point-set-topology

[gritty widget]

This is more direct

Your proof doesn't do it formally

It's not wrong, we basically had the same idea

The only informal step is not showing that any neighborhood crosses a line segment.

But to justify it you'd have to do what I just did anyway

What for?

If you're content with informal (visual?) proofs then feel free to use yours

Not a big deal

[formal tide]

How to show that any covering of a rectangle is trivial?

[gritty widget]

It is formal, I just didn't write it out. I only asked if the argument worked...

[gritty widget]

What's a trivial covering?

[formal tide]

p:Y->X is trivial if Y is homeomorphic to X x D where D has the discrete topology

[gritty widget]

Then we're not talking about the same thing?

[unreal stratus]

Oh I completely got the wrong end of the stick of that question with rectangles lol

because of all the point set topology questions

[gritty widget]

Precisely: you asked if your reasoning is wrong or not

I provided you improvements to your reasoning

Nothing to argue here

[formal tide]

it seems obvious geometrically to me but idk how to tackle the problem

[gritty widget]

No need to be rude but okay.

[unreal stratus]

Basically comes down to simply connectedness

[marsh forge]

This seems like the type of thing that you might want to post on MSE (failing that, MO)

I wish I could help but I don't know the answer

[gritty widget]

I wasn't rude (I'd agree I was unrefined or blunt, but definitely not trying to be rude)

[white oak]

Hausdorff's 4th axiom for neighbourhoods is cyclic!

[formal tide]

geometrically, a homotopy beween the border of the rectangle and any point in the interior has as image all the space

so it makes sense that the preimages can be adjoined

but again this is very informal, how could I make this more formal

also thanks

[white oak]

<@&286206848099549185> or anyone else, can you kindly look into my thread above? https://discord.com/channels/268882317391429632/1004072710609055785

[gritty widget]

Helpers are for the help channels, not for pinging people in channels here

[white oak]

Oh, I see.

• These channels are for pre-university homework-type questions. Broad conceptual questions, as well as questions in university-level topics — are more appropriate in topic-specific channels (Pre-University, Early University, Advanced Mathematics) and in

I do not see how my question can be seen as fitting this condition. I do not think it will fit in a help channel.

[marsh forge]

Yes you got it the other way around

not that you should be posting in the help channels but instead that you should be posting here

its just that helpers won't be of much help here

[marsh forge]

So, what you have described here I think is usually called the weak-f topology

and indeed surjectivity isn't necessary

But when describing quotients we want surjection

just because we want to think of the quotient as coming from gluing parts of X together, essentially

[hollow basin]

The notation is confusing because discord screwed me

hence the compilation error

this is the correct one:

[gentle osprey]

TheZachMan

[gentle osprey]

Andrew071

[young lantern]

wouldn't the signed distance for x in M just be 0?

[hollow basin]

No

[young lantern]

x in S*

d(x,S) = 0 if x in S

[hollow basin]

it's a little hard to see, thanks latex bot, but that's S complement

[young lantern]

if it's easy for you, can you edit it and change your notation to d(x,S)?

it'd be a lot easier to read

and then signed distance can be like d*(x,S)

[hollow basin]

the bot won't pick up on the edits at this point, but yeah that'd probably be better

[young lantern]

and S^c is S^c not bar S which is typically the closure of S

[hollow basin]

ok I realize now that in a topology context S-bar is a bad choice of notation, but in general it's a reasonable way to write complement

[young lantern]

Reading on stack exchange, someone claimed the result is actually false, and limit point should be defined as the one given in wikipedia and then the proof would be correct

[unreal stratus]

With the discussion of local degree in Hatcher and in my uni's notes, it seems if one makes consistent choices of isomorphisms then one finds the degree of a map f: S^n -> S^n is equal to the sum of local degrees (where one picks some y with finite fibre under f) .

But our uni notes define the local degree by itself and basically seem to imply that you can get different signs for different choices of isomorphisms between homology groups. Is there a problem here, or is the point just that we have to calculate local degree after already having chosen isomorphisms H_n(S^n) \cong Z so that everything is consistent?

[marsh forge]

Should be the latter

[unreal stratus]

Sure thanks

[golden gust]

Let $X$ be $\bbR_{\geq 0}$ with some nonstandard topology, and let $Y$ be $\bbR_{\geq 0}$ with the standard topology, so that $\mathrm{id}\colon Y \to X$ is discontinuous. now let $\bullet$ be a group operation on $X$ which pulls back under $\mathrm{id}$ to a metric $d$ (which of course \emph{is} continuous, as all metrics are)
[\begin{tikzcd}
Y \arrow[r, "\textrm{id}"] & X \
X^2 \arrow[u, "d"] \arrow[ru, "\bullet"'] &
\end{tikzcd}]
can $\bullet$ be continuous? I think no, right?

[gentle osprey]

Average J∘du=du∘j enjoyer

[marsh forge]

I dont have an intuition either way

wait

What is this metric product thing

[golden gust]

I can tell you it for $\mathbb{N}[\frac{1}{2}]$, because for $\bbR_{\geq 0}$ it's not constructive

[gentle osprey]

Average J∘du=du∘j enjoyer

[marsh forge]

I did not even know such a thing existed

thats wild

[unreal stratus]

Silly question but is X even a group like

[golden gust]

it's bitwise XOR, so like 101.01 * 11.101 (in binary) is 110.111

then you extend this to R_>=0 with axiom of choice or whatever

[unreal stratus]

Oh okay fair you've put some weird topology and weird operation on it ig

[golden gust]

yeah

btw potato I'm writing an invariant article about this so this is a sneak peek lol

[unreal stratus]

Lol nice

[gritty widget]

So • is a group operation and a metric? That's impossible

[golden gust]

why?

[gritty widget]

A group operation must be onto, but a metric must be non-negative

It's R_+ and not R. Never mind

[golden gust]

yeah lol

[gritty widget]

|a-b| <= a • b is a curious inequality here

[gritty widget]

Doesn't this inequality lead to a contradiction with assumption that Id is discontinuous?

How I arrived at it:
a • a = 0 for all a, so 0 is the identity element
0 • b <= 0 • a + a • b so that
b-a <= a • b by triangle inequality

So such situation is impossible, Id must be continuous

This only uses that • has identity element and is a metric (no associativity or other group axioms)

[golden gust]

if $X = \mathbb{N}[\frac{1}{2}]$ and $\bullet$ is the bitwise XOR described above, then the open ball around 1 of radius 1 is
[{x \in \mathbb{N}[\tfrac{1}{2}] : x \bullet 1 < 1} = \mathbb{N}[\tfrac{1}{2}] \cap [1,2)]
which is not open in $Y=\mathbb{N}[\frac{1}{2}]$ with its standard topology. so the identity $Y \to X$ must be discontinuous, no?

[gentle osprey]

Average J∘du=du∘j enjoyer

[golden gust]

I know that you can bound it below by |a-b| and above by a+b, but I don't see how this leads to contradictions @gritty widget

[gritty widget]

The inequality implies that if y_n converges to y in Y, then y_n converges to y in X. But this is precisely what means for Id to be continuous

[golden gust]

isn't the inequality the wrong way round for that? seems like it implies X —> Y is continuous, but not necessarily Y —> X

[gritty widget]

If y_n is convergent to y in Y, then Id(y_n) = y_n converges to Id(y) in X

X has Euclidean topology here

|Id(x)-Id(y)| <= x • y = d(x, y)

It's in fact Lipschitz continuous

[gritty widget]

If you calculated this right then you got it wrong

If this set, say A, would be open in Y, then using Id_2: X to Y, Id_2^-1(A) isn't open

So it's Id_2 that's not the continuous one

[gritty widget]

You switched them around

[golden gust]

no no, X has the topology induced by this weird metric, and Y is the euclidean one

[gritty widget]

you said here that d is continuous

as all metrics are. This made me think that Y is generated by topology from metric d

[golden gust]

metrics are continuous as functions to R with its standard topology, right?

and the domain space has topology induced by the metric (which is the coarsest topology that makes d continuous)

[gritty widget]

yes

[gritty widget]

you mean X^2?

the domain space

[golden gust]

X^2 is given the product topology I guess

[gritty widget]

I think $\bullet$ can be continuous, I don't see why not

[gentle osprey]

Blitz

[gritty widget]

unless d is somehow a quotient map

It wasn't said, but I'll assume d is a metric on X, as otherwise d might not be continuous

[golden gust]

it would be nice if it is, so that this object is a topological group, but I'm not sure either way

[gritty widget]

I don't know. I think this is the place that we can't get too far by abstract considerations and need to get "nitty gritty" into the analysis

[wise coyote]

is it the case that every subspace of a t1 space containing all its limit points is closed?

[unreal stratus]

That is true for all spaces, not just T1 ones

The closure of a subset A of a top space X is equivalently A union its limit points

The proof should be in any good topology textbook or you can try it yourself

[swift fjord]

This is vacuously true

[unreal stratus]

what lol

Yeah sure so any textbook that doesn't have this is bad

[swift fjord]

I was implying there are no good topology textbooks

[unreal stratus]

Pain

What about Rudin chapter 2

[gritty widget]

There's no good topology textbooks that you know of

And that's different

[swift fjord]

There's no good topology textbooks because point set bad

[gritty widget]

nice

For topology textbooks I like Dugundji. It's not perfect, but it did prepare me for all the basics of topology that I'll probably ever need

I don't like the part about the simplicial approximations though, I'd skip that part of the book, Dugundji has a bad approach to simplexes, too informal and using old nomenclature

It doesn't appeal much to intuition and often doesn't really justify concepts, or why they are, but I don't mind that approach

Figuring out why something is myself was good enough for me

I didn't read Munkres but introduction to AT in the last chapters sounds promising

Willard has cool contents, I've only really read the proof of Hahn-Mazurkiewicz theorem there. It didn't make a good impression though, since Willard makes a crucial error in the proof

Not sure how much that speaks for the rest of the content in his book, they probably don't contain many errors (or so I hope)

[pallid lion]

how the hell do you guys read 20 topology books when it takes ages to just finish a single one

[gritty widget]

I only really read one. Why read the same thing over and over again?

[pallid lion]

I mean why fly over the others then

I get that they may have different content, but there is often a heavy overlap that is redundant

[gritty widget]

I had doubts about the proof of Hahn-Mazurkiewicz book presented in one book (not Dugundji but something that touches on different fields, mainly separable metrizable spaces), and I was looking for something which wouldn't have an error

in the end, I had to fix the proof myself, since Willard was the only book I found with a similar proof, but that one was also wrong

[surreal lantern]

lol

[gritty widget]

anyway, I am reading few different books which are also technically topology, but they go further in one direction

[pallid lion]

ok I guess, but do you really feel qualified judging the other books if you just went to search this specific thing in them?

[gritty widget]

but I haven't finished them yet

[gritty widget]

I feel qualified by looking at the table of contents

Willard touches on some continuum theory which is pretty neat

[pallid lion]

it's one thing for a book to have bad content, it's another to have bad content explained in a terrible way

[gritty widget]

bad content is something subjective

[pallid lion]

you could say the same about writing styles

[gritty widget]

I wasn't judging Willard on the level of content, rather that the error might mean that there's more errors in the proofs out there

That's a maybe of course as I didn't read it

The direction the book is going is pretty cool though, as it doesn't go in the "standard" algebraic topology direction, but it touches on some continuum theory

[pallid lion]

cool ig, no idea about it though

[marsh forge]

No

I think there are good examples with the wedge of two circles in that section

But I haven’t looked in a bit

Huh?

The idea is that the deck transformation preserves the fibers/preimages over points in the base space

So if I have an n-sheeted cover

My deck transformations can permit r the even lifts

Permute

Yes

Same thing

[pallid lion]

I think the source of your confusion is that just because the composition is p, it doesn't uniquely determinte tau.
I mean p doesn't have to be a homeomorphism.

This example is not a covering, but to illustrate
Imagine a linear map from f R^2 to R
where you map (x,y) to x
if I compose f with the map
g(x,y) = g(x,n*y)
then f composed with g is still the same map f
even though there are many choices for g

[marsh forge]

I mean X

The point is like

Over each point x in X

There’s like n points x’ living in the cover over x

And the deck transformation is allowed to permute these n points

Now, not all permutations will be valid for other reasons

[pallid lion]

you aren't really allowed to just swap though, more like sliding since you'd have some continuity issues if I remember correctly, I'd have to look it up again

[marsh forge]

It would be ideal if you could type it in latex...

[gritty widget]

explain to me what you're doing when you're calculating interior of B

Let's start with $B^\circ \neq 0$

[gentle osprey]

Blitz

[gritty widget]

I don't get it

[gentle osprey]

Blitz

[gritty widget]

better?

[plain raven]

nlab diagram of models for homotopy theory in increasing order of convenience

boy i sure find it convenient to have to establish what a cohesive infinity topos is before i can do basic topology.

[cedar pebble]

[pallid lion]

the euler characteristic of a torus minus a disk is just -1 right?

the red lines are the 1-cells
the red dots the 0-cells
and yeah the blue one I can't be ask to draw properly but you can probably imagine how the 2-cells are supposed to glue in the end

(I tried)

the two points close together at the left are supposed to enclose the hole of the missing disk with their connection

[surreal lantern]

this kinda looks like some physics picture from electromagnetism

[pallid lion]

it's as good as it gets for 1 minute of effort with screenshot mouse paint

though in hindsight

drawing the blue lines that way was stupid

this might be better

https://cdn.discordapp.com/emojis/887072458803404890.webp?size=160&quality=lossless

[gritty widget]

There is this standard arguement that if X=U_1 union U_2

and there is some element s1 that kills H^*(U_1) and s2 that kills H^*(U_2) then s1s2 kills H^*(X)

and its easy to induct this up for a finite cover of X

is there a version of this for paracompact X

this is the result for a finite cover

[forest sky]

Is that Gage "Nhymnim" Poljack of the series which progresses?

[pallid lion]

The goat

[cedar pebble]

Yes, since a torus minus a disk is homotopy equivalent to a wedge of two circles so the Euler characteristic is 1-2=-1

[pallid lion]

Just for feedback. Is it in any way recognisable how I tried to construct this thing as a CW complex?

[marsh forge]

Yes

your CW complex structure looks correct if I am interpretting the image properly

and then you have the right euler char from the alternating sum on the number of cells

[near trench]

I'm not too deep into topology so I'm sorry if this is a bit of a simple question but is there actually a good reason to define neighbourhoods? It seems like every proof using neighbourhoods would be simpler just using open sets. So what is the intuition behind understanding neighbourhoods?

[cedar pebble]

open neighborhoods are literally just open sets containing a given point

[near trench]

I know what they are. They just seem pointless.

[cedar pebble]

you can't imagine any situations where you would only want to consider open subsets that contain a given point?

[near trench]

I can. I just can't imagine why I would want to define neighbourhoods after already defining open sets which fill that role.

[cedar pebble]

I mean you can call it what you want

you're not really defining anything new

[near trench]

I know.

[cedar pebble]

it's just giving a name to something is all

[dim meadow]

The word neighborhood also conveys intuition about how to think about a neighborhood of a point

imo

[cedar pebble]

that and it's just convenient to not have to say "open subset containing" every single time you invoke this notion

[near trench]

I guess..

[cedar pebble]

idk you're free to not use it if you feel it's redundant, and you're free to use it if you feel it conveys some useful meaning

[dim meadow]

Anyway I think the idea that notation should be concise to this extreme is a bit weird

By which I mean not having names for things that you can define in terms of other things

[near trench]

Oh, I did read there were 2 definitions for neighbourhoods. One was open, the other just contained an open set containing the point. I guess the first one makes sense, my gripe was probably more with the second one which was what I was reading about.

[cedar pebble]

well so there's a notion of open neighborhood, and a notion of neighborhood

[surreal lantern]

you also have a convenient abbreviation with nbhd.

[wise ruin]

Not sure if this belonged in #top or #diffgeo, so top it is.

I am being asked to make sense of a classification of boundaries in the case of a closed manifold M. The classification relies on being able to construct, at a point x on the boundary E of M, a basis for the tangent space T_x M given a vector of T_x M and a basis for T_x E.
How is such a definition sensical for closed manifolds? Namely, I'm not sure what a basis for the "tangent space" of a nonexistent point on its boundary would look like.

[gritty widget]

I found a defn of a boundary that might help

[wise ruin]

I appreciate this, but I’m not looking for the definition of a boundary, this is already known. I’m trying to understand how an empty boundary would be classified when, in order to do so, I need a basis for a tangent space in the boundary

However, after looking back at the “definitions”, it seems like the best interpretation is that the empty boundary vacuously satisfies the criterion for both “types” of boundaries. So I will just go with this, thanks though!

[marsh forge]

I’m not sure if I am misunderstanding

Oh sorry never mind I was

It’s been a long day

Wait yes okay I was right the first time

I don’t understand the problem you are talking about

The classification of boundaries for a closed manifold is just that they are all empty

[wise ruin]

I’m not classifying boundaries by their cardinality, I’m classifying boundaries by whether they are “in” or “out”.
In fact, the problem I was wrestling with is exercise 1.2 of that TQFT book you posted in Ivory.

[marsh forge]

I’m still a little lost

Are you asking which closed manifolds can be boundaries of what higher dim manifolds

I was reading it as like

Which manifolds appear as the boundary of a closed manifold

Which ofc is easy to classify

[plain raven]

Let f : X -> Y be a function between topological spaces. Then f is said to be continuous at x if, for every neighborhood V of f(x), f^{-1}(V) is a neighborhood of x.

Equivalently, for every open set V containing f(x), f^{-1}(V) is a neighborhood of x.

I think this is a good motivation for neighborhoods.

We say f is continuous if it is continuous at every point in its domain.

From this you can recover the usual definition.

So my short answer is, neighborhoods are useful when you want to talk about functions that are continuous but some points, but not all.

When talking about functions that are continuous everywhere, it becomes easier to use open sets.

[surreal lantern]

[unreal stratus]

Also perhaps worth points out that so many properties of functions and spaces in general are dependent on local behaviour - e.g.. it suffices to check continuity (or smoothness etc as it may be) of a function on a neighbourhood of each point. I'd argue it's much more elegant linguistically too in many cases

[wise ruin]

For what it’s worth, here’s the context and the problem was “Make sense of the definition of in-boundary and out-boundary in the case of a manifold M with empty boundary.”
Like I said though, the empty boundary is then both an in-boundary and and out-boundary.

[gritty widget]

Why is this called this name? what relation does this have with chains

I don't know, alternatively you could just say it has countable cellularity

from set theory https://en.wikipedia.org/wiki/Countable_chain_condition

if we consider inclusions of non-empty open sets, then this is exactly countable chain condition from set theory

[wise coyote]

why can we not make a countable basis for an arbitrary metric space (X,d) by the following logic:
-take any point x in X
-take the set of open balls with rational radius as a basis B(x,r)
this would be countable since the rationals are countable, and for any point y in X, take some rational radius ball such that r > d(x,y)
and this basis is closed under finite intersection

i know that not every metric space has countable basis, so what is the flaw in this logic?

the definition i use for basis is the following from munkres:

[gritty widget]

can you describe your proposed basis more precisely?

what it's supposed to be isn't very clear from your description

[wise coyote]

take an arbitrary x in X

then the proposed basis is the set of all open balls B(x, r) where r is a rational number

sorry, i dont know how to use texit

[gritty widget]

thanks

i just can't read lol

[wise coyote]

so just all rational radius open balls centred at the same point where the point is arbitrary

[gritty widget]

one more clarification

is the proposed basis $${B(x, r) : x \in X, r \in \bQ},$$ or is it $${B(x, r) : r \in \bQ}$$ for some $x \in X$?

[gentle osprey]

TTerra

[gritty widget]

presuming the second, but i just wanna be sure

[wise coyote]

the second, yea

i mean, the first could easily be uncountable if X is uncountable

[gritty widget]

yeah

[marsh forge]

The latter won’t be a basis

[gritty widget]

the second is clearly closed under intersection, but that's not what (2) in the definition of a basis says

[marsh forge]

You can think about R to see a counterexample

Ie might as well choose x=0

Find an open set that doesn’t contain any ball around 0

[wise coyote]

wait so what is the difference between 2 and closed under intersection?

[marsh forge]

2 is just slightly weaker

You don’t need the intersection to be a basis element

[gritty widget]

ah i am drunk. this is a basis for a topology on R, but not the standard one

replace R with any metric space

the point is that not every metric space has a basis for its metric topology

[young lantern]

Because this is equivalent to requiring the space be separable

[gritty widget]

read their next message lol

[young lantern]

If you look at for example L2, you can see that it’s not seaparable

[marsh forge]

You really don’t need to go as far as l2 to understand why this construction fails lol

Also, you really aren’t answering the question at all other than saying it’s wrong lol

Which the question asker is already aware of

[gritty widget]

the open sets in the topology generated are just gonna be open balls at the point you picked, which makes it pretty clear that you won't get the original topology

[marsh forge]

@wise coyote maybe this is a point of confusion: just because some collection of open sets inside of a topology T satisfy the axioms of a basis, does not mean that this collection of basis open sets is a basis for T

You can potentially get a much coarser topology

[gentle osprey]

Rishi

[dusty talon]

why can't i delete this

[unreal stratus]

Probably left it too long

[boreal solstice]

Im reading a book and It says all finite subcoverings of are open coverings open But that would mean compact spaces would be impossible because every subcovering would need another subcovering and so on... What am i missing?

[hollow harbor]

If the cover you start with is already finite, you're done. A "finite subcover" can just be the cover itself, if the cover is already finite.

[boreal solstice]

Mhm, but then the closed interval [0, 1] wouldnt be compact

[hollow harbor]

The problem is to reduce infinite covers to finite covers: once that's done, we have enough to get a lot of stuff done. That's why compactness is a useful notion.

Why not?

Which open cover can you give me where I cannot find a finite subcover?

[boreal solstice]

Because to make a subcover of the infinite open cover of open sets only using the open sets you need infinite open sets

[hollow harbor]

I guarantee you that I don't need infinite open sets.

[boreal solstice]

But the subcover of an open cover needs to be open right?

[hollow harbor]

Open in the sense that each set in the subcover should be open itself, yeah

The subcover just consists of a bunch of sets that I already had in my bigger cover.

All those sets are open cause they were open before.

As elements of the bigger cover

[boreal solstice]

And then how can you make a finite subcover of the open cover that corresponds to infinite open intervals approaching 0 or 1

[hollow harbor]

That is not a cover, it does not contain the endpoints

That's why (0, 1) is noncompact

But once you add the endpoints, you need an open set around each

And that open set will be able to replace all but finitely many of your intervals approaching the endpoints.

[boreal solstice]

Ohhh, yeah, it clicks now

Thanks

[hollow harbor]

The more clear case of this would be to try to prove that the set {0} union {1/n : n in N} is compact

[gritty widget]

I don't understand the confusion here

[hollow harbor]

That gets rid of the messy interval stuff

[boreal solstice]

Me not getting the concept of cover

[hollow harbor]

And lets you see how an open set around 0 contains all but finitely many of the points 1/n

But the idea is the same for the unit interval (it's just harder to prove properly)

[gritty widget]

A cover is just something that sums to the whole space. In this sense, covering it

Because every element of your space is in one of the sets from the cover

[boreal solstice]

Yeah, i get that but i was not getting that an infinite union of open sets that approaches 0 or 1 doesnt cover [0, 1]

[hollow harbor]

That's a key to compactness.

Compactness is all about holes being filled in and ends being capped off

And too many things not being too far apart

(what you would expect "compact" objects to be)

[boreal solstice]

Yeah, the problem is that i try to visualize it in my head and It makes everything x1000 harder

[gritty widget]

Holes being filled in hmm

[hollow harbor]

Here i'm referring to the fact that a metric space is compact iff it is complete and totally bounded

This isnt catch all intuition-wise for topological spaces

But it's still usually a pretty good picture

[gritty widget]

{0, 1} has a pretty big hole (0, 1) here

But the rest of the intuition seems alright

[hollow harbor]

I suppose you have to be clear about what kinds of holes are a problem

Leaky ones

[gritty widget]

Yeah. But it's because arbitrary open covers are impossible to visualize

So you want to visualize [0, 1] itself and the process of how would you go about to extract a finite subcover from an arbitrary one instead

[plain raven]

Observation -

Let Y be a fixed space.

There's a monad on Top/Y which sends each f : X -> Y to the mapping cylinder M_f , equipped with the obvious projection onto Y.

The unit is the usual inclusion of X.

What I find interesting about this is that the unit of the monad splits iff Y embeds in X in such a way that f is a strong deformation retract.

An algebra of the monad is some kind of ... very strong deformation retract, haha.

It's cute if you want to work out what the extra condition is to be an algebra. Or I'll tell you. Whatever.

[gritty widget]

what's the property "nets have convergent subsequences" equivalent to

[unreal stratus]

subsequences or subnets?

for subnets that's just compactness

[gritty widget]

I know

[unreal stratus]

mb then lol

[gritty widget]

say a subsequence of a net is a subnet which happens to be a sequence

such space would have to be compact and sequentially compact

perhaps that's the characterization of it?

[rancid umbra]

if i have a manifold M, is the usual way of defining a basis for a topology on M to use the coordinate charts for M?

this seems like the topology on M would depend on the coordinate charts

[gritty widget]

A topological manifold can be defined as a topological space with few properties

But it can also be defined using charts

I'm guessing former would be preferred in the study of them since it's simpler. But probably both are beneficial

See Lee, either topological or differentiable manifolds

It should be in both of them

[gritty widget]

do carmo does this in his book on riemannian geometry. see chapter 0, remark 2.3

at least for differentiable manifolds. but just remove the smoothness assumptions present and you'll get what you want for topological manifolds

some people define differentiable manifolds this way to try and write introductions to differential geometry without much prerequisite topology knowledge

but no one defines "topological manifold" this way

relevant parts so you don't have to set sail

[fading vale]

I like that he uses phi for the empty set in (2) lmao

[rancid umbra]

thanks! doesn’t this depend on the charts tho? for example, the circle has two atlases, one parametrized by (t, +- sqrt(1 - t^2)) and the other by (cos t, sin t)

this should be independent of the charts used correct?

[gritty widget]

as long as two atlases are compatible (transition maps are homeomorphisms, diffeomorphisms, whatever), the topology induced should be the same

side remark: (cos t, sin t) is not an atlas, you need at the bare minimum two charts to cover the circle

[rancid umbra]

thanks for clearing that up

[gritty widget]

I have this weird question, dunno if it's more topology or linear algebra. Suppose I have a group $G$ and I construct $V=\oplus_{i=1}^\infty Lin(G)$, the linear spans being over $\mathbb{R}$, so a direct sum of real vector spaces; introduce the weak topology on $V$. Denote the $i$th copy of $G$ in $V$ with $G_i$. The claim is that $H=\cup_{i=1}^\infty G_i$ is discrete in $V$. Is there an obvious way of seeing this?

I tried going the rout of no accumulation point of $H$ is in $V$. To this end I can reduce the problem to no accumulation points in any finite dimensional vector subspace $W$ of $V$ (because of the weak topology). But besides this, I can't really see what I else I could do. Any ideas?

[gentle osprey]

miqq

[gritty widget]

Linear span where?

Weak topology? You mean if you map V to Lin(G)? So Lin(G) is a topological space now?

It's not a weird question but as it is, doesn't make a lot of sense

[gritty widget]

The weak topology is on V wrt the finite dimensional vector subspace of W, and the linear span is over R. Maybe it's a terminology issue, Lin(G) means the real vector space with an abstract basis in G

So C in V is closed iff C cap W closed for all f.d.v.ss of V

Yeah. Normally you specify that you're taking the free R-vector space with generators G

Lin(G) would be the linear span of G

That doesn't tell you what space we are in or if G are independent

So H is a linearly independent set

Your claim boils down to: if V is a vector space with weak topology and H is a linearly independent set of vectors in V, then H is discrete

In fact, H is a basis here

For v in H, let U(v) = {x in V : x expressed in basis H has non-zero v coefficient}

This is an open set in V and U(v) cap H = {v}

So H is discrete

@gritty widget

[gritty widget]

Of course! I actually got to the point that H is a basis for V but it didn't even cross my mind that it must be linearly independent.

Thank you very much!

[gritty widget]

How do I show there is no continuous bijection from (0,1) to S^1? Intuitively it's quite clear but not sure how to show it properly.

you mean, from (0, 1) to S^1?

"between" sounds a bit weird, because relation of there existing a continuous bijection from one space to another isn't symmetric

Yeah

Then there'd have to be a continuous bijection from union of two open intervals to an open interval

On both of those intervals it would be a continuous injection, so it'd have to be strictly monotone

[gritty widget]

How do you conclude that? 🤔

So without loss of generality there'd exist a continuous strictly monotone bijection from (0, 1) union (1, 2) to (0, 1)

[gritty widget]

By removing some point from (0, 1), say 1/2

removing a point of S^1 gives us space homeomorphic to (0, 1)

Oh I get it. Thanks

[gritty widget]

Say this function is f. Then A = sup f(0, 1) and B = inf f(1, 2) would exist and be some elements of (0, 1)

Moreover A <= B

Does there exist t in (0, 1) union (1, 2) for which f(t) = A then? No, there can't

this is a contradiction

[wary depot]

Is it true that the number of (not necessarily connected) n-sheeted coverings of S^1 is equal to the number of partitions of n?

up to isomorphism

[plain raven]

Hm. Are you accounting for the fact that an isomorphism may switch two components around?

Like consider the covering space over S^1, which is the coproduct of the identity map and the map z |-> z^2.

This has three sheets

but i could write it as 0 | 1 2 or 0 1 | 2

I think

If i'm understanding you correctly you would have to identify those partitions.

[wary depot]

yeah, those should be identified

[plain raven]

So maybe we could come up with a rephrasing which involves a choice of each equivalence class under this bijection. Example - all partitions of n where the partitions are in monotonically increasing order of size.

[wary depot]

my bad, I probably should have clarified that by "partition" I meant without regards to ordering

[obtuse meteor]

generally this is what is meant by a partition

[plain raven]

Hm ok thanks. Not familiar with this terminology i guess

[obtuse meteor]

quick proof: Two coverings are iso if and only if they have the same set of connected components (as always up to iso)
For each m, there is only one connected m-sheeted covering of S^1

[obtuse meteor]

this is why for the purposes of classification we only need to consider connected coverings

[wary depot]

alright, and then the connected components are themselves coverings

i see, thanks

[obtuse meteor]

I suppose you need to also know that sheets sum additively over components when they're finite but...obvious.

this means a covering of S^1 is classified by how many copies of R it has, and a partition of n where n is the number of sheets once you remove all copies of R, as the only infinite quotient of pi1(S^1) is pi_1(S^1)

er, assuming non-infinite connected components

[gritty widget]

Let $A \subset \mathbb{R}^n$ such that $A^c$ is unbounded. I want to show that A compact iff $d(A,B) >0 $ for any closed disjoint set $B$. I think I managed to show compact implies the other condition, kinda stuck on the other one and correct me if Im wrong but I think this other direction is not true in R (so need to assume $n>1$) since you can take say (-infty,0]. So how can I show that holds for $n>1$?

[gentle osprey]

Catematician

[gritty widget]

If d(A, B) > 0 for all closed disjoint sets B, then d(A, x) = 0 implies x in A, so A is closed

Yeah closed is easy, but the problem I have is how to show A bounded

Sry should've said that

It definitely uses that we are in dimension n > 1

Yeah

I think an illustrative example is something like (-infinity, 0] x R in R^2

Yeah I get why it breaks, we can take some kind of a graph of a function in those simple cases

Was thinking about something like R \ A_eps but it doesnt work unless maybe there is a way to make it work lol

(To construct a counterexample if A is unbounded)

If A is unbounded, we want to extract a sequence in A^c which would be unbounded and at the same time, get closer and closer to A

So we probably would like to, say, divide the whole space into rings {x : a <= | x | <= b}

And I think here is why n > 1, we use that those rings are connected

If there's no ring for us to choose a point from A^c, then well, this would contradict that A^c is unbounded

Kinda confused whats your idea, what are a and b and by sequence in A^c you mean a sequence of closed sets?

If there's no ring for us to choose a point from A, then this would on the other hand contradict that A is unbounded

[gritty widget]

a sequence of points, of course

a and b are constants

they will be chosen for every ring in the construction

because you want the rings to get more and more far from 0

and at the same time, you want them to contain both points from A and from A^c

Since such ring is connected, the intersection of A with it and A^c with it are 0 distance apart

because otherwise those intersections would be non-empty, closed and disjoint

so we can always take a point in A^c in such ring, which would be close to A

and we want to take them closer and closer to A

This sequence of points will be disjoint from A, closed, and would be 0 distance apart

contradiction

Sorry I dont get it but even if I did why is this sequence closed?

Doesn't it converge to something in A

no, it converges to infinity

so has no limit points in R^n

How are the rings chosen? I dont get the construction and still what are the a and b in the rings youre choosing.

numbers

so I choose 6 and 8

they are constructed recursively

Say R_n = {x : a_n <= | x | <= b_n} = R(a_n, b_n) is going to be the n-th ring

For R_0 we can take a_0 = 0 and b_0 to be large enough so that R_0 has non-empty intersection with A and A^c, and |b_0-a_0| >= 1

Then take x_0 in A^c intersection R_0 so that d(A, x_0) <= 1

For a_1 we can take something like b_0+1

Then choose b_1 so that R_1 has non-empty intersection with A and A^c and |b_1-a_1| >= 1

Then take x_1 in A^c intersection R_1 so that d(A, x_1) <= 1/2

and in general, take R_n so that it has non-empty intersection with A and A^c, |b_n-a_n| >= 1
a_n < b_(n+1)

and take x_n in A^c intersection R_n so that d(A, x_n) <= 1/2^n

Then {x_n} is closed and disjoint from A, but d(A, {x_n}) = 0

Hmm I think I see it now, and is |b_n - a_n| >=1 so that you choose different points or doesnt matter as long as the rings are increasing?

[gritty widget]

it's so that {x_n} doesn't have a limit point (in R^n), and converges to infinity

Okay yeah I see it now

R_n were taken to be disjoint also because we want {x_n} to have no limit points

Yeah I get it

Thanks, thought it's going to be some kind of one liner proof im missing

actually if you choose a_(n+1) = b_n+1 then you probably don't need that condition

[gritty widget]

if your proof depends somehow on dimension, then it's probably not a one-liner

Yeah, the problem stated implied it works in R which clearly doesn't

yeah. Here R^n was crucial to say that the rings R_n are connected so that d(A cap R_n, A^c cap R_n) = 0

We can see where the proof breaks using your example (-infinity, 0]

How is connectedness needed for that?

well you see that it doesn't work for R

If d(A cap R_n, A^c cap R_n) > 0 then A cap R^n and A^c cap R_n would separate R_n

so R_n would be disconnected

here it's also crucial that A cap R_n and A^c cap R_n are non-empty

[gritty widget]

I'm not sure where I should post this, but what is the purpose of dictionary orders?

[steel glen]

if a function has compact support on a space (X, tau), does that mean $f(\partial X)={0}$?

ok that's not what i meant to ask

if we have a set X \subset U, U being the universe, and f has compact support in X, does that mean f(x)=0 for all x in the boundary of X?

i think that's what i meant to ask

[golden gust]

yes, because every neighbourhood of a point in the boundary intersects U

for p in the boundary, if f(p)>0 there would be a neighbourhood of p where f is positive, which intersects U

I don't even think the support has to be compact

maybe I'm accidentally assuming some nice things (separation axioms) about U but I don't think so

[steel glen]

for my case that's nbd since i'm really asking about X \subset R^n, with the usual topology

but thank you. i'll try to wrap my head around this

[golden gust]

feel free to ping me if you have questions about anything I said

[unreal stratus]

Rephrasing, { x | f(x) > 0 } is an open subset of X and hence contained in the interior of X

[dry jolt]

Given the action of a group on a space, there is an induced action on (co)homology. How should I think of this induced action intuitively?

I guess right now, the picture I'm starting with is to take a simplicial structure and see how the group action affects it, but I'm not sure how to visualize what the fixed points and orbits in homology correspond to

[boreal solstice]

What does the "real projective plane" has to do with the real plane? Im reading about it and the book im reading describes it like "an homeomorphism from the mobiüs strip with a disc attached to its boundary"

[gritty widget]

You mean the lexographical orders? They're the most natural way to totally order cartesian product of (totally ordered) sets

They're useful for examples and counter-examples in topology. Some people study order topologies and they'd obviously be useful there.

[gritty widget]

It's a 2-dimensional real manifold

Projective spaces are pretty big but I have yet to see them applied, someone else could expand on this

They're a special case of the Grassmannian

[raven spire]

Yes

[gritty widget]

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

see "Puzzle solutions"

[hardy robin]

thanks!

[winged viper]

can anyone explain in a simple way how dehn surgery is possible? so from my understanding, I take the 3-sphere, remove a solid torus T, then reattach a solid torus T' so that the boundary of T and T' agree. How is it possible to reattach T' in a way that is "different" from how T was originally attached?

[cedar pebble]

what do you mean how is it possible?

you're choosing a homeomorphism T^2->T^2 to glue along so to speak

[winged viper]

like is there a simple example

im just imagining taking R^3 and removing a torus, and the only way i can think of putting it back is in the same way. but this is probably just cuz i cant visualize 4 dimensions?

or maybe im misunderstanding the definition

[cedar pebble]

do you know what a Dehn twist is?

[winged viper]

hmm ive had it explained to me before, i can try to look it up again

[winged viper]

ah yes i think i remember

[cedar pebble]

you're cutting an annulus from a surface, modifying by an element of the mapping class group, and then gluing back in

Dehn surgery is the same thing just a dimension higher

[winged viper]

ohh okay i see

[winged viper]

ok dehn twists make sense ill just have to think about the higher dimensional one more lol

[cedar pebble]

this is sort of the picture of how you're doing a twist in Dehn surgery

[winged viper]

wait sorry in this case, you're removing an annular region from the torus and gluing it back, but isn't the resulting manifold still a torus

[cedar pebble]

yes

[winged viper]

in dehn surgery is it possible to glue back the torus and get a different manifold?

[cedar pebble]

yes this is the point, you remove a solid torus, and glue it back in along a homeomorphism of the boundary

that homeomorphism is this Dehn twist

you typically end up with a different manifold than when you started when you do this procedure

[winged viper]

is this possible in a dimension lower? like is there a way in the 2-dimensional case where you can remove an annulus, glue it back and you get a different manifold?

[winged viper]

since in like this example you get the same manifold back

[cedar pebble]

the lower dimensional analogue would be like, cutting a disc from a surface and gluing it back in along a degree n map on the boundary S^1

[winged viper]

ohh i see

ok that kinda makes sense thanks

[gaunt linden]

For a worse example that has no lower-dimensional analogy (I think), the homeomorphism of the torus boundary could be one that interchanges "latitude" and "longitude". One kind of great circles on the torus can contract through the solid torus itself; the other cannot. And it's the opposite way around for contracting through the complement of the solid torus. So if we stitch the solid torus back in after that twist, we get something that is not simply connected -- some of the circles contract neither on the inside nor on the outside.

[gaunt linden]

Dehn twists and Dehn surgery are separate things, right?

[silk tapir]

Hello all, stuck at a fairly basic question. I am trying to prove the fact that if f maps convergent sequences to convergent sequences, then the preimage of an open set under f will be open. Not sure how to go about it. Any hints? I thought maybe I ought to try showing the contrapositive?

[gritty widget]

Try proving that the preimage of a closed set under f is closed maybe

also I don't think this is equivalent in general, unless you're maybe, in a metric space

[unreal stratus]

Equivalent for first countable spaces I suppose

[rancid umbra]

anybody have any hints on how to show that GLn/On is homeomorphic to [0,infty)?

[cedar pebble]

GL_n(R)/O(n)?

that's not homeomorphic to [0,\infty)

[rancid umbra]

i’m trying to do 3-24

not quite seeing how the hint helps atm

[pastel linden]

O(n) is inner product preserving, thus norm preserving on R^n. The orbits of this action are thus spheres of fixed radii

[rancid umbra]

i think i’m confused a bit

how is the quotient space of R^n by On defined?

[cedar pebble]

okay this is R^n/O(n), not GL_n(R)/O(n)

the points of the quotient space are orbits of the action of O(n) on R^n

O(n) acts on R^n by rotations, so the orbits are spheres of a fixed radius as bacono said.

[rancid umbra]

i’m confused on how R^n/On is defined

[cedar pebble]

two points in R^n define the same point in R^n/O(n) if they are in the same O(n)-orbit

[rancid umbra]

sry, i was confused on notation. A/B has a lot of different uses and i was messing them up

thanks

[unreal stratus]

Q/Z or R/Q has caught me out a couple of times aha

[cedar pebble]

Q/Z is wack

[hidden crag]

Would that mean that they’re equivalent iff they have the same norm?

[rancid umbra]

yea

that’s why the hint helped

[hidden crag]

Alright neat

That’s kinda cool

[rancid umbra]

right

[cedar pebble]

(I complained earlier because GL_n(R)/O(n) is something different: it's homeomorphic to the space of upper triangular matrices with positive diagonal entries, by the QR decomposition)

[hidden crag]

Oh makes sense

Learned those decomposition in numerics some time ago, didn’t think I’d see them again

[rancid umbra]

so would u map an orbit OnM to the unique upper triangular matrix in the QR decomposition for the homeomorphism?

[cedar pebble]

yup I believe that's how the identification goes

this is kind of a general pattern when you describe some quotient of matrix groups G/H in terms of some matrix decomposition

[tawdry widget]

This is the definition of classifying space on fiber bundles by husemoller . And wiki gives classifying space of U(n) like this. Are they homeomorphic? How?

[cedar pebble]

idk if they will be homeomorphic but they should be homotopy equivalent, which is what matters for the construction

for what it's worth the infinite Grassmannian definition seems to be the more modern definition that shows up everywhere

[chrome ridge]

This is a construction of a 4-manifold with a given group presentation. Why does $\pi_1(X \setminus int(N_j))$ is still the free group on generators $a_1, ..., a_{|S|}$ ?

[gentle osprey]

ru0xffian

[cedar pebble]

so there's a kinda subtle difference between retraction and deformation retraction but roughly speaking you think of these as continuously shrinking a space into a subspace

e.g. a cylinder S^1x[0,1] or a Möbius strip both deform retract to a circle S^1

they're important because these are examples of homotopy equivalences, so all the different invariants you study in algebraic topology (homotopy groups, (co)homology groups, etc) are invariant under deformation retract

Other way around (the example with S^1 and R^2-{0} is extremely similar)

But yeah this map from the cylinder to the circle that collapses the interval is a retract

[steel glen]

sorry for the necro, but im guessing its to construct ordered topologies on set products

[uncut surge]

My instinctive thought was "okay, but order topologies are kinda pathological anyway" when thinking of something like the lexicographic order on the unit square, but I suppose this is useful for lattice things anyway

And I guess generally lexicographic orders are just a thing in the real world that people sometimes actually work with, so even without the topology in the background, they are useful and good to think about

[gaunt linden]

It's the simplest way to make a total order on a cartesian product of totally ordered sets, and total orders are often useful to have.

(It's an additional bonus that it preserves the "well-order" property of the original orders if they have it).

[swift fjord]

This is one way to define ordinal multiplication

[chrome ridge]

This is a construction of a 4-manifold with a given group presentation. Why does $\pi_1(X \setminus int(N_j))$ is still the free group on generators $a1, ..., a{|S|}$ ?

[gentle osprey]

ru0xffian

[chrome ridge]

Why doesn't removing the tubular nbhd change the fundamental group ?

[shadow charm]

For any space with the discrete topology I can define a delta complex structure on it by letting every point be the image of a 0-simplex right?

[plain raven]

Yes.

[uncut stratus]

Hi everyone! I'm currently writing a thesis on simplicial homology groups, and I wanted to incorporate a chapter fully dedicated to examples and computations of homology groups. For now I've included zero-dimensional homology and the homology of surfaces (I'm reading Munkres' Elements of Algebraic Topology), and I saw some yt videos titled "homology on graphs". Does anyone know any book which talk specifically about homology on graphs? It would be fantastic if it used just the definition given for simplicial homology (for example I've found a book that talked about homology on graphs but used terms like "cell complex" which I've not treated in my thesis, and I don't want to develop a whole new theory for just a section of a chapter, it should be like a class of examples).

(I tried asking on #book-recommendations and I was redirected here)

[lunar yoke]

graphs are 1-dimensional simplicial complexes / cell complexes, so it's not really a surprise that this term pops up

you define homology for these complexes, and graph homology follows as a special case

there is not much to it, so I would be surprised if there is a whole book on homology of graphs unless it goes into heavy graph theory

maybe also check out the wiki article https://en.wikipedia.org/wiki/Graph_homology

[uncut stratus]

Thank you. I've already introduced complexes and the general homology groups, I thought there was something more to discuss about the 1-homology of graphs (like, idk, some special results, for example with the fundamental group one shows that every graph can be seen as the product of n circles glued in one point, so once you compute the fundamental group of those n circles glued together you are done, I was thinking about putting a computational example like that for homology)

[lunar yoke]

yes the same happens for homology with basically the same proof

[uncut stratus]

But I don't want to go too deep on graph theory since it's not the aim of my thesis. Thanks

[lunar yoke]

because everything is homotopy invariant

you quotient out a maximal subtree and get a wedge of circles

then the fundamental group on that is free on the number of circles

and likewise the 1st homology is free abelian on the number of circles

generally you have H_1(G) = Z^(|E|-|V|+1) for a connected graph G

as you can see on the wiki article

[uncut stratus]

I haven't shown that $H_p(K)$ is homotopy invariant, indeed that could be a problem, i don't want to use results I've not proven 😅 . I'll definitely check the article

[gentle osprey]

Mòrag

[swift fjord]

The proof of homotopy invariance for simplicial homology is an incredible drag

What people usually do is prove it for singular and show Equivalence for triangulable spaces

[lunar yoke]

not if you use the method of acyclic models

[swift fjord]

I'd still call that a drag phil

[lunar yoke]

no way

[uncut stratus]

I know, I was trying to avoid it like hell, it's a mess with simplicial approximations and stuff

[lunar yoke]

its so much more straightforward

[swift fjord]

I think something like this which will take a while to develop can be stated without proof

It's pretty much the main reason one cares about homology at all

[uncut stratus]

I was thinking about that. Is singular homology hard to introduce? Could it be done in a single chapter, with the proof of its equivalence to simplicial homology (and its topological invariance)?

[lunar yoke]

the definition is easy, but verifying the axioms takes some work

by axioms i mean like homotopy invariance, excision, etc

[uncut stratus]

Oh, those things I've skipped, like... Eilenberg-Steenrod axioms are they called?

[lunar yoke]

yes

these make homology fairly easy to compute

as compared to (higher) homotopy groups for example

[uncut stratus]

Really? My professor who is supervising me just said "ye you can skip those"

[lunar yoke]

well maybe its out of the scope of your project

[uncut stratus]

I'll check them out nonetheless, thank you

[plain raven]

Proof of the excision axiom using acyclic models

[marsh forge]

Not everything that satisfies the ES axioms (minus dimension) is easy to compute!

In fact, stable homotopy satisfies them 🙂

[lunar yoke]

fair enough, i was only thinking about singular homology really

[swift fjord]

I somehow doubt that they want to get into acyclic models in their paper

[lunar yoke]

btw this is some of the coolest stuff i've seen in a while, thanks!

really nice and short

[plain raven]

yeah i am like

a collector of all things acyclic models

i hoard everything i come across

[lunar yoke]

nice

[plain raven]

Barr wrote a 200 page book on applications of acylic models

[lunar yoke]

yeah i saw that but never really made the time to look into it

[plain raven]

it's not bad it's like

pretty self contained

it's like 120 pages of general hom alg to just get all the groundwork laid for the main theorem and applications

dude really likes acyclic models

[lunar yoke]

you too apparently xd

i dont know much about them except the few applications i've seen, but they are all really neat

[plain raven]

i just found a new version of the theorem last night

and i've been thinking about it this morning

it's in an old paper of borel

[lunar yoke]

maybe you could actually introduce singular homology like this in an algebraic topology course if there was a course on homological algebra and some category theory the semester before

we only got to see acyclic models very briefly in the follow up course sadly

[plain raven]

just start the algebraic topology class with like 6 weeks of category theory and ignore the kids who keep asking about """"" spaces """""""

whatever those are

[lunar yoke]

our class literally started with the eilenberg steenrod axioms lol

[marsh forge]

Just start with spaces, define homotopy types, and then stabilize them

then you never need to define singular at all

Just provide the t-structure on spectra and note that the heart is equivalent to Abelian groups

then you get singular homology for free

[plain raven]

nice

[lunar yoke]

true just start teaching from blue book

why not

[marsh forge]

this is way more insane than the blue book

which is altogether pretty reasonable

[lunar yoke]

is this understandable without too much knowledge on infnity cats and if so do you have a paper / source recommendation for this

[marsh forge]

hm

[lunar yoke]

i know how the stable category is triangulated

but ive never worked with t structures

[marsh forge]

So this is just a very dumb way of saying

that the spectra with homotopy concentrated in degree 0

are the EM spectra

which are equivalent (as a subcategory) to abelian groups

[lunar yoke]

oh i see

yeah ok that was actually mentioned in our course

but not shown

that EM spectra are "the way to port algebra into stable htpy theory" or something like that

[marsh forge]

Right

So the construction is usually not so bad, but is model dependent

You start with EM spaces and just apply the \Sigma^\infty functor

[pale frost]

i dont mean to interrupt but ive heard about homotopy having a meaning in other categories. are there examples of uses of this?

[marsh forge]

this is just model category theory and infinity category theory

the basic example is the derivated category

where "homotopy" is really just homology

[pale frost]

derivied category

[lunar yoke]

oh yeah we also had a construction of these, but we didnt show that the stable htpy groups are concentrated in deg 0

[pale frost]

oh

okay

[marsh forge]

Oh

I mean thats a very simple computation

Since the EM spaces have homotopy concentrated in a single degree

you just write down the relevant colimit

and see that its identity in degree 0 and 0 everywhere else

unless you are saying you didn't show that EM spaces have homotopy only in one degree

which follows from their construction

[lunar yoke]

no yeah ik that

[pale frost]

i wonder if there is an inbetween where homotopy isnt either homological algebra or the usual theory

[plain raven]

EM spectra makes me think "electromagnetic spectrum"

[lunar yoke]

i think its just that we had a different construction

we didnt work immediately in the stable category

[marsh forge]

There are all sorts of model categories and infinity categories out there

[lunar yoke]

but worked with the model of orthogonal spectra

[marsh forge]

ah

Orthogonal spectra are

yucky

[lunar yoke]

well they're certainly better than symmetric spectra imo

[marsh forge]

many things are better than garbage

[lunar yoke]

xd

[marsh forge]

no i mean

at the time

[lunar yoke]

what other model do you prefer then

[marsh forge]

these ideas were super revolutionary

[lunar yoke]

or do you just not care at all

[marsh forge]

but the infinity categorical approach to spectra is just miles better

[lunar yoke]

true, that was also mentioned in the introduction

[pale frost]

this has to be related to spectra right? or dont you need talk about spectra when talking about homotopy in other categories?

[marsh forge]

you dont

[lunar yoke]

but you cant just require that ppl know infinity category theory in their 2nd master semester

[marsh forge]

Spectra end up being integral to the theory of stable homotopy categories

[marsh forge]

True, I would use sequential spectra

[lunar yoke]

oof

but then what about the symmetric monoidal structure

[marsh forge]

Assert it exists

[lunar yoke]

hm

[pale frost]

maxj how did you fo about learning AT for your degree?

or do you do AT?

[marsh forge]

I'm a stable homotopy theorist

im getting my phd at the moment

[pale frost]

what made you go this path?

since start of phd?

[marsh forge]

its neato mosquito

Oh I was on this path way before phd

[lunar yoke]

stable homotopy theory is just pretty cool

[marsh forge]

i'm one of the most like

specialized people I know

[pale frost]

bruh moment

[marsh forge]

in early grad school anyway

[pale frost]

i want to study AT but i only took one course

[marsh forge]

I worked with Peter May at UChicago as an undergrad

[pale frost]

and i think exploring is good

[marsh forge]

and that basically set me on this path

[pale frost]

woah!

[lunar yoke]

cool

[marsh forge]

Did his REU twice and mentored for it once

[pale frost]

oh jesus

[lunar yoke]

i heard later on you dont really care much about the actual model as you said, and the phd student i talked to also said that they think of spectra as abelian groups

[pale frost]

are u phd there?

[marsh forge]

I probably would not have gone to grad school if I didn't already know what I wanted to research, and before I clicked with AT I was sort of getting bored with math

I'm a PhD student at UCSD

working w Zhouli Xu

[lunar yoke]

i guess this will get more clear at some point but rn it just confuses me

[pale frost]

ive heard omega spectra being something about abelin groups

but thats all i know

[marsh forge]

The first statement is correct

[lunar yoke]

i guess thats from the E_infinity stuff

[marsh forge]

the second statement is unhinged lol

I think the right way to think about it is like

Stable homotopy categories (triangulated categories) are something in between

a "space like" category

and an abelian category

But there are fundamental differences

[lunar yoke]

hmm ill learn more about that next semester

we're doing a seminar on enhancements of triangulated categories

with lots of infty stuff in it

[marsh forge]

nice!

Yeah currently AT has a pedagogy problem

where to read basically any paper written in the past 5-10 years

you need to know lurie's approach to infinity categories

but this stuff can be hard to learn and its rarely taught

[lunar yoke]

and that takes a long ass time to get used to

[marsh forge]

I think graduate schools just like

need to offer topics courses on them at this point

[lunar yoke]

there was a 3 semester course on infinity categories here in bonn before i came here, and since then nothing of the sort

[marsh forge]

oh! Youre at bonn

Thats a great place to learn some homotopy theory

[lunar yoke]

eh sadly its kinda dying out rn

at least it feels like it

schwede is basically the only one doing this

rest is geometric topology

or at least the people at the max planck institute doing this stuff are not giving lectures

it is what it is

[marsh forge]

You should try to meet Dominic Culver too

And Bastiaan Cnossen

[lunar yoke]

yeah i know him already

[marsh forge]

oh fuck lol

how'd I miss good old toby

Tobias Barthel is like

one of the pre-eminent people in my field

[lunar yoke]

hes the guy who told me spectra are like abelian groups lmao

[marsh forge]

lol

oops

[lunar yoke]

he doesnt teach and doesnt do master theses sadly

[marsh forge]

Yeah but maybe you can still corner him or get coffee or something haha

It would be a shame to be at Bonn and never meet with him

[lunar yoke]

most of the people i know from the weekly topology seminar, but he's never been there afaik

[lunar yoke]

well i still have a while to go

like 1 or 1.5 years for master

[marsh forge]

I hear he might be a little on the private side

[lunar yoke]

planning to start thesis early next year and do something like global homotopy theory

[marsh forge]

Nice!

[lunar yoke]

but idk yet

for now next semester is equivariant stable htpy theory

[marsh forge]

Ugh I have so many projects I want to do

[lunar yoke]

and this semianr

[lunar yoke]

relatable

also since i got here ive basically only been doing abstract stuff, and i feel a little sad that i never really did a lot of computations

especially with cohomology operations and the like

or spectral sequences

which apparently everybody dreads but i've never actually done them

[marsh forge]

They are not so bad

[lunar yoke]

i think i saw your video on the serre spectral sequence on yt a while ago

i really liked it

[marsh forge]

hahah like you found it independently?

[lunar yoke]

yeah it was recommneded to me actually lol

[marsh forge]

oh wow almost 500 views

im basically famous

holy shit my hair is so short

If you want to read some like

stuff at the edge of abstract homotopy theory and computations

[lunar yoke]

rn i also have an exam coming up in one week on L^2 invariants, its a good refresher to do some more geometric type topology

[marsh forge]

work by my advisor Zhouli Xu, Dan Isaksen, Bogdan Gheorge, and Guozhen Wang might be of interest

[lunar yoke]

aight i shall add that to my readlist

[marsh forge]

this is the computational motivic homotopy story

super fascinating stuff

[lunar yoke]

oh motivic

[marsh forge]

and what most of my current research is related to

[lunar yoke]

i have absolutely no idea what that is

[marsh forge]

There are a few different ways to think about it

one is as a homotopy theory for schemes

[lunar yoke]

but its a word i've seen a lot also on the AT server

oh no, AG is my nightmare sadly

[marsh forge]

The other way to think about it is as a deformation of spectra with respect to the Adams-Novikov

I also don't love AG

I think about it as the latter thing

[lunar yoke]

i tried it when i came here but the course was just straight up insane

bonn ag is not to be messed with

[marsh forge]

I think AG requires a particular brand of persistence and interest to really get invested in

and if it doesn't click with you its hard to force it

[lament needle]

I've been stuck on this for a while. for b id think That it's similiar to tychoffs theorem and I choose each of these sets in the family to be compact.

I know for the product topology on an infinite family for any open set O there can only be a finite number of preimages $\pi_i(O) \neq X_i$ but this restriction doesnt seem to appear here

[gentle osprey]

matthew

[lament needle]

here it would appear that each for $(X_i,i) \subset \coprod X_i$, $X_i$ is an open set.

[gentle osprey]

matthew

[gritty widget]

Assuming all X_i are non-empty, there needs to be a finite amount of them for their disjoint union to be compact

Because X_i form an open cover of the disjoint union

And are disjoint

[lament needle]

Right, that is the problem that was scarying me

But that is a condition on the size of A and not $X_a$

[gentle osprey]

matthew

[gritty widget]

Next, each X_i needs to be compact, because they are closed in the disjoint union

[gritty widget]

Too bad, it'll depend on A

Now what's left is to prove that a finite disjoint union of compact sets is compact

[lament needle]

That Im comfortable proving

But what you are saying is that we can't treat infnite families

[gritty widget]

No. All but finite amount of X_i need to be empty

[lament needle]

ok, then I know what to do. Thanks

[gritty widget]

Np.

[lament needle]

I just thought I had to treat infintie families and I went crazy

[gritty widget]

Well. You do, but it's not compact there

Because if X_i is non-empty for infinite amount of i then non-empty X_i form an open cover without any subcover

So their disjoint union isn't compact

[bitter smelt]

Does anyone have a SVK diagram for more than 2 open sets?

I would like to know if it is just the SVK diagram for two with another branch off the bottom

nlab has the statement for two open sets, but hatcher has it for any number so long as the intersections are "nice" https://ncatlab.org/nlab/show/van+Kampen+theorem

wolfram has this as well https://mathworld.wolfram.com/vanKampensTheorem.html

(though it cites hatcher lmao)

[marsh forge]

Not sure what you mean by diagram?

in the category of groups you should get the colimit of the diagram where you have all the three-wise intersections including into all the cover components

[bitter smelt]

I mean the pushout diagram

The statement for hatcher implies we do not consider three way intersections, which is part of what makes creating the diagram a little confusing to me

elements from N come from pairwise intersections

[cedar pebble]

For three open sets you should get a sort of pushout cube

[pure hornet]

The iwoat seminar has started! The recording is at https://youtu.be/HRsDJPzl_Hg . Please admire and appreciate the teaching of one of our lords and saviors, prof. Jacob Lurie!

[shadow charm]

In Hatcher it is said that the torus forms a two sheeted covering of the Klein bottle by looking at the Klein bottle as the orbit space of R^2 through a group of the form (ZxZ) x< Z/2Z, which has ZxZ as an index 2 subgroup (here x< indicates a semidirect product).

Is this a general theorem? How would it be rigorously stated?

(That subgroups of deck transformations correspond to orbit spaces that cover the original space?)

[gritty widget]

What is a borel action?

[gritty widget]

They either mean that every g represents a Borel map or that it's jointly Borel

Most likely the latter

what borel structure on G?

G is a discrete group

Which means it's homeomorphic to natural numbers

Being a topological space, it has a Borel sigma-field

yeah but it can be homeomorphic in many different ways, its not clear that those ways preserve the group strucutre or something like that

What's your point

i dont see why it doesnt matter which homeomorphism to natural numbers you take actually nvm

its discrete so doesnt matter

i understand now

That was just to help you imagine what that is topologically

i was thinking about 2^N instead of discrete for some reason

The group structure is irrelevant to me

since its discrete it doesnt matter, i was thinking about non-discrete topologies and those could be put in different ways in G

and that could perhaps matter

Idk. No one says it's a topological group

But ig that's implied. But doesn't matter

We were talking about in what way the action can be Borel, and giving G a topology answers the question (+- it's not really "Borel" but more like measurable because X doesn't have a topology but that's like, can be ignored)

yeah but if the topology on G wasn't discrete, then different ways to put the topology on it could make the map stop being measurable

wait

if G is discrete then isnt any map measurable ?

[swift fjord]

Doesn't standard borel imply a measurable space structure

[gritty widget]

You have a map from G x X to X

[gritty widget]

ah ok

[swift fjord]

I think they're then asking for the action of g as a map X\to X to be measurable

[gritty widget]

Yes

[swift fjord]

Idk if they mean that the action is measurable as a map G x X\to X

What's the context Carla

[gritty widget]

borel equivalence relations

[gritty widget]

If you google Borel actions then they mean for the whole action to be measurable

Not just the maps from X to X corresponding to elements of G

[swift fjord]

Alright then

[gritty widget]

Anyway I think those might be equivalent but I don't want to bother proving it

(in this case of G discrete)

it is equivalent cuz inverse image of open open iff of form whatever x open in X

replace open by measurable

huh

Yeah actually it is simple

$$h:G\times X\to X,\ A\subseteq X$$ with $A$ measurable, then $$h^{-1}(A) = \bigcup_{g\in G} ({g}\times X \cap h^{-1}(A))$$ is a countable union of measurable sets.

[gentle osprey]

Blitz

Blitz

[fading vale]

Im not 100% sure im clear on what you're asking but here the neighborhood being evenly covered means that its preimage under p: X tilde -> X is a union of disjoint opens mapped homeomorphically onto N_t x (a_t, b_t)

and the point is that because such U cover X, you have F(y, t) in some U evenly covered by p. then the preimage of U under F in Y x I will contain such a N x (a_t, b_t), since the N and (a_t, b_t) cover Y and I respectively

tilde F isnt being used at all in this part of the proof

(it might help to like, draw a commutative diagram and look at it while thinking about this stuff because its kinda easy to mix up maps)

[wise ruin]

Oh you’re right, I was just mixing up maps. Thanks for pointing that out

[lament needle]

Not sure how to approach this. I know it's easy to map X to a line segment, and X is a collection of line segments of length 2 B_i so I tried to think of f as a family of functions {f_i}_{i \in I} each defined by the preimage of f on a closed line segment B_i, and I know I can take the extension of each of the f_i's and that i can define f by each of the f_i's but it didnt seem like i could define an extension of f from this family of extended functions.

Another idea I had was to compose f with a map $g: R^n \rightarrow R$, and then I can take the continous extnesion of $g \circ f$ but I know g cant be a continous injection so that seems like a problem

[gentle osprey]

matthew

[gritty widget]

look at the components of your map from A to the ball

you can apply tietze to each of those

you'll just have to make sure, either in the application of tietze or after the fact, that the extension takes values in the ball

forgot how to english words

[lament needle]

what do you mean by components

[gritty widget]

f maps into (a subspace of) R^n

f = (f_1, ..., f_n)

f_i are the components

[lament needle]

oh geez I should have thought of that. oof

empty brain moment, ok then i know how to proceed

[plain raven]

I came across an old notion of factorization system in the literature and it's all spelled out somewhat "equationally"

I am posting this here because i am hoping that more recent literature has given an equivalent formulation of the notion more .. categorically

like

idk factorization systems are a kind of algebraic structure right

So let C be a category and M be a subcategory of C. (Faithful but not necessarily full)

Our factorization will, for each u : m -> c, where m is in M and c is in C, return a kind of 'terminal factorization' of u through morphisms in M; the factorization gives a decomposition

u = beta(u)\circ alpha(u) where alpha(u) is a morphism in M.

It should satisfy the following axioms:

1. alpha(id_m) = id_m; beta(id_m) = id_m.

2. For any u, alpha(u) is a morphism in M; for any u : m -> m' in M, alpha(u) = u, and beta(u) = id_m'

3. For any u : m -> c, alpha(beta(u)) = id_m' and beta(alpha(u)) = id_m'

4. alpha(alpha(u)) = alpha(u); beta(beta(u)) = beta(u)

And lastly

~tex [\begin{tikzcd}
m && c && {c'} \
& {m'} && {m''}
\arrow["{\alpha(u)}"', from=1-1, to=2-2]
\arrow["{\beta(f\beta(u))}"', from=2-4, to=1-5]
\arrow["{\alpha(f\beta(u))}"', from=2-2, to=2-4]
\arrow["u", from=1-1, to=1-3]
\arrow["f", from=1-3, to=1-5]
\arrow["{\beta(u)}"{description}, from=2-2, to=1-3]
\end{tikzcd}]

[gentle osprey]

diligentClerk

[plain raven]

in the diagram above,

$\beta (f \beta(u)) = \beta(fu)$ and $\alpha(fu) = \alpha( f\beta(u))\alpha(u)$

[gentle osprey]

diligentClerk

[plain raven]

I have tried a bunch of different ways of taking a crack at this but to no avail

[void gazelle]

I am not sure if this is the right place to ask, but if I want to define limit of sequence of non-closed subsets, is it possible? can they form a Hausdorff space?

[plain raven]

Can you give more detail about the scenario? If X is a space and we take an ascending chain of subsets, we could take their union or their direct limit

[void gazelle]

oh, sorry, If X is Riemann sphere, can we define limit of sequence of non-closed subsets of Riemann sphere

[gritty widget]

There are ways to give the topology to all subsets of a topological space but they are almost always not Hausdorff

And not very interesting

Try book hyperspaces by Nadler

[shadow charm]

Am I correct in computing the universal cover of a sphere with a diameter as an infinite alternating chain of spheres and intervals

[void gazelle]

Thank you!

If Dn and D converges in Hausdorff distance, do their complement also converge

[shadow charm]

And for a sphere with a circle intersecting in two points x_0, x_1, I’d get this infinite pattern of helices tangent to spheres?

Pictures of the two. First is easy to see from drawing but the second is whacky to draw

[gritty widget]

I don't think so. Consider R(t) = {r exp(2pi is) : 1 < r < 2, 0 <= s <= t}, then R(1-1/n) should converge to R(1) but closures of R(1-1/n)^c are continua in the Riemann sphere while closure of R(1)^c isn't, but this impossible

The space of subcontinua of the Riemann sphere is compact

You can probably see this explicitly too

The point 1.5 exp(2pi i (1-1/(2n)) should be the culprit here

[void gazelle]

Thank you

[obtuse meteor]

I agree with this but the way I’d compute it may be different

A sphere with a diameter is homotopy equivalent to a wedge sum of a sphere + a circle

And wedge sums behave quite nicely with universal covers

[obtuse meteor]

Great blog post abt this: https://www.math3ma.com/blog/a-recipe-for-the-universal-cover-of-x-y

[shadow charm]

Thanks I’ll give that a look later!

[shadow charm]

Yeah alright I see, this is how I approach wedge sums too so good to know it’s not unjustified. And I’m pretty sure the same method also confirms my second drawing so that’s cool, thanks 👌

I just always feel a little unrigorous about making universal covers because it feels very intuitive I kind of just imagine starting from a point and extending the space in local steps while keeping stuff simply connected, but then I never bother getting a explicit form of the space or showing that you actually do get a covering space (at least most of the time it’s obviously simply connected)

[gritty widget]

Sin(1/x) continuum and {0} x [-1, 1]

[unreal stratus]

Yooo

[gritty widget]

Graph of sin(1/x) for 0 < x <= 1 say and the interval I wrote

That's called sin(1/x)-continuum

Yes

[unreal stratus]

(Silly example but also like a circle retracting to a point for example)

[gritty widget]

Something something, sin(1/x)-continuum is not path connected