point-set-topology

[unreal stratus]

Idk like you have given a good argument for why this is the case aha

[tribal palm]

[unreal stratus]

But you can also just use the metric to see why this is the case if you want a lil more intuition

That probably helps more generally w knowing the opens lol

[tribal palm]

oh? yeah i should probably do that, thanks

honestly my intuition tells me no subset of A should be open (because none are in R, all being unions of isolated points)

but of course that is opennes in R, not in A

[unreal stratus]

Yeah

[torn jungle]

can someone help me with the definition of group amalgams?

we have that A is an "alphabet", A-1 the set of ther inverse elemets of A

is A supposed to have some operation we take the inverses in terms of?

if A is a group is A-1 just A \ {0}?

I really don't get this i feel like there's something missing

[umbral panther]

A-1 is in bijection with A. You don’t exclude the identity

[torn jungle]

not the identity the zero elemnt as it wouldn't have an inverse

also is A a set, a group

[umbral panther]

The identity is its own inverse

A-1 is usually formal inverses. It’s just symbols whose job is to be the inverse of the elements of A. But it’s just a set. They don’t do their job until they are put in a group

[torn jungle]

So very arbitrary

we can construct a set of letters say A = { a, b,c }

and claim that the inverses are the capital letters corresponding to each letter and construct a group this way Fr(A)?

and is the group operation on the free group concatanation?

professor fucked up this lecture, so unclear 😭

A has to be a subset of a group?

[umbral panther]

If you’re talking about free groups, A can just be a set

If you’re talking about amalgamation, then you start with groups

[torn jungle]

btw in the lecture all he wrote was A set, A-1 inverses of A, A+- disjoint union

and a word is the product of the elements?

[torn jungle]

hmmm

just trying to understand the statements first which are missing some context i feel like?

[tawdry widget]

Your messages are chaotic. What exactly are you asking

[umbral panther]

Words in the set A form the free monoid over A. If you want a group, you need inverse. Words in A, A-1 modulo cancellation form the free group

[torn jungle]

I tend to do this sorry

[tawdry widget]

According to what I found on the internet amalgam of groups has nothing to do with what you sent. Your messages looks like ordered groups, the things you discussed are about free groups

So chaotic…

[torn jungle]

During the last lecture we looked at the pushout of groups. We started with some "definitions" of

1. the alphabet set A (a set)
2. A-, A+- (inverses of A and disjoint union on A,A+-)
   my first question is if there are any conditions on the set A. I assume it must have some operation possibly with some rules?
   and would words (ai in A, a word is a1 a2 a3 a4 a5) be the product of the elements? or just the order the elements are placed in?

Why I need this is because the pushout is defined as:

<A | R>

= Fr(A) / <R>ncl

for A the disjoint union of groups G1 and G2 and R the set of the following (relations(?))

{abc s.t. for some i, a,b,c in Gi and abc = e} U {f1(a) f2(a)^{-1} for a in H }

[torn jungle]

where G1 and G2 are subsets of H

[tawdry widget]

Still isn’t clearer.
About push out of groups I know that push out of f:G->H and g: G->K is free product of H and K, over normal subgroup generated by {f(x)g(x)^-1: x from G}.

So when you have G=<A | R>, H=<B | S>, K=<C | T>, then push out is <B disjoint union C | S union T union {f(a)g(a)^-1: a from A}>

I still can’t understand your context. Better wait for others

[torn jungle]

just can't understand some parts of the lecture notes

mostly what things are denoted as

wikipedia answered my question

[tribal palm]

simply {{1/n} : n in N} ?

[unreal stratus]

That is part of the basis i had in mind

[tribal palm]

wait, no some open sets contain 0

[unreal stratus]

Well in fact they must be part of any basis

[tribal palm]

the other half being {0} u { 1/k : k > n} for every n?

i suppose i have the tools to check

[unreal stratus]

yup

[tribal palm]

it clearly covers A

so i just need to show any finite intersection in the basis is also a union in the basis

well in fact every finite intersection is in this case either empty or itself a basis element

[heady skiff]

oh wait

so each component of a topological space is closed

wait hold up i'm confused now

because each component of a topological space is a closed set

so under the finite complement topology

every component is a closed set

and hence a finite set

well just R i guess

well it wouldn't be connected by the argument i just proposed

[ebon galleon]

In the cofinite(/finite complement) topology, the finite sets are closed and the entire space is also closed.

[heady skiff]

oh so i guess the components couldn't be the finite sets and the entire space cuz that's not a partition of the entire space obviously

now i'm confused

are the components finite sets?

[ebon galleon]

No.

[heady skiff]

why not?

component

=> closed

closed in cofinite topology => finite

?

[ebon galleon]

No.

[heady skiff]

?

and then i said the finite sets

[ebon galleon]

Ah, I mispoke there. The finite sets and the entire space are your closed sets.

My bad

[heady skiff]

all good

[ebon galleon]

Also, the finite sets are not connected in the subspace topology (since they inherit the discrete topology) unless they are just a single point

[heady skiff]

i can't be the only one that things that armstrong is not a good book

munkres seems so much better

so uh

it would just be the finite sets and R then i guess

lol

[ebon galleon]

Think about it this way. Let's suppose that you have an infinite space X with the cofinite topology. A space is disconnected if it can be written as a union of two nonempty, disjoint, closed sets (this is equivalent to other characterizations of disconnected)
Now, if a subset of your space proper and closed, then it must be finite (since the only infinite closed set is the entire space and proper rules that one out)
Can you write X as a union of two finite sets if X is infinite?

[heady skiff]

no

[ebon galleon]

Right. So then X cannot be disconnected. Or in other words, X is connected.

So since X is clearly maximal among subsets of X, that means X is the only connected component of X

(because remember, the components should partition X. So we won't include any smaller connected sets)

[heady skiff]

i see

that makes sense, thanks

i didn't even know that the components partitioned the whole space

it was briefly mentioned in my book but wasn't a definition

would the closed sets of the half-open interval topology be ones of the form $(-\infty, a] \cup (b, \infty)$? my logic was that if $A$ is closed then $\mathbb{R} - A$ is a half open interval so I kinda worked backwards

[gentle osprey]

okeyokay

[heady skiff]

so if $A = (-\infty, a] \cup (b, \infty)$ then $\mathbb{R} - A = (a, b]$

[gentle osprey]

okeyokay

[heady skiff]

and same with switching the closed brackets around to b

nvm

how does this show that x and y are not in the same component?

[patent bloom]

because y is in [x,y)^c ?

and x is in [x,y)

[tribal palm]

god it’s gotten to the point where i keep doing topology in my sleep

i just woke up with the idea: let f:X->Y be bijective and fancy B_X a basis for X, if f(fancy B_X) = { f(B) | B in fancy B_X } is a basis for Y, what can we conclude about f? (my hope is that it will be a homeomorphism but i’ve yet to get out of bed and actually check)

[warm meteor]

Bonjour,

je suis à la recherche d’une suite de cauchy divergente biensure dans un ensemble à dimension infinie. Mais j’arrive pas à en trouver au moins une . Quelqu’un a une idée ?

Good morning,

I am looking for a sequence of cauchy divergent course in an infinite dimensional set. But I can't find at least one. Any ideas ?

[paper wedge]

wdym course

are u asking for an example of a cauchy sequence that doesnt converge in an infinite dimensional space?

if so then i think the set of all continuous functions under integral norm is not complete

consider C[0,1]

and then consider f_n(t) = t^n

[tribal palm]

is the hilbert cube $[0,1]^\omega$ with the sup norm complete

[gentle osprey]

Jens

[tribal palm]

?

[opaque scroll]

The sup "norm" would just be the discrete metric, so yeah

[tribal palm]

whaat

then we’re thinking of different things… probably me who’s using the terminology incorrectly

[opaque scroll]

Right, sorry I think I misread what you said

Anyway, if something is Cauchy in the supnorm it's Cauchy pointwise, and then just show that the pointwise limit is a limit

[tribal palm]

an element in the cube is a sequence x = (x_n)_{n in bbN}, with the sup norm of x being the supremum of these xn, then the induced metric is the sup of the absolute value of the componentwise difference

is that not standard terminology and definitions?

[warm meteor]

yes

[opaque scroll]

Yes, it is. I just misread which space you where working with mb

[tribal palm]

thanks

[paper wedge]

yea then read what i said

or wrote whtver

[warm meteor]

to prove that it’s a cauchy series i have to take the norm(t^p+n - t^n ) and since 0<t<1 it tends to 0 when n goes to infinity ?

[paper wedge]

this is not a series

[warm meteor]

i don’t know how to say suite in english

😂😂😂

[paper wedge]

its a sequence

[warm meteor]

ohh yess thank you

[paper wedge]

and u have you show that for all epi >0 there exists N such that for all n,m >= N norm(f_n-f_m) < epi

[warm meteor]

yes that’s what i did no ?

[warm meteor]

here

of course i will do it with epsilon

[paper wedge]

idk i just got confused by what ur saying

so i wrote it out

but yeah

idk what do you mean by letting it to infinity

compute the integral

and then u will get some term that goes to 0 as the ns and ms get very large

probably something of the form 1/something

[warm meteor]

I‘m going to write it a language that we all will understand

[paper wedge]

money?

btw

this is not really #point-set-topology

you should ask this again in #real-complex-analysis

or maybe this is more functional analysis but meh

[warm meteor]

this the first step

now we have to take another norm

@paper wedge

[paper wedge]

no

ur done

this goes to 0

As those get large

now this is cauchy

now why would this not be convergent?

remember we are in the space of continuous functions

[warm meteor]

yes

because when n goes to infinity it will not be continuous anymore

the limit will be 0 if t in [0,1[ and 1 if t = 1

[tribal palm]

[warm meteor]

so it’s divergent ?

like it doesn’t converge in the same space

OHH yes

😂😂😂😂😂

so it is divergent

thank you @tribal palm

saiki helped me

@paper wedge thank you too

[paper wedge]

sorry im afk im watchinf tje most boring anime so i can fix my sleep schedule

[warm meteor]

tell me what’s the name of the anime yoûre watching

[tribal palm]

i was reading this wikipedia article about comeagre sets and was struck by the genius idea that we should rather be calling closed sets, co-open

[paper wedge]

monster

[warm meteor]

hehehe

[warm meteor]

co- open

co-open 🧐

I can’t really see the joke

[opaque scroll]

Nah, open sets are closed under union which feels like a colimit thing. So we should be calling then coclosed

[tribal palm]

making the closed sets cococlosed

[opaque scroll]

🥥closed

[queen prism]

and a coconut should just be called a nut

[tribal palm]

[red yoke]

1 am coco

[tribal palm]

when i have some function from A to B, and i want to refer to the obvious function P(A) to P(B) induced by f, namely f(C) = {f(c) | c in C} for all C subset A, what should i call it, "the image function of f (or induced by f)?" or "the image function under f" ?

[limber ravine]

what is P?

[queen prism]

means the function that sends subsets of the domain to subsets of the codomain
so normally you'd write f(x) = y for a point x in A and a point y in B but now we want to talk about the function sending a subset C of A to its image f(C) in B

I don't think there's really a standard way to call it
maybe just write it out in words
"... and now consider f as a function f : P(A) -> P(B) sending subsets to their images"

[tribal palm]

yeah i think i've seen my prof write something similar

[lime sable]

maybe direct image and inverse image, or pushforward and pullback, but that doesn't fit perfectly here

[unreal stratus]

You can be fancy and call it P(f) i think lol

Oh nvm that is ambiguous

Calling it the image function would make sense to me

And what nlab does I think

[tribal palm]

something like this, then

oops i meant to add "under f" after the "P(Ninfty) to P(A)"

and i suppose when considered on all of P(Ninfty) it is not a bijection, only when restricted to Binfty

ah, right, this is what my lecture notes has

[tawdry widget]

Yes f is a homeomorphism

[tribal palm]

first thing i did aftr getting out of bed was verify this

if f were not a bijection

but it is

[tawdry widget]

He said f is a bijection so continuous

[ebon galleon]

wait yeah bijection i guess should give continuity then

Since both sides would be bases

[tribal palm]

yes, i think this is good enough for a solution

for context:

and bbN_\infty is just bbN u {\infty}

[tribal palm]

oh oops i forgot to show fancy B actually is a basis for A

[tribal palm]

if x is some point in the topological space X, is there a name for the smallest open set containing x?

sounds like a set of some importance

[ebon galleon]

There typically isn't such an open set

For example, consider the real line: no points have a smallest neighborhood

[tribal palm]

right right, but there sure are spaces in which such a set does exist

[ebon galleon]

When every point has a smallest neighborhood, the topology is called an "Alexandrov" topology: https://en.wikipedia.org/wiki/Alexandrov_topology

Simplest examples: indiscrete and discrete topology (also every finite topology)

[tribal palm]

ooh

øystein ore mentioned here worked at my uni

i wondered what happened if i modified the usual top on R so that {0} was open in the topology, and found if i took the set {0} u {(a,b) | a < b} as a basis this does make a top finer than the usual, in which {0} is open

and in this the only seq converging to 0 is the constant 0 sequence

i’m not sure what other questions i could ask about it

i suppose i should try and see if i can find some homeomorphic space

[narrow cairn]

hmm

well, its still second countable

and hausdorff

and locally euclidian

so i think it should still be a manifold

however it is not connected

it has 2 components

[umbral panther]

It is locally Euclidean, but the dimension varies from point to point. This is usually excluded from manifolds, but it should probably be allowed

It has 3 components

[narrow cairn]

right duh

[narrow cairn]

oh right 0 pnly has a 0-euclidian nbhd

my bad

R is homeomorphic to (0, infty) right?

if so this space should be homeo to the disjoint union of two copies of R and a single point

whats the homeo from (0, infty) to R?

[umbral panther]

Log

[narrow cairn]

ahhhh

thats epic

[tribal palm]

how do you define a component?

[umbral panther]

A component is a maximal connected subset

[tribal palm]

we haven’t even gotten to connectedness yet, but i’ve only sneeked a short peek

[umbral panther]

A subset which is both open and closed is a union of components. Anything bigger can’t be connected because it is the union of the clopen and the complement

[narrow cairn]

connected space is a space that cant be the union of two nonempty disjoint open subsets

connected subset of a space is a set connected with the subspace topo

[queen prism]

you should probably try to understand subspaces before talking about connected spaces

[narrow cairn]

connectedness is so easy compared to compactness/paracompactness bullshit

[tribal palm]

yeah i’m very uncomfortable with subspaces… product spaces as well honestly

[heady skiff]

^ same

what's the usual text for topology

because we're doing armstrong for our class and it's not that helpful lol

Munkres?

[queen prism]

idk about product spaces

subspaces go like this
consider the set Y = [0, 1] as a subspace of X = R
the open sets in Y are defined like this: take any open set in X and consider the part of it that overlaps with Y
for example, (1/3, 2/3) is open in Y because (1/3, 2/3) n Y = (1/3, 2/3) is open in X
on the other hand, consider the set formed by overlapping (-1/2, 1/2) with Y, which is [0, 1/2)
[0, 1/2) is now open in Y, because it was formed by overlapping Y with an open subset of X
but it is definitely not open in X

you can check that these kinds of sets form a topology on Y
and if you're not convinced, maybe look at it from the metric-space perspective

[heady skiff]

so whenever we're considering say a neighborhood of some point x in X, and we want to figure out the open/closed sets of the neighborhood, we do so under the subspace topology?

[queen prism]

if you take a neighborhood of a point and you want to consider it as a topological space in its own right, you would most likely use the subspace topology

you could put a different topology on it, but when you're talking about subsets of a larger space, subspaces just make the most sense

you can try reading munkres

another book, probably on the easier side, is https://www.topologywithouttears.net/

[heady skiff]

like i'm confused as to when to consider it a subspace and when not to

[heady skiff]

oh ok thanks

[queen prism]

it's just the most natural candidate
take the part of an open set that happens to lie inside your subset and call that open

[heady skiff]

well what if you have a subspace of a subspace lol

say

i'm considering an open ball V of x contained with U and I want to talk about clopen subsets of V, call it A

then i would have to decide whether to write A = V \cap C for some closed set of U or some closed set of X

but i guess my question is I have that freedom to do so right

[queen prism]

here's a cool fact

[high hill]

you can show subsubspace topology is a subspace topology

yh that

[queen prism]

subberspace

[high hill]

now what about an infinitely descending chain of subspaces...

the thing that lies at the end...

[queen prism]

zorn's lemma wizardry?

[narrow cairn]

well idk what the question would even be then

[queen prism]

o I remember now
for product spaces you need to understand topological bases

[narrow cairn]

yes

[heady skiff]

here does letting V = B(x, \eps) contained in U work for a locally connected neighborhood?

i need a hint for this shit idk how the fuck to do topology lol

i've tried decomposing B(x, \eps) into two disjoint nonempty open sets A U B and tried to get at a contradiction or some shit

[narrow cairn]

wait, what part of the problem are you working on?

the last part?

[heady skiff]

i was working on 1 but i got frustrated and looked up the answer lol

[narrow cairn]

lol

try 2

[heady skiff]

i'll probably just go back and start reading munkres, this book armstrong is terrible

[narrow cairn]

read Lee

Lee is efficient

[heady skiff]

isn't that a grad book

[queen prism]

graduate's just a label

[narrow cairn]

well, its technically GTM but the first 4 chapters are definitely not grad level

idk about chapter 5

are CW complexes generally grad level?

[queen prism]

i don't think they're undergrad

is that part of point-set

[narrow cairn]

ive heard its usually taught in algtop

[queen prism]

ok probably not undergrad then

[narrow cairn]

im doing CW complexes so i should get G+

[heady skiff]

yeah lee looks good actually

[narrow cairn]

lee is very good

[heady skiff]

idk i'm skipping and probably making assumptions but looks a lot less hand-wavy then armstrong lmfao

so many more examples too

[narrow cairn]

yea the proofs are thorough

its thorough but also efficient, does all of pointset in like 150 pages

[heady skiff]

whereas armstrong is just like "glue this side to that side and warp it a little bit QED"

damn ok

[queen prism]

is lee top good? i was going through it but i remember backreading and someone liked smooth more than top

[narrow cairn]

ITM is good so far

some of the problems are fucking hard but you can just skip those ones

[lyric iron]

yooooooooooo

[narrow cairn]

like "show that if every open cover of a space admits a partition of unity subordinate to it then the space is paracompact"

didnt even try that one sounds hard

[lyric iron]

am new

[narrow cairn]

hello

what topology are you doing?

[queen prism]

that sounds painful

[lyric iron]

what is topology i am in 7th grade

brooooooooo

[narrow cairn]

lmfao

[fickle elm]

You can prove the open ball in R^n is path connected by explicitly constructing a path between 2 points.

[narrow cairn]

wrong channel

[fickle elm]

usually proving path connectness is easier

[narrow cairn]

check #discussion or #❓how-to-get-help

[lyric iron]

can u clear my doubts in maths???

[narrow cairn]

someone can but not here

[lyric iron]

so where

[narrow cairn]

i just linked 2 channels

[lyric iron]

k thank uuuuuuuu

[narrow cairn]

(this is very easy btw, remember that balls are convex)

[heady skiff]

Wait is munkres point set or algebraic topology

[fickle elm]

Both i remember

but people seldom use Munkres for the AT part

AFAIK

[narrow cairn]

lee ITM seems to have some good algtop stuff

idk tho havent got there

[fickle elm]

Quickly checked the content of Munkres's topology.

He includes Jordan curve theorem and classification of compact surfaces, but only fundamental groups and some covering spaces.

Can be a supplement to a topology course but not as a standalone AT course IMO

[tribal palm]

introduction to topological manifolds?

[queen prism]

ye

[narrow cairn]

yes

[queen prism]

i wanna go through all 3 of his manifold books, i'm just lazy

[heady skiff]

a set that is not open is not necessarily closed right?

trying to show that S a subspace of X and B closed in S implies that B is an intersection of a closed set of X w/ S

so I got S - B = X \cap O for some open set O of X

[fickle elm]

That is the definition of subspace, right?

[heady skiff]

nah not how lee defined it, he left it as an exercise actually

to characterize the closed sets

[high hill]

yh the open defn and closed defn are equivalent

its a short exercise to show this

[heady skiff]

not short for me sadly!

[high hill]

well short in the physical proof needed to be written down

its purely symbol manipulation with sets

Let (X, T) be your space. Now, (S, {O n S : O in T}) is your subspace topology

Then you wanna end up showing {O' n S : O' in T'} = {O n S : O in T}' I think, notated loosely

(where T' := {O' : O in T})

[heady skiff]

what does the ' mean

[high hill]

complement

[heady skiff]

o ok

[high hill]

im also using it loosely in 2 places

so ' on a set of sets

is the complements on the inside

{O' n S : O in T} = {(O n S)' : O in T} I think reads a bit better

i also wrote that from the top of my head, so there may be errors. But the proof of what u wanna show really is a "follow your nose" proof

[heady skiff]

oh okay thanks

wait i'm confused

open set U of X implies X - U is closed?

wait i need to get my quantifiers right holy shit

bc A closed iff X - A open

but the negation of A closed is not necessarily A open

and same w/ X - A open is not necessarily closed right

[ebon galleon]

X - (X - A) = A

[heady skiff]

right

i'm high lol

[narrow cairn]

all of the notation here sucks

[narrow cairn]

that is the definition of a closed set, yes

[tribal palm]

blessed notation

[narrow cairn]

wouldnt T' be the complement of T in P(X), so all of the sets that arent open

[ebon galleon]

this was what Shuwi was using for T'

[narrow cairn]

sure but thats why this notation sucks for complements

i hate A' instead of X \ A or even X - A since it doesnt make explicit which set the complement is in

[ebon galleon]

I don't disagree.

[high hill]

well the more implicit notation is the quickest to hash out ideas in plaintext

[narrow cairn]

sure but X \ A takes marginally more time than A'

[high hill]

i usually use -

[narrow cairn]

thats good too

[high hill]

its less readable to me at least as plaintext

because its more characters non formatted

{(X - O) n S : O in T} = {X - (O n S) : O in T}

thats what u wanted right

[narrow cairn]

that seems less ambigious yes

i do like \ more since - can get confusing in algebra

[high hill]

id argue \ is more easily confused

is why i avoid it

[ebon galleon]

I have never actually had that cause any confusion

[narrow cairn]

like for groups/fields A, B ive seen A - B := {a - b | a ∈ A, b ∈ B}

[high hill]

the reverse slash being the quotient

[narrow cairn]

ah thats true

Lee uses a half titled backslash

[high hill]

i feel like i wrote my algebra wrong rereading it

{(X - O) n S : O in T} = {X - (O n S) : O in T}

oh yeah ur right, i confused my complements

{(X - O) n S : O in T} = {S - (O n S) : O in T}

caught me red handed

[novel ember]

send help

i suck ass working with bases

[high hill]

@heady skiff important point raised ^ pay attention to complement within which set u want

otherwise u end up with funny things

thanks henry

[narrow cairn]

use the based criterion

probably idk

[novel ember]

what

[narrow cairn]

based criterion is very useful for basis problems usually

(a basis B generates X iff for all open subsets U of X and all points p ∈ U there is some open V in B with p ∈ V subseteq U)

usually called the basis criterion but its funnier to call it the based criterion

[heady skiff]

what do we mean by "pasting"? what's a mathematically precise way to describe it? i mean intuitively it makes sense i guess, but other than that it's not really rigorous or i can't see it as being mathematically rigiorous

rigorous

like this looks like an origami book or some shit lol it doesn't make sense mathematically to me

[viral atlas]

Quotients

You "identify" points in your space through some suitable equivalence relations

The construction for the torus above quotients a square by identifying opposite edges with each other, i.e., pairs of points collapse into one in your quotient space

[heady skiff]

huh okay... i think that makes more sense then

thanks

[viral atlas]

Writing out details of these intuitive constructions made me appreciate the origami more than the rigour at times

[limber ravine]

If I have a set which is open, then for each p in U there's another open V such that p in V subset U. When can I ensure that the closure of this V is compact?

[high hill]

"glue" is the word i usually see

[narrow cairn]

what are some interesting properties of $\mathbb{S}^{\infty}?$ is it locally euclidian of dimension $\aleph_0$?

[gentle osprey]

most likely to honorable

[ebon galleon]

It's contractible.

[narrow cairn]

no idea what that means

[ebon galleon]

Homotopy equivalent to a point.

[narrow cairn]

well that sucks doesnt it

so it doesnt have onteresting topological properties?

[ebon galleon]

R^n is also contractible. Does that mean R^n doesn't have any interesting topological properties?

It's just interesting that it is, since S^n is very much so not contractible for any n. But in this sense, you can kinda just think of it as "you always have a higher dimension that you can contract a lower dimensional sphere through"

[unreal stratus]

It is a model for E(Z/2Z)

lol

[narrow cairn]

what is E there

[feral copper]

E(Z/2)->B(Z/2), like the classifying space

Yeah so I didn't read it was about S^oo

[feral copper]

Its quotients by finite groups are the infinite lens spaces L(oo,q), and those are K(Z/q,1) spaces

[feral copper]

The case q=2 is what potato said: B(Z/2)=RP^oo, and E(Z/2)=S^oo.
More generally, B(Z/q)=L(oo,q), and it is covered by S^oo

Z/2-principal bundles over some space X are very interesting, and they 1:1 correspond to homotopy classes of maps from X to RP^oo, or of fiber-preserving maps from the total space to S^oo

[narrow cairn]

curious tho, like i asked earlier, is it locally euclidian of dimension |N|?

[narrow cairn]

is it just me or does Lee ITM have a really big jump in density when it gets to CW complexes

[umbral panther]

It is homogeneous. Once you get to infinite dimensions, “local Euclidean” is not precise and there are many model spaces

[narrow cairn]

is it not precise? by locally euclidian of dim |N| i mean that every point has a nbhd homeomorphic to an open subset of R^|N| (with the product topology0

[umbral panther]

Infinite product of non compact spaces are nasty

This is modeled by the vector space with a countable basis. Finite sequences of real numbers, followed by all zeros

Manifolds modeled on Hilbert spaces or Banach spaces are common

Manifolds modeled on the Hilbert cube are rare, but have good technical properties

(The Hilbert cube is homogeneous!)

[swift fjord]

A fun way to see this is that you can construct the homotopy explicitly using a convergent series. So for example in the first half a second you contract S^1 (through S^2 say), then in the next quarter second you contract S^2 etc.

[tawdry widget]

I remember someone told me a categorical way to prove it’s contactable in this channel, nerve of a category of two isomorphic objects (which is equivalent to category of one object) stuffs. I forgot to screenshot it. I swear I can remember his name if he appears again.

Maybe it was potato

[gritty widget]

What could this surface be?

[red yoke]

Torus

[quiet thorn]

🍩

[gritty widget]

How do you know this? What is your thinking logic

[red yoke]

Visualize

A punctured torus is just a tube with a thin strip attaching two ends

Alternatively move the two discs of attachment close together

Then you can cut off a torus

[bright acorn]

can someone clarify to me what is meant by this statement?

[gritty widget]

Would you be able to draw any pictures? 🙂

[distant lichen]

I believe this refers to the approach to SW/Chern classes viewing them as elements of H(BO/BU, Z_2/Z) respectively

If I remember correctly the general case comes from the Chern-Weil homomorphism

Arun Debray has a good document on this

https://web.ma.utexas.edu/users/a.debray/lecture_notes/u17_characteristic_classes.pdf

[feral copper]

This, and we mostly care about Banach manifolds (model spaces are Banach spaces) @narrow cairn
The sphere S^oo is modeled by a space with a countable basis indeed, and functional analysis tells you it cannot be a Banach space! So it's more interesting as a topological space than a manifold

[tribal palm]

should i look into “pointless topology” (locales) ?

or would that be pointless

[tribal palm]

genuine question

• bad pun

[hidden crag]

do you have a good reason to look into it

[tribal palm]

well i’ve just noticed that a lot of the topology we’ve done so far could be phrased without reference to any point, and i usually find those formulations a lot more elegant

[tribal palm]

i should certainly get settled with point-set topology first though

[hidden crag]

then it depends on what your goal is

if you're gonna study it for its own sake because you like it then sure why not

but it's most likely not going to be useful for further studies of subjects that involve topology

[lime sable]

stone duality is fun

but yeah

[tribal palm]

girls just wanna have fun >:S

[ebon galleon]

No reason to do so for most purposes

[scarlet turtle]

how would i prove this?

f|A and f|B is the function f restricted to A and B respectively, and continuous wrt the respective subspace topologies

[coarse night]

U is closed if A ∩ U and B ∩ U are closed

[bright acorn]

sorry, but by / you don't mean a quotient right? but "either"? like elements of either H(BO,Z), H(BO,Z_2), H(BU,Z) or H(BU,Z_2).

[umbral panther]

It looks like it’s talking about Chern-Weil which shows that real characteristic classes of a Lie group G are symmetric polynomials in the characteristic classes of the maximal torus

But that doesn’t cover Stiefel-Whitney, which are torsion. Grothendieck 1958 gave a unified treatment of Chern and SW, but that doesn’t draw attention to the torus

[distant lichen]

I do not, I mean respectively Stiefel Whitney and Chern

[distant lichen]

Ahh, thank you. Does this apply to integral Stiefel Whitney classes at least?

[umbral panther]

Does what apply to SW?

Chern-Weil is about connections and differential geometry. It is completely incapable of dealing with SW

You could do something more subtle, like the Oliver transfer that says that the integral cohomology of BG is a summand of the cohomology of BNT, where NT is the normalizer of the maximal torus. But NT has a lot more cohomology than the invariant classes H(BT)^W. (Whereas rationally it’s all the same)

There are definite analogies between Chern and SW which drive a lot of treatments, such as in Milnor. But only Grothendieck’s treatment applies formally to both

[tidal lynx]

This still holds if ദ and ദ ' are (either or both) subbases too, right?

[queen prism]

mogus

[tidal lynx]

The proof should go the same, except replacing an occurrence of B or B' with B_1 \cap ... \cap B_n or B'_1 \cap ... \cap B'_m ?

[tidal lynx]

hi

does this mean smthn

[queen prism]

ඞ

[tidal lynx]

Wait I just realized I'm being dumb

the set of finite intersections of the subbasis is itself a basis

so the same statement applies

[narrow cairn]

how can we guarantee that such a W exists

i get that since its locally finite theres a W with a finite intersection with each e but why does that hold for the closures?

,rccw

[feral copper]

Because the cell decomposition is assumed to be locally finite

[narrow cairn]

sure but isnt the cell decomposition just the open cells?

[feral copper]

They are glued to other cells along their boundary

[umbral panther]

I would just take that as the definition of locally finite

But it probably follows from the hypothesis about the interiors. The whole cell is compact, so the boundary can only add finitely many more

[feral copper]

But it probably follows from the hypothesis about the interiors. The whole cell is compact, so the boundary can only add finitely many more

[narrow cairn]

hmm i dont follow

[tidal lynx]

For this question

I'm wondering why in the above solution, we can guarantee the second equality on the second line (that B cap A = (B cap Y) cap A)

Wouldn't that mean B is a subset of Y ?

[queen prism]

more explicitly:

B n A = B n (Y n A) = (B n Y) n A

[tidal lynx]

Ok that makes sense

But the conclusion that B n A = (B n Y) n A doesn't make sense

because that implies B is a subset of Y

And by construction this should hold for all open sets B of X ?

[ebon galleon]

No it doesn't. It's just that B and BnY agree on their parts contained in A

[tidal lynx]

oh wtf yea

thanks

[narrow cairn]

still dont understand this

[feral copper]

The boundary of the cell glues to at most finitely many others. If locally finite means a point has a neighborhood that intersects the interior of finitely many cells, then it also intersects finitely many closed cells (might be more, but still finitely many)

[narrow cairn]

hmmm, but what if other cells that would usually be disjoint have closures that end up intersecting W

[feral copper]

Now that I think about it, it's obvious but the proof not so much. You'd need to use something like the Levesgue covering lemma (some version of it)

[feral copper]

Doesn't matter, still gonna be at most finitely of them, because where they meet, use local finteness again

[narrow cairn]

oh, because the boundary is closed and therefore compact?

[feral copper]

Nah you can just cover the boundary with finitely many opens and that's good actually!

[feral copper]

What are you asking about?

[tidal lynx]

How do I show that in the subspace topology of [0, 1], if E ⊂ [0, 1], E does not contain 1, and sup E = 1, then E is not open in the subspace?

[queen prism]

what about E = (0, 1)

[tidal lynx]

U suck man

Ty

[tribal palm]

what book is this?

[queen prism]

lee
prob

[narrow cairn]

lee ITM chapter 5

[narrow cairn]

well, i was thinking for each point in the boundary you could take some locally finite neighborhood, those cover the boundary, then take a finite subcover and you have a finite cover of the boundary each element of which intersects with only finitely many cells so the boundary only intersects with finitely many cells

err, wait, isnt that just the definition of the (C) property? i think im dumb

[narrow cairn]

well, it only is if we suppose W is contained in a cell, but i suppose we could intersect it with some n-cell's interior

wait, no i still dont understand

W isnt an n-cell but even if it was we couldnt say that it intersects only finitely many cells' closures just that its closure only intersects finitely many cells

[heady skiff]

doesn't this proof only establish $q^{-1}(V) \implies V$ open in $p(A)$ but not $V$ open in $p(A)$ implies $q^{-1}(V)$ open in $A$? or am I missing something I'm rather tired

[gentle osprey]

okeyokay

[narrow cairn]

well isnt that just a consequence of continuity of q since q is just a restriction of p (which is continuous) to a subspace

[heady skiff]

ohhh right forgot quotient maps imply continuous oops

lol

alr thanks

[rancid umbra]

is a wedge of circles homotopy equivalent to a circle?

like, im pretty sure an 8 is homotopy equivalent to an O by linear interpolation

[umbral panther]

No 8 is not equivalent

[rancid umbra]

but i can deform O to an 8

just pinch

[umbral panther]

You need maps in both directions. Pinching gives a map from 0 to 8. What is the map back?

[lime sable]

a "homotopy" is between maps, where you play a movie deforming from one map to another

a "homotopy equivalence" is when you have maps f: X -> Y and g: Y -> X such that fg and gf are homotopic to identity maps

so when two spaces are homotopy equivalent, there is no movie playing that turns one into the other (automatically)

[rancid umbra]

hmm. okay. im trying to do something in complex analysis concerning the integral of some rational function over a circle

thought that i could get loops around each of the poles and then integrate and add

but i need the contours to be homotopy equivalent

back to the drawing board. thanks

[umbral panther]

Contours don’t have to be homotopy equivalent

[rancid umbra]

is it true that if they are, then the integral of any analytic map over the curves are the same?

[umbral panther]

Yes, but there are more contours that give the same integral, like two disjoint loops that can be dragged together and combine in an 8

[opaque scroll]

The integral along two paths are the same if the paths are homologous, so they don't have to be homotopic.

The figure 8 is homologous to the sum of the two circles that make it up.

[narrow cairn]

im still super lost

[red yoke]

If U and V are disjoint and V is open then cl(U) and V are disjoint

[narrow cairn]

i cant believe its that simple

god damn it

[red yoke]

Point-set be like:

[narrow cairn]

half of the proofs take 2 seconds and the other half take 20 pages and 7 years

[fickle elm]

Is the map defined by z --> z^2 between complex planes a proper map?

I know that z--> z^n can define n-fold cover on unit circle, so is it true that the preimage of a compact set in C for z-->z^2 is its double cover？ How can I see this?

[red yoke]

How about preimage of R

[red yoke]

Yes

[fickle elm]

R is not compact in C, right? I need to look the compact subset of C, then look at its preimage under this map. I am quite convinced (it is wrong, I am just an idiot) it should give back its double cover in general but I do not know how to prove this?

[red yoke]

I meant it's not a double cover

Then consider [-1, 1]

[fickle elm]

Ah you are right.

[opaque scroll]

Consider {0}

[fickle elm]

Its preimage has only a point 0, it is still compact, right?

A compact subset A of C is a closed and bounded subset of C. Square map is continous so preimage of A is closed, the square root of a bounded complex number is still bounded, so square map is proper.

Is it a convincing argument?

[unreal stratus]

Yes that works, though you need to say like "all" square roots or something but yes

[arctic island]

I had my first topology lecture today and i have a question. My professor said that in the definition of "Topology (topologic structure)"

(1) Both the empty set and X are elements of τ.
(2) Any union of elements of τ is an element of τ.
(3) Any intersection of finitely many elements of τ is an element of τ

you don't really need the (1).
He said that (2) and (3) imply (1), why is that?

[queen prism]

maybe he’s talking about the trivial intersection/union?

[limber ravine]

I think you mean that the topology is closed under complements instead of (3)

[queen prism]

that’s a sigma algebra

[limber ravine]

fk ahahah I am studying analysis sorry

[unreal stratus]

Idk I mean empty intersection is a bit like

Just a convention in this case

I would usually say 1) is necessary for clarity lol

[arctic island]

Oh i see it follows from the interesction of empty family of sets...

But how about the empty set? Can you do the Union of "no sets"?

oh i guess its again the same just the union a the quantifier changes to "Exist"

[queen prism]

when we consider intersections they are normally taken with respect to the overall set X
so a given element x in X is in the intersection if x is in A_i for every i in (empty)
but this is vacuously true

[arctic island]

I misunderstood your reply before...
Why is this vacuously true? Why not just true?

[queen prism]

an implication, in this case, “x is in A_i for every i in (empty),” is vacuously true if the hypothesis never holds
we can rephrase the statement as “for every i in (empty), x is in A_i,” or, “if i is in (empty), then x is in A_i”
i is never in (empty) so x is in A_i

[tribal palm]

yeah it's a bit set-theoretically shaky, but people typically just take the union of the empty family of sets to equal the empty set, and the intersection of the empty family of sets to equal the entirety space in question (

[silver umbra]

so this paper im reading casually mentioned that if 3-manifold M is a fiber bundle M -> S^1, it can be expressed as a mapping torus M_phi

but its not clear to me why this is the case

[coarse night]

think of M = Σ × [0,1] / identification on 0 and 1 via some diffeo of Σ

[fading vale]

let F be the fiber of your bundle. pick some point x in S^1 and take a bicollar of its fiber, which is F. the complement of this bicollar is the preimage of a subset of S^1 homeomorphic to R^1 and hence can be trivialized as F x I. Thus this is two copies of F x I glued by some homeomorphism

technically you are gluing along two places but it is not that hard to see that the gluing maps must be the same so you get some automorphism h of F, and M is the mapping torus of h

[narrow cairn]

wait, so there exists a family of sets A_i such that the intersection across A_i is strictly bigger than the union across A_i? wack

[tribal palm]

it's a bit like 1/0, it doesn't really make sense, but sometimes it's useful to just declare it equal to something

[unreal stratus]

I mean you can think of it as like

Fix some set X which we are considering subsets of. Then we can partially order P(X) by inclusion and the union of a collection of subsets of X is the "least upper bound" and the intersection is the greatest lower bound

[silver umbra]

what is a bicollar?

[unreal stratus]

Then it makes sense why you'd want the union of the empty set to be empty and the intersection to be X

[fading vale]

like take F x I and imagine your original F is embedded as F x {0}

[silver umbra]

so by complement of the bicollar

do you mean the complement of p^-1(x) x {0} in M x I?

im a bit confused sry

[feral copper]

That's a collar. A bicollar is F×I and F embeds as F×{1/2}

(Sorry :P)

[fading vale]

Yeah sorry i have no idea why i was imagining 0 centrally

i guess i was thinking F x [-eps, eps]

[fading vale]

You have a 3-manifold M with a circle fibration M -> S^1. the fiber of this is F, which is some surface. Fix some point x in S^1. if you imagine taking a small neighborhood of radius epsilon about x, its preimage in M is going to look like F x [-eps, eps], where F x {0} represents the fiber of x. The complement of this bicollar is the preimage of S^1 with a tiny interval removed. But S^1 with a tiny interval removed is trivial, so we can trivialize the fiber bundle over it. Thus the complement of the bicollar is also isomorphic to F x [0, 1]

The 3 manifold M is given by gluing the copy of F at -eps to the copy of F at 0, and the copy of F at eps to the copy of F at 1

[feral copper]

If you think of the bundle over S¹ as gluing the two ends of the bundle over the open interval, then this gluing map (called the monodromy) uniquely characterizes the bundle. The ambient space is the mapping torus of this map

I think there's a notion of monodromy for more general fiber bundles F->E->X which involves a morphism \pi_1(X)->Diff(F)

[feral copper]

As principal stick bundle said, the bundle over the interval is trivial because the interval is contractible

I see them typing, but that was also explained earlier by S³/W 😉

[limber ravine]

Where do we use Hausdorfficity to show that a compact subset of an Hausdorff space is closed?

[queen prism]

hausdorfficity

[limber ravine]

The definition, is that for any two distinct points we can find two disjoint neighbourhoods, one for each point

xd

[gritty widget]

are you asking where it comes up in the proof, or why it is necessary?

[limber ravine]

oh wait

[gritty widget]

it's necessary because of, say, something like the line with two origins. take the "closed unit interval" of R but replace the origin by one of the new origins. this will be compact but not closed

[limber ravine]

oh interesting

[gritty widget]

could argue compactness by saying it's the image of the compact [0, 1] in R under the map taking R to the line with the "bottom" origin

[ebon galleon]

In a Hausdorff space, you can separate a compact set and a point not in the set with disjoint neighborhoods, from which it follows that compact sets are closed. This is not true in nonHausdorff spaces

[limber ravine]

Need an hint for this

[ebon galleon]

Think about how you can use compactness of the set, combined with the Hausdorff condition, to get this intermediate lemma

[ebon galleon]

Explicitly, if a space X is Hausdorff and K is a compact subset, for any point x in X \ K there are neighborhoods U of K and (this is the important part) V of x such that UnV = Ø

[limber ravine]

that means that each open (in the ambient space) covering of the set has finite subcovering

well K can be covered by a finite subcovering (U_i) of opens in X. For x in X\K and y in K there will be opens U_j (one that covers K) and V subset X such that V n U_j = {}

[ebon galleon]

How do you use Hausdorffness to get those U_i's? This is where that assumption should be used

[limber ravine]

I would say they need to be disjoint

but it can't be because for two points in such U_i, then we wouldb't be able to find another two disjoint neighbourhoods

[limber ravine]

wdym?

[ebon galleon]

So compactness isn't just that there is a finite cover of K (this always holds: just take X lol), it's that any open cover of K can be reduced to a finite one. So it's a useful property, because it essentially allows us to prescribe certain properties to the open sets.
So for this, you need to find the correct open sets (using the Hausdorff condition, probably something to do with x not being in K) to cover K and take a finite subcover of. Hint 1: ||Think about the Hausdorff condition applied to x and y for each y in K||. And hint 2 for the important part: ||You'll get corresponding neighborhoods of x, and finitely many of them. Think about what you can do with finitely many open sets||

[limber ravine]

we can cover K with the sets that appear from the hausdorff condition for each y in K

and all these will be disjoint of a fixed neighbourhood of x

ok, now I saw the hints

with finitely many open sets, we can intersect them

which will be open again

there's something missing, how exactly are we proving that K is close with this approach?

[ebon galleon]

Since U and V_x are disjoint, and K is a subset of U, we have K and V_x are disjoint too.

||union of open V_x||

So it's actually a slightly stronger condition than you need, but a similarly-flavored proof allows you to show that disjoint compact sets in Hausdorff spaces can be separated by open neighborhoods. Which is how you prove that compact Hausdorff spaces are normal (something to think about later on perhaps)

[limber ravine]

Thank you very much!!! I was studying differential geometry and ended up re-doing topology exercises. Nice 😄

[silver umbra]

i see. but how do we define this gluing in such a way that we get M back?

[fading vale]

the fact that M is the total space of the bundle means that the automorphism of F that prescribes the gluing recovers F by definition

[silver umbra]

hmm wait ok

[silver umbra]

i understand everything here so far, modulo the fact that fiber bundles over contractible spaces are trivial

[silver umbra]

but im not rly understanding this gluing

so after we remove the preimage of the interval (-eps, eps)

we're left with the trivial bundle F x I

where is the copy of F at -eps in F x I?

or am i misunderstanding

is it because we have two copies of F x I now?

like, the strip we removed is isomorphic to F x I

but then the complement of that strip is also F x I

[fading vale]

Yes

and then M is given by gluing them together

[silver umbra]

sure but then

how is that a mapping torus

like you remove the preimage of the interval

then you get F x I

and then there needs to be some kind of identification of

F x 0 with F x 1 directly no?

is the quotient like

ok so we have our two copies of F x I

we glue F x 0 to F x -eps

and F x 1 to F x eps

and then we collapse F x (-eps, eps)?

[fading vale]

Kind of

[silver umbra]

like i entirely understand how you recover M

by gluing F x I with F x (-eps, eps)

i just dont understand how that gives M a mapping torus structure

[fading vale]

the point is that this is given by two automorphisms h_1 and h_2 of the genus g surface F. if i first glue F x {-eps} to F x 0 by h_1, before i glue F x {eps} to F x 1, the resulting space is still homeomorphic to F x I because i can just untwist the F x [-eps, eps] piece by applying h^{-1} to it. then i can glue F x {eps} to F x 1 by applying h_2 circ h_1^{-1}

this means that i am constructing M by taking F x I, and then gluing the two ends by the automorphism h_2 circ h_1^{-1} of F

this is the mapping torus of h_2 circ h_1^{-1}

[silver umbra]

waittt ok so

by untwist, do you mean the fiber isomorphism that identifies the complement of the collar with F x I?

and so when you untwist the complement of the collar

when you want to glue F x (-eps, eps) back to it to get M back

you have to twist that by the same isomorphism

[gritty widget]

Hi, I am wondering about "vertices" and gluings, I am confused how they just merge the two edges into 1 when there is supposed to be a verticy, would anyone be able to help, thank you:

[heady skiff]

what do they mean by constant on each p^{-1}(y)? that they're all sent to the same set consisting of one element?

[gritty widget]

for each y, constant on the set p^{-1}(y)

[heady skiff]

tteppa being cryptic as usual

yea my question was what does that mean 😹

[gritty widget]

i didn't actually read what you wrote, i just looked at the underlined "constant"

[heady skiff]

lmao u chillin

[gritty widget]

for every y, g(x) = g(x') for each x and x' in p^{-1}(y)

or, as you wrote, g(p^{-1}(y)) is a singleton for each y

the element of this singleton will be f(y)

[heady skiff]

oh aight

thanks ur the goat

[silver umbra]

is there an easy way to see that the mapping torus of a closed orientable surface always has nontrivial first homology?

[umbral panther]

A mapping torus is a bundle over a circle. It has the homology of the circle

[silver umbra]

wait what? the torus is a mapping torus over the circle and H_1(T) = Z^2 while H^1(S^1) = Z

[umbral panther]

It has at least the homology of circle

If X is connected then any mapping torus of X has the circle as a retract

[silver umbra]

oh wait could you also say

you have a projection p: M -> S^1

which induces p_*: H_1(M) -> H_1(S^1)

which is a surjection by functoriality

and since H_1(S^1) = Z

H_1(M) is at least Z

[umbral panther]

It is a surjective if the fiber is connected. If the fiber has finitely many components, then it has nontrivial image, but not necessarily surjective. If the fiber has infinitely many components, it could fail. The mapping torus of the shift of Z is R

[silver umbra]

wait im confused so does the proof i just gave not work in general? which part fails when we dont assume connectedness of the fiber

[umbral panther]

What did you mean by “by functoriality”?

[silver umbra]

so the projection is surjective

so it has a right inverse q: S^1 -> M

so then q_* is the right inverse of p _*

[umbral panther]

If the fiber is not connected, it does not have a right inverse

[silver umbra]

ahh okay

[umbral panther]

Say, the fiber is two points and the map is switching then

[silver umbra]

i see

but then the rest of the proof i wrote works, no?

assuming connectedness

the mapping torus i'm looking at is a fiber bundle over the circle with the fiber being a closed orientable surface

wait actually nvm im still confused

any fibering M -> S^1 needs to be surjective to begin with

[umbral panther]

The argument is fine

[silver umbra]

another question if thats okay

so if we have a fiber bundle F -> M -> S^1 where M is a 3-manifold, we have that M is a mapping torus of F wrt a homeomorphism f: F -> F

i want to show now that f is in the image of the monodromy representation pi_1(S^1) -> MCG(F)

i have no idea how to approach this tho

[umbral panther]

What is the monodromy representation? How is that different been the previous statement?

Gotta go

[silver umbra]

np, thanks for the help so far!

[silver umbra]

actually, a more approachable result that would imply this would be

are the orientation preserving diffeomorphisms of the fiber F of a mapping torus M_f always homotopic to f?

given that F is a closed orientable surface

[narrow cairn]

https://cdn.discordapp.com/attachments/359052581022203914/1148316776497676379/image.png?ex=651ef9d6&is=651da856&hm=b88210b12aa466d5089f0b79b996616cd26781dc70a63e54f32c739c49b3beb7&

sorry didnt mean to ping

[narrow cairn]

started reading about simplicial complexes and got hit with a bunch of annoying linalg :(

[arctic island]

I am thinking about this one for too long now...

I have a space M with a metric d. Let S be a subset in M.
Now i have this function
d(_,S) : M -> R
defined as
d(_,S) = inf{ d(x,s) | s from S }

Show that this function is continuous.

My idea that seems obvious to me is this:
to prove continuity at a point a I forumulated this:

for a epsilon > 0 i need to show that there exists a delta > 0 that:
d(x,a) < delta => |d(x,s) - d(a,s)| < epsilon

and this makes perfect sense like if two points are close to each other then their distance to the "closest point of the set S" should be close to equal thus the difference approaches 0... But idk i need help proving it...

[trail charm]

reverse triangle inequality maybe

yeah it should work

[trail charm]

try reverse triangle inequality

[heady skiff]

would this constitute a rigorous proof?

[ebon galleon]

Yes.

[heady skiff]

ok cool

i was told that topology is a very visual subject

should make tests easier

[ebon galleon]

Memes are acceptable proofs in topology.

[heady skiff]

when coffee mug becomes donut 😹

i'm having trouble seeing how to show it's continuous - i know that we have to map some o pen set that looks like this (obviously just the intersection of the circle and the lines) back to an open set in X, but idk how to see this rigorously

well i guess intuitively it makes sense, you just "straighten" out that line

rigorously tho idk

could I use the epsilon-delta method?

[gritty widget]

the components are continuous functions

(x, n) -> x is continuous and (x, n) -> x/n is continuous

[heady skiff]

oh that's a thing

i didn't know that

well that's handy

[heady skiff]

also by "the quotient space X* whose elements are the sets g^{-1}({z}) is simply the space obtained from X by obtaining the subset {0} x Z_+ to a point" do they mean that since g is injective for every point not of the form (0, x) that all those points remain the same but everything of that form gets identified to a single point?

or rather in general, when they say "let X* be the quotient space obtained from X by identifying the subset A to a point b" do they mean that we just consider the complement as the rest of the partition

[limber ravine]

proving that each proper map of a Hausdorff space to a Hausdorff locally compact space is closed

I am not seeing where locally compactness is used

i know that any compact subset of a Hausdorff space is closed

and that continuous maps from compact to Hausdorff are closed maps

For each V subset Y compact we know that its pre image in X is compact. But both of them are compact, so both sets are closed. So we get f^{-1}(V) closed in X and V closed in Y.

[umbral panther]

What is your definition of proper? If the target is locally compact, universally closed is equivalent to preimage of compact is compact

[limber ravine]

a continuous map f : X -> Y is proper if for each A subset Y compact, its pre image in X is compact

[heady skiff]

here do we not need the requirement that A and B are open?

[queen prism]

neither of which contains a limit point of the other

[heady skiff]

ah ok

wait what's the indiscrete topology

open sets are empty set and whole space?

[gritty widget]

yes

[obtuse meteor]

annoying linalg my brother in christ the point is the linalg

[narrow cairn]

just so many random sums and shit

annoying really

[heady skiff]

does a path have to have domain [0, 1], or can it have domain of an arbitrary length, say [a, b]?

[novel acorn]

You can always rescale your interval

[ebon galleon]

By convention it's often [0,1] but it doesn't really matter

Since they're all homeomorphic

[heady skiff]

ah yea i guess that makes sense

[heady skiff]

dumb question but why is this set closed (underlined in blue)? it's not necessarily the preimage of a closed set, right?

[gritty widget]

f^{-1}({0} x [-1, 1])

[heady skiff]

O

i'm high

thanks

[narrow cairn]

f is continuous, and {0} x [-1, 1] is closed since it is the product of closed sets

well actually i have no idea what space youre dealing with

but presumably {0} x [-1, 1] is closed

[heady skiff]

uh the closure of the topologist's sine curve

considered as a subspace of R^2 i believ

e

[narrow cairn]

right which is a subspace of a product space

[heady skiff]

ye

and singleton sets are closed

and similarly closed sets

lol

[narrow cairn]

so {0} x [-1, 1] must be closed since it is the intersection with {0} x [-1, 1] with cl(T)

closed sets do happen to be closed

p implies p

[heady skiff]

lol

well i mean i just saw it as f being continuous

so preimage of closed set is closed

[narrow cairn]

right i was just saying why the set is actually closed

[heady skiff]

oh yea my fault

[ebon galleon]

"closed sets are closed"

[mint pollen]

At this point I wouldn't be surprised if someone tells me something like
"Closed sets are closed... in finite dimensional cases"

[coral pawn]

Can someone help with a?

What ever the homotopy is, at each point it probably is a conjugation since it fixes I_n

[umbral panther]

Right, just find a path in U and use conjugation by each element of the path

[coral pawn]

The only path I can think of is tI_n + (1-t)P but i don't think it lies in U

[gritty widget]

unitary matrices are like elements of the unit circle - how is the unit circle parametrized?

[coral pawn]

[0,2pi] with end points glued?

[gritty widget]

something more complex number-y

[coral pawn]

e^ix

But for matrices we would have to define that using a taylor expansion won't we?

[gritty widget]

a taylor expansion which always converges. the matrix exponential is a thing

[coral pawn]

And showing that thing exists is not something we have done

How would we write P as e^ix?

[gritty widget]

if you don't want to use the matrix exponential then just diagonalize

[coral pawn]

For the identity we just take x = 0

[gritty widget]

well, i guess showing that you can write a unitary as e^{iA} might go through diagonalization anyways

[coral pawn]

Let's just do exponential because I don't want to think about diagonilzation and other matrix operations much

So the goal is now to write P as e^ix

Wait

[coral pawn]

Yeah idk how to do this

[gritty widget]

just diagonalize

[coral pawn]

Ah I see

Can unitary matrices always be diagonlized?

[umbral panther]

You could diagonalize to find a fairly explicit path, but you could also just prove U is connected

[gritty widget]

https://en.wikipedia.org/wiki/Spectral_theorem#Normal_matrices

bw how do you prove that U(n) is connected without doing something like this

[coral pawn]

Okay so I diagonlize and get BDB^{-1} = P where D is diagonal with entries of the form e^ix

I still don't see how to go from a diagonal matrix to the identity without leaving U

[gritty widget]

B diag(exp itx_1, ..., exp itx_n) B^{-1} will remain unitary for any real t (you may wish to specify what kind of matrix B is)

[coral pawn]

Won't that leave U?

Oh wait

[umbral panther]

For most groups, in particular the classical families of groups, it is easy to understand the orbit of a generic vector in the standard representation and the stabilizer. In the case of U(n), it’s S^(2n-1) and U(n-1). If they are both connected, then the group is

[coral pawn]

Will B end up being in U(n)?

[gritty widget]

did you read the article?

[coral pawn]

No. I'll read it now

Thanks

I understand now

[coral pawn]

Is it true that if A is a unitary matrix and A = BDB^{-1} = B' D' B'^{-1}, then B D_t B^{-1} = B' D'_t B'^{-1} where B and D are unitary, D is diagonal, and D_t is the matrix we get by raising each entry in D to t?

Is this true?

Because D and D' have the eigenvalues on the diagonal

So B and B' must "differ" by row/column switches, which should be unaffected by our operation

Is this the correct thing to do?

[coral pawn]

I'll post my solution for b if anyone could look at it

I am not 100% convinced that this is well-defined

[tawdry widget]

I think it’s good

BDtB * is irrelevant with choice of B and D indeed

[west brook]

What about the 5D irrep of SO(3)?

[tawdry widget]

Now I think about it again, it doesn’t look well-defined. Anyway I come up with a workable one: maps A to (cos(πt/2)I, sin(πt/2)I; -sin(πt/2)I, cos(πt/2)I)diag(A, I)(cos(πt/2)I, -sin(πt/2)I; sin(πt/2)I, cos(πt/2)I)

I wrote the case when r=2 but you know what I mean

[umbral panther]

Why have you introduced the auxiliary matrix A? Are you trying to form a homotopy between A in one position and A in another position? That is not uniform in A

[arctic island]

Is that really just it? Because im comparing a function thats defined like this: d(_,S) = inf{ d(x,s) | s from S } So idk i guess there could be problems with that infimum

[trail charm]

don’t think it should be an issue since reverse triangle inequality holds for inf

unrelated but i dont see how this factors?

intuitively the pikj = 0 for i ne j makes sense but i dont see how it factors through Hn(Uj, Uj)

[trail charm]

for S^n times S^n as a CW complex (one 0-cell, two n-cells, one 2n-cell), is its homology group just Z for 0, 2n and Z (+) Z for n? and then just 0 for everything else

[fickle elm]

Yes. By Künneth formula, and the fact that homology groups of sphere are torsion free.

[trail charm]

ngl i have no idea what kunneth formula is

and i did not know that homology group of sphere is torsion free but that is good to know

i kinda just guessed bc CPn has cells of dimension 2k =< 2n and so Hk(CPn) = Z for even and 0 otherwise

[feral copper]

Basically computes homology of product spaces

This isomorphism is not natural

[gentle osprey]

Matplotlib

[feral copper]

If you look at homology over a field, then you don't need the second big thingy with torsion

But here, to compute the homology of a product of two spheres, it's better to just come up with a cellular decomposition and go for cellular homology

(so what you did)

[fickle elm]

Maybe I am studying too much algebra and forgetting about topology...

[feral copper]

Don't worry, I realized recently that some people have intuition for the algebra (like juggling between four long exact sequences) and not as much for the geometry

That's not a bad thing, it's just different

I'd need to be better at these shenanigans

[trail charm]

i like algebra more than geometry :)

[trail charm]

ty

makes sense

is there any good way to visualize S2 x S2

or is there not bc it’s a 4 dimensional object

[feral copper]

There are ways to try to visualize it, but essentially, no, because it is indeed 4-dimensional. Its homology is easy to describe, because there are two big 2-spheres looking at you. The intersection form is pretty easy too, for this reason as well

The fact that it is also CP(1)xCP(1) is interesting from a geometric point of vue (if you care about Kähler stuff and complex geometry that is); that's a quadric surface embedded in CP(3)

So, again: the best way to think about it, purely from the topological point of vue, is its CW decomposition

[fading vale]

i often draw it as a square with sides identified (like a torus) where you just tell yourself that lines represent spheres instead of circles

[rustic lava]

I'm so stuck on what A' is in this

I realise it's pretty basic topology

[gentle osprey]

JJP

[iron pewter]

Right, it is closed

A' is - assuming the book is using the notation I'm familiar with - the complement of A, so R - (0,1].

[rustic lava]

Nah it's the derived set

[iron pewter]

Love me some different notations, huh

[chrome ridge]

Is there a way to do the local homology of an n-manifolds without excision thm?

[heady skiff]

how does this definition of the quotient topology imply that it's the largest topology for which pi is continuous? if we're defining the quotient topology as U open iff p^{-1}(U) open then I can see that, but i'm not sure how to see it with this definition

[iron pewter]

Sorry for delay, let's unpack some definitions. I'm sleepy and thus I don't think well.

Derived set is, iirc, the set of all accumulation (limit) points of A. Since A is closed, A' is a subset of A.

The point x is an accumulation point of A, if every open neighbourhood of x contains an element of A other than x.

Now consider any point x \in A. The two-point set {{x},{pi}} is an open neighbourhood of x not containing any other element of A other than x.

Thus there are no accumulation points and so A' is empty.

[rustic lava]

Thank you :). Wrote that then scribbled out about 8 times lmao

[heady skiff]

anyone?

[limpid fern]

finest topology?

[limpid fern]

any partition also defines a quotient space anyway

pi is certainly surjective

in fact this construction is the quotient space construction anyway

[chrome ridge]

Is there a way to do the local homology of an n-manifolds without excision thm?

[novel acorn]

Uhhh Mayer vietoris sequence iirc?

I think it's doable with just the long exact sequence of homology iirc

[chrome ridge]

how tho?

[novel acorn]

Here local means H_n(M, M-{x}) I assume

[chrome ridge]

yup

[heady skiff]

yea i guess

[novel acorn]

Now iirc there's some fucked up way to do it via cones of spaces

[heady skiff]

i like

forgot all my trig

is this roughly where (1, 1) would get mapped to if we're considering the unit square of R^2?

ignore that top comment i'm a dumbass

wait now i'm confused

i wish i paid attention in high school maths

[limpid fern]

maybe some graphing software might help

[heady skiff]

wait now i'm confused as to what e^2\pix gets mapped to on the unit circle

oops

[heady skiff]

like what

wait so (1, 0) and (1, 1) would get mapped to the same point on the torus then

right?

Something tells me complex analysis is not going to be a fun time

[silver spruce]

i like plotting these kind of things in matplotlib but there's plenty of other tools you could mess around w/

[silver spruce]

e^(ix) = cos(x) + isin(x)

[silver spruce]

and yes

[queen prism]

https://www.mathsisfun.com/algebra/trig-interactive-unit-circle.html

[ebon galleon]

mathisfun

[chrome ridge]

any hints/ references of this method?

[heady skiff]

is the easiest way to really check that this is continuous is to let some open set in the codomain and show that it's preimage is open? if so that seems like a pain....

there's gotta be an easier way right?

in fact i'm not sure how to check most of the continuity for constructions like these

because some of them involve an abstract topological space which has no notion of distance

and if you have a subspace of a topological space you have to consider intersections

and the preimage of those intersections which is hard

[unreal stratus]

How is the geometric cone defined here since it seems to be as a subspace of Euclidena space anyway, hence you can use standard facts about reals

[heady skiff]

so the epsilon-delta way here is the best method?

[unreal stratus]

namlye that multiplication and addition are continuous

[heady skiff]

oh

[unreal stratus]

i wouldn't reprove it, this is topology and you can usually just assume basic things like that

[heady skiff]

ugh yea i gotta remember this thanks

it was a mistake to take topology before real analysis

well

[queen prism]

lol

[heady skiff]

i don't know how the continuity of multiple variable real-valued functions work

i would assume it's just like

phrased in terms of balls

[unreal stratus]

yeah in terms of a metric typically

[heady skiff]

so when you say multiplication is continuous, you mean that the function f(x, y) = xy is continuous for all x, y in R^2 (if we're working in R^2). just to be clear lol

and similarly for all the other elementary operations

[unreal stratus]

ye

[heady skiff]

what's the point of this argument underlined in blue right here? it's not injectivity right, since f(x', 1) = f(x, 1) and (x', 1) \neq (x, 1)?

oh is it just showing that it's well-defined?

wait no

oh nevermind

it's about the quotient space in the next sentence i'm dumb

[sullen bear]

I'm having a very tough time proving that a function mapping out of a quotient space is continuous. Let $X = \mathbb{R} \sqcup \mathbb{R}$ be the disjoint union of two real lines and define the equivalence relation $\sim$ such that $(x, i) \sim (x, j)$ if $x \neq 0$. The quotient space $X/{\sim}$ is a line with two origins. I want to prove the function $\varphi: X/{\sim} \to \mathbb{R}$ that maps $(x, i) \mapsto x$. How do I prove that this function is continuous?

[gentle osprey]

srhoosteen

[sullen bear]

maybe i should have an intermediate function that goes from X/~ to X and then from X to R?

[ebon galleon]

The map R U R --> R given by [ (x,i) |--> x and (x,j) |--> x ] out of the disjoint union is clearly continuous. So since you are quotienting, show that this is constant on equivalence classes of the relation, from which it will follow that it factors through the quotient (and from this description, the map from the quotient is phi)

[sullen bear]

aaahh thank you ill give this a shot