point-set-topology

[marsh forge]

pairs (a,t)

so im mostly askin that t is here

[pearl holly]

yeah okay to here t = 0

[marsh forge]

great okay so my final question

does f_1(a)=f_2(a,0)?

[pearl holly]

yes

[marsh forge]

So can we build f:XuCA->X?

[pearl holly]

yeeee

[marsh forge]

Okay now final question before you go to bed

is X->XuCA->X a retract?

(remember the first map is just the inclusion)

[pearl holly]

ye it is by construction

[marsh forge]

amazing

now think deeply about this construction

and reversing it

is basically the exact same concept

For after you get it, Exercise: ||Prove that a map f:X-> Y is nullhomotopic if and only if there is a map g: CX->Y such that the composite X->CX->Y is equal to f (where the map X-CX is the inclusion at time 0)||

[pearl holly]

yeah okay I will try that one out tomorrow. Thank you so much for the help! Tbh I was kind of lost at the very start but then I started to get the idea, thank you!

[marsh forge]

I think that a big problem in topology is finding maps with certain properties and the solution is almost always to like, partially define the map in some easy way and bootstrap up to make the full map you need

And this gets more complicated the more topology you study

So my advice is whenever you feel totally stuck try to write down what you can guess and what the problem is giving you and see if together that can get you to where you need to be

[fickle marsh]

Is (1,1) a valid basis of the standard Topology and is it the empty set or {1}

[ivory dragon]

the standard topology on what? ℝ? if so, certainly not

[fickle marsh]

Yeah on R

[ivory dragon]

also i'd interpret that as the empty set but like

i wouldnt use that notation in general

{} or {1} are both more clear for whatever you want

[fickle marsh]

Basically I'm trying to determine if there are finite elements

And that's the only thing I could think that would be finite

[ivory dragon]

there are no finite open sets in ℝ except the empty set

(under the standard topology)

there are various ways to prove this

one way: prove that singletons are closed, and that closed sets in ℝ are not open unless they are {} or ℝ itself

[that second statement is equivalent to ℝ being connected, but you probably havent learned that yet]

[fickle marsh]

Nope

[ivory dragon]

a more direct way: let U in ℝ be a nonempty open set, so it has an element u. Then there is an interval (u - r, u + r) that is a subset of U for r > 0, since such sets collectively form a basis of the standard topology. but that would imply U has infinitely many elements.

intervals of the form (u, u) are not part of the standard basis for the standard topology

only (u - r, u + r) for some r > 0

as i said above, thats how id interpret it

but i would never use this notation in practice

i suppose if you interpret it like that, the empty set could be a part of the basis of the topology (though it need not be)

but in any case, its... well.. empty

so it doesnt contain u

hence doesnt present a problem to my above argument

i dont see what youre adding.

[marsh forge]

Namington let me explain

Typically

There are no numbers between 1 and 1

[ivory dragon]

thanks for contributing max.

the world is enriched by your presence.

[marsh forge]

Tru

[fickle marsh]

Well I've seen problems with calling the closed set infinite or finite

Most my professors leave it as ambiguous

[ivory dragon]

closed sets can be finite or infinite.

{1} is closed, as is [1, 2]

[plain raven]

apparently there are some physicists whose convention is that the word "finite" is not used to describe 0

[fickle marsh]

Sorry I meant empty set@ivory dragon

ex. my analysis professor considered N to be {1, ....} and thus there was no natural number that represented the carnality of the empty set => empty set it not finite

and in terms of if it was then infinite he said that there were a lot of complexities and issues with assigning finite or infinite to the empty set

he ended up saying that (1,1) wasn't valid

[plain raven]

it's fine to disqualify that notation, it's confusing

but you could also just define (a, b) : { x | a < x & x < b }

and then (1,1) would denote the empty set

[marsh forge]

I guess when you asked if there were finite open sets I assumed you meant non empty

I think it is reasonable to call the empty set finite

[gusty flame]

Does anyone know where I could find a proof for this claim?

[plain raven]

i think this is a theorem of topological dimension theory

maybe try Nagata's Modern Dimension Theory

or Connections, Curvature and Cohomology by Greub, Halperin and Vanstone

not a very useful reference, sorry, these are big books

[tough imp]

Nagata wrote a book on topological stuff?

[tight agate]

yeah it's called counterexamples in topology

[tough imp]

-_-

[eternal nimbus]

"closed sets of R by the usual topology are NOT the segments [a,b]" WAIT WHAT

[gritty widget]

not all closed sets are closed intervals

[eternal nimbus]

OHHHH in that sense

okey okey okey

[ivory dragon]

intervals [a, b] are in fact closed sets

but there are more closed sets than that

for example, (-infty, a] U [b, infty)

[eternal nimbus]

Yup yup, or simply the union of two segments that don't overlap right?

yeah yeah, i interpret it as [a,b] are not closed and i was like "whuuuut"

[ivory dragon]

well its fine if they overlap, itll just give you a single segment

but yeah

[eternal nimbus]

(yeah was thinking a closed thing that is not segment)

[ivory dragon]

finite unions of closed sets are closed

in general

but yeah i understand what youre saying

you just read it backwards

[raw sedge]

is every closed set of a Hausdorff space the preimage of 0 under some continuous map into R?

[true robin]

This property implies the space is regular, so no

[empty grove]

No, R_K is a hausdorff space in which K is closed but is not the preimage of 0 under any map

[raw sedge]

Do you have a sufficient condition?

okay, so it's equivalent to saying the space is perfectly normal?

Is there anything particularly special about R here? What other spaces could we replace it with?

[true robin]

R appears in the def of perfectly normal, I assume that is what makes it special in this case

[gentle osprey]

Nobody

[plain raven]

i think it's a diferent nagata. the commutative algebra guy is Masayoshi Nagata, the author of this book is Jun-Iti Nagata

[fair idol]

In a general topological space how would one show that a set is open?

[gentle osprey]

Nobody

[marsh forge]

(This strategy works particularly well when you know a basis for the topology)

[tough imp]

This makes a lot more sense

[hollow harbor]

Good morning!

[pulsar lynx]

Trying to understand the comb space as an example of a connected but not path connected space

Does A ⊂ B ⊂ C where A,C are connected imply B is connected? I don’t understand this chain of logic

Intuitively I understand it like “any open ball around (0,1) has to contain another point in D, so separating into two open sets is impossible”

[unreal stratus]

Not in general, no

[pulsar lynx]

Ok

[ivory dragon]

[1, 2] subset [1, 2] U [3, 4] subset R

[pulsar lynx]

Ah yeah that makes sense

But still, why is that screenshot valid

[ivory dragon]

because its specifically the closure of E

what would it mean if D could be disconnected?

(if you want a formal proof of this fact, https://proofwiki.org/wiki/Set_between_Connected_Set_and_Closure_is_Connected)

[pulsar lynx]

It wouldn’t be a closure, right

Ok yeah that makes sense, thank you

[ivory dragon]

the formal proof is a bit finnickier than you might expect

but the fact should be intuitive enough

(unfortunately thats a theme for a lot of point-set topology...)

(thankfully because this is a "familiar" topology you dont need the proof of the statement in full generality)

[pulsar lynx]

Ah

Yeah I looked at it like … what

[ivory dragon]

well proofwiki has a habit of constructing weird maps without explaining what its doing

lmao

if you unpack what that map is actually trying to communicate, it should match your visualization

[unreal stratus]

(Sometimes proofwiki seems to write in more detail than most humans reasonably would imo too or is that just me?)

[ivory dragon]

alternatively you could just remark that its obvious since you know how ℝ² works

whatever you prefer ¯_(ツ)_/¯

[unreal stratus]

Proofs almost always seem much longer and often more tedious than textbooks I mean

[ivory dragon]

absolutely yeah

it likes to cite hella prereqs too

its a... weird organization

were i to make a collection of proofs, i would not structure it like proofwiki

but eh, can still be handy

[unreal stratus]

Ye I agree

[ivory dragon]

the nice part is that proofwiki's proofs are completely unambiguous

you very rarely have to fill in details

its just... you might have to hop back 3 steps of hyperlinks to figure out how they proved some random fact theyre citing that any reasonable human would bypass with an "obviously..."

or perhaps a "because ___"

actually thats another style thing with proofwiki: they rarely say "because _ is _" or whatever

its always

"By definition of bleurgh:

and therefore:

and so from [proposition]:"

which makes it look (and read) a lot more intimidating than it ought to be lmao

[unreal stratus]

yeahh

[crimson path]

Let (X,tau) be a topological space.
I want to show from first principles that a subset A is in tau (namely, open)
iff each point x in A is inside a open set U contained in A.

I have managed to show this, but I had to go via A = Int(A) iff A is open.
Can one circumvent this sidestep and directly link the iff-statement I want to show?
Notice I am not referring to any basis or such, rather working directly from and only with
the definition of a topological space.

@gritty widget yes I feel so too but it's not clicking for me. The converse direction
is straightforward I feel, since the topology is closed under arbitrary unions.

But the other direction. Let A be open, that is, in tau. Now let x be in A.
"open" now just means "a member of tau". I don't see how to proceed from
here to exhibit a set O in tau that x lies in, and also such that O is a subset of A.

I am probably missing something here but I can't spot it

Oh lord I am dumb, thank you

[lean marten]

Hey looking for some help on a problem from "Topology and Groupoids"

I'm most of the way there but having some trouble constructing a pushout in the form:

[\begin{tikzcd}
{\pi (X,(B_1\cup B_2))} & {\pi(X,B_1)} \
{\pi(X,B_2)} & {\pi(X,B_1\cap B_2)}
\arrow["{s_1}", from=1-1, to=1-2]
\arrow["{s_2}"', from=1-1, to=2-1]
\arrow[from=2-1, to=2-2]
\arrow[from=1-2, to=2-2]
\end{tikzcd}]

[gentle osprey]

Oatman

[lean marten]

Where $s_1$ and $s_2$ are retractions given by 6.7.4

[gentle osprey]

Oatman

[lean marten]

Actually they're deformation retractions

[tough imp]

I won’t lie, Bourbaki is honestly like this lol. When I have a specific result I need a proof of Bourbaki has consistently been the easiest proof to read as long as you can find it lol

This is in contrast to Stacks Project where it still has this reference web but all the proofs still require a ton of filling in

[ivory dragon]

well you see

you dont read stacks projects for the proofs

you read it so you have something to cite and can sidestep the proof

[tough imp]

I do

[ivory dragon]

if there is an error in the stacks project's proof network, no one will notice for 7 decades

and then itll suddenly break 5000 papers

[tough imp]

Maybe you but

I instead spend 193949293 hours

Poring over some dumb detail

And there’s still 1 proof I gave up on after like actually 2 days

Torsors am I right

[coral pawn]

Can someone please help me answer this question from my homework?

How would one answer this?

What exactly is it even asking?

Polar coordinates don't really work if we include 0, so I'm guessing it wants us to consider R^+ x (-\pi,\pi)

But then the inverse map of the chart would require us to use atan2 function

And I don't think that's a commonly used function

Also, does this chart give the same smooth structure as the cartesian coordinates?

My hws due tonight and this is the only problem I need to do

Ping me if you have any ideas

[trail tiger]

It's asking if chart transition function is smooth between the canonical polar and Cartesian charts.

That's what it means for a chart to be smooth.

[wet flame]

Quick topology question bc I have done it in a while lol

Just about topology of R^n

So for a subset E of R^n, if E is compact and E consists of only isolated points, then I am trying to prove E must be finite. So since E is compact, it is closed and bounded. I think I can approach this by contradiction. So assuming E is infinite, so it has infinitely many isolated points, clearly no open cover will have a finite sub cover

But like

How do I formalize that

[rancid umbra]

@wet flame if E consists of infinitely many isolated points, then for each x in E, there is an open neighborhood U_x of x so that U_x cap E is {x}. the collection of all such U_x's gives you an open cover of E with no finite sub-cover

[wet flame]

Brilliant

Thanks

Wdym by cap E btw?

Set of isolated points of E?

[rancid umbra]

E is your set

[reef shore]

cap means intersection

[rancid umbra]

i made a slight typo, U_x cap E is U_x intersected with E. And since x is an isolated point, it should mean that U_x intersect E is just the singleton set {x}

[wet flame]

Oh that’s what you mean by it lmao

Got it

[rancid umbra]

i dont think you need contradiction either, but it works

just take the same open cover of U_x's from above

it has a finite sub-cover by compactness

each U_x contains only one point from E (namely x)

E is finite

[wet flame]

Ohhhh

[wet flame]

This is beautiful omg

The simplest things amaze me

[rancid umbra]

lol same haha

and thanks

[plain raven]

this trick is useful very often

good one to keep in your pocket

[rancid umbra]

which trick

[plain raven]

it can be hard to figure out a good open cover for a problem sometimes, it seems so silly to take an open neighborhood around every single point (isn't that way more than you need?) and then apply compactness

so you might miss it at first

[rancid umbra]

yea good trick

[plain raven]

this is only tangentially related but like

if X is a space

then for any open cover of X

we can associate to it the Cech cochain complex of the cover, with coefficients in some Abelian group.

and there's a category of open covers and refinement maps

now, this category is not filtered, so colimits are poorly behaved.

you wouldn't want to take the colimit of all cech cochain complexes over all open covers here

because the answer would be nonsense

but if you pass to cohomology first, then the colimit kind of becomes filtered

this is a terrible explanation and probably doesn't make sense

but

where i'm going with this is

if instead of considering all open covers of X

you consider just those open covers of X indexed by the points of the space

with x\in U_x for each x

then you do get a filtered poset

and you can take colimits at the level of cochain complexes and get a reasonable answer

which is nice

this trick blew my mind when i first saw it lol

[rancid umbra]

diligentClerk about to send me down a rabbit hole of wikipedeia articles at three in the morning with all this vocab

[plain raven]

i'm sorry haha

[rancid umbra]

lmao its cool. wish i could appreciate it, i just have no idea what most of those terms mean

[plain raven]

suffice it to say that there's a common form of argument where you are dealing with open covers in a space, and it's always ok to replace the open cover you're working with with a refinement, so you'll say something like "we need property P to hold for the cover, and U doesn't necessarily have that property, but that's fine, we can replace it by a sufficiently fine refinement V such that P holds"

often this is phrased in a way where it's understood that U, V are arbitrary covers, but it turns out that a lot of these arguments still work when you let U, V be open covers indexed by the points of the space with x in U_x for each x in X, and they're much better behaved in some technical ways (although perhaps not as computationally friendly because then the cover is necessarily infinite if X has infinitely many points)

[rancid umbra]

i see. i didnt realize that covers came up that much later on. only places ive really seen them be used is for compactness and total boundedness and some uniform continuity

[plain raven]

yeah I think i first started to understand the importance of covers when i saw this picture

give me a sec

idk it was a blog post on shape theory

but like

take a circle or something

and cover it by like, short segments of arc

say, connected open sets

and cover it in an efficient "minimal" way so that no point lies in more than two open sets

now you can associate to this open cover a combinatorial data structure

this is going to be a simplicial complex

it has one vertex for every set in the cover

and it has one edge between i and j iff U_i \cap U_j is nonempty

if you draw a picture of this simplicial complex you get exactly like, a regular n-gon, where n is the number of open sets in the cover to start with

so from this nice cover of the space you can extract a simplicial complex homeomorphic to the original space

and you can imagine doing this with a sphere or something as well

[rancid umbra]

whomst in the fvck was big brained enough

[plain raven]

or just like, an arbitrary nice topological space

idk. cech? maybe mardesik?

say like

a torus or another compact manifold

you should often be able to, from a nice enough open cover, extract a simplicial complex which is a triangulation of that space, or at least homotopy equivalent or something

https://upload.wikimedia.org/wikipedia/commons/2/21/Cech-example.png

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

oh that's interesting

so in this example the cover is somewhat "bad"

and there's an example of a point with a triple intersection

but you can see that the new thing is still homotopy equivalent to a circle

just not homeomorphic

the triple intersection means you add a 2-simplex connecting the points i,j,k

does that make any sense lol

[rancid umbra]

yes the picture and extracting the simplex made sense

same with the example of the triple intersection, at least intuitively. two lines kind of got identified as one

[plain raven]

shape theory apparently was like, an independent branch of topology that kind of existed independently of algebraic topology for a while. it was combinatorial but not algebraic if that makes sense

in retrospect that distinction doesn't seem well founded because you can associate interesting algebraic invariants to the cech complex

but

at the time they were considered two different branches of topology

and like, two different proofs of the same theorem by two different methods (shape theoretic vs algebraic) would be published in separate journals

https://wildtopology.com/category/shape-theory/

aha this was the bllog post

https://wildtopology.com/2012/10/30/the-cech-expansion-i-nerves-of-open-covers/

[rancid umbra]

wait i feel like im failing to see something important here. so, different open covers might give you different simplexes, right?

wait, but big picture right, so if you take nice enough open covers, you get back something thats possibly easier to work with and still preserves some of the important properties of the original space, right

[reef shore]

In a compact, totally disconnected, separable metric space with cardinality less than that of the continuum, given an uncountable closed set, can it always be partitioned into 2 uncountable closed sets?
Context is these Stone spaces, and proving what I said will be enough. Any other suggestions would be welcome

I think whatever I said is all we know about the space S_n(T) topologically, but if it helps it is the space of complete types of over a theory T over a countable language (I don't think that would matter though, looking at the language used here. I have seen a proof of this before which doesn't use metrisability but uses the elements of S_n(T) itself, and I want to see how to get a proof from metrisability)

[plain raven]

yeah this is right, different open covers means different simplicial complexes, taking refinements of the open covers gives a more detailed, more accurate representation of the original space, like increasing the number of triangles in a rendering.

[rancid umbra]

cool thnx

[shy moss]

Hi

Let X a space. How to prove that if every map f:S^1->X extends to a map f':D^2->X then the fundamental group of X is 0?

Using functoriality of fundamental group i found that $\pi_1(f)=\pi_1(f')=0$

[gentle osprey]

Or x1

[gritty widget]

there is a very nice picture you can draw for this

i think this might help you, to show that the fundamental group is trivial its enough to show that any map is homotopic to a point

D^2 is contractible

can you put together

[crimson path]

Hello.
I have made an attempt to exhibit a topological space (X,tau) with a
subset A such that lp(A) = {all limit points of A} is not closed.
Is it correct?

Take X = {a,b,c} and tau = {null, X, {a}}
Consider the subset A = {b}.
c is the only point in X such that each neighbourhood of c
intersects A. Hence lp(A) = {c}.
But the complement of {c} is {a,b}. And this is not in the topology, that is,
it is not open. So lp(A) = {c} is not closed.

I have a slight suspicion it is spurious, since I can't readily
spot an error. Does anyone see any?

[reef shore]

Yeah seems legit as long as your definition of limit points doesn't make all points of A limit points of A

[crimson path]

@reef shore The definition I used is that p is a limit point of A iff
if U is an open set containing p then U \ {p} must intersect A non-trivially.

"all points of A limit points of A", hmm well here A = {b}
and I verified b cannot be a limit point of A. This is because
X is the only open set containing b. And X \ {b} = {a,c} intersects
A trivially, so b is not a limit point.

Where could it go wrong, do you mean?

[reef shore]

Yeah so sometimes limit points are defined with just U instead of U \ {p}, which would make this counterexample not work

Here it is fine, I was just making sure

[crimson path]

Oh ok interesting, I am working with the "punctured" definition, that is, we remove the limit point when checking intersection.

@reef shore big thanks for the help!

[pearl holly]

Wait, does $X \times S^n$ always retract to $X \times {x_0}$?

[gentle osprey]

Tokidoki ✓

[pearl holly]

ye it does, never mind

[pearl holly]

I mean just a retraction

ye I forgot about that lmao

[reef shore]

empty spaces

sorrt

space*

[pearl holly]

sorrt space

[reef shore]

isn't that when the first oog homology is 0

[pearl holly]

wtf is oog homology?

[marsh forge]

empty set isnt a space because real spaces have basepoints

[reef shore]

Then I shall define a pointed empty set

[shy moss]

what is the difference between deformation retract and strong deformation retract?

[marsh forge]

like 3 hatcher exercises

[reef shore]

The homotopy to identity in a strong deformation retract must fix the subspace being retracted onto throughout the homotopy

[marsh forge]

i actually do not know any important results that depend on the distinction

[reef shore]

I only googled this because someone asked me a related question when I was TAing their topology course

[marsh forge]

yeah i mean

there are literally a handful of hatcher exercises where it matters

but idk if it comes up at all after

since like

strong deformation retract is kind of a weird notion for a homotopy theorist to care about

maybe its nice to know when homotopies are homotopies of pairs

that could come in handy

but i think CW replacement makes this unnecessary for practical purposes

id have to think a little more

[shy moss]

in spaces of homomorfism

[marsh forge]

huh

wdym by this

[shy moss]

i see it in a talk

[marsh forge]

ah no, you can't make it fix A by replacement

orx1 i really have no idea what you mean by "spaces of homomorphism"

[shy moss]

for example Hom(Z^n,G) with G a topological group

[quasi forum]

So I'm having a bit of a hard time being able to concisely describe these topologies. I am ending up with a notational mess and it's making it harder for me to understand them.

Can someone help me boil this down a bit.

[marsh forge]

Okay but how is this related?

[marsh forge]

Can you start by giving me a basis for each

[pearl holly]

I'm sorry to disturb here but Max, I have to pick a subject to write about. I want to write about algtop but I don't know what I want to write about in that field. Any suggestions on cool stuff that I can write about?

[marsh forge]

Hm what have you learned that you like most

[quasi forum]

Sure. The first basis is open boxes and boxes with one of the boundaries included.

[marsh forge]

There are a bunch of things I could suggest depending on how much time you have and what you like

[quasi forum]

So something like Ax[0,b) or (a,1]xB for open intervals A and B, and real numbers a and b

[gritty widget]

do max's research topic

[pearl holly]

tbh I kind of want to explore something new in AT that I haven't learned yet and write about that. Like cohomology and the rings sound really cool but I don't know if I will be able to write a project on that topic lmao

[marsh forge]

I would be happy to help you learn about stable homotopy theory

[pearl holly]

tterra I'm too noob wtf

[marsh forge]

You could write about equivariant topology

Maybe that’s too much idk

[quasi forum]

Should I wait?

[reef shore]

Thread perms when

[pearl holly]

yeah I will take a note of that. I will go now lmao, sorry for the disruption

[marsh forge]

I’m waiting for all three dackid but the first looks good

Except you of course also have just AxB

[quasi forum]

Okay cool. Wanted to make sure we were okay

[shy moss]

let me search the paper

[quasi forum]

Yea, that was the first part (open boxes)

[marsh forge]

Gotcha

@pearl holly we can talk in ur special server

[quasi forum]

Okay, the dictionary order is tough to explain, but I'll try. but if you have two points (a,b) and (c,d) with $a<c$, then $a_b<c_d$. And ((a,b),(c,d)) is an open set.

[gentle osprey]

dackid

[quasi forum]

In contrast to the product topology, we have open intervals of the form [(0,0),(c,d)) and ((a,b),(1,1)].

If the notation is confusing, there is ordered pairs, like (0,0) and the outside paranthesis refers to the interval

[quasi forum]

There is also the situation in which a=c, so we need to check if b<d. If so a_b<c_d

[quasi forum]

Finally, the last one is defined as this on the reals intersected with the set [0,1]

Okay, I think I'm done

[solar shore]

Is it true that in a Hausdorff space X, for A⊆X and x∈X a limit point of A, there is an infinite sequence of points of A \ {x} converging to x?

@gritty widget In your example, x is not a limit point of A because the neighbourhood {x} of x doesn't intersect A \ {x}.

[marsh forge]

Sorry was watching a YouTube video

[empty grove]

You can take X = omega_1 + 1, A = omega_1 ⊂ X, x = omega_1

[marsh forge]

This is not quite accurate you’re doing everything squared

My advice is to first look at the first and third examples and write down the basis sets explicitly

[solar shore]

@empty grove My definition of a limit point: x is a limit point of A iff every neighbourhood of x intersects A \ {x}

[marsh forge]

Hint: || they are the same ||

[empty grove]

Ye, then my example works

Assuming you're familiar with ordinals

I'm putting order topology on X

[marsh forge]

I think hausdorff second countable is the assumption you want

[quasi forum]

I'm not sure what you mean

[solar shore]

I am familiar enough with ordinals. Give me a minute, I'll try to understand your counterexample.

[quasi forum]

About first and third examples

[marsh forge]

You did not work very hard on the third examples basis

You can write down a basis for all of R^2 and use this to write down a basis for I^2

[marsh forge]

The intuition is that if you space has too many open sets you might need more than a countable amount of points to properly converge

[solar shore]

@marsh forge @empty grove what if I allow not only infinite sequences, but transfinite sequences?

[marsh forge]

You get to the theory of nets very quickly in which case I think stuff works out how you want

But I don’t think about spaces that aren’t second countable often

[quasi forum]

Yea, it's pretty much open intervals intersected with I.

something like ((0,-1),(0,2)) is an open interval, and that intersected with I would be [0,1]x{0}

[marsh forge]

No

It is open sets of R^2 intersected with I^2

[quasi forum]

Yea I know

[marsh forge]

[0,1]x0 is not open

[quasi forum]

That is the intersection of the open interval I mentioned on I^2

[marsh forge]

Oh I completely misread

I’m very sorry

I assumed they meant the usual subspace top

[quasi forum]

Oh no. It wants dictionary order

[solar shore]

@empty grove Indeed that works, thanks. So I guess in topology sequences are not good objects for defining stuff. Idk if transfinite sequences are though.

[marsh forge]

Well

Most decent spaces

Are second countable

In which cases sequences are fine

[empty grove]

They're not, I remember a counterexample but I can't remember its name

[solar shore]

So naturally I wonder if I can redefine a limit point in terms of transfinite sequences and their limits 🙂

[empty grove]

Looking it up lol

[marsh forge]

Wait maybe in weak* stuff this doesn’t suffice

[empty grove]

Ah Aren's space. I think it works, will have to check

[quasi forum]

So now back to the OG question. Is there a better way to express these?
My explanation for dictionary order was not great and I'd really like to know how to improve it

[marsh forge]

No your explanation of the dictionary order was fine. My guess would be that the second and third spaces are homeo and have the same basis

Maybe try to prove that and see if something goes wrong

[quasi forum]

Actually, the page right before this gives an example in the subspace that is not in the I^2 dictionary topology

The set was {1/2}x(1/2,1]

This is because the only time we will have a half closed interval is when we include the least and largest elements. In this case (0,0) and (1,1) respectively.

So that is not included in the dictionary topology for IxI

[marsh forge]

Oh then that is a pretty good answer I think for those two

I don’t think this problem is very good

It’s worth thinking about a little

[quasi forum]

Eh, compare in these classes tends to refer to try and see if there is some sort of inclusion

[marsh forge]

Were you given it as an assignment

[quasi forum]

We know those two are different, but is one finer than the other? Or are they completely separate topologies?

[quasi forum]

Yessir. Last question for it

[marsh forge]

bad instructor

Okay

I think one of those two should be finer

The only reasonable inclusion is the identity

Maybe you could scale everything

And get an inclusion

But again this question is too open ended

[quasi forum]

Well, with the tools we have now, compare only really means is it coarser or finer. We don't have many tools in our kit right now

[marsh forge]

Yeah then one of the latter two is probably finer, whichever one you just showed can’t be coarser

And then the first should be incomparable to the others

[quasi forum]

Actually pretty sure Product top equal to relative top

[marsh forge]

Oh maybe

Sorry I’m not thinking very hard about this but it also feels like you know what you are doing

No

[empty grove]

@solar shore ok I'm pretty sure the Aren's space works. The reason is that (infty, infty) doesn't have a nested neighborhood basis (I hope) and that should suffice. For a net of a certain order type to converge to a point, the net of open neighborhoods of that point should have a cofinal subset with the same order type and that is a neighborhood basis of that order type. So to converge to a point, a net must mimic the order type of some neighborhood basis of that point, and to say that transfinite sequences (or even totally ordered nets) suffice would be to say that all spaces have nested neighborhood bases. I'm actually not sure about a couple details here, but this is what my intuition says. Of course that doesn't go very far in point set topology

[marsh forge]

There is an open set between (x,y) and (x,y’) where y’<y and everything is inside the interior

This is not open in the product topology

@quasi forum

Although you should be correct that these two are not incomparable

[empty grove]

hmm what I said about some subset of neighbourhood basis having the same order type as the net is not true, the converse is

[quasi forum]

Ah true, because {x} is not an open set

But in contrast, that set is indeed in the dictionary topology on RxR

Okay, so I believe the inclusion is I^2 dictionary topology is included in product and relative top and product is included in relative top

[empty grove]

I didn't follow the discussion so far but that doesn't sound right. The I² dictionary top should contain open sets which are not there in the product topology

In particular the open vertical lines

[marsh forge]

I was starving for that whole convo

So my brain was just

Not working

[solar shore]

@quasi forum you might be interested to know that there exists a solution book for Munkres' first few chapters

@empty grove Do you mean this https://en.wikipedia.org/wiki/Arens–Fort_space space? And is your (infty, infty) the same as (0, 0) in the wikipedia article?

[solar shore]

@empty grove

I think I have figured it out.

1. Suppose we are given an infinite sequence s of points that converges to this special point (∞, ∞).
2. Then s has a point in an infinite number of columns, for otherwise there would be a neighbourhood of (∞, ∞) not intersecting the columns s is contained in, and no tail of s would lie in that neighbourhood.
3. Let t be an infinite subsequence of s such that t has no more than one point in each column.
4. Clearly, "all points except those in t" is a neighbourhood of (∞, ∞). There is no tail of t that lies entirely in this neighbourhood. Therefore, there is no tail of s that lies entirely in this neighbourhood either. Hence, s doesn't converge to (∞, ∞).

[quasi forum]

Got my answer. Prod top and dict top are subsets of the relative topology inherited by the dictionary top on R^2, but they are NOT subsets of each other in any way

[quasi forum]

I do not. I'd rather figure it out on my own, even if that means losing a couple points on my homework.

[frigid patrol]

Why is it true that Symm^g(C) surjects onto Jac(C)?

[frigid patrol]

Why is this as variety and can we give algebraic equations for it?

[empty grove]

Ah yeah the one I had seen before in some blog post was slightly different but captures the idea yeah (there it was a counterexample to "sequential closure is sequentially closed" but needs slight modification for that)

[empty grove]

Only detail that is missed I think is that you need t to be a cofinal subsequence. Infinite subsequence need not be cofinal of your sequences are transfinite

[tough imp]

Wtf is Symm^g

[gritty widget]

Can any one help me with this, I have two modules A and B I want to know that A and B have the same free parts

A and B are cohomology groups, the map between then is the map induced by the inclusion of B into A

I know that the support of the kernel and cokernel of this map is contained in some set K

[gritty widget]

Moreover I know that the kernel and cokernel of the map are torsion

Is this enough

[fair idol]

Does anybody have some tips on how to demonstrate two different sets of topological bases generate the same hotel sigma algebra? For instance the closed sets and the open rays

I'm just wondering on general tips

[tight agate]

@gritty widget what ring are the modules over

you said free part, so I assume they're over a PID?

or dedekind

[gritty widget]

Modules over C[x1,..,xn] , Specifically A and B are equivariant cohomology groups

And the map between A and B is the map induced by an inclusion

[tight agate]

so youre saying you have some map B ---> A such that the kernel and coker are C[x1,..,xn]-torsion?

[gritty widget]

Yes

[tight agate]

just tensor with the fraction field?

[gritty widget]

Sorry this is very rude of me I have to run off and help a friend

I’ll reply later

[gritty widget]

ahh

and that’s is free

and tebsoring with free things is exact

[tight agate]

localization is exact, yes

but you dont really need that

tensoring still gives you a complex

ker --> B ---> A --> coker

tensor that with the fraction field and ker and coker become 0

it still doesnt make sense to talk about a "free part" without more information

[gritty widget]

But is the tensor if B with the fraction field not just exactly the free part of B

[tight agate]

free part doesnt make sense unless it's a direct summand of B

tensoring with the fraction field kills torsion

but this doesnt imply that it is free

in your particular case if you somehow know that B/B_{tors} is projective, then it works

but projective implies free for modules over C[x_1, ..., x_n] is a very deep theorem

it's called the Quillen-Suslin theorem

[gritty widget]

Hmm okay

It’s this highlighted line I’m trying to make sense of

[tight agate]

okay yeah if you only care about rank

the thing above works

[gritty widget]

From the line before having support in that union implies that the module is torison

Is the rank not just the number of generators of the free part?

Ooh no

[tight agate]

what do you mean by free part

[gritty widget]

It’s the number of free generators

[gritty widget]

Okay I think this is where I was going wrong

I thought we could decompose our module into a free part and a torsion part

[tight agate]

yeah that doesnt work in general

you always have an s.e.s 0 --> B_{tors} --> B ---> B/B_{tors} --->0

but it doesnt always split

[gritty widget]

And that the free part was just generated by the free generators and the torsion part was generated by the torsion

But yeah I see now that you can have some mixing

Thanks very much

That cleared up a lot

[tight agate]

[gritty widget]

And just to double check if we have something like

C[x,y,z,w]/(y^2,z,w)

y, z and w all have torsion

x is the only free generator

So the rank of this thing should be 1?

[tight agate]

rank over what

[gritty widget]

Okay I need to go read the definition of rank

[tight agate]

the definition of rank is simple

if R is a domain

and M is an R-module

then the rank of M is the dimension of M (x) F over F

where F is the fraction field of R

[gritty widget]

So then above since A and B are isomorphic after tensoring with F they must have same rank

Thank you very much king

[tight agate]

yup

[gritty widget]

Excellent

[frigid patrol]

C^g modded out my action of Symmetric group on coordimates

[gritty widget]

What does it mean to say that a map is fibering?

[empty grove]

Do you mean fibration

[meager python]

Stupid question, does anyone have an example of an etale morphism with infinite fibres?

[tight agate]

take an infinite disjoint union of spec(k) and map to spec(k)

[meager python]

I thought of that but couldnt see why it is etale, and I still can’t

I see that it’s relative dimension zero so that checks at least

[tight agate]

O rip I see the problem

yeah idk

[meager python]

My guess is that etale morphisms must have finite fibres but dunno how to see that

[winged badger]

Yes, it is etale, at least according to Vakil's first defenition.

The map Spec(k)->Spec(k) is etale, and that's enough because being etale is a local condition on both the source and the target

[meager python]

I’m a bit confused, is the original map quasi finite?

Or how is k -> \prod k defined

[plucky verge]

<@&286206848099549185>

if B is the basis for a topology

is B in T ?

does this even make sense as a question ?

[meager python]

@winged badger I don’t think this example works

[plucky verge]

@gritty widget thanks

[meager python]

How do you define Spec(\prod k) -> Spec k?

[winged badger]

No, it's not Spec(\prod k)-> Spec(k). Let X=union of countably many Spec(k). It's not equal to Spec(\prod k). The map is from X to Spec(k)

It's not true that the disjoint union of Spec(A_i) is Spec(\prod A_i) when the union is infinite. Only for finite unions is this guaranteed to hold

[meager python]

you're right it's not affine

gimme one sec I need to pay the bills :U

[pearl holly]

Okay so Hatcher writes this: "Another sort of process we have encountered is the transformation of one functor into another, for example: Boundary maps H_n(X, A) -> H_n-1(A) in singular homology, or indeed in any homology theory". I don't really understand this. How is this a transformation of one functor into another?

[novel acorn]

I assume it's in the vain of the n-th homology group and the n-1th are different functors because one looks at C_n and the other one at C_n-1

[empty grove]

Transformation of the functor (X,A) → H_n(X,A) into the functor (X,A) → H_n-1(A)

He should define what he means by this precisely later

[pearl holly]

yeah okay I see

[meager python]

@winged badger you're right this works, I was being confused by some formulations of etale morphism requiring quasi finite but missed the local part of quasi finite

[empty grove]

Of course I've only told you how the functor acts on the objects here, but it should be easy to see how to turn this into a functor by defining a map on the morphisms

The functors are from Top-2 to Ab, Ab being the category of abelian groups and Top-2 being the category of space-subspace pairs with maps of pairs as morphisms

[pearl holly]

ah yeah okay I see

Hatcher also defines a functor and then says "strictly speaking, what we have just defined is a covariant functor". So wtf is the difference between a functor and a covariant functor?

[empty grove]

A functor can also mean contravariant functor, but usually you'd say if that is the case

Contravariant functors are arrow reversing and covariant are the ones he defined. If you just see "functor" written anywhere, it means they are talking about contravariant functors

[pearl holly]

yeah okay, got it! Thank you so much!

[solar shore]

I am not sure what cofinal means. But maybe the observation that we only have countably many points in the space will give you what you think I missed?

[empty grove]

Here cofinal = unbounded

Yeah it might, I'll have to think

Don't see it directly at least

[marsh forge]

Virgin covariant functor Chad contravariant functor chadder presheaf chaddest copresheaf

[solar shore]

Well, since we have countably many points, if a transfinite sequence converges to some point, then there's a (the usual kind, countably infinite) subsequence of it converging to that point

[empty grove]

:catThink:

[empty grove]

How would you argue for this

Like it might be immediate lol I'm probably just missing something

[marsh forge]

You should be able to use transfinite induction

To construct such a subsequence

I think

Actually not even

You can do it more simply

Oh no nevermind my idea requires a second countable assumption

I think you can still do it transfinite my

Transfinitely*

[solar shore]

actually now I am thinking about this and i'm unsure so let me think

[empty grove]

I feel like ordinals with cofinality ≠ omega might mess things up in some way but not sure

[marsh forge]

So I mean

You only need continuum many points to pick one for each possible open set

Then I think you can refine that

Set theory isn’t my strong suit I might be messing something up

[empty grove]

Hmm so you're saying take the original sequence and you can assume that it is at most continuum size?

[marsh forge]

Yes I believe so

Actually

Eh

You probably need some directed system of neighborhoods

To make this choice

Which is i think what nets are

Yeah actually how do you even phrase the notion of convergence without a net converging to you point for a transfinite sequence

You can define limit points

But a directed limit point seems harder

[solar shore]

For every point x of the space except for maybe the limit of our transfinite sequence, there exists an ordinal α s.t. x doesn't occur in our transfinite sequence after α (because the space is Hausdorff and hence t1 and hence everything-except-x is an open set).

Now what do I do next ...

[empty grove]

So if you have cofinality = omega, you can take a cofinal (usual) subsequence and that also converges to the point. Now if cofinality > omega, then you can do the following: suppose no point in the space occurs cofinally in s. Then since you have countably many points, the union of the indices corresponding to each point is a countable union of non cofinal subsets and because cofinality is uncountable this can't be the size of domain of s, which is a contradiction. Hence there's a point which appears cofinally in s, so you can just take t to be the constant sequence at this point and this should converge to the same point

Cofinality < omega case is trivial

[empty grove]

You can still say that for every neighborhood there is some index N such that for every n > N...

[solar shore]

Yeah actually how do you even phrase the notion of convergence without a net converging to you point for a transfinite sequence

I have no idea what a net is. So...

We say a point L is a limit of a transfinite sequence of points s iff for every neighbourhood of L, there exists an ordinal α s.t. all points of s after α lie in that neighbourhood.

[empty grove]

Which is the same definition as in the case of nets actually

I used cofinality above, it just means " size of smallest cofinal subset"

Cofinality < omega implies cofinality = 1, ie the ordinal is a successor ordinal so any notion of convergence is trivial

Or cofinality = 0 I guess

Which means ordinal is 0

There might be a less set theory brained way of doing this but I can't think of anything else

[solar shore]

Continuing, how about this?

So, from our transfinite subsequence, take the last occurence of each point and throw away everything else. Done, we've got either a finite or a countably infinite convergent subsequence.

[empty grove]

throw away everything else
By everything else do you mean everything upto this last occurrence of x?

Ah I see

Only keep the last occurrences

[solar shore]

We take our transfinite sequence. Throw away all its elements except those that are the last occurences of their respective points.

[empty grove]

Right that makes sense

[plain raven]

is there really such a thing as "set-theory brain" beyond the set-theory brain that we all have because this is the dominant working language for mathematics

[empty grove]

Perhaps not

[novel acorn]

👉 👈 I need a bit of help with this one

Tbh I'm not sure how to even show what is created when I identify the edges

[novel acorn]

So if I mold it I imagine I'll get a 2-simplex with 3 edges connecting to a point in it's interior

Two triangles with an edge connecting the diagonal?

Oh yep I forgot about the second one

I feel like I'll get this

Oh wait no

I'm stupid

wrong one

Gimme a sec I'll draw it

Now I just have to somehow turn this into the sheet for the Klein bottle

Maybe I could cut it along the diagonal

Oh yeah I seee it

You have the [v1,v2] and you cut along that

Hm I thought of something like

rotating it by 60° but they I'll have to identify [v0,v1] with [v1,v2] instead of [v1,v3]

Hmmmmmmmmm

AAA I keep getting a torus

Cut along the [v1,v2] edge and then invert the lower segment so that the edges of the same color are parallel

Like "flip it" would be a better term although that doesn't seem allowed

Oh yeah I see where I fucked up
I basically managed to replace v1 and v3 when I did that

OK OK

OOp

I had to

do another thing

sorry

So this is where I got to

Now

if I were to flip the lower part about the red axis
I'd just get the torus
However if I put it back together that way the vertices get bungled around

so there's probably some other way to rotate this that I can't see cuz it's 12AM

OK so
Flip in the way I'm thinking i.e. the same as rotating 180° around an edge or some other kind of flip

OK yeah

it was easy

I was just being stupid

OK now sleep and then I'll start computing the homology groups of this thing tomorrow

thanks a lot btw

[obtuse meteor]

@sleek thicket snake snake snake

[sleek thicket]

🐍

Snake lemma best lemma

[obtuse meteor]

we also proved that extraordinary homologies play well with suspensions on our homework from today

[sleek thicket]

Nice!

[obtuse meteor]

Ellenberg-Steenrod OP

[sleek thicket]

Is this advanced AT?

Or whatever it's called?

[obtuse meteor]

yeah it's like a second course

[sleek thicket]

nice

[obtuse meteor]

there's 3 courses on Algtop in the department

592 (qual course), 695 (Algtop 1), 696 (Algtop 2)

I took 592 last semester and am taking 695 now

[sleek thicket]

Pog

[obtuse meteor]

the lecturer's notes are awful

[sleek thicket]

I'm thinking I might review complex analysis today

I want to take complex geometry in winter

[obtuse meteor]

http://www.math.lsa.umich.edu/~ikriz/2021695/20216950908.pdf

I am readable

[sleek thicket]

But I took one quarter of complex analysis like 2 years ago

[sleek thicket]

Oh dear

Well it looks pretty

[obtuse meteor]

what is this word

[sleek thicket]

Lmfao

[obtuse meteor]

please

[sleek thicket]

hougued?

bougued?

[obtuse meteor]

I didn't go to class today bc my allergies are acting up so I got a covid test just in case

it's in the context of a wedge sum

[sleek thicket]

....based??

No way

[sleek thicket]

No there's t oo many letters

What the fuck

[obtuse meteor]

involves the bouged?

bouged?

[sleek thicket]

I recognize cohomology

[obtuse meteor]

bouged???

[sleek thicket]

Bouquet!!!

That's it

[obtuse meteor]

bouquet!!!

lel

[sleek thicket]

Lmfao

I am trying "what was the bus drivers name" Twitter rn: https://twitter.com/grassmannian/status/1435739700623396865?s=19

Brendan Murphy (@grassmannian)

Let k be a field and R a countable product of copies of k. What's the global dimension of k?

Twitter · Sep 8, 2021, 10:59 PM

the global dim of R = 2 iff CH holds but the global dim of k is just 0

[obtuse meteor]

lel

[gritty widget]

have you tried abandoning a physical form?

that way, it cannot physically hurt.

idk try harder

[sleek thicket]

Same

Also bored

Good new profile pic tTerra

[gritty widget]

thanks

[shy moss]

if G is a reductive algebraic group and K a maximal compact subgroup of G, then Hom(Z^n,K) is a strong deformation retract of Hom(Z^n,G)

[marsh forge]

Oh okay I was looking for theorems that use this in the hypothesis

That is a cool result tho

[shy moss]

there are also more results that relate strong deformation and spaces of homomorfism

[shy moss]

what are lagrange manifolds?

[gritty widget]

lagrangian submanifolds?

[native raptor]

hey so if i'm trying to show that certain projective spaces don't admit fields of tangent 2-planes, i can do that by showing that their total steifel-whitney classes don't factor into a degree 2 polynomial and a degree n-2 polynomial in the Z_2 cohomology ring right?

[coral pawn]

Can someone help me with this?

[winged badger]

Probably more suited for #advanced-analysis ?

[gritty widget]

one word: chain rule

[ivory dragon]

hint: read the problem

[shy moss]

why there isn´t a homotopy between f and id which is stationary in both boundary circles?

[reef shore]

can you visualise what f is doing?

[shy moss]

yes

[reef shore]

could you describe it in words?

ok actually idk if that will be productive

look at the subset {1} x I

[shy moss]

f is something like twist the points which aren in the boundary circles

[reef shore]

think of what its image looks like under f, and under id, and why they couldn't be homotopic intuitively

[shy moss]

i found a homotopy between f and id

[reef shore]

right yeah

I twists the cylinder

[reef shore]

Yeah they are homotopic

Just that the homotopy cannot fix the boundary circles throughout

[shy moss]

this is a line right?

[reef shore]

yes

[shy moss]

i made a drawing

[reef shore]

yep perfect

so now if you have a homotopy of f with id which fixes both boundary circles, you could restrict it to this line x [0,1] to get a homotopy between f restricted to this line and id restricted to this line (which would fix the end points)

are you able to visualize this?

like visualize it like some continuous deformation

So if you can prove that such a restricted homotopy cannot exist, then you are done

[shy moss]

continuos deformation from the twisted line to the line?

[reef shore]

Yeah

[shy moss]

the image of f_0 should be twisted line and f_1 a line

[reef shore]

Yep

But it should fix the end points because the whole homotopy fixes the boundary circles, and the end points lie on the boundary circles

Do you see this? And also that it is impossible?

[shy moss]

i dont see why is imposible

[reef shore]

one of the lines is twisted around the cylinder

and the homotopy wouldn't be able to undo this twist

To prove this formally you can use fundamental groups

[shy moss]

ooh i see, is because the f_0 is a line twisted

[reef shore]

||project the cylinder onto S^1||

Yep

[shy moss]

like brouwer fixed point theorem of D^2?

[reef shore]

I don't see how you could use that

[marsh forge]

Probably means a similar contradiction

[shy moss]

yes, i mean, the proof of brouwer fixed point theorem is by a contrdiction and using functoriality of fundamental group. I think this proof is similar to prove that there is no homotopy f_t between f and id which fixes boundary circles

[ivory dragon]

how is ℝ's topology defined for you?

[hollow harbor]

i would pick a point in (0, infinity)

and then i would find a ball around that point which is contained in (0, infinity)

so if i have x in (0, infinity), then if i pick say r = x/2, we have the interval (x - r, x + r), and this is (x/2, 3x/2). since x/2 > 0 and 3x/2 < infinity, this interval is a subset of (0, infinity).

i could have picked other r's too

(and r can depend on x as you can see)

[marsh forge]

There are a few equivalent ways to define the topology on the real numbers

depending on what your definition is

the proof will look different

[marsh forge]

Well, if you don't know how to define the topology on R then it will be impossible to prove a set is open

just like as a matter of the question making sense

It's sort of like me asking you if apples are foobars without telling you what a foobar is

[pearl holly]

gib me foobar

[marsh forge]

or gib me deth

[wheat gulch]

does a clifford torus have a pair of antipodes where the villarceau circles meet?

is that the correct terminology?

[little hemlock]

Ye so that’s what it means for a set to be open in a metric space (such as the real numbers w/ the usual Euclidean distance )

So for this question, it suffices to show that given an arbitrary point x in (0, infty), you can find an open ball centered at x contained in (0,infty)

[quasi forum]

So I'm having a hard time with this one. Neither A^CxB or AxB^C are open, so I am not entirely sure if this is achieving anything

[bitter yoke]

Remember that the open sets in the product aren't only products of open sets

The product of open sets just gives you a basis for the topology

[quasi forum]

Yea, open sets are unions of open boxes in the product topology.

Yeah, sadly it does not

[hollow harbor]

this is what i was just about to say

[gentle osprey]

slimvesus

[quasi forum]

Le slimsavage

Thanks, I was thinking of that earlier, but I second guessed myself since I was struggling to justify it.
This gets the job done easily, so I appreciate the help.

Ya know, for a grad class, some of these questions are super free.

Half of it. The second half is algebraic

[ivory dragon]

My grad algebra class had a problem of showing an action A to an ordering of S_|A| could be lifted to another ordering of S_|A|

You might realize that this is not only trivial

It's like

Super trivial

...I was trying to think of a stronger word than "trivial" but failed

[quasi forum]

So jumping back into this: when might this not be equal in an order topology?

Firstly, not interested in the sass.
2nd) this inclusion seems super trivial to me and I feel like it should always be an equality.

However, this does not appear to be true and I would like to mold my intuition into the correct one.

But yes, the first part is very simple

[rancid umbra]

think about when X is discrete, like X = Z

[quasi forum]

Ohhhh! So for instance, (1,2) is just the empty set, so the closure is also the empty set.

More generally, if (a,b) is open, then [a+1,b-1] is the closure

I think what you're getting at is that a and b have to be limit points of the set in order for the equality to hold.

[rancid umbra]

no, in Z, every set is open and closed... its just that cl(a,b) = (a,b) != [a,b]

like, cl(1,2) is empty, but [1,2] = {1,2}

[quasi forum]

Yep, that's what I was getting at earlier.

Hmmm, so how to generalize this?

[marsh forge]

So the issue is clearly the end points right

So your question is “when are the endpoints not limit points”

Er I guess any point can be a problem

Mainly the issue is that R is not a good example to keep in mind

Because it’s so connected

But Z is a better example as above

[quasi forum]

So yeah, I am pretty sure this claim holds, since 1 and 2 are not limit points of (1,2) (the empty set).

[marsh forge]

That is necessary but not sufficient

For example, 2 is not a limit point of [1,3] in Z

(1,3)*

[quasi forum]

Why?

[marsh forge]

Let me explain better when I get home

[rancid umbra]

i wanna say equality holds if and only if X contains a dense proper subset

if not that, then something like that.
assuming cl(a,b) = [a,b] for all a,b in X, then for any two points a,b in X, a and b will be limit points of (a,b)

it seems like you need some global property of X to hold

[marsh forge]

@quasi forum exercise: every one element subset of Z is open in order topology

[quasi forum]

(a-1,a+1) is open in the order topology

(a-1,a+1)={a}

If there is a smallest element, let's say a, then [a,b) is open for any b. So [a,a+1)={a} is open

Largest element has a similar approach

The above is using the fact that these are basis sets of the order topology

[marsh forge]

Okay so now convince yourself 2 is not a limit point of (1,3)

In Z

[quasi forum]

Oh geez...this is super obvious! Cl(A)=AUL, where L is the limit points of A

[rancid umbra]

yea

you can characterize closed sets by limit points too. A is closed if and only if (something about its limit points)

[quasi forum]

I can't, cl((1,3))={2}

[marsh forge]

I agree that is the closure

[quasi forum]

And the closure is all the elements of A and it's limit points.

[marsh forge]

Yep

That is not true

[quasi forum]

I guess what you're trying to get at is because {2} is a Singleton, there is no neighborhood U/{2} that meets {2}

And so yeah, that means {2} is not a limit point of itself

[rancid umbra]

@quasi forum
for cl(a,b) to be equal to [a,b] for all a,b in X, you need that for every x,y in X, if x < y, then there is a z in X such that x < z < y

these are necessary and sufficient conditions

[frigid patrol]

Can someone explain how divisors correspond to line bundles?

[cedar pebble]

a divisor D corresponds to a line bundle O(D)

conversely a line bundle L corresponds to a divisor D(L) of zeros minus poles

[tough imp]

[gritty widget]

Is the classifying space of products the product of classifying spaces?

B(G\times H)=BG\times BH

[surreal rain]

How option 3) is true?

[empty grove]

You can write every element of S_1 as a union of elements of S_2 and vice versa

Thus T_1 contains S_2 and T_2 contains S_1 and therefore T_1 and T_2 contain each other

[surreal rain]

How you write elements of S_1 as union of elements of S_2

[gritty widget]

countable union

[empty grove]

Well you can write elements of S_3 as union of elements of S_2, and elements of S_1 are special cases of elements of S_3

Do you see how to do it for S_3?

You might have seen this as "S_2 forms a countable basis for the standard topology on R"

[surreal rain]

Ok

[shy moss]

what is equivariant homology?

[sharp frost]

Hi how can I find the vertex of the following paraboloid?

[hollow harbor]

#multivariable-calculus #geometry-and-trigonometry this channel is for higher geometry (upper undergrad / grad level)

[sharp frost]

if you answer in another channel I'll move there but until then I don't believe you when you say it's easy

this is for a diff geom class

[hollow harbor]

fine, ask it here and see if someone answers.

[reef shore]

ryc if you are so good why are there unsolved problems in math????

[hollow harbor]

the worst part is i don't even know how to do this

and i really should

[sharp frost]

owned

[reef shore]

isnt vertex where derivative is 0

[hollow harbor]

what do you mean

[pseudo crane]

Then you’re dumb. This is easy stuff

[sharp frost]

no that's the centre

[hollow harbor]

it's implicit in z

[sharp frost]

but paraboloids dont have centres

[reef shore]

oh epic

[hollow harbor]

my strategy here would be to do changes of variables to get rid of the xz and yz terms

and then to complete squares

actually you can just complete squares from the start with the xz and yz terms to see what the change of variables should be

[reef shore]

maybe linear change of basis so that the axis is parallel to one of the axes

[sharp frost]

the strategy that I have managed to catch onto is

1. determine asymptotic directions of quadric surface
2. the vertex is where the gradient vector is equal to the asymptotic direction vector

[reef shore]

and then partial derivative of 3rd coordinate wrt other 2 = 0

[sharp frost]

but idek how to find the asymptotic directions

[sharp frost]

yeah the aim of the question is actually to find canonical orthogonal coordinates such that the equation reduces to the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=cz$

[sharp frost]

but to do that I think I first need to know the position of the vertex so I can translate the axes accordingly

[reef shore]

maybe you can figure out direction of axis without exact line?

[sharp frost]

I know that the asymptotic direction can be found by ignoring the lower-order terms so you can reduce it to finding a nonzero vector $v=(x,y,z)$ such that $$4x^2+2y^2+3z^2+4xz-4yz=0$$

[sharp frost]

but a method for actually finding such a vector escapes me

[flint cove]

Can „f: X→Y is a quotient map“ be checked locally? i.e. if we have an open cover Uᵢ of Y, and we can show that every f⁻¹(Uᵢ) → Uᵢ is a quotient map, does that suffice?

[reef shore]

are you allowed to use wolfram

[sharp frost]

haha I have been using it for some factoring

[unreal stratus]

Not sure what I'm overlooking, but can we just not complete the square to get it into a quadratic form and then diagonalise?

Meh no, completing the square would leave mess still, ouchies

[reef shore]

I think so, because some set V is open iff V cap U_i is open for all i, so it suffices to check f(S) cap U_i is open whenever S is a saturated open set, and that works

[flint cove]

What does „saturated“ mean?

of the form f⁻¹(f(S))?

[reef shore]

inverse image of some set ie union of fibers

yeah

[flint cove]

ah ye

so „saturated wrt f“

makes sense, thank you

[reef shore]

yeah

[flint cove]

context: I was looking for a short argument that a locally trivial fiber bundle π: E→X always is a quotient map. Surjectivity is clear because Eₓ ≅ F for every fiber, and now we can see that π clearly is a quotient map when restricted to E_U whenever U is an open set over which our bundle is trivial

and since these E_U are saturated open we get that π as a whole is a quotient map

[reef shore]

Neat

[flint cove]

Not so neat: I have like a week left to finish my B-thesis and I'm still digging in the basics lmao

[reef shore]

all the best

[flint cove]

thanks, should be just the right amount of pressure to not fuck around with perfectionism

[sharp frost]

in case anyone is interested in the solution to the question I asked earlier: here is an answer I got in another discord

[sharp frost]

and one of the columns of P turned out to be the solution to this equation

[shy moss]

i´m trying to understand why loops in X x {y_0} and {x_0}x Y conmutes

the isomorfism between $\pi_1(X \times Y),(x_0,y_0)$ and $\pi_1(X,x_0) \times \pi_1(Y,y_0)$ is defined by $\varphi([f])=([p_1f],[p_2f])$ with $p_1,p_2$ canonical proyections right?

[gentle osprey]

Or x1

[reef shore]

yes

[shy moss]

so if i have paths $f:I \rightarrow X \times\lbrace y_0 \rbrace$, $g:I \rightarrow \lbrace x_0\rbrace \times Y$ then $\varphi([fg])=\varphi([f])\varphi([g])=([p_1f],[p_2f])([p_1g],[p_2g])=([f],[e])([e],[g])=([f],[g])$ right?

[reef shore]

yep

except there should be square brackets in the later expressions too

[shy moss]

ooo right

[reef shore]

here too, the right side should have square brackets

[gentle osprey]

Or x1

[shy moss]

but if $[f]$ and $[g]$ conmutes then $\varphi([fg])=\varphi([gf])$, $([f]),[g])=([g],[f])$ and this implies $[f]=[g]$, is this right?

[gentle osprey]

Or x1

Coldilocks ✓

[reef shore]

Because the projection onto the first coordinate is still f

and similarly for the second coordinate

[shy moss]

now i see why loops in X x {y_0} and {x_0}x Y conmutes

[pearl holly]

yo I'm reading the first page of "The Idea of Cohomology" and I'm already lost lmao. Hatcher writes "let X be a 1-dimensional \Delta complex. For a fixed abelian group, the set of all functions from vertices of X to G also forms an abelian group...". How does that form a group tho? With what operation? If it's just composition then you can't just compose X->G with X->G so what is going on here?

[tough imp]

You just add them component wise

Think about maps to R first (the Reals)

If you want to add f,g: X -> R

Just use the rule that f + g is the function sending x to f(x) + g(x)

The identity is the constant 0-map

And the inverse to f(x) is given by g(x) = -f(x)

[marsh forge]

Chm alg top arc

[pearl holly]

oh lmao so it's just "normal" addition?

[tough imp]

Yee

[pearl holly]

oh okay yee then I see it lmao

[tough imp]

You can do this to turn the set of functions to any sort of algebraic structure with that sort of algebraic structure

(Okay I’m not sure any, but it works for like groups, rings, modules, etc)

[pearl holly]

yee okay I see. Thank you so much!

[tough imp]

[wise sigil]

Hey! a soft question: I'm self studying AT from tom dieck. Is understanding all the "side constructions" necessary to becoming fluent in AT? For example, the build up towards homotopic map => same induced map on homology groups. Also, I've just spent about one and a half weeks on covering spaces. I'm worried avoiding too many constructions will trip me up later but I'm getting a little impatient too

[reef shore]

Reading tom Dieck as a first intro to AT

I guess it should be fine to skip examples and come back later if you feel the need, or decide for each topic based on how well you think you understood it

[wise sigil]

Hatcher was a bit too informal, but yeah i think ive made a mistake

[wise sigil]

How do uni AT courses usually go? straight to thm or no?

[reef shore]

Yeah Hatcher skips some important details and I feel that it is inefficient if you already know cat thy. I am planning to read from JP May because it was recommended to me by the people here. I was trying dieck earlier, went very poorly

[reef shore]

Usually profs tend to give examples, some give them after theorems and some before

but books would give more examples than you would get in lectures

Not enough time in lectures

So should be safe to skip some annoying ones

[wise sigil]

Thanks thats reassuring

[marsh forge]

So

AT courses in general

normally dont focus

on nitty gritty proofs

every class ive ever had skipped the proof of singular = simplicial = cw, for example

A lot of intro AT is building the tools you will need for real AT

Understanding certain constructions is important

and other less so

one of the hard parts of self-learning AT is knowing which is which

and most books don't do a great job of distinguishing

[wise sigil]

yeah thats exactly my problem

[marsh forge]

also

i dont suggest tom Dieck's book as a first pass

[wise sigil]

what would you recommend?

[marsh forge]

I like Hatcher more than most, but I also would say a lot of people seem to like Rotman

I would disagree with your comment that hatcher is not rigorous

and in general say that omitting certain verifiable details and being unrigorous are very different concepts

[wise sigil]

Mmm alright ill have another look at them

[reef shore]

Yeah true, it's rigorous except maybe in ch0 which is a quick review of prereqs

[marsh forge]

I would say that peter's book is not great for learning AT without like

[reef shore]

but it leaves a lot up to you to check I feel

[marsh forge]

a heavily involved mentor

the biggest issue is the lack of exercises

but also just the writing is like, very concise, shockinglky

Peter's book also is really really written for aspiring algebraic topologists

like homotopy theorist style topologists

and not great if you are interested in e.g. knot theory

[wise sigil]

But will going through dieck after familiarizing myself with AT be fruitful?

[marsh forge]

ive never done it and dont plan to really

let me scan the TOC

[reef shore]

I should mention that I am doing a quick pass of Hatcher with someone else rn anyway I read the first 3 chapters in May so far and I plan to read the rest later so it will be my second pass

[marsh forge]

yeah no tom Dieck is like

a massive reference text

i dont think i would suggest reading this unless you are trying to read some specific part of it

also why the fuck does it introduce spectra before homology

[reef shore]

I remember That was where I started learning AT

[marsh forge]

but then not use the stable perspective for homology

I would think there is little to be gained in trying to read tom Dieck as a book

[reef shore]

I quit and moved to Hatcher when he defined the universal cover of a space to be a space such that the associated coverings functor becomes an equivalence or something

😵‍💫

[marsh forge]

My like, suggested course would probably be