho. I don't get why. Is it because we are specifically talking about 'sub covers'? That's how I made sense of it before. But I am not sure anymore.

[umbral panther]

What’s easy for one person is difficult for another... Consider the fiber product or pull back. Given two spaces Xt and Y mapping to X, consider the subspace of the product YxXt consisting of pairs of a point in Y and a point in Xt that map to the same point of X. Call it Yt. If Xt is a covering space of X, then Yt is a covering space of Y. If Y->X is a homotopy equivalence and Xt a covering space, is Yt homotopy equivalent to Xt? Is your original space covering Y homeomorphic to Yt?

[unreal stratus]

Yes

It's not wrong

Are you basically wondering why we care about compactness?

[hot locust]

No, I am wondering why some things are compact while others are not.

[gritty widget]

compactness is "every cover has a finite subcover," not "there is a finite cover"

[hot locust]

Oh my god, that makes so much sense

Thank you

[hot locust]

Like, we haven't done connectedness yet, so so far it's just a bunch of sets for me which I can use other sets to stick on and cover so far. 💀

Though I proved that a Torus is compact, idk what exactly is compact about it.

I don't see the picture.

[hot locust]

But yea ig, it's a related question. Like, if I know why certain things are compact and others are not, I can maybe make sense of why we need it?

[unreal stratus]

I do think it is sort of hard to visualise what it really means hm

I think it makes more sense just to think about why it's useful and stuff

Often you only want to work with finitely many things, so that for example you can stitch stuff together nicely from each open set

[hot locust]

Makes sense

[narrow cairn]

take for instance R, and take the open cover {(n, n+2) | n ∈ Z}

this has no finite subcover

i always just think about compactness intuitively as "closed and bounded" but obviously that only makes sense in metric spaces (and isnt always true in metric spaces)

[queen prism]

compactness takes local properties to global properties
if every point in a compact space has a neighborhood with property P, then so does the entire space (klaus janich)

[narrow cairn]

uhh, [0, 1] U [2, 3]

compact and locally connected but not connected

(that is locally connected right)

[hot locust]

Ahh right

And we can say this for every cover? That's neat now that I think about it.

[narrow cairn]

no, R does have finite covers

it just has some open covers without finite subcovers

which means it isnt compact

remember that the compact subsets of R^n are exactly the closed and bounded ones

[hot locust]

What kind of metrics does it not work for?

[queen prism]

ok I think I mischaracterized the statement, hold on

[narrow cairn]

discrete on an infinite space

like take a countable discrete space

[queen prism]

https://math.stackexchange.com/a/1409858

[narrow cairn]

the whole space itself is closed and bounded

since every point is contained in the ball of radius 2 around any point

[hot locust]

Ah okay so one way to show that something isn't compact is to find one cover such that it doesn't have a finite subcover

?

[narrow cairn]

but the open cover { {a} | a ∈ A} has no finite subcover

[narrow cairn]

open cover specifically but yes

another easy way to show it is to find a continuous map f: X -> Y with f(X) not compact in Y, then X cant be compact

[hot locust]

Ohh

[hot locust]

Ooh thanks

[hot locust]

And hence it's not, compact?

[narrow cairn]

yes exactly

[narrow cairn]

for instance if A is a countably infinite discrete space, since every map from a discrete space is continuous, enumerate the elements of a as a_1, a_2 etc and let f: A -> R by f(a_n) = 1/n. then f(A) = {1/n | n ∈ N} which is not closed (it has a limit point at 0 that isnt contained in the set) and therefore isnt compact

[hot locust]

Wait, cts maps which map to a compact set, their preimages are open in a compact space. And so is this the contrapositive of the prev. Theorem?

[narrow cairn]

this is the contrapositive of "the continuous image of a compact set is compact" which is the main theorem on compactness

[hot locust]

Ahh that Makes sense

Wow

[narrow cairn]

(and as a corollary compactness is a topological property)

[hot locust]

So it's preserved under homeomorphism

[narrow cairn]

yes

[hot locust]

So if we find a homeomorphism between a space that is not compact to a space that we want to find if it's compact

That would mean that the given space isn't compact either

[narrow cairn]

right

[hot locust]

That makes sense

[hot locust]

Thanks, I learned a lot.

[merry geode]

What was first, simplicial homology or singular homology?

[umbral panther]

Simplicial

[tidal glen]

Does anyone have recommendations for reading material on hyperbolic manifolds?

[merry geode]

Oh, the simplex one is first

[tawdry widget]

Probably better in #diff-geo-diff-top ? Since there is curvature involved

[red sage]

Yes

[narrow cairn]

nice

abbott based

[tidal glen]

oh sure

[sudden flower]

if f and g are maps X -> S^n with f(x) + g(x) ≠ 0 are they always homotopic by the homotopy given by H(x,t) = (1-t)f(x) + tg(x)/| (1-t)f(x) + tg(x)|?

[unreal stratus]

Yes

[knotty vine]

No

[unreal stratus]

Why not lol

[knotty vine]

Take the (perturbed) constant map and the identity S1 -> S1

[unreal stratus]

What do you mean by perturbed here

That seems to be doing the heavy lifting

[knotty vine]

So f(x) + g(x) avoids 0

[unreal stratus]

How are you gonna do that

You have to perturb it a lot to avoid that

I think

[knotty vine]

Try it. Im trying to come up with an easy argument

[unreal stratus]

Hm why are you telling me to try it when they seem to have written a perfectly good homotopy and I have done this question before too

[knotty vine]

Maybe im mistaken

[torn jungle]

What's the 0th Homology of a simplicial 1-sphere in F2?

is it 0?

[knotty vine]

Oh I think I see now. You're right!

[unreal stratus]

Cool

[unreal stratus]

Do you mean the simplicial set or simplicial complex?

[torn jungle]

ASC

[unreal stratus]

Sure

[knotty vine]

Doesnt really matter what kind of homology

[torn jungle]

Should be zero if im not mistaken?

[unreal stratus]

Well it shouldn't matter since it should agree with the homology of the realisation

Which doesn't vanish in 0th degree

[knotty vine]

Unless you mean reduced homology

[torn jungle]

This is what I have

im lazy so I didnt write span when computing the quotient

[unreal stratus]

Oof

[torn jungle]

:p

[unreal stratus]

Yes you are working with reduced homology it seems

Since otherwise delta_0 is just the zero map

[torn jungle]

yes you're right

[knotty vine]

Yah then it's 0

[unreal stratus]

Usually one would write tildes for reduced homology

[torn jungle]

trying to prove/disprove claim a

[torn jungle]

we also work on F2

[unreal stratus]

Okay sure

Hm why is that

[torn jungle]

it is a 3.5 week crash course on homology so nothing too advanced

[unreal stratus]

Or just for simplicity

Okay sure

Seems somewhat idiosyncratic lol

[torn jungle]

yes

we also know how to work on Z etc. but ptofessor doesnt wanna make out lives too difiicult

I also have this

too small but

attempting to prove the statement

might be completely bogus

[knotty vine]

Not every ASC is equivalent to the wedge sum of circles right?

What about S2

What about the projective plane

[torn jungle]

i only consider the 1-skeleton as H0 depends on the boundary of the 1 skeleton and the boundary of the 0 skeleton

of our ASC K

[knotty vine]

That sounds good

Why is the intersection of loops always connected?

[tawdry widget]

What book is this from?

[torn jungle]

not a book

not sure if my lecturer ripped it off somewhere or formulated his own questions

[torn jungle]

the intersection of loops cannot itself be a loop

so its equivalent to a point

[knotty vine]

But cant it be 2 points?

[torn jungle]

yes

[knotty vine]

But then H0 isnt 0?

[torn jungle]

FUCK

[tawdry widget]

I see, thanks

[torn jungle]

i think the intersection always was a line in it

[torn jungle]

or

rather

there exists another distinct loop/loops such that the intersection also contains lines

dk how to formulate this exactly

[knotty vine]

Could we transform the ASC so that all loops connect in only a single point

[torn jungle]

thats also true

[knotty vine]

If you have to start thinking about the combinatorics of these kinds of hellish diagrams then I think you're on the wrong path 🙂

[torn jungle]

I agree

made it a biiit too complicated

my main idea was to use mayer - vietoris

to show its always zero

[knotty vine]

I think the idea is good

Maybe you could also do without (a la https://topospaces.subwiki.org/wiki/Zeroth_homology_is_free_on_set_of_path_components) but never mind that

Mayer vietoris is usually stated for two subspaces. How were you planning on dealing with a load of loops? What if theres transfinitely many?

[torn jungle]

we define ASCs as strictly finite

what I was planning to do is work in pairs of loops

and combine what I get for each pair to get some notion for the Union of all loops

ie

[knotty vine]

Ok then you're good

[torn jungle]

l1 intersection l2 ---> l1 U l2
l3 intersection l4 ---> l3 U l4

and then combite what I get'

i think there should be a simpler way

[knotty vine]

Just do induction

[torn jungle]

yeah

something along the lines

[knotty vine]

So then you really want all loops to connect in only 1 point since otherwise you also have to think about the intersection of a loop with another union of loops

Not just the intersection of two loops

Is the empty ASC connected?

[torn jungle]

i think i managed the proof

i showed we can get all the loops to intersect at at most 1 point with each other loop, and then inductively showed that H0( l1 U .. U ln ) = 0

[grim knot]

guys, does somebody know what are the most important theorems of algebraic topology?

[hidden crag]

up to what point

[grim knot]

I have a book, algebraic topology by allen hatcher

and I was trying to find the most important things about it, since I have to write a "summarizing paper" about the most important results, eventhough I didn't do the course yet

https://pi.math.cornell.edu/~hatcher/AT/AT+.pdf

[hidden crag]

You have to summarize things that you don't know?

[grim knot]

[hidden crag]

[hidden crag]

You could do that for one chapter of the book

[grim knot]

I wanted to write something about simplicial homology

[hidden crag]

but for the entire one?

[grim knot]

I thought about simplicial homology, homology group and homotopy groups and showing the connection through the Hurewicz theorem

[hidden crag]

you could show that it satisfies ES axioms and compare it to other homologies

[grim knot]

but I don't know how relevant this theorem is

[hidden crag]

how long is the paper supposed to be

[feral copper]

Hurewicz, as in relating the first nonzero homotopy group to the corresponding homology group? Maybe that's too much, and something more entry-level would be cellular spaces and their homology (cellular homology)

[grim knot]

two pages 😄

[grim knot]

yes, that is what I was thinking about

[hidden crag]

two pages might be enough for the special case pi_1

[feral copper]

If you really want to go into homotopy groups, why not, but that's not as basic as homology

[hidden crag]

you probably don't wanna go into higher homotopy groups for that one

[feral copper]

Tbh, cellular homology is much more helpful to give tons of examples of computations

[grim knot]

I thought about the special case for n=1, otherwise it's gonna be too much

[hidden crag]

however: funny triangles

[grim knot]

I haven't really heard of it

lemme take a look

[feral copper]

A very neat application is the homology of lens spaces: that's the perfect example of what can differ through different xoefficient rings

[feral copper]

It depends. For CW pairs, there's this "disagreable lemma" (that's what my old German teacher called it), which states that a CW pair (X,A) is n-connected iff A contains the n-skeleton of X.
If you can prove this, then Hurewicz for CW spaces is an application of the case A={*}!
(that's for culture and if you're interested for your own sake; don't talk about this in your "summarizing paper")

(and ofc, there are easier and more direct proofs of Hurewicz!)

[unreal stratus]

It's really cute how it follows easily from the π1 Hurwicz along with the serre exact sequence in homology (from the Serre spectral sequence)

Unfortunately that requires formalism one probably hasn't done when you first learn hurewicz lol

[distant lichen]

Spectral sequence arguments and "cute" does not compute

There we go

Haha

[unreal stratus]

I think they can be very cute

Well like a nicely interlocking puzzle

Tickles a similar part of my brain to, say, doing character tables

:)

[distant lichen]

Ahh

[unreal stratus]

Except when it doesn't work

😭

[distant lichen]

Well, to each their own

[unreal stratus]

e.g. a preprint i am working on

[distant lichen]

I tend to find them very useful but rarely aesthetically pleasing

[unreal stratus]

Fair enough

I think a nice example I had in mind was like

Stuff like cohomology of K(Z,2) where you get nice periodicities and stuff

or like can completely work out the differentials "in reverse" hehe

[feral copper]

Working on spectral sequences?

[feral copper]

But, don't you have a somewhat-explicit description of K(Z,2)=CP^oo?

[unreal stratus]

Definitely yes, the cohomology groups are immediate and not too hard to compute the ring structure directly ig

Just I think it's nice that you can also do all those computations without even having a nice model

[unreal stratus]

More or less

[feral copper]

Yeah makes sense. Ig it also helps in more general situations where you don't have such a nice description of the space

[radiant cedar]

what is the fundamental group of the space X where X is the subspace of R^3 in which at least one of the coordinates of (x,y,z) is an integer?

[muted arrow]

Try sketching part of it and what you think a generator might be

Note that X is spanned by the planes {x=i}, {z=i}, {y=i}

For i an integer

[radiant cedar]

i've tried and seems like this space consists of parallel planes in the xy, yz and xz directions. it's not very intuitive to figure out what the generators should or if the space is contractible

[muted arrow]

For an idea of what the generators "should" be, consider the problem in R^2 first

[radiant cedar]

in R^2 we end up with a "grid" right?

[muted arrow]

Right so what does a generator look like there

[radiant cedar]

just to be clear if i consider the base point the origin the loops in this space should be "stuff going along" the grid lines and not actually in R^2?

[muted arrow]

Yes

And since the space is path connected your base point doesn't matter, you can forget about it

[radiant cedar]

we dont have any other trivial loops except the constant one right?

[muted arrow]

Not up to homotopy, no

But that's always the case

Draw the R^2 case and an example of a non trivial loop

(An easy one)

[radiant cedar]

for example the one that goes around the square with vertices (0,0), (1,0), (1,1) and (0,1)?

[muted arrow]

Right, so you're bounding a square

Now what do you think it should be in R^3

[radiant cedar]

a cube

[muted arrow]

Yep. But a sphere is simply connected and a cube is your username to a sphere

[radiant cedar]

wait is the loop then nullhomotopic

[muted arrow]

Yes

The space in your example is simply connected

Draw one of the cubes in R^3 and a curve to convince yourself

[radiant cedar]

i see pi_1(X) = 0 in that case

thanks for helping out this was enlightening

[muted arrow]

No problem!

[heady skiff]

could i have a hint on this question pls?

[plain quiver]

prove the contrapositive using the fact that you can divide by f(x) + x

[heady skiff]

is it just that if we assume that $f(x) + x \neq 0$ for all $x \in C$, $F(x, t) = \frac{(1 - t)f(x) + tx}{\norm{(1 - t)f(x) + tx}}$ is a homotopy between them?

[gentle osprey]

okeyokay

[plain quiver]

yeah, something like this will worm

[heady skiff]

cuz denominator is never zero

[plain quiver]

work*

[heady skiff]

oh okay sick

what do they mean by effect on the square? shouldn't it be on G since F is a map from the square to G?

[limber wren]

I assume they mean draw a picture of what f does to [0,1] x [0,1]

like the square being [0,1] x [0, 1]

[heady skiff]

wait i'm confused how does it affect [0, 1] x [0, 1] tho if that's the domain

[gaunt linden]

It's a bit handwavy language.

[limber wren]

haha yeah it's weirdly worded

[heady skiff]

classic armstrong

ohh

i see now

like the effect "restricted to this square"

even tho there's no restriction

[gaunt linden]

The idea something like imagine the map physically lifts each point on the square out of R^2 and puts it down somewhere in G instead.

[limber wren]

lol really it's like "just draw a picture"

[heady skiff]

lol yea makes sense

[limber wren]

oh maybe by the square they mean the boundary?

[heady skiff]

also i'm assuming they want us to show that the fundamental group of G relative to e is ableian

[limber wren]

like f restricted to the boundary

[gaunt linden]

Just showing the effect on the boundary is probably easier than trying to depict the whole square, yes.

[limber wren]

since G is path connected you can always look at e, since all fundamental groups on G are isomorphi

isomorphic*