point-set-topology

[frosty sundial]

sorry

wasnt a valid CW complex

just a single n-cell floating in space

[pearl holly]

lmao lonely n-cell

okay but anyway, thank you so much!

[frosty sundial]

haha np

also the closest thing to just "an n-cell flaoting in space" would be "an n-cell and a 0-cell, and the boundary of the n-cell is all glued to the point"

and that is S^n

which does have homology in degree n

(as your original argument showed)

[pearl holly]

right right I see

[gritty widget]

in its generality or specifically when topology is a manifold

im assuming second case for obv reasons

[pearl holly]

Are there some spaces, apart from projective spaces, that have a cohomology ring isomorphic to some polynomial ring?

Say like Z[alpha] or something, not truncated by anything

Or are the projective spaces the only spaces with these kinds of cohomology rings? That would be kind of cool tbh

[honest narwhal]

@pearl holly lemme process this

Do you mean direct limit of CPn?

[pearl holly]

Ye I think so, I’m referring to like RP^infrty, CP^infty and so on

[honest narwhal]

Okay yeah because CP^7 definitely isn't it lmao

[pearl holly]

The only examples I’ve seen of such spaces are the ones that I mentioned and it kind of feels like that’s the only class that have these cohomology rings but I’m only guessing

[honest narwhal]

Here's an interesting point

CP^{\infty} is a K(Z,2)

RP^{infty} I think only has a clean cohomology ring mod 2

And it's a K(Z/2,1)

I wonder if there's something going on there. Fwiw anything with some cohomology ring as you describe is an infinite CW complex and how many of those do we know

So maybe there are a metric fuckton of examples

[pearl holly]

Yee lmfao

[honest narwhal]

But this is just an idea that gets the noggin joggin

[pearl holly]

I think there’s something like HP^infty as well and it has a similar cohomology ring

Hatcher says some stuff about it but I haven’t read it

Like octagonic projective plane or some shit like that

No that was like O

Nah it does feel like there should be a ton of infinite CW complex spaces with these cohomology rings

But it would be cool to classify those I guess

But cohomology doesn’t tell us a lot of the actual space we are working right?

[gritty widget]

Complex grassmannians also have cohomology as a polynomial ring

[pearl holly]

oh daim

I think Hatcher talk a little about grassmanian stuff but I didn't read it

okay so like projectives planes, Complex grassmannians but is there anything else?

I honesly hoped that only projective spaces have thsi property, it would be cool lmao

[gritty widget]

Oh now I should state that

It’s not a polynomial ring in one variable

[pearl holly]

oh daim

back at square one then

but daim this is kind of a cool thing insn'y it?

[gritty widget]

So you want a space X

With H*(X;Z)=Z[u]

[pearl holly]

yes exacjtly

no truncated version or no bullshit

just straight out polynomial

[gritty widget]

Is it obvious that u has to have degree 2

[pearl holly]

yeee or degree 4

that's a cool theoreom btw

[gritty widget]

What theorem is this

[pearl holly]

it's in Hatcher

[gritty widget]

Very cheeky of me

[pearl holly]

if H^*(X; Z) is Z[\alpha] then alpah has degree 2 or 4

[gritty widget]

You hardly have Hatcher open do you

[pearl holly]

you prove it using steenrod operations

[gritty widget]

Oh okay

[pearl holly]

no I read the proof like 2 days ago I have it fresh in my mind

for HP^\infyy it is \alpha = 4

but for like RP^\infty it is 2

so we know that it has to be infinite dimensional and alpha has to be a generator of degree 2 or 4 but that's like it, no?

[gritty widget]

Okay well

Stupid answer

Because it should be homeomorphic

But

Oh nevermind

[pearl holly]

no what were you going to say?

[gritty widget]

This is wrong but I was going to say

[pearl holly]

it's always nice having at least some argument thingy

[gritty widget]

You can realise CP infty as some infinite complex grassmannina

[pearl holly]

daim I don't even know what the grassminnian is

[gritty widget]

And then I was thinking about using how G(k,n) is the same ask G(n-k,n)

If you look in lee smooth manifolds

It explains these

G(k,n)

[pearl holly]

huh okay I see

gotta read up on my manifold thingies

[gritty widget]

Is essentially all k dimensional substances of an n vector space

So G(1,n) is all lines

So this is projective space

[pearl holly]

okay

hmm

I asked another member of this server btw (he left) and he said that if there were such a space, it would be reall weird because the fundamnteal group if either perfect or zero

but this is getting quite cool right?

[gritty widget]

Yeah

[pearl holly]

I will ask my prof about this

[gritty widget]

I suspect this will have a slick answer

[pearl holly]

I almost belive it's true

yeeeee

[pearl holly]

but I have only seen examples so idk lol

THIS SI SO FUCKING COOOL

this whole AT thing is so cool

[gritty widget]

Also can you please @agile anvil when you get an answer

Opps

[pearl holly]

yeee I will

lmao

imagine having me as your name

pain

I will hit my prof up tomorrow and I will ping you if I get an anwer

yo anyone say cool stuff about AT

I'm sorry but like I'm very bored

[gritty widget]

Okay

http://www.map.mpim-bonn.mpg.de/Fake_complex_projective_spaces#References

[pearl holly]

lmao fake

[gritty widget]

Can we just take a limit of these

To get some infinite version

[pearl holly]

The classification of these spaces was one of the early milestones in surgery theory.

daim that's cool

but I don't know what sugery theory is about lol

sounds like some doctor phd in phramacy stuff

yo lime_soup what have you read btw?

like AT stuff?

or just anything else?

[gritty widget]

Hatcher and concise

Sorry very sleepy will reply properly tomorrow

[pearl holly]

ye I'm sleepy too tbh

but have a great day

and goodnight

💤

[hexed holly]

For a space A, what does the map -id: ΣA -> ΣA induce on homology?

We know it's an isomorphism and its square is the identity, but is it always -id?

I should say the map on the level of spaces is given by f(a, t) = (a,1-t) followed by the appropriate identifications

[hexed holly]

nvm I've figured it out

[gritty widget]

dawg wut

are those chain maps

[hexed holly]

No it's a map of spaces, which induces a map on homology

[gritty widget]

wtf summation symbols for

[hexed holly]

It's a suspension

[gritty widget]

oh

[hexed holly]

Reduced suspension in this case

[gritty widget]

ok

[jolly garden]

Does anyone know how to visualize the Smash product [0,1]∧[0,1]=([0,1]x[0,1])/([0,1]∨[0,1])? [0,1]x[0,1] is the unit square obviously, the wedge product is gluing two intervals at a certain point. But what would the quotient look like?

[glossy pine]

there's a smash product? damn

[jolly garden]

Yeah, look at the Smash product S^1 ∧ S^1 for example: https://en.wikipedia.org/wiki/Smash_product#/media/File:Visualization_of_the_smash_product_of_two_circles.gif

pretty cool

S^1 ∧ S^1 = S^2

[glossy pine]

see this is the kind of topology im I interested in

not the boring product topology and whatever

[empty grove]

It is the square but you collapse 2 of the sides

It becomes homeomorphic to the square itself I am pretty sure

Like if you collapse a side you could imagine that you get a triangle

And then another side and you'll have 2 sides remaining but there's content inside so you'll get 👁️

And 👁️ is homeomorphic to square

[jolly garden]

Supposing we glue the one interval to the center of the other intervall, we'd get a "T" basically. Let's imagine [0,1]x[0,1] to be a sheet of paper. Wouldn't collapsing the sheet of paper along the T result in simply folding the paper once and being left with a rectangle that has to be collapsed to one edge?

I'll draw a picture

nevermind

([0,1]x[0,1])/([0,1]v[0,1]) basically means that we identify (a,b) with (x,y) iff (a,b)=(x,y) or {(a,b),(x,y)}⊆[0,1]v[0,1].

I feel like you could fold the paper once and be left with a half square with a half red line at the top

[hexed holly]

Why would you get any folding?

The resulting space of your diagram after the collapse would be simply a bouquet of two disks

[jolly garden]

🤨

[empty grove]

Depends on what your basepoint is

I assumed basepoint is 0

[jolly garden]

All points on the red intervals are considered the same in the quotient, right? I can't imagine that to be honest

[hexed holly]

All points on the red intervals are considered the same, right?
Yes

In the quotient space that is

[jolly garden]

nvm, forget the folding

[hexed holly]

You can collapse in turns. first collapse the top edge to get a triangle. Then collapse along the middle line to get two disks joined at a point

This shows the smash product is highly dependent on the basepoint that you choose

Different basepoints result in different smash products

Not even homotopy equivalent

[jolly garden]

Ahh, that's clever.

I see, thanks

[hexed holly]

Now exercise for you: can you show that the square with one edge collapsed is homeomorphic to a triangle?

[jolly garden]

I'll give it a try

[hexed holly]

You should construct an explicit homeomorphism and prove that it's a homeomorphism

[sturdy notch]

This is a somewhat elementary question but I just wanted to make sure I understood
If $X$ is a set with two topologies $\tau_1$ and $\tau_2$ such that $\tau_1 \subset \tau_2$, if $(X,\tau_2)$ is path connected, then $(X,\tau_1)$ is too right?

[gentle osprey]

dadaurs

[sturdy notch]

Because for $\gamma: [0,1] \to (X,\tau_2)$ a continuous path, $U \in \tau_1\implies U\in \tau_2 \implies \gamma^{-1}(U)$ is open

[gentle osprey]

dadaurs

[empty grove]

yes

[sturdy notch]

great thanks

[west spindle]

so apparently this exercise which i thought was easy to solve isn't

i was trying to come up with a space consisting of three points

[empty grove]

{1,2,3} discrete metric, ball of radius 1.5 at 2 and ball of radius 2 at 1

[west spindle]

discrete metric

more like social distancing metric

[empty grove]

[west spindle]

also completely unrelated

fund group of products equals product of fund groups

right?

[empty grove]

yes

[west spindle]

well not equals but iso

ok

so then the fund group of R^3 \ R^1 is just Z

[empty grove]

yes

[west spindle]

because it's the punctured plane \times R

what's the fund group of a double-punctured plane? is it Z^2 or Z freeprod Z?

[empty grove]

F_2

[west spindle]

[empty grove]

freeprod

[west spindle]

is the fund group of a plane with n holes free of rank n

[empty grove]

because it deformation retracts to figure 8

yes

that deformation retracts to wedge of n circles

[west spindle]

right

wedge means they're joined together at one point

[empty grove]

yes

[west spindle]

okay

another one just to make sure

[empty grove]

at the basepoint specifically

[west spindle]

the fund group of an indiscrete space is trivial, right? bc everything is homotopic to everything else and all that

[empty grove]

just that here it doesn't matter

yes

[west spindle]

ok good

[west spindle]

full disclosure i was assigned these problems a few days ago and am soon going to be in a teams call with the geo/top professor discussing them and showing that i actually know how to solve them, i.e. that i know my shit

[west spindle]

wait does this actually work?

both of these balls are the entire space are they not

oh but i suppose you could inherit the metric from Z instead and have the ball of radius 1.5 at 2 and the ball of radius 1.9 at 1

[empty grove]

Ball of radius 2 at 1 doesn't contain 3

Open ball

[west spindle]

ah

[empty grove]

oh true

[onyx crow]

https://wikimedia.org/api/rest_v1/media/math/render/svg/3db92acff9f0a685c8a7580c5afa4bfed93075fe

What does the “lim” here signify?

I understand the definition otherwise

[empty grove]

colimit of the diagram they give

[onyx crow]

ohh

[empty grove]

But the diagram they give is weirdly described

It is X x [0,1]

and then 2 branches pointing into it

[onyx crow]

hmm I didn’t realize there was a diagram involved

[empty grove]

[onyx crow]

(I mean I didn’t see it as a diagram)

[empty grove]

the diagram is shaped like
X x [0,1] ← X x {1} → {v_1}
↑
X x {0}
↓
{v_0}

I see

lol

[empty grove]

or they might not be connecting the 2 diagrams I am not sure actually lol

but if this is the suspension I think they should

[onyx crow]

Hmm I can’t really see why exactly colimit means “sewing” all those together

[empty grove]

You have a map from X x [0,1] to the colimit (which turns out to be the quotient map) and the inclusions just tell you how those 2 boundaries sit inside the colimit, but because their maps to the colimit factor through the singletons, they both collapse to a point

So it is just a way to say that the 2 boundary circles have their "inclusions" as constant maps by saying that they factor through singletons

[onyx crow]

Ooo nice

So the universal property is just to make sure the map is the quotient map?

[empty grove]

yes

it ensures that you have no extra points, and nothing else is identified, and that the topology is the quotient topology

It is the free-est way to sew these things together while satisfying the commutativity of the diagram condition

[onyx crow]

Hmm this is cool

[pearl holly]

okay so apparently you can take the cohomology of the Steenrod algebra. How tf does that work tho, the Steenrod algebra isn't a CW complex, right? So how does it make sense to talk about cohomology?

[pearl holly]

ok max helped nvm

[frosty sundial]

Just fyi there are lots of different kinds of cohomology

You can take@cohomology whenever you have a chain complex

Chain complexes can come from CW complexes but they can also come from lots of other things

E.g. group cohomology

[pearl holly]

oh ye I was stuck thinking about generalized cohomology

andt ere you use cw complexes I think

[hexed holly]

Quick question: For a space A, what does the map -id: ΣA -> ΣA induce on homology?
The map sends the equivalence class of (a,t) to that of (a, 1-t)
Where ΣA= Ax[0,1]/ (Ax{0} cup Ax{1})

More precisely, on reduced homology

Is it always -id, or sometimes id, or sometimes neither?

[flint cove]

Hm, for A=S¹, it induces the inversion map, right? Because it's the map S²→S² switching north to south pole

[hexed holly]

Yes for A=S^1 it induces -id

[flint cove]

And for A=∗ it should be id because every map is homotopic to the constant map

[hexed holly]

By Hurewicz

[flint cove]

Oh, wait, reduced

[hexed holly]

reduced it's zero, so doesn't apply

Or vacuously -id

[flint cove]

ah right I failed to see that this is not a contradiction x)

[little hemlock]

more explicitly than what's written here, how do you construct this lift? I don't quite understand this explanation

[hexed holly]

@little hemlockthis is May's Concise.

It's a very evil book

don't read it

read something else

[flint cove]

I don't think it gets any more explicit than that in the general case
If you have a C¹ map you can construct the lift by integrating over the angle derivative, so the lift would tell you the accumulated angle (i.e. possibly surpassing 2π or going under 0)

The argument of the general case is basically saying „we can construct this lift locally and stitching the lifts together works because of compactness“

[hexed holly]

It also has incorrect proofs. Do yourself a favor and pick something else.

[little hemlock]

sure okay, but like what is the open cover of [0,1] that we are taking a finite subcover of here?

[flint cove]

That depends on your map. Because your map puts constraints to how large you can take your interval around your point to which you can extend your lift without looping around

e.g. if I→S¹ goes linearly from [0, π] then you only need one interval
If it goes linearly from [0, 50π] (i.e. looping around a few times then you need more

And if it does horrible shit like “most” of the continuous functions do then you can't say in general

[hexed holly]

You're using compactness indirectly, through the Lebesgue number lemma

https://en.wikipedia.org/wiki/Lebesgue's_number_lemma

[cobalt sonnet]

How do you do this arranging?

[honest terrace]

For any k, you can't have M_k cap (cup[i > k] U_i) = emptyset for any i > k (connectedness).
So there is some i > k such that M_k cap U_i is nonempty. Renumber your sequence so that U_i = U_k+1

@cobalt sonnet

[gritty widget]

hey, does anyone have an example of a cover that does not have lebesgue number?

if possible, a cover of subset on R that does not have lebesgue number?

[hexed holly]

Yes

Take [0, infinity)

Cover it by intervals (n, n+1/n)

Plus the interval [0, 2)

That's a cover of the open space, but the overlaps get smaller and smaller

For every number d>0 there is some overlap small enough so that at each of its points the d-neighborhood is not contained in any of the sets of the cover

@gritty widget

[gritty widget]

@hexed holly i dont see it

[empty grove]

why not just take the subset to be union of (n,n+1/n)

[gritty widget]

if you take a point from [0, 1'9)

any ball with radius less than 0.1 will be contained in this set

[empty grove]

wait it has a lebesgue number

how about union of (log n, log (n+1))

[gritty widget]

if you take a point somewhere else

it will be contained in one of the other neighborhoods

[hexed holly]

What is its lebesgue number?

[empty grove]

wait

[hexed holly]

It doesn't

you're mistaken

[empty grove]

does (0,1) union (1,2) work

[gritty widget]

what do you mean by (n, n+1/n)

omg

[hexed holly]

No, finite unions obviously don't work

[gritty widget]

i though it's (n+1)/n

[empty grove]

why

[hexed holly]

Nevermind, they do

[empty grove]

given any r, you take (1-r/4, 1+r/4)

and this is not contained in either

[hexed holly]

You can take in my example the union of sets for even n to be one open set and that for odd n to be the second

That example still works and has only 2 open sets

Finite union of bounded sets wont work

[empty grove]

doesn't my example work

[gritty widget]

but how is 3.9 covered in your set @hexed holly ?

[empty grove]

(0,1) union (1,2)

[hexed holly]

3.9 is covered by (3, 4+1/3)

[gritty widget]

you wrote "(n, n+1/n)"

[hexed holly]

Shit I meant (n, n+1 + 1/n)

But (n, n + 1/n) also works

[empty grove]

this works too though

[hexed holly]

When it covers their union

[gritty widget]

(n, n + 1/n) doesnt cover it?

[hexed holly]

Yes

[empty grove]

you can take small enough neighbourhoods of 1

[hexed holly]

But it's more satisfactory when the space covered is an interval

Or the whole real line

my example is very easy to modify so it covers the whole R

Just do the same thing at the other end

[gritty widget]

ok thanks i like this example

[cobalt sonnet]

@honest terrace But how can you ensure that every $U_k$ is present in the renumbered sequence.

[gentle osprey]

IlIIllIIIlllIIIIllll

[hexed holly]

@cobalt sonnetYou ensure it by always picking U_(k+1) to be the first one that intersects M_k in the remaining list

You can prove that with such a construction it exhausts all elements of the cover by the fact that M is connected

Suppose the union of the resulting M_k's is not M

Then for all of the U_k's that have been left out, they don't intersect the union of M_k's, since otherwise it would have been chosen at some point

That means the union of M_k's and the union of all the U_k's that have been left out form two disjoint open sets whose union is M

But this contradicts the fact that M is connected

Therefore none of the U_k's are left out

[cobalt sonnet]

@hexed holly Then for all of the U_k's that have been left out, they don't intersect the union of M_k's, since otherwise it would have been chosen at some point

@hexed holly Why is this true

[hexed holly]

If it intersects the union then it intersects some M_k

[cobalt sonnet]

Actually the premise is not right

You can't assume that the union is not M.

The union can be M even if not all U_j are exhausted

[hexed holly]

Good point

[cobalt sonnet]

though it is not relevant for my application

meaning, it is okay if we toss out some U_j s

[hexed holly]

So if U_i hasnt been chosen and it intersects M_k, it must be chosen after a finite number of steps

Do you understand?

[cobalt sonnet]

yeah I think U_j not chosen implies it cannot intersect any M_k

[hexed holly]

Im on my phone

Will explain more in 10min

[cobalt sonnet]

because if U_j intersects M_k_0, then it intersects M_k for all k >= k_0

Ah I see

"So if U_i hasnt been chosen and it intersects M_k, it must be chosen after a finite number of steps" this is correct

if we pick the smallest index that works at each step

[hexed holly]

I am back

You can ensure all U_i's are exhausted as follows

Instead of picking just the first U_i that intersects M_k, at step n, pick the first n U_i's that intersect M_k

If there's less than n U_i's, pick just those, but don't pick more

That way they're all exhausted

@cobalt sonnet

[cobalt sonnet]

Product of connected spaces is connected?

It's easy for path connected

[shy moss]

Yes

[gritty widget]

one would hope it's t rue

[cobalt sonnet]

my only intuition for connected is from path connectedness

[brisk horizon]

Does homology theory have other application other than counting n-dimensional holes?

[cedar pebble]

sure I mean if you're happy to dualize this to cohomology then there are more applications that are easy to name

homology isn't just about counting holes but also about like, giving homology classes of cycles that you can e.g. integrate over

[shy moss]

There are algebraic applications of homology

For example R and C are the only finite dimensional division algebras over R which are conmutative and with identity

[gaunt linden]

Whenever you have two nonempty open subsets of a product space, there is a point in each subset that differ only in finitely many coordinates. That means you can make one point into the other in a finite number of changing only one coordinate. If the two original open sets partition the product space, then at one of those coordinate changes takes you from one partition to the other, and therefore their intersections with to a single line along that coordinate (homeomorphic to the relevant factor) partitions that line nontrivially too.

[dim meadow]

What do you mean by "there is a point in each subset that differ only in finitely many coordinates"?

[gaunt linden]

A point in the product space $\prod\nolimits_{i\in I} X_i$ is a function defined on $I$ such that $f(i)\in X_i$ for all $i\in I$. I'm saying that if we have nonempty open $A,B\subseteq \prod\nolimits_i X_i$, then there is $f\in A$ and $g\in B$ such that $f(i)=g(i)$ for all but finitely many $i$.

[gentle osprey]

Troposphere

[gritty widget]

hey, i'm stuck at this statement in my topology book: "If X is locally compact Hausdorff space with no isolated points, then every point in X is a closed subset with empty interior." Why is that true?

Then it says "We also notice that Int(A) = Ø iff X-A is everywhere dense in X"; why is that true?

[empty grove]

if a point has non empty interior then it would be an isolated point

every point is closed because it is contained in a Hausdorff neighbourhood, and in a Hausdorff space singletons are closed

wait that doesn't quite work

but you can go to the one point compactification, and there singletons are closed

[gritty widget]

is that the definition of isolated point? i realise i can't find the definition anywhere in the book

[gritty widget]

hmm we haven't covered compactification yet and it's covered later in the book

[empty grove]

x is an isolated point if {x} is open

hmm let me think then

[gritty widget]

it's in the chapter on baire category theorem if it helps

[empty grove]

wait

locally compact hausdorff is hausdorff in particular

singletons in a hausdorff space are always closed

Because you can separate them from all other points, and then the union of all those open sets not containing this point is the complement of the singleton

[gritty widget]

ok i see yes

[empty grove]

This is also kinda the defintition of dense. Int(A) = complement of Closure(complement of A)

[gritty widget]

why exactly is this true "if a point has non empty interior then it would be an isolated point"

[empty grove]

The interior of {x} is either empty or {x} itself

because Int(A) is a subset of A

if Int({x}) = {x} then x is open

because interior of anything is open

[gritty widget]

ok thank you!

[idle night]

If a fundamental group is equal to 0 does that mean that it's trivial?

[shy moss]

Yes, 0 is the trival group

[wintry heart]

very basic topology question i have:

are all metric spaces with the same set homeomorphic? (if not, for which metrics are the corresponding metric spaces homeomorphic?) (is that extra question even answer-able?)

and what about product metrics?

[drifting sundial]

R with the discrete metric and the usual metric are not homeomorphic

[wintry heart]

ohh

so you can apply it to any set?

i mean

can you apply the discrete metric to any set and have the resulting metric space not be homeomorphic to (other metric spaces)

you should be able to do that, right?

[drifting sundial]

yes, if your other metric isn't some kind of discrete metric then usually they won't be homeomorphic

[wintry heart]

wait a second, is there another metric like the discrete metric?

nvm yes there are

rubber duck thing got me

[sturdy notch]

I got a basic question too, in a metric space, $cl(A) $ can be defined as the set of all limit points of $A$, but we saw in class this isn't necessarily true for topological spaces.

[gentle osprey]

dadaurs

[sturdy notch]

is there some basic counterexample in a topological sapce

[empty grove]

It's always the set of all limit points of A, but it may not be the set of sequential limit points of A

As in there may not be a sequence in A converging to every point in the closure, the closure can be larger

An example is all ordinals up to the first uncountable ordinal under the order topology

Then the first uncountable ordinal is in the closure of its complement but there is no countable sequence converging to it (you need uncountable sequences)

And then there are examples where even longer sequences don't work so you gotta use nets

One example is the Aren's space (which is a counterexample to almost anything about sequences that you can think of)

[sturdy notch]

wait how do you define limit points as opposed to sequential limit points?

[empty grove]

x is a limit point of A if every neighborhood of x contains a point of A

Some books say distinct from x at the end of that sentence

[sturdy notch]

ah right that makes sense

[empty grove]

But you shouldn't do that if you want closure to be the set of limit points

[sturdy notch]

yeah so given a set A, we always have that the set of all sequential limit points of A is included in cl(A) right?

[empty grove]

Yes because sequential limits are always limits

[sturdy notch]

but the other direction doesn't necessarily hold in all topological spaces

[empty grove]

Yes

[sturdy notch]

that makes sense, thanks a lot

I'll also ask one quick question regarding lebesgue numbers, given a metric space such that each covering has a positive lebesgue number does not imply that the metric space is compact right?

because i can always take some infinite set with the discrete metric

[empty grove]

Yeah that seems right

[hexed holly]

Can someone do a sanity check for me?

pi_2(S^2) = Z and is generated by the homotopy class of the identity map [id], correct?

[jolly garden]

Yes.

I guess you should understand it without translation, the words are pretty similar.

[pearl holly]

I guess this follows from Hurewicz

[long hornet]

I'm struggling with this proof from tom Dieck. More precisely, I only understood what we're trying to do, not how.

The second sentence feels strange because I thought if we have two trivializations s, t over an open set W, then they agree on the first component, i.e., we only get an interesting map on F (the fiber space)

[tame lotus]

A sphere over a vector space is the set of vectors of unit norm S(V) = { v in V | ||v||^2 = 1 }. Geometrically speaking what does it mean to have a sphere of m by n matrices S([R^n, R^m]) using the spectral norm? I have a hunch all spheres ought to be decomposable to combinations of simpler n-spheres but IDK where to start at all. It might be something like S(V) -> X ought to be decomposable to primitive homotopy groups.

[long hornet]

It is a norm on R^(m x n), so I think this is topologically the (mn - 1)-sphere?
I don't know about the geometry though.

[tame lotus]

Yeah you would assume it would be related to just S_(mn - 1) right? but I don't really get the spectral norm stuff.

[long hornet]

It's a theorem that all norms on R^N are equivalent, or something

[tame lotus]

I know there's a basis theorem for vectors so you can probably encode it as S_(mn - 1)

But IDK what that means for adding it together and stuff

I don't think you even need AC for finite dimensions

[rancid umbra]

do you mean like as a product decomposition?

im pretty sure u can decompose every n sphere into n circles

[tame lotus]

Yeah I would assume but it would have to be wedge product or something right?

Product of n circles is a toroid not a sphere I think

[rancid umbra]

yea

[tame lotus]

And that makes sense for S(R^n) and S(R^(m * n)) but more complicated examples like S([R^n, R^m]) feel awkward

I just don't have an intuition for normed vector spaces

[rancid umbra]

is [R^n,R^m] the space of m by n matrices or something diff?

[tame lotus]

Yeah aren't matrices exponentials in Vect?

Not sure

I've heard them treated as spans too.

But you might expect sphere as a functor to be monoidal and maybe closed wrt to pointed product and exponential

Presumably theres a map S(U) /\ S(V) -> S(U * V)

[long hornet]

Not just that, but actually given any two norms v, v', then every v'-sphere is bounded within two v-spheres, and the ratios of the radii can be chosen uniformly.

[tame lotus]

Huh oh wow. Vector spaces are funky

[long hornet]

I meant norms on R^n

[tame lotus]

So if you have like the complex numbers that's not true anymore?

I was playing with real matrices because they're simpler but maybe they're too simple

[long hornet]

Actually it's still true

Just saw it in Wikipedia

[tame lotus]

I think this mostly answers my question.

[spiral stratus]

That exact same proof is in Hatcher, Vector bundles think
With slightly more detail
Maybe try reading that one

[long hornet]

Oh, thanks!

[tame lotus]

I asked in another forum and I think I have an idea
https://math.stackexchange.com/questions/4344644/whats-a-sphere-in-the-space-of-matrices/4344710

Interpreting singular values as scaling factors then the ball of matrices ought to be the set of matrices that rotate, reflect or shrink

Does this make sense as a geometric interpretation of the "ball of matrices" { A in R^m*n | || A || < 1 } under the spectral norm

[glass ermine]

I've been stuck on this question for a while now. I've managed to split M into multiple parts. I've shown that if x, y, z are all rational then that subset of M isn't connected. You can also have combinations of two real numbers that multiply to give a rational and a rational number and that isn't connected, but I'm stuck on the case where x, y, z are all irrational but multiply to give a real. Specifically I can't show that the set {(x^n, x^m, x^k) such that n+m+k is an integer and x is rational} isn't connected. - is this even the right approach?
Sorry if I've explained it poorly lol

https://cdn.discordapp.com/attachments/233997066890772483/926088016500523008/IMG_0545.png

[glass ermine]

OHHHHHHH

yeah ok that's way easier lmfao how did I not think of that

[pearl holly]

Omg

It’s wew

[glossy pine]

Omg

It's wew

[flint cove]

Not sure if this is a stupid question, but is there a cohomology ring isomorphic to the complex numbers? i.e. is there H*(X, ℝ) such that there is a 1-cohomology class ω squaring to -1?

Just viewed from abelian groups this would mean H*(X, ℝ) = (ℝ, ℝ, 0, …), but I have no idea how the cup product behaves (I just know it exists and it is a natural transformation)

[flint cove]

Nevermind I'm stupid, the only trivial degree-2 thing is 0, -1 explicitly has degree 0

[gritty widget]

silly question

when is it possible to determine all cts maps between two spaces like done for Sn->Sn

i find that fascinating

[long hornet]

Up to homotopy, I think

[gritty widget]

um

ig an example

S1->S1

theta -> e^i(theta*n)

damn thats not the best way to write it

[long hornet]

But they aren't all of that form.

[gritty widget]

oh i guess up to homotopy

you are def right

because any rotation of circle is cts function

[tame lotus]

IIRC there's an algorithm to calculate for some classes of stuff homotopy groups

Not sure how fast or general it is

[idle night]

Can someone tell me why ∂_1 = 0 here? This is for a delta^2 complex with its 3 vertices identified to a point

[pearl holly]

Because the verticies are identified

I think idk I’m drunk

[gritty widget]

lol

but yea

[gritty widget]

All the points are identified so when you compute the boundary of any given edge its going to be 0

[idle night]

Ah that makes sense

I just got confused by the i and j indices

[gritty widget]

because boundary of edges is [vi]-[vj]

for i,j being endpoints of a given edge

[spiral stratus]

In general completely impossible

We probably wouldnt be using co/homology for many things if we could calc homotopy classes for anything

I mean you cant even calc homotopy classes between spheres

[gritty widget]

oh wow

is this what homotopy groups do?

how do i phrase this to make sense

[spiral stratus]

Yeah homotopy groups are just homotopy classes from spheres to other spaces

And have a group structure because spheres are H'-spaces

But they're not nice functorially

So co/homology is a replacement for that which is nice

But cohomology is actually also just homotopy classes into certain spaces called spectra

(At least for CW complexes)

[shadow charm]

CW complexes are whack

I might start regretting asking for Hatcher for Christmas

[spiral stratus]

CW complexes are based

[empty grove]

That's not the default convention though

[spiral stratus]

based? based on what point?

[gritty widget]

For me, homotopy groups "clicked" when I saw how the structure is co-represented by the point-preserving "equator collapse" map Sn --> (Sn V Sn), ie the cofiber of the equator S(n-1) --> Sn. A pointed space (b : X) represents a functor [ - , (b : X)] : PointedSpaces^op --> PointedSpaces, and the value of this functor applied to the equator collapse map of the pointed n-sphere is the multiplication map for the n'th homotopy group of X at basepoint b. The contravariant hom-functors [ - , (b : X)] turn colimits (taken over pointed maps of pointed spaces) into limits, and in particular [ Sn --> (Sn V Sn) , (b : X)] is naturally homeomorphic to [Sn , (b : X)] <-- [Sn , (b : X)] × [Sn , (b : X)]. When confronted with the question of what homotopy groups "do", this is the picture in my head!

[gritty widget]

i lost some of the details after you introduced the functor

you are saying this functor takes equator collapse to multiplication for n’th homology

i also dont see how they take colomits into limits either

ill try and ponder this though, thanks a lot

[long hornet]

Related question: If X is a CW complex, does the knowledge of the homotopy groups of X give us a description of [X, X]?

[stone cipher]

If we have an equivalence relation $\sim$ on $X$ and $A\subseteq X$ then is it correct to say that $[A]\sim=\cup{a\in A}[a]\sim$?
This question was motivated by quotient map of a topological space. Say $q:X\to X/\sim$, and if we pick an open set $U$ in $X$ then $[U]\sim$ means $\cup_{x\in U}[x]_\sim$?

[gentle osprey]

RaD0N

[honest narwhal]

Time for cohomology of arithmetic groups

[long hornet]

Whaat

[honest narwhal]

Changing gears a bit (I was gonna start with some general stuff on root systems but that's very algebra this channel is more topology)

[gentle osprey]

several people

several people

several people

several people

several people

several people

[honest narwhal]

So this gives us a cochain complex

[long hornet]

Is there a good example of E to keep in mind?

[honest narwhal]

The cohomology is called the group cohomology of Gamma with coefficients in E

Benedict: so... I'm pretty much learning this now but one thing that I know people care about is Galois cohomology

So you might have, for example, Gamma a Galois group of a field extension and E is the non-zero elements of the larger field, or the roots of unity, etc

Oh btw

So what's d_0

d_0:C^0\to C^1

[gentle osprey]

several people

several people

[honest narwhal]

In hindsight I should prob prove stuff about the chain complex

So first off why does d_q land in C^{q+1}?

Well that's clear just read it off lmofa

[long hornet]

I assumed it can be proved by manipulation

[honest narwhal]

Put a gamma in each slot

And omit one variable

You've just put a gamma in each slot of f

So you multiply by gamma on each term gg

Slightly less obvious (until further notice) is why this gives us a chain complex

[gentle osprey]

several people

[honest narwhal]

So you have smth like

$\sum_{j < i} (-1)^{i+j} f(x_0,...,\widehat{x_j},\ldots,\widehat{x_i},\ldots, x_{q+2}) + \sum_{i \le j} (-1)^{i+j} f(x_0,\ldots, \widehat{x_i}, \ldots, \widehat{x_{j+1}},\ldots, x_{q+2})$

[gentle osprey]

several people

[gritty widget]

what is [X,X]

[long hornet]

Mapa from X to itself modulo homotopy

[gritty widget]

oh

thats how its related

So S1 -> S1 is the hom set i was talking abojt?

[long hornet]

Yeah, but in general it's not a group

[gritty widget]

what is first example?

[honest narwhal]

Oh yeah forgot to finish off the thing. So being slightly more careful about the double sum, it's

[long hornet]

I don't know honestly, but it's "naturally" a group iff X is a group object or a cogroup object in hTop*.

[gentle osprey]

several people

several people

[honest narwhal]

Note the d^2 = 0 computation didn't rely on the fact that things live in C^q

So L^q forms a chain complex as well

And apparently this is an injective resolution of E

[long hornet]

Isn't it exactly like how we prove it for singular cohomology?

Also, just realized that I don't know whether L^q is about set maps or G-morphisms

[honest narwhal]

Pretty much

I thought about this for a minute and I think set maps tbh

Because you're not really thinking of Gamma^q as a "Gamma-module" I feel

Unless you wanna just say multiplication but idk feels off

Though the description that will be important is different

[honest narwhal]

I guess keep these facts in mind

So this seems kiiiiinda wrong

I think he meant that Gamma\X satisfies the properties in line 1 rather than just X

[long hornet]

I don't see why they are injective, then

[honest narwhal]

Hmm

[honest narwhal]

Going here then

[stone cipher]

(a) So far I know that for every open set of H at origin, there exists a circle contained in that open set. Suppose that we have a loop based at origin of winding=1. How do I prove that this is not path-homopotic to trivial loop based at origin, in H?

[empty grove]

Any one of the circles C_n is a retract of H, so the inclusion of that circle into H induces an injection into the fundamental group

[stone cipher]

what is the retraction?

[empty grove]

Map that circle to itself, all the other circles to 0

[stone cipher]

LMAO

[empty grove]

Why are you lyao

[stone cipher]

cause I'm done with topologuuu

[plain raven]

feels goo

[empty grove]

[gritty widget]

woah looks sussy

[rancid umbra]

is it a differentiable manifold?

[empty grove]

Not even a topological one

[rancid umbra]

is it because it doent look like a graph when u zoom in on zero?

[empty grove]

0 has no simply connected neighborhoods, all points in all manifolds do

[rancid umbra]

is there an easy way to fix it? like removing 0?

[empty grove]

Yeah removing 0 just makes it a disjoint union of (circle - point) s

I think

And that's a manifold

[rancid umbra]

is it true that a continuous injection between totally ordered sets must be monotone?

[gritty widget]

yea

[stone cipher]

I'm having trouble finding a counterexample of a universal covering space not contractible

[gritty widget]

im thinking of proof but it should be true

[rancid umbra]

i think you need domain to be connected dallas

[gritty widget]

yeah

probably

[rancid umbra]

nvm my c.e. wasnt continuous whoops

[gritty widget]

wait did toy delete example

-x

[rancid umbra]

i can fix it though i think

[gritty widget]

we want to show that given cts injective function between two totally ordered sets that the function has connected fibers

every fiber of injective function is a singleton

which is connected

or wait

lol

i think thats it

[rancid umbra]

why does that show its monotone?

[gritty widget]

unless we working with different definition of monotone

we want to show it preserves order correct?

[rancid umbra]

yea

[gritty widget]

so if aR_1b then f(a)R_2f(b)

[rancid umbra]

yea. we can just use < and assume wlog its increasing. easier for readability

[gritty widget]

yea

uh they are total orders

[rancid umbra]

yes

using order topology

[gritty widget]

let a < b, then f(a) < f(b) or f(b) < f(a) is true

[rancid umbra]

we're going to need to compare three points

[gritty widget]

it could flip the relation also

[rancid umbra]

wdym

[gritty widget]

a<b implies f(a)< or > f(b)

nah im curious what definition being used

[gritty widget]

lol

[rancid umbra]

this is my c.e.

lol

continuous injection that is not monotone

[gritty widget]

under standard topology how is this cts

[rancid umbra]

wdym lol its three line segments

[gritty widget]

so where does f(-1) go?

idk domain or range

[rancid umbra]

my b. domain is X = [-6,-3] U [0,4] U [6,12]

its a continuous injection from X to R

[gritty widget]

f^-1(0)?

just empty?

[rancid umbra]

emptyset

[rancid umbra]

not defined

pretty sure you need connected domain

[gritty widget]

maybe?

and ur using subspace topology ig

[rancid umbra]

ye

once u have connected domain u can use IVT

i think the proof is just going to be hella case work tho

[gritty widget]

wait

[rancid umbra]

because you have to show that for any three points a < b < c, you have f(a) < f(b) < f(c) or f(a) > f(b) > f(c).

then you have to show that for any four points if a < b and c < d, with f(a) < f(b), then f is increasing, meaning f(c) < f(d). (if f(a) > f(b), then you would have to show that f is decreasing)

[gritty widget]

lol

ur right ig

i wish it was true but it isnt

thats good

[empty grove]

Need connected order for that

Oh wait you already got it

[karmic hornet]

can every metric induce a topology?

[frosty sundial]

every metric induces a unique topology

[karmic hornet]

is it possible for a topology to induce a metric?

[frosty sundial]

you can ask if a topology is metrizable

i.e. "is there a metric on this set which induces the given topology?"

sometimes there is and sometimes there's not

e.g. the indiscrete topology is not metrizable (on a set of more than 1 element)

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

[karmic hornet]

oh so that's what it means for topology to be "metrizable"

[obtuse meteor]

I would definitely not say the word "induce" here, bc induce usually implies it's unique IMO, and even when you're metrizable the metric is not unique

Ex. The Euclidean vs. taxicab metric on R^n both induce the same topology

[karmic hornet]

got it

kinda

lol

[wanton marsh]

A sphere

[swift fjord]

Suppose I have two different triangulations $K,L$ of $\mathbb S^1$. I have a continuous map $g:\mathbb S^1\rightarrow \mathbb S^1$, that is induced by a simplicial map $\phi:K\rightarrow L$. Is it always the case that the degree of $g_{\ast}$ will be the degree of $\phi_{\ast}:H_1(K)\rightarrow H_1(L)$? What if $K,L$ have an identical geometric realisation (That is then hoemo' to $\mathbb S^1$). Even in this case, I find it hard to reconcile that I can essentially choose two different generators for the domain and codomain and the degree will still be well-defined

[gentle osprey]

Kosher Nostra Mashgiach ShiN

[swift fjord]

like even if K,L have the same realisation and are homeomorphic to S1 by the same homeomorphism, the isomorphism in homology will still not be identical by the definition of the induced map of a general continuous map in simplicial homology

I find that all of this 'up to homeomorphism' identification kinda obfuscates what you actually can and can't do.

[stone cipher]

How do I prove that?

[wanton marsh]

well if you know that a sphere is not contractible

[stone cipher]

I don't know xd

[wanton marsh]

ah

[empty grove]

use proof by obvious

[stone cipher]

Sadge

[empty grove]

I think wedge of 2 comb spaces is not contractible but is simply connected

base point being the pathological point

and proof of not contractible should be elementary

but then its not locally path connected etc

oh they don't require X locally path connected

nice

[stone cipher]

Comb space?

https://en.m.wikipedia.org/wiki/Comb_space

So do u mean wedge sum of this?

[empty grove]

ye

[stone cipher]

And the base point?

[empty grove]

top left

[stone cipher]

(0,1)

Oops

[empty grove]

ye

oh

(0,1)

[stone cipher]

Thx. I will try

[shy moss]

You can prove it with the second homotopy group or with second homology group

A sphere has a hole so it can’t be contractible

[stone cipher]

@shy moss I didn't study these things :|

[hollow harbor]

why is moldilocks being so mean...

[empty grove]

who me?

[swift fjord]

This is drivig me insane

[odd flame]

what is the relationship between isometries and metric spaces

[shy moss]

An isometry is a function between metric spaces that preserve the distance

[idle night]

Is it accurate to say that singular homology, practically speaking, can only actually compute the homology groups of simplicial complexes?

[stone cipher]

is any simply connected space, semilocally simply connected?

seems immediate to me, by choosing U to be X itself as X is simply connected...

[spiral stratus]

i mean thats asking for spaces that arent homotopic to simplicial complexes/admit triangulations
which is relatively pathological but definitely a thing, e.g. https://en.wikipedia.org/wiki/E8_manifold this piece of crap
But I honestly have no clue what that thing's homology looks like

[shy moss]

Yes

[plain raven]

find a common refinement of the triangulations?

it's a good question

i'm really not in the mood to think about this rn but like, i will talk to you about it later if you don't figure it out

[idle night]

I mean that (this is what I've heard, I'm trying to figure out if it's correct) computing homology groups with singular homology is practically infeasible due to the amount of maps from n-simplexes -> x, and homology groups are actually calculated in practice using simplicial homology (which then has a canonical iso to singular homology groups)

[plain raven]

This is mostly true, certainly what you said is obviously correct, you can't program a computer with an uncountably infinite number of simplices so having a practical representation of the chain complex by finitely generated means is like, impractical

but also it's easier to prove that singular homology is homotopy invariant and then you can just freely deform your space into a simpler one without worrying about respecting the triangulation

so i think in practice you'd want to switch back and forth between these approaches pretty frequently, you use the homotopy invariance of singular homology to deform your shape into a simpler one that admits a triangulation and then use the equivalence of singular/simplicial

[idle night]

Thanks for the answer. What would this part consist of mathematically? Is it just saying "X and Y are homotopic, so singular homology tells us H_n(X) is iso to H_n(Y) and Y admits triangulation", so we calculate H_n^delta(Y) and get H_n(x) from that, or is there some more mathematical machinery involved than that?

[plain raven]

yeah that's all

I don't have a good example off the top of my head, i wonder if there is any like, contractible space which is not a CW complex

[idle night]

That simplifies things a ton, tyvm

[tame lotus]

IIRC every space is "weakly equivalent" to a CW complex.

But weak equivalence is clunky? This stuff is mostly beyond me.

I mostly come at topology from category theory and group stuff.
IIRC most notions of infinity categories only correspond to a "well behaved" subcollection of spaces.

Part of why I'm interested in learning more about topology I guess.

[spiral stratus]

Depends on the space
Eg for E8 we dont know if it has the homotopy type of a CW
So you cant apply Cellular homology either
Same for triangulation etc
Im honestly not sure how you'd approach calculating that homology, perhaps through a bunch of bundle sequences

[gentle osprey]

mns

mns

[honest terrace]

This is equal to (1/2,1), if I'm not reading it wrong

[limber ravine]

indeed

[empty grove]

If n starts from 1, it is empty

Second one is true

[swift fjord]

suppose you have some arbitrary triangulation of S^n. Is it always the case that the generator of H_n(S^n) will contain all of the n-simplices of this triangulation?

this feels true but also feels like there should be some counterexample for n>=2

or, at least, the simplices generating the homology contain every vertex

[swift fjord]

More generally, is the generator for the top homology of a sphere always going to be the sum of the simplices?

Regardless of triangulation

We gave a similar argument for the top.homology of the torus under a specific triangulation

I guess

You could always argue that the sum of simplices is a cycle

And it has to generate the homology since it's a nontrivial cycke

So you can choose it as a generator

Right

Oh

Hmm

Right

But it'll always contain all.the simplices

Ok thanks

Ok stupid question but at most 2 n-simplices can share an n-1 face right

In a simplicial.complex

I don't mean explicitly

I guess i'm just thinking of simplices in 3d space

and there only 2 simplices can share a 2-face

but that might not generalise

just cuz like, any other simplex would either identify with one of them or not intersect them at a face

ok how about for n>1

like of course there can be arbitrarily many lines coming out of a point and not intersecting

I don't see how you can even produce 3 tetrahedrons that all intersect in shared faces and share a 2-face

Oh ok, yea I guess that's true

Then what's going to be the problem if some simplex is missing from the generator

Like why would it prevent it from.being a cycle in that case

Huh

Is this something that's easy to show?

At least, in the case of a sphere

Or is it something I should take as a fact

Problem is we haven't formally talked about manifolds in this course

I'll talk to my prof about this and see what says

[limber ravine]

the first, would $\bigcap(0-\frac{1}{n},1)$ work?

[gentle osprey]

mns

[limber ravine]

Now I need to show that every open set of the real line is a union of disjoint open intervals. We take $(a,b) \in R$ and claim $(a,b) = (a,\frac{b-a}{2}) \cup (\frac{b-a}{2},b)$. After showing the equality holds would it suffice to finish the proof?

[gentle osprey]

mns

[swift fjord]

You just showed every open interval is a union of disjoint open intervals

You need to show it for a general open set

Do you know that open sets in R are the (not necessarily disjoint) union of open intervals?

[limber ravine]

yes

the standard topology in R is the set of unions of all intervals (a,b)

with a,b in R

[blissful cedar]

can anyone help me understand why a line in R^2 is closed? I was looking at the proof and it's because R^2 - L is open and therefore the compliment has to be closed. Which makes sense, but why wouldn't taking a finite amount of unions of n-balls along the line work?

[fading vale]

How would a finite union of 2-balls cover a line in R^2?

[blissful cedar]

define an infinitesimally small point and small radius and just plot along the line i guess?

[frosty sundial]

[fading vale]

Thats not going to be a finite union though

[blissful cedar]

ok but if its an infinite union would that make a difference to its open property

[fading vale]

I mean even if you take open 2-balls their union will be an open set containing L but it wont be L itself

[frosty sundial]

(sorry my sweat was just because balls have positive radius by definition, you can't have "infinitely small radius")

[fading vale]

It looks like this

[blissful cedar]

ahh right

so then that answers my next question, so if we have an open disc in R^3 it is neither open nor closed because we can't hit the boundary points with our 3-balls?

[fading vale]

well its an open disk so it has no boundary points

[blissful cedar]

🤦‍♀️

so cant we cover the open disc with just one open disc?

namely itself?

[fading vale]

But yeah neither it nor its complement can be represented as the union of open balls

"open" here refers to open in R^2

[blissful cedar]

ohhhh because we dont have a z-direction

[fading vale]

An open disk would be a 2-ball, not a 3-ball

Yeah

[blissful cedar]

Okay that makes a lot of sense now thx

[fading vale]

Happy to help

[empty grove]

That's [0, 1)

@limber ravine

[stone cipher]

,tex It is known that $C=\left{(x,y,z):x^2+y^2=z^2\right}$ is not a manifold. But finite union of manifolds is a manifold and $C_+=\left{(x,y,z)\in C:: z\ge 0\right}$ is a manifold. Then the union of $C_+$ and $C_-$ would give $C$ a manifold?

[gentle osprey]

RaD0N

[shy moss]

Union of manifolds not necessarily is a manifold

[stone cipher]

Let me see where did I get that idea

[frosty sundial]

it's true for disjoint unions of manifolds (although you might want to be careful about your definition and if you would allow unions of manifolds of different dimensions or not)

[hollow harbor]

[stone cipher]

Hmm ok. Maybe I got it wrong. Thanks.

[hollow harbor]

R x {0} and {0} x (0, infinity) are disjoint manifolds

their union is not a manifold

[frosty sundial]

that's not a disjoint union

[hollow harbor]

yes, it is

[frosty sundial]

"disjoint union" is not just "union of disjoint sets"

no

[hollow harbor]

ok fine

i see what you mean

you're tagging them

sure

[frosty sundial]

the disjoint union of A and B is (A x {0}) U (B x {1})

i mean that is just the definition of disjoint union of topological spaces haha

manifolds don't come embedded in some larger euclidean space where "union" could actually make sense

[hollow harbor]

ok, i'm with you on that, but i do think it bears clarifying if you're mentioning it to someone

[frosty sundial]

ok sure

[hollow harbor]

cause they often are embedded and then you need to be careful

*someone who is learning, specifically

or just someone who forgets like me 😵‍💫

[frosty sundial]

😵‍💫

[stone cipher]

Ohh.. now I see. In fact disjoint union of countable "manyfolds" ( ) is a "onefold" ( ). They all need to be of the same dimension.

[rancid umbra]

wait why

this seems really manifold-y

[hollow harbor]

the union in R^2 is a T

well, it's upside down

[rancid umbra]

right but it’s separated

there’s no corners

[hollow harbor]

how is it separated

wut

i mean i agree if you do a disjoint union and tag them

then they live in totally different R^2s

but if you union them in the plane the origin is not locally euclidean

[rancid umbra]

the intersection of the two sets is empty right?

[hollow harbor]

yes, so they're disjoint

[rancid umbra]

am i having a blonde moment?

[hollow harbor]

"separated" and "disjoint" are not the same things

[rancid umbra]

right, so i don’t see the issue with the union not being a manifold

[hollow harbor]

do you see what the union is?

[rancid umbra]

my bad, i didn’t mean separated in the technical sense

[hollow harbor]

the union is a T but with the ends all being infinitely long

like maybe if i write this it's easier

R x {0} and {0} x (R \ {0})

now the union is just both the x and y axes

(it's {(x, y) : x or y = 0})

this is not a manifold because at the origin, you're always going to end up with these 4 branches no matter how far you zoom in. it's a bit of an exercise, but you can show that no neighborhood of the origin is homeomorphic to an open interval in R.

(and it's certainly not a higher dimensional manifold).

[rancid umbra]

no yea i got that

but u said that the upside down, disjoint T was not a manifold

[hollow harbor]

my other thing is the same but it's missing the bottom leg

it's still got 3 prongs coming out of the origin

(important to note that the origin IS in the union).

[rancid umbra]

i could see it not being a manifold if the union wasn’t disjoint because of the intersection at the origin

[rancid umbra]

oh my god

bruh

[hollow harbor]

lol it's ok, buncho got me with a similarly brainfarty mistake

[frosty sundial]

^

[hollow harbor]

when i see a disjoint union of sets that are already disjoint, i just immediately parse it as union. analysis brain

like you do this in measure theory, you don't tag stuff

it's a bad habit

rly they should never be using sqcup or anything in measure theory i guess

[rancid umbra]

@hollow harbor sorry for the ping. i’m still trying to get the hang of manifold stuff. but if you remove the origin from the upside down T, that should turn it into a 1 manifold, correct?

and the reason it’s not a manifold rn is because the component functions derivatives approaching the origin are zero, so the rank of the derivative isn’t 1 at the origin

[hollow harbor]

yep that's fine!

exactly

that's a good reason as to why

[rancid umbra]

cool. thanks

[gritty widget]

is it locally euclidean tho

or is just a disconnected manifold

[hollow harbor]

wut

[gritty widget]

how would i go about showing that upsidedown T with origin removed is locally euclidean

[plain raven]

a disjoint union of bananas

[finite heath]

@rancid umbra aren't u on winter break my guy?

test

[rancid umbra]

lol yea but i’m trying to get over my fear of manifolds and differential forms

[gritty widget]

holy hell

i was thinking of it without point removed lol

but its still 3 connected components

doesnt change much ig

also wdym idenity x idenity

[hollow harbor]

woah, i was given honorable?

[orchid forge]

grats

[long hornet]

I want to prove the fact that if X is compactly generated Hausdorff (GCH) and Y is locally compact, then X x Y is compactly generated, in a "categorical" way

I know that, since Y is locally compact, then Hom(Y x A, B) = Hom(Y, Hom(A, B)), so Y x -- is a left adjoint, and since X is GCH, it is a colimit of its compact subspaces, so these two operations commute: If X = lim K_i then X x Y = lim (K_i x Y).

If we can do (2), then X x Y will really be compactly generated, because K_i x C_j is compact, and we will be done.

[empty grove]

2 seems legit idk about 1

1 also seems like it would be true

[long hornet]

The issue is that we haven't use Hausdorffness at all, and it's unlikely that it is unnecessary.

[empty grove]

I'm just commenting on the categorical bit, that seems fine to me

Topology idk enough for this lol

That could be it

[long hornet]

Even the lim K_i x lim C_j = lim (K_i x C_j) step?

[empty grove]

Ye limits do that

You can take limits of limits

Or limits directly

Wait

[long hornet]

I'm taking the limit over the "product" of two posets

[empty grove]

Ye true that might be false

Wait that is is legit because you can say that product is a right adjoint functor from Top × Top to Top and you are taking the limit of the diagram (K_i, C_j) then the product and the limit of this diagram is the pointwise limit and product is a right adjoint functor so it commutes with limit

[long hornet]

I see! I am worried about the maps in this "product" diagram. They are just the products of the morphisms?

[empty grove]

Yep

And there's a general theorem that says that if D is complete and D^C is the functor category for some C, then D^C is complete with limits computed pointwise

And you can identify Top × Top with Top² (lmao) where 2 is the 2 object 2 morphism category

@long hornet

[long hornet]

I'm willing to believe that

I originally wanted to prove that if X, Y are CW compelxes and Y is a locally finite (= locally compact) then X x Y is a CW complex with the product topology (equivalently, its compactly generated).

The other result of this type is that the same holds if X, Y both have countably many cells

So again X and Y are limits of locally finite (in fact finite) CW subcomplexes, and these limits are indexed by integers

But it's not obvious how to generalize this proof

The one given in Hatcher relies on induction, so :(

[empty grove]

Categorical induction lessgo

[long hornet]

Oh yeah I totally see it!!!!

[empty grove]

[finite heath]

wahoooo!

[blissful cedar]

hey guys im getting a little confused with the notion of manifolds if anyone could help me that would be great :). The cross (or just simply R^2 with the origin) is not a manifold because it isn't locally homeomorphic to a Euclidean space right, but how comes R^n is a manifold (presumably this contains the origin)?

[empty grove]

It is homeomorphic to itself via the identity map

The origin is not the problem point

The "self intersection" is the problem

[blissful cedar]

whats the difference?

between the origin and self intersection

is it because we're creating a new origin w.r.t. our open balls?

[empty grove]

The cross is lower dimensional because it is like a line everywhere else

so if it is locally like some R^n then n=1

but near the origin it is not like R^1

[blissful cedar]

that makes sense

ahhh

so once we pass the boundary of n=3, then the intersection part doesnt matter

because we're in a higher dimensional space, and we can represent the intersection with some n-ball where n>=3

am i thinking about this correctly?

What if you had two manifolds with different dimensionality intersecting each other, this would presumably be not a manifold right?

[empty grove]

nothing to do with dimension

if you have the xy plane and the zw plane intersecting at a point in 4d space

That would also not be a manifold for a similar reason

and same in all higher dimensions

[fallow venture]

[blissful cedar]

err

when we talk about R^n being homeomorphic to itself (via the identity map), is n finite?

so R^n is a manifold because the identity map which is a homeomorphism that just takes a point on R^n back to a point R^n is a manifold. But we can't take an identity map on R^2 because everywhere apart from {0} looks like R^1?

im confused trying to relate the two :((

[fallow venture]

I think there's a lot of people as confused as you in the world

Including me

[empty grove]

Any space is homeomorphic to itself via the identity

[empty grove]

To prove that something is a manifold, you need to show that every point has a neighbourhood homeomorphic to R^n for some fixed n

In the case of R^n, you can take the neighbourhood to be the whole space for any point

For the cross, 0 has no neighbourhood that looks like R

(any neighbourhood V of 0 in the cross contains a connected neighbourhood U such that U - {0} has 4 connected components, whereas there is no point in any connected subset of R such that removing that point gives you 4 components, so 0 can't map to anything under this homeomorphism from V to R)

[blissful cedar]

ahh right i think i understand the subtlety behind fixed n

[empty grove]

The cross wouldn't be a manifold even if you could choose different n for different points

[blissful cedar]

So we can have it being homoeomorphic to different dimensions of Euclidean sspaces?

[empty grove]

no neighbourhood of 0 in the cross is homeomorphic to the plane

[empty grove]

no, but even if you allowed that, the cross is not a manifold

[blissful cedar]

so here, when you say every point in the space has a neighbourhood homeomorphic to R^n, every point has to be of the same dimensional Euclidean Space right?

[empty grove]

yes, in the standard definition

[blissful cedar]

and in this case? why does it fail? is there a 4+1 dimensional intersection or something?

[empty grove]

I don't think there is an elementary argument, you would have to use homology or something of the sort

[flint sonnet]

Hi, just passing by ; can't you say that a cross minus the central point is not connected, but removing any point from a connected n-dimensional manifold (n > 1) does not disconnect it ?

[empty grove]

They are asking about the higher dimensional version of the cross example

that you can have 2 planes in 4d space intersect at a single point

and their union is then not a manifold