point-set-topology

[urban zinc]

not every nbhd, there exists a nbhd

[trail charm]

oh right

oops

[urban zinc]

yea otherwise a local homeomorphism would just be a homeomorphism (onto the image) haha

[trail charm]

yea

ig revised would be smth like

since X is locally n-Euclidian, for all x in X, there exists nbhd U_x ≈ open ball of R^n

and then rest of the proof as above?

[bitter smelt]

You still need to consider that your U_x is perhaps not a neighborhood for which f is a homeomorphism

[trail charm]

aw shid

[next crystal]

why is U \cap V_\alpha nonempty for some alpha and how does p restricted to this V_\alpha being a homeomorphism make p(U \cap V_\alpha) open in B?

[bitter smelt]

b is in p(U). U_b contains b. Thus, the pre-image of U_b must have some intersection with U. The preimage of U_b is composed of disjoint union V_alpha, so at least one of these must intersect with U

Intersection of open sets are open, homeomorphisms are open maps

[next crystal]

so just to make sure im understanding, lets say e \in U and p(e)=b. Since b is in U_b, then the preimage of U_b must contain e, so p^{-1}(U_b) intersects U at e?

[bitter smelt]

Correct

I don't think this proof is quite complete though.

4.5/5 stars. Said "NTS p(U) open," yet never concluded p(U) is open

We're on pedantic mode now though

[next crystal]

well b contained in p(U \cap V_\alpha) contained in p(U) shows that right

since b was an arbitrary element of p(U) and p(U \cap V_\alpha) is open

so it shows every element in p(U) has an open neighborhood contained in p(U)

[bitter smelt]

Yep

[trail charm]

would i consider the product topology of two spaces of continuous functions?

something like f_1 in F_1 and f_2 in F_2, then f_1+f_2 is a mapping F_1 x F_2 -> F?

where i guess F is the space of cts functions X -> R

[obtuse meteor]

No. In General the set of functions X -> R does not carry a natural topology (for bad spaces X)

[trail charm]

ah

would i consider the product f#g: A -> R x R via (f1 # f2)(x) = (f1(x), f2(x))?

and then R x R -> R via f1(x) + f2(x) or f1(x)f2(x)

[obtuse meteor]

Yes!

[trail charm]

ty :)

[next crystal]

So we have $U \cap V_\alpha$ is open in $E$\
But the homeomorphism $p|{V\alpha}$ sends open sets in $V_\alpha$ to open sets in $U_b$\
but $U \cap V_\alpha $ isnt necessarily open in $V_\alpha $ so how can we say $p|{V\alpha}(U \cap V_\alpha)$ is open in $U_b$, let alone $B$?

[gentle osprey]

michαel

[lofty rain]

hey, im not sure if i have the right understand because i think all of them should work unless im incorrect?

since open discs are a basis, and we can say that every point has an open disc centered around this point, we can have a square, rectangle, triangle, whatever else, that contains this centre?

which would also make them a basis

but to have them all be a basis seems a bit wrong, at least given the nature of the question

[coarse night]

yes they all are basis

[lofty rain]

alright alright thank you

[stable kite]

In a book on Riemann Surfaces I'm reading the author constructs a covering space from a presheaf over a space by taking the disjoint union of all stalks and then inducing the topology from subsets of the form [U,f], (where U open in X and f a section) containing all germs of f in u, for u in U.
Then, the author goes on to proof that for X a Riemann Surface and the sheaf of holomorphic functions, this is a Hausdorff covering space.
I haven't been able to find this construction elsewhere so far. Does someone here recognize this? Thanks in advance. 😊

[arctic relic]

Whats a good introduction to minimal models

My end goal is to be able to go through sullivans thing on infintesimal computations in topology

Im good with twisted (co)homology

[umbral panther]

It’s called the espace étalé

[stable kite]

Oh, so those notions coincide? I've heard about étalé, but read somewhere it's not T2 so suspected it's a different space.

[umbral panther]

For holomorphic functions it’s Hausdorff because they are determined by their germ. But for the skyscraper sheaf it isn’t. Nor for the sheaf of smooth functions

It’s not a covering space. It’s always a local homeomorphism, but a function that doesn’t extend globally (which is all non constant holomorphic functions) shows that it isn’t a cover

[stable kite]

Oh yeah, it shouldn't be covering space, but complex analysis has a slightly relaxed notion of covering it seems.

[stable kite]

Thank you kindly, that clears up my confusion!

[bitter smelt]

It is. Subspace topology for first question and both V and U_b are themselves open so openness in the subspaces is equivalent to openness in the bigger space (assuming the set is a subset of the smaller space) for second

[urban zinc]

Wait Van Kampen's theorem is just that the fundamental group(oid) is the pushout

neat

[bitter smelt]

SvK for groupoids is a little different than ususl SvK

Groupoids goes harder

[umbral panther]

The Hawaiian earrings are homeo to the wedge of two copies. But not uniquely. Is it a universal property? Are they somehow a universal space with a map from the circle and a homeo to the self wedge? Is the fundamental group a universal group with a generator and an iso to the free square?

[trail charm]

does “prove directly from definition” just mean to construct an explicit quotient map, like (cos 2πx, sin 2πx)?

[unreal stratus]

Okay this is Lee right so if I'm remembering correctly how he does it, the point is like

[novel acorn]

I think they mean construct an explicit homeomorphism?

[trail charm]

yea

[unreal stratus]

Okay, you can just write down the map I -> S^1 you wrote and show it induces a continuous bijection I/~ -> S^1

[trail charm]

ah kk

[unreal stratus]

But showing it's a quotient map is tedious without using the extra technology Lee develops later

Namely, I don't think he's mentioned compactness at this point (?)

[trail charm]

ohh gotcha

nah not yet

[unreal stratus]

Yeah so the easiest way to show that the map I/~ -> S^1 is a homeomorphism is that uh

Well, spoiler alert lol but any continuous map I/~ -> S^1 is a closed map, so if it's bijective it's a homeo

As you'll see later on

:)

[trail charm]

closed map lemma?

[novel acorn]

showing bijection isn't too hard (I think)
The part part is showing that the map is open I'd say

[unreal stratus]

Yes lol

[trail charm]

ive read through a few chapters lol im just going back and making sure i understand it all

[unreal stratus]

That's what I'm saying basically lol

[trail charm]

kk ty

[unreal stratus]

Noice

[trail charm]

i just wanna get to diff top 🤬🤬🤬

its like how real analysis is the prereq for everything cool

gen top is prereq for everything cool

[solemn oar]

Good medicine tastes bad, you know

[trail charm]

yeah yeah

this tastes better than real analysis tho

[trail charm]

theres a similar question, which is like “X is hausdorff iff the diagonal, {(x, x)}, is closed”

does it suffice to just take q: X -> X to be the identity map?

under the equivalence relation x1 ~ x2 iff x1 = x2

then q(x1) = q(x2) iff x1 = x2

[unreal stratus]

Well, that is a special case sure

[plain raven]

When you learn some thing and start to understand it better then you rearrange things in your mind and restructure them.
Intuition grows and develops. Perhaps it is better to say that topology will force you to grow and develop your intuition

(than to say topology is unintuitive)

[sonic warren]

I do feel like the subject gets considerably more digestible as soon as you add in some algebraic or differential structure

[bitter smelt]

it doesn't, it just allows you to brush your weak understanding of the fundamentals under the rug since there is more structure

[next crystal]

haha this is so true my intro topology class was like 60% point set and 40% algebraic and once we got to the algebraic part i did exactly that

[bitter smelt]

Black boxing is a time honored tradition among mathematicians

[stable kite]

Why is there no local homeomorphism from S2 to the plane?

I guess from one of the theorems on maps S2 to the plane one can show local injectivity fails?

[quiet thorn]

i don't think i have ever seen a super reaction get so many stacks

[coarse night]

local homeomorphism is an open map, can you get some contradiction from this?

[stable kite]

Oh, image of S2 is compact as cont. image of a compact space, so bounded and closed by Heine-Borel, so not open in the plane per connectedness, but S2 open in itself?

[coarse night]

yeah

[coarse night]

spit out potato

[unreal stratus]

Eh I was gonna say like calling it Heine Borel feels odd when compact subspaces of Hausdorff spaces are always closed

But like nothing wrong w just saying Heine Borel especially if we are mostly concerned w subspaces of R^n here or metric spaces

Lol

Hence the deletion

[coarse night]

yeah it's not the most straight forward way to argue but whatever works works

[stable kite]

I agree, the argument with closed in Hausdorff is more precise, as it's exactly the property that matters here. Thanks for the comment. 😇

[unreal stratus]

Np

But yeah nice pf guys

[plain raven]

i prefer to call it abstraction

[trail charm]

when we say "identify X with q(X)", does that just mean X ~ q(X)?

[gritty widget]

context?

[trail charm]

where X U_f Y is adjunction space

[gritty widget]

i hate this new discord thing where i have to open each image to see what's in them lol

i miss when they were just stacked

[trail charm]

and q : X \amalg Y \to X U_f Y is associated quotient map

agreed

on one hand it's nicer because it doesn't take as much space for a message w/ 4 images

but it's annoying

[gritty widget]

"identify X with q(X)" means you don't make a distinction between X and q(X), you pretend they're the same

like if you write x you're thinking of x in X and q(x) in q(X) at the same time

[trail charm]

ah gotcha

[trail charm]

is this proof sufficient?

i took some ideas from here: https://math.stackexchange.com/questions/183319/showing-the-adjunction-space-d2-cup-f-d2-is-homeomorphic-to-s2

[unreal stratus]

I don't think q is open

Ah yes, you claim the iota are open. They aren't - in particular their images aren't open

@trail charm

Try showing q is closed - generally this is easier (why?)

[trail charm]

aw shid

ok gimme sec to write out

[trail charm]

tbh im stuck on this

[unreal stratus]

There is a general lemma for showing maps are closed

[trail charm]

closed map lemma: since B^2_a \amalg B^2_b is compact and S^2 is hausdorff, q is continuous implies q is closed?

this feels a bit overkill (if it's even correct), considering this is in a later chapter

[unreal stratus]

Yes and in practice this is certainly how I'd do it

But hm is there a more nice / elementary way uh

So in this case it's actually possible to just create an inverse lol

and show it's continuous

Another (slightly fancier) way would be to show that S^2 actually satisfies the same universal property as that adjunction space but that's basically the same proof

[trail charm]

ah yeah i was thinking about finding an inverse but tbh this isnt an exercise, more of an example in lee's itm in the adjunction spaces chapter

and it's more of the form "you can verify this is a homemorphism using techniques from the previous section" i.e., the quotient space section

oh well i suppose this will suffice for now, i will think about it another day maybe

ty

[unreal stratus]

Sure

But yes in terms of satisfying the universal property, hopefully it is clear why like

the datum of a map S^2 -> X is the same as the data of maps S^2_+ and S^2_- compatible on their intersection

[trail charm]

wdym compatible?

[unreal stratus]

Well tbh just mean agreeing here lol

[trail charm]

ah

yeah that makes sense to me

kinda gross how lee calls them characteristic properties instead of universal but oh well

dont sully me >:( universal sounds cooler

[unreal stratus]

I think the way he presents them is slightly different to universal properties but it's been a while hm

Idk been too long

[trail charm]

oh oops

i take it back

[unreal stratus]

Actually I may be wrong lol dw

[craggy sage]

Carried over from help channels

[solemn oar]

Here is an example of a T6 space which is not metrizable: https://topology.pi-base.org/spaces/S000043/properties

In fact, all of these are also examples of such: https://topology.pi-base.org/theorems/T000268

[craggy sage]

Thank you very much for the links

[craggy sage]

Holy shit this is the lower limit Topology

This is what my project instructor gave as an example of an non-metrizable Topology but I never knew it was T_6

[balmy field]

Looking for some help: Let $f: M \to N$ be so that if $U \subset M$ is open then $f(U)$ is open. Is $f$ continuous?

[gritty widget]

you mean "f(U) is open"?

[balmy field]

Yeah my bad

[cosmic socket]

No

[gentle osprey]

Ultimate Chad

[gritty widget]

it's not true

it is false

[balmy field]

I figured... what does the counterexample look like?

[gritty widget]

there are many

[cosmic socket]

See https://math.stackexchange.com/questions/75589/open-maps-which-are-not-continuous

[gritty widget]

one is the angle function on the whole circle

as a map from S^1 to [0, 2π)

more boring examples can be constructed by taking maps into discrete spaces

e.g. the floor or ceiling functions from R to Z

[balmy field]

Just to check my understanding, are the open sets in S^1 just arcs (and unions of arcs)

[cosmic socket]

Yea

[unreal stratus]

Take any bijective continuous map which isn't a homeo and consider its inverse

So some boring examples will be like, take a set X and equip with topologies τ and τ' with τ' strictly containing τ

Then consider the "identity" (X,τ) -> (X,τ')

[tidal lynx]

almost all open mappings are not continuous (probably)

[spice basalt]

this problem asks to prove that any pairwise disjoint set of open intervals on R is countable

we can do this by showing that each interval contains at least one rational number and the rationals are countable

does this require the axiom of choice?

[languid patrol]

Just countable choice

@spice basalt which is also necessary to prove, eg, that the countable Union of countable sets is countable

[umbral panther]

No, the rationals can be well ordered without choice. Thus every open set contains a canonical rational, the simplest rational in it

[spice basalt]

ah ok that makes sense. thank you!

[cosmic socket]

are there phantom maps between eilenberg-maclane spectra

(I don't know anything about phantom maps so maybe this is a naive question)

but I guess can I treat maps between eilenberg maclane spectra as just some families of cohomology operations?

[cosmic socket]

if the answer is yes what if I restict to only field

and if that is also yes, what about the case for F_2

The reason I asked is because I read something that says the steenrod algebra can be viewed as the homotopy endomorphisms of $H\mathbb{F}_2$. But does that mean there are no phantom maps between $H\mathbb{F}_2$ and itself?

[gentle osprey]

bigradedSphere

[thorny agate]

Is there any way to describe which neighborhood of x it's talking about in that second paragraph?

Because obviously not every neighborhood works (like take just the whole space X for example)

[gritty widget]

@thorny agate if φ: [0, 1] -> [0, 1] is the map described in the second paragraph, then φ o f is zero on f^{-1}([0, 1/2))

[winged viper]

my professor defined the projective resolution of a chain complex C as a quasi-isomorphism P -> C where P is a projective chain complex, and he then went on to define Tor between two chain complexes C and D as H_*(P \otimes D) where P -> C is a projective resolution (or equivalently H_*(C \otimes Q) where Q -> D is a projective resolution)

does anyone know any reference that defines things this way? i've only seen projective resolutions/Tor of modules being defined, but these definitions completely generalize that

like one interesting consequence is that the universal coefficient theorem is stated as H_*(C; R) = Tor(H_*(C), R) for a projective chain complex C. The proof of this is to view H_*(C) as a chain complex (whose differential is 0) and show that C -> H_*(C) is a quasi-isomorphism, hence a projective resolution. Then H_*(C; R) = H_*(C \otimes R) = Tor(H_*(C), R) is true by definition

and this agrees with the usual universal coefficient theorem (I think at least) if you break down what Tor(H_*(C), R) means in terms of Tor's between modules

but i think you lose naturality since Tor(H_*(C), R) involves a direct sum which comes from splitting the usual universal coefficient theorem

anyway would love if someone knew a reference that did things this way since it seems quite clean

[empty grove]

Probably look at Gelfand-Manin, it has a chapter on derived categories so should cover this

A more modern approach is using homotopy theory. You could look at the first couple chapters of Riehl's categorical homotopy theory for that, though she gives more topological motivation

There should be some more homological resources but I haven't read any

Actually Gelfand-Manin does later talk about the homotopy theory and triangulated categories, so should be fine. I just found it difficult to read

@winged viper

[coarse night]

[empty grove]

[coarse night]

seeing you after a lot of time

btw what would be an example of a module which is locally finitely generated but globally not?

[empty grove]

So excited to see me that you posted in the wrong channel

Is that even possible though

Like if M is locally finitely generated, then for each p, there is a surjection (R_p)^n → M_p for some n, which lift to a map R^n → M. The direct sum over p of all of these lifts should be a surjection because it is locally a surjection

Does this not work?

I might be misinterpreting "locally finitely generated"

[coarse night]

oh lol it's topology

[coarse night]

idk actually, some prof asked this question

but the n is Rⁿ_p may not be uniformly bounded right?

[urban zinc]

What does the K stand for in K(G,1)?

[coral pivot]

hmm these are also classifying space

so i imagine the K is from this in some appropiate language

[hidden crag]

Klassifizierend

[solemn oar]

The original reference is a paper by Eilenberg-MacLane (1945). They introduce the notation but do not explain it.

[hidden crag]

[coral pivot]

lol

[solemn oar]

They were studying homology groups, so maybe K was just the next letter in the alphabet that wasn't I or J

I was taken, btw

[coral pivot]

I the additive group of integers....

[unreal stratus]

Lol

[urban zinc]

Wait seriously

and no one changed the notation to something better!!!

[unreal stratus]

Presumably lol

[solemn oar]

Eilenberg-MacLane "RELATIONS BETWEEN HOMOLOGY AND HOMOTOPY GROUPS OF SPACES" (1945)
Earliest reference I could find.

[urban zinc]

And they were both American 🤔

brb just gonna start calling them EM spaces

The universal cover of a connected graph is a tree
Is this just because every cover of a graph is a graph, and a simply connected graph has to be a tree?

[coral pivot]

yes

[unreal stratus]

yes

[urban zinc]

Ah okay

wtf S^infty is contractible

[unreal stratus]

Yeah

[urban zinc]

Actually I guess this makes sense because it can't have a hole in any dimension

[unreal stratus]

Yeah you can think about it heuristically by like the homology/homotopy groups being pushed up into higher and higher dimensions

as n -> infty

Ofc that doesn't show contractibility at least without more

But yeah there's the cute proof with shifting lol

[urban zinc]

I hate infinity, glorified 8 symbol fr

Wow Hatcher sure does like to... use words he's never mentioned before

But it's okay because he tells us that later he'll define it in Example 2.43 🥲

It is not hard to construct an example using this notion DEFINED IN THE NEXT CHAPTER

[coarse night]

[solemn oar]

Yeah but you have to work a little bit for it unless you use Whitehead's theorem.

[urban zinc]

This is the proof Hatcher gives

[rough cedar]

RP^inf

[solemn oar]

Note that as opposed to finite dimensional euclidean space, linear maps need not be continuous in infinite dimensions. Hatcher is skipping some steps here.

[jovial lance]

ayOO @urban zinc did you become hatcherpilled

[coarse night]

this reminds me of this ex: Can you show there is no spade X s.t. ℝ → X is a 2-fold covering

[urban zinc]

It needs to be bounded too right?

[urban zinc]

Lol idk what that means but I'm reading it

[empty grove]

What 💀 is this about quasicoherent sheaves?

[winged viper]

Thanks!

[empty grove]

Ouch that is true

Rookie mistake

[winged viper]

Can you just use the fact that such a space is a manifold and then the classification of 1-manifolds

[coarse night]

too fancy

[coarse night]

idk she randomly asked in class

[winged viper]

Hmm somehow use the fact that the fundamental group must be Z/2?

[coarse night]

already answered in #algebraic-geometry

[coarse night]

ye

fancy way || X must be K(Z/2, 1) which is contradiction ||

[winged viper]

Oh lol

[coarse night]

this argument works for R^n -> X as well

[winged viper]

“Too fancy”

[coarse night]

yeah I admit it's a fancy way but there's an elementary way to do this

[winged viper]

What’s that

[empty grove]

Any action of C_2 on ℝ has a fixed point

[coarse night]

what can you say about the ||deck transformation group||

[solemn oar]

Yeah, essentially. A linear operator between normed spaces is continuous if and only if it is bounded.

[winged viper]

Ohhh okay that’s smart

[thorny agate]

this proof does not make sense to me

1: how is it clear that f is one-to-one?
2: why does showing that f is a closed map suffice for showing that f is a homeomorphism

[gritty widget]

1. if x and y are distinct points, take C = {y} as in the statement of the lemma and you'll see that f(x) does not equal f(y)
2. saying f is closed is the same as saying f^{-1} is continuous

[frosty needle]

Could someone help me understand "sequentially closed"? The definition I'm seeing is a bit confusing to me because I thought convergent sequence meant it had exactly one limit.

I'm trying to prove this but I'm struggling to understand the definitions of closure.

I'm taking a prelim exam in topology at the end of the summer but need to learn it from the ground up to get the necessary vocabulary.

[thorny agate]

so iirc limits are unique in a topological space iff that space is Hausdorff

metric spaces are Hausdorff, so limits in them are unique

[frosty needle]

So, I'm kinda a total topology noob

I'm pretty good with algebra and fairly familiar with analysis

but I've got next to no experience in topology.

I'm trying to build up a vocabulary with it so that I can work with the hypotheses directly.

[thorny agate]

Uhhhh

are you working through any sort of text?

[thorny agate]

since you have a metric space you don't have to worry about this but this generalization of sequences to nets is how you talk about topologies outside of metric spaces

[frosty needle]

So is sequentially closed just the same as the regular limit point definition of closure from real analysis?

I'm just not used to looking at things from a topological lens. This is a crash course intended for people who've taken the prelim sequence but I'm taking the exam at the end of the summer because my algebraic reasoning is really strong, but my topological vocabulary is laughable.

I took the algebra exam halfway through the course and passed it.

[thorny agate]

yea

[frosty needle]

Sounds good.

[thorny agate]

just in general topologies

[frosty needle]

And regular closed just means open complement?

[thorny agate]

you need Hausdorffness for unique limits

yes

(you should know what Hausdorff means, comes up alot)

and other types of spaces

as that's part of the topological vocabulary

[frosty needle]

Right. I would like to learn those things as thoroughly as possible given time constraints.

[median sand]

Varying definitions aside, a normed vector space is locally compact iff the closed unit ball (and hence all closed balls) is compact, right? IIRC this isn't true for general metric spaces, no?

[gaunt linden]

Yes and no. There are two ways to define "closed" in good old real analysis, and they happen to be equivalent as long as we're talking about metric spaces. However, once you consider general topological spaces, they are not equivalent anymore, and then "sequentially closed" is explicit about which one you're using.

[frosty needle]

I mean the one that says the set contains all its limit points. Is that not the same as sequentially closed?

The other definition of closure meaning it has an open complement. Right? Or not right?

[gaunt linden]

Oh right, that's a third definition. :-D

"Limit point" is itself a bit ambiguous in the same way (but the "sequential" version of that is used seldom enough that it's rare to see "sequential limit point").

You can either say:

"p is a limit point of A" means "there's a sequence of points in A that converges towards p"
or:
"p is a limit point of A" means "every open set that contains p also contains at least one point of A"

[novel acorn]

Isn't the second one more common (in general topology)

[gaunt linden]

Yes, the second one is the commonly used one, but I'm suspecting the first might be what Amizar has internalized.

There are weird topological spaces where these concepts are not the same.

[novel acorn]

Measure spaces are too nice we should ban them

Screw you being T6

[median sand]

Anyone?

[gaunt linden]

That's not true, for example, for an infinite discrete metric space where all nonzero distances are 1.

[unreal stratus]

For general metric spaces I assumed they meant iff all closed balls are compact right since otherwise no good notion of the unit ball

Hm

[gaunt linden]

(Discrete spaces are still locally compact, though).

Hmm. @median sand, could you clarify? I understood you as focusing on how you in a vector space can choose a single radius for all the balls you consider.

[median sand]

Clarify what? My question is about normed vector spaces specifically. The way I understand it there are multiple potentially non-compatible definitions of local compactness, but for normed vector spaces these all agree and come down to the closed unit ball being compact.

Wanted to check if that's true.

My 2nd, offhand, question was whether this (locally compact iff all closed balls compact) holds for general metric spaces (which it probably ought to, since a compact neighbourhood must contain a compact closed ball and a closed ball is a compact neighbourhood).

[median sand]

I'm not sure what this is a counterexample of.

[gaunt linden]

It's a metric space which is locally compact, but where closed unit balls are not compact.

(It is locally compact because every point p has a compact neighborhood, namely {p}).

[median sand]

Oh right, since the entire space is a ball, my bad.

[median sand]

But this was correct, right? Your counterexample is not a normed vector space (a NVS can't be discrete, since 1=|2x|\neq2|x|=2).

[gaunt linden]

I think it's true for normed vector spaces, but I'm not strong enough in functional analysis to rattle off a proof.

[median sand]

Does this need actual functional analysis though?

Whatever the definition of local compactness, it presupposes the existence of a compact neighbourhood, no?

This then contains a closed ball, which becomes compact, and all balls in an NVS are homeomorphic.

[gaunt linden]

It's just the heading I file "weird facts about the topology of infinite-dimensonal vector spaces" under.

[ocean narwhal]

yes

[thorny agate]

I think I'm undervaluing the visual parts of topology especially with finding examples and such.

Like some of this stuff is visual but my notes have very few pictures / diagrams

Anyone have a good source which has these diagrams for topological things

[thorny agate]

where does this T_ notation come from?

like T1 spaces, T2, etc

the numbering seems arbitrary-ish (like I guess bigger number = more powerful guarantee)

[urban zinc]

T = trennungsaxiom (separation axiom)

[urban zinc]

T_n implies T_m for n>m

[urban zinc]

You might be need to be more specific, diagrams of what?

Also it might help you understand better if you draw your own diagrams

Topology is very visual but visual aids are also often deceptive

[coarse night]

cool

I used to think T stands for Topology

[next crystal]

topologyaxiom

[unreal stratus]

Tbf visual aids for topology can be funny imo lol

Like with common constructions just drawing ur space as essentially a disk lol e.g. when people draw suspension

[lime sable]

most spaces are homeomorphic to disks, so it's fine

[trail charm]

are separation axioms (eg T_1, T_2, T_4, etc.) still often used in papers and textbooks and whatnot?

[plain raven]

yes

[trail charm]

aw man

more notation to memorize

[plain raven]

oh

no

you don't have to memorize the notation lol

[trail charm]

oh gotcha

[plain raven]

i mean some have names. T2 is called Hausdorff and T4 is called normal. That's more common in my experience. T3 and T1 are less common.

you should memorize T2. the others you should pick up as you go. And Urysohn's lemma has as a corollary that all compact hausdorff spaces are T4. but you should just unfold that and learn "all compact hausdorff spaces are (definition of T4)" rather than memorizing T4

there's even a T3.5 axiom that comes up when discussing the stone cech compactification lol

[urban zinc]

tbh it's not that hard to memorize T0, T1, T2, T3, T4; they're mostly the same

and the numbering system is actually pretty logical

they're all just about separating things from each other topologically

T0 – If you give me any two points, is there an open set containing one but not the other? (In other words, can I tell them apart topologically at all?)
T1 – If you give me any two points, is there an open set containing the first one but not the second one? (This is equivalent to singleton sets being closed)
T2 (Hausdorff) – If you give me any two points, can I draw non-overlapping neighborhoods around them?
T3 – If you give me a point and a closed set, can I draw non-overlapping neighborhoods around them? (Usually also required to be T1 so that single points are closed sets)
T4 – If you give me any two closed sets, can I draw non-overlapping neighborhoods around them? (Usually also required to be T1 so that single points are closed sets)

They're all things you should be aware are able to go wrong in topological spaces

Also yeah, Hausdorff is important

[coarse night]

there's also like 3½

T_π

[urban zinc]

I thought you were joking but Wikipedia actually lists that as another name for T3½ wtf

[coarse night]

yeah there's 4½ also

upto T6

[solemn oar]

Then there is R0 and R1, but I never saw those used for anything.

[empty grove]

Why did the highly separated space go to the dentist?

T_30

[hidden crag]

lmfao

[empty grove]

[urban zinc]

Is the pun here teeth dirty

I'm trying to figure it out 😭

[hidden crag]

you're close

[urban zinc]

Teeth hurty???

I finally know what singular and simplicial homology are 🥺

This is neat

[quasi forum]

Question about two spaces being locally homeomorphic. Do you need a function f:X->Y that is a local homeomorphism, or can you have different functions for each point and its corresponding neighborhood that are homeomorphisms?

[hidden crag]

the latter

see manifolds and charts

[quasi forum]

Okay good! I was hoping so!

[empty grove]

But then locally homeomorphic spaces may not have a local homeomorphism between them

[quasi forum]

Yep, I'm 100% with ya. As an example, I believe there is no local homeomorphism between R and S1, but they are certianly locally homeomorphic

[hidden crag]

There is one from R to S^1

since R is a covering of S^1

[quasi forum]

Oh really? I didn't realize that

Oh yeah, you can just wrap it around infinitely many times

Just the typical map (cos(theta), sin(theta)) does the trick I think

[grizzled ibex]

If you really think about it alg-top is really really really weird

[quasi forum]

I haven't thought of coverings in so long, so if I said something silly, let me know

[grizzled ibex]

i don't know about the advanced topics, but the introductory ones focus too much on computing holes

[hidden crag]

no that's pretty much it

[grizzled ibex]

isn't this too much restrictive? why do i feel like holes and bordism isn't everything?

[quasi forum]

Okay rad. Thanks Timo.

[obtuse meteor]

It turns out once you start trying to compute (higher dim) holes you have to understand a whole lot about your space so you’re led to a lot more concepts

Also there’s like shape theory in pure topology land or <waves hands at anything at all geometric> when you’re dealing with manifolds

[urban zinc]

Omg if I finish reading this section I can finally prove invariance of domain

[solemn oar]

Very nice

[coarse night]

can you show there's no continuous bijection from R → R² || without using invariance of domain?||

corrected

btw it's not something I need help with, just ex for the readers

[languid patrol]

hint: baire category theorem

[coarse night]

shh

[unreal stratus]

Lies

[gritty widget]

call her peano the way she curve until i fill her space

why the fuck did that super react

[unreal stratus]

Lmao

[next crystal]

no please stop

[empty grove]

Turns out T_7 to T_23 are all equivalent via a Urysohn like argument and T_24 onwards is just discrete

[warped rover]

T_23.5

[empty grove]

That is totally disconnected to this discussion

[next crystal]

do those actually exist

[empty grove]

do you care

idk if they do I am sure someone's made some stupid definitions past T_6

[next crystal]

you definitely did 🧙‍♂️

[empty grove]

https://tenor.com/view/walter-white-walter-falling-breaking-bad-dm4uz3-gif-18078549

[plain raven]

let (X, A) and (Y, B) be two topological pairs. define their product by
(X x Y, X x B \cup A x Y)

1. does this monoidal product have interesting categorical properties (a universal property)
2. if both inclusions are cofibrations is their product a cofibration?
3. are there other model categories where something similar holds?

[median sand]

If X is Hausdorff, how does 2''=>3?

[untold lily]

use this (source willard)

[gritty widget]

nhood

pronounced "nude"

[untold lily]

mood

[median sand]

you in the wrong nhood nword

[gritty widget]

[quick bough]

how does one show that this map is continuous

the claim here is that X is contractible if it has a generic point

the problem here that i have is that if you consider the preimage of an open subset in X, you either get U x {0} if \eta does not contain U and otherwise, X x (0,1] \cup U x {0}

unless i made a mistake

and U x {0} doesn't necessarily have to be open (?)

or is there another way to define a homotopy

[unreal stratus]

Consider preimages of closed sets instead

[quick bough]

ugh, i was gonna do that and decided not to

okay, wait

but hmm

you will get C x (0,1]

no?

similar problem

[unreal stratus]

Hm no

If it contains eta, then it is X

So preimage is X x I

And if it doesn't contain eta...

[quick bough]

wait, why is it the whole space if it contains eta

am i missing something here lol

[hidden crag]

generic

[quick bough]

oh

oh god..

yeah

okay, got it

if it doens't contain eta, is it C x {0}?

[hidden crag]

yes

[quick bough]

okay, just to make sure

[unreal stratus]

Yes

[hidden crag]

sry for butting in potato

[unreal stratus]

Lol np

I am walking anyway

[quick bough]

also, another question

we don't have an explicit representation for the induced maps, right

[unreal stratus]

Not sure what you mean

[quick bough]

like

[unreal stratus]

Like we have the definition lol

[quick bough]

how does pi_0(f) look like

[unreal stratus]

Sk like what is the definition

[quick bough]

tbh, we didn't rlly have a definition for the map itself

[unreal stratus]

Sends path component of x to path component of f(x)

[quick bough]

oh

wow

okay, thanks

[unreal stratus]

Np

[cursive tendon]

Is this true? On a open and bounded subset of ℝⁿ, functions that vanish near the boundary are compactly supported.
Specifically. Claim: $Ω ⊂ ℝⁿ$, $Ω$ is open and bounded. Let $𝑓: Ω → ℝ$ be such that $∃ε > 0. ∀𝑥∈ Ω. d(𝑥, \partial Ω) < ε ⇒ 𝑓=(0)$, then $𝑓$ is compactly supported.

where

$$d(𝑥, \partial Ω) := \inf_{y∈\partial Ω} |𝑥-𝑦| $$
and
$f$ is compactly supported when $\operatorname{cl}_Ω{x∈Ω| 𝑓(𝑥) ≠ 0}$ is compact.

[coarse night]

Where’s the bijection then

[gentle osprey]

Mattuwu

[coarse night]

I mean you get that the support is closed and being subset of Omega itself, it’s bounded so compact

[cursive tendon]

the support is closed in Ω, and it's not necessarily closed in ℝⁿ

[coarse night]

Ah I see what you mean

But it will turn out to be

[cursive tendon]

yep, "support" is defined by taking the closure in Ω as a topological space of the set-theoretic support. For example, Let Ω := (-1, 1) the restricted bump function 𝑓: Ω→ℝ, 𝑥 ↦ exp((x²-1)⁻¹) has a closed support (-1, 1), which is not closed in ℝ, and is thus not compact.

[coarse night]

You can, say take eps/2 and make the support sit inside a closed subset of omega

*closed subset of R^n

Let’s see if it’s possible to write a proof

look at the set Γ:={ x ∈ Rⁿ| d( ∂ Ω, x) ≥ ϵ/2 } then it's closed subset of Rⁿ and your support sits inside this closes set

so your suppf is closed subset of Rⁿ which is bounded so compact @cursive tendon

[coarse night]

she curved so much she crashed into herself

[cursive tendon]

right...it's 𝑑⁻¹([𝜀/2, 𝑀]), where 𝑀 is the maximum distance, now because 𝑑(əΩ, -) is continuous, the inverse of a closed set is closed and thus it's compact

[coarse night]

ye

you can just take M= ∞

[cursive tendon]

why, isn't [𝜀/2, ∞) no longer closed?

[coarse night]

you don't need bdd

just need closed, bdd comes from Ω being bdd

[cursive tendon]

sorry I edited, it's no longer closed

[coarse night]

it's closed

[ ϵ/2, ∞) is closed subset

[cursive tendon]

OH, yes

thanks!

[quiet thorn]

uwu

[plain raven]

I see that the typical woke media has been characteristically silent on the hard questions i'm asking

[languid patrol]

1.) This product is a kind of pushout so I suppose it does have a universal property, but I don't know anything deeper than the obvious one.

2-3.) In a cofibrantly generated model category it suffices to check this on generating cofibrations. So on simplicial sets whether or not the product operation preserves the cofibration property can be checked for the inclusion of S^{n-1} \to D^n on, say sSet or compactly generated hausdorff spaces. There it's easy to see that the answer is yes

in Top with whatever model category structure you want to put on it (say the Strom one) I think the answer is no.

[plain raven]

yeah ok. i'm interested in a different model structure on simplicial sets so i'm just trying to get a grasp of when this fails and why it's true in certain places.

Top satisfies that if (X,A) is a cofibration and Y is a space then (X x Y, A x Y) is a cofibration. https://math.stackexchange.com/questions/381527/the-product-of-a-cofibration-with-an-identity-map-is-a-cofibration
That's pretty interesting. I wonder how general that is

https://mathoverflow.net/questions/158929/is-a-join-of-two-cofibrations-a-cofibration

This is helpful

"This axiom is exactly what's needed to make sure that the monoidal structure on the point-set level descends to a monoidal structure on the homotopy category (more generally that the monoidal structure on a model category descends to a well-defined monoidal structure on the homotopy category)."

Huh

[empty grove]

It's a closed monoidal product so that's one universal property

Hom((X, A), (Y, B)) = (all maps of pairs, Hom(X, B))

Seems to work

[lunar yoke]

Surprisingly (at least to me st the time) it is generally not enough at all to require the monoid axiom when you want to have a right induced model structure on categories of commutative algebras, i.e with underlying weak eqv and fibrations. The problem is that symmetric powers are generally not homotopically well behaved. In spectra this is usually circumvented by replacing your model structure with a „positive“ version of it, where in degree 0 you require cofibs to be isos. Magically this turns out to make symmetric powers work well, and you get all the nice model structures you want

[lofty rain]

would i be correct in saying that if some T1 is a basis and it is homeomorphic to some T2 then T2 must also be a basis?

given that T1 and T2 are on the same set

[novel acorn]

sorry here you're treating the bases T1 and T2 as subspaces?

[plain raven]

Great. cool. had not thought of that.

[plain raven]

do you mean the subspace of maps carrying A into B

[empty grove]

Nope

Wait for the big space?

[plain raven]

I'm wondering if you meant
(All maps from X to Y, all maps X to Y carrying A into B)

i guess not but like

it's surprising that A does not come up

that's why I'm asking

[empty grove]

The subspace is correct

The big space might need to be maps that take A into B

Subspace is maps that take all of X into B

[plain raven]

ohhh

interesting

[empty grove]

Ye I thought about it a little and you want the mapping space to be space of maps of pairs and subspace to be that

And the subspace should be too surprising because of analogy with pointed spaces

There you want the basepoint to be the map that maps all of X to the basepoint

Rather than just the basepoint

This tensor hom adjunction is probably just a restatement of the universal property you'd get from this being a pushout as the e-girl mentioned

[plain raven]

yeah i guess in retrospect the universal property doesn't matter. i was trying to understanding why it was important and whether there were common reoccurrences of this in other model catd

[empty grove]

It takes boundaries to boundaries

Like (I^n, ∂I^n) × (I^m, ∂I^m) = (I^(m+n), ∂I^(m+n))

And often we work with pairs where the subspace is the boundary

It also generalizes the smash product if you take the functor (X, A) ↦ X with the class of A as the basepoint

As in this functor becomes monoidal

From the category of pairs to Top_*

[plain raven]

yeah. that's a good observation

[thorny agate]

How important is Baire category stuff?

only asking since it came up in a proof (I was trying to find examples of products of normal spaces that aren't normal)

and I couldn't follow it and it used Baire and I've never run into it before

[languid patrol]

@thorny agate for abstract point set it is often very useful. It’s a good statement to spend a day or two seriously trying to understand it’s consequences, but maybe not too much more than that

Shows up in a lot of weird places e.g. p-adic geometry/analysis, theory of toplological vector spaces, etc.

[thorny agate]

hmmmmm

ok I'm not gonna look into it unless it comes up again

ok second question

What's a good example of a space which is the product of normal spaces but itself is not normal

OTHER than the Sorgenfrey square

[hidden crag]

why exclude that one

[thorny agate]

is there a proof that doesn't require Baire category stuff

only because I want to focus on learning more fundamental topology stuff (don't really have alot of time to do topology hence wanna focus on the essentials)

or maybe I'll just leave that as a todo in my notes

[untold lily]

baire category is a super powerful theorem

a lot of things I've seen proven with it I think you have to use Baire

[thorny agate]

hmmm ok

any good references?

Bredon doesn't mention it

[untold lily]

well, pretty much any other topology book should, like munkres has it I believe

willard obviously does

[trail charm]

lee ITM

[thorny agate]

I'll add it to the ever expanding list of TODOs in my notes

[untold lily]

you'll be forced to learn it if you ever wanna do functional analysis

[trail charm]

something something meager

[empty grove]

Are you having difficulty with the statement or the proof?

Or the applications

Or did you just see the word category and run

[thorny agate]

None of the above

I saw the statement used in a different proof and was wondering if it would be worth the time to learn it

And I guess it is

[fading vale]

Let f: M -> R be a morse function on R and take a critical point x of f and a morse chart U about x in which f locally looks like f(x) + Q(x). write x = (x-, x+) where x- and x+ are the components in the negative definite and positive definite parts of Q on f

[nimble portal]

This is the appropriate response

[gentle osprey]

Moth

Moth

[fading vale]

appropriate image

[white oxide]

Why does it have category in its name actually

[fading vale]

Can anyone pin down why that's true?

[urban zinc]

What's the boundary and interior of a 0-simplex? Math SE suggest that the boundary is empty and the interior is just the point, but is this standard?

[unreal stratus]

Well that's true topologically and I guess necessary for it to be true that every simplicial complex is (as a set) a disjoint union of insides of simplices

[urban zinc]

Wait why is it true topologically? The standard 0-simplex is a point in R right? So I feel like topologically the boundary should be the point and the interior should be empty

I get the second thing you said (it was one of the reasons I asked originally)

[umbral panther]

A simplex is a manifold with boundary. The zero simplex has empty boundary

[unreal stratus]

Okay sorry yes as a subset of R sure

But aren't the interiors etc defined such that they match up with how they look lol e.g. 1-simplex has boundary/interior corresponding to it being [0,1] in R

[urban zinc]

Oh wait that makes sense

Ty!

[empty grove]

Category appears here as the English word. Certain spaces are called "first category" or "second category" (if I recall correctly a second category set is one which is not a countable union of nowhere dense subsets) and Baire's theorem is a theorem about these (eg it says that complete metric spaces are second category)

Ye looked it up, first category sets are meager, second category the non-meager. We have better terms for these now so they see less use I suppose

[tulip bluff]

Reading an intriguing blog post about cohomology at https://grossack.site/2021/03/01/cohomology-intuitively.html, but I'm left wondering what on earth is meant by contradicting the intermediate value theorem when we are on a complex plane?

[white oxide]

True, but the name still slaps so...

[empty grove]

In the case for √i, consider the small circular arc connecting 1 to i centred at the origin. If Re √i < 0, then Re √z = 0 for some z on this path by the intermediate value theorem, but then z = √z² must be real which is a contradiction.

[quiet thorn]

Is category in baire category theorem related to category of category theory?

[novel acorn]

no

Baire sorted sets into two categories

the first category are today called meager sets

and the second category are non meager sets

[quiet thorn]

Yeah

Sad

[novel acorn]

(note this was long before the term category was a thing in math)

[quiet thorn]

I see

[white oxide]

Let's change it, someone make a ncatlab page about the very boring corresponding subcategories of the category of subsets of some top space and call it the first and second baire category associated to the top space /s

[late iron]

This is a bit late but for $\partial_+ U$, observe that this set is equivalent to $U(\epsilon, \eta) \cap {Q(x) = \epsilon}$.

Since $Q(x) = \epsilon$, we can write $$\epsilon = -||x_-||^2 + ||x_+||^2$$, then substituting it in place of $||x_+||^2$ in the inequality $$||x_-||^2 ||x_+||^2 \leq \eta(\epsilon + \eta)$$, which turns out to be equivalent to the inequality $||x_-||^2 \leq \eta$

[gentle osprey]

ScarletScorch

[late iron]

Just read the picture lol

In general, you can think of the 3 components of the boundary of U(e, \eta) as each maximizing / minimizing one component of U(e, \eta)

[late iron]

Recall that U(e, \eta) is defined using $$U(\epsilon, \eta) = {x \in \R^n | -e < Q(x) < e, |x_-| |x_+| \leq \eta(e + \eta) }$$

[gentle osprey]

ScarletScorch
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[late iron]

There are 3 inequalities here, and the equality at one of them corresponds to one of the boundary components

[late iron]

You can think of $\partial_+ U$ and $\partial_- U$ as the bounds on the sublevel sets of Q(x), and the third piece $\partial_0 U$ makes up the "integral curves" of the gradient of Q that connects the first two.

[gentle osprey]

ScarletScorch

[late iron]

Audin and Damian is a really terse book, good luck 😂

[fading vale]

I understand the construction of the positive and negative boundary components, I think. What I'm confused about is why the third component in question corresponds to the trajectories of the gradient of Q

[gentle osprey]

Moth

[echo oyster]

For two spaces to be homeomorphic, their dimensions must be same?

[gritty widget]

what notion of dimension do you have in mind?

[echo oyster]

The basics as R2 is 2 dimensional,R3 is 3

I haven't learnt about topolgical dimensions

[gritty widget]

well, i can tell you at least that no R^n and R^m are homeomorphic for distinct n and m

this is a consequence of the invariance of domain theorem

this also implies the statement that homeomorphic topological manifolds must have the same dimension

[echo oyster]

So homeomorphism implies same dimension, about the converse ?

[gritty widget]

what do you think about the converse?

[echo oyster]

Not true

[gritty widget]

why?

i agree with you

but let's see an example

[echo oyster]

Any (x,y) & [x,y]

[gritty widget]

what does that mean

oh intervals

[echo oyster]

Open & closed interval

[gritty widget]

that's an example

[echo oyster]

I was working on Rn & R1 if homeomorphic

[gritty widget]

you don't need heavy machinery such as invariance of domain for this one

think about something you can do to one but not to the other

[echo oyster]

I can't think of any

Both are connected

[gritty widget]

connectivity is a good thing to think about

they're both connected

[echo oyster]

Does it require compactness ?

[gritty widget]

R is easy to make not connected

[echo oyster]

R \ {0} not connected

[gritty widget]

true

can you do the same thing to R^n for n > 1 to disconnect it?

removing one point, that is

[echo oyster]

We can do it with path connected

[gritty widget]

"connected" and "path connected" are equivalent for all R^n so i'm not sure what you mean

[gritty widget]

can you?

[echo oyster]

I can remove the origin

[gritty widget]

does removing a point from R^n for n > 1 disconnect the space

you know it does for R

what about R^n for n > 1

[echo oyster]

I don't know

I'm trying to think

[gritty widget]

picture

take a point out of R^2. is it still connected?

[echo oyster]

No

[gritty widget]

why?

[echo oyster]

No it's connected

[gritty widget]

yes

[echo oyster]

Its the whole plane

Except the origin

[gritty widget]

the origin isn't special, any point will do

[echo oyster]

Yes true

So it works same for any n , Rn minus a point is connected

[gritty widget]

n > 1

[echo oyster]

Yes

[gritty widget]

but for n = 1 removing a point gives you a disconnected space

[echo oyster]

Yes understood

[gritty widget]

so why does this imply that R and R^n for n > 1 cannot be homeomorphic

[echo oyster]

Connectivity is topolgical invariant, we showed it for the spaces after removing point

So it's a restriction which is not homeomorphism

[gritty widget]

could you be more precise

[echo oyster]

I'll think about it, sorry need to go or I'll be late

[late iron]

Oh so if you solve the system of ODEs for the integral curves of Q(x) , they will be of the form ||x_-||^2 ||x_+||^2 = constant. If this helps, try to think of the example when you have a quadratic form of 2 variables.

[gritty widget]

backslash your norm brackets

so that discord doesn't think you want to spoiler whatever's inside

[late iron]

Ahhh

It works sort of

[gritty widget]

asterisks and underscores should also be backslashed

[late iron]

Ahh lemme see * _

Perfect

Thanks

[rough cedar]

wait wtf I actually thought this lol

[urban zinc]

😭

https://youtu.be/58B5dEJReQ8?t=1422 @rough cedar

This is so funny

[rough cedar]

lmaooo

he has a nice handwriting

wtf is his left brace compared to his right one though lol

also #category-theory message lol

[coarse night]

no that's αC_n ζ_n

[languid patrol]

syntax error

[thorny agate]

Drawing left braces is hell

[thorny agate]

Very confused at the construction of g_1

namely what is g_1(x) for x not in F?

relevant lemma

[languid patrol]

Something in between [0, 1/3], you don't get full control from Urysohn

[thorny agate]

what about for x in F but f(x) is in (1/3, 2/3)?

[languid patrol]

The closed set is $F \cap f^{-1}((-\infty, 1/3])$ and the $U$ is $f^{-1}((-\infty, 2/3))$

[gentle osprey]

Topos_Theory_E-Girl

[thorny agate]

ohhh

that was not explained at all

[languid patrol]

true!

[thorny agate]

this is just "Tietze Extension but apply it to each projection?"

or is there some subtlety I'm missing?

[unreal stratus]

sounds right

[thorny agate]

neato

is there a reason Tietze requires that we use R?

specifically I don't see where we need completeness

like wouldn't any ordered metric space suffice?

[unreal stratus]

the proof i know uses convergence of series

[thorny agate]

oh so does this one

ok yea there you go lol

[plain raven]

the real numbers are honestly dope

bless up

[urban zinc]

YAYYYYYYYYYYYYYYYYYYYYYYYYYYY

Good theorem

[thorny agate]

It seems like one of the theorems of all time

[urban zinc]

I love Urysohn

[thorny agate]

I'm struggling to prove that open subsets of locally compact sets are locally compact

Not really sure where to start tbh

[thorny agate]

nvm (this is an awful proof)

[gritty widget]

[neat stag]

does this mean $X$ and $X'$ are just some sets in the topologies respectively, or does it mean $\tau$ is a topology of $X$ and the same with $X'$? I think the statements are wrong with the first interpretation

[gentle osprey]

darylgolden

[gritty widget]

you should read it as: there's a set X and two topologies T and T' on it which we care about, and were going to write X for the topological space (X, T) and X' for the topological space (X, T')

[neat stag]

thanks!

[wicked yew]

how does the hint given help?

[unreal stratus]

Well you're meant to use i) to show that V(K,U) is open iirc

[wicked yew]

I think I solved it by just saying f(K) is compact in U so about every point in f(K) we can take a ball of radius epsilon for every y in f(K) which lies in U

[unreal stratus]

And then take ball radius epsilon about f right

[wicked yew]

yep

[unreal stratus]

Yeah so I remember doing this question and had like the same as you, lol

[unreal stratus]

I think whoever wrote this didn't expect you to use this

And they basically reproved it in a different way

But it certainly is in the notes

[wicked yew]

oh

[unreal stratus]

Given that this question is 10 marks though, I'd reprove this

[wicked yew]

can you give me a hint on how they wated me to prove this

[unreal stratus]

Yeah I remember the way I did it was reproving that and then it easily follows as you said lol

But yeah dw your way is probs ideal lol

Uhh

[unreal stratus]

So iirc you can define a function which does this for you

and use that

[unreal stratus]

Like assuming compactness you can prove this in a couple of ways and one uses like a function and lemmas about compact spaces and the other is working directly with open covers

P sure it's again slightly quicker just to work with open covers lmao

[wicked yew]

oh did the question want us to work with open covers to solve the prolem?

[unreal stratus]

iirc they did it using functions in some slightly odd way lol

There are solutions available for this year though p sure but ye

[wicked yew]

ah fair I'm not very good with convergence of functions in spaces so I'll check it out

Thanks for the help and gl with the rest of your exams

[unreal stratus]

thank

[urban zinc]

What is a linear map from the n-simplex to Euclidean space? I though the n-simplex wasn't a vector space?

[unreal stratus]

An n-simplex is naturally embedded in R^{n+1}

[urban zinc]

I see

So it's a map that extends to a linear map from R^{n+1} to Euclidean space

[plain raven]

yes

[urban zinc]

What exactly is a chain homotopy? I understand the equation that defines it, and I understand that chain homotopic chain maps induce the same homomorphism on the homology, but is there a more intuitive/geometric picture I should have for them?

[coarse night]

there is, let me find

[languid patrol]

An actual homotopy induces a chain homotopy

in the obvious way

[coarse night]

Keep this picture in mind

[urban zinc]

ohhh okay

ty!

Can you get an actual homotopy from every chain homotopy?

Or does it only work the one way

[coarse night]

by "actual homotopy" do you mean the one we have in topology?

[urban zinc]

Yeah

If you have two chain complexes made via chains of singular simplices and you have a chain homotopy, does it correspond to a homotopy between two continuous maps?

[coarse night]

then no, in a sense chain homotopy is more abstract and non unique. However every homotopy induces a chain homotopy

[urban zinc]

Ah okay

Ty!

Darn I got so excited when I finished reading the proof of the excision theorem and then I realized that I still have many more pages to read before I finish this section... back at it

The excision theorem is cool though! Tbh the details were interesting

[coarse night]

Just accept as axiom

[urban zinc]

The proof details were cool!!!!

this on the other hand... wtf

[coarse night]

(spoiler: you can actually do entire homology theory just using axioms, excision is an axiom there)

[urban zinc]

No... you can't do this to me... you can't spoil section 2.3 of Hatcher while I'm on 2.1!!!!!

[coarse night]

I just did

[white oxide]

Eilenberg-Steenrod pilled

[white oxide]

Warm-up for spectral sequence diagrams

[plain raven]

#math-discussion message

@urban zinc

[urban zinc]

WOAHHHH

[thorny agate]

How does this prove continuity in either direction

relevant proposition

[gritty widget]

a function f is continuous at a point x if and only if f(x_α) converges to f(x) for every net x_α converging to x

[thorny agate]

is it just "a function is continuous iff limit of f(x)_a = f(limit of x_a)"

ok

that's what I thought but wanted to make sure

[gritty widget]

you would hope it would hold for just sequences but it doesn't so we make nets to make it work

[thorny agate]

Here beta is the Stone-Cech compactification

How is mu(f) a real number?

[gentle osprey]

spamakin

[thorny agate]

so mu is a tuple of values between [0 ,1] as far as I can tell

[gritty widget]

[0, 1]^F is the same thing as functions from F to [0, 1]

[thorny agate]

ohhh that's how they're treating it

that would have been good for them to clarify since in the prior proof they use the exact same notation to mean a tuple 🙃

ok thanks

[plain raven]

cofibrations are a kind of good subspace embedding

are there any interesting model categories where an interesting class of cofibrations is closed under pullback along an interesting class of maps

i am aware that cofibrations are closed under pushout but that is not what i am asking

[urban zinc]

tuples and functions are the same

i.e. tuples indexed by X of elements of Y are the same as functions from X to Y

[thorny agate]

Ye that fact just slipped my mind in the moment

[tidal lynx]

but the iff holds if we use just sequences and not nets for functions R -> R… is the function being between metric spaces enough for this to hold?

[gritty widget]

it works for functions between metric spaces

try to prove that sequential continuity implies regular (i.e. open set) continuity and see what property of the space you need

do it for metric spaces and then try to rewrite the proof without referencing the metric explicitly. see if it generalizes

answer: ||sequential continuity at a point x implies regular continuity at x when the space admits a countable basis at x||

||when you have a metric d, a perfectly good countable basis at x is the set of open balls around x of radii 1, 1/2, 1/3, ...||

[tidal lynx]

wdym by “basis at x”

oh I see

[gritty widget]

a basis at x is a collection of open sets such that every open neighborhood of x contains one of them

[gritty widget]

https://en.wikipedia.org/wiki/Lazard's_universal_ring

[unreal stratus]

Informal group laws putting on suits to be more formal

[neat stag]

Reading munkres and trying to understand the difference between product and box topologies, am I correct to say that in $\mathbb{R}^{\omega}$, $(-1, 1) \times (-\frac{1}{2}, \frac{1}{2}) \times (-\frac{1}{3}, \frac13) \ldots$ is open in the box topology but not in the product topology?

[gentle osprey]

darylgolden

[next crystal]

yes

each (-1/n, 1/n) is open in R so its open in box
but the terms arent eventually all R so its not open in product

[fickle hamlet]

To show that product of 2 compact sets, (X and Y) is compact, can I just let C be a collection of open sets that cover the finite product, then each element x in X, there is an element S in C, where S =OxV, where O is open in X and V is open in Y and x is in O. We do this for every x in X and get a subset M of C whose union is XxZ, where Z is some subset of Y. then apply compactness with the subsets in X to get a finite subset of M_f of M. Do the same process with Y and get P_f a collection of open sets that is finite and union to RxY, where R is some subset of X. So then take the union of these two collections M_f and P_f and get a collection which covers XxY that is at most the size of M_f+size of P_f. Does this work?

[unreal stratus]

such S needn't be in C

But I think the idea is roughly correct hm

[tacit canopy]

Are the isometry groups of two metric spaces over the same set with the same induced topology, the same group?

[umbral panther]

No. Consider a metric space with three points. You could call this a triangle. Or you could look at triangles as subsets of the plane. Different triangles have different isometry groups. S3, Z/2, trivial

[tacit canopy]

What if the space is simply connected

Nevermind

[umbral panther]

A useful example is the Euclidean plane vs the hyperbolic plane

[tribal palm]

what’s a book that’s in style similar to bourbaki’s General Topology but actually digestible

[pseudo coral]

Idk about that text but the two that work for me are intro to topology by Gamelin & Greene and Topology by James Munkres these got me through point set topology

[plain raven]

kelley

?

just a guess

there's also Dugundji.

I don't know much about those books i'm just hoping they're similar in style to bourbaki. I would normally recommend munkres to a student

[abstract saffron]

Munkres is great tbh

I'd do the same

[tribal palm]

i’ll have a look at all of these

thank you peeps

[urban zinc]

Can anyone give a fun application of mapping cylinders/mapping cones

Just wondering if anyone has one ready

[bitter smelt]

Fundamental group of real line with two origins

See hatcher's solution here: https://mathoverflow.net/questions/25472/fundamental-group-of-the-line-with-the-double-origin

[plain raven]

The mapping cone can be viewed as a kind of homotopy-theoretic cokernel.

If f : X -> Y and C_f is the mapping cone of f, with i : Y -> C_f the canonical inclusion, then i\circ f is homotopic to a point - it is homotopically killing off the image of f, similarly to how you kill off generators for the fundamental group in a graph by adjoining 2-cells as in chapter 1.3 of Hatcher.

[bitter smelt]

(I leaned on the fun word here since this is practically worthless but neat)

[trail charm]

proving the dunce cap is contractible, see Bredon I.14 Problem 6

[plain raven]

The mapping cylinder M_f of a map f : X-> Y is important in part because M_f is homotopy equivalent to Y, and moreover, homotopy equivalent in a way that commutes with f and the inclusion i : X -> M_f.

Even if the map f is pathological, the inclusion X -> M_f will be a subspace embedding and probably a "good pair" in Hatcher's sense. Thus in some sense the map f :X -> Y is being replaced with a homotopy equivalent pair (M_f, X) where X is a subspace; now we can talk about its relative homology. H(M_f, X) = H(M_f/X) = H(C_f)

there are probably some technical caveats i'm omitting here.

[urban zinc]

Woahhh I like that!

[hidden crag]

what is a good pair in Hatchers sense

[urban zinc]

Oh wow that's cool

[urban zinc]

(X,Y) where Y is a nonempty closed set in X that has a neighborhood in X that deformation retracts to Y

[hidden crag]

ah

[plain raven]

It's similar to a cofibration, but not exactly the same.

[hidden crag]

ok nvm it's not equivalent to cofibration

yeah

[plain raven]

This was a fucking pain in the ass for me to figure out

[hidden crag]

that's what i was gonna mention

[plain raven]

There's a good stack overflow post about this which I posted here a few months ago

[urban zinc]

Oh huh, I see

[plain raven]

https://math.stackexchange.com/questions/854281/good-pair-vs-cofibration

[urban zinc]

ty clerk!

[hidden crag]

but being a cofibration is sufficient

[plain raven]

Well, ...

[hidden crag]

for the homology thing

don't mind me it's 3 am

[plain raven]

Ah, yes. For the homology thing.

[hidden crag]

my thoughts are not coherent

[plain raven]

I thought you meant cofibration is sufficient for good pair.

They're not equivalent in general but they both have the homology property yes.

[hidden crag]

no for the relative homology is reduced homology of quotient part

[obtuse meteor]

Oh this is so cool

[plain raven]

https://math.stackexchange.com/questions/3547820/neighborhood-deformation-retracts-vs-cofibrations/3550166#3550166

i'm only posting these because i don't want others to suffer the same headache as i went through

I thought they were the same for a hot minute

[urban zinc]

Oh wait Hatcher mentions this for spaces other than good pairs

Htilde(X U CA) ≈ H(X, A)

What's a cofibration?

[cosmic socket]

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

[urban zinc]

Oh God

[winged viper]

it's just a slight generalization of a good pair

the definition of a cofibration on wikipedia is equivalent to being an "NDR-pair"

[urban zinc]

Ahhh okay I see

[unreal stratus]

is good pair common language? or is it just from hatcher (or maybe it is popular now because of hatcher)

[trail charm]

is there a good "idea" behind the notion of a "saturated set?" i keep having to look back at the definition because i cant draw a connection between the word "saturated" and the definition itself

[unreal stratus]

Is this in the context of quotient spaces?

I guess the way I think about it is that if f:X -> X/~ is the quotient map then a subset of X is saturated iff it's a union of fibres of f (i.e. of the form f^-1(A))

So like there's a sense in which these are the most "full"/ complete subsets of X (in view of f), since otherwise you're missing out some elements of the fibres

[urban zinc]

Oh neat, a map from the sphere to itself with no fixed points is homotopic to the antipodal map

Omg I get to learn the hairy ball theorem

Wait wtf this is so cool

Just to check understanding, does z \mapsto z^n have degree n? For S^1 → S^1

[gritty widget]

it would suck if it didn't

[coarse night]

But doesn’t not

[trail charm]

yea

[trail charm]

ohhh this makes sense

[plain raven]

What Frank said is highly sensitive to precise choices of definition. Please read the stack exchange posts I linked just before

[gritty widget]

Hello people. I have a question: I have a complete and seperable metric space D (with metric d), and a specific subset S. I want to show that S has a compact closure. For this specific S, we can always find a set S_k that has a compact closure, st. for any x€S we can pick a y in S_k such that d(x,y) =< 1/k. Does it follow now that the closure of S is compact? If yes, why?

Unfortunately i have never really studied topology, but intuitively it sounds good. Currently im studying the Skorohod metric

[viral atlas]

Are there any conditions on the subset S? As I understand it, the claim shouldn't even be true. Take D to be the real numbers with their usual metric, and set S=D for a silly example.

[gritty widget]

Not really. But for the real numbers we cant find such S_k right?

[viral atlas]

Oh, okay, so S is constrained by the existence of such sets S_k as you described

[gritty widget]

Ah yes sorry, i wrote it poorly

[umbral panther]

Take the circle and delete one point. This is R with a weird metric. Set Sk = everything farther than 1/k from the deleted point

Oh, that isn’t complete

[umbral panther]

Is the infinite product of [0,1] in the box topology a (noncompact) complete metric space with Sk = product of first k factors?

No, I probably have that backwards. That’s probably the metric for the product topology and L-oo for box

[trail charm]

"Consider the directed set $D'$ consisting of ordered pairs $(\alpha, U)$ where $\alpha \in D$, $U$ is a nbhd of $x$, and $x_\alpha \in U$, ordered by the $D$ ordering and inclusion."

So is this ordering something like $(\alpha, U) \le_{D'} (\beta, V)$ iff $\alpha \le_D \beta$ and $U \subset V$?

[gentle osprey]

anamono for anamono 🍓

[trail charm]

actually it should be U \supset V

i think

yeah V \subset U makes sense because later he says (\gamma, U \cap V) \ge (\alpha, U)

oh man now im confusing myself lmao

wait no yeah (a, U) \ge (b, V) iff U \subset V

[coarse spade]

Can anyone help me out? I've got no clue how shapes can be described like "ABD^{-1}C^{-1}BACD"

Sorry I have no clue how to use the bot

These questions I've got derive the euler characteristics from the shape above and similar ones, X(M) = F - E + V and I'm completely lost

[gritty nest]

$ABD^{-1}C^{-1}BACD$

[gentle osprey]

bee [it/its]

[gritty nest]

just put dollars around the part that's latex

[umbral panther]

Yes, it is compact

First use the open cover characterization of compact to replace Sk by a finite set. Now take a infinite sequence in S. We want to prove that it has a sub sequence that converges in D. For each k, find a point in ask such that infinitely many elements of the sequence are in the 1/k neighborhood of the point and throw out the rest of the sequence. The sequence that remains after doing this the all k is Cauchy, so has a limit in D. (To prevent the sub sequence from being empty, there’s some trick. Maybe when we’ve done the first Sk, keep the first k elements and only throw out later ones)

[umbral panther]

What kind of shape? A curve or a surface?

[gritty widget]

Nice thanks!

[coarse spade]

A surface I believe, I don't understand how vertices can be found from pairwise identification and then the euler characteristic

[umbral panther]

It’s not the vertices, but the edges

Take an octagon. Label the edges ABDCBACD. Glue A to A. But which orientation? Inverse means to switch from your default connection

[coarse spade]

Ok but then whats up with finding the euler characteristic? Where does the V come from in X(M) = F - E + V

[umbral panther]

You start with an octagon. You glue some edges. This causes some vertices to be glued. The end of one A gets glued to one of the ends of another A. After you trace through the connections, you wind up with 1-4 vertices left

[coarse spade]

I've got this representation in a mark scheme of a question asking for the euler characteristic of the polygon $ABD^{-1}C^{-1}BACD$, how can V = 3 be found from this?

[gentle osprey]

hondo_ohnaka

[coarse spade]

( expert drawing )

I've seen that the number of vertices can be found by coutning equivalence classes but I can only see 2 classes here not 3 unless I've missed something

[umbral panther]

One class is the orange line. The other two classes are the triangles

[coarse spade]

Ok thats, that's clear now

Thanks*

topology is hard

[pseudo coral]

Specially algebraic you can visualise things but writing down the rigorous proof is a whole other beast

[pseudo coral]

Hi, for space that isn’t 2nd countable but Hausdorff can I use uncountably many copies of the reals

[empty grove]

Yes

[pseudo coral]

Cool thanks

[unreal stratus]

Discrete space on uncountably many points lol

[brittle rapids]

why are separable spaces important in constructive mathematics

[urban zinc]

Because uncountable things are scary, and countable things are nice

It's the same reason why the rationals are useful when talking about the reals

[brittle rapids]

what sense of useful?

i know practically nothing btw

[urban zinc]

the real numbers are full of nasty numbers which you can't explicitly construct, but the rational numbers are nice and you can fully describe each rational

Therefore it's helpful that the rationals are dense in the reals

[queen prism]

every countable subset of R has zero measure

[urban zinc]

It's a very similar situation with separable spaces in general

It's like studying something by looking at its skeleton

Or a tree by looking at its branches

You can't describe each leaf, but maybe describing its branches is enough

[brittle rapids]

thank you

so a bit like finitely generated things in algebra

[grim knot]

hey guys, I was trying to prove that, if we have a compact topological space X, then every closed set is compact. I was looking at these two proofs and I do not see, how this can be a covering of K. Do we not need to have an equality?

I guess that it is enough to have inclusin, or am I completely off topic?

[abstract furnace]

closed \implies compact isn't even true in \R^n

[grim knot]

the topological space has to be compact of course

[abstract furnace]

oh right misread

ok so let $X$ be a metric space and let $K\subset X$ be compact. take some closed $F\subset K$. you want to show $F$ is compact. So take any open cover ${U_\alpha}$ of $F$. Now, since $F$ is closed, it must be $X\setminus F$ is open, so ${X\setminus F}\cup {U_\alpha}$ covers $K$. But $K$ is compact so take a finite subcover, and since $F\subset K$, you've obtained a finite subcover

[coarse night]

U_alpha need not be open in X

Also why’s there K X Y F?

[abstract furnace]

xd

it shouldn't matter because F is compact in K iff it's compact in X

at least i think maybe that's just for metric spaces

[gentle osprey]

mim

[abstract furnace]

the same idea works though in a topological space you just have to use relative openness instead of openness

in answer to your question though an open cover of $V\subset X$ is a collection of open subsets $U_\alpha\subset X$ so that $$V\subset\bigcup_{\alpha\in A} U_\alpha$$

[gentle osprey]

mim

[abstract furnace]

that is, no you don't need it to be an equality @grim knot

and you can just remove F^c if it's still in your finite subcover afterwards if you want