point-set-topology

[odd flame]

ohhhhh wait is this just bc any neighborhood on a circle is homeomorphic to R^1

[shadow charm]

Rereading through chapter 0, I don’t quite get why the last sentence of example 0.15 works here

In particular, I don’t quite see why we can say that on the boundary of N, the homotopy we get by extending the homotopy at A agrees with the constant homotopy outside of N

I have a very poor picture of this example, ig the general idea is that if we can “thicken” A then (X,A) will have the HEP but I have little intuition for why it’s true

[marsh forge]

depends what you mean by any neighborhood

its untrue in most definitions

but it is true that like, if i take a connected open set around a point that is not all of S^1

then i get something homeomorphic to R1

[odd flame]

ok ok sorry to change gears rq

question about this picture

im reading the wiki and it says that a transition map can be defined only if there's an overlap in the charts

now im still definitely not comfortable with the terminology so i could be wrong in what im thinking here, but doesnt this map always exist just bc R^n is obv homeomorphic to itself?

[marsh forge]

You should read a real textbook on this stuff

[odd flame]

probably but munkres is checked out at my library

[marsh forge]

download it

[odd flame]

pdf's

im usually reluctant to read off a screen but i'll try lol

[marsh forge]

Why are you okay with reading wikipedia off a screen then lol

[odd flame]

good point

i know im groping i just havent met w prof in a while

[marsh forge]

I meant to say

You should read a textbook bc i think wikipedia is omitting details

that are confusing you

[odd flame]

ah

[crude vector]

I wanted to prove that compact subsets of a metric space are bounded (A subset of a metric space is defined to be bounded if there exists an open ball in that space containing the subset), so I came up with the following outline and was wondering if anyone could look it over and let me know if it works, cause the proof I found for it used a different open cover:

Let K be a compact subset of a metric space X, and pick some $p \in K$.
Consider the collection of open subsets {$V_n$}, where $V_n = B_n(p)$, for each n $\in$ N [Open ball of radius n, centered at p]. Since for all $q \in K$, d(p, q) < n for some n $\in$ N, {$V_n$} is an open cover of K.

Since K is compact, there exists a finite set of indices M = {$n_1, ..., n_m$} such that {$V_i$} for i $\in$ M is a finite subcover of {$V_n$}. Let s = max{$n_1, ..., n_m$}. Then $V_i \subset V_s$, for all i $\in$ M. Therefore, $V_s$ alone is a subcover of {$V_n$}, i.e. K $\subset V_s = B_s(p)$.

Hence, K is bounded.

[gentle osprey]

strad

[gritty widget]

@crude vector yes

[crude vector]

It's correct?

[gritty widget]

You should work on your LaTeX

It's correct

[crude vector]

Yeah I definitely just kinda winged it here, no clue how to use LaTeX correctly yet

And awesome, thanks a bunch 😄

[crude vector]

In the proof of (a), he shows that every point of the complement of E-bar has a neighborhood (open ball) that doesn't intersect E and concludes from that the complement of E-bar is open, but how can I be sure that there exists a neighborhood (open ball) of p that doesn't intersect E', the set of all limit points of E? Did he skip a step?

Oh wait if every neighborhood had a limit point of E

Then every neighborhood would have a point of E

Right?

[vocal wharf]

yes

[crude vector]

Thank you, I was about to go insane

[vocal wharf]

every open neighbourhood containing a limit point is an open neighbourhood of that limit point

[crude vector]

Ahhh I see, that I did not know

But I can see why it makes sense now

Thanks a bunch

[celest flare]

What does it mean for something to be "strong"? In particular I sometimes hear homeomorphism is too strong for xyz

what does that mean?

[gritty widget]

@celest flare depends on the context

you can think of homeomorphisms as sort of equivalence relation between spaces, for me too strong would mean that it makes too many spaces equivalent to each other

basically, it forgets too many structure

[wanton marsh]

that would be the opposite

a "strong" equivalence relation is a one that can tell many things apart

so the equivalence classes are small

a "weak" one cannot see as much and its equivalence classes are larger

so a point is not homeomorphic to a segment

homeomorphisms see the difference

but they are homotopically equivalent

[gritty widget]

depends on the perspective
while the set of equivalence classes might be larger or smaller, the equivalence classes might do the opposite

that's why I said things such as
depends on the context and for me

there's no formal take on this (at least I don't think so), so no real point in arguing

[marsh forge]

fwiw I think Zef's usage is a lot more common, especially in topology

Where most of our tools are too weak to even start to talk about detecting homeomorphism

[gritty widget]

yeah but I think they asked this because they saw my math.se answer I wrote online

maybe it's just a coincidence, but timing was too perfect 🤔

[limber ravine]

Hey guys

I have two topological spaces, X and Y

X a discrete and Y an indiscrete

suppose they're topological spaces of {1,2}

what would be inside the balls ? all the subsets of {1,2} in X and the empty set and {1,2} in Y?

[little hemlock]

what are the balls supposed to represent?

|| ||

[limber ravine]

xd

nvm I got it now

[gritty widget]

you can only really speak about open and closed balls in a metric space or a gauge space

and not in topological spaces like here

[lunar yoke]

two neat problems from my recent exam:

show that RP^n cannot be covered by n open contractible subsets

show that there exists no covering map from the compact orientable surface of genus 47 to the compact orientable surface of genus 11

[kind depot]

Hey yall so i was workign through hatcher and was stuck on this prob; i have the presentatin version of this gorup but am not sure how to show its isomorphism to integers as shown below

$$ < a, b, c | [a, b], [a, d] >$$

how do i show that this is the same as $$( \mathbb{Z} \star \mathbb{Z} ) \times \mathbb{Z}$$

[gentle osprey]

Optimism

Optimism

[kind depot]

sorry there was a typo

[sinful cloak]

$$\mathbb{Z} \star \mathbb{Z} = \langle a, b \rangle$$, is this clear to you?

[gentle osprey]

M8732

[sinful cloak]

Wait, d? Are you sure it's nto a typo now?

[kind depot]

Oh haha still a typo and yes that’s clear

Ye should be d

[sinful cloak]

<a,b,c | [a,b], [a,c] >?

[kind depot]

Or thag

Ye but like how do I go from free product to croess

I’m not very comfortable with groups

[sinful cloak]

You can define a group isomorphism.

Due to universal property of free groups it is easy to map out of them, so...

[kind depot]

Mm group isomorphism how??? I’m confused a little

[sinful cloak]

$$\Theta\colon \langle a, b, c \rangle \rightarrow (\mathcal{Z} \star \mathcal{Z}) \times \mathcal{Z}$$ $$ a \mapsto (0,a)$$ $$b \mapsto (a,0)$$ $$c \mapsto (b,0)$$

[kind depot]

Ohh ok

Ty ic now

[gentle osprey]

M8732

[kind depot]

Should prob learn group theory haha

[sinful cloak]

This is obviously surjective.

So you just figure out the kernel and make it an isomorphim.

[kind depot]

Mm I’m a little confused by the map so b gets mapped to a pint (a,0)?

Or like I can see that the kerned of this free product is empty and conclude it’s an isomorphism?

[sinful cloak]

The kernel is going to be all the elements in the normal hull of [a,b] and [a,c]

this is because they are the element that commute in the group product.

[sinful cloak]

This is using the above identification of <a,b> with Z star Z.

[kind depot]

Mm thx I’ll look at it

[shadow charm]

Is figuring out how R^3- a knot/ a link deformation retracts usually a hard problem?

Hatcher gives deformation retracts for simple examples and sure they’re intuitive but it seems like a pain to actually prove these deformation retracts

[tardy topaz]

Does anyone want to study Hatcher's algebraic topology book together?

[gritty widget]

Hatcher is too hand wavy for me, I prefer something more formal

[tardy topaz]

Do you have a specific book in mind?

[gritty widget]

There's Spanier, Dieck

Some people recommend to learn manifolds first

From Lee for example

[sleek stag]

Is Bredon good? It covers diff. manifolds also in the second chapter.

[tardy topaz]

Have you read these books or would you want to study them?

[gritty widget]

I've read some of Dieck, like the first two chapters.
right now I'm studying something topology-related already + exercises, so that takes time
So I don't really want to read algebraic topology, but only for right now.
Good luck finding someone, for me studying with other people was always more of a mental pat on the back though

[tardy topaz]

Thanks

[gritty widget]

the book of Dieck can be pretty technical ig, but that's what I like, so... I'm not everyone, you know

[gritty widget]

I think I read on math.se that it's good for learning both manifolds and alg topology at the same time

but I haven't read it so I wouldn't know

[hasty pasture]

Im trying to show that $(\mathbb{Z}\mathbb{Z)}/[\mathbb{Z}\mathbb{Z},\mathbb{Z}\mathbb{Z}]\cong \mathbb{Z}\times\mathbb{Z}$ and my first thought was to use the first isomorphism theorem. I've been trying to use the homomorphism $\phi:\mathbb{Z}\mathbb{Z}\to \mathbb{Z}\times \mathbb{Z}$ defined by $\phi(a^{n_1}b^{m_1}\cdots a^{n_k}b^{m_k})=(a^{n_1}\cdots a^{n_k},b^{m_1}\cdots b^{m_k})$ Where we are considering $\mathbb{Z}*\mathbb{Z}$ to be the free product of two infinite cyclic groups generated by a and b respectively. I have shown that this is a homomorphism and is surjective, but i am really struggling to show that the commutator is the kernel. I have that the commutator is contained in the kernel but i cannot show the other containment. I am beginning to suspect that it is in fact not equal to the kernel, but if that is the case I am stumped on how to proceed. I was hoping someone would be able to shine some light on the problem.

[gentle osprey]

llspacebarll

[coarse night]

what are your preferred book for algebraic topology (other than Munkres because that's what I'm following rn)

[empty grove]

Munkres is for point set

Rotmanstan has been stanning rotman

Unsurprisingly

learned from spanier

[coarse night]

it has some 'homotopy' stuffs

[empty grove]

It barely introduces fundamental groups

[coarse night]

yes that's why I'm asking lol

[empty grove]

@swift fjord tell us why rotman good

[lunar yoke]

moldi tell us why may good

[empty grove]

May not good

[lunar yoke]

[empty grove]

It is so difficult

[lunar yoke]

yeah it skips a lot of details

[empty grove]

I am moving along but very slowly

It's better than Hatcher though for sure

[lunar yoke]

i recently tried to fill in the details in one claim in the cofibrations chapter

[empty grove]

which claim was it

[lunar yoke]

something about products of ndr pairs being ndr pairs again

[empty grove]

oh god ndr pairs

[lunar yoke]

and he gave the construction

but he said its obvious that its continuous

but it wasnt

by far

[empty grove]

lmao

[lunar yoke]

i had to hunt down various statements in dieck and brown

[empty grove]

imagine checking continuity in 2022

[lunar yoke]

fair enough

[high hill]

moldi #help-18 can u help make sense of the pinned quotient group pl

[swift fjord]

clear and concise writing which doesn't leave you in the dark with hazy geometric intuition, but doesn't completely hold your hand either

good chapter order

just overall excellent writer with supreme pedagogical sense

no, it just actually does algebra unlike some authors

hatcher be like
proof:

look I drew a pretty picture

[lunar yoke]

reading 400 page book in 1 summer

[swift fjord]

summer is long

[empty grove]

😌

[swift fjord]

munkres be like:
S I M P L I CI A L H O M O L O G Y

[empty grove]

Simplicial homology

[pearl holly]

Does munkres do simplicial stuff?

[swift fjord]

elements of algebraic topology

[pearl holly]

Oh right

[swift fjord]

the driest book i've ever read

[limber ravine]

you know elementary topology by viro?

[empty grove]

Wait hold on I just realised this simplicial homology is not the homology of a simplicial object

[empty grove]

Simplicial homology

[coarse night]

can't even distinguish which one is a suggestion and which one is a joke

[lunar yoke]

id say singular homology is actually more in that direction

[empty grove]

Read Rotman then come back and tell us why shin is wrong

[lunar yoke]

singular set is kan complex etc

[empty grove]

Ye

I just forgot that simplicial homology was a thing in topology too

[lunar yoke]

i think its good as a first introduction to homology

very visual

[empty grove]

True

[swift fjord]

yea except proving anything in it is awful

[lunar yoke]

easy to compute for small examples

[swift fjord]

so you want to immediately move to singular

[empty grove]

But isn't simplicial homology also the homology of a simplicial abelian group

[lunar yoke]

cellular is also nice, but it sucks when you have to prove that certain maps have certain degrees to compute the boundary maps

[empty grove]

Or are you just saying no kan condition

[lunar yoke]

much nicer if you can just say "obviously this is degree ..."

[empty grove]

so true

[lunar yoke]

hmm

you mean you take the chain complex via dold kan and homology of that?

[empty grove]

No lol that would be stupid

I wouldn't say stupid things would I

[lunar yoke]

why would that be stupid

[empty grove]

kinda redundant

[lunar yoke]

its what you do for singular

[coarse night]

[empty grove]

Wait isn't that redundant

Idk dold kan well enough to say

[lunar yoke]

well ok which simplicial abelian group are you talking about then

[empty grove]

You have singular simplices still but only the specified ones

So viewing it as a simplicial subgroup of the singular simplicial group

[empty grove]

specified ones as in the ones that appear in the delta complex structure

[lunar yoke]

ah yeah ok

[coarse night]

I'm already liking rotman

[empty grove]

noooo you were supposed to tell us why shin is wrong

[swift fjord]

Moldi why do u hatr me

[coarse night]

@swift fjord want me to say it?

[empty grove]

I went to ivory once and everyone was bullying you

I thought it was cool

[swift fjord]

It's all part of my plan

I have the entirety of ivory in the palm of my hand

Wait were u there for CP^infty

[empty grove]

No

[swift fjord]

Oh

Don't look it up

[empty grove]

brb

[swift fjord]

[empty grove]

I don't get why that was bullyworthy

[swift fjord]

Me neither tbh

At least to that extent

[empty grove]

I guess it is just cool to bully you

[swift fjord]

It's not just me

Everyone there bullies everyone

[empty grove]

oh

Good thing I don't go there

😌

[hollow harbor]

[hollow harbor]

it's because he's so smug about it

it's too easy to bully some people. but it's a challenge to bully shin

[empty grove]

That justifies bullying? Wow ryc

[swift fjord]

I am the most well-liked member of the ivory tower

[hollow harbor]

would you bully someone you didn't like? no of course not

you only bully the best people

[empty grove]

Yasuke would never have let this happen

[hollow harbor]

guess that's why i haven't been bullying moldilocks as of late... 😮‍💨

[empty grove]

[gritty widget]

I realize i think I have misunderstood the definition of a vector bundle

can someone clear this up for me

[hollow harbor]

sure

[gritty widget]

you don't actually have to specify the "transition" maps

you just specify the local trivializations

[hollow harbor]

yes

[gritty widget]

and the transition map cocyle conditions are automatic

[hollow harbor]

well

yeah

you specificy local trivializations and then you need to check that the transition maps are linear isomorphisms on fibers

and like

the smoothness stuff you can check either on transitions or on local trivs

[gritty widget]

but do you though?

[hollow harbor]

i guess not

[gritty widget]

the definition doesn't seem to mention transition maps

[hollow harbor]

you can also check that on the local trivs

yeah

there are two ways to construct these things

once you're constructing things from coordinates you often have to refer to the transition map defns anyway

but

yes, you can do either and you don't need to do both

[gritty widget]

this seems very strange

it doesn't seem to suggest any compatibility on interesections?

[hollow harbor]

you just don't need it

the compatibility on intersections all comes from composing things you know

you already know E is a manifold

[gritty widget]

ooooh

[sweet wing]

https://media.discordapp.net/attachments/787355364017700884/942451842665807924/223340518619217920.png

https://media.discordapp.net/attachments/787355364017700884/942463721945563136/223340518619217920.png

i feel like im not seeing something obvious but not sure where could anyone help :p

(also this is an exercise from bredon i finally decided to actually learn nets LOL)

[gritty widget]

hmm do we actually need that E is a manifold

[hollow harbor]

Well

Then you start to need more stuff

Vector bundles are manifolds

[gritty widget]

a finite dim vector bundle over a manifold is definitely a manifold

but when you want to talk about classifying vector bundles

you need to introduce slightly more general spaces

maybe i am wrong

but it seems like h being a vector space isomorphism on the fibers should give the glueing data?

yeah okay we don't need them to be manifolds

[shadow charm]

Here X_m is the mapping cylinder of the map z -> z^m on S^1

I’m having a lot of trouble seeing how letting x vary allows us to get X_m

It’s in the context of the torus knot K(m,n)

Ie that winds m times longitudinally and n times meridianally

[empty grove]

You are actually reading about the torus knots

[shadow charm]

Why, is that a meme

[empty grove]

I thought it was universally skipped

[shadow charm]

Lmao why

I would only skip trivial stuff

[empty grove]

who cares not like I need donuts to untangle my earphones

[shadow charm]

I mean i don’t care all that much about those knots but it’s a working example

And if I don’t understand the method there’s something I need to fix

do you think you might be able to explain how the radial lines trace the required mapping cylinder in the end

[empty grove]

,rotate

[gentle osprey]

Couldn't find an attached image in the last 10 messages.

Feb 13, 2022, 7:41 PM

[empty grove]

bruh

[shadow charm]

Oh I think I might have understood it

The center of the radial segments is S^1 x{0}

And the outside is S^1 x {1} with the required quotient

Because as x varies the center loops once but the edges loop m times

That’s such a weird way to see it

There is nothing cylindrical about this

Aight now I just need to get how they make the deformations agree on the boundary torus

New problem if I view it this way the intersection of X_m and X_n isn’t S^1 x {1/2} like it should be

Ohhh i think it’s the other way around

The « circle » on the boundary of the torus is the intersection of the two « cylinders » and corresponds to S^1 x {1/2}

The core circles of the two torii are S^1 x {0} and S^1 x {1} with the proper identification

That’s mind bending ngl

[gritty widget]

I'm struggling to understand why we can find these U_x,j for each x

for every (x,t) in X \times I we can find a U_{x,t} such that the bundle is trivial

[gritty widget]

Does anyone see why this is true?

I'm neither sure why the projection is a homeomorphism

nor why it would lift to the bundle

[novel acorn]

I read it and I was given therapy afterwards

It's the worst construction he does

it doesn't even showcase Van Kampen that well anyways??

[empty grove]

Hatcher bad 🤝

[novel acorn]

Nah
I read like half of Hatcher

it's it's OK

not that I could do any exercises 😔

Also the only other one I know is Dieck
But Fiona stans that one

[empty grove]

Damn

[bitter smelt]

i have a question in #diff-geo-diff-top which should have been in here, but i misplaced it

[gritty widget]

Is there any result about fiber bundles that says something like

The nth homotopy groups of sections of the bundle

Is equivalent to the path components of some maps?

Since R^n+k is contractible, the space of bundle mono morphisms is homotopy equivalent to space of sections of this fiber bundle

I don’t get why this is true

[shadow charm]

https://m.youtube.com/watch?v=nq8QBdP_N_o&list=PLpRLWqLFLVTCL15U6N3o35g4uhMSBVA2b&index=2

Found this playlist to use alongside Hatcher

Anyone know if it’s any good?

[viral yoke]

Let $f : (X, x_0) \to (Y, y0)$ be a pointed continuous map, and
consider the induced homomorphism $f_* : π_1(X, x_0) \to π_1(Y, y0)$.
(1.) If $f$ is one-to-one, is $f_$ also injective?
(2.) If $f$ is onto, is $f_$ also injective?
(3.) If $f_$ is injective, is $f$ also one-to-one?
(4.) If $f_$ is surjective, is $f$ also onto?
For those that are true, give a proof; if not, give a counterexample.

[gentle osprey]

eM

[viral yoke]

I think I have an idea on how to approach this, but it feels like I'm bouncing between proving injectivity but needing surjectivity, and vice versa, which violates by intuition that all of these should be true.

[dense flame]

he's great

[empty grove]

Circle, line, disk should probably be enough to do all of them

where I am saying that all 4 should be false

Also why sometimes injective and sometimes one-to-one

Ye all 4 are false

[viral yoke]

I have no idea why it is worded like this lol

[gritty widget]

because homomorphisms can be injective or surjective, but pointed continuous maps can be one-to-one or onto

clearly!

[midnight echo]

quick question, is the empty set and the set of real numbers open?

i'm trying to prove that the cofinite topology on the reals is a topology

[dense flame]

yes

axiomatically, the empty set and the whole set X are open for any topology on X

[midnight echo]

cool cool cool, thank you

[swift fjord]

The cofinite Topology is usually explicitly defined so that it includes the empty set

Because the empty set is not cofinite

[midnight echo]

ahhh thank you i've written this so far, but I can't see how the cofinite topology will include R itself since R is not finite

[swift fjord]

R is cofinite

The complement of R in R is the empty set

Which has cardinality 0 by convention is you wish

[midnight echo]

Ohhhh yeah, gotcha cheers

[swift fjord]

Np

[midnight echo]

i'm trying to to prove the statement of arbitrary open unions are open, however it seems like what i've written is a contradiction as the complement of an open set is closed

[sleek thicket]

What part of this gives a contradiction?

[midnight echo]

the final line I didn't finish writing it but it says R\u is the complement of u. However I defined u as open, so the complement is closed, therefore the intersection of them will be closed as well

[sleek thicket]

Yep, that's all true

But none of that is contradictory

All you need to show is that the intersection of all those complements is finite, that would imply that the union of all the Ui is open. Does that make sense?

[midnight echo]

i'm slightly confused as to how it would imply it's open? Is it by definition of the cofinite topology?

[sleek thicket]

Yep

"U is open" means "R \ U is finite (or U is empty) " when we're talking about the cofinite topology

When you write τ = ..., you're defining the collection of open sets (which is τ)

[midnight echo]

Ahhh brilliant, thank you so much

[cosmic beacon]

Would the homology of a space with coefficients in:
-Q as a ring
-Q as a Z-module
-Q as a Q-module
always be the same for any space?

As I understand, homology with coefficients only cares about the abelian group structure, which begs the question, why do we care about homology with coefficients in a general ring or module?

[boreal plover]

good evening

can anyone here help me understand the difference between an identification and a homeomorphism?

A bijective function f between two topological spaces X and Y is a homeomorphism if and only if both f and f^(-1) are continuous

A function f : X -> Y is an identification if and only if U \subset Y is open if and only if f^(-1)(U) is open```

I was thinking that if, for example, X is a space with a lot of points i a single open set, then f is not bijective but is an identification

[sleek stag]

Those definitions are equivalent

if f is a bijection

[boreal plover]

ahhh

[sleek stag]

but the requirement is dropped in the definition of the identification

[boreal plover]

so the only difference is that f need not be bijective?

hmmmm then why do we choose a homeomorphism rather than an identification to be the statement of "these are the same topological space"

wouldn't, for example, connectedness or compactness be preserved by an identification?

[sleek stag]

Yes, but if you don't have a bijection between the spaces then you really can't say the spaces are the same.

Note that a topological space is a set + a topology on the set. To say topological spaces are the same, we want a one-to-one correspondence between the underlying set AND the open sets.

[boreal plover]

hmmm wait the very next example seems to clarify things

In this case, an identification isn't a "homeomorphism but not bijective", is it?

for e.g. the equivalence relation could cause the sets in X/~ to be coarser than the topology on X

[empty grove]

Hausdorffness isn't

I'm not sure about the other 2 either, I don't think an identification needs to be surjective?

[sleek stag]

At least not in the definition given above

[empty grove]

Ye I was just thinking if maybe that forces surjectivity somehow

But doesn't seem to

[sleek stag]

Yea, I've only dealt with quotient maps which are by definition surjective

[sleek thicket]

the ring structure of coefficients matters because it gives an algebra structure on cohomology

the module structure of a coefficient group means the homology (and cohomology) will be a module over the same ring

(for Z, Z-modules are just abelian groups so there's no extra structure)

[gritty widget]

I think it has to, but even if it isn't, the other part becomes discrete or something

So doesn't matter much

[empty grove]

I was thinking of an inclusion of the point space into a 2 point set

You could define a topology on the 2 point set using this definition

You should get the discrete space yeah

So it's a non surjective identification and if you generalise this to 1 point space including into however many point set, you lose both connectedness and compactness

[gritty widget]

Something like "homeomorphism but not bijective" would be a perfect map

https://en.wikipedia.org/wiki/Perfect_map

This is a closed continuous surjection with compact fibers

[boreal plover]

Ah so what is an identification? 😅

[gritty widget]

Identification/quotient map is a (surjective) map q:X to Y such that U is open in Y iff q^-1(U) is open in X

Closed/open continuous surjective maps are quotient maps

They have this useful property that to prove f:Y to Z is continuous, if f o q^-1: X to Z is well-defined, then it's continuous iff f is continuous

So it's enough to show f o q^-1 is continuous (I'm abusing notation a bit here)

So for example, say you have a map f:R to R

And q:R to S^1 is given by q(x) = exp(ix)

Then this is a quotient map

And you want f o q^-1 to be well-defined, this means f has to be 2pi periodic

So we get a correspondence between continuous 2pi periodic functions on R and continuous maps from S^1 to R

Just a basic application

Does it make more sense now?

[prisma seal]

what do you get from the result of the quotient map on this complex? Munkres says it's not a Torus, and I think I get why - you have the repeated ad edge, the be edge, and the cf edge, but I don't know what you would actually get from this

[marsh forge]

i dont think it is necessarily a nice space

i mean i guess theres a classification

but the best way to figure out what surface it is is probably computationally

[prisma seal]

so it's not necessarily something I can just draw/picture easily?

[marsh forge]

Well I didn't think about it hard but in general these things can be like, arbitrarily complex

so easily no

in theory you can figure it out with enough effort

[limber ravine]

So basically a continuous function is such that an open set in the counter domain is open in the domain. Then an homeomorphism is like a overpowered continuous function where an open set in either side is open in the other

[marsh forge]

codomain

not counter domain

basically mns yes

homeomorphism should say "these two topological spaces are the same"

so of course they should agree on which sets are open

[limber ravine]

Right?

Gotcha

Thanks

[errant stump]

yeah like max said quotient spaces formed like this can be crazy. i think it's like every compact riemann surface* or something like that has a corresponding fundamental polygon

for example you can write RP2 as one of these, and you can't even put it in R^3 without it self intersecting

[marsh forge]

It's easy to describe all orientable closed manifolds this way

I think you also get all nonorientable ones

You also get a lot of stuff that is not a manifold

[errant stump]

also of genus >0 ig lol

[marsh forge]

You can build spheres this way

[errant stump]

yeah good point. just identify every point on the edge together and glue two copies of the shape together along their boundaries and you get a 3 sphere. so good luck visualizing that

[marsh forge]

Oh i meant the normal sphere

Just identify two pairs of consecutive edges of a square

[prisma seal]

I guess I expected this to be nice since it was relatively "simple"

[gritty widget]

A homeomorphism is a bijection $f:X\to Y$ such that it induces a map between topologues $f^{-1}:\tau_Y\to\tau_X$ which is also a bijection.

[gentle osprey]

Blitz

[gritty widget]

So I'd describe it as a bijection which is also a bijection between the topologies

[marsh forge]

Sure, but I dont think this really captures the intuition

[gritty widget]

It captures the fact that they are essentially the same

[gritty widget]

here

the only real difference between homeomorphic spaces is underlying sets

[marsh forge]

Well... yeah

my point is that you've simply rephrased the textbook definition

I was just trying to explain how one thinks about them

I would not say "these two spaces are the same" is a suitable formal definition

But if one has a homeomorphism and asks whether it should preserve property P

The intuition I gave is normally helpful

[gritty widget]

rephrasing makes it clear what they are, regardless of your intuition

[marsh forge]

[gritty widget]

you've said, more or less, "homeomorphic spaces are essentially the same as topological spaces" and I answered "in what way are they the same?"

[digital peak]

a topology is not the ultimate way to look at a space

[marsh forge]

lmfao

[gritty widget]

They are one way to look at topological spaces

[fading vale]

They already know the formal definition because they stated it though max was just explaining the intuition

[marsh forge]

the intuitive way to think about Definition X is by restating definition X

[gritty widget]

of course it is

[digital peak]

different people have different intuition btw

[gaunt linden]

They can have intuitions that are not different while not being the same either.

[digital peak]

you could describe topology as algebras of a certain profunctor, and then yes this is what you would get for isomorphism of algebras

[marsh forge]

This convo went from bad to terrible

[digital peak]

don't thank me

[light rivet]

If I had a top space X has k path connected components, how can I show its zeroth homology group is isomorphic to Z^k? I saw the proof where X is path connected, so H_0 is Z, but that used an trick of a surjective hom f: C_0 \to Z and showing the ker f = Im \partial_1, and doesnt seem like it generalizes

[fading vale]

have you seen that H_n(bigcup X_i) = bigoplus H_n(X_i)

@light rivet

[light rivet]

Nope, not yet

[fading vale]

okay

we can do it pretty directly anyway

[light rivet]

The trick I have seen for the X path connected case is the following: Take the surj hom f: C_0 \to Z that just maps the formal sum to the sum of its coefficients in Z

Then C_0/ ker f = Z and then we could show ker f = im partial_1 which gives H_0 = Z

But I don't see how to extend this idea to path connected components

[fading vale]

The way i would do it is that if you have two vertices in C_0, H_1 = C_0/im d_0 basically identifies any two vertices connected by a one chain right

because if i have a one chain, thats a path between two vertices v and w, and its image is v - w so that v - w = 0 and hence v = w in H_0

[marsh forge]

An instructor has marked this as a good answer!

except its H_0

not H_1

[fading vale]

Oops

Typing is hard

Also is the instructor you

Who is watching me

[fading vale]

anyway like @light rivet does this make sense?

[marsh forge]

im trying to outline a talk so I didnt want to type this out and am glad you are doing it instead

[fading vale]

You're legally obliged to answer any AT questions

[marsh forge]

Yes

It’s why I get the yellow name

[fading vale]

If this isnt clear think of it like this: Since ker d_0 = C_0, H_0 is obtained by taking the free group C_0 generated by the vertices and quotienting it out by saying that any two vertices in the image of d_1 (i.e connected by a 1 chain) are identified

[marsh forge]

(equivalently a-b in img d_1 iff theres a path connecting them, and there are no other relations)

this trick also works for the case of one connected component, of course

[light rivet]

hmm, gimme sec to digest this

[fading vale]

it might help to draw some stuff out

[rancid umbra]

if i have a set that is diffeomorphic to a convex set, is it simply connected?

[lunar yoke]

homeomorphism should suffice

it gives an isomorphism on fundamental group

which detects simply connectedness

by being 0

(since convex sets are simply connected)

oh even more its already conctractible

[rancid umbra]

what is contractible

[fading vale]

homotopic to a point

[marsh forge]

very simply connected

[rancid umbra]

alr

thanks

[lunar yoke]

btw is there a nice example for a space that is not contractible but simply connected

[marsh forge]

sphere

[fading vale]

the sphere

[lunar yoke]

oof

[fading vale]

Max

[marsh forge]

im here for the low hanging fruit moth

[lunar yoke]

why am i like this

[marsh forge]

u can do the hard work

[rancid umbra]

i see. thats interesting. its fundamental group is trivial tho...

[fading vale]

Yes thats what simply connected means

Its just not homotopic to a point

[marsh forge]

it has a handful of nontrivial higher homotopy groups

[rancid umbra]

i dont really have any background in this type of topology yet. but i kinda dont like this

[lunar yoke]

its the best kind of topology

[fading vale]

just a few

[marsh forge]

if you believe that circles are not contractible, spheres dont seem much worse

[fading vale]

Dont like it in what sense

[rancid umbra]

i need a result from complex analysis actually, and it says that any holomorphic function on a simply connected space has a primitive. but my professor phrased it in terms of being diffeomorphic to a convex set, since we havent covered the algebraic topology part of topology

[marsh forge]

ah yeah the result is just topological

phrasing it as diffeomorphic just seems silly

[rancid umbra]

i uh. also dont know how to prove this result but i need it for something else...

i have some form of cauchy's integral theorem

[rancid umbra]

but im not sure how to get this from the thing i have

oh wait nvm i think i got it

[drifting sundial]

I don't get this part in May's proof of van Kampen. doesn't the existence of a path x -> y follows from the fact that x,y in U and U is path connected?

why do we need to assume that the cover is finite?

[fading vale]

We dont just need to assume there exists a path from x to y in U, we need to choose paths from x to y for every y in X such that if y is in U, the path is also contained in U

So like

if y is contained in U_1 and U_2, we need the path x to y to be contained in U_1 and U_2, eg U_1 cap U_2

since the cover is closed under finite intersection we know that y is in some finite collection of elements of the cover, and the intersection of those elements is also in the cover, and then we can take a path from x to y in that intersection

if the collection is infinite then if take some cover U_alpha and write y in bigcap U_alpha, then there is no guarantee that bigcap U_alpha is in the cover, and in particular not even a guarantee that its path connected, so there might be no path from x to y in bigcap U_alpha

[drifting sundial]

ah yeah this was the part I missed

[fading vale]

I honestly cannot think of an example where finiteness is crucial off the top of my head lol you need something weird and fucked up where you're taking like infinitely thinner slices so that when you take the infinite intersection its a point or some shit

and that disconnects it

maybe the comb space would provide something concrete

But theres really no need to worry about that

This is just being very careful

[fair idol]

How can I show the interior of a set is nontrivial?

[empty grove]

Find a non empty open subset

[true robin]

prove that the complement is not dense

[gritty widget]

equip the set with the discrete topology

*i should add that if you are to use this argument, it is essential that you state you can do this without loss of generality

this is actually a really nice trick to use in general

you can you use it to show things like

every manifold is connected

every manifold is infact not a manifold

I think this should be true but im not sure

M an manifold, N a closed submanifold, U a tubular neighborhood

the closure of U should deformation retract to N?

[fading vale]

Theres a general fact here where the total space of a vector bundle deformation retracts onto the 0 section

locally the total space is B x F so if we take local coordinates x, v for B and F then H(x, v, t) = (x, (1 - t)v) is well defined and works

[gritty widget]

this works for any bundle?

[fading vale]

Yes

The only important bit is that its locally trivial

[gritty widget]

im confused about one part

why do we need to closure?

[fading vale]

It comes from like, the formal way that your tubular nbhd works

Like

a tubular neighborhood U of a closed submanifold S of M is equivalent to like

you want a bundle E -> S and a map f: E -> M

such that S -> E -> M (f composed with the section S -> E) commutes with the embedding S -> M and f has closed image in M

Like if U is a tubular neighborhood regarded as a subset of M containing S then it corresponds to a bundle with total space E which deformation retracts onto S

E embeds into M though, and its image is closed so its the closure of U, not U itself

[gritty widget]

this still confuses me

[fading vale]

@gritty widget how are you defining a tubular neighborhood

(if you see this sometime soon you should probably ping me if/when you respond )

[fair idol]

If I could ask a vague question... How can I decide whether a topological space can be the covering space of another?

[marsh forge]

Homotopy groups can be a great test

You need to be able to realize pi1 of the cover as a subgroup of pi1 of the base

And pi_n has to agree

[supple locust]

Can I just pick a generator $z$ of $C_n(D^n)$ such that $\partial z \in C_n(S^{n-1}) $ (so it becomes cycle in relative homology $H_n(D^n, S^{n-1})$)? Seems obvious but I can't really justify why.

[gentle osprey]

bert

[shadow charm]

Wow, I think van kampen’s theorem has finally made me intuit pullbacks and pushouts

[umbral finch]

what does it mean for sth to be "identified"?
like the definition of a torus where the left and right side are identified and the top and bottom sides are also identified.

[empty grove]

Quotient by equivalence relation which says that those corresponding things are equivalent

[fair idol]

Thank you so much!

[gritty widget]

I'm confused about this deifnition again

it seems there is compatibility required

it seems that there is no compatibility required

[fading vale]

What kind of compatibility do you want

You usually also want pi circ h to be the usual projection U x R^n -> U i guess

[long coyote]

Show that deleting a point from a manifold of dimension greater than 1 does not affect orientability of the manifold.

[gritty widget]

This definition makes it seem like

Every bundle is a trivial bundle

[fading vale]

Why would it?

Trivial bundle would imply that E is globally B x R^n

is there something here that implies that

[fading vale]

If you ask a question its probably best to elaborate on what definitions of orientability you're working with and what ideas you have/what specifically you are unsure of

[long coyote]

i am reading hatcher's book, my current idea is to apply canonical isomorphisms. not sure i am on the right track

[fading vale]

So have you defined it as a choice of generator of H_n(M; Z) where M is ur manifold of dimension n?

Or is orientability = admitting an oriented atlas?

these are equivalent but id just like to know which one you're working with

[long coyote]

first one

[fading vale]

Oh okay

so basically your goal is to show that if M has dimension n > 1, H_n(M; Z) cong Z implies that H_n(M - p; Z) cong Z

right?

or i guess since you want to show it preserves non-orientability too you want an iff

[long coyote]

yes, the orientability won't change after removing a point, if M is orientable, M-{x} is still orientable vice ver sa

[fading vale]

Yeah thats what you want to show

[long coyote]

either i am missing something or i overthink the problem, will work on it after a break

[fading vale]

Ok tbh i was kinda busy and idk where i was going with this proof i think its kind of hard to do it with the global homology

But im going to leave these msgs up in memorial to my shame and just delete the actual thing or something

[shadow charm]

In defining the CW complex, the 0 skeleton is only ever defined as some discrete space

Does this mean we can take it to be as disgusting as we like, like R with the discrete topology?

And all theorems about CW complexes with such a disgusting 0 skeleton would still be true?

[empty grove]

Yes but how can discrete be disgusting lol

[shadow charm]

I mean it’s disgusting in the sense that it doesn’t really fit with the nice intuition of a 0 skeleton as a “discrete set” whatever that means in my mind of points

I guess it’s because I’m viewing R as a continuum but with a discrete topology I should view it as separated points, just uncountably many of them

[empty grove]

Yes

[shadow charm]

I just have trouble imagining an uncountable amount of separated points

[gritty widget]

That's not what continuum means

A continuum is a compact connected space, so more like [0, 1]

[shadow charm]

I wasn’t taking a formal notion of continuum

Rather the idea of a connected “continuous” line which fair enough a discrete topology also contains (it’s still a connected space) but it doesn’t have the same qualifies

Qualities*

This is all in an informal sentiment

[empty grove]

Compactness isn't part of it

Or at least linear continuum is ordered set with archimedean property and suprema/infima

Connectedness is a consequence, compactness is one if it has global supremum and infimum

It would be weird if ℝ weren't a continuum lol

Oh apparently compact is part of continuum

[gritty widget]

Yes

[empty grove]

And it's different from linear continuum

Bruh

[arctic relic]

Given a free group F, why can I find a graph X st the fundamental group of X is F? Does it have to do with being able to write any graph as a wedge of circles that correspond to generators in F?

[empty grove]

There doesn't exist a graph

Oh wait free group

Yes

Wedge of κ circles has fundamental group on κ generators

And every graph is homotopy equivalent to such a wedge

And every such wedge should be homotopy equivalent to some graph if there is justice in the world

[arctic relic]

Sweet thank you

[empty grove]

Well if by graph we mean a 1 dim CW complex then every wedge of circles is a graph

[empty grove]

Also connected graph oops

[arctic relic]

Yeah that’s what I kind of realized just now since it’s all just boundaries of 2 discs

Not sure if that’s a good way to think about CW complexes though

[shadow charm]

In general you can count the number of edges of a graph - its minimal spanning tree

Fundamental group of the graph is then free group on as many generators as such edges

[coarse night]

I get f(X) has to finite in Y so g(f(X)) is a finite set in X, but idk how to show it's not homotopic to 1_X, or f(g(Y)) ~ 1_Y

continuity in spaces other than lR feels impossible to me

[empty grove]

Think of them as embedded inside ℝ

[coarse night]

no

[empty grove]

I'm almost saying that unironically

Ok here's another hint in addition to the one given, you might want to first think about what homotopies in either space look like

[coarse night]

that I am unable to visualize

[empty grove]

Hint 3 a homotopy is a bunch of paths next to each other

[coarse night]

I will be able to extend the finite set slowly to whole X, is that what you are saying

[empty grove]

No, I'm saying that to understand homotopies in a space, you first have to understand paths in that space

[coarse night]

true

[coarse night]

what about arbitrary space, f:X-> Y. Thinking about paths will be of any help?

[empty grove]

If you know about the structure of Y to some extent

[coarse night]

hmm

thinking of F(x0, t) could be helpful

[empty grove]

Ok so you have 2 spaces X and Y here, and you wanna see if there's a homotopy equivalence between them

Which means you want to check if there exist functions f: X → Y and g: Y → X

Such that fg ≈ 1_Y and gf ≈ 1_X

And this is why you want to understand homotopies in X or homotopies in Y

So you can see what a function homotopic to the identity would look like

[coarse night]

ye

[empty grove]

Now do you see what paths in either space look like?

[coarse night]

yes

[empty grove]

What are they?

[coarse night]

singletons

[empty grove]

Yep

Now what do homotopies look like?

[coarse night]

the same function

[empty grove]

Yes, constant homotopies

[coarse night]

nice

[empty grove]

So now every ≈ so far becomes a =

The question is then asking if these spaces are homeomorphic

[coarse night]

homotopic eqv*

so I get if gf ~ 1 then g must be constant function from f(X)

[empty grove]

No

Actually homeomorphic

Because ≈ became =

[coarse night]

oh that's what you meant by it

thanks

[shadow charm]

Anyone can provide a little hint on showing that if convex sets in R^m intersect non trivial in triples then they must have a common intersection?

I’m trying to set up some kind of function of the distance of a point in a k-th set to other convex sets to show it intersects with their intersection somehow so that the result would follow by induction but struggling

I feel I’m missing something super basic

[fading vale]

What does intersect non trivially in triples mean

[shadow charm]

Every triple of sets intersects

[fading vale]

Like we have a collection of convex sets and any 3 have non trivial intersection

[shadow charm]

Yeh

[fading vale]

and you want to show that the entire thing has non trivial intersection

[shadow charm]

Yup

[fading vale]

https://www.cut-the-knot.org/pythagoras/ConvexSets/HellysTheorem.shtml @shadow charm heres a proof

this is called hellys theorem

This is only R^2 though hm

[shadow charm]

Thank you 🙏

[fading vale]

https://en.wikipedia.org/wiki/Helly's_theorem

[shadow charm]

Should be adaptable lemme check

[fading vale]

Heres R^m

it seems to require compactness

[shadow charm]

Hmm this uses radon’s theorem which I don’t really have access to I don’t think

I feel like I might be approaching this problem the wrong way

The original problem is showing that the union of open convex sets in R^m with non trivial intersection in triples is simply connected

If I can show they all intersect I can just use van kampen

But maybe there’s another way to go about it

[shadow charm]

Wait this doesn’t even really work because it considers intersections of more than 3 sets as soon as we are above R^2

Alright I’ll assume then that there is a counter example and that I must approach this differently

Wait I’m an idiot

Okay I’ve got another way at this

Thanks for the help

[fading vale]

i didnt do very much but im glad it workedo ut

[arctic relic]

Is it true that D^2/{p} where p is not in the boundary is homotopic to a wedge of circles?

I’m trying to compute the fundamental group of the complement of a point in the torus (S1 x S1) and reasoned that D^2/{p} is homotopic to S1 but I’m not sure if this is enough to conclude that the punctured torus is homotopic to a wedge of circles ie the fundamental group is generated by <a,b|. >

I guess the punctured torus is homotopic to it’s 1 skeleton which is a wedge of circles is good enough?

[marsh forge]

To one circle

Do you mean T^2?

[marsh forge]

Ah I see I’m not sure how you plan to use the fact about D2

[marsh forge]

Depending on who you are proving this to yes this is enough

[arctic relic]

By D^2 I mean the 2 disc

[marsh forge]

Yeah, the first sentence is just confusing

It is of course homotopic to a wedge of circles in the sense that one circle is a wedge of circles

[arctic relic]

I think it’s 2 circles since the 2 disc has gluing instructions for a torus now that I think about it

My bad question was poorly asked

[gritty widget]

what's an example of a limit that the pi functor doesn't commute with

I feel like it doesn't preserve all limits

[shadow charm]

Okay I think I’m getting the hang of this type of argument I just need someone to check that what I’m doing makes sense

So I want to show that the complement of a finite set of points in R^n with n>=3 is simply connected

The case n=3 is trivial and I’ll be doing the rest by induction

So I suppose it’s true for n=k-1 and consider the case n=k: let X_k be the complement of k points x_1,…,x_k and X_k-1 of the first k-1 of these points

I consider a loop f in X_k based at some point x_0 in X_k

As a loop in $X_{k-1}$ this is hullhomotopic so there is a homotopy $F: I\times I \to X_{k-1}$ between $f$ and the constant loop $\gamma_{x_0}$

[gentle osprey]

𝓛ittle ℕarwhal ✓

[shadow charm]

Now consider an open ball neighborhood B of x_k that doesn’t contain any of the other x_i

Its inverse image is open in $I \times I$ so a union of possibly infinitely many open rectangles

[gentle osprey]

𝓛ittle ℕarwhal ✓

[shadow charm]

The inverse image of $x_k$ is compact in $Int(I\times I)$ and so contained in finitely many of these rectangles $(a_i, b_i) \times (c_i, d_i)$

[gentle osprey]

𝓛ittle ℕarwhal ✓

[shadow charm]

For each of these rectangles consider a map $G_i: I\times I \to \overline{B} - {x_k}$ that maps $\partial (I\times I)$to $\partial ((a_i, b_i) \times (c_i, d_i))$after reparameterizing the open rectangle

[gentle osprey]

𝓛ittle ℕarwhal ✓

[shadow charm]

Then the restriction of $F$ to the closure of the open rectangle is homotopic to $G_i$ rel $F(\partial( (a_i, b_i) \times (c_i, d_i)))$

[gentle osprey]

𝓛ittle ℕarwhal ✓

[shadow charm]

Hence doing this for each rectangle homotopes $F$ to $F’$ in $X_k$ where $F’$ is a homotopy in $X_k$ between $f$ and $\gamma_{x_0}$

[gentle osprey]

𝓛ittle ℕarwhal ✓

[shadow charm]

And so we obtain simple connectedness of X_k

Is this a good argument?

[empty grove]

Looks fine

You could also use van kampen for the induction step

Cover the space with 2 open sets, both having fewer holes

[shadow charm]

I knew there was some way to do van kampen but couldn’t figure it out

Since it’s the van kampen chapter

Like you would take X_k union X_k-1?

Wait no that wouldn’t help at all lol

[empty grove]

Without loss of generality there is a point with highest x coordinate

Because you can rotate

[shadow charm]

Oh right and cut with a vertical line

And just make them overlap a bit

[empty grove]

Yep

[shadow charm]

Yeah makes sense thanks

[crude vector]

fellas, if I have Y, a subspace of a metric space X, do I say a subset of Y is closed relative to Y if it is the intersection of Y with some closed subset of X?

or does that definition only work with open sets

[empty grove]

This works with closed

Easy to prove 😌

[crude vector]

Perfect, many thanks

oh maybe I should try and prove it myself then

[fair idol]

Can someone explain to me the difference between the homotopy lifting property and it's special case, the path lifting property.

Homotopy lifting property
if f:Y->X and X has a covering space p:Z->X then there exists some g:Y->Z s.t. pg=f.

If Y is a point this is called the path lifting property. What does this have to do with a path?there's got to be some definition I'm not thinking about. I would assume if Y=[0,1] it would be called the path lifting property

[empty grove]

The definition of HLP is wrong

f needs to be a homotopy

So f: Y × [0,1] → X

[fair idol]

Oooooooooooo

I'm insane. Thank you so much ...

[crude vector]

As an alternative, does this suffice as a definition for "closed relative to Y"? $\
\text{E is closed relative to Y if whenever p is a limit point of E and p} \in Y \text{, then } p \in E \text{is also true.}$

[gentle osprey]

strad

[empty grove]

Yes

[crude vector]

Merci beaucoup

[gritty widget]

what do people use symmetric products for

also idk any visualization for it

make new tensor from old tensor

oh lol

[pearl holly]

Wtf is a symmetric product

[gritty widget]

i didnt mean that one

i meant one from normal topology

apparently symmetric product in diff top is just Sym(T tensor U)

symmetrization of any tensor product

apparently there is a notion in topology where given topology X and S_n symmetric group
SP^n(X) = X^n/S_n where Sn x X^n -> X^n by (permutation, x)->permuation(x)

https://en.wikipedia.org/wiki/Symmetric_product_(topology)?wprov=sfti1

im trying to visualize this

0 avail

its like its treating X^n as a tensor

[marsh forge]

I think it appears in the construction of the free commutative monoid on X

[bitter smelt]

point set question (its been a while):
Are pull backs of covers of a quotient space (w/ quotient space toppy) also covers of the base space?

(with pullback being pullback of the canonical map)

nvm thats a silly quesiton

[gritty widget]

Let f: X --> Y be a continuous surjective map of a compact space X onto a Hausdorff space Y. Then Y has the quotient topology Uf.

So fat I have that Y is compact Hausdorff.

[arctic relic]

Suppose f from S^1 to S^1 is null homotopic. Show f has a fixed point and maps atleast one point x to its antipode -x.

I already know of a way using the inclusion of s^1 into D^2 to force Brouwer fixed point theorem and the borusk ulam lemma but is there a more obvious way to do it?

[plain raven]

maybe apply the intermediate value theorem?

to the function f(x)-x

[gritty widget]

what is Uf

[gritty widget]

Quotient topology

Fancy-U-sub-f

[arctic relic]

Yeah that’s what I ended up doing since the induced homomorphism from fundamental group of S^1 has image of 0, using the real numbers as a covering space on to the other S^1 we also get 0, and then there exists a lift from S^1 to R and the IVT solves it from there

[gritty widget]

what is the relation? open in U_f iff open in Y implies open in X?

or wait bad phrasing

[gritty widget]

Any continuous f.

And surjective.

[arctic relic]

I think the definition of quotient topology is whatever topology makes f a quotient map

A quotient map takes preimages of open sets to open sets

Since y is hausdorff, pushing a compact set into Y via a cont map means it’s a compact set in hausdorff -> f(E sub X) is closed

[gritty widget]

idk exactly how Y is compact with quotient topology

i think i have idea

[gritty widget]

Yeah?

I think I have that ao far. Surjective f => Y is compact.

what is E_X?

[arctic relic]

A compact set in a Hausdorff space is closed. A continuous map takes closed sets to closed sets. Then, try to partition into Y into some union of sets.

[gritty widget]

open set in quotient?

[gritty widget]

Union of open points?

what open points?

[arctic relic]

Do you know what it means for a set to be saturated?

[gritty widget]

arent all points closed in Y

[gritty widget]

No.

wikipedia gives cool definition

C = f^-1(f(C))

for C a subset of domain

so like restriction of X to a saturated set is injective

[arctic relic]

X is compact. Write it as a finite sub collection of an open covering. Each one of these is closed by argument above. Then, there’s a natural partition of Y into disjoint subsets whose pre image is X

[gritty widget]

nvm thats false

is preimage of every cell a saturated set?

[arctic relic]

You know what I’ve confused myself I need 10 minutes to google some stuff

[gritty widget]

why would the partition of Y be natural?

cant the open covering map to subsets that meet?

or do we use hausdorff here?

[arctic relic]

Yes

[gritty widget]

1. a continuous map from a compact space to a Hausdorff space is closed
2. a closed continuous surjection is a quotient map
   1 and 2 imply it's a quotient map

[gritty widget]

How is it closed?

And thanks!

[gentle osprey]

Blitz

[gritty widget]

Ohhh wow

yeah. I usually work with metrizable spaces so for me it's enough that domain is compact, but this is how you do it in generality

[arctic relic]

How does this imply that Y has the quotient topology?

[gritty widget]

what is "this"

[arctic relic]

Showing that f is a quotient map

[gritty widget]

well, it doesn't exactly have quotient topology, but it's homeomorphic to a quotient of X

[gritty widget]

Does closed = open for maps?

[gritty widget]

nope

Then how does this make f open?

it doesn't, it makes it a quotient map

quotient maps don't have to be open

Hmm

For example, $x\mapsto \exp(2\pi i)$ with $x\in [0, 1]$ is a quotient map from $[0, 1]$ to $\mathbb{S}^1$

[gentle osprey]

Blitz

[gritty widget]

it's closed, surjective, continuous, but it's not open

[gritty widget]

Are you a grad student?

to give you an example of a map that's an open quotient but is not closed, we can cook up an example using products and projection maps

I think $\pi:\mathbb{R}^2\to\mathbb{R}$ given by $\pi(x, y) = x$ should work

[gentle osprey]

Blitz

[gritty widget]

for example, consider graph of tan in (-pi/2, pi/2), then it's closed and its projection is an open interval

so that's a map which is open continuous and surjective, but not closed

[gritty widget]

Thanks for all the help!

Could I ask another question?

sure

Thx

So, if I take S1 x I over the eq. relation (x, y) ~ (s, t) iff xt = ys, how would I show this is homeomorphic to the unit disc?

And why would I need the quotient at all?

I can get it visually.

Take the map $f:\mathbb{S}^1\times \mathbb{I}\to B^2$ given by $f(x, t) = xt$

[gentle osprey]

Blitz

[gritty widget]

B^2 is the unit disc? 👀

this is a quotient map, and from the fundamental property of quotient maps, $B^2$ is homeomorphic to the quotient of $\mathbb{S}^1\times \mathbb{I}$ by the "kernel" of $f$, so precisely by the relation $\sim$

[gentle osprey]

Blitz

[gritty widget]

2-dimensional unit disk, yes

[gritty widget]

Hmmm

It's (S1 x I)/~

where kernel of a map $g$ means something like $\text{ker}(g) = {(x, y) : g(x) = g(y)}$

[gentle osprey]

Blitz

[gritty widget]

so it's a little different than in linear algebra

Okay...

So basically take (x, y) in the S1 x I and map it to xt in the unit disc?

How does that work?

S^1 x I is compact, and the map is continuous and surjective

so it's a quotient map

But you have to use (S1 x I)/~, right?

use it?

Ohh, show S1 x I ~= (S1 x I)/~?

[gritty widget]

The problem is to show that (S1 x I)/~ ~= unit disc

that $f$ is a quotient map implies that $((\mathbb{S}^1\times\mathbb{I})/\sim)\ \cong B^2$

[gentle osprey]

Blitz

[gritty widget]

Huh...

How is thay implied?

it's a general fact

if $f:X\to Y$ is a quotient map then $Y\cong X/\text{ker}(f)$, and the homoemorphism is given by $\bar{f}([x]) = f(x)$ for $[x]$ an equivalence class of $x\in X$

[gentle osprey]

Blitz

[gritty widget]

I see.

I have never been introduced to that concept.

you should read a book about general topology

[gritty widget]

I'm in my first intro to topology class.

Sorry.

So

Bascially

I can use that map to show (S1 x I)/~ ~= (S1 x I) to start?

use LaTeX, I don't understand

The quotient is homeo to the space?

depends on what space

(S1 x I) is homeo to (S1 x I)/~

nope

B^2 is contractible while S^1 x I isn't, which proves the two spaces aren't homeomorphic

[gritty widget]

I'll have to think about this some more and read over what you've wrote to understand it fully, but thank you for your help.

S^1 x I/~ is homeomorphic to B^2, this was settled already

but B^2 is not homeomorphic to S^1 x I

the reason is basically that, we can shrink B^2 to a point, but we can't shrink S^1 x I to a point, the reason begin we can't do it with S^1

[gritty widget]

How can I prove it wiyhout using kernels?

lol

We haven't used that in class.

I just defined it

Can I use compactness or Hausdorffness somehow?

[gritty widget]

I mean, I don't think that's the solution he wanted.

why don't you write your own solution

[gritty widget]

Yeah...

Sorry, I'll work on it some more. Thanks for giving me ideas.

[junior coral]

Hi, does anyone know how to compute $\pi_1(S(O(p) × O(q)))$?

[gentle osprey]

sheeppunk

[junior coral]

For context, I'm trying to find the fundamental group of the Lie group $SO(p, q)$ by examining the fundamental group of its maximal compact subgroup$S(O(p) \cross O(q))$.

[gentle osprey]

sheeppunk

[gritty widget]

Homotopy equivalence preserves the amount of path-components of a space

Is it true that any two subspaces of R are homotopic iff they have the same amount of path-components

[empty grove]

Should be, any connected subset of ℝ is contractible

[shadow charm]

Would the space {1,1/2,…,1/n,…} be homotopic to {1,2,3,…} though

[empty grove]

Ah right it's not

Nice

[shadow charm]

Maybe they were asking for finitely many path components though

[hollow harbor]

hmm? do you not need to add 0 to the first one for them to be nonhomotopic?

uhhh

[empty grove]

Ye

[hollow harbor]

i'm confused

ok

yeah

[shadow charm]

Yeah should’ve specified oop

[hollow harbor]

they're homeomorphic even

but then yes definitely

[shadow charm]

Yeah I mean I think homotopy equivalent and homeomorphic are the same in these cases no?

[hollow harbor]

yes i suppose so

idk when these are the same

presumably for totally disconnected spaces?

[empty grove]

Ye since no non trivial paths

[hollow harbor]

mhm

ok then i should have said totally path disconnected i guess

those might be the same thing

[empty grove]

I doubt it

Totally disconnected seems stronger

[shadow charm]

Why do we need a notion of connectedness and path connectedness

Is connectedness useful?

[empty grove]

Yes

[shadow charm]

(As in not path connectedness)

[empty grove]

It's a property preserved by continuity and lets you examine functions out of a space

[shadow charm]

Yeah but it feels like connected spaces that aren’t path connected exist for counter example’s sake

[empty grove]

Functions out of a space correspond to pairs of functions out of any separation of it

[shadow charm]

I probably ought to read up on connectedness before opening my mouth lol

[empty grove]

Because in ℝ^n these turn out to be equivalent for open sets and most of the time when we work with functions we take their domains to be open subsets

[empty grove]

Responding to this

[hollow harbor]

There are plenty of things which should obviously be connected which are not path connected

And which are relevant outside of point set

E.g. functions with bounded / unbounded variation come up all over analysis

[marsh forge]

on the flip side if all you care about is algebraic topology, path connected and connected tend to coincide

so you are not entirely off base

I would agree both notions are important, though

[shadow charm]

I see

[gritty widget]

I've used this recently to show that the topologists sine curve is not an absolute neighbourhood retract

[empty grove]

What's an absolute NR?

[hollow harbor]

And you still want to talk about their connectedness properties a lot. E.g. sometimes I want to say that a weierstrass function has the intermediate value property to pick a relevant point even if its image is not path connected

Oh yeah moldi theres our example

Image of a sufficiently nasty weierstrass function is connected and totally path disconnected

[shadow charm]

Why is it’s image not path connected

[hollow harbor]

It wiggles too fast

[gritty widget]

It's a space which is a neighbourhood retract of any space it embeds to as a closed subspace

[hollow harbor]

Depends on the weierstrass function

[shadow charm]

But isn’t the restriction of the function to an interval a path between the two points