point-set-topology

[thorny agate]

Yea but isn't the scaling different on one side to the other?

[feral copper]

No biggie

Just take a translation of the map $$x\mapsto\frac{x}{|x|}$$

[gentle osprey]

Matplotlib

[thorny agate]

oh wow yea I was overthinking it

[feral copper]

And using the fact that you actually take a small circle around (-1,0) so you'd rather take like 1/4x the map above

Yeah basically your map can be $$x\mapsto\frac{x+(1,0)}{4|x+(1,0)|}$$

[gentle osprey]

Matplotlib

[feral copper]

And I realize the 1/4 is useless too x')

[heady skiff]

could i get a hint for E? I'm unsure whether it's open in Y; for example, I know that for points x with |x| > 1/2 we're good, and 0 is not included so we're also good, but for points that get super close to the origin i'm wondering if any open interval around that point necessarily intersects some point of the form 1/k

i do know that for every epsilon positive, (x - e, x + e) contains a point of the form x + 1/n but i don't really think that tells me anything either

ohw ait

is it just that if |x| < 1/2, then by the density theorem for e > 0 small enough we have x - e < r < x + e and since |x| < 1/2 it follows that this rational number must be of the form 1/k

yea you just choose |x| < e < 1/2 right

wait no i don't think that makes sense 💀

oh 0 < e < 1/2 - x since 1/2 - x is positive implies x < x + e < 1/2 there we go

talking to myself!

[brittle rapids]

density theorem?

{x | 1/x in Z+} divides 0 < x < 1 into countably many open intervals

and the union of open intervals is open

[heady skiff]

wdym

i just said that if $0 < x < \frac{1}{2}$, $\frac{1}{2} - x > 0$ so we have $\epsilon > 0$ such that $0 < \epsilon < \frac{1}{2} - x$. Also $-\epsilon > x - \frac{1}{2} \implies x - \epsilon > 2x - \frac{1}{2} > -\frac{1}{2}$. now $x < x + \epsilon < \frac{1}{2}$, and so we can use the density theorem to assert the existence of $r \in \mathbb{Q}$ with $-\frac{1}{2} < x - \epsilon < r < x + \epsilon < \frac{1}{2}$, and since $|r| < \frac{1}{2}$, it follows that $r = \frac{1}{k}$ for some integer $k$?

[gentle osprey]

okeyokay

[heady skiff]

wait am i trippin

😭

oh wait i was thinking about limit points and stuff

😭

it's too late at night

[brittle rapids]

is it okey if i say i don't understand what you wrote

sorry

[heady skiff]

i don't know what i was doing

💀

[thorny agate]

this would be the way to go

[heady skiff]

now that i think about it

yeah

wait wtf was i doing

💀

that's so funny

LMAOO

[brittle rapids]

it's alright

[heady skiff]

hey it was a fun exercise tho

even tho i don't know what i was proving

but i proved something i think

i think i proved that if |x| < 1/2 and x is nonzero, then we can find an epsilon neighborhood that contains a point of the form 1/k

yeah

oops

if 0 < x < 1/2

[brittle rapids]

hm

but your proof looks sussy

because your epsilon can be really small

well whatever, have a good rest

[heady skiff]

i guess so but i'm hoping the proof proves that that doesn't matter?

lol yea i think that's a sign to stop w/ math fortn

oh wait i just realized that the first part is asking if they're open under the subspace topology

welp time to redo this exercise

[thorny agate]

bruh

[heady skiff]

well i guess that B is the only open set under the subspace top

bruh

well for the first part i guess i was discretely finding the open sets considered in R

so we chillen

final answer: B open under subspace top, A E open in R

ok no more math

[brittle rapids]

kind of a naive question but here we go

[swift fjord]

Are you going to go?

[brittle rapids]

i got too scared

[swift fjord]

Don't be

We're not going to judge you

[potent sky]

If i have a knot compliment M and I find a disk D where the boundary of D is an essential curve on the boundary torus of M, how exactly does this mean that the knot I was looking as is the unknot?

[potent sky]

^ ping me if you have an answer im not always checking this

[feral copper]

Doesn't that mean you have a heegaard splitting of genus 1 of the 3-sphere then? In which case of course both tori are unknotted

[feral copper]

Ping

What's important here of course is that the curve bounds a disc

Which means this disc is a cut system for the complement of the knot

[potent sky]

hi

one sec

idk im just not getting it

like i understand why the unknot is the only knot which bounds a disk

[potent sky]

im having a really hard time visualizing the disk that is bound by a longitudinal curve on the boundary torus

[feral copper]

You're not asking that the knot itself bounds a disc though

[potent sky]

yes that's true

im asking sort of the reverse question, i have a disk with essential boundary

[feral copper]

Also, wouldn't that be a meridinal curve rather?

[potent sky]

this is somehow supposed to tell me that the knot is therefore the unknot

well both curves are essential

on the torus

it's easiest for me to picture a disk on the inside hole of the torus

i can visualize why that's the unknot, but im not sure how it's a theorem

[feral copper]

But the disc lives inside the solid torus

[potent sky]

yes

of the knot compliment

[feral copper]

But yeah you can think of the solid thing to lie 'outside' for the drawing

[potent sky]

anyway, can you tell me why this is a theorem

[feral copper]

So the meridian of the knot bounds a disc in the complement. I'm sure this implies that the fundamental group agrees with the first homology, and the only abelian knit group is that of the unknot

[potent sky]

the fundamental group of what?

the boundary?

a yes, because it'd be one of the generators of the fundamental group of the torus

but what about the boundary torus of a non-trivial knot

can the same thing happen?

i guess you can't really get a disk that lies entirely in the complement

but what would the fundamental group of the boundary of that disk be?

wait isn't this just the loop theorem?

[feral copper]

Lemme be more precise: M is the complement in S^3 of an open regular neighborhood of a knot K
I'm saying that pi1(M) should agree with H1(M), because a meridian of the knot (living on the boundary of M) bounds a disc in M. I'm not sure how you'd go proving that though, I was just spitting out random ideas!

[potent sky]

oh

[red yoke]

(New to knots so maybe someone could help proofread this)
By Meyer-Vietoris, the 1st homology generators for the knot and its complement generate that of the torus, so since the essential curve is trivial in the knot complement, it must be a nonzero multiple of a generator for the knot.

[feral copper]

If you want an actuel theorem, you might want to look into https://en.wikipedia.org/wiki/Gordon–Luecke_theorem

[potent sky]

i dont like these

i feel like the reason is supposed to be simple

for one it's easy to visualize

i just want to understand why

[feral copper]

Yeah, well, sometimes that's also why it's hard to prove x)

[potent sky]

i dont think it is hard to prove

[feral copper]

I'd google stuff like "esential disc in knot complement" to try to look for cleverer people than me! Also, you can ask your question here again, and some knot theory pros can answer

[potent sky]

ok thank you for trying at least

[red yoke]

But homology is visual

[tulip bluff]

If A is a closed subset of X and j : A -> X the inclusion map how can I show that if j is a cofibration, then X x {0} U A x I -> X x I is a retraction?

[red yoke]

Consider X → X × {0} and A × I → A × I

[red yoke]

Point-set be messy

Why does every alg top book have false claims

[tribal palm]

lol

that is interesting

[uneven bronze]

Does the closure of a set always exist?

[ebon galleon]

Yes.

In a topological space, ofc.

[uneven bronze]

hmm, so even an unbounded set has a closure?

[ebon galleon]

Correct.

You can define it in terms of limit points, or in terms of an intersection of closed sets.

[feral copper]

Closure is the intersection of all closed sets containing what you're interested in

That exists since the whole space itself is closed

[uneven bronze]

ok

[ebon galleon]

(and "unbounded" is more a term for metric spaces, I think. Altho there is a related notion of compactness)

[half creek]

Is U equal to the disjoint union of the empty set and itself?

[knotty vine]

Not literally

[empty grove]

And since the intersection of any family of closed sets is closed

[empty grove]

They mean take U ⊂ X

And empty as a subset of Y

And vice versa with V ⊂ Y

[grizzled socket]

\pi:\Tilde{M}\rightarrow M is the universal cover. I have a subset homeomorphic to \mathbb{R}^2 inside of \Tilde{M}, and I want to know what will its image by \pi be. I suppose it is either again a plane, a cylinder or a torus, but is there a way (looking at the specific problem which I am omitting for simplicity) to figure out which one I get?

[empty grove]

There are more possibilities. You'd have to look at the specific problem.

[half creek]

So the contain here does not mean that the topology on X is a subset of the coproduct topology. Is it right?

[knotty vine]

Yeah

[empty grove]

Yes it's not quite a subset but X under the subspace topology is the same as the original X

[knotty vine]

If you take $X \subseteq X \sqcup Y$ and equip it with the subspace topology $\mathcal{T}S \subset \mathcal{T}{X\sqcup Y}$ then there is a homeomorphism $(X,\mathcal{T}_S) \iso (X, \mathcal{T}_X)$

[gentle osprey]

Semer

[empty grove]

The homeomorphism being equality

[knotty vine]

Well, the T_S thing consists of pairs of an open U of X and the empty set

If youre being pedantic

[grizzled socket]

Ok, but what kind of information should I be looking in my specific problem to figure out to what this image is homeomorphic?

[knotty vine]

I guess if were really being pedantic then this is still an abuse of notation... since X is not a subset of X u Y

[empty grove]

In what sense? Are you using some specific construction of the disjoint union of sets?

Lol

[empty grove]

smh

[grizzled socket]

\Tilde{M} is actually \mathbb{R}^5, foliated by 2-planes. I am projecting it onto a quotient. I need to know how a specific leaf gets wrapped.

[empty grove]

Depends on your map lol idk

[normal umbra]

Can I ask about "It's image is given by [x_0] and thus implies that g* is surjective?

Nevermind, it's by definition

[red yoke]

Does the n-fold torus retract to any non-separating circle?

I think I can prove it doesn't retract to any separating circle using homology

And I can draw a retract for this particular case (C')

[cedar pebble]

if it retracted to any circle what would that say about the homology?

[red yoke]

Its H1 contains Z

And the inclusion maps the circle to a generator of some Z in H1

[scenic pecan]

Someone help me prove this. I am not able to prove the basis criterion for part b. If x belongs to the intersection of two basis elements then there is a basis element in B1 \cap B2 that contains x.

[red yoke]

x is some distance away from the boundary of each of the two balls

Then you can find a ball with center sufficiently close to x such that it contains x yet is contained within the boundaries of both larger balls

A useful fact is that if y is in Br(x) and r' < |x-y|, then B(r-r')(y) is contained in Br(x)

[scenic pecan]

1st and third are topologies. 2nd one is not. Am I right?

[unreal stratus]

Sounds good yes

What was your proof for T_2

[scenic pecan]

Arbitrary intersection of infinite sets may not be infinite.

[unreal stratus]

Yes

[scenic pecan]

For T3 I am stuck a little bit to prove that arbitrary intersection of countable sets is countable. Do you have an idea how to prove this?

[unreal stratus]

Subset of a countable set is countable

Ofc here they include finite sets as countable

Or it doesn't work

[scenic pecan]

Oh since X-(union of arbitrary open sets) is a subset of a countable set?

[unreal stratus]

yes

[scenic pecan]

????

[unreal stratus]

wdym lol

some authors take countable to mean countably infinite

others include finite sets (which is more sensible imo lol)

but here we need the latter

otherwise like X isn't even open

[scenic pecan]

Okay so you mean if we take the former defn then phi is not countable? I was thinking countable means either finite or countably infinite

[ebon galleon]

Last time I checked I can count finitely many things.

[alpine nest]

Yeah, it seems to me that whether finite sets count (heh) as countable is not entirely standardized

Some authors treat them as such, others don't

[tidal lynx]

Wikipedia does, Rudin doesn’t

[tulip bluff]

Any good books that go over fibrations, cofibrations and higher homotopy groups in detail? Hatcher and May seem both a bit dull.

[hidden crag]

Check out Davis and Kirk

https://www.maths.ed.ac.uk/~v1ranick/papers/davkir.pdf

[last topaz]

Sometimes "countable" means "infinite but countable". Other times it means "finite or infinite but countable".

[coarse night]

skill issue

[radiant cedar]

Why is the map (x,s) -> (sqrt(1-s^2)x, s) from S^{n-1} x [0,1] -> S^n in S^n? I don't understand what this first component does.

[balmy field]

well the first issue is that if x is a unit vector (in S^n) then sqrt(1-s^2)x has norm sqrt(1-s^2) which is not 1 for s not 0

[white oxide]

Behold! An uncountable set:
{1,2,3}

[gentle osprey]

Living Hydrogen Cyanide

Living Hydrogen Cyanide

[normal umbra]

Nevermind, I figured it out

[meager frigate]

I have a topology final(intro topology) soon and I was wondering if anyone had any tips or important stuff I should be focusing on. I haven’t been attending my lectures since after quotient topology nothing made sense at all (connectedness, homotopy, homology)

i sorta gave up and i’m trying to teach myself this stuff now and i just wanna know the best way to do it or most important topics to start with

[normal umbra]

Is my working correct?
(R³, w = xdx-y,dy)

Also d theta

[last topaz]

In response to "important stuff I should be focusing on" I would say the Van Kampen theorem for homotopy and its homology counterpart, the Mayer-Vietoris sequence (also the Long Exact Sequence associated to a Short Exact Sequence) because that is how you compute stuff. By decomposing a space into parts that you already understand, you learn about it. And remember that the key to computing with an exact sequence is the fact that whenever a part of an exact sequence looks like 0 --> A --> B --> 0 then the arrow A --> B is an isomorphism.

[last topaz]

Your d(theta) is good.

[last topaz]

Seems like there's something your misunderstood about how to compute the exterior derivative of a 1 form. In general, a 1-form on R^3 will be of the form

f(x,y,z)dx + g(x,y,z)dy + h(x,y,z)dz

and you're suposed to do

f_xdxdx + f_ydydx + f_zdzdx + g_xdxdy + g_ydydy + g_zdzdy + h_xdxdz + h_ydydz + h_zdzdz

Instead you've got anxiety inducing expressions like d(xdx -y)/dy that don't mean anything.

[normal umbra]

Then what should I write for that

My brain is small sometimes so it doesn't get through

[last topaz]

Well, how does your

w = xdx - ydy

fit into the general expression of a 1-form that I wrote? I.e. what are the f, g and h?

[normal umbra]

w = xdx-y,y
F(x,y,z) = xdx - y
G(x,y,z) = y

dg/dx, dg/dz and dy wedge dy = 0, G redyces into nothing

I don't know how to compute dF/dx and dF/dy thougj

[last topaz]

f, g h are functions. They can't have dx, dy, dz in them!

Hence f is not xdx - y. In fact this expression has no meaning.

[normal umbra]

I have no idea

[last topaz]

Then I'll give you the answer.

f(x,y,z) = x

g(x,y,z) = -y

h(x,y,z) = 0.

[normal umbra]

Huh???

w = xdx-y, dy

not w = xdx-y

Or is the -y unimportant, but if so f should be x and g should be -1???

Idk

[normal umbra]

Can I ask how did you get this? I'm like, not thinking well rn

[last topaz]

Ok well let's back up because you keep writing w=xdx-y,dy. What is that comma supposed to mean?

I thought it was a typo but you seem to insist on it being there.

[normal umbra]

It's either my professor's typo and I'm just ranting on this bullshit when it's trivial

Cuz if the comma is gone then x and -y is trivial

[last topaz]

Well, what meaning does the comma has in your eye?

[normal umbra]

I have no idea

😭

[last topaz]

That's problematic.

[normal umbra]

I never seen a comma in k-forms...

[last topaz]

I think it's safe to assume it's a typo.

[normal umbra]

Welp, then it's x and -y

I wasted like 30 minutes on a typo

[tribal palm]

they meant to type \, for a small space

[normal umbra]

Ohh, does that even exist

[tribal palm]

as they have just before

[normal umbra]

You learn more everyday

Never knew \, existed

Also, is poincare lemma alg topo or dg

[last topaz]

Yeah, if you want your integrals to look good, write \int f(x) \ , dx

[swift fjord]

Also use \mathrm{d}

[dusty talon]

If I want to show a space is totally disconnected, is it sufficient to prove that every set in the base is disconnected

[umbral panther]

You can give a basis for the usual topology in the real line in which every basic open set is disconnected. Just take pairs of open balls

[dusty talon]

i don't get why ^ works

[steel glen]

which part of the basis definition do you think would fail

[dusty talon]

wait just to clarify

we're saying to form a base where every element in the family is a pair of open balls

[steel glen]

open balls in the usual sense

[dusty talon]

ok sure but then like

how do I form the interval (0,1) as a union

if I have to deal with pairs

if they intersect then i need a smaller set in the base contained in that intersection

and then either that set wouldn't be a pair or the problem is going to repeat itself

and if they don't intersect then you can't form the interval

unless im completely misunderstanding

[steel glen]

take the basis members (0,.5) U (.5,1) and (.25,.75) U (.9,1)

[dusty talon]

oh ig so

thank you

[merry geode]

I want to like algtop but prof made me dislike it with meh teaching and heavy exams

How do I recover?

[quiet thorn]

self study it

[feral copper]

What specifically did you dislike about the teaching? Maybe you simply don't like the material, not the lecturer? (Genuinely asking, although maybe I shouldn't because that's not #point-set-topology per say)

[merry geode]

Well, I felt the lectures were not structured enough

I guess part of it stems from hatcher's style, but it was harder to follow the professor's lecture than reading through the hatcher book.

[feral copper]

That's a point raised by lots of students across lots of unis tbh
Maybe it's a mixture of 1) alg top teachers aren't as rigorous as you're to in analysis classes for instance, and 2) drawing proofs is a first

Yeah Hatcher never helped me

[merry geode]

The problem is

[feral copper]

What were the lectures like?

[merry geode]

Reading hatcher, I got what it was trying to say right away

Lectures were harder to follow

The professor also prefers to write in small letters as well.

[feral copper]

Not a real issue; ask them to write bigger

[merry geode]

I don't need the rigor per se, but when she introduced e.g. barycentric subdivision, I did not understand why she was trying to do that other than it is in the proof of excision.

[feral copper]

Isn't that plenty enough reason to use them? (Also, Whitehead proved lots of things, and one of them uses them too)

[merry geode]

I mean I had hard time seeing the connections to the proof.

It was kind of introduced out of the blue, with a brief mention that it would be used in the proof.

[hidden crag]

That’s not uncommon

[merry geode]

I didn't like how the connections wasn't made effectively

[feral copper]

Typically, the theorems you see proved took months and months, or even years, to be proved. You're introduced to them in 20mins... ofc you're not gonna grasp it all at once

[merry geode]

Indeed, I did not expect to grasp it at once

[feral copper]

What I'm saying is true in general, but particularly hurting in topology

[merry geode]

In this case tho, it didn't help that the professor did not complete the proof.

[hidden crag]

There’s always other sources to read it from

[feral copper]

May sound stupid, but actually works: try googling it!

[merry geode]

Introduced the barycentric subdivision and linear chains, and kind of directly to "so let's say you can take singular homology where the map is restricted to open cover"

[feral copper]

Math SE and Overflow are great tools for that

[merry geode]

Yea I mean, I manage to understand it reading the textbook - I think it is greatly written

[hidden crag]

Where’s the issue then

[feral copper]

Anyways; I guess you don't really like alg top in general, not necessarily the lecture(er)s

That's fine, you'll find something else you'll enjoy!

[merry geode]

I doubt that

[merry geode]

Well, the issue is that it left some sour taste in my mouth regarding algtop.

[hidden crag]

If you decide to be pissed at the very subject itself even though you want to like it because of one course

Idk what to tell you

[merry geode]

Hmmm

[hidden crag]

[merry geode]

I don't think I "decided" to be pissed, maybe I did some overstatements?

[hidden crag]

The course is over right

[merry geode]

Yea it just ended today

[hidden crag]

So staying pissed about the professor instead of just reading the textbook that you said you like seems like a decision

[merry geode]

Well idk, maybe it was just the exam that was too hard for me

Yea, sorry that I was a bit ranting due to the exam

Ultimately if the prof did not expect much in e.g. exam, I would not have complained I guess.

[merry geode]

Ah I recalled, it was because the professor will be teaching algtop 2 next semester

Guess I can simply read Hatcher instead of taking the class?

[feral copper]

No, do attend the lectures

Did you discuss your complaints with other students? With the teacher? Because they also attended the courses and we didn't.
And yeah, go to the classes next semester. You'll be in no position to rant about failing if you don't

[merry geode]

Ugh

No, I didn't, because I am largely solitary there

But why go to the classes next semester?

[feral copper]

Then next semester, try working with other students. Working in group does help, really, it's not even one of those '10 weird tricks to pass exams'.

[feral copper]

Well, you asked whether you should go or skip them, I said don't skip them. Do you have a choice of courses to attend to make? Like, does the uni make you choose between alg top and something else?

[merry geode]

I mean it will help, but I don't think I can justify taking time for classes.

[merry geode]

Oh

I need to look into it, but I think algtop is not one of the necessary courses.

At least algtop 2 should not be

[feral copper]

Then try to review the material of sem1, and see if you fancy attending. If you end up realizing you dislike alg top as a field of maths, then don't ofc.

[merry geode]

(Algtop 1 could be necessary for the qualification exam)

Ah, sorry for not making it clear. Yea, I am not planning to stay absent in class or simular.

• I am interested in algtop for the knot theory, which drags my attention

[feral copper]

Yeah that's what I thought, sorry for me misunderstanding.
Then if it's a non-manfatory class, it's ezpz: go and see if you like it. If you don't, just stop going?

[merry geode]

Yea, I guess I shall read through Hatcher again. Thanks!

[feral copper]

There are plenty of books on knot theory, which are waaaaay better than Hatcher. So give those a read instead

[merry geode]

Ah trueee

[feral copper]

(And honestly, drop Hatcher altogether, it's bad especially for newbies to algtop)

[merry geode]

Oh?????

Hmm, I guess I will look for another material then. Thanks!!

[hidden crag]

It always surprises me anew that you don’t like Hatcher matplotlib

[feral copper]

Yeah ikr! It's a pretty unpopular opinion (despite somebody telling me otherwise someday, but it was based since they disagreed)

I feel like it's too handwavy for it to be the reference. There's nothing else out there which is as complete, but whenever I open Hatcher it's a nightmare for me, it's poorly-written and I do it as a last resort

[hidden crag]

I get the Hatcher hate in general but it just seems so up your alley

[feral copper]

Also, esthetically speaking, it's blocky and ugly, which doesn't help

Ah! Sorry you got it this way! It's still a reference to have on the shelf, and I do (it was one of the first books I bought with my own money)

It's funny because if somebody asked me what they could read, I'd still say Hatcher as there's nothing else

[hidden crag]

No worries it’s just funny to me because like

[feral copper]

there's nothing else
If I don't want to give them 12 references that is

[hidden crag]

I struggle with Hatcher sometimes because it’s too geometric

But then a geometric person doesn’t like it either

[feral copper]

Yeah for me it's too algebraic

[hidden crag]

May

[feral copper]

Concise?

[hidden crag]

Yeah

[feral copper]

Dang I forgot that existed, I never opened it! Gonna correct that mistake

[hidden crag]

Or don’t

If you don’t like things being too algebraic you won’t like it at all

[feral copper]

Yeah chap2 categories

[hidden crag]

I just brought it up as a joke because it’s a death sentence as a first course in alg top

define colimits
leave every single example and statement as exercise

[feral copper]

Perfect use case of :gigachad:

[merry geode]

Excuse me what

[hidden crag]

Don’t get me wrong I love the book but the pedagogy is questionable to say the least

I’ve also really been enjoying Davis and Kirks book

It has so much good stuff

[feral copper]

Looks like it's on the algebraic side of the force (that's my 2 cents from reading the toc)

For pedagogical purposes, I think it's best to start with \pi_1 before homology, since you can make drawings and start grasping how the algebra plays a role. Then homology can be seen as an abelian version of \pi_1 (except it's not, but that's a good enough motto for newbie students), and then do things like CW spaces and connect that with homology and \pi_1

Also, historically, wasn't \pi_1 considered before homology? (I remember Poincaré thought that all homology groups were equal to all homotopy groups)

[hidden crag]

I agree

[feral copper]

this way you can give a course which more 'tells a story' rather than delves into algebraic details

And if people think they like the algebra more, then they can move on from there

[hidden crag]

pi_1 + coverings into homology

Is the call

[feral copper]

Yeah coverings ofc that's a must

Also, this helps dissolve manifolds into the world of general topological spaces, and see in what ways they are special

Because I feel like lectures either only deal with manifolds, or not at all

[hidden crag]

My alg top class barely covered homology

[feral copper]

[hidden crag]

Because we didn’t get very far

Hm but what’s something geometric one could read after the standard content in alg top

I only know the stable homotopy route

[feral copper]

The Whitehead theorems for all CW-shenanigans is a possibility ofc. But then you could have (in no particular order, and some may be too difficult for students too, and most, if not all propositions are 101% based):

• characteristic classes (still decently algebraic too), and intersection/obstruction theory
• knot theory (e.g., in French: Knot theory through the Alexander polynomial)
• branched covers
• geometric structures: handle decomposition (Morse/Cerf theory are king), and then Heegaard splittings and their classification
• surgery theory and Kirby calculus

Oh sorry, something to read! Uhm, Seveliev's Lectures on 3-manifolds is nice!

[hidden crag]

All good, thank you for all the suggestions!

[white oxide]

Is there a way that given some base A and another base B with some nice property (say low cardinality) to replace its open sets of B with ones from A without invoking the axiom of choice? Rn I am looking at a hypothetical countable base for the power set of some uncountable cardinal equipped with the product topology and I want to replace the open sets with elements from the canonical basis and then do some careful diagonalizing

[coarse night]

this is nice

read h-cobordism after reading handles

[tulip bluff]

Let X and Y be topological spaces and f :X \to Y a homotopy equivalence, x_0 in X and y_0 = f(x_0). Show that if {y_0} -> Y is a cofibration, then f has a homotopy inverse g for which g(y_0)=x_0.

I thought this would have come from the cofibration condition by choosing the homotopy inverse g of f as the "test map" and the homotopy H : f o g \simeq id_Y as the homotopy, but I can't show that g(y_0) = g(f(x_0)) = x_0?

[unreal stratus]

i think the point is just that you can define a homotopy {y_0} x I -> X in the obvious way and take a map Y -> X to be your original homotopy inverse a, and then use HEP

like if a is a homotopy equivalence and b is homotopic to a then b is also a homotopy equivalence

so you are just using the fact that Y is well-pointed to modify a

[tulip bluff]

Can you elaborate on this? I'm quite new to this stuff so I don't really understand what you mean.

[unreal stratus]

No worries! So the point was basically that if $y_0 \hookrightarrow Y$ is a (Hurewicz) cofibration, then to define a homotopy $Y \times I \to X$, you need only define compatible bits on $y_0 \times I$ and $Y \times {0}$ and then it'll extend to a homotopy $Y \times I \to X$.

[gentle osprey]

homotopy coherent potato

[unreal stratus]

Now pick any homotopy inverse $a: Y \to X$ of $f$. This will define smth on our $Y \times {0}$ in the above notation. Our aim is to make the map on $Y \cong Y \times {1} \to X$ send $y_0 \mapsto x_0$. So we can just pick a path $\gamma: I \to X$ from $a(y_0)$ to $x_0$, and define a map $y_0 \times I \to X$ by $(y_0,t) \mapsto \gamma(t)$.

[gentle osprey]

homotopy coherent potato

[unreal stratus]

Now we use the homotopy extension property to get a homotopy $h: Y \times I \to X$ where $h(y,0) = a(y)$ and $h(y_0,1) = x_0$. So just let $g: Y \to X$ be $y \mapsto h(y,1)$

[gentle osprey]

homotopy coherent potato

[unreal stratus]

Now $g$ sends $y_0 \mapsto x_0$ by construction, but it's also homotopic to $a$. Since $g \simeq a$ - by construction! - we see that $g$ is a homotopy inverse to $f$ since $a$ is one too.

[gentle osprey]

homotopy coherent potato

[unreal stratus]

Basically the argument is: we take our map a and use the HEP to homotope a to a map which sends y_0 -> x_0.

Here I assumed a(y_0) and x_0 lie in the same path component. This holds because a(y_0) = af(x_0) is connected by a path to x_0 (since af is homotopic to the identity)

Is that clear?

[tulip bluff]

Thanks! I'll try to digest this.

[tulip bluff]

I'm a bit lost in the beginnig, why do you want to define a homotopy Y x I -> X?

[unreal stratus]

Well basically we have to modify a to find g, but we know that g (if it exists) must be homotopic to a

since (homotopy) inverses are unique (up to homotopy)

So it makes sense to build a homotopy in order to produce g

But also this is just an element of experience - often you can produce maps by taking one map and using HEP or something and then taking t = 1

[tulip bluff]

Oh okay. I am still a bit lost on when you say that "our aim is to make the map on Y -> X send y_0 to x_0"? The way I see this is that you are defining some conditions on a map Y x {0} -> X and {y_0} x I -> X that then play nicely when we use HEP.

[unreal stratus]

Well the aim is just exactly what you're doing

And then yes I'm sort of saying that the implies we need to define those maps in the way I did

[muted arrow]

https://mathoverflow.net/questions/459281/examples-of-counting-holomorphic-curves-in-cylindrical-reformulation-of-heegaard
Any knowers in chat?

note the comments

My theory here is that somehow the index of that one I drew is -1

so despite there being a disk in the image, there cannot be a holomorphic curve

But if that's the case, I still don't see a coherent way of determining when this is the case without using the disks in the standard Sym^g setting

[normal umbra]

Can I ask how does the connectedness of the open covers imply that the differential form is a constant function?

(Proving the sequence 0 -> A^k(M) -> A^(k,0)(U) -> A^(k,1)(U) -> ..., where the A^(k,n) is the nth de rham cohomology class)

[shadow charm]

Cause its exterior derivative vanishes

[normal umbra]

o

Oh yeah I'm stupid

Thanks

[shadow charm]

Np

[normal umbra]

Can I ask why delta(r(w)) = 0 and delta(c) = 0?

Also I'm also not that sure on phi(x) = [c], the notation doesn't ring a bell or I just don't work with cochains enough

Can I ask about where the \gamma_0 thing came from?
The author just jumped into "Indeed, there exist" without explaining 😭

I feel so dumb asking so many questions but my small brain isn't big enough to figure most of them out

[merry geode]

Woah that’s such a huge diagram

[fickle elm]

Not very sure about the notation. I have to guess what is this double complex A^{i,j}(M). But I guess that delta(r(w))=0 because on your third picture, if i=0, delta is replaced by r, so delta(r(-)) is the composition of two differential, it is 0 by definition of a complex.

And what is this curvy big U, is it an open cover of M? how many open sets does it have? l open sets?

[normal umbra]

U is open cover of M, All intersection U_alpha is contractible

[normal umbra]

If this helps

[merry geode]

.. De rahm cohomology is just Cech cohomology?

[normal umbra]

???

There exist an natural isomerism between them

[merry geode]

Yea I mean it seems interesting

[normal umbra]

isnt that a well known result anyways

[merry geode]

Idk algtop

[red yoke]

What are the A^p,q?

Is it computing Cech cohomology by adding more contractible sets

[fickle elm]

It is the cech de rham complex, i guess.

the curvy A

The normal A seems to be an index set.

[merry geode]

This seems like definitional

The author did not delineate it clearly, but I think they are talking about two representative of the same class

[fickle elm]

Wait a min. It is diagram chasing, most of them. It should not be difficult but may be hard to navigate.

The vertical direction is the de Rham differential, the horizontal direction is induced by the inclusion of each degree of open covering.

From the notes, i think A^l(M) denote the l-forms on M.

[normal umbra]

Yes, A is an index set and the complex is cech derham with d representing de rham ans delta cech

Yesss

The notes goes from easily comprehensible to an insane elysium

You know what, I'll just send the pdf here

It's only 11 pages long

I've been staring at this piece of notes for at least 25 hours these past 3-4 days so yeah...

[merry geode]

Are you having hard time with diagram chasing?

[fickle elm]

The reason for delta(c)=0 is such. If you calculate d(delta(c)), by commutativity of diagram (1), d(delta(c))=delta(i(c))=delta(w_l)=0. delta(w_l)=0 because this is how you construct w_l. You can look at all previous w_0, w_1,..... So delta(c) is d-closed, note that the vertical line is exact by proposition 3, so delta(c) is also d-exact, meaning it comes from the lower row, but it is all 0 for the rows lower than C^l(U), so delta(c)=d0=0.

[fickle elm]

Diagram chasing for this is always hard to navigate if you do not often do this.

[fickle elm]

You could try asking someone more experienced in this in real life, TA maybe. Sometimes you will go into a dead end without knowing and waste a lot of time. It is much easier to explain with chalks and blackboard, you can draw a large diagram and actually see how you track an element.

BTW, in Bott & Tu's differential forms in algebraic topology chapter II, he also has discussions for cech and de rham cohomology. It may be helpful but it uses new notations and some definitions are a bit different.

[normal umbra]

Kind of...
The w_0 w_1 ... w_l and eta_l part I understand but the rest I'm just sort of, "Ok it clicked and sure ig"

[normal umbra]

Why is delta(r(w)) 0 though

I don't see where it came from

[normal umbra]

I see, thanks for the advice 🙇🙇🙇

Partially because I sort of like, am new to alg topology ig? So I'm not like, that familiar to all the stuff

[fickle elm]

w is in A^l(M), delta (r(w)) is just the row A^l(M)-->A^{0,l}(U)-->A^{1,l}(U)

[normal umbra]

Also when did I get active role

[normal umbra]

Yeah

[fickle elm]

i dont know why he wants to change notation, but basically it is just delta^2

[normal umbra]

Oh...

[fickle elm]

And for a complex, the differential squared is always 0

[normal umbra]

Isn't r mapping from w to w_0 tho

[fickle elm]

Yes. w in A^l(M) and w_0 in A^{0,l}(U)

[normal umbra]

Ahhh

[fickle elm]

Look at the big diagram (1). Write element in the correct place and see where it comes and goes.

[normal umbra]

Oh...

I know what it's doing now

[merry geode]

Hmm

[normal umbra]

The c' = c + gamma part is just because that they are in the same eq right

[merry geode]

Same equivalence class?

[fickle elm]

You mean w=w'+dgamma_0?

it is because w and w' are both closed forms representing x in cohomology, so they differ by a boundary.

[merry geode]

Yea, they in the same class

[fickle elm]

This is how we define the map phi. We start with a de rham cohomology class x, we want to find a cech cohomology class. So we find a closed cochain c in cech complex, and we define phi maps to the cech cohomology class represented by c.

The next part is to show that this is well-defined. On the level of closed forms, everything just differs by some boundary, so we indeed define a map between cohomology.

[fickle elm]

Not in general. Note here in the proof the paper chooses a special open covering U of the manifold M. All the intersections are contractible. It is not possible for a general space.

[merry geode]

Ah I was only thinking of C^infty manifolds.

Thanks for clarification

[normal umbra]

How do you proof that it differs by a boundary tho

Conceptually I understand but algebraicly I don't seem to be able to wrap my head around those symbols

I self learnt algebraic topology so I might be lacking on some skills

[fickle elm]

By definition of cohomology, it is closed forms modulo boundary forms, so if two closed forms represents the same cohomology class, their differences must be in the boundary.

[normal umbra]

Yes

But like, how do you proof that they differ by a boundary

(Algebraicly

[merry geode]

You mean the step (II)?

[normal umbra]

The author just said "Indeed, c' = c + g0" then directly goes into induction

[normal umbra]

Yes...

[merry geode]

Ah they introduced c'

Welp. If you look closely, they explained why the claim holds

[fickle elm]

More generally, the closed forms are an abelian group G and the boundary forms are a normal subgroup H (because differential^2=0). The cohomology is just the quotient group G/H. Consider the standard projection G-->G/H. Two elements g1, g2 mapping to the same elements if and only if g1+H and g2+H are the same cosets, meaning g1-g2 in H.

This is completely algebraic.

[normal umbra]

Like, I understand that c' and c are cochains generated by the process but how do they "differ by g0" and how do you prove it???

It's like, counterintuitive to look at

[merry geode]

See the arrows on the left side of each line? Look at the 6~7th arrow

It says "On the other hand, since ... indeed homologous".

[normal umbra]

How is w'_0 = w_0 + dg_0 though

[merry geode]

That one is because of this, I believe.

[normal umbra]

Because dr = 0?

[fickle elm]

It is an algebraic fact about quotient group as I describe. g1=w and g2=w', so w-w' must in the boundary form since cohomology is closed/bounadry

[normal umbra]

Does the quotient group contain that element?

I guess

[fickle elm]

Which element?

[normal umbra]

The inductive proof and everything carrying on until the end of the proof of theorem 2 I can understand, I just had no idea why "Obviously \gamma_0 has to exist you little bitch"

I think the reasoning should be trivial

[fickle elm]

Let me say it in this way. Now you start with two closed forms w and w', you want to know if there exist a form gamma0 such that dgamma0=w-w'

[normal umbra]

Yeah

[fickle elm]

And this is the same as w-w' must be the boundary

Do you agree?

[normal umbra]

Yeah

[fickle elm]

OK. So we need to show w-w' is the boundary. We now use that they represent the same cohomology class

Meaning they maps to the same element in the quotient group closed forms/boundary forms

Now you can use the algebraic fact I mentioned about quotient groups.

[normal umbra]

AHHHH

Yeah

I understand now

Thank you so so sooo much 😭

Pov third day of actually learning algebraic topology

[regal garnet]

Hello , can you please help me with this exercise ? I don't know how to think for the problem 5.3 . If it's possible to give me some hints?

[tiny obsidian]

for compactness: the polygon is compact
for without boundary: interior points should be clear, think about how neighbourhoods of points on the polygon fit together after identification (two cases, vertices and edges)

[knotty vine]

I think the question was about the second exercise

[tiny obsidian]

oh whoops

[hexed steppe]

maybe use euler characteristic

as Q is not a sphere, chi(Q) = V - E + F <= 1.

and you want V=1

[tiny obsidian]

hint for the right question this time (sorry for some reason I read 5.3 and 5.4): somewhere in P are two adjacent vertices that are not identified to the same point, say identified into u and v. If more than one point is identified into u, show that there is a polygon with the same number of edges but one fewer point goes into u, and otherwise (or when you've gotten down to one vertex) show there is a polygon with fewer edges that gives the same surface

(squares dealt with separately)

[gritty widget]

https://math.stackexchange.com/questions/524492/proof-that-two-homotopy-inverses-are-homotopic

[knotty vine]

How does this work on the following polygon? It forms the torus but with an extra point v on one of the two loops

[tiny obsidian]

it's homeomorphic to this, same six sides and one fewer v

hopefully it's clear how to get something with fewer edges and only u

colours and shape are an extra hint I guess

[knotty vine]

I see that it is homeomorphic, that is one fewer v, and that we can remove edges to only leave u. I just don't see what the process is going from my polygon to yours and why that would be homeomorphic in general.
Couldnt we just contract any edge that joins two different vertices and skip the "keep the same number of edges" step?

[tiny obsidian]

I chose the blue edge as going from v to u, then cut the triangle with the edge before the blue one, choosing before to be the one further clockwise for my drawing convenience, and glue it back along that edge

contracting the edge...

I'm convinced it works

and it probably is easier

[knotty vine]

Cool!

[tiny obsidian]

I was going off of 'well when I did classification of surfaces, we did cut and glue'

[verbal sage]

Not sure if this is the right place to post this, but I am running a very casual reading group that is going through Hatcher and we're looking for new folks. We're on section 3.2, Cup Product. Folks can message me if they're interested. (I can totally appreciate the complaints. Some of his proofs can feel stream-of-consciousness and imprecise to me)

[scenic pecan]

Why do we require Quotient topology induced by a map to be a subjective map? I don't see it is necessary.

[prisma garnet]

you want X to be in the induced topology, don't you?

[scenic pecan]

But won't X always be there? As every x has an image. I may be missing something

[prisma garnet]

I mean the quotient space when I said X

[scenic pecan]

Oh yeah for Quotient space it is nice but I was asking in general from X to a set Y with Y having the quotient topology

[prisma garnet]

you mean to say for two topological spaces X and Y, why do we require a quotient map $q: X \to Y$ to be surjective?

[gentle osprey]

DarQ

[prisma garnet]

coz you want the topology induced by q on Y to be the same topology Y has already

@scenic pecan

[scenic pecan]

Yes

[scenic pecan]

But aren't we defining what topology we want on the set Y. I don't think it is already a topological space.

[prisma garnet]

I'm confused

[scenic pecan]

Take a look here. Y is any set it doesn't have a topology on it already 😭

I showed that the quotient topology is indeed a topology but there was no need for the surjectiveness

[red yoke]

Did you show that Y is in the topology?

Oh wait that's included

[prisma garnet]

ig you want the induced topology on Y to be related to X through q.

[red yoke]

But anyway the quotient space is important because we can turn continuous functions on a space into continuous functions on the quotient space

[prisma garnet]

and the point in Y that're outside the image of q just have the discrete topology

which isn't very useful

[scenic pecan]

Ig the defn is constructed to give nice properties to the quotient map?

[scenic pecan]

Yes this makes sense. Too many open sets would be there

[wispy veldt]

surjectivity is used to show that any cont function f:X-->Z with pie(a)=pie(b)--> f(a)=f(b) must factor through Y in a unique way , namely we have a unique cont map h from Y-->Z such that f=h o pie

iirc

[red yoke]

If ~ is a relation on X, then any continuous function on X that is compatible with ~ can be turned into a function on the quotient

Where compatible means it takes the same values on "equivalent" points

E.g. If your continuous function is sin: R → R, then since this is compatible with the relation x ~ x+2π, you can turn this into the continuous function S¹ → R.

[scenic pecan]

I don't get it.

[red yoke]

Where the circle S¹ has the quotient topology induced by the relation x ~ x+2πn

[scenic pecan]

Makes sense. 👍

[red yoke]

This is basically saying the same thing

[scenic pecan]

Okay..I will think more about this. Thanks 👍

[prisma garnet]

can someone give any feedback on this proof?

this is trying to solve 18.a fore reference

[prisma garnet]

I'm especially concerned about how I defined A_i, it feels hard to read

[red yoke]

Is q: X × I → SX?

Why are the preimages open

[prisma garnet]

oh, yes, sry

actually, I have one improvement one sec

[prisma garnet]

does it make sense now?

[red yoke]

Why does it deformation retract to S¹

[prisma garnet]

the preimage of $A_i$ is ${1/i} \times I \cup X\times ([0, .25) \cup (.75, 1])$ ofc

[gentle osprey]

DarQ

[prisma garnet]

oh wait I made a mistake

[red yoke]

Why is it open

[prisma garnet]

[red yoke]

Particularly at (0, 1/2)

[prisma garnet]

huh

wait

[broken nacelle]

Grr, that point at 0 is annoying

[red yoke]

You can adapt SvK to work

And take X × [0, 2/3) and X × (1/3, 1]

[red yoke]

Or even this works actually

[prisma garnet]

huh

but the intersection wouldn't be path connected

besides, we want the fundamental group of each A_i to be \bZ

[red yoke]

Which is where adaptation comes in

[red yoke]

Or you can do this and adapt SvK to work without openness

(To work specifically for this construction)

Hm maybe you can also map the CW complex with 2 vertices and countable edges to SX

And show it induces an isomorphism

[prisma garnet]

how could I adapt SvK?

[prisma garnet]

I still got no clue how to solve this

[prisma garnet]

@wispy veldt did you solve this one?

[wispy veldt]

not yet

[red yoke]

Any loop can be broken into a finite union of arcs with endpoints in the intersection

[red yoke]

Then you can use this to identify with paths in this complex

Or directly get group structure from combi

I guess the hardest part is just showing existence and non-existence of path-homotopies

[prisma garnet]

but like, why?

[red yoke]

To this or other methods

[red yoke]

For existence you can slide the arc towards 0 or 1 while fixing endpoints

(I don't think I went into this much detail when doing Hatcher's exercises, I was just like "oh this looks good enough so it's probably true")

[prisma garnet]

[prisma garnet]

I shouldn't have asked for feedback on my proof here

[red yoke]

Actually I think I have a better method

[red yoke]

You can do this to turn any loop into a union of loops with a common basepoint at 0

Then each piece will be determined (up to homotopy) by which two values of X it travelled along

Non-existence of path-homotopies can be shown by considering the collection of winding numbers around the points (1/(n+0.5), 1/2)

[supple cloak]

Has anyone here done knot theory?

is there anyone I can ask for some help regarding finding oriented knots as a sum of prime knots

[knotty vine]

[supple cloak]

trying to do this question here, and I have got this step

nvm Ive done it

[heady skiff]

i'm confused

who's DarQ

[broken nacelle]

that's a weird question

it's me obviously!

[heady skiff]

@prisma garnet how do you feel about this

you have an impostor

[alpine nest]

For twelve years you have been asking: Who is DarQ?

[fair idol]

If I have a manifold how do I characterize the sub manifolds without boundary?

So for instance the open disk doesn't have a boundary and it's a sub manifold of R2

[red yoke]

All open subsets are submanifolds of the same dimension

[unreal stratus]

Pretty sure this is bad in general, since e.g. every smooth m-manifold embeds in R^2m, which embeds in any (non-empty) 2m-manifold

I guess it depends what you mean by "characterise", sorry, lol, i guess i was thinking more if you wanted to classify all possible submanifolds or smth

[fair idol]

Well what I'm really interested in is why d^2=0 is a natural condition for a chain complex. My understanding is the historical context is that sub manifolds without boundary that aren't boundaries of other submanifolds detect holes. Manifolds without boundaries are called closed submanifolds right? Is the definition of a closed submanifold just something without a boundary?

[languid patrol]

It’s because a boundary doesn’t have a boundary (think about it!)

d = 0 is saying that you are a cycle (boundaryless) d^2 = 0 means all boundaries are cycles

[tulip bluff]

How can I show that the map r : R^2 \ {0} -> S^1 is a fibration? I though about using the fact that r is a homotopy equivalence with homotopy inverse the inclusion, but I can't find any properties relating homotopy equivalences and fibrations.

[unreal stratus]

I would use the fact that it is a fibre bundle over a compact space

Those are always Hurewicz fibrations

[lime sable]

i think "closed submanifold" here means "closed embedded submanifold" aka a closed subset which inherits a manifold-with-boundary structure

not "compact and without boundary"

[tawdry valve]

the (deg 0) derivations on C^infty(M) are exactly vector fields. Is there a nice characterizations of deg -1 on Omega(M)? I know that interior multiplication by a vector field is a derivation, but are there more?

[hexed steppe]

what is degree

[tawdry valve]

$D(\Omega^p(M)) \subseteq \Omega^{p-1}(M)$

[hexed steppe]

what

[tawdry valve]

I mean like how interior multiplication i_X(omega), for omega a p-form, is a (p-1)-form

[hexed steppe]

what is degree

of a derivation

[tawdry valve]

okay let me restart

[hexed steppe]

a definition would suffice

[tawdry valve]

yup one sec my b

[merry geode]

Maybe they mean derivations which are like, sending Omega^n to Omega^(n+d)? Where what they want is d = -1 case

[hexed steppe]

oh

[tawdry valve]

a degree $p$ derivation on the graded $\mathbb R$-algebra $\Omega(M)=\bigoplus_{k=0}^\infty \Omega^k(M)$ is a $\mathbb R$-linear map $d: \Omega(M)\to \Omega(M)$ such that $d(\Omega^k(M))\subseteq \Omega^{k+p}(M)$ and $d(ab) = d(a)\wedge b + (-1)^{p \cdot \mathrm{deg}(a)} a\wedge d(b)$

[hexed steppe]

like degree of a map on the graded

yeah

ok

[merry geode]

(Yea, derivation as in maps btwn dgca with certain grade change)

[gentle osprey]

Joseph

[hexed steppe]

do you suspect vector fields being deg 0 derivations on C^infty is relevant?

or was it just a comment

[tawdry valve]

I'm not really sure, mostly just a comment

[hexed steppe]

because interior prod is like dual

like

on 1 forms

i_X is identified with X

[tawdry valve]

right, i_X(df) is the same as X*f

[hexed steppe]

well

more than that

for any 1form w, i_X(w) = w(X)

like a vector space V embeds canonically in its double dual V**

it is the same type of thing here

X \mapsto i_X

is an injection

and i think conversely

if a derivation D of deg -1 behaves this way on Omega^1, then it should be i_X for some X

[tawdry valve]

right so maybe our goal is to first recover a vector field X from looking at the action of D on Omega^1

and maybe define it pointwise using that sort of duality

and making sure it assembles smoothly using bumps

or something

[hexed steppe]

hmm

actually intuitively

should all such D be given by i_X for some X?

i guess there are some infinite dimensional spaces here

so maybe not clear that this is invertible

[tawdry valve]

yeah, but maybe we can use bumps/local frames to reduce to pointwise stuff

say x is in M, and we want to find X(p) given the action of i_X on Omega^1

choose a coordinate chart (x1,...,xn) at p

then we can write X = X1 d/dx1 + ... + Xn d/dxn locally for some functions X1,...,Xn: M -> R

so to recover X(p), we need to recover X1,...,Xn at p

i_X(dxi) at p = dxi(X) at p = dxi(X1(p) d/dx1 + ... + Xn(p) d/dxn)

which is just Xi(p)

so X(p) = i_X(dx1) d/dx1 + ... + i_X(dxn) d/dxn

so maybe given deg -1 derivation D, try to define a vector field X at p by X(p) = D(dx1) d/dx1 + ... + D(dxn) d/dxn

and prove that this doesn't depend on charts?

unless there's a more intrinsic way of doing this

oops maybe i should've put this one channel below

[red yoke]

The interior product iX is the unique degree -1 graded derivation such that iX(ω) = ωX

So ig you can just take the dual

Are degree 0 derivations exactly vector fields

[tawdry valve]

yeah actually I'm not sure exactly if on the entire de Rham algebra if degree 0 derivations have to be vector fields? I know a degree 0 derivation restricted to just 0-forms (i.e. C^infty(M)) are just vector fields (more or less by definition, depending on the author)

yeah maybe not? could you just define a derivation which is 0 on all p-forms for p>1, and then a vector field for p=0?

[red yoke]

What if you take a vector field X and do D(ω^k) = 2^k LX(ω)

[tawdry valve]

right I think that also works

or does it lol idk

I think it meets the axiom

cause every time you do Leibniz don't you get a 2 factor outside?

[red yoke]

Nvm it fails

[wind bronze]

Is there any reason for which a topology on a set cannot have the difference of any two members ?

[steel glen]

what do you mean

did you consider the powerset by chance

[wind bronze]

The topology on a set is a subset of its power set such that the set itself, the empty set, the intersection and the union of any two members belong in it.

The question is, why can't we have the difference too ? There is no intuitive reason for this.

[steel glen]

we can though

that’s my point

[wind bronze]

Why isn't it required to have the difference too ?

If we can always tell if a point is in a neighbourhood, why shouldn't we always be able to tell if it isn't in a neighbourhood too ?

Because the union basically tells that we can tell whether a point is in any of these two neighbourhoods.

The intersection basically tells us that we can tell whether a point is in both of these two neighbourhoods.

[steel glen]

by difference

[wind bronze]

And the difference would tell us that we can tell whether a point is NOT in a neighbourhood. Why isn't this a requirement ?

[steel glen]

are you talking about symmetric?

or literally A \ B

[wind bronze]

A \ B

Or, at least the complement set (that is, the set the toplogy is defined on, minus the neighbourhood).

[steel glen]

well consider (0,1) \ (.5, infty)

why should that be open

what makes you say that should be open

[wind bronze]

It wouldn't be open. But why do we define open this way ?

[steel glen]

i know you know it’s not open

im more asking

why do you think it should be

i think it’s a fair question to ask “why” instead of “why not”

[wind bronze]

The closed intervals of R satisfy all the properties of a topology, thus, the closed intervals are open too so-to-speak. The reason we don't have the difference is that it would allow, in both cases (either if we take open sets as being literally open intervals, or closed intervals), for intervals which are open on a side, and closed on another side. But what is wrong with this ?

[wind bronze]

Because it is a neighbourhood in R as well.

Isn't this what the elements of the topology are meant to represent ?

[steel glen]

they are meant to represent open neighborhoods

your intuition for open may vary

i don’t know exactly why we don’t require difference, but it clearly already creates problems

i always understood topologies to generalize metric spaces, and having one of the most canonical examples for a metric space break seems discouraging

intuitively,

if you take x = 0, and then consider (-1,1) \ (0,1)

it feels as if you are removing the neighborhood of 0

[wind bronze]

But how does it break the metric space ? As I said, closed intervals form a valid topology too, so in this regard they are open too. What happens if we allow the difference too is that we get intervals which are closed on a side and open on another side. But I don't see any problem with this, because we can already reason both about closed and open intervals as being open sets. What is the problem with something being both of these things ?

[wind bronze]

I see. I guess I'll think about it more.

[steel glen]

if you required differences you would quickly get the discrete topology, which is too weak

we’re not trying to generalize to an arbitrary topology

the canonical example is R with the metric topology

[wind bronze]

In a lot of cases, yes, but not always.
For instance, take a set P, and define a topology T which contains P, the empty set, and any subset of P. If we allow differences, it won't be the discrete topology, but rather we get the awesome property that, for any element x, we can tell both what neighbourhoods it is in, as well as what neighbourhoods it is NOT in. It is a "complete" topology, so-to-speak.

WAIT, I THINK I FOUND A REASON WHY (nevermind*)

[steel glen]

a good way to look at it is this maybe:
if you require set differences, then the space (R, d) would simply not generate a topology. so you immediately lose the thing you were trying to generalize

[steel glen]

this example doesn’t make sense to me

did you mean any subset of P

[wind bronze]

Yes, my bad. Sorry for the typo.

[steel glen]

if so, i don’t see how that isn’t the discrete top

it is by definition t = P(X)

[wind bronze]

Take, for example the set P = {0, 1, 2, 3, 4, 5}, and the toplogy starting from {P, empty set, {2, 3}}.
After we complete all unions, intersections and differences, we get {P, empty set, {2, 3}, {0, 1, 4, 5}}.

This is not a discrete topology.

Wait, I made a mistake.

Done, it has even less neighbourhoods than I thought.

[steel glen]

that’s still a topology in the usual sense

i don’t think you’ll get illustrative examples in the finite sets

if you go about this in the “closeness” route, then differences would break things by “taking away” elements close to x

much like in the example with (-1,1) \ (0,1)

finite intersections and arbitrary unions still keep this sense of “closeness”

[wind bronze]

I see, I see. I'll think about it more.

[lime sable]

they do not, an infinite union of closed sets can be not closed

[wind bronze]

😮

I did not think of this.

And the reason we care specifically about open intervals is that they are a neighbourhood of every point they have, meanwhile closed ones are not a neighbourhood of their extremes (if we assume the topology induced by the metric space).

IT MAKES SENSE NOW!

Thank you both @steel glen and @lime sable!

Now, I have just one more question - in the definition of a manifold, why does each point need to have at least one neighbourhood homeomorphic to an open interval of R^n, and not EVERY neighbourhood ? I don't see any reason for the quantifier being existential rather than universal.

[lime sable]

well closed sets are precisely those sets that contain their extremes, if by "extremes" you mean "limit points"

[wind bronze]

Yup.

[lime sable]

the entire manifold is a neighborhood of any point

[wind bronze]

Because having an existential quantifier literally means that there are neighbourhoods of a point which do NOT behave like R, which I don't think is exactly what we want from manifolds.

[steel glen]

because you again lose the usual examples

[lime sable]

yeah, they are locally like R^n, but perhaps not globally, which allows interesting stuff to happen

[steel glen]

for example, R would no longer be a manifold

[steel glen]

this is the intuition why

think of requiring S^1 to be homeomorphic to R

[wind bronze]

Ugh, what is S ?

[steel glen]

S^1 is the circle

[wind bronze]

Oh, okay.

[steel glen]

but yeah, the circle is different topologically from R, or any interval. but locally, if you “zoom in” enough, it looks like R

that’s the global v.s. local stuff

[wind bronze]

I see, so if we required every neighbourhood to be homeomorphic to R, it would mean that the manifold would be globally homeomorphic to R* too, meanwhile local-ness is one of the main motivations of the manifold.

Thank you both once again!

[steel glen]

yes exactly

if you require the manifold to be homeomorphic to R, you’re just working in R lol

[wind bronze]

Yup. It makes sense.

[steel glen]

yeye just thought that remark was funny haha

[wind bronze]

So, basically, each point in a manifold has a chart.

[scenic pecan]

Is the author talking about the subspace topology in S1? I am not sure how the image of [0,1/2) is not open in S1?

[red yoke]

It's not open at a(0) = 1

It's an arc that's open at one endpoint and closed at the other

Yea he means S¹ with subspace topology from R²

[scenic pecan]

But how to prove this rigorously like the map of [0,1/2) maps to a subset of R^2 so do I have to show that there is no basis element contained inside some point of this subset?

[red yoke]

There is no basis element of S¹ that contains 1 and is contained in the image

[scenic pecan]

Ok Thanks

[scenic pecan]

let U be a saturated open set in $R^{n+1} \setminus {0}$ $x \in$ $q(U)$ Note that for each x we can construct an open ball "B" containing $q^{-1}(x)$ $\in$ $R^{n+1} \setminus {0}$ contained in U as U is open. since $S^n$ is the subspace of $R^{n+1} \setminus {0}$ implies that $B \cap S^{n}$ is an open ball containing x and contained in $S^{n}$ which implies $q(U)$ is open. Is my proof correct? Please correct me if you feel it is not rigorous enough.

[gentle osprey]

NotAntiMatter

[tiny obsidian]

U isn't open though

[scenic pecan]

Sorry. edited it

[tiny obsidian]

there also isn't a single ball containing the fiber of x, because the fiber goes off to infinity and no ball can

[scenic pecan]

Okay. So can we take fibres that intersect S^n? I am not sure how to prove this.

[tiny obsidian]

I think if you show there is exactly one point on a ray also in S^n, then U n S^n will do

[tulip bluff]

In the definition $$(f+g)(t_{1},t_{2},\ldots ,t_{n})={\begin{cases}f(2t_{1},t_{2},\ldots ,t_{n})&t_{1}\in \left[0,{\tfrac {1}{2}}\right]\g(2t_{1}-1,t_{2},\ldots ,t_{n})&t_{1}\in \left[{\tfrac {1}{2}},1\right]\end{cases}}$$ of composition for higher homotopy groups why is the $2t_1$ and $2t_1 - 1$ in the $t_1$ coordinate? I'm trying to compare this with the binary operation in $\pi_1$ which is given by concatenating loops, but this seems wildly different.

[gentle osprey]

rrrembrandt

[unreal stratus]

This is actually the same formula

for loops, we usually define $(\alpha \star b)(t) = \begin{cases} \alpha(2t) & t \le 1/2 \ \beta(2t-1) & t \ge 1/2 \end{cases}$

[gentle osprey]

homotopy coherent potato

[tulip bluff]

Is there a geometric way to picture for this?

In the case with n > 1

[unreal stratus]

Yes, it is uhhh

1 sec

Well no point drawing it ig, but for n = 1 this corresponds to shrinking two copies of the interval and placing them beside one another

(so one is now traversed twice as fast, then the other twice as fast after that)

For n=2 this means you put two squares side to side, with one next to the other along the first coordinate

Same for n > 2 as well

So like [] [] lol

Of course, you could also stack them like
[]
[]

which would use a different formula, but it turns out that the way you put the squares/cubes doesn't matter at all for n > 1 :)

(Basically, you can always "shrink" the squares/cubes and then move them around however you like to get the same map up to homotopy)

[brave geode]

i love eckmann-hilton

[obtuse meteor]

for @tulip bluff this is a general argument and a good thing to look up. The wikipedia page is pretty decent.

it comes up a lot more often than one might think

Challenge: Once you understand this proof, show that any topological group has abelian pi1 using the same technique

[umbral panther]

Challenge: prove the etale fundamental group of a group scheme need not be abelian. What went wrong?

[gaunt linden]

(responding to Faye) Hmm, I can come up with a proof that sort of looks somewhat like the diagrams in Wikipedia's 2D argument for commutativity if I squint a lot. Does "using the same technique" run deeper than handwaving similarity?

[knotty vine]

The first connection is that pi_(n+1) is the same as pi_n of the loop space. Loop spaces are topological groups via loop concatenation

[gaunt linden]

Hmm, I'll confess I'm not entirely clear on what "the loop space" is in this context. If it's a group via concatenation, wouldn't it need to be "up to homotopy" and thus discrete?

[umbral panther]

Loop spaces are up to homotopy, but that’s sufficient for Eckman-Hilton
Moore made them strictly associative by keeping track of the length of the loops.
Kan made them groups, but that’s not necessary for most purposes

[knotty vine]

Oh wait youre right, I guess I shouldve said h-space

[gaunt linden]

I'm more confused about how a space whose points are counted up to homotopy can itself have an interesting pi_n.

But perhaps I should first have asked: "Loop space" sounds like pi_1, but if that were true it would probably not have a name of its own, so what is it actually?

[knotty vine]

Sorry, the loop space isn't loops up to homotopy, it's all loops

Then its set of path components is pi_1

[gaunt linden]

Okay, so by keeping track of lengths and acknowledging a trivial loop of length 0, I can see we get a monoid, and thus Eckmann-Hilton is within reach.

[unreal stratus]

hm tbf what i said wasn't quite eckmann-hilton

i was more thinking in terms of operads i.e. space of embeddings of 2 cubes inside R^n is path-connected

[unreal stratus]

There is a formal eckmann-hilton argument, not sure if you've seen it

(As in, if you have a monoid with two compatible operations and same units then the operations coincide and are commutative)

Then for a topological group you can put two operations on their pi1 and hence ...

Oh it seems people were doing the homotopy commutativity on the level of spaces rather than pi1 lol

But then yeah sure it still works

[gaunt linden]

Ah, I think I see it -- an element of pi_2 is (a homotopy class of) loops in the identity component of the loop space, so everything in pi_1 outside the identity is irrelevant!

[unreal stratus]

I'm not too sure what you mean here when you talk about stuff outside the identity

[knotty vine]

Outside of the path component containing the base point in the loop space

[unreal stratus]

Sure but I didn't see why that was relevant to anything

[gaunt linden]

Hmm, do you mean something like composing two loops in one order or the opposite? And then somehow in the case of a topological group there's an argument that the compatibility assumption of E-H is satisfied?

[unreal stratus]

No, the point is that can compose the loops as normal (in one way) and then for a topological group there is another way to "multiply" loops

and then you can just check they are compatible in the E-H sense

[knotty vine]

(I think Troposphere was just clearing something up for themselves)

[unreal stratus]

Oh okay sure

I guess the nice way to think about it imo is that we have this suspension-loops adjunction

[gaunt linden]

Yeah, I'm not convinced Semer's point about higher homotopy groups is directly relevant to Faye's challenge, but it looked interesting all the same.

[unreal stratus]

But then you need to check compatibility with group structures

[unreal stratus]

Oh I mean it kind of it if you want to do stuff on the level of spaces

I went straight to pi_1 so you get monoid (well, in fact group) structures on the nose

[gaunt linden]

Just to be explicit, I suppose your "another way" is, given loops a, b, consider the loop f(t) = a(t)b(t)?

[unreal stratus]

Yes.

This is hopefully quite nice and clear for, say, paths on the circle

You either go round the circle n times then m times, or you sort of go round at a faster pace the whole time by multiplying paths

[knotty vine]

I guess also any topological group is an h-space, so if you can do it for them...

loop spaces are h-spaces

h-spaces are topological magma up to homotopy

magmas? magmai?

idk greek

[unreal stratus]

i think magmas but yeah lol

[gaunt linden]

Right. This seems connected to what I cooked up myself, namely given a and b, consider the map from [0,1]² given by f(t,u) = a(t)b(u).
Then the ordinary composition goes along the bottom then right edges of the square, and the other possibility goes diagonally through the square.
But following the left then top edges gives the ordinary composition of a and b in the opposite order, so this square directly says that pi_1 is abelian without needing to do E-H.

(It confused me for some time that I could also have taken f(t,u) = b(u)a(t) and get a different homotopy between a+b and b+a, and I was trying to somehow interpolate between them for no good reason).

But okay, I see the E-H argument now, thanks!

[knotty vine]

What is this diagonal in the case of a loop space?

I guess this part

But in Tropospheres proof we seemingly don't need the intermediate equalities

oh I guess we use that a(0)b(u) = b(u) and a(t)b(1) = a(t)

[gaunt linden]

I think now that my proof is not what Faye had in mind.
(And yeah, we do need at least a topological monoid unital magma structure on the original space -- for either proof).

[knotty vine]

No I think your proof is pretty much E-H

except specialized ofc

[gaunt linden]

Hmm, are we in agreement about which proof "my proof" is? I'm thinking about the one that's based on "bottom-then-right and left-then-top are homotopic in [0,1]²".
The E-H based proof Potato hinted at doesn't need any reasoning about the topology of [0,1]², and instead just uses the purely algebraic fact that (a+b)*(c+d) and (a*c)+(b*d) are identical curves, not merely homotopic (where + is compress-and-concatenate and * is take-the-diagonal). The only topological facts it needs is, first, that * produces a curve at all, second, that + has an identity once we identify homotopic loops.

Oh, and third: that * (which was defined to work on curves) lifts to a well-defined binary operation on homotopy classes. There may be more footwork swept under the carpet there than I thought at first.

[obtuse meteor]

sorry I don't know what this proof is

[gaunt linden]

What I called "my proof" there was this one (but with no particular focus on the diagonal).

[obtuse meteor]

This seems EH’y just more direct

[knotty vine]

Okay heres maybe an idea.
Let me first try to set (sketch) the stage. E-H says a one-object monoidal category is the same as a commutative monoid. Hence, a one-object monoidal groupoid is an abelian group. Another consequence: for the unit I of a monoidal category, End(I) is a commutative monoid and Aut(I) is an abelian group.
The homotopy hypothesis says that spaces are infty-groupoids. What's an infty version of E-H? I.e. if a monoidal infty-groupoid is a one-object (infty,1)-category, then what is a one-object monoidal infty-groupoid?
Troposphere's argument seems to hinge on the (obvious, i.e. easier than E-H) fact that Paths( (0,0), (1,1) ) in [0,1]^2 is contractible. Can we make an infty version of this fact that helps us prove E-H using Troposphere's argument?

[umbral panther]

It’s not true in the infinite context. You can have n commuting multiplications. The space of based maps from S^n to X has that. It’s only when you take components that n=2 implies commutative

[radiant junco]

Hi all, could somebody explain me the proof that R with the k topology is strictly finer than R with the usual topology?

To me both are equivalent I cannot find a set that belongs to Rk that doesn't belong to Ru

I'm using B= {(a,b), a<b, a,b€R} U {(a,b) - {1/n}n€N*, a<b, a,b€R} as a base for the k-topology in R

[plain raven]

What is N*?

@radiant junco

Ok, I understand. The set (-1, 1) \ { 1/n | n \in N, n > 0 } is open in the K topology (obviously) but not open in the classical topology, because it contains zero but does not contain all points in a ball of radius epsilon around zero for any epsilon strictly greater than zero.

[radiant junco]

@plain raven tyy

[hidden crag]

Does someone know where i can find the computation of the integral homology of a K(Z/4Z,1)? Is this done via model or is there a nicer way? Hatcher states this without proof in his document on spectral sequences

it's Z/4Z for odd n and 0 for even n

apparently this is the group homology of Z/4Z but this isn't very helpful to me

it doesn't seem like the specseq computation via path space fibrations works here

[dry jolt]

This is the integral group homology of G = Z/4Z so you could take a free resolution of Z over ZG, tensor with Z, and take homology that way. A nice topological way to do this is to put a CW structure on S1 that Z/4Z acts freely on. Then the cellular chain complex consists of free ZG modules whose homology is isomorphic to homology of S1. In particular, you can patch this resolution to itself to get rid of H_1, and do this repeatedly to get a free ZG resolution of H_0 = Z. This yields a periodic resolution of period 2, hence the periodicity in the homology of Z/4Z

I can describe this in a bit more detail a little later, but hopefully that makes some sense. This is also described in Brown's book on Cohomology of Groups in 1.2 and 1.6 I think

[hidden crag]

Thank you Walter

I'll come back to this another time i think

[dry jolt]

You might also be able to apply Leray-Serre to 0 -> Z/2 -> Z/4 -> Z/2 -> 0

[hidden crag]

does this coincide with the specseq associated to K(Z/2,1)->K(Z/4,1)->K(Z/2,1) after realizing Z/4->Z/2 with a cont. map and replacing that by a fibration

because that's the example i'm reading rn where the fact is used

ok yeah i didn't know the serre specseq is sometimes also called leray serre

[dry jolt]

Yeah, it should be

Well it's funny because I've also seen it called Lyndon-Hochschild-Serre specifically in the context of group cohomology

But yeah, I'm not sure how that computation will actually work out but I think it's worth trying

[white oxide]

Given two bases B0 and B1 so that B1 has less cardinality, is there always a way to construct a subset of B0 which is still a basis of cardinality at most B1.

[tiny obsidian]

I think so?

I don't see why the proof for second countable spaces (every basis has countable subset that is also a basis) doesn't generalise

at least for infinite cardinals obvs

[lunar yoke]

Its essentially spelt out here, using the serre spectral sequence, but its pretty easy so try it yourself maybe https://people.math.wisc.edu/~lmaxim/753f13w7.pdf

[hidden crag]

Right, l(\infty,4) is a model for K(Z/4,1)

Thank you Phil

[gentle girder]

yeah this should be true, if you have two bases $B,C$ such that $B$ has cardinality $\kappa$, then for each pair $\alpha, \beta$ where it's possible, you choose $C_{\alpha, \beta}$ such that $B_{\alpha} \subseteq C_{\alpha,\beta} \subseteq B_{\beta}$, This is a family of cardinality at most $\kappa \times \kappa = \kappa$, and forms a basis (it's the same as the proof for second-countable spaces, as @tiny obsidian said) https://math.stackexchange.com/questions/65425/bases-having-countable-subfamilies-which-are-bases-in-second-countable-space seems like the first answer on this page generalizes, since it doesn't use any assumption about the cardinality of the basis

[gentle osprey]

smay

[gentle girder]

oh texit allows you to delete the source

neat

[bronze wind]

Hello ! I am trying to compute the homology groups of the Klein bottle using the Mayer-Vietoris exact sequence. My covering is the following : U is a small inner disk, V is the whole klein bottle minus a small closed disk in U. I have that U is contractible, U cap V is homotopic to S^1 and V is homotopic to the wedge of two circles. Using Mayer-Vietoris, the only bit of the sequence that is not zero is (I am using reduced homology groups) between H2(K) and and H1(K). I am struggling to calculate the induced map in homology (that would help me to to calculate H1 and H2). I tried to compute the induced map of the inclusion of U cap V in V by taking a generator of H(U cap V) (the class of a circle) and mapping it to the boundary of the square. Howhever I don't know how to continue from here

This leads me to a more general question : what are the methods to calculate the induced map of the canonical inclusions when using Mayer-Vietoris. I know that this really depends on the space you are studying, but maybe you guys have some general advice

[tribal palm]

i just formally withdrew from my topology exam, i haven't at all found the time to work through the curriculum

now i have until probably sometime in the middle of january

[uneven bronze]

I have a basic question. Suppose there are two open sets A and B, and B contains A. Must the closure of A also be contained in B?

[gentle girder]

no, A and B can be the same open set for example, and B would contain A

uh if you want a more satisfying example, you can take A to be (0,1) and B to be (0,1]

then B contains A and also an extra point, but still the closure of A contains 0

you can generalize this, take A the open unit ball, and then take B to be A with the upper hemisphere attached

[outer harness]

B is not open though

[gentle girder]

oh open

lol

just attach a neighborhood of the upper hemisphere

[outer harness]

Oh true

[gentle girder]

so A = (0,1) and B = (0,2)

sorry yeah I missed "open"

[uneven bronze]

ok, thank you. Just to be sure. If we suppose A and B would be two, different open balls, then my statement would be true, right?

[gentle girder]

the thing that you should be visualizing for this @uneven bronze is that you have A some open set, then you can imagine the "boundary" of A (which is a notion that does make sense in topology in general), then you can take B to be A unioned with some open neighborhood of a subset of the boundary, then B would contain A but not the closure of A.

[gentle girder]

that depends on the open balls that you choose and doesn't work in general

[uneven bronze]

ok

[uneven bronze]

this helped, thanks

[uneven bronze]

I'm reading through the following proof, and I have a basic question about it.

Why is f continuous? This seems to be the characteristic function for B, and it is only continuous if B is both open and closed, which we...don't know, or?

The theorem I'm referring to is this:

[coarse night]

this certainly does it

[gentle girder]

if A, B separate X, X\B = A, so B is closed

[coarse night]

but you don't really need a "proposition" to see that is true

[uneven bronze]

ok, now I see, got it, grazie!

[tidal lynx]

you can also directly see that the f^-1 of {0}, {1}, and {0, 1} are A, B, and X, all of which are open

you don’t even need to check the last one actually, cuz inverse of unions is union of inverses

[whole matrix]

for 2.3 i got that none of those collections is a topology, is that right?

[tribal palm]

no

[whole matrix]

oh :/

i think i found my mistake

[tribal palm]

that’s why we consider the complement instead

[whole matrix]

the set is of the elemtents U not X\U

[tribal palm]

[whole matrix]

so if S_1 and S_2 are in T_1, X\S_i has to be finite, which means S_i is infinite, but since S_1 ი S_2 can be finite, X(S_1 ი S_2) would be infinite so it wouldnt be in the topology

@tribal palm is that logic making sense?

i don’t really like straight up asking for the answer, but which ones are topologies (no proof needed i want to work on that myself)

[whole matrix]

nvm i understand why the collection is a topology now

[whole matrix]

this just clicked

[next crystal]

this isnt right, lets say X\S_1 and X\S_2 are finite. Then we do indeed have S_1, S_2 being infinite, but we actually have the stronger statement that S_1, S_2 include all but finitely many elements of X. So lets say S_1 = X \ A and S_2 = X \ B where A, B are finite. Then S_1 \cap S_2 = X \ (A \cup B), which includes all but finitely many elements of X, so X \ (S_1 \cap S_2) is finite.

[whole matrix]

i see, that makes a lot of sense, thanks

that stronger statement was what i was missing

[next crystal]

this also kind of hints at how to prove it, specifically the formula S_1 \cap S_2 = X \ (A \cup B), which is demorgans law

[whole matrix]

yea, thats what i had it when the answer clicked

i was already using that when i did the question wrong for the first time, i had misunderstood which sets were in the collections

[blissful yoke]

i'm trying to understand this covering map, but i don't really get why f~ and g~ start at the origin

like shouldn't going around A or B correspond to going around some A_i and B_i? but f~ and g~ dont go around any of those circles

if i'm understanding it correctly, each of the tangency points (n x 0) and (0 x m) are mapped to x_0, and B_i is mapped to B, A_i is mapped to A, but then what are the actual x-axis and y-axis mapped to?

unless its each interval [n, n+1] x 0 being mapped to B, but then the circles in the covering space don't seem to be doing anything lol

[tulip bluff]

Is there some relationship between mapping cylinders and cofibrations?

[umbral panther]

The mapping cylinder of a map X->Y is a factorization if the map X->C->Y such that the first map is a cofibration and the second is an equivalence. And it is a natural factorization

[tulip bluff]

By factorization you mean what here specifically?

[umbral panther]

I mean that the composite of the two maps is the original map

[hidden crag]

Replacing maps by fibrations/cofibrations is a neat concept

[tulip bluff]

Thanks this was helpful 🙂

[edgy tree]

Hi, have the following assertion:

If A and B are subsets of (E,d) a metric space then diameter( A U B ) <= diameter (A) + diameter (B)

I feel this assertion is false but i can't find a counter-example. if someone can help me, it would be nice

[limber wren]

think about two sets with small diameter that are far apart from one another

[edgy tree]

My idea (didn't find anything else):
Let (E,d) = (R, | - | ),
A = ]0,1[, his diameter(A) = 1
B = ]2,3[, his diameter(B) =1
A U B = ]0.1[ U ]2,3[, his diameter(A U B) = 3 ? (im not sure here)

[limber wren]

sure, that seems good

[edgy tree]

i had a doubt about the diamater of A U B. looks like i was correct.

ty!

[limber wren]

no worries, you had the right idea

[gentle girder]

the x-axis is mapped to A, and the y-axis is mapped to B (in the "usual" covering space sense of R covering S^1), Notice that each time we map something on that axis back to x_0, we have to attach another circle corresponding to the "other loop" (since x_0 has an evenly-covered (or elementary idk what munkres uses) neighborhood)