point-set-topology

[ornate berry]

(At least, this is my justification)

[gritty widget]

$\pi_i:\mathbb{R}^n\to\mathbb{R}$ given by $\pi_i(x_1, ..., x_n) = x_i$

[gentle osprey]

Blitz

[candid hedge]

So we're modeling each variable as a projection right?

[unreal stratus]

So you're saying $\pi_i(\mathbb R^n) = \mathbb R$? Scandalous

[gentle osprey]

potato

[unreal stratus]

[jk]

[gritty widget]

This is what it is

And a polynomial is obtained from those and constant functions using multiplication and addition

[candid hedge]

On another note i asked a question in help and a helper gave me this answer. I am not convinced. can someone weigh in and tell me if this argument is flawed or not

the question is f(x+y)=f(x)+f(y) and f continuous at 0 show that its continuous everywhere

[gritty widget]

This should probably be in #real-complex-analysis

[candid hedge]

oh

[candid hedge]

Okay got this

thanks

[bitter smelt]

Should the homology groups of the klein bottle be anything recognizable when the ring is not Z?

Simplicial homology btw

[small hemlock]

There is an uncountably infinite amount of possible topologies one could define on R, right?

[tidal cedar]

yes

[junior flower]

something something even P(P(R)) topologies

[long hornet]

Check out the universal coefficient theorem. It boils down to whether 2 = 0 in the coefficient group. So for Z2, the groups are Z2, Z2 x Z2, Z2, but for, say, Zp (with p odd), they are Zp, Zp, 0 iirc.

[bitter smelt]

We haven't gotten to any theorems yet, so I'm just trying to prove what it is from definitions alone

For a generic commutative ring R

[hard wind]

I’m beginning Singular Homology in Hatcher and he says this, but doesn’t clearly explain what a “singularity” means

Is this a region of the map that isn’t injective? Or something else

I just want to figure out what intuition he’s getting at before I move on

[ornate berry]

yeah p much

It is as he says: its image may not look like a simplex, insomuch as it is squished beyond recognition

(he doesn't say the second part, but this is what the first part means)

[tawdry valve]

This is also in contrast with simplicial homology, where we had characteristic maps sigma_alpha which were actually embeddings (in the delta complex structure). Now we’re looking at arbitrary continuous maps

[coarse night]

In Hatcher, he says that a CW pair (X, A) is a cofibration and for this he shows Dⁿ×I (def) retracts to Dⁿ×{0} ∪ ∂Dⁿ ×I. And uses it to show that Xⁿ ×I retracts to Xⁿ×{0} ∪ (X^{n-1}UAⁿ) ×I. I don't understand where this is coming from

Is it like we are shrinking every n-cell that are in Xⁿ but not in Aⁿ?

[plain raven]

yes

I think one way to get intuition for the proof is that

if (X,A) is a pair and you have some space Z, and a homotopy h : Ax I -> Z and an extension of h at time t=0 to a map f : X ->Z
then using quotient topologies you can glue h and f together to give a map X x 0 \cup A x I -> Z

and it's not hard to see that the problem of extending the homotopy to X is equivalent to extending the map h \cup f : (X x 0 \cup A x I) -> Z to a map X x I -> Z along the inclusion (X x 0 \cup A x I) -> X x I

if you want to do this for all Z it suffices to give a retraction of the map (X x 0 \cup A x I) -> X x I

[coarse night]

yes, that has been discussed in the previous section

Btw any resources I can read cofibration and fibration theorems from?

[candid hedge]

he said there exists a unique map from a set T to a cartesian product (A*B) such that these 2 triangles commute. i have no idea what triangles hes refering 2, ig its the projections since he dre them in a a triangular form but i dont know exactly what he means by they commute

anyone can recognize the property hes talking about and help me out

then he wrote this is what he meant

with u defined as u= {ta(x),tb(x)} such that ta and tb are 2 maps from T to A and B respectively

[ornate berry]

These are the two "triangles" in the diagram

A commutative diagram is one in which if you choose two points and any two paths between those points, the composition of the functions in those two paths are equal

[ornate berry]

So he is saying exactly this.

[candid hedge]

Ohhh oh

got it

sorry it seems obv now

[ornate berry]

Dw

[candid hedge]

thanks a lot

[ornate berry]

No problem

There's no way you'd know the terminology without learning it so don't sweat it

[candid hedge]

yepp. The more i know💫

[icy schooner]

when you cover P^n with say U_i in k^(n+1), k a field

U_i is all stuff in k^(n+1) with x_i \neq 0

and you get a map (x_0,... , x_n) to (x_0/x_i, ... , x_n/x_i)

why is this bijective

it is onto

but if I have a homog coordinate in P^n I can multiply it by anything nonzero and let that factor be x_i

[icy schooner]

nvm it is mapping from U_i in P^n to k^n

can indeed take any nonzero x_i

doesn't matter in P^n

[candid hedge]

Limit points and sequence limits have the same definition?

A limit point is a sequence limit, and a sequence limit is a limit point?

Can i take the isolated point to be the limit of a constant sequence?

[gritty widget]

are you talking about limit points of a sequence

limit point of a sequence is a limit of some subsequence and conversely

[candid hedge]

yo

I ma trying to remember

ig not

Yeah i cant see exactly how i would go about doing this

Ah yes makes sense

[unreal stratus]

connectedness is very common too

[candid hedge]

Okay yeah i can see how. Theres a lot of talking about holes and shapes in algtop

Right

i have videos on this topic but they arent part of my required studies

so i skipped them

Okay so if a homeomorphism exists ebtween the two no?

Is it a very long argument the relation between morphisms and continuous functions?

or are we almost there?

okay

[unreal stratus]

they are the same thing in this context lol

[candid hedge]

in what context

when are they different

[unreal stratus]

A morphism X -> Y in the category of topological spaces is precisely a continuous function X -> Y

[candid hedge]

right

i mean homo

this is no no homo territory

right

yeah

dying to know

gooooood

wtf is this sorcery

[unreal stratus]

Well homomorphisms also come into the picture with fundamental groups - a map X -> Y of spaces induces a homomorphism between fundamental groups

[candid hedge]

Ohhhh

[unreal stratus]

Yeah sure

[candid hedge]

this isnt necessary

i def should learn this on my own

just wanted this

Yeah lol xd in uni we're still defining closure

in a normed vector space

but i am moving at a faster pace

I think i am done with general topology concepts

but havent touched any of the algtop since it isnt required

wdym

i am reading wikipedia and watching videos and reading books

its fine

i undestand. quality vs quantity

yeah

thanks a lot for the explanation i rly needed it

mainly rn i am struggling with the 10 definitions of limit points lol

and closure

ik theyre all equivalent but fuck they are many

theres definitions involving every concept, one definition for distance, 2 for sequences, one for closed sets intersection, one for the interior and the frontier, one for balls

[candid hedge]

whats the difference between compact and complete spaces?

As ive seen on wikipedia, both seem to be defined as spaces that have no holes or missing endpoints

[lunar yoke]

First off complete spaces make only sense if you have something to complete, like a metric or a norm

whereas compact is a notion that makes sense for arbitrary topological spaces

If you have a compact metric space, then it is also complete, but not every complete metric space is compact, like e.g. the real numbers with the euclidean distance

[candid hedge]

For cauchy sequences to exist you mean

[lunar yoke]

well you need some notion of "distance" for the definition to make sense yes

in a general space you only get to know which subsets are open

[candid hedge]

real numbers with euclidean distance are complete or compact iif its a closed interval no?

[lunar yoke]

real line is complete (but not compact) and a subset is compact iff its is closed and bounded, by the Heine Borel theorem

it doesnt have to be connected, so e.g. [0,1] u [2,3] is also a compact subset of the reals

and is hence complete

[candid hedge]

so ig if its bounded on a complete sspace then its compact?

completeness just makes sure there is no holes

[gritty widget]

or a vector space

[candid hedge]

without a norm?

[gritty widget]

yes

[candid hedge]

how would you do it?

whats a complete space on a vector space?

[lunar yoke]

well a metric space is compact iff it is complete and totally bounded, which is a bit different from just bounded

[gritty widget]

Complete vector space

[lunar yoke]

you can find lots of equivalent statements on wikipedia or the like

[gritty widget]

We say it's complete if any Cauchy net in it converges

[lunar yoke]

lmao

and how do you define cauchy

without any norm or metric

[gritty widget]

a Cauchy net on a topological vector space is a net x_a such that for any neighbourhood of zero U, x_a - x_b is in U for big enough a, b

[lunar yoke]

alright

makes sense actually

[gritty widget]

i. e. it's a Cauchy net using the standard uniform structure on a topological vector space

[swift fjord]

No, for example (0,1) in R is bounded but not compact. In general even closed and bounded isn't enough

[gritty widget]

But when we think about Cauchy or complete, we usually want some kind of metric.
This can be defined more generally for so called uniform spaces (it just happens that topological vector spaces have a natural structure of a uniform space)

[candid hedge]

niiice

[candid hedge]

0,1 is not complete tho

I mean the set itself is complete

ill fix my question then a sset that is bounded and complete is compact

[swift fjord]

Still no

[candid hedge]

counter example?

[unreal stratus]

Complete and totally bounded right

How about Z with discrete metric

[gritty widget]

yes, it needs to be totally bounded

[swift fjord]

For any metric, you can create a metric that is topologically equivalent but bounded, so take eg [0,1]^infty, this has a standard metric d that makes it complete, then you can take the metric d_b(x,y)= max{d(x,y),1}

Then this space is complete and bounded but not compact

[unreal stratus]

The ball radius 2 about any element is Z (hence it is bounded) and any Cauchy sequence is eventually constant (i.e. it's complete), but it's obviously not compact

For what it's worth, totally bounded means that for any epsilon, you can cover the space by finitely many epsilon balls

[candid hedge]

yeah i just looked it up on wiki

or its called precompact

[unreal stratus]

Which is certainly stronger since for example you can't cover Z by finitely many balls of radius 1/2

[gritty widget]

Yep. Z is pretty standard because for (metric spaces I think), being non-compact is equivalent to having Z as a closed subspace

which is a very cool property

you can have something similiar for the Baire space

[unreal stratus]

I suppose non-compact is equivalent to not being limit point compact for metric spaces and it follows from that right? Roughly

[gritty widget]

that;s called sequentially compact, and yes, it should follow from that

[unreal stratus]

Well limit point compact is different to sequentially compact in general

[swift fjord]

By having Z do you mean a countable discrete subspace

[gritty widget]

I don't know that property then

[unreal stratus]

Every infinite subset has a limit point

[gritty widget]

yes, a closed subspace homeomorphic to Z

[unreal stratus]

Implied by compactness for any space and equivalent to compactness and sequential compactness in metric spaces

if memory serves

[gritty widget]

I see. I've learned something today

[unreal stratus]

Noice

Yeah in my view a nice way to prove that compact => sequentially compact for metric spaces is through limit point compactness

[gritty widget]

It's in Munkres, makes sense why I didn't know about it

[unreal stratus]

Ah okay

[unreal stratus]

But yeah this style would be to say take a sequence (s_n), if it's only got finitely many distinct terms then extract a constant subsequence, otherwise the set of elements has a limit point, then constant subsequence converging to that

Actually that argument will work for any first countable space I suppose

Nice

[candid hedge]

does bounded make sense without a metric?

[gritty widget]

depends on the context, sometimes it does

for a topological space in general, no

[candid hedge]

totally bounded does tho no?

does make sense everywhere

[gritty widget]

I don't think so

[candid hedge]

they're 2 different concepts ig

[candid hedge]

its only defined in terms of open subsets so i dont see why not

tho bounded is defined in terms of distance

[gritty widget]

no, because the subsets need to be small enough

[candid hedge]

okay so its balls not open sets

[gritty widget]

it makes sense for uniform spaces though

[gritty widget]

well, spaces of small enough diameter

[candid hedge]

ok

[gritty widget]

either way, it uses the metric in crucial way

but can be generalized to uniform spaces, which are sort of inbetween metric and topological spaces

like, how topological spaces tell us about all the continuity properties, uniform spaces tell us about all the uniform continuity properties

but people don't really study them anyway so they're not that interesting ¯_(ツ)_/¯

[candid hedge]

okay

[gritty widget]

For me they're kind of unintuitive, hard to imagine.

[candid hedge]

tho whats the difference between compact and precompact

so compact and totally bounded

[gritty widget]

precompact means the closure is compact

[candid hedge]

GOOOOD so many fucking concepts

[gritty widget]

precompact is a property of a subspace, compact is a property of a space

[candid hedge]

so precompact is not defined for a space?

i dont see how subspaces and spaces are different

[gritty widget]

nope

[candid hedge]

Ohhh

[gritty widget]

subspace means it's a subset of some space

space is just a space

that's the difference

in other words, the property of being precompact is relative to what space are we in

the property of being compact is a property of a subset taken as a topological space itself

no matter what space it's contained in

this can be seen because in the definition of precompact we use closure which means we need to be in some topological space in the first place

equivalently, a subspace is precompact if any net in it has a convergent subnet

the limit here doesn't have to belong to the subspace

[candid hedge]

I still dont intuitively understand compactness

what is it really

Its excruciating to memorize this

i need to understand so i can tell on my own what is what

[gritty widget]

a space is compact iff any net has a convergent subnet

[candid hedge]

gooooood whats a net lol xD

ill wiki it

[gritty widget]

a generalized sequence

[candid hedge]

how is this supposed to be intuitive

i mean how would it let me take a look at something and know if its compact or not

[gritty widget]

it wouldn't

[candid hedge]

or whats the properties of compactness

Then its not helpful really just one more thing to memorize

[gritty widget]

it's helpful sometimes

[candid hedge]

Okay

ill see on youtube if i can find somethings that explain it in a memorizable way

[gritty widget]

you check a property of a space on many different ways and no way is better than the other

sure, they lie on youtube but feel free to

probably the easiest definition of compactness to memorize, and the easier to check in most contexts, is the one with covers

a space is compact iff any open cover has a finite subcover

[candid hedge]

finite subcover that is open right?

[gritty widget]

to check a space is compact is even easier once you know Alexandroff subbase theorem

[gritty widget]

a subcover of an open cover is open itself

[candid hedge]

so how is [0,1] compact in this definition?

[gritty widget]

we don't need to signify that it's open

[gritty widget]

Alexander, sorry

[unreal stratus]

A standard proof is to take an open cover U of [0,1] and let s be the supremum of all t such that [0,t] is contained in a finite union of elements of U

Which may sound slightly complicated but it has to be since we need completeness of the interval

[candid hedge]

open cover of [0,1] contained in [0,1] right?

[gritty widget]

0, 1?

[candid hedge]

sry

[gritty widget]

it doesn't matter if you consider open covers of [0, 1] as a subspace of R, or open covers of [0, 1] as consisting of open sets in R

[candid hedge]

so is ]-1,2[ an open cover?

[gritty widget]

Sure, under the right definition

[unreal stratus]

The set containing that is an open cover in one definition ye

[candid hedge]

right

okay

[gritty widget]

because all open covers of [0, 1] can be translated into open covers of a subspace [0, 1] in R

and conversely

[candid hedge]

so can i just say that U containing ]-1,2[ covers it then its compact?

[gritty widget]

huh?

[candid hedge]

Or is it that any subset of [0,1] needs to be covered also

[gritty widget]

(-1, 2) won't be in the cover in general

[candid hedge]

why?

[gritty widget]

because there are covers which don't contain it

[candid hedge]

So open covers are the smallest covers than contain [0,1]

[gritty widget]

no

open covers are covers consiting of open sets

[gritty widget]

cohomology is a ring with the cup product

does reduced cohomology still have a ring structure?

if so what is the unit

[steep parrot]

@ebon stream

[ebon stream]

hello i am skin, anyone know how to use fundamental groups in knot theory?

[gritty widget]

knots are images of S^1

so if you have a shape

and you calculate its fundamental group

it you get Z

then its a knot

[ebon stream]

ok

what is z

[gritty widget]

so knot is continuous right?

[bitter smelt]

This seems suspicious

[bitter smelt]

What do you mean "continuous"?

A knot is an embedding of S^1 into R^3

[rain stratus]

Anyone here once told me Willard was a bad book or at least had some wrong theorems/definitions. Does anyone else find this too?

[rain stratus]

I say this because I feel he skipped on proving something very interesting maybe because he tripped on his words; or that I am not seeing that it applies to the specific case too.

My problem is that in chapter 2 section 4 (Neighborhoods) he never actually talks about the neighborhood system of a single point in a topological space X being a topology for the whole set X. He doesn't even leave it as an exercise.

He instead points that if you get a collection N_x of subsets of X for each x in X, which fulfill the properties of neighborhood 1 to 4, then if you define open sets as in 5 you get a topology in which N_x is a neighborhood system for each x... That is just saying that a collection of neighborhood systems (one for each x in X) defines its own Topology in X; but not saying that a single neighborhood system can define its own (maybe) smaller topology for X.

Is it that this is not true in general or that he just didn't want to prove it? I'm actually a bit angry because I feel Willard actually thought he proved what I said with different words, but I'm thinking maybe this book just doesn't want to prove the things I expect

Jesus Christ I hope to be wrong on this one because this school has smugly been using this book as an intro to topology for a while thinking Topology is just hard and students are bad when the book might actually be the problem

[gritty widget]

I never said it's a bad book. Students most likely are bad/not dedicated enough though, in my experience a lot of them are

classic victim-blaming

This isn't a result used in standard cirriculum anyway

[rain stratus]

I wanted to use what I said for an example of a singleton in which its closure is an infinite set.
So since I want a Topology in which the only closed set which contains this singleton, is the whole space X, then using the neighborhood system at the point p of some infinite space would do if it formed a Topology in its own right. Because all open sets are empty or contain this point, it means the only closed sets are X or don't have p (*since they are the complements of open sets which contain p or of the empty set), so the closure of the singleton {p} would be the whole space X

[timid snow]

for that specific problem you could just define the topology as any set containing p + the empty set

[rain stratus]

non-trivial, that's literally the trivial topology

[timid snow]

huh

it has nontrivial open sets

and unless i am incredibly bad at topology im pretty sure it isnt the discrete topology either

[swift fjord]

that is definitely not the trivial topology lol

[rain stratus]

maybe you're right then

[timid snow]

for your original question i guess its just because it isnt really that useful

[swift fjord]

I've never seen anyone define that or feel the need to

in regards to your original question

that makes sense maybe in topological groups where the topology can often be defined by a nbhd system of the identity

[timid snow]

like i think every topological space can be defined by its neighborhood system at each point but thats definitely not true if you only take one

[rain stratus]

define what

[swift fjord]

a topology using the nbhd system of a single point

[rain stratus]

But even if that's not by definition that's still quite trivial, did you mean from an infinite set X, the empty set, a set containing the point, and X? that has barely any open or closed sets, X, empty set, that set is open, its complement is closed. I thought you meant something else

[timid snow]

i meant empty set, X, and any set containing p

also triviality isnt really a subjective thing

[rain stratus]

I find it the exception that it's used to define something, I usually see it when the author wants to skip a proof he thinks it's easy. So it's subjective in the sense I'm using it. I'm using the word in two senses because the word is not a single thing

[timid snow]

like if they say "nontrivial topology" they specifically mean any topology that isnt the trivial topology (and maybe the discrete topology)

[rain stratus]

Well true that you gave me a non-trivial topology but that's a very easy thing to do, so that in the non-topological sense to me that's a trivial answer. Not the topology you're giving. If I have the trivial topology, I just add a proper subset which contains the singleton then it's not technically the trivial topology

[timid snow]

yeah sure that works

i think youre making this too hard for yourself

[swift fjord]

authors note stuff they think is important or useful, and i'd generally trust their judgement over yours, especially when it comes to introductory books

this is not to berate you to be clear, this is just something that evidently comes from experience

[rain stratus]

At the same section the book gives us the definition of the Sorgenfrey line, and since asking for a singleton which is dense in an infinite space was asked I find it natural to try and give an infinite topology rather than a finite topology

I said I think the same at the beginning if you read what I said. If the book means what it means then he's just not proving or talking about the things I expected

[arctic relic]

I was trying to show that the category consisting of one object X and one morphism, the identity on X is equivalent to the category with 2 objects, Y and Z, and 2 morphisms, f and g st fog=idY and gof=idZ. I defined 2 functors, one from cat(1) to cat(2) and the other viceversa, took the composition, and showed that the composition of the 2 functors were naturally isomorphic to the identity functor on the category. Is this the right idea for equivalent categories?

[timid snow]

#category-theory but i think thats right if you took both possible compositions of the two functors

[candid hedge]

so whats the easiest way to think about quotiened spaces?

Is it collapsing some space around certain points?

[golden gust]

sometimes you can think of it as "gluing" certain subsets together

like [0,1]/0~1 is a circle because you glue 0 and 1 together

[candid hedge]

Okay

[graceful sequoia]

Define an equivalence relation on [0,1] so there is a simple bijection b/w the set of equivalence classes and the unit circle.

Can anyone give me any hints?

[gritty widget]

find a piece of string in your house and make a circle out of it

now turn that into a topology proof

[graceful sequoia]

Am I correct in thinking that only 0 and 1 will be equivalent while all other numbers will just be equivalent to themselves?

Like [0] = [1] = {0,1}, while [a] = {a} for all a != 0,1

[gritty widget]

you are correct

[graceful sequoia]

Ah okay, but finding the relation is difficult xD

But thank you for the hint!

[gritty widget]

i think you described the relation just fine there

[graceful sequoia]

Dont I have to describe it mathematically?

[gritty widget]

what's not mathematical about what you just wrote?

[graceful sequoia]

Like I was wondering it should be of the form..

aRb iff (some expression/equation involving a,b)

[gritty widget]

it doesn't HAVE to be

[graceful sequoia]

Okay

[gritty widget]

you essentially wrote down a partition of [0, 1] here, which is equivalent to specifying an equivalence relation on it

[gritty widget]

you also described it just fine in words here

[graceful sequoia]

Ah okay thanks!

[gritty widget]

Yes. Collapsing spaces but with respect to an equivalence relation. And they can be pretty pathological.

Or you can think like. You already can consider neighbourhoods of subsets of your topological space

So it all happens in it already, sorta

The sets of relevance being equivalence classes.

So quotiening is like forgetting all the information about the points and making equivalence classes those points, based on how they behave with each other as before

That's why separation properties are usually bad to preserve with quotients

[gritty widget]

Hi for some reason Im stuck with one step in the problem. I need to show that if $f,g : X \to R$ cts where X compact and $I_x$ is a closed line segment in $R^3$ with endpoints $(cosf(x), sinf(x),0)$ and $(0,0,g(x))$ then $Z= \bigcup I_x$ is compact. I'm not really sure how to properly show it's closed. (bounded follows from g being cts) I could probably use the argument 'complement is open' by handwaving, but maybe there is some argument using general results?

[gentle osprey]

Catematician

[gritty widget]

Define G:X x [0, 1] to Z by G(x, t) = t(cos(f(x)), sin(f(x)), 0)+(1-t)(0, 0, g(x))

G is continuous, onto, and X x [0, 1] is compact

@gritty widget

Oh that's pretty cool

Thanks

[pearl holly]

can someone please help me understand the concept of a (B, f) structure/tangental structure on a manifold as in https://ncatlab.org/nlab/show/tangential+structure ? I really don't understand where this definition comes from

Like is it the case that BSO(n) structures on a manifold correspond to orientations on the tangent bundle? Because I can't see this at all

[dusk heron]

Assume that we have a fiber bundle $S^1\to E\to B$, where $E$ is simply connected. Does it follow that $B$ is also simply connected?

[gentle osprey]

gustavn64

[hidden crag]

i think so

by the long exact sequence of htpy groups induced by the fiber bundle

you get $0\to \pi_1(B,x_0) \to 0$

[gentle osprey]

Timo

[hidden crag]

So the Kernel of the second map is all of $\pi_1(B,x_0)$ which is also the image of the map that sends everything to 0

[gentle osprey]

Timo

[hidden crag]

the zero on the left comes from the fact that E is simply connected and the one on the right comes from the fact that S^1 is path connected, hence pi_0(S^1) vanishes

[coral pawn]

If the closure of A is contained in the closure of B, then A is contained in B

I think this is true, but I can't prove it

Can someone verify this?

[gritty widget]

the closure of [0, 1] is contained in the closure of (0, 1), but...

[coral pawn]

Nevermind not true

Yeah

[unreal stratus]

Converse true tho

[echo dove]

Yes, that's right. Here's a sketch... The tangent bundle of an n-dimensional smooth manifold is classified by a map $M \to BGL(n)$ which is basically the same as saying that, if we choose a nice open cover ${U_\alpha}$ of $M$, and trivializations (frame fields) of $TM$ over each open set in the cover, then on intersections we get maps $U_\alpha \cap U_\beta \to GL(n)$, called transition functions (giving the change of basis from one frame field to the other).

[gentle osprey]

daveamayombo

[echo dove]

We can always choose a metric on $TM$, and choose our trivializations to be orthonormal... in which case the transition functions land in the orthogonal group $O(n)$ instead of $GL(n)$. That data (those transition functions) then correspond to a map $M \to BO(n)$ that lifts the map to $M \to BGL(n)$ mentioned above.

[gentle osprey]

daveamayombo

[echo dove]

If $TM$ admits an orientation, we can choose all the trivializations to be oriented as well as orthonormal. The transition functions then naturally land in $SO(n)$, and correspondingly we get a lift $M \to BSO(n)$. (Conversely, if such a lift exists we can choose trivializations for which the transition functions land in $SO(n)$, which implies $TM$ is orientable).

[gentle osprey]

daveamayombo

[pearl holly]

okay I see, but I don't really understand how those transition functions correspond to a map M --> BO(n). Could you elaborate on this?

[pearl holly]

nvm lol. Thank you so much! This makes a lot more sense now

[shadow charm]

Someone have some intuition for what this prism operator is computing?

[ornate berry]

It's splitting Delta^n x I into a simplicial complex

[shadow charm]

I get that decomposition at the top

But I’m not sure what exactly the prism is

It’s like going through these simplices in the simplicial complex in alternating directions but I’m not sure what the intuition for considering that is

[ornate berry]

You should think of it as making the faces that touch each other cancel out

[shadow charm]

Ah yeah that makes sense

And so like building delta x I bit by bit

[ornate berry]

This is related to the "basic relation" hatcher presents

Yeah

[shadow charm]

Okay yeah that makes more sense

Got to say homology is pretty lit

[unreal stratus]

It is

Also wow physical hatcher

What a madman

[ornate berry]

Did you print it out yourself?

[coarse night]

why not I also have a physical copy

[unreal stratus]

Moneys

[coarse night]

is it expensive you you live?

[ornate berry]

Hatcher is free. Printing is not

[coarse night]

it's <10$ here

[shadow charm]

This actually isn’t physical Hatcher but a pdf on my tablet

However I also have a physical copy that I got last Christmas and occasionally touch

[ornate berry]

Oh my it's e-paper

lovely

[shadow charm]

e-paper is so useful

[ornate berry]

I someday wish to have an e-paper tablet so my eyes don't hurt as much 🥲

[shadow charm]

Cause you can actually write on the book

[shadow charm]

Understandable

[unreal stratus]

Dw I was joking

Though on amazon for example it's like not cheap

[gaunt linden]

Is there an established term for a metric space where any two points are connected by a rectifiable curve?

[swift fjord]

wtf

where are university books that cheap

[coarse night]

how much does it cost for you?

[swift fjord]

I can't go to a bookstore physically and get it

so the prices online

[unreal stratus]

what's the easiest way to see that the inclusion is an isomorphism on homology? I guess it suffices to show that H_n( X+,X) = 0 for all n using the LES on homology

[plain raven]

You can use CW homology. Then the inclusion induces an isomorphism of chain complexes, not just homology, in degree n >= 2

because it sends cells to cells

[unreal stratus]

Hm I don't see how it induces an isomorphism of chain complexes (with CW homology) since isn't C_3(X)= H_3(X^3, X^2) free with basis the 3-cells of X, and since X^+ has (potentially at least lol) more 3-cells than X we can get non-isomorphic groups C_3(X), C_3(X+)?

[plain raven]

Ah fuck

ok

[unreal stratus]

I think the long exact sequence of the triple does it though, though I may be wrong

[bitter smelt]

Im trying to determine how perumutation of vertices in a simplex affects orientation

if I force v_0 to be fixed (since standard orientation is [v_1-v_0, ... , v_n-v_0]) I believe I have proven that the sign of the orientation being positive implies orientation preserving, the other is orientation reversing

just using a permutation matrix and taking its determinant

I thought to precompose with another permutation but I'm not sure how to determine orientation when v_0 <-> v_i

I thought maybe rewrite as such

\begin{align*}
\tau: (v_1 - v_0, \ldots, v_n-v_0) &\mapsto (v_1-v_i, \ldots, v_0-v_i , \ldots, v_n-v_i) \
&= ((v_1 + v_0-v_i )-v_0, \ldots, (v_0+v_0-v_i)-v_0, \ldots, (v_n + v_0-v_i )-v_0
\end{align*}

[gentle osprey]

Migillope

[bitter smelt]

But I'm not sure where this gets me

[echo dove]

Have you looked for patterns in low dimensions? See how freely permutting the vertices affects orientation in dimensions 1, 2 and 3?

Also, you're free to put the vertices wherever you want, so long as they're in general position, right? So you could arrange xi = -x0 (and x0 non-zero), with all the other vertices in the plane orthogonal to x0. Then swapping x0 and xi is just a reflection across that plane... so the effect on orientation is easy in that case.

[bitter smelt]

we require that the sum of the v_0 is leq 1 and each one positive

for the standard n simplex

So first I would show that that transformation is orientation preserving

multiplying every vector by some scalar or adding some vector to all of them ought to be orientation preserving, but I'm not sure about that sort of rearrangement

morally yes, I'm not sure it would be so easy to prove?

I guess now that I am examining specific cases directly I'm not so sure I see how [v_1-v_0, v_2-v_0] induces the normal orientation on the two simplex

[echo dove]

Y'know I think it's easier even than I said above. For the std 2-simplex, for example, you have three vertices v0, v1, v2. Normally those are taken as the three std basis vectors in R^3, right? The orientation of the simplex is such that (v1-v0, v2-v0, v0+v1+v2) forms a positively oriented basis for R3. So just show that the sign of permutation of those basis vectors affects orientation of R3 and correspondingly the simplex.

[bitter smelt]

it seems as though this would suggest the orientation is v_0 -> v_1 and v_0 -> v_2

[echo dove]

Right. That's an oriented basis in the plane containing the simplex, for example.

[bitter smelt]

So how is that permutation orientation preserving, my picture must be wrong somehow

should I not also be swapping vertices in the picture

yeah that would always preserve orientation then

and im not actually moving the vertices im just reordering okay

so actual picture should be

[bitter smelt]

huh, so you're suggesting going up a dimension and checking there

[echo dove]

Yeah, that's at least one option -- if you know how orientations of subspaces and their orthogonal complements relate to orientations of the ambient space. Another way to think: the simplex is defined to have the orientation associated to the ordered basis (v1-v0, ..., vn-v0) and you want to know how that orientation compares to the one given by (v0-vi, ... vn-vi). Both of those are bases... so if you can find the matrix that sends the first basis to the second, the sign of its determinant will tell you whether they give the same orientation or not.

[bitter smelt]

This second option was my strategy the first time

but I couldnt easily see what the matrix would be in the case that v_0 is movd

I guess I can actually do the change of basis formula and write it all out

but that seemed hard(er than necessary)

Maybe I will try your method to avoid having to do that

[echo dove]

Yeah, going up a dimension seems easier.

[bitter smelt]

very clever, thanks for this hint!

[echo dove]

👍

[odd flame]

how are g and h countable here

[gritty widget]

(h) is a countable union of sets like (g), so you can just worry about (g)

for (g), it might be helpful to use when the function becomes constant to rewrite the set

like "union over n in Z_+ of sets of functions which are 1 at and after n"

basically just push "countable union of countables is countable" and "finite product of countables is countable" as far as you can

also, not a topology problem (despite you finding it in a topology book)

[coarse night]

There's another way to do (g) (which was given by someone here)
you can biject sequences of Z+ to Z≥0 by subtracting 1 from each term. Then you can send that sequence (a_n) to ∑ (2^n)^(a_n) which gives you an injection to Z≥0.

nvm too complicated, basically those seq correspond to finite subsets of Z≥0 and you need some binary rep to produce a unique number from it.

[shadow charm]

So in Hatcher he gives a long exact sequence of homology groups for good pairs $(X,A)$ ($A$ is a deformation retract of some neighborhood in $X$) that involves the homology group $H_n(X/A)$. To prove this he instead works with relative homology groups but never comes back to prove that this is a specific case of that. I figured all you’d need to prove is that $C_n(X,A)$ is iso to $C_n(X/A)$. My first naive idea was to send any simplex $\sigma:\Delta^n\to X$ in $C_n(X)$ to the simplex $\sigma\circ q$ with $q:X\to X/A$ the quotient map, but the kernel of this morphism isn’t quite $C_n(A)$ but the subgroup of it consisting of sums with net 0 coefficients

[gentle osprey]

𝓛ittle ℕarwhal ✓

[shadow charm]

And of course none of my attempt was using deformation retracts

So I’m thinking I need to send simplices in A to 0 and somehow take a certain set of simplices in the inflated neighborhood to send to the constant simplex sending Delta^n to the point A collapses to

(Meant to say q o sigma)

[shadow charm]

I see it’s not that simple: the proof I require is actually given a few pages later

[unreal stratus]

Yeah I thought that one was slightly odd in how the proof and statement were so far apart

[shadow charm]

It’s normal to be stumped by the proof of excision on a first reading right? It’s mad

[unreal stratus]

It is probably like in the top 3 longest proofs in the book oof

[shadow charm]

Okay oof good to know

[shadow charm]

Hatcher says $H_n(X\cup CA,CA)\cong H_n(X\cup CA \setminus{p}, CA \setminus {p})$ by excision, for $A$ a subspace of $X$, but $p$ is not in the interior of $A$ (it’s the tip of the cone)

[gentle osprey]

𝓛ittle ℕarwhal ✓

[echo dove]

But the tip of the cone \emph{is} in the interior of $CA$ as a subspace of $X \cup CA$, which is what's needed for excision.

[gentle osprey]

daveamayombo

[shadow charm]

Gaahhh I don’t know what I was thinking thanks

I guess I still have the naive idea of thinking about my topological spaces as embedded in something bigger

[echo dove]

No worries! 🙂

[shadow charm]

Would there be any use in constructing homology groups by using formal sums in Z/nZ instead of Z (so not using the free abelian group as a staring point but adding in some cyclic relations)

For that matter you could probably use any group G 🤔 by just imposing more relations

I suppose you’d lose a lot of the geometric interpretation though…

[echo dove]

You gain quite a lot, actually. This is called homology with coefficients -- Hatcher covers it at some point.

[shadow charm]

Ah yeah that sounds vaguely familiar

Coolio

[lunar yoke]

when you do it with a field, the dimension of the homology are the betti numbers

so in some sense you could say that initially homology was computed with coefficients in fields

[shadow charm]

What are Betti numbers again?

[lunar yoke]

well you could use that as the definition

alternatively, the rank of homology as Z-module

b_n(X) = rk H_n(X; Z)

b_n(X) = dim_C H_n(X; C)

where C are complex numbers

[shadow charm]

What do we mean by dimension here then

[lunar yoke]

as vector space over C

[shadow charm]

Ah right

Why go to a field if this is already the case for Z 🤔

[lunar yoke]

so you can basically do the whole thing with homology and chain complexes and so on over R-modules

[echo dove]

You don't get the same Betti numbers over different fields tho.

[lunar yoke]

for many things you want your ring to be a PID though

[shadow charm]

I mean Z is already a PID so C doesn’t tell us more at least for the numbers no?

[lunar yoke]

right

but sometimes its easier to compute with C

[shadow charm]

Ah okay

[lunar yoke]

for example you get that over fields homology and cohomology coincide

for finite dimensional things

betti numbers generally have lots of nice properties

[shadow charm]

Still don’t know cohomology so can’t appreciate that but neat

[lunar yoke]

homotopy invariance, künneth formulas, poincare duality, relation to euler characteristic, etc

[shadow charm]

Is it nice also for finite fields or only infinite?

And does the characteristic matter

[echo dove]

Over fields, modules have very simple forms. Over PIDs it's still easy to classify and deal with modules.

[shadow charm]

Fair enough

[lunar yoke]

Yeah, I think thats where daveamayombo's remark form above comes in

[echo dove]

Yup. Field characteristic is what matters for Betti numbers, iirc.

[lunar yoke]

mhm makes sense

[shadow charm]

I see

[lunar yoke]

if you wanna get an even cooler invariant you can look at the universal cover and incorporate the action of the fundamental group

[shadow charm]

How so

[lunar yoke]

this leads to L^2-betti numbers

pretty cool stuff

those have the nice properties of betti numbers and even more, like multiplicativity under finite coverings (much like euler char)

but also you have to get functional analysis involved

at least a bit

[shadow charm]

Wait so what are the L^2 betti numbers?

Betti numbers of certain covering spaces?

[echo dove]

The alternating sum of Betti numbers is the Euler characteristic, which is the same as the alternating sum of the counts of simplices in a finite complex.

[lunar yoke]

L^2 betti numbers are defined for spaces with group actions

and they're a bit more complicated to define

[shadow charm]

Ah okay

[lunar yoke]

the most basic situation is you have a connected finite cw complex X

then you look at the universal cover

[echo dove]

And Betti numbers participate in some nice inequalities with the numbers of critical points of Morse functions (smooth functions that are reasonably nice).

[lunar yoke]

and that has an action by the fundamental group

[shadow charm]

Right

[lunar yoke]

the problem is that usually the universal cover can be huge even if you base space is small, like S^1 only has one 0 cell and one 1 cell, but its universal cover R already isnt a finite cw complex anymore

so normal betti numbers arent really useful anymore there, as they give you infinity

so the idea is to incorporate this group action we have

to reduce the size again in some sense

[shadow charm]

Normal Betti numbers of what space

[lunar yoke]

the universal cover

R

[shadow charm]

Isn’t that 0?

[lunar yoke]

well yeah

[shadow charm]

It’s contractible so homology groups are trivial

[lunar yoke]

but the method of how you go about computing it is what i mean

[shadow charm]

Ah right okay

[lunar yoke]

R was just an example to demonstrate that the universal cover may be "huge" in some sense

[shadow charm]

Yeah

[lunar yoke]

but you can view it as a Z-cw complex, which has one equivariant 0-cell, and one equivariant 1-cell

and this gives you a tractable chain complex again

[shadow charm]

I’m not sure I know what a Z-CW complex is

[lunar yoke]

if you look at the universal covering, the preimage of the 0-cell is Z right

[shadow charm]

Yeah

Oh i seee what you mean

[lunar yoke]

the action of the fundamental group pi_1(S^1) = Z on this is free and transitive

so we can interpret as one equivaraitn 0-cell

[shadow charm]

Yeah okay

So like you take the fibers of each cell

[lunar yoke]

yeah basically

[shadow charm]

Alright

[lunar yoke]

and in this situation of finite connected cw complexes and their universal coverings, this is the method you use to get tractable chain complex for the universal cover again

[shadow charm]

So how are you getting your chain complex?

[lunar yoke]

like the cellular chain complex

but you count equivariant cells

[shadow charm]

Oh I haven’t done cellular homology yet

[lunar yoke]

well the n-th chain module is free on the number of n-cells

and the boundary maps are a pain in the ass to compute

lmao

[shadow charm]

Oof

[lunar yoke]

but yeah so after all of this you bring your nice chain complex into the world of modules over the so-calles von Neumann algebra over your fundamental group

in that world you have a very nice dimension function coming from a nice trace

[coral pivot]

did someone say von nuemann algebras

[lunar yoke]

and this you use to compute your l^2 betti nunbers

[shadow charm]

I see, well I mean I don’t cause half of the stuff I haven’t heard of but it sounds pretty neat

[lunar yoke]

anyway this is all a bit far out if you havent seen cellular homology yet lul

its honestly also not really something you come across randomly in most algebraic topology books

its more geometric topology

[echo dove]

Can you name an example of a space where the L^2 Betti numbers differ from the usual ones? Is S^1 already an example?

[shadow charm]

What’s the difference? (Geometric vs algebraic)

[lunar yoke]

yeah for example all the L^2-betti numbers of spaces that cover themselves vanish

by multiplicativity under finite coverings

[echo dove]

I.e. simply connected, yeah?

[lunar yoke]

no, not necessarily

you also dont have many relation with normal betti numbers

[coral pivot]

btw phil you got a reference on the topic of, is this L^2 homology? I am intruiged by the connection to W* algebras.

[lunar yoke]

for any finite sequence of rational numbers and natural numbers, you can find a finitedimensional cw complex that has l2 betti numbers the rational sequence and normal betti numbers the natural sequence

[lunar yoke]

There is Kammeyer's book (a bit shorter and newer) and Lück's book (a bit older but basically a tome, the main reference)

also Lück gave a nice introductory talk on it here https://www.youtube.com/watch?v=gdj5Vyh6xBc

he was my prof for this course

nice guy

funnily enough this area has lots of applications to things you would never expect

like the kaplansky conjecture, that for torsion-free group G and field K of char 0, the group ring K[G] has no zero-divisors

[coral pivot]

interesting

will give these a read, thanks!

[echo dove]

Yeah, thanks! Curious too now, will take a look.

[coral pivot]

(also btw you should apply to the grad+ role, you will be good for our grad channels)

[shadow charm]

Thanks for the answers though the last stuff was way out of my league

[lunar yoke]

i saw that but cba to apply lmao

[coral pivot]

lol fair

[lunar yoke]

also this was unusual for me in that i basically only do abstract nonsense otherwise

very nice for a change though

[coral pivot]

i see, what math are you into?

[lunar yoke]

well homotopy theory

(stable)

[coral pivot]

makes sense

[lunar yoke]

also trying to learn infinity categories rn

[coral pivot]

nice, i learnt stable homotopy from adams book, so havent seen the infty category treatment yet

[lunar yoke]

me neither

our course on it focused on the model of orthogonal spectra

[coral pivot]

i see

[lunar yoke]

i mean in a course setting you cant really expect people to know infinity categories

at least not yet

lmao

[coral pivot]

yeah a friend ran a seminar on it this summer and a bunch of us learnt from that

(on adams book that is)

[shadow charm]

Max’s course?

[lunar yoke]

oh was it max

[coral pivot]

yep

[lunar yoke]

neat

next semester ill have a course on equivariant stable homotopy theory

thats gonna be fun

[coral pivot]

oh yeah sounds fun

[unreal stratus]

Btw what is needed to start working on stable htpy theory from a books like Adams - can you go straight from a book like Hatcher? (having read chap 4 i mean lol)

[coral pivot]

yeah thats what i did

[unreal stratus]

Noice thanks

Homotopy theory seems p dope

[coral pivot]

https://people.math.rochester.edu/faculty/doug/otherpapers/Adams-SHGH-latex.pdf

dont forget the tex'd up version :)

[unreal stratus]

Lol yeah dw I was definitely planning on using that :)

[coral pivot]

im into operator theory btw

I am interested to see how much of homotopy theory can be extended usefully to the category of C* alg for example

[unreal stratus]

Oh that's cool

Hadn't realy thought of links between operator theory and homotopy theory, not that I'm surprised

[coral pivot]

yeah like C* algebras is really non commutative topology, and i know there are model category structures on C* too

[silk tapir]

Hello everyone, I'm trying to construct an explicit deformation retraction between the doubly punctured closed disk and the figure eight, dumbbell and theta each. I know visually how it's supposed to go but I'm quite clueless about how to write it down formally. Can anyone help?

To be precise, the spaces in question are $\Theta = S^1\cup {0}\times [-1, 1]; \infty = (S^1+(1,0))\cup (S^1+(-1, 0)); O-O = (S^1+(1.5, 0))\cup (S^1+(-1.5, 0))\cup [-0.5, 0.5]\times {0}$

[gentle osprey]

adi

[long grail]

In my minds eye I see why it's true but I'm having trouble getting a proof started

Is my topology {N, empty, N-F_i} where F_i is just an arithmetic progression of length N or less?

[gritty widget]

N, empty set, and all the sets which for every N contain an arithmetic progression of length N

The point is to prove that it is a topology

[long grail]

I'm super new to topology except for on R^n and find the proofs awkward.

[gritty widget]

You just need to show that it's closed under all the properties that topologies are closed under. Which is finite intersections and arbitrary unions

[gentle osprey]

Blitz

[gritty widget]

For intersections you want to take two sets A, B with this property and show that the intersection also has this property

[long grail]

Thanks

[coarse night]

why is union of finite sets also have this property? ain't {1, 2,..., N-1} U {N, N+1} an CE?

[gritty widget]

does it contain an arithmetic progression of length N+2?

[coarse night]

I see the wording now.

[hard wind]

Got another question about Hatcher, he makes the following proposition about relative homology.

For good pairs $(X,A)$, the quotient map $q:(X,A) \rightarrow (X/A, A/A)$ induces isomorphisms $q_{*}:H_n (X,A) \rightarrow H_n (X/A, A/A) ≈ H^{~}_n(X/A)$ for all n.

[gentle osprey]

Espio

[hard wind]

Basically he’s connecting relative homology to actual homology of a quotient space. What I’m confused about is the word “good pairs”… what does that mean exactly?

[swift fjord]

Page 114

[hard wind]

Thank you! I spent a while flipping around looking for a definition and resigned myself to thinking it was a handwaved term

[swift fjord]

No problem, that is definitely a possibility with Hatcher lol

I recommend whipping out the ctrl+f on the PDF for those moments

It's free on his website

[hard wind]

Yeah good call

I’m pretty sure the pdf is somewhere in my google drive

[queen shale]

ok so im curious about "gluing" manifolds together

A section of my paper explores this concept via the "paddlewheel"

https://www.overleaf.com/8397941641bvjmggsxwvmk

what do you guys think?

[lament needle]

The open disc is a 2-cell but is it a cw complex. It seems to me that the answer is no but this seems a little funky to me. Because the X^0 skeleton has to be nontrivial for a non empty space, then any way you attach a 2 cell you wont get an open disc since you are mapping it's boundary onto the X^1 skeleton

[tawdry valve]

a finite CW complex is compact. trying to give a CW structure to a non compact space using only finitely many cells isn’t going to work. Try using infinitely many cells

[formal flint]

is it fine to skip chap 1 in munkres if im familiar with most things in there already

it seems like an introduction to naive set theory, logic, etc.

[gritty widget]

yeah it's fine

you can go back to it when you need

[formal flint]

👍 thanks

[gritty widget]

the most important stuff from there is probably the stuff on countability

[formal flint]

ive only seen countability in baby rudin chapter 2

[gritty widget]

i dunno what rudin does but you're probably fine skipping the first chapter of munkres

[formal flint]

rudin chapter 2 is basically all i know about topology at the moment lol

[formal flint]

alright, got it. thanks

[gritty widget]

since you've read the topology part of rudin, you should try comparing what you see in munkres to what you've seen in rudin

[plain raven]

it's a cofibration

i think he just doesn't want to overwhelm the reader with terminology or like

he doesn't want to prove that a good pair is equivalent to a cofibration at that point or

idk

[candid hedge]

are convergent sequences dense in all sequences?

[plain raven]

what's your metric on all sequences?

I think it's obviously true in the product topology on R^N

[candid hedge]

infinity norm

also is there some sort of guidebook on all the dense and open and closed subspaces of familiar spaces

like sequences, functions, matrices, real numbers...

[candid hedge]

Whats the product topology on R^n exactly

[plain raven]

Ok. So you didn't mean all sequences, you meant all bounded sequences?

[candid hedge]

yes

i realized this now

[candid hedge]

But what does the product topology mean

[plain raven]

Hmm. If {a_n} is a sequence then there is an open neighborhood U_k, epsilon of {a_n} consisting of all sequences {b_n} such that |a_i - b_i | < epsilon for all i < k

these open neighborhoods form a basis for the topology

In this topology it's obviously true

[candid hedge]

why?

(1,0,1,0,1,0,1,0,1,0...) whats the convergent sequence closer than epsilon to this one?

[plain raven]

not just epsilon but within epsilon for the first k

so if k = 5 take

(1, 0, 1, 0, 1, 1, 1, 1, 1, ... )

[gritty widget]

No, this space is complete (assuming you're asking if c is a dense subspace of l^infinity)

sadly c doesn't inherit the topology as a subspace of R^N

Here denoted by s

[small hemlock]

Is this proof valid? I'm to prove that every open subset of R is a union of open intervals. So, suppose A is an open subset of $\mathbb{R}$, and let $x \in A$. Since A is open, there exists $\epsilon > 0$ such that $(x - \epsilon, x + \epsilon) \subset A$. We can write A as $\bigcup_{x \in A} (x - \epsilon, x + \epsilon)$, for $\epsilon > 0$. Hence, A is a union of open intervals? I just feel I may be lacking a little rigor here

[gentle osprey]

J.Ross

[gritty widget]

Note that the epsilon depends on x

but yes that's correct

that's the only not rigorous thing I see here

[small hemlock]

Alright, nice. So since epsilon depends on x in each subset of the union here, I should designate it as $\epsilon_x$ ?

[gritty widget]

yes

[gentle osprey]

J.Ross

[small hemlock]

Okay that makes sense, thank you!

[long grail]

One inclusion is obvious, I'm having trouble with the other

[swift fjord]

Which direction is the one that's obvious to you

[long grail]

Fancy A subset of the intersection of topologies containing fancy A

[swift fjord]

the topology generated by A you mean. Well, for the other direction, notice that the topology generated by A is a topology containing A, and therefore is also being intersected over, so what can you conclude

[gritty widget]

Isn't a subbasis by definition a subfamily of a topology that generates it

[long grail]

Oh duh, of course it is

[long grail]

Thanks!

[swift fjord]

np

[long grail]

@gritty widget from Munkres

[gritty widget]

I see. So they chose to define it in terms of an explicit description of what it is

[candid hedge]

Whats the product topology i still dont understand it

[candid hedge]

this convergence definition is really weird

Is there a guidebook on the dense-open-closed-compact subspaces of all familiar spaces?

[gritty widget]

Product topology on $\prod_{i\in I} X_i$ where $X_i$ are topological spaces, and $\pi_j:\prod_{i\in I} X_i\to X_j$ is the projection onto the $j$-th coordinate is the topology with a subbasis consisting of sets of the form $\pi_j^{-1}(U)$ with $U\subseteq X_j$ an open set, that is ${(x_i){i\in I}\in \prod{i\in I} X_i : x_j\in U}$

[candid hedge]

on the space of sequences- matrices-functions-real numbers...

[candid hedge]

i mean most important ones

[gentle osprey]

Blitz

[gritty widget]

it's the smallest topology which makes all the projections pi_j continuous

[candid hedge]

So the collection of all combinations of open sets from the respective spaces right?

[candid hedge]

how do this topology lead to this definition of convergence?

[gritty widget]

I'm not sure what you mean by this.

The basis of topology here consists of sets $${(x_i){i\in I}\in\prod{i\in I} X_i : x_{j_1}\in U_1, ..., x_{j_n}\in U_n}$$ where $U_k\subseteq X_{j_k}$ are open

[gentle osprey]

Blitz

[gritty widget]

You can think of it as "the topology of pointwise convergence"

because a sequence (or net) in the product being convergent to some element is equivalent to it being convergent for each i

[candid hedge]

what i know to be the product topology its {U1xU2xU3.... with U1 open in X1,U2 in X2} etc

[candid hedge]

but its def the same thing

[candid hedge]

how does this lead to this result?

is there an intuition?

why is it a pointwise convergence?

so for a function it would be matching up a few points to the function and the rest can do whatever the fuck it wants?

[candid hedge]

So we consider each element in the sequence to be a projection on real numbers line?

[long grail]

When we say "Y inherits its topology from X" what do we mean? I can't find it in the book

NM I think I get it

Y is made up of open sets in X

[surreal lantern]

this is probably referring to the subspace topology

where Y has to be a subset of X ofc

[bitter smelt]

This almost always refers to subspace topology, or sometimes quotient topology if Y is a quotient of X

[long grail]

Quotient topology

I'm not there yet. What conditions on the subspace are required for it to be modded out?

Don't tell me I'll find it

[gritty widget]

What does modded out mean here?

[woven sundial]

i think they are asking what you can take a quotient by

[long grail]

Yes

[gritty widget]

everything

[surreal lantern]

you take the quotient w.r.t an equivalence relation

[gritty widget]

any equivalence relation you can think of, and any subspace

quotiening by a subspace is defined as quotiening by an equivalence relation which identifies all points in that subspace

[long grail]

Ok I understand

Thanks

[gritty widget]

but the quotient might not have the separation properties we would like to have, so we might restrict to certain types of equivalence relations

[long grail]

There's one where if you have the ordinary topology on R, if you quotient with respect to ~ where x~y iff x-y is an integer, the quotient space is homeomorphic to the unit circle

That's kinda interesting

[surreal lantern]

yeah that's the same as saying R/Z where Z is seen as a subspace of R

this is how you generally quotient by subspaces

[gritty widget]

no unless you interpret the quotient as a quotient by a subgroup

in topology R/Z would be interpreted as x ~ y iff x and y both belong to Z or x = y

[surreal lantern]

oops you're right

got it mixed up mb

[gritty widget]

Any ideas how to find homeomorphism between these 2 spaces? $\mathbb{N} \times X$ and $\mathbb{N} \times Y$, where $X = {0} \cup {\frac1{i} : i = 1,2, \dots }$, $Y = X \cup \mathbb{N}$

[gentle osprey]

Catematician

[gritty widget]

So the limit points have to map to limit points, I was thinking of a map like (x, 1/2i) -> (x,2i), (x,1/(2i+1)) -> (x,1/2i+1), but that doesn't work I think. No other obvious mapping comes to my head.

you can just map (n, 1/i) to (n, 1/(i-1)) for i > 1 and N x {1} to N x {2, 3, 4, ...}

Oh okay yeah, thanks.

Bijection between two discrete spaces of the same cardinality is homeo right?

Yeah it is

if you're not sure then prove it

Ya just did

[unreal stratus]

Yes, you are just renaming the elements essentially

[surreal lantern]

im not sure what you mean

but the way you phrased it doesn't make sense

[long grail]

I think it must be. Sorry the question is dumb

[surreal lantern]

you're taking the pre image of something that is not in the codomain

[long grail]

so if f: X×Y -> X, then the pre-image of an subset of Y is empty

Here I'll give context

[surreal lantern]

the pre image of a subset of Y is not a thing here

because your function is mapping to X

[long grail]

Oh I transcribed incorrectly

It would have been convenient for the proof I'm working on if it is empty

But I copied something down wrong

[surreal lantern]

can you give proper context

what are you trying to prove

[long grail]

Sure

4

And I'm trying to use this theorem:

I can include my sloppy work

but i notice my mistake, it should be pi_2

*pi_2^{-1}

[surreal lantern]

the first line doesn't quite make sense

you want W to be an open set in X x Y right

so you'd have to take the pre images of pi_^and pi_2 in the first line

but yeah that theorem works for proving that the projections are open maps

[surreal lantern]

but it still wouldn't be entirely correct

[long grail]

Ok

That union in the theorem

I think I understand but I was having trouble

In the def of the subbasis

[surreal lantern]

yeah i was about to suggest to review the def of a sub basis

[long grail]

Ok

I'll do that

[surreal lantern]

i don't think i can be of more help rn it's pretty late for me

[long grail]

Well you were a big help

Sincerely

Good rest

Thanks

[surreal lantern]

Think about the fact that every open Set in X x Y can be written as union of finite intersections of sets in the given subbasis

and then how pre images are compatible with unions and intersections etc

that should give you your result if you work the details out

[long grail]

All right that helps

[candid hedge]

is there an example of an open hyperspace subspace?

[candid hedge]

hey, sorry to bother you again. But can you explain this i didnt get it yesterday

whats convergence in the product topology

[candid hedge]

How is this convergence?

[gritty widget]

We say that a sequence or a net $x_\alpha$ converges to $x$ if for any neighbourhood $U$ of $x$ we have $x_\alpha\in U$ for all big enough $\alpha$.

[gentle osprey]

Blitz

[candid hedge]

i mean whats the open ball in product space

[candid hedge]

and can you answer this?

[gritty widget]

Product space doesn't have a metric in general

[candid hedge]

Ah

[gritty widget]

As for $\mathbb{R}^\mathbb{N}$ you could define it as $$d(x, y) = \sum_{n=1}^\infty 2^{-n} \frac{|x_n-y_n|}{1+|x_n-y_n|}$$ but it's not the only metric that you could define on it

[candid hedge]

well?

[gentle osprey]

Blitz

[crisp bough]

https://math.stackexchange.com/questions/4526397/please-help-me-evaluate-the-following-integral-analytically
could someone please clarify the derivation steps suggested in this solution to the integral
and i misunderstand the concept of dilogarithms; please explain what they are and their application in the solution.

[candid hedge]

just to conclude then

[crisp bough]

sorry for posting in the wrong channel

and interrupting the convo

i urgently need some clarification with the derivation of this integral, if anyone can help
many thanks

[gritty widget]

stop spamming

[crisp bough]

sorry, my folio is due today and thought if i posted in multiple channels i would receive more help

i will refrain from doing so in the future

[gritty widget]

Are you asking what's a sequence in R^N convergent to this point?

[crisp bough]

can u pls clarify the derivation process?

[gritty widget]

oh my god shut up

wrong channel and stop pinging me

[crisp bough]

next time, be polite...

k, i will delete my posts

[candid hedge]

yes

no

i am asking if convergent sequences are dense in the product topology of sequences space

[gritty widget]

Yes. You can put 0 after finitely many terms

[candid hedge]

why??

this is what i dont get

can you please write an intuition or sketch a proof

[gritty widget]

If $x = (x_n)_{n\in\mathbb{N}}$ then $y_n = (x_1, ..., x_n, 0, ...)$ is a sequence convergent to $x$ in the product topology

[gentle osprey]

Blitz

[candid hedge]

howww?

in what sense?

[gritty widget]

In the product topology

[candid hedge]

you're no texplaing

not*

[gritty widget]

You asked in what sense is it convergent

But since we have topology here, the way in which it converges is well-defined

[candid hedge]

whats a neighbourhood of x in this topology?

[gritty widget]

I don't know how to answer this question other than give you a tautological answer.

It's a set which includes an open set which contains x

[candid hedge]

okay

[gentle osprey]

Blitz

Blitz

Blitz

Blitz

Blitz

[candid hedge]

AHHH

fucking hell

i got it

thanks

[candid hedge]

how can a hyperplan be dense in a given vector space?

and do you have an example of a non closed hyperspace?

[gritty widget]

if can if it's given by a functional which isn't continuous

[candid hedge]

ohhh

[gritty widget]

in fact it's equivalent

a functional f on a locally convex topological vector space is discontinuous iff f^-1(0) is dense

and any hyperplane is of the form f^-1(0) for some linear functional f

[gritty widget]

note that open isn't equivalent to not closed

If a hyperplane f^-1(0) were open, then it'd contain x+V where V is a neighbourhood of 0, hence span(V), which is the whole space

so it wouldn't be a hyperplane

[candid hedge]

yeah

no i mean not closed

[gaunt linden]

One concrete example: Let V be the vector space space of all real sequences (a_n) with finite support (that is, only finitely many entries in each sequence are nonzero).
Equip it with the sup-norm.
Then the set of sequences where a1+a2+a3+...=0 is dense in V.

[gritty widget]

like I said, just take any functional that isn't continuous

[candid hedge]

why?

is it dense

[gaunt linden]

Suppose you have a sequence and you want to approximate it to within a distance of espilon>0 by a sequence that sums to 0. Just take your target sequence and add or subtract something smaller than epsilon to sufficiently many of its entries to make its sum 0.

[dry jolt]

I'm reading that if G acts freely on the cells of a compact CW complex, then G is finite. Is this because a compact CW complex is finite, hence the free action on the cells determines an injective homomorphism from G to a finite group (namely the product of the symmetric groups on the cells)?

[shadow charm]

Hatcher proves the first and fourth maps are isomorphisms and then says we can assume so are the second and fifth by induction on k to then prove the third is with the five lemma. I’m not sure I see what the inductive argument on k is, though I assume it must be obvious

Nvm I’m an idiot I see it

[shadow charm]

Just to be sure I’m not misinterpreting Hatcher, if you’ve got a delta complex and take a quotient by identifying simplices in a way that preserves order you automatically get a delta complex structure on the quotient space by composing the simplices with the quotient map right?

So for example if I’m asked to realize a delta complex structure of RP^n as a quotient of S^n after giving S^n a delta complex structure with vertices at +- the unit coordinate vectors, I just need to give that complex an order where two vertices on one axis are consecutive and then we follow some arbitrary order of the axes correct?

And then identify all simplices by exchanging antipodal vertices

[median sand]

Does anyone here know about crossed modules (as related to the 3rd group cohomology) or is this not the place to ask (I figured the people here would be more in touch with homological stuff than the algebra channel)?

[bitter smelt]

[median sand]

OK, so I'm reading about H^3(G,A) in Brown, here's his definition of a crossed module. According to him H^3(G,A) corresponds (up to equivalence) to exact sequences 0->A->N->E->G->1, where N is crossed over E. Now when reading about H^2(G,A) and extensions 1->A->E->G->1 in Jacobson, first thing he does is show how the extension induces a G-module structure on A, so I tried to show how such a crossed module sequence does the same thing and I ran into some problems.

Namely, given a sequence with N crossed over E, I can show that G acts on A, but I'm not sure how to show it acts by homomorphisms (i.e. that A becomes a G-module). So my question is: i) in the definition of crossed modules, is it necessary to assume E acts on N by homomorphisms or not (perhaps Brown neglected to mention this) ii) if not, how to show that G then acts on A by homomorphisms.

[cedar pebble]

yeah I think you need to assume some particular action by automorphisms

[median sand]

So just to clarify, here's what my thought process was: if i is the injection of A in N, lift \sigma\in G to s\in E, then by (5.2) in above picture \alpha(s\cdot ix)=s\alpha(ix)s^-1=1, so s\cdot ix=iy\in iA and then you can check that the y is independent of the choice of s, so you set \sigma x=y and then you can check the usual action identities. Only thing I had trouble showing was, as I mentioned, that s\cdot i(xy)=(s\cdot ix)(s\cdot iy). Another consequence of (5.2) is that \alpha(e\cdot(nm))=\alpha(e\cdot n)\alpha(e\cdot m), so it's like the action of E on N is "almost" by homomorphisms, but actually up-to elements of iA (the kernel of \alpha). I got stuck here.

[small hemlock]

I'm really struggling with a problem that I feel should be very basic: Let X be a topological space and U a subset of X, where U is open. (So U has the subspace topology). Then a set V a subset of X is open in U iff V is open in X and V is a subset of U.

For the first direction, let V be a subset of X and suppose V is open in U. Then V=U∩A for some set A open in X. Since V=U∩A, V is a subset of U. But from here, I simply cannot think why V is open in X.

For the other direction, suppose V is an open subset of X and V is a subset of U. Is a set that's open in a main topological space always open in a subspace topology of that set?

Any help would be greatly appreciated

[gritty widget]

V is intersection of two sets open in X

[small hemlock]

Oh my gosh

That's straight from the axioms, okay

I think I've figured out the other direction as well, thank you for that

[coral pivot]

I am reading something that states Ext is dual to topological K theory

in what sense is this true?

[cedar pebble]

I can't think of a single sense in which this is true

where is this?

[coral pivot]

https://www.math.umd.edu/~jmr/algtopK.pdf

page 5 paragraph 2

[cedar pebble]

oh this is C* algebra stuff lol

[coral pivot]

figured if this were true in C* it would be true in top but i guess not oof

[cedar pebble]

I mean the setup with KK-theory is a lot more complicated

it's worth reading this page https://ncatlab.org/nlab/show/KK-theory

[coral pivot]

i see, will do thanks

[shadow charm]

Is this simplicial homology computation correct? The space I’m working with is $n+1$ copies of $\Delta^2$, $\Delta_i^2$ for $i=0,…,n$ with all edges of $\Delta_0^2$ identified, the top left edge of $\Delta_i^2$ identified with the bottom edge of $\Delta_{i-1}^2$ and the two other edges of $\Delta_i^2$ identified, with $i=1,…,n$

[gentle osprey]

𝓛ittle ℕarwhal ✓

[shadow charm]

All in an order preserving fashion ofc

I feel like I might be missing a Z term or smth

[shadow charm]

Am I being dumb or is the homology of the delta complex with all faces of the same dimension identified just 0

[shadow charm]

Consider the space $X \cup_F (D^2\times I)$ for $F:S^1\times I\to X$ a homotopy between $f$ and $g$. This space deformation retracts onto both of the subspaces $X\cup_f D^2$ and $X\cup_{f’} D^2$ so they are homotopy equivalent

[gentle osprey]

𝓛ittle ℕarwhal ✓

[unreal stratus]

Also this is covered in chap 0 Hatcher I fink

[gritty widget]

you should not be spending a whole third of your time checking edge cases, in my opinion

[gritty widget]

get the main idea on your first read, and do the annoying parts (trivial stuff like empty set cases) later if you really need to

if you're really checking empty set cases for a whole third of your time, you're probably not doing it right

as for actual important details in proofs (ones that aren't waved away by declaring a set nonempty or something), don't skip those

[shadow charm]

Too true

[gritty widget]

Omega is generated by open sets

f^-1(sigma(open sets in Y)) = sigma(f^-1(open sets in Y)) which is a subset of sigma(open sets in X)

where sigma means a sigma algebra generated by

[cerulean oriole]

I think if one of the sets needs to be non-empty, a good book should state so rather than making you figure it out.
(Or introduce a convention that “set” means non-empty set .)

[gritty widget]

no reasonable mathematician wants to think about empty set cases for everything they do

[cerulean oriole]

Or perhaps conventions in specific places, such as “subsets of X that come up might be empty, but we always assume X is nonempty because we don't want to study the empty topological space”.

etc.

[cerulean oriole]

Sure, but then you should say you're excluding it and where.

[candid hedge]

example of 2 disjoint closed sets A,B in a normed vector space but the distance(A,B)=0

i cant think of any

the only way 2 disjoint sets is enough to say the distance is >0 is when one is compact and the other is closed

so i am looking for a counter example

thatll open my eyes

[gritty widget]

just think of something in R^2

[candid hedge]

the distance is always greater than 0 no?

[gritty widget]

no

[candid hedge]

if they dont intersect

and are closed

[gritty widget]

no

[candid hedge]

howww? All sequences that converge will converge inside the ball

[gritty widget]

there can be sequences a_n in A and b_n in B such that d(a_n, b_n) converges to 0, but A, B still have all of those properties

[candid hedge]

An example would be incredible

[gritty widget]

exercise for you

[candid hedge]

cause i cant see how

First if we're in R^2 then closed means a closed ball. so any sequence inside a closed ball has a limit point because its bounded. so if the d(an,bn) converges to 0 then an and bn will have the same limit point which is a contradiciton

[gritty widget]

you can't disprove the truth

[candid hedge]

i am not trying to i am trying to see where my logic is failing

[gritty widget]

then write out your logic in clear to digest steps

[candid hedge]

is it not clear?

[gritty widget]

it's not

[pallid lion]

closed does not mean closed ball

[candid hedge]

can we agree that an,bn have limit points?

[candid hedge]

even in R^2? arent they equivalent?

[gritty widget]

no, we cannot agree on that

not all sequences in R^2 have limit points

[pallid lion]

it doesnt even mean that in R^1, this is not how topology works. closed sets are not just closed balls

[gritty widget]

this is because R^2 is not compact

[candid hedge]

all bounded sequences do

[candid hedge]

they're equivalent. Open and closed balls form a basis

[gritty widget]

well, yes. So try and come up with something unbounded

[candid hedge]

Something unbounded in a closed ball?

[gritty widget]

R^2 is not a closed interval

nor it is a closed ball

[pallid lion]

nonono, you cant just reduce to the case of a closed ball, the line x=0 is closed in R^2 but not a closed ball

[candid hedge]

I am not talking about R^2 i am taking about our choices of sets A,b

[candid hedge]

AH fuck