point-set-topology

[silver umbra]

this obviously doesn't generalize to higher dimensions

[red yoke]

I'm assuming 1.17 is "{gm} a sequence in Γ"

If for each C you draw the disk E in the region bounded by the sphere

The region of H bounded by the E's and the complement in S² of the D's seems to be a fundamental region for H

So the limit points should correspond to the sequences of shrinking circles similar to ones in figure 4

Which is a Cantor set

One way to think of this is that two copies of the fundamental region are adjacent iff they are related by an inversion, so the dual graph (with copies of fundamental region of H being vertices and adjacent regions beings edges) should be an r-regular tree

(idk this is based on intuition, maybe someone can verify)

[tough jay]

Hi, in my teacher's solution for computing the homology groups of the 2-torus using the decomposition with two cylinders I don't understand how he concludes that H1(T)=Z+Z given the fact that neither the map j1_* + j2_* nor the boundary map ∂ are isomorphisms (∂ is not even injective). Did I miss something in his arguments ?

[silver umbra]

wait sorry could you clarify what you mean by this

[red yoke]

Like the disk in H bounded by C

Rather than on S² bounded by C

[silver umbra]

okay well we can just take S^2 to be the boundary of H^3

how does a circle C in the boundary of H^3

bound a disk in H^3?

[red yoke]

Well H looks like a 3-ball

[silver umbra]

right

[red yoke]

So you draw the disks E like you normally would in R³

[silver umbra]

ah so a flat disk

or would it have to be flat in the hyperbolic sense

[red yoke]

Actually ye

[silver umbra]

okay i get what you mean

so then what..

[red yoke]

So a corresponding inversion should flip across this disk

[silver umbra]

im having a little bit of trouble visualizing

[red yoke]

Imma draw

[silver umbra]

i guess im not fully grasping how a circle inversion (a mobius transform of C) acts on elements in H^3 treated as the 3-ball

[silver umbra]

^

oh also

here's the general definition of schottky groups

here we can restrict jordan curves to just circles

so you take pairs of circles with disjoint interiors C_1, C_1',..., C_g, C_g'

then take the maps T_1,...,T_g

which are mobius transforms mapping the outside of C_i to the inside of C_i'

this resource gives this very vague explanation as to why the limit points is totally disconnected

[red yoke]

Something like this, for H² and generated by 3 disks

[silver umbra]

okay so the three big curves you've drawn are the disks bounded by three circles on the boudnary

so how do points in the ball

get corralled

into those smaller and smaller circles?

[red yoke]

And then reflecting them across each other generates more curves

If you take the fundamental region and keep reflecting across one of the curves on its boundary then you can get to any copy of the fundamental region

And since the orbit of a point intersects each copy of the fundamental region exactly once, this determines the limit set

[silver umbra]

i'm not sure im following those steps

what do you mean by reflecting the disks across each other?

[red yoke]

Inverting the curve across another curve

[red yoke]

Say, inversion by large black curve takes the large blue curve to the small blue curve on top

[silver umbra]

so is the fundamental region the interior region cobounded by all three of those curves?

[red yoke]

Ye

[silver umbra]

okay that makes sense, since inversion is an involution

why are we taking the images of each of the curves

with respect to the inversions

[red yoke]

They divide H² into copies of the fundamental region

[silver umbra]

wait, so actually each bounded region is a fundamental region? hmm hold on

right because the orbit of a point would be

a point inside curve 1, curve 2, curve 3, and in the inside part

ahh okay, so then i see how reflecting all the points inside the black curve

wrt to the blue curve

results in them all being placed inside

the smaller black curve that you've drawn inside the blue curve

okay im beginning to understand your picture at the very least

[silver umbra]

what exactly do you mean by this though?

the first part is obvious

[red yoke]

The limit points can be obtained by picking any point and reflecting it closer and closer towards the boundary

The limit point you end up with only depends on which sequence of reflections you use, but not the starting point

Since the curves shrink to 0 anyway

[silver umbra]

wait i think limit sets of kleinian groups in general

only depend on the sequence of transformations

and not on what you start with

i think because they always act transitively?

unless im very mistaken here

[red yoke]

[silver umbra]

wait but why does the fact that the curves shrink to 0 mean that the starting point doesnt matter?

[red yoke]

Cuz the sequence of fundamental domains shrinks to the same point

[red yoke]

Maybe this follows from some hyperbolic space metric argument

[silver umbra]

ok wait which transformation

is sending that point to the other

?

[red yoke]

Is actually precomposition by large blue curve

[silver umbra]

arrows here indicating reflection wrt the blue curve

ahhh okay i see how you got it

take middle point, reflect across blue curve, reflect across black curve, reflect across blue curve again

[red yoke]

Ye

[silver umbra]

i understand the visual intuition but

how do you actually prove that the limit set is totally disconnected

like the curves remain disjoint

and go to zero

[red yoke]

Well each limit point corresponds to a shrinking sequence of curves

So it's a Cantor set

Or at least very close to the Cantor set construction

[red yoke]

Ye ig

[silver umbra]

i mean yeah i see the cantor set construction

you take the interior of a curve

then you remove the two other curves inside it

then you keep going

that's convincing enough i suppose

[red yoke]

Imma go hunt dinner soon
Thx for the fun read

[silver umbra]

yw! i am super new to kleinian groups but am trying to get my knowledge up to speed for related work. ty for the help

[somber phoenix]

I cannot get through this thing, for like 2 weeks...

Why is $L(S) = L(\overline{S})$? where S is a subspace of $C([0,1]^n)$, and L some functional.

[gentle osprey]

Elin

[coarse night]

because L is continuous

L(S)=0

[somber phoenix]

maybe it's measure theory... for $\mu [(0,1)] \neq \mu ([0,1])$, but I guess measure theory says they are the same, so there must be somewhere there.

[gentle osprey]

Elin

[somber phoenix]

I guess the idea is that the measure of a single point is mu null, and I guess the collection of these points are also mu null.

[coarse night]

?

[somber phoenix]

and how does that change it? the distance to the limit point is guaranteed to be less than epislon, but not zero.

[coarse night]

what is your question exactly?

you asked something about L and commented something about μ

[somber phoenix]

Yeah, something has to say that L(S) = L(clsr(S)), but just continuty doesn't really say that

[coarse night]

why not?

[somber phoenix]

at least I've never seen anything about it

[coarse night]

Exercise

Also use proposition 5

[somber phoenix]

but this is only the case if L(R) = 0? does not seem to hold otherwise, I'm thinking you need to use a clopen set in "Y"?

[rancid umbra]

what?

[alpine nest]

L is defined to be zero on R ,which is a superset of S

so it's zero on both without the need to invoke topology

but also it's true that a continuous function that's zero on a set is also zero on its closure

[alpine nest]

ok ,topology is needed to know that the closure contains the original set

[heady skiff]

this is probably a trivial question lol, but if all the vertices of a simplex sigma lie in a simplex A_k, do we have sigma in A_k since we can consider those vertices a subset of the vertices of A_k?

[knotty vine]

In a simplicial complex, yes. Every simplex is uniquely defined by its vertices and every subset of vertices of a simplex defines a face

[heady skiff]

cool, thanks

ok, this is probably another trivial question, but can we consider the union of two simplices to be another simplex? or is this something we just choose to distinguish as a complex

wait do we just take the two vertices of the simplices and then let their union be the simplex spanned by their vertices

or rather, it ends up that they're equivalent

gah

[strong swift]

Hello, I'm stuck with these two affirmations of Hatcher's book, can someone explain?

[heady skiff]

wait... is the converse of this (intersection of all topologies on X that contain A contained in the topology generated by A) literally just that that the topology generated by A contains A, so it's a subset of it

💀

i deadass spent over an hour thinking about the converse when it's literally just ohhh intersection small so intersection in everything

[ebon galleon]

The topology T_A generated by A (the explicit description) is contained in every topology containing A, and contains A, which I think is what you mean

[heady skiff]

yeah, i think?

Let $\mathcal{T}\mathcal{A}$ be the topology generated by $\mathcal{A}$, and let ${\mathcal{T}\alpha}$ be the collection of all topologies on $X$ which contain $\mathcal{A}$. Let $O$ be an open set of $\mathcal{T}\mathcal{A}$. Then for each $x \in O$, we can find a basis element $A_x$ such that $x \in A_x \subset O$. Since $\bigcup{x \in O} A_x = O$ and $O = \bigcup_{x \in O} A_x \in \cap \mathcal{T}\alpha$ ($\mathcal{T}\mathcal{A}$ contains $\mathcal{A}$ and is a topology) we have $O \in \cup \mathcal{T}\alpha$. Hence $\mathcal{T}\mathcal{A} \subseteq \cap \mathcal{T}\alpha$. Conversely, if $O$ is open in the intersection of all topologies containing $\mathcal{A}$, then $O$ is open in every topology; in particular, it is open in the topology generated by $\mathcal{A}$ (since $\mathcal{T}\mathcal{A}$ contains $\mathcal{A}$). Hence $\cap \mathcal{T}\alpha \subseteq \mathcal{T}\mathcal{A}$.

[gentle osprey]

okeyokay

[heady skiff]

sanity check: the closed interval [0, 2] in R would be open under the lower limit topology, but not open under the K- topology right?

and then to show that the K-topology is not contained within the lower limit topology, it suffices to just take (0, 1) - K right

where K consists of all 1/n such that n is a positive integer

[red yoke]

[0, 2] is not open in lower limit topology

(0, 1) - K is open in the lower limit topology

[heady skiff]

Wait how

[red yoke]

What's an open neighbourhood of 2

[heady skiff]

Oh ya true lol

[ebon galleon]

That's kinda the point of the lower limit topology btw, you get the half open intervals [a,b) containing the lower limit, but not the other half open intervals (a,b]

[heady skiff]

No ya idk what I was thinking

[eager vigil]

Hey guys, Idk if any of you have worked with Armstrong's textbook before, but I'm slightly confused here: here L is a simply connected, path connected subcomplex of path-connected K and we want to compute the edge group E(K,v). He says that we can therefore ignore simplexes of L and thus says the generator g_ij = 1 if v_i v_j span a simplex of L.
However, isn't this only true if we take say v_k v_i v_j v_l and replace it with v_k E_k^{-1} E_l v_l where E_n is a path in L connecting the basepoint v_0 to v_n?
Idk if my question makes sense

[eager vigil]

Hmm, or I guess maybe it doesn't matter after all...

[coral pawn]

Is there a way to do this without using Kunneth theorem?

[red yoke]

By cellular homology

[coral pawn]

What cellular structure would be the least painful here?

[hexed steppe]

…how many different ones do you know

you can basically choose at random

[coral pawn]

let's take the usual one on S^1 (one 0-cell e^0 and one 2-cell e^1) amd RP^2 (one 0-cell j^0, one 1-cell j^1, and one 2-cell j^2)

with the attaching map on RP^2 being the double loop around S^1

The cells for the product is the product of the cells

So you'd get something like 0 --> Z(e^1 x j^2) --> Z(e^1 x j^1) (+) Z(e^0 x j^2) --> Z(e^0 x j^1) (+) Z(e^1 x j^0)

(don't need to compute the first homology because of the fundamental group)

How would you figure out the maps here in terms of the generators?

[hexed steppe]

by thinking about it idk

[coral pawn]

[hexed steppe]

how do you compute them in general

in terms of the attaching maps

[coral pawn]

Yeah but attaching maps now aren't so straightforward

[hexed steppe]

why

[coral pawn]

What is the attaching map of e^1 x j^2?

For e^1 its the constant map

And for j^2 its the map that goes twice around S^1

[hexed steppe]

ok

i agree

[coral pawn]

What will it be for the product tho?

You can't just take the product of the attaching maps

[hexed steppe]

it will be the product?

why?

what else could it be

[coral pawn]

Because the boundary of the product is not the product of the boundary

[hexed steppe]

idk i dont feel like thinking about this but you can see hatcher

the attaching maps are the products

[hexed steppe]

appendix A i think

[coral pawn]

You're confusing characteristic maps for attaching maps

[coral pawn]

@red yoke Do you have any suggestions?

[red yoke]

Ch.0 should have covered product complex constructions iirc

E.g. ∂(A×B) = A×∂B ∪ ∂A×B for cells A, B

[hexed steppe]

you are right

though i am guessing in this case it will work

[coral pawn]

This seems way too painful

at least to write down

[red yoke]

There's 6 cells

[coral pawn]

Can someone help me write down the boundary maps?

Like I'm not sure what e^1 x j^2 maps to

[coral pawn]

I computed the homology groups and I'm getting H_2 = 0 and H_3 = Z

Using Kunneth, this is definitely not the case so I messed up but I cannot figure out why

[unreal stratus]

why avoid Kuenneth 😭

But fr I'll have a think

I think a nice way to do this is to decompose $\mathbb{RP}^2$ using the affine charts, so like opens $U,V \cong \mathbb R^2$ with $U \cap V \simeq S^1$. Then Mayer-Vietoris tells you $H_n(S^1)^{\oplus 2} \to H_n(S^1 \times \mathbb{RP}^2) \to \tilde{H}{n-1}(T^2) \to \tilde{H}{n-1}(S^1)^{\oplus 2}$ is exact

[gentle osprey]

Potato E-Girl

[unreal stratus]

In particular, if n >= 3 then H_n(S^1 x RP^2) = \H_(n-1)(T^2)

n=1 is easy by pi_1, for example, so that reduces you to H_2

But this thing tells you H_2 is a submodule of Z^2, so in particular free, and then the rank can be read off from the sequence I've given

I think that works nicely

[hot locust]

What is an example of a metric space whose closed and bounded subset isn't always compact?

[tiny ridge]

Take open disk of radius 1 with the Euclidean metric. The ball of radius 2 is bounded, as well as closed.

[hot locust]

but shouln't it be compact according to heine borel theorem?

[red yoke]

Heine-Borel works on R^n

In general totally bounded and complete iff compact

[hot locust]

oh wait euclidean metric isn't the usual metric?

wait wait

[red yoke]

Open ball is not complete

[hot locust]

ahh

makes sense

thanks

[valid zenith]

more generally, take any noncompact metric space (X,d) and equip it with the metric d'(x,y) = min{d(x,y),1}. you should check that d and d' both generate the same topology on X.

[gentle osprey]

DarQ

[red yoke]

No

The requirement is just that n-dim cells stick to the (n-1)-dim skeleton

Through any continuous map from its boundary to the skeleton

[broken nacelle]

I know but I can't imagine the continuous map attaching to only part of a cell in the skeleton

[red yoke]

The map need not be injective

You can have it be constant

[broken nacelle]

Oh wait bruhhh

So if your 1-skeleton is just a 1 cell you can attach a 2-cell's boundary along any point in the 1 cell?

[red yoke]

Ye

[broken nacelle]

[red yoke]

Then you have added a sphere

[broken nacelle]

You can't do that with delta complexes, can you?

[red yoke]

No

You rigidly attach faces to faces for delta complexes

• orientation too

[broken nacelle]

Yea

I see

Aight

Thx

[merry geode]

Where does S^1 \times RP^2 come from there, may I ask?

[unreal stratus]

Just product the stuff I said w S^1

[merry geode]

Oh. Sorry I was stupid

[unreal stratus]

No dw

[narrow cairn]

i dont really understand the setup here, what is X?

[trail charm]

likely just an arbitrary space

[narrow cairn]

why does it need to exist

[trail charm]

because that's how a path is defined? not quite sure what the question is

[narrow cairn]

are we just like embedding delta^2 into X in some fashion for some reason?

[trail charm]

there's a name for this kind of map but my mind is blanking gimme a sec

[narrow cairn]

like im really confused as to if i can just say "the space is contractible so all closed paths are nullhomotopic" or not

[trail charm]

this kind of map is called a "singular n simplex"

[narrow cairn]

i dont understand why we use X and not just consider the maps in delta^2

seems weird and arbitrary

[fading vale]

What do you mean the maps in Delta^2?

[trail charm]

https://math.stackexchange.com/questions/3699502/intuition-behind-singular-n-simplex

delta^2 is a space

[narrow cairn]

also these maps seem not nullhomotopic, they seem like full loops

[trail charm]

studying maps Delta^n -> Delta^n isnt really useful because they're homeomorphic

[narrow cairn]

the ones the problem uses eps for

[fading vale]

The purpose of this construction is to use the n simplices to study an arbitrary topological space

[narrow cairn]

ah so its studying the way that this "loop" gets mapped into X?

how would i even start to show that its nullhomotopic?

oh it gives a hint

let me look at th 1.6

issue is that theorem mostly regards free homotopies

this is th 1.6 (f is any map from S^n to a space Y and its "the following are equivalent")

[heady skiff]

I’m working through Armstrong rn, and it’s quite terrible

[heady skiff]

wait sorry how is this the case

because if you have any interval [a, b) contained in (0, 1) - K, don't you have points of the form 1/n for positive integers n

[heady skiff]

is there any characterization of the open sets of the cofinite topology? i suspect that it's sets of the form R - O where O is finite, and of course R and the empty set

[umbral panther]

That’s the definition. What other definition do you have you want to compare it with?

[heady skiff]

well a set O under the cofinite topology is open iff R - O is finite

so I'm saying, that we can take the sets to be R - U themselves + R + empty set

Because R - (R - U) = U is finite

[hot locust]

Why is arbitrary product of discrete spaces not necessarily discrete?

[knotty vine]

cantor space

[hot locust]

I was trying to prove it by showing that singleton basis won't be a basis for it but idk if it makes sense.

[hot locust]

Wait, what about it?

[knotty vine]

Cantor space is 2^omega, ie an infinite product of discrete spaces

but it's not discrete

[hot locust]

What kind of discrete spaces can you please elaborate a bit?

[coral pawn]

What charts are you using?

[unreal stratus]

standard ones

as in like U_i = { [x_0,...,x_n] : x_i not 0}

[knotty vine]

2 is the discrete space of two points

[unreal stratus]

(not necessarily discrete)

[knotty vine]

More general discrete spaces will result in the cantor space as well but I forget the details

[coral pawn]

The standard one requires 3 copies of R^2 doesn't it?

[unreal stratus]

Uhhh yes I messed up lol

My bad

😭

[coral pawn]

Can your solution be salvaged?

Well

[hot locust]

Edited 👍

[unreal stratus]

well you still only need two of those charts to cover ig right

but i didn't think about what the intersection looks like

[coral pawn]

It's gonna be R^2 - a line I think?

[unreal stratus]

Also if you're interested in a more like abstract reason why you shouldn't expect arbitrary products of discrete spaces to be discrete: given a set $X$, the discrete topology on $X$ is the largest topology you can put on $X$. But on a product of spaces, you put the smallest topology such that projections are continuous. So these two notions compete in some sense

[hot locust]

Ahh

[gentle osprey]

Potato E-Girl

[coral pawn]

the line where one coordinate is 0

[unreal stratus]

In particular, note that the product topology is strictly contained in the box topology generally - let alone the discrete topology

[coral pawn]

Since it's [x_0,x_1,x_2] where two of them are non-zero

And this is homotopic to the two point discrete space

This could work now

[unreal stratus]

uh

Sorry what i said was right lol

[coral pawn]

well removing a line from R^2 leaves us with two copies of R^2 basically

[unreal stratus]

Okay true lol

[coral pawn]

and these are disjoint

[hot locust]

But sometimes it can be discrete right

The product

[unreal stratus]

I'm pretty sure it is never discrete if you have an infinite product unless like lots of the spaces are singletons or empty sets lol

Because uh if $X$ is a space and $\emptyset \subsetneq U \subsetneq X$ any proper open subset, then like e.g. $\prod_i U \subseteq \prod_i X$ isn't open

[gentle osprey]

Potato E-Girl

[unreal stratus]

And then this generalises to, say, if every space in your product is discrete and has cardinality > 1

For finite products, yes it'll still be discrete

And also silly cases, like an infinite product of singletons lol

but you can always pull out the singletons into a single point and then be left with a product of bigger sets, and if that product is still infinite then argument above still works

[knotty vine]

Any product of compact spaces is compact, but the only compact discrete spaces are the finite ones

[unreal stratus]

[i should say proper, non-empty in words, but the notation is clear]

Lol was gonna say your solution requires choice / more tools, but in fact so does any proof aha

since ofc otherwise the product could be empty and hence somewhat vacuously discrete

nice

[coral pawn]

actually i don't think this is true

[unreal stratus]

i agree

[coral pawn]

you miss a point

[unreal stratus]

i was thinking too mucha bout RP^1 today lmao

i think also like

lol okay this is nice

well yeah so mod 2 cohomology of RP^n is F_2[x]/x^(n+1) so in particular RP^n can't be covered by < n contractible opens

¯_(ツ)_/¯

[coral pawn]

Wait

S^1 is RP^1

We could try doing the same thing with that

Something seems wrong here

[hot locust]

Is the converse of Product of T1 spaces is T1 true?

[coral pawn]

Okay so let U_0 and U_1 be the usual covers of S^1

[hot locust]

If you have a product of spaces that is T1, then is it necessary for the components to be T1 too?

[coral pawn]

Then U_0 x RP^2 and U_1 x RP^2 cover S^1 x RP^2

U_i is homotopic to a point

U_0 cap U_1 is homotopic to the two-point discrete space

This gives an exact sequence H_n(RP^2 U RP^2) --> H_n(RP^2) (+) H_n(RP^2) --> H_n(S^1 x RP^1) --> H_{n-1}(RP^2 U RP^2)

[heady skiff]

is this correct?

[coral pawn]

But using the disjoint union of homology thing, this would imply that the homology of RP^1 x RP^2 is trivial

[coral pawn]

@unreal stratus Do you see anything wrong here?

[heady skiff]

here, could you just take an irrational number p, so that [p, p + e) is open under the lower limit topology but not open under the topology generated by C

[coral pawn]

nvm i figured it out

[heady skiff]

what does this rigorously mean? does this just mean, say let L = [0, 1], so there's a simplical approximation s: |L^m| -> |K| of the path f: |L^m| -> |K|, and we let this "edge" path be s?

other than that, i can't really make sense of this garble since a path is a function and an edge path isn't...

also can somebody help me make the last two sentences precise

i can't believe this is in a formal proof

so we would consider |T'| to be |T| union the simplex spanned by v_i and v_i + 1 right?

[knotty vine]

I think so? A space is T1 if every map from the Sierpinski space S is constant. Suppose for contraposition that one of the factors is not T1, then there exists a nonconstant map from S to it. You might be able to construct a nonconstant map to the product from this and hence conclude that the product is not T1. I think this will use choice again

[ebon galleon]

Yeah that works

Just need a choice of element in each component

[knotty vine]

Could we do something more general using the lifting property formulation of the seperation axioms and show that this works for all of them?

How often does it happen that a property which is preserved by limits is also "reflected" by them?

I never ever think about this, odd

[knotty vine]

Would just LEM be enough? If a component is empty, then the product is empty which is trivially T1

[ebon galleon]

Well

Okay

You need all the spaces nonempty

Otherwise take a non-T_1 space and product it with the empty set

[knotty vine]

oh right

[knotty vine]

Every edge path can be made into a path by linear interpolation and every path can be made into an edge path by simplicial approximation. Furthermore, turning an edge path into a path and then back into an edge path like this yields the original edge path and turning a path into an edge path and then back into a path yields a homotopic path to the original path.

So we can safely conflate them if we only care about paths up to homotopy

[ebon galleon]

I'm not sure whether there's a general reason for when this happens

[knotty vine]

Ya

[knotty vine]

This is not a formal proof and almost all proofs that mathematicians write are not formal either

[heady skiff]

i thought proofs are formal by nature

anyways you know a lot more than me, so yeah you're probably right

[ebon galleon]

Nope

Formalizing things is a bitch

So usually it's not done like purely formally

[heady skiff]

hmm interesting

[knotty vine]

In addition, most formal proofs are unintelligible

[heady skiff]

yea i'm sure producing explicit homotopies or whatever is a pain

so then like

if that's the case, how can we verify something is true

wait that sounds dumb

[knotty vine]

no its a good question

[heady skiff]

because don't you need some level of rigor to verify things

is it just an experience thing

[knotty vine]

If a piece of text convinces most mathematicians (that understand that piece of text) that some fact is true, then you could say that it's a proof

Or maybe something is a proof if the act of understanding it also means being convinced of its truth?

Anyways, we're not really doing philosophy here, just mathematical practice

[ebon galleon]

There's a reason that a lot of details are stated informally in some stuff: we know they're true because someone checked them, but it's a pain to reproduce again and again, and provide no insight. So you end up with more "handwavy" proofs like this.

[knotty vine]

(we hope someone has...)

[ebon galleon]

True

Sometimes people claim things and thorough proofs have never been published

[heady skiff]

i see

[knotty vine]

The "not providing insight" is important: if your proof is completely bloated with boring details then not many people will understand it and hence we might even say that it's not a proper mathematical proof

[ebon galleon]

cough cough page long logic symbol proof cough cough

[heady skiff]

isn't that just a conjecture lol

[knotty vine]

The line gets blurry with things that seem obvious (at first sight)

[knotty vine]

[ebon galleon]

No I mean they claim that they have a proof but it never sees the light of day. Such things end up as like "folklore" results: they're well known, but nobody has an actual reference for them

[heady skiff]

oh yeah i've heard of those

[ebon galleon]

An example I came across: the fact that the category of delta-generated spaces is locally presentable was a folklore result for a few years: Smith claimed it when he first suggested them (in a paper which literally doesn't exist), but no proof existed for a few years

[knotty vine]

From Barwick "The future of Homotopy Theory"

[heady skiff]

that is very interesting yeah

i always thought the reason that like

proofs of some results were hundreds or even thousands of pages long is because they involved breaking the problem into small steps

coupled with tedious verifications

[ebon galleon]

Yeah I've read that note... Note sure I've done quite enough homotopy theory yet to fully understand this "culture" referenced, but I thought there were some valid complaints

[heady skiff]

ryx this is kinda of a vague question but what subject do you think interests you the most

you seem to be a big fan of topology and category theory lol

[ebon galleon]

Well, my work so far has mostly been in topology and category theory (in particular, homotopy theory), so that's why lol
But I generally find most things interesting

[heady skiff]

oh cool

[ebon galleon]

I'm not "oh I only do algebra I hate analysis" no I quite like my analysis courses too

[heady skiff]

based

real analysis is interesting but i'm sure there's a lot of other cool stuff involved in analysis too

[unreal stratus]

Analysis is fun

[unreal stratus]

what are delta-generated spaces again lol

hm

oh fair enough

[heady skiff]

how did they get that the fundamental group of X is a free group on n generators?

[ebon galleon]

X is delta generated if it has the finest topology with respect to continuous maps from all simplices to X

(In fact, I believe it turns out to be sufficient to look at maps from \bbR to X)

[heady skiff]

anyone?

[unreal stratus]

ye nice

characterising them as small colimits of simplices is cool

[plain raven]

Whoa. Small?

Hm.

[ebon galleon]

Yes

[unreal stratus]

yeah according to nlab

[ebon galleon]

This is due to Vogt

[unreal stratus]

Oh really

[ebon galleon]

I believe

[unreal stratus]

Okay cool I wasn't sure if it'd be some trivial consequence

[plain raven]

Wait, I knew this.

[unreal stratus]

of your definition

but doesn't seem it

[ebon galleon]

Vogt has a paper "Convenient Categories of Topological Spaces for Homotopy Theory "

[plain raven]

Did I ? I proved something similar to this independently like a year go

[unreal stratus]

What are delta-generated spaces used for? Me new to this

[plain raven]

before i knew about vogt's paper

[unreal stratus]

I imagine yeah a convenient category

Or do they have any particular applications in homotopy theory

[ebon galleon]

In which this is shown in generality, for a notion of (insert subcategory here)-generated spaces

The smallness here comes from Delta being small

[unreal stratus]

vogt and boardman feel too underated

[ebon galleon]

(blah blah small dense subcategory or whatever)

[unreal stratus]

I more meant that you get all small colimits rather than just simpler ones

[plain raven]

Ok nvm. The thing I figured out was that the unit interval suffices.

You don't need higher simplices

[unreal stratus]

Oh sure

[ebon galleon]

Oh yeah, that just comes from being coreflective and full

Like, knowing something is (co)reflective in a bicomplete category is very strong for this reason

[hidden crag]

This is a standard exercise, van Kampen/ convince yourself by looking at concrete loops

[ebon galleon]

Yeah, I forget exactly who this is due to. Some combination of Haraguchi, Kihara, or Christensen-Sinnamon-Wu, I think.

[ebon galleon]

Anyways, these ones being locally presentable is very nice. Other convenient categories such as k-spaces or CGWH are nice in their own regard, but they are not locally presentable

[merry geode]

Why do we have both homology and cohomology

When one can be defined as a dual of the other.

Cohomology has cup product so why not just go with that?

[hidden crag]

Why discard the other

[merry geode]

Idk, too many things to learn?

[plain raven]

Can you prove this as follows:
Let X be Delta-generated.
You have a comonad G which spends each X to \coprod_n\coprod_{f : \Delta^n -> X} \Delta^n
You have a canonical map from G(X) -> X
Then take the kernel pair of this, call that Y. You have a canonical map G(Y) -> Y
Now you want to prove that X is the coequalizer of the two maps G(Y) -> G(X)

Which i guess is equivalent to saying it has the quotient topology

i.e. iff the map G(X) -> X is a quotient map.

Ok nvm i'm convinced this works.

Am I?

Yes

[knotty vine]

Really?

[merry geode]

I hate endless learning before ever getting to research level

[plain raven]

Absta there's something interesting called the cap product which describes a relationship between homology and cohomology. It shows up in the theory of Poincare duality.

If you are studying continuous maps from a simplicial complex X into a more pathological space Y, it is easier to use simplicial homology than cohomology.

[merry geode]

Ah I see, thanks

[ebon galleon]

Vogt shows that there's a canonical way to identify that coequalizer so that it has the underlying set of X, so then it really just becomes a question about topologies on X

iirc, at least

[merry geode]

Perhaps I would never get to the research level

[plain raven]

Yeah that makes sense.

It is definitely clear by construction that the underlying set agrees with X.

I had to think about it for a minute to be sure that the quotient topology is correct, but after thinking about it for a minute i think that it just follows immediately from the definition of delta generated.

[ebon galleon]

yup

[knotty vine]

Homology groups are the homotopy groups of chain complexes (i.e. simplicial R-modules, which are Kan complexes)

[ebon galleon]

Anyways, yeah they're a pretty cute category of spaces. I know there are some more specific reasons to prefer categories such as CGWH (something about quotients of spaces in CGWH being in CGWH again? As opposed to modified topologies on them), unsure if this also holds in DTop

[merry geode]

Huh, homotopy groups?

[ebon galleon]

But I know a bit too much about specific types of spaces lol, I gotta start learning other things

[merry geode]

Oh. So simplicial complex is just one of Kan complex

[knotty vine]

Simplicial sets are a nice kind of topological spaces, if they have the Kan property, there is an easy definition of homotopy groups for them which corresponds to the standard one. If all the sets are actually abelian groups (or R-modules) and the boundary/degeneracy maps all homomorphisms, then it is always Kan. In addition, these simplicial R-modules correspond precisely with chain complexes. The homotopy of the simplicial R-module is the homology of the corresponding chain complex.

[knotty vine]

A simplicial complex is usually something different entirely.

[merry geode]

Hmmm, never faced that notion

As I am just beginning to peek into algtop

[ebon galleon]

it's okay, I still don't know what (co)homology is

[merry geode]

Ah so simplicial complex is just too simple in this area

[knotty vine]

Yeah, it's more a bit of motivation than anything precise

[knotty vine]

Not simple per se, just not that common in homotopy theory

[merry geode]

Hmm

[plain raven]

It's not that important to distinguish between a simplicial complex and a simplicial set until you start working with concrete theorems about them. At the level of intuition they're basically the same, it's just slightly different formalizations.

There's a fully faithful embedding of simplicial complexes into symmetric simplicial sets, so I think of simplicial complexes as nice symmetric simplicial sets.
A symmetric simplicial set is not a simplicial set but it's very similar.

A simplicial set you can think of as being like. Just a response to the fact that there's not a sensible Cartesian product of simplicial complexes.

[knotty vine]

And we get function spaces for free!

[plain raven]

Yeah, that's also a nice property

[ebon galleon]

cartesian closure my beloved

[merry geode]

Hmm

[unreal stratus]

is the point of the "symmetric" part basically to forget the ordering you have with simplicial sets?

[plain raven]

Yes.

[unreal stratus]

or am i missing something

not heard of this notion actually

I'll google

[plain raven]

It's not the greatest language.

[knotty vine]

You can do without, but you get a way bigger space (which is still weak htpy eqv)

[unreal stratus]

this reminds me that i meant to learn smth about simplicial modules stuff today lol

Well, just there's the general fact that the π_* functor sset -> graded abelian groups commutes with filtered colimits

i guess this is basically compactness in some form

Oh nice in Goerss-Jardine (and now sorry for talking to myself lol)

[plain raven]

Homotopy equivalent because they're both simplicial complexes right

[knotty vine]

right

Theres an adjunction between them and the counit (unit?) is an equivalence

[plain raven]

Ok. Yeah.

[knotty vine]

but i forget the details...

[plain raven]

It's alright I'm not in the mood to grill you on it

[hot locust]

Ooh yea okay, gotcha 👍

[hot locust]

Can a non-metrizable space be totally bounded or is it a definition specifically associated with metric spaces? I think it's the latter but I am not sure.

[umbral panther]

Maybe you can generalize it to bornological spaces (ie, spaces with a notion of bounding)

[languid patrol]

good intuition!

[coral pawn]

Can someone explain why they are looking at the homotopy class of the braiding map?

Also, what do they mean by "exchanging the top left" here?

[unreal stratus]

Lol barely, I mean this is more or less the definition of compact objects right lol

Hehe

[trail charm]

what does $S^2 \vee S^1 \vee S^1$ look like

[gentle osprey]

anamono

[trail charm]

i.e. can anyone draw me a picture lol

[fading vale]

Like a sphere and a circle and a circle all touching at one point

[trail charm]

like this?

[quiet thorn]

S1 is a unit circle though(?)

[trail charm]

yea but u get the idea

i tried to draw it like one circle in front of the other

like a butterfly's wings

this is why i do math and not art

[cedar pebble]

doesn't need to be no

[trail charm]

oh ok

i win

[quiet thorn]

Oh

[tribal palm]

is it possible to define S1 without every mentioning any metric space?

[red yoke]

CW complex with a 0-cell and a 1-cell

1d compact Lie group

(tho metric spaces are hidden in defn ig)

[tribal palm]

cw complexes are defined in terms of Dk so wouldn't that be circular

[cedar pebble]

I mean you can just define this as gluing the two endpoints of a closed interval together, without necessarily specifying which interval this is

you can define the homotopy type of S^1 in a much more universal way than this without specifying a space

$S^1\simeq*\coprod_{\amalg}*$

[gentle osprey]

nGroupoid

[cedar pebble]

(as a homotopy pushout this gives you the homotopy type of S^1)

it's probably hard to construct the actual topological space S^1 without using some metrizable space like an interval or R, but it's not like the space remembers any of this metric information when it's considered as a topological space alone

[tribal palm]

waw

[rancid umbra]

fr

[tribal palm]

yeah i was wondering if there were some characterizaton in terms of connectedness or something, but i guess that would only be a partial characterization

[rancid umbra]

its the only compact, connected one-manifold without boundary

[tribal palm]

genus is probably the word i was thinking of

[tribal palm]

waw

is the one point compactification of R homeomorphic to S1 ?

[rancid umbra]

yes

[tribal palm]

but not the extended real numbers

[rancid umbra]

the extended reals are already compact
no, the one point compactification of R is not the same as [-infty, infty]

the extended reals are homeomorphic to [0, 1]

[tribal palm]

oh right

[ebon galleon]

Adding two points is more than one!

[tribal palm]

ngl that's where i often go wrong, basic arithmetic

i felt the need to write explicitly 1+1 = 2 < 3 in my complex analysis exam the other day

cannot be too careful

[hidden crag]

It’s even true in greater generality for R^n and S^n

Makes sense when you look at the stereographic projection and what prevents it from being global

[quiet thorn]

i should bring an abacus to my next exam so that i never end up doing 2 * 3 = 5 again

[tribal palm]

i was familiar with the n=2 case and generalized from one example

[tribal palm]

been there

god i really want an excuse to look into "pointless topology," the name is too funny, and i really don't like bothering with points

.<

[alpine nest]

point-based topology vs point-cringe topology

[tribal palm]

i am looking to do more logic in the future, and i hear there is some applications there

but yes i really should take the advice of my seniors and get comfortable with the standard approaches to things first

but i really like approaching things historically and peeking into the paths people did not take

anyone know a introductory book on point set topology with historical commentary along the way?

[unreal stratus]

good

[unreal stratus]

honestly just try alg top :)

Well okay you needn't, but I think it's a lot more fun for most people after doing pointset stuffs

[ebon galleon]

Locales ("pointless" spaces) have connections with topoi, which have connections with logic

[tribal palm]

i will

assuming i will survive life outside of studies, which is not at all given, ehehe

also aggresively Bourbaki examples are always a trip

[half creek]

I have trouble in proving arbitrary union of open sets is open for this topology. My proof is, excluding the empty set and R, arbitrary union would be (−∞,n) if we could find a largest n among the open sets, otherwise the union would be R. Can someone check this for me? Thanks.

[unreal stratus]

yes, this works just fine.

[half creek]

Thank you!

[unreal stratus]

Of course if you really wish you could prove it more rigorously but yes i'd be happy with this

[pulsar lagoon]

Hello

I was curious if the map x -> -x was nullhomotopic

Here x is in S1

Intuitively it feels as tho it should be

[half creek]

Do you mean I can write it more formally or there is a more rigorous way to prove it?

[unreal stratus]

oh i mean you could write it more formally but depends on your courses' preferences

this is basically the only way to prove it though

[winged viper]

This map is just rotation by 180 degrees, so yes it would be nullhomotopic

[umbral panther]

It’s a homeomorphism, so it’s not null

It’s a homotopic to the identity, not to a constant map

[winged viper]

I meant homotopic to the identity 😭

Oops

[pulsar lagoon]

Ah I see

But if I compose it with a nulhomotopic map

You get another nullhomotopic map

[umbral panther]

If you compose anything with a null map, you get a null map

[junior adder]

I have a topology question that needs answer for my thesis. In the hit documentary Attack on Titan there was a popular theory that Eren had actually stayed on Paradis Island by remote controlling the Founding titan via a power of the Warhammer Titan, which allows a shifter to control their titan outside of the nape by encasing the user in a diamind external to the titan and connected via a flesh cable. The question is, how would Eren have accomplished this without having a very visible and very detacheable (if the cable detaches he loses control of the titan) cable dragging across hundreds of miles of surface land as a weakspot? . Image of remote controlled WH for reference

[umbral panther]

When did they change the channel description?

[safe torrent]

[quiet thorn]

can someone ask an actual question so I can get this whatever thing off my screen

[white oxide]

Take your lorentzian manifold of choice. Deform two separated flat region into half sphere shapes. Cut off the interior, take the quotient space identifying the boundaries and make sure the metric is compatible with the new structure. Pull the wire through the shortcut.

Only do this with an adult present. For the quotening use good glue, none of the cheap stuff.

[ebon galleon]

Topological glue is the good stuff!

[pseudo ocean]

the strong stuff

[narrow cairn]

okay so i had an idea... is the rough idea for the first problem here to collapse the loop in Delta^2 to e1 (ie, show that eps_0 * eps_1^-1 * eps_2 is nullhomotopic rel I dot) and then compose that with sigma?

[grim gull]

From Loring W. Tu's Intro to Manifolds book:

I'm having trouble understanding in what sense the fibers are preserved. Is φ(Ep) a fiber of π'?

[gritty widget]

no but it's contained in a fiber

[grim gull]

that's my understanding as well, which leads to confusion. clearly fibers arent mapped to fibers, so what's preserved?

[gritty widget]

the base point?

with only this definition i would agree with you that it seems more natural to call a map fiber-preserving if it maps whole fibers to whole fibers. but if you go on to see some examples you'll see that this is way too restrictive

[brittle rapids]

the relevant diagram commutes

[\begin{tikzcd}[column sep=0.3em] E \ar[rr, "\phi"] \ar[dr, "\pi" swap] & & E' \ar[dl, "\pi' "] \ & M & \end{tikzcd}]

[gentle osprey]

quasi_semi_group

[heady skiff]

in order to show that some topology T is contained with the topology generated by a subbasis S, does it suffice to show that for every x in a basis element B of T, there exists S' as an element of S with x in S' contained in B? i'm a little bit confused since i'm not entirely sure this criterion is valid since i'm not sure S is a basis

actually let me just ask a more straightforward question: is the topology generated by a subbasis S just defined to be the unions of all finite intersections of elements of S

[red yoke]

Yes

[heady skiff]

ok cool

[red yoke]

"Being in the fiber"ness

Maybe this is skill issue but T-invariant subspace also sounds weird to me

[heady skiff]

i'm a little confused, is this just saying that the order topology is one of the canonical choices of topologies to place on R, or is it saying that the order topology and the standard topology coincide

[ebon galleon]

The order topology and the topology induced by the metric coincide.

[heady skiff]

hm ok

[urban zinc]

@coral pivot what's your opinion on that?

[coral pivot]

lol

[ebon galleon]

lol

[heady skiff]

why is it that this set isn't open in the order topology? given $x \in {\frac{1}{2}} \times (\frac{1}{2}, 1]$, $x = {\frac{1}{2}} \times x_1$ for some $x_1 \in (\frac{1}{2}, 1]$, and wouldn't $U = (\frac{1}{2} \times \frac{1}{2}, \frac{1}{2} \times 1)$ be a basis element such that $x \in U \subseteq {\frac{1}{2}} \times (\frac{1}{2}, 1]$?

[gentle osprey]

okeyokay

[heady skiff]

too lazy to add \bigl(

[brittle rapids]

definition of open is "every point in the set is a point in a basis element contained in the set"

and here the issue is the point (1/2, 1)

by the way, the same phenomenon happens for subspaces of R

[fervent root]

can i get some help showing R^n is not homeomorphic to R^m using connectedness

i’ve shown that R is not homeomorphic to R^n but i can’t adapt the proof

[rancid umbra]

what was ur proof for this?

[rancid umbra]

to lazy to add big but not lazy enough to write x/y instead

[umbral panther]

That’s a hard theorem. I don’t think you can do it with just connectedness. You can use pi_1 to distinguish R^2 from higher

[fervent root]

i defined a homoemorphosm from R to R^n and then restricted its domain to like R - {0} and one was connected and the other wasn’t

[rancid umbra]

okay. so u poked a 0-dimensional hole in R

what if you tear a “one-dimensional hole” in R^2 if you have a homeomorphism R^2 —> R^3

R^2 cut the real line perhaps

[umbral panther]

R^n is homogeneous, so no matter which point you remove, the complement is homeomorphic. But there are lots of ways of embedding a circle in R^n. There’s only way to embed it by a lipschitz map, but there are crazy fractals with crazy complements

[tiny ridge]

Good puzzle: Can you find a pair of embedded RP^2's in RP^3 which (necessarily non-transversely) intersects at a single point?

[fervent root]

so i should restrict it to R^n-1?

[hot locust]

Never heard of them but I'd check those out

Thanks

[tribal palm]

god i’m looking back at the notes i took in the lectures we covered quotient spaces and i don’t understand any of it

three different diagrams with different types of functions and i’ve no idea what they’re saying

i probably did not understand when i wrote this down either, two months ago now, should’ve asked someone about this then, oh well

[rancid umbra]

no, i’m just saying look at this specific case for some more intuition

[tribal palm]

i know the intuition of gluing together, and taking equivalence classes makes perfect sense, but these factorings of different kinds of functions did not at all to me

[white oxide]

Like, the universal property of quotient spaces?

[tribal palm]

i’m not sure if just one or all of the diagrams were the universal property

certainly one of them

[gentle osprey]

Rishi

[dusty talon]

ignore ^^

[coarse night]

There is also another way by using lifting lemma

Look at map G’xG’->GxG and lift to G’xG’

[dusty talon]

hm

[dusty talon]

I figured out my issue here ^

[dusty talon]

how would you do it this way?

im a bit confused

[coarse night]

remember something about lifting
X'
| p
Y → X
if π_1(Y) ⊂ p* π_1(X')

can you do something similar for this diagram?
G'
/ |
(m'?)/ | p
/ v
G'xG' → GxG → G
pxp m

[dusty talon]

oh ic

I think i get it

tysm

[fringe cypress]

If X is a separable, are there no functions f : X → X continuous only on a countable dense subset of X? Also, are there non-Hausdorff separable spaces? If there are, assume also that X is Hausdorff

I half remember that you can't have functions f : ℝ → ℝ continuous only on ℚ, so I'm trying to see how much this generalises

[silver umbra]

if p: Y -> X is a covering map, and f: X -> X is a map

is there a lifting criterion on f so that there exists some lift f: Y -> Y?

[umbral panther]

I think the criterion is if f preserves the conjugacy class of the fundamental group of Y in the fundamental group of X

[alpine nest]

In a complete metric space, if the set of continuity for some function is dense, it can't be countable.

It follows from Baire's theorem

[silver umbra]

so, it the lift exists if f(p(pi_1(X))) = g p(pi_1(y)) g^-1 for some g in pi_1(X)?

[red yoke]

fp(π1(Y)) ⊂ p(π1(Y)) should be an equivalent condition

[gritty widget]

Can you just share these diagrams there

[umbral panther]

This condition is not invariant under charge of base point, which is why you need conjugation

[silver umbra]

does that matter if all the spaces involved are path-connected?

also, another general question

i've got a normal cover p: Y -> X

since its a normal cover, that means that p(pi_1(Y)) is a normal subgroup of pi_1(X)

which means that its the kernel of some group homomorphism pi_1(X) -> G

how can i determine what G is?

[unreal stratus]

Well G is the cokernel of p_*

So basically you have to compute what p does on fundamental groups

[normal umbra]

$\Sigma_{k} c_{k}(V)t^{k} = [I + i\frac{tr(\Omega)}{2\pi}t + i\frac{tr(\Omega^{2}) - tr(\Omega)^{2}}{8\pi ^{2}}t + ...]$?

[gentle osprey]

det(\frac{it\Omega}{2\pi} + I)

[normal umbra]

Is this directly expanded from the $det(\frac{it\Omega}{2\pi} + I)$ form using the identity tr(log(A)) = log(det(A))

[gentle osprey]

det(\frac{it\Omega}{2\pi} + I)

[obtuse meteor]

yes

[unreal stratus]

Also remember you can use \sum

Seeing sigma like that in a topology context confused me aha

[normal umbra]

Thanks, I suck at latex

[obtuse meteor]

It pays to consider how you do it in the reverse! It might be easier to see after you do some things like that

A common type of technique/question is, given a finitely presented group G, you can realize it as the quotient of a free group on the generators. Try to realize this in terms of regular covering spaces

[tiny ridge]

More easily, it's the characteristic polynomial of iOmega/2pi

[normal umbra]

Thanks

[tiny ridge]

The coefficients are then computed by "diagonalizing" and using Newton's formulae

[scenic pecan]

I have a silly question regarding quotient topology. I understand that it can be thought of as gluing points to together which belong to the same equivalence class so [0,1] can be mapped to S1 but I don't get why we want open sets to map to open sets only from the definition. I can't intuitively think of what will go wrong if we don't have a quotient topology and if there are other ways to "glue" the points.

[knotty vine]

We want the quotient map to be continuous, and nothing more

There are other ways to glue points, for example by gluing each to one of the endpoints of an interval

That's not homeomorphic to the usual quotient, but it is homotopy equivalent

[red yoke]

More precisely, the quotient topology "inherits" as many open sets as possible

[scenic pecan]

So we are defining a homeomorphism?

[knotty vine]

We have the function already, we just define the quotient topology so that the function is continuous

[steel glen]

we're not asking for open -> open. [0,a) does not map to an open set, for example
what we want is for a neighborhood of a point to be open if the preimage under the canonical surjection is also open, which is like preserving open sets like arki said

[red yoke]

Moreover, if q: M → N is a quotient, S is a space and r: N → S is a function of sets, then rq is continuous iff r is continuous

[scenic pecan]

What function are we talking about here?

[steel glen]

q(x) = [x]

if your canonical map is not continuous, then something probably went wrong. topology is the study of continuous maps, after all

i think you might be looking for a more geometric answer though

[red yoke]

So say you have a continuous function S² → R and you realize that antipodal points have the same value. Using this property, you can turn this into a continuous function RP² → R.

[knotty vine]

That's the "universal property" of quotients

[scenic pecan]

P2 is the set of antipodal points right?

[red yoke]

Ye

So it "inherits" continuous functions from the original space

Which is what you care about RP² in geometry

[scenic pecan]

Now it makes sense. Thanks.

[potent sky]

Does alexander's trick (https://www.wikiwand.com/en/Alexander's_trick) prove that the only knot which bounds an embedded disk is the unknot?

[eager vigil]

Yeah, I'm just waiting for the semester to be over so I can switch books... found both Hatcher and Armstrong terrible... hoping that Rotman, which seems my best option, will be good

Hey, I'm slightly unsure about this finishing argument here... it seems that Armstrong is arguing that if we have an element whose abelianization is the identity, then the element must be in the commutator subgroup. Is this true?...

[dusty talon]

I'm having trouble showing that the operations on a simply connected covering space of a topological group are continuous (I've shown the covering space is a group, that the projection is a group morphism, etc.). I've tried a lot of things but they haven't worked

[heady skiff]

Out of curiosity why do you not like Hatcher? This semester, the textbook was Armstrong and it was so bad that I started studying from another book entirely, but I was planning on reviewing point set and going w Hatcher

[eager vigil]

The problem I experienced with Hatcher was that it made very little use of point-set topology. Hatcher seems to be hand-wavy on a lot of things which I didn't like, so I'm planning on going with Rotman who, I've heard, is a lot more rigorous. From what I've heard, a lot of people have the same experience with Hatcher and say that it's rather supposed to be a second-course book on algtop which is also how I'm planning on approaching Hatcher in the future - mb also as a reference sometimes for material from chapter 0 and 1 which I covered. Though this is just my experience and other people might be completely fine with Hatcher

[heady skiff]

I see, yeah I hate hand waviness so I’ll give Rotman a look then

[eager vigil]

As an alternative or just a future supplement, I've noticed that my university has notes written for Hatcher which, from a quick skim-through, seem to write a lot of the stuff out rigorously. I've attached it here.
I'll still go with Rotman, but for anyone looking at Hatcher, I think it'd be a nice supplement

[tiny ridge]

Not really. It is quite easy to prove that the only knot which bounds an embedded disk is an unknot: Suppose K is a knot and D is an embedded disk bounding it. Then K = delD is a circle of radius 1 in D. Push K along D and isotope it to a circle of tiny radius around the center. This is obviously an unknot. So you have isotoped K to be an unknot.

There is some extra care required in this argument if D is not smoothly embedded and only topologically embedded (one must replace the top embedding by a smooth one). But these are mostly non-issues

[red yoke]

What are some knot theory book recommendations for someone who is familiar with the main sections of Hatcher 1-3? Are there other prerequisites one should be comfortable with?

[potent sky]

thanks!

[tiny ridge]

Rolfsen perhaps. I like Saveliev's 3-manifolds book

Rolfsen is very vivid and easy going

[potent sky]

Wait isn't that just the alexander's trick isotopy?

[tiny ridge]

Not really, it's the isotopy which radially shrinks the unit circle to a circle of radius epsilon > 0 in the disk

Alexander trick is about isotopy of self-homeomorphims of the disk. Yes, a radial homotopy is involved, but that doesn't make this simple straightline homotopy the "Alexander trick"

[potent sky]

i thought you said it pushes along D, like outwards

[red yoke]

Thanks! ❤️

[dusty talon]

bump

[tiny ridge]

what is your definition of the operation on Gtilde

[dusty talon]

you take the paths from a fixed base point in G tilde to the points tha you want to multiply

actually

let me just send a ss

[tiny ridge]

so points in Gtilde are homotopy classes of paths with a fixed starting point in G, to you

[dusty talon]

yes

(because simply-connected implies we only care ab endpoints ?)

[tiny ridge]

what youre doing is sort of messed up

[dusty talon]

huh

[tiny ridge]

once you have xtilde, ytilde in Gtilde, why on earth bother about paths to them starting from identity

[dusty talon]

^^

wait wait

not in G

xtilde ytilde are points in G tilde

[tiny ridge]

typo, edited. my statement stands

[dusty talon]

oop

wdym like

I've defined multiplication the way the problem wants me to define it

[tiny ridge]

what is your definition of Gtilde

write it down for me

[dusty talon]

simply-connected covering space of G

[tiny ridge]

thats it? no explicit definition?

[dusty talon]

like

[tiny ridge]

aka apriori you do not know that it exists?

[dusty talon]

define the points in G tilde?

[tiny ridge]

im just asking what you know

[dusty talon]

im given that it exists

if thats what you mean?

[tiny ridge]

OK good

do you know the map lifting lemma for covering maps

[dusty talon]

yes, roughly

[tiny ridge]

state it for me

[dusty talon]

if you have a path in the base space then there's a unique lift to a path in the covering space by giving a fixed point for the path in the cover

[tiny ridge]

thats the path lifting lemma, which is a special case of the map lifting lemma

i was asking about the latter

[dusty talon]

oops

I do not know the general case then

[tiny ridge]

ok thats fine

I was trying to understand what you have in your repository

Do you know an explicit construction of a simply connected covering space of a space

[dusty talon]

no

oh wait

there is something for this in our notes

[tiny ridge]

That would help

[dusty talon]

wait

nvm

its something ab the fundamental group i misread while skimming my bad

[tiny ridge]

It’s going to hurt to prove the map you constructed is continuous without knowing some of these

Because you just have some abstract definitions available to you.

Nothing concrete

[dusty talon]

Yeah... the only idea i've had so far is like

the only notion of openness that we have

is that I'm given that there are continuous maps in the base space

and that p is a local homeomorphism

[tiny ridge]

Yup. So you really want to lift that map in the base to the cover

Continuously.

[dusty talon]

right

[tiny ridge]

This is what the map lifting lemma enables you.

[dusty talon]

but then I don't know how to show like for products

that if I have U V in the base space (i get these by projecting down W tilde)

then i want my U tilde V tilde to be the lifts of those sets

but then I'm not sure how to show that those U tilde V tilde actually multiply into W tilde

[tiny ridge]

Let me tell you something a little more concrete. You can incorporate this in your solution to make life easier.

Before I start, what is G here? A topological group such that X Y Z…?

X = Path connected, anything else?

[dusty talon]

uhh

could i get a hint

[tiny ridge]

Its a genuine question. Its not true that every topological group has a simply connected cover.

You need hypothesis

Just trying to understand what hypothesis you are given

[dusty talon]

I'm given that G is a topological group, G is path-connected, G has continuous maps (inverses and products) (I mean isn't this part of what it means to be topological group lol?), and it has the simply-connected covering G tilde

that's it

[tiny ridge]

OK, that’s part of the hypothesis!

No, I meant that it has a simply connected cover is part of the hypothesis.

[dusty talon]

oh ok

also

does path connected imply locally path-connected?

[tiny ridge]

Very strange homework, I hate it. Its needlessly abstract.

[dusty talon]

💀

[tiny ridge]

Not in general, no.

[dusty talon]

ok im also given locally path-connected

[tiny ridge]

In general thats also not enough to admit a simply connected cover.

[dusty talon]

o

[tiny ridge]

A space admits a simply connected cover if and only if it is path connected, locally path connected and “semilocally simply connected”

[dusty talon]

ahhh I've heard that before

[tribal palm]

when my prof writes "let i: A subset X" it means i is the the function that takes each a in A to itself in X, right?

[dusty talon]

we've never talked ab it in class though

yea inclusion map

[tribal palm]

wonderful, thank you

[dusty talon]

I also don't know what it means

[tiny ridge]

Semilocally simply connected means for any point p in X, there exists a neighbourhood U of p, such that any loop in U starting and ending at p, is nullhomotopic in X

[dusty talon]

whats nullhomotopic?

[tiny ridge]

Homotopic (rel p) to the constant loop at p.

[dusty talon]

o

ok i kinda see how that makes sense

[tiny ridge]

Given all the hypothesis above, here’s a construction of a simply connected covering space of X. Fix a basepoint p in X. Define Xtilde to be the set of all equivalence classes of paths starting at p, f : [0, 1] -> X, f(0) = p, where you say two such paths f, g are equivalent if f(1) = g(1) and f, g are path-homotopic.

This is a set. We need to give it a topology.

We do it by specifying the basic open sets. The basic open sets around a particular point f in Xtilde looks like {g in Xtilde : g(1) is in a neighbourhood U of f(1)}

U varying over all possible open neighborhoods of f(1) in X.

There is a map ev : Xtilde -> X (ev for “evaluation”) given by ev(f) = f(1). Check that wrt the above topology ev is continuous, and in fact is a local homeomorphism, and moreover in fact a covering map.

It remains to check Xtilde is simply connected. This is not too hard, and is a good exercise.

Use this description in the exercise above. Define the multiplication on Gtilde directly, using similar idea as in your write-up. You’ll see that the problem has simplified considerably.

[dusty talon]

because I have more concreteness with what is considered open?

[tiny ridge]

Yup

[dusty talon]

ok

I will try it 🙂

[merry geode]

I forgot how to compute again

So the basic example stopped computable for me

Say, X = [0, 1] and A = {1/n : n \in N} \cup {0}

How do I compute H_1 (X / A) ?

My computation gives that it should be countably generated since X/A is practically countable wedge product of S^1's

But apparently this does not hold

Is this another of the Hawaiian ring moment?

[unreal stratus]

i mean, it's not

[unreal stratus]

basically yes

[merry geode]

Gah, Hawaiian ring

[unreal stratus]

you can take paths in X and jump from point to point within A sort of

if you think about paths that way then you can actually do infinitely many jumps in one path

since points in A get really close etc

Pretty sure you can basically encode infinite binary sequences this way

[merry geode]

How do you do infinitely many jumps, and why is this impossible for simple wedge product?

[eager vigil]

Any hints for this one?

[normal umbra]

Why is the k-cochain formed?

I don't really wrap my head around the idea

[unreal stratus]

Can get Infinitely many by gluing together short paths

Think binary expansions

Then for normal wedge product you can just use compactness of I ig

[merry geode]

Ahh, as we have many distinct open sets in normal wedge product

I still do not get how the binary expansions are linearly independent, or at least that you cannot reduce the generating set.

Wait, is countable direct sum of integers countable?

[red yoke]

Yes

Wedge sum of circles has fundamental group being direct sum of integers

[red yoke]

Does binary expansion refer to a sequence indicating whether a path loops around each circle?

Notice the "number of loops" around each circle is invariant under path-homotopy

[heady skiff]

i don't really see how we necessarily need Y to be convex in either for a to be a lower bound or upper bound for Y?

[red yoke]

And addition of paths leads to addition of number of loops

So the fundamental group of earring is at least as large as direct product of integers

Which is uncountable

[red yoke]

X = R, Y = (0, 1) ∪ (2, 3), a = 2.5

[ebon galleon]

R - 0 as a subset of R

[heady skiff]

wait convex sets in R are just intervals right

i guess that's how intervals were defined in my real analysis course

[ebon galleon]

Yes, since they're (path) connected, and the only connected subsets of R are intervals

(open, closed, half-open, rays)

[merry geode]

I think so

[heady skiff]

oh so it's just that if Y is not convex, there exists a1, a2 in Y with a1 < b < a2 but b not in Y, so in general it's not true that if a is not in Y and Y is not convex then a is a lower or upper bound for Y?

i guess that's contradiction

[merry geode]

Yea, I was curious if it was uncountably generated.

[red yoke]

Ye it is

[merry geode]

How do you show that?

I see that direct product of integers embeds into the homology, but I don't see how the direct product is uncountably generated.

[red yoke]

Since a countable direct sum is countable, an uncountable free module is uncountably generated

[heady skiff]

do bases of a topological space always exist? can we just take X?

[merry geode]

I see. But is countable direct product of Z free?

[red yoke]

The topology is a basis of itself

[red yoke]

Yes

[merry geode]

Hmm, this article indicates it is not free @red yoke https://en.m.wikipedia.org/wiki/Baer–Specker_group

[red yoke]

Wait what

[merry geode]

Am I misinterpreting the article?

[red yoke]

But the same argument should work

[merry geode]

Same argument?

[red yoke]

Countably many elements only generate countably many elements, so uncountable Abelian groups is not countably generated

[merry geode]

Ah, indeed.

Since an abelian group cannot be bigger than free group of its generating set

Well btw, so is countable direct product of Z a free abelian or no

[red yoke]

Apparently not

Oh I guess I assumed it was finitely generated

[umbral panther]

Free abelian groups are reflexive !

[merry geode]

I think I am going to screw up final too

I forgot Regularity of deck transformations so I have to review that, and need to go through homology exercises as well..

Homotopy lifting theorems as well

[merry geode]

Algtop is so tricky

[silver umbra]

is the hopf fibration the only way to express S^3 as the total space of a fiber bundle?

[unreal stratus]

||S^3 x {*} -> {*} and S^3 x {*} -> S^3 ||

But fr I mean there are other examples - for example S^3 is the universal cover of RP^3

[silver umbra]

ah right ofc

but are there any examples that are not covering spaces?

[red yoke]

S³ → S² → RP²

[rigid path]

Hi, anyone can suggest me a good source to better understand compactifications? (Base level)

[umbral panther]

Exercise: use Poincaré duality to show that if a closed manifold is the total space of a fiber bundle, the base and fiber satisfy Poincaré duality, which really limits the options

[tiny ridge]

You can do silly stuff like S^3 -> S^2 and then compose with S^2 -> RP^2

This gives a fiber bundle with disjoint union of two circles as fibers

[unreal stratus]

lel

[tiny ridge]

But essentially that's the only thing yes

[unreal stratus]

Yeah

[tiny ridge]

As bw said, if X -> S^3 -> Y is a fiber bundle, X and Y are Poincare duality spaces. All PD spaces of dimension 1,2 are homotopy equivalent to manifolds

For manifolds, you can now go case by case. If Y is a surface, homotopy LES tells you pi_3 Y is nontrivial forcing Y = S^2 and RP^2 and the maps being either Hopf or Hopf composed stupid

If X is a surface, S^3 must be a mapping torus which is nonsense

Ok, there are fringe cases where either X or Y are 0-dimensional, where the answers have already been discussed

For completeness one should also show that X, Y are upto homeomorphism manifolds. If not you'll get something like: R^3 has a factorisation R^3 = U x V where U, V are not manifolds

I don't think this can happen but I do not have a proof

[coarse night]

[tiny ridge]

(It can happen in higher dimensions)

What a time to enter, Mr S^3/W

[coarse night]

Lol sure

[tiny ridge]

Let's try a special case. If X x R is homeomorphic to R^3, I conjecture X is homeomorphic to R^2

[coarse night]

Why do you think that?

[umbral panther]

If it weren’t true, you’d already have heard the counter example

[tiny ridge]

"No funny business in low dimensions"

[coarse night]

Not convinced till I see a proof

[tiny ridge]

That's the point of a conjecture

You don't have to be convinced

[coarse night]

What about X\times R homeo R2 does this also say X homeo R?

[tiny ridge]

If yes, X x 0 sits inside R^2 as a separating subspace, thus bounds a domain above

Riemann mapping theorem should tell you X is a quotient of R

Sorry, only true if X is locally connected

The topologists sine curve also disconnects R^2, but isn't a quotient of R

But if X is not locally connected, neither is X x R = R^2 so that rules those out I suppose

So X is a quotient of R by Caratheodory's Riemann mapping theorem

[coarse night]

cool

[tiny ridge]

Start listing Hausdorff quotients of R which do not have any pi_1

There's only one, R, I am sure

[coarse night]

sounds correct but hard to justify rigorously

some kind of bing shrinking seems like

[tiny ridge]

This is nonsense, take R with a stick poking out of the origin

Clearly a quotient of R, fold an interval down the middle

These cannot occur because they have cut points of order > 2, X disconnects into more than 2 pieces if you remove these. So R^2 would disconnect into more than 3 pieces if you removed an embedded line from it, which is nonsense

So what are the quotients of R st all points are cut points of order 2?

https://academic.oup.com/plms/article-abstract/s2-41/1/191/1483886

That it is a Hausdorff quotient of R implies the space is metrizable, connected, locally connected. That it is a subspace of R^2 implies it is separable.

Above applies

[tiny ridge]

So the answer to this question is YES

[tiny ridge]

Good luck with this 🙂

[silver umbra]

thanks guys! ill have to take some time to read over all this

[umbral panther]

A good technique for this is local cohomology H^n(X,X-pt)
This satisfies a Kuenneth formula, so if AxB = R^2 and neither is a point, they must be disconnected by removing each point. No Riemann mapping

[tiny ridge]

How do you prove they're connected/loc connected

Connected is easy

Locally connected obvious because I already said so OK

Good argument

Sounds like this might be useful for R^3

[umbral panther]

Local homology makes this a lot easier. It makes it a local argument

[thorny agate]

How do I show that $S^1$ and ${(x, y) \in \mathbb{R}^2 | x < 1/2 } \setminus (-1, 0)$ are homotopy equivalent?

[gentle osprey]

Spamakin🎷

[thorny agate]

Just struggling to come up with a map

[feral copper]

Did you make a drawing?

[thorny agate]

ya

[feral copper]

Then you have a map looking at you on the drawing

[thorny agate]

no but like the actual map >_>

my issue is the scaling

[feral copper]

Take a circle around the point (-1,0), and contract the space around it to that circle