point-set-topology

[unreal stratus]

Lol sounds like you're using the phrase 'topology of _' in two different ways

which topologies do you mean exactly, because if you mean those generated by open balls vs open rectangles then they are the same thing

[odd flame]

oh that's what munkres is saying

yeah that's what i meant

i can see visually why theyre the same thing ofc but why do we say theyre the same if they technically have different elements

[unreal stratus]

The topology don't have different elements

[odd flame]

or is that a silly thing to ask

[unreal stratus]

You're generating the same topology using two different bases

[odd flame]

hmmmm

[unreal stratus]

Well, what do you think 'being the same thing' means there?

[odd flame]

well i was thinking it meant having the same elements

[unreal stratus]

Perhaps 'it' is the problem

[odd flame]

but i think it might be better to think of it as covering the same space/set?

[unreal stratus]

The topology is the same whether you generate it with open balls or open rectangles

[odd flame]

i know covering might not be the most ideal word though

[golden gust]

you can make any rectangle out of balls, and any ball out of rectangles

[odd flame]

but exactly?

idk if it matters that theyre exactly the same though just contained right

[golden gust]

what do you mean exactly

[unreal stratus]

and by 'they'

[odd flame]

big fivehead moment on my part, but im imagining like interior rectangles approximating the area of a circle and vice versa

[golden gust]

yeah and because we're allowed arbitrary unions, you can union a bunch of interior rectangles to get an open disc

not something that roughly looks like a disc, an actual disc

[odd flame]

ahh that was it

https://tenor.com/view/patrick-star-dumb-gif-20952040

my b, thanks for patience

[unreal stratus]

These pointset topology questions are giving me flashbacks to recent exams

[odd flame]

im preparing for my fall semester topology class

definitely not avoiding studying for my algebra class too

[golden gust]

really? we must have taken different exams

[unreal stratus]

Well

Pointset topology reminds me of our topology course which reminds me of the exam

Even though the exam didn't end up being pointset

[golden gust]

biggest bait of 2022

[unreal stratus]

Yeah I kinda dissociated

Idk

Anyway

[gritty widget]

Overkill proof is that any two norms on R^n are equivalent. Now consider the Euclidean norm and supremum norm

[cobalt sonnet]

that's not really overkill

It's equivalent to what he wants to prove isn't it

[livid narwhal]

Super niche thing I'm curious if anyone has any resources for. Is there a unique representation up to rigid transformations for a complete distance matrix on the surface of a flat torus?

So example I've got a mapping of a flat torus to a square with wrapping edges, I've got the complete distance matrix (along the surface), is there an arrangement of 4 points that has the same distance matrix but cannot be mapped onto the others by a rigid transform?

[odd flame]

are the elements of a subbasis open sets in the topology it generates? no right

that’s probably phrased weird

[unreal stratus]

What's the definition of the topology generated by the subbasis?

[odd flame]

unions of finite intersections of elements of the subbasis

oh

[unreal stratus]

Yeah

[odd flame]

i guess intersect an element w itself right

[unreal stratus]

Well just intersection of a single set works fine too

[odd flame]

very general but what’s the point of a subbasis

[unreal stratus]

So one useful thing is you might want to have the minimal topology such that e.g. certain functions are continuous, for example

Then you can just consider those sets that have to be open and let that generate the topology

For example if $(X_i){i \in I}$ is some collection of spaces, if we want to endow $X = \prod{i \in I} X_i$ with the smallest topology such that the projections $\pi_i: X \to X_i$ are continuous we can just consider the subbasis given by $\bigcup_{i \in I} {\pi_i^{-1}(U) \mid U \text{ open in } X_i}$

(since we require those sets to be open)

[gentle osprey]

potato

[unreal stratus]

Then in fact this produces the standard product topology

There are also other theorems which involve playing around w subbases

[odd flame]

ok sorry i’ll come back to that in a sec but i was reading about the order topology - what does it mean to say that “open rays form a subbasis for the order topology”

moreso the “for” part, bc i was seeing bases and subbases as separate things

[unreal stratus]

S is a basis (resp. subbasis) for a topology T if S generates T (as a basis resp. subbasis)

[odd flame]

resp.?

[coarse night]

respectively

[gaunt linden]

The naming is somewhat confusing because a basis for T is always a subbasis for T, but not the other way around.

[golden gust]

honestly I'd never heard the word subbasis until a couple days ago. as shin said it's just an open cover

[gaunt linden]

No.

{R} is an open cover for R with the usual topology, but definitely not a subbasis for it.

[unreal stratus]

Shin said cover and that was talking about when you don't already have a topology on the set (or at least not in reference to that)

[golden gust]

oh my bad

[gaunt linden]

Wikipedia states that books disagree about whether a subbasis should be required to cover the entire space in addition to generating the topology,

[gritty widget]

huh, never seen anyone to require that the subbasis covers the whole space

[gaunt linden]

Apparently it's about whether you dare to appeal to an empty intersection in order to conclude that the entire space is in the topology generated by the subbase.

[gritty widget]

I see. You could argue that we can define subbasis as any family generating the topology of our space, in the sense that it's the smallest such family.
Imo that's more natural, you always define "closure" of something, and then try to describe this "closure"
So just alternatively define the topology to be arbitrary unions of finite intersections of sets from the basis and the whole space

If you're scared of empty intersections, that is

otherwise it's just an artificial requirement that it covers the whole space

closure not in the topology sense, but in like, something generated by something

like in https://en.wikipedia.org/wiki/Closure_operator

[empty kettle]

I am somewhat not great at topology. Suppose there is a torus which is modified so that there is a tube connecting the torus to itself. How would you go about classifying it.

My guess is its the connected sum of the torus and another shape however i am not sure what that tube would be

[ivory dragon]

the 2-torus?

i.e. the connected sum of two tori

assuming you mean "classifying it up to homeomorphism"

(which is usually what is meant when talking about classifying closed surfaces, but your wording makes it sound like you don't know these definitions)

[golden gust]

in general, adding a "handle" is the same as connected summing with a torus

[empty kettle]

Ok thank you!! @golden gust could you look at my attempt

The figure is two Klein bottles connected to two torus with two open disks removed. I need to identify the type of shape. My guess is that the disks can be homeomorphically moved to create a "hole" with the hole in the Klein bottle resulting in a 4 torus.

This is my attempt for this figure

[golden gust]

the klein bottle is not orientable, so the final surface shouldn't be orientable, and in particular can't be the 4-torus

do you know the classification of closed surfaces?

[empty kettle]

They have to be either a torus or projective plane

[empty kettle]

Tbh my prof didn’t have a great explainatiin and I’m kinda piecemailing it together

[golden gust]

yeah, either the connected sum of tori or the connected sum of projective planes

[empty kettle]

So how would I see this turned into projective planes

[golden gust]

the relation $\mathbb{RP}^2 # \mathbb{T}^2 \cong \mathbb{RP}^2 # \mathbb{RP}^2 # \mathbb{RP}^2$ might be useful

have you seen this before? would you know how to prove it?

[gentle osprey]

Average J∘du=du∘j enjoyer

[empty kettle]

We didn’t really do proofs for this. I’m good with point set proofs

[golden gust]

but would you be allowed to use this statement in your solution?

[empty kettle]

Yep I know that one haha

[golden gust]

(use \#)

[empty kettle]

Yes

[golden gust]

yeah and $\mathbb{K} \cong \mathbb{RP}^2 # \mathbb{RP}^2$

[gentle osprey]

Average J∘du=du∘j enjoyer

[empty kettle]

How do I handle the removed disks

I remember that connected sim proof!

Thank you so much btw. If you need any analysis help lmk

[golden gust]

you can just say the final surface is "<whatever the connected sum of surfaces is> minus two open discs"

like, they don't really interact with the connected sums at all

[empty kettle]

Is that acceptable?

Bc the question only wants me to be like Torus sphere or projective plane

Or will the cancelation then allow what the removed disks to make sense?

[golden gust]

if you remove two discs then the boundary of your surface is the boundary of those discs. but the torus and projective plane are both closed (they have no boundary) so the final shape cannot be a torus or projective plane

can you send the question?

[empty kettle]

Yeah sure one sec

this is the question

@golden gust for the first part i said it is an 8-genus torus

[golden gust]

oh I see, the boundaries are identified. my bad I didn't realise

[empty kettle]

How does that change it?

[golden gust]

because then the surface is closed

[empty kettle]

Ok I see

[golden gust]

semi-important note: the orientations of the circles are such that you are connect-summing with a Klein bottle, not a torus (these are a priori different, but it turns out that since your surface is already non-orientable, it's the same. still, important to keep in mind)

[empty kettle]

so its 4 Klein bottles put together

[golden gust]

because of the circle things, the final shape is actually 5 klein bottles

[empty kettle]

Where does the 5th Klein bottle come from

I see the 2 main ones. Then the two holes result in two more?

[golden gust]

I'm sorry I misunderstood (but luckily it doesn't change the answer). I thought that we were gluing the two circles together, but actually we're just identifying the antipodal points of each one individually

yeah ok so

each of these circle-things is the same as connect-summing with a single RP^2 (this is called "adding a crosscap", if you've heard of it)

[empty kettle]

Ok so its basically $\mathbb{RK}#\mathbb{RK}#\mathbb{RP}^2#\mathbb{RP}^2$

[gentle osprey]

Lord Mind

[empty kettle]

Then I substitute in $\mathbb{RK}=\mathbb{RP}^2#\mathbb{RP}^2$

[gentle osprey]

Lord Mind

[empty kettle]

So the final answer is a 6-projective plane?

[golden gust]

you forgot about these two holes

i.e. connected sum of two tori

you need to add those in too

[empty kettle]

Ok so it is summed with the tori

Ok let me try again!

[golden gust]

yeah

[empty kettle]

I think im getting the hang of it

[golden gust]

you're nearly there

[empty kettle]

$\mathbb{RK}#\mathbb{RK}#\mathbb{RP}^2#\mathbb{RP}^2#\mathbb{T}#\mathbb{T}$

[gentle osprey]

Lord Mind

[empty kettle]

So basically it is 2 + 2 + 3 + 3 = 8 projective plane?

[golden gust]

um

2 + 2 + 3 + 3 = 10

but yes

[empty kettle]

oh lol

[unreal stratus]

Mod 2 though

[empty kettle]

yes

[golden gust]

yeah exactly

[empty kettle]

whats the mod 2 for?

[golden gust]

or in other words, this

[empty kettle]

ok

Could you check if I am right that the other question it is an 8 genus tori

[golden gust]

can you explain how you got 8?

[empty kettle]

Sure, so there is currently a 6 tori with a hole in the middle and a tube. As discuss we can move that tube to become another torus. And the center hole can be shrunk to be another torus

[golden gust]

sounds good to me

[empty kettle]

Ok yay!! I have another question to do but I am going to try to do it myself!!!

Thanks so much!!!

[empty kettle]

Ig I am confused on how the loops can be orientation preserving when the Klein bottle is non orientable

[golden gust]

orientability is a global property

you can imagine that a tiny loop can be orientation preserving if it stays within a small open set homeomorphic to R^2

[gaunt linden]

Is it just me, or is C two separate loops?

[golden gust]

looks like it lol

[empty kettle]

C is

I got two mobius stripes for that one

@golden gust could you explain how I could figure out if something is orientation preserving or reversing

[empty kettle]

Also I was able to successful do the word question yay!

[golden gust]

this is the picture

follow an oriented disc around the loop, and if it has a different orientation on its way back, the loop is not orientation preserving

[empty kettle]

Ok that makes sense. @golden gust for the 1st on how do I glue it after I make the cuts. My idea is to create a new square but that seems wrong

[gritty widget]

stuck on this topology problem

the closure of A is the union of A and its limit points

i need to show that the closure of A is equal to the intersection of all closed sets containing A

where are you stuck

for one direction i have two cases

that an element x is in A or x is in the set of limit points

im stuck on the limit points part

i assume some j in I and by contradiction that x is not in V_j

so x in A/V_j

[empty kettle]

@gritty widget for any closed sets containing A it must contain the limit points of A

So the intersection must have the closure of A in it

[empty kettle]

@golden gust @gaunt linden is this the way to do it?

[gaunt linden]

The upper and lower halves of the vertical sides are not interchanged by the gluing!

[empty kettle]

What does that mean sorry

[gaunt linden]

If I understand your symbolism right, your drawing says that the upper half of the right side should be matched with the lower half of the left side.

[empty kettle]

I’m trying to do A on this

[gaunt linden]

But it really should be: upper half of left side to upper half of right side.

Yes, I recognize it as A.

[empty kettle]

Ok so flip those?

[gaunt linden]

But the sides you have marked up as c and d have been matched up wrongly.

[gaunt linden]

Yes.

[empty kettle]

So this labeling?

[gaunt linden]

The lines with with three red chevrons are cuts. They become boundaries after cutting, not something to glue together again.

(And you forgot to cut the lower one).

[empty kettle]

Ok let me try this again

Lol I thought I didn’t have to

[gaunt linden]

Yes. This should now be recognizable as a well-known surface.

[empty kettle]

So like that

@gaunt linden is this right?

[gaunt linden]

Again the edges you have labeled as c are cuts. They are the boundary of your new manifold and should not be glued together again.

[empty kettle]

Ok so I can’t do words

[gaunt linden]

I'm not familiar enough with alg topo that the words thing makes sense to me in the first place, but that's probably just me.

[empty kettle]

Ok so without using words what do I do

Bc I’m just confused what that shape is

Or is still Klein?

[gaunt linden]

A rectangle with two opposite edges left unglued and the others glued together with the orientation flipped ...

[empty kettle]

Möbius strip?

[gaunt linden]

Yes.

[empty kettle]

Ok yay!!

Let me try b

@gaunt linden is this right?

[gaunt linden]

Yeah.

[empty kettle]

So it turns into two surfaces?

Ok yay I’m getting somewhere @gaunt linden

[gaunt linden]

Looks like it.

(That's not surprising in itself -- if you cut along a loop in a plain old sphere, it will fall into two parts too).

[formal tide]

Is there any work on n-dimensional paths? The homotopy groups are defined in terms of maps from S^n, but I think they could also be defined as closed n-dimensional paths, i,e, maps from D^n somehow

[surreal lantern]

They can be defined as maps from I^k (k- times the cross product of the unit interval) into your space (X,x_0) such that the boundary of I^k is mapped to the basepoint x_0

(that would be the k-th homotopy group)

Now from this you get to your definition over S^n by collapsing I^k and the boundary of I^k along the boundary itself

collapsing I^k along the boundary gives you S^k and collapsing the boundary gives you a point that has to be mapped to the basepoint

so you can also think of maps from (S^n, s_0) in terms of maps from (I^k, del I^k) here

this is pretty chaotic but i hope it makes sense

[formal tide]

thanks, I do get it! wondering how it'd work for other paths though, not just the closed ones

so for example, I could have a usual homotopy H between to paths p1 and p2 (with same endpoints)

The (image of) the homotopy H is homeomorphic to I^2

if I had another homotopy G between to paths p1 and p2, it'd also be homeomorphic to I^2

[surreal lantern]

I don't think isomorphic is the right word here

[formal tide]

I want to say what a path between H and G is

oops homeomorphic*

too much algebra (;p

[unreal stratus]

isomorphic in Top

[formal tide]

^^

[surreal lantern]

hmm to be honest im not quite sure what you're getting at shiranai

[formal tide]

I'm basically trying to formally state this part of the HoTT book

sorry for long wall but I can't seem to explain it well myself

I want to rigorously define what a "3-dimensional homotopy between homotopies" is

(third line of 2nd pic)

[surreal lantern]

I'm not familiar with some of the terms in there

but a homotopy between two dimensional homotopies sounds like a map I^2 x [0,1] into your space that preserves the basepoints to me

making it just a map from I^3 into your space

those classes are maps are used to define the third homotopy group of a space

Maybe look into higher homotopy groups, i might be missing your point tho, if that's the case im sorry

[formal tide]

np, thanks timo; the higher homotopy groups are the equivalence classes of the n-dimensional paths, as homotopies of the (n-1)-dimensional constant paths over the base

hmm I think I'm getting close to what I need, thanks again

[surreal lantern]

You're welcome. I think it will be helpful to see elements of those groups as maps from (I^k, del I^k), might make that "recursive" relation easier to formulate

[shadow charm]

Why do they stress on semi locally connected here? Every point x has arbitrarily small neighborhoods U that admit lifts projecting homeomorphically to U by p if I’m not mistaken, so the same argument could be used to say that a necessary condition is that X be locally simply connected no (which is stronger)?

[empty kettle]

Im still confused on the orientation reversing curves. My guess is B,C are because they basically show that the shape contains a mobius band. Not sure on A tho

[flint sinew]

I remember someone saying that topologies (or was it top. spaces) are the among the most fundamental mathematical objects there are and that many other things/objects/structures can be described as topologies. Is this right?

[surreal lantern]

That's kind of a vague statement

It does generalise a lot of terms from other subjects like analysis tho for example

[sturdy notch]

This is something that bothered me too. First off, locally simply connected is not stronger than semiloc simply connected as per wikipedia. For a universal covering p:E->X, the ley thing to notice is that for some trivializing open set U , i the inclusion of U in X and j an inclusion into E we have pi1(i)=pi1(p)opi1(j) where the lhs is just the 0 map as pi1E is trivial.
Thus for a universal covering to exist pi1(i) must always be trivial which coincides with the definition of semiloc simply connected. I dont have any good intuitive reason for why this maut be the case though if i recall correctly the condition does appear somewhat naturally in the construction of the universal cover.

Tbf though these are somewhat technical conditions which are satisfied for most(if not all) reasonable spaces, my prof (and presumably your future prof too ;)) didn’t stress this too much

Oh abd for a space which is not locally simply connected but is semiloc simply connected the cone over hawaian earring works (as per wikipedia)

[shadow charm]

Yeah so locally simply connected is stronger then

I can’t quite understand what you’re saying with the universal covering quite yet but I’ll give it another look later thanks

[surreal lantern]

cone over hawaian earring

[shadow charm]

This example was in Hatcher too

[surreal lantern]

sounds funny

[sturdy notch]

Oops read that wrong sorry, im tired. Yeah i dont know the order in which hatcher does things, if you take the definition for granted it should appear rather naturally afterwards

[shadow charm]

See the thing is in principle I agree with you (if I understood what you were saying with the universal cover correctly) I just don’t get why you have to restrict yourself to semiloc

Since you can take U arbitrarily small if im not mistaken

[sturdy notch]

Well locally simply connected is a stronger condition so youre not restricting yourself by looking at semiloc sc

[shadow charm]

No but I mean saying semiloc is necessary is weaker than saying loc is necessary

So if you can show loc is necessary why bother with semiloc

So I’m guessing my argument fails somewhere

Just not sure where

[sturdy notch]

You should soon see why semiloc is the right condition for covering spaces, loc is not a necessary condition for the exostence of universal covers

Ie the cone over the hawaian earring admits a universal cover (itself) but isnt loc

[shadow charm]

Oh I think I see where my argument fails

You can have arbitrarily small neighborhoods U such that pi(U,x)->pi(X,x) is trivial but that doesn’t imply the neighborhoods themselves are simply connected

Okay that makes more sense

Thanks for the help 👌

[sturdy notch]

Yes

[sturdy notch]

Np hope i didnt confuse you as much as i just confused myself

[shadow charm]

Lol relatable

[empty kettle]

Could someone walk me through how to tell if a curve is orientation reversing

[broken nacelle]

@crude vector

[grave maple]

Anyone familiar with Foundations of Topology by Preuss?

[stark creek]

Favorite Anime? (This is allowed according to the channel description):

[gritty widget]

it will probably be chainsaw man whenever mappa finally releases the adaptation, but for now i think that attack on titan takes that spot

i have generic anime taste

it was beastars s1 until the recent season of aot dropped

[gritty widget]

I can see that

i don't watch a lot of anime to begin with

odd taxi was really good

[gritty widget]

is this the part where you tell me to watch whatever obscure late 90s/early 2000s "art house" almost-hentai deconstruction anime film you have on your mind?

No. But I admire your creativity

Mushishi is really good one

[gritty widget]

Is $I_+\wedge X \cong I \times X$?

[gentle osprey]

lime_soup

[empty grove]

Not quite, the left side collapses x × I where x is the basepoint of X

I × x rather

So a homotopy from X to Y in Top_* is exactly a function I_+ ∧ X → Y

Smash replaces product, and you add disjoint basepoints to specific spaces

[gritty widget]

yes thank you

another thing I am confused on

if X is some pointed space, the reduced cohomology of agrees with the regular cohomology everywhere except degree 0

If (X,A) is a good pair then H(X,A)=H(X/A)

the reduced cohomology of X is H(X,x_0)

if the base point of X is nondegenrate then (X,x_0) is a good pair

So this should say that for spaces with non degnerate base points

the reduced cohomology is always the cohomology of X/x_0

but now where I am confused is I think X/x_0 is just X? so this should mean that the reduced cohomology and regular cohomology agree in every degree

ah

[gritty widget]

this one also requires degree greater than zero

[empty grove]

Ye the left side is unreduced the right is reduced

[cobalt sonnet]

@stark creek Death Note is a good one

[odd flame]

depending on the day either the original dragon ball (the tournament sagas in particular) or samurai champloo

in any case maybe a weird question but i left a note to myself in my notes asking "on what sets is it useful to create an order topology"

so uhhhh... on what sets is it "useful" to create an order topology

[gritty widget]

Any ordinal. R. Long line iirc

Hopefully this will cease your thirst for examples

[odd flame]

https://tenor.com/view/miku-insatiable-hunger-gif-25085127

ngl i dont know what past me meant by useful but i think i get it anyways

how can we talk about an open set of X if we dont necessarily know what the topology on X looks like

[gritty widget]

"let X be a topological space"

[odd flame]

but X can have multiple possible topologies

[gritty widget]

"let X be a topological space" presumes we have picked one and are talking about it

[odd flame]

unless this lemma is more about "given a topological space with this particular topology, we want to check if a collection is a basis"

i'll assume we kinda said the same thing there so merci

[surreal lantern]

A topological space is a tupel (X, tau)

with a base set X and a topology tau

[gritty widget]

the phrase "topological space" does not mean "set which can be given a topology", but "set which is given a topology"

[surreal lantern]

so when speaking about a topological space the topology is already determined

[odd flame]

understood, probably shouldve realize that sooner

so if i say consider the topological space R, it's assumed that that's (R, T) with the obv ordering topology on it right

and not "R with some topology"

[surreal lantern]

I mean

you usually just say consider R with it's standard topology or something along the lines of that

[odd flame]

or i guess talking about that would really be consider the topological space R with topology _______ right

yeah ok

im fixating on language lol i'll stop thinking too hard abt it

[surreal lantern]

[plain raven]

a monotonic function between posets is continuous wrt the order topology if it preserves... ?

i think directed suprema but i could be wrong

but yeah like my guess would be to give an order-theoretic definition for the continuous monotonic functions

and that's when the topology is useful

when you care about monotonic functions satisfying that extra hypothesis

[swift fjord]

Where's nikita when you need him

[plain raven]

Ha!

[rancid umbra]

if i have a collection of indexed top spaces X_i each with a base B_i for the topology on X_i, is there a way to construct a base for the disjoint union of the X_i’s using the B_i’s?

something like every disjoint union you can make with each set from the B_i’s if that makes sense?

[cedar pebble]

yes

[flint cove]

Does anyone have an overview of how homotopy equivalence interplays with the separation axioms?
Since any indiscrete space is contractible, this suggests to me the following particular questions:
✅ Is any space homotopy equivalent to its T₀ quotient?
❔ Is any T₀ space homotopy equivalent to its T₁ quotient?
❔ Is any T₁ space homotopy equivalent to its T₂ quotient?

[rancid umbra]

do u mind elaborating?

[flint cove]

Homotopy equivalence vs. separation properties

[cedar pebble]

if you have bases B_i for X_i then the union of the B_i's should be a basis for the disjoint union of the X_i's

[rancid umbra]

oh shoot, i thought i was going to have to do some weird indexing and set manipulations to get all the required properties. thanks

[unreal stratus]

I think you probably need to close under finite intersections though too right?

[gritty widget]

When they say "union of B_i" they mean union of their images under the canonical map from X_i to their disjoint union

You don't need for them to be closed under finite intersections because they're disjoint if from different indexes

Slight neglect but a forgivable one

[unreal stratus]

True sorry yes i was being silly

[gritty widget]

This is from fomenkos homotopical topology

I can't see what the "obvious" homomorphism should be

[swift fjord]

Personally I wouldn't forgive it

[pearl holly]

I know that if E is a ring spectrum then E^* is a multiplicative cohomology theory. Does the converse also hold?

[odd flame]

reading munkres' section on a subspace topology, i'll assume here by subspace he really just means subset? or is using the word subspace already implying that it's a subset as a topological space

idk if i phrased that well

[gritty widget]

He means that Y has subspace topology

[odd flame]

[manic garden]

a multiplicative cohomology theory gives rise to a ring spectrum (up to phantoms) via brown representability

but you often can't really say anything about how structured the multiplication is

[pearl holly]

okay I see

[pearl holly]

but how is this true?

[odd flame]

here does he mean the same as in both R and Y have an order topology? so like the same kind of topology as opposed to the same open sets?

[gritty widget]

"its subspace topology (as a subspace of R) and its order topology"

[gritty widget]

he means that for [0, 1], the subspace topology and order topologies coincide

the open sets in the subspace topology for Y come from R, but they are not the open sets in R

they are the open sets in R intersected with Y

[odd flame]

ohhhh it means order topology on Y

language tripping me up a lot lately for no reason

thanks yall

[vague brook]

Why can't you use van kampen on this. Intersection of every spheres is just (0,0,0) which is path connected, fundamental group of spheres is trivial, so by van kampen there is a surjection from free product of trivial groups to fundamental group of space, or fundamental group is trivial, or simply connected

math stack exchange ppl said there was no way they could see of doing it so idk whats wrong with this

[gentle osprey]

c squared

[rancid umbra]

if each of the X_i are hausdorff then the disjoint union is also hausdorff

if we pick distinct points a and b in the wedge sum, then there are essentially two cases: a and b are both singletons, or one of a or b is equal to {p_i}_{i\in I}

[cerulean oriole]

I think that's for the Scott topology, I happened to read about it a bit recently.

[cerulean oriole]

++++++

[cerulean oriole]

Also I think what's usually called order topology is only applicable to tosets.

[odd flame]

mfw my pdf is just randomly missing a page

[vague brook]

lol

[gritty widget]

what do you think a neighborhood basis of a point should look like

what are the simplest open sets in a metric space you could possibly use?

no, you can't just use every open set

here's a more precise hint: think about open balls

[gritty widget]

that is what i meant by this, which you didn't answer

[rancid umbra]

try drawing concentric circles

there might not be a smallest open ball

not necessarily

[rancid umbra]

just bumping. having a pretty hard time with this. any suggestions would be awesome

[unreal stratus]

So if neither of the points is the point of gluing then it should be fairly straightforward (if both in the same X_i just take open nbhds not touching the identified points ; try to fill in the same where they're in different ones)

If one point p is that basepoint and the other is xbin, say X_i, then we can pick a nbhd of x and a nbhd of (the preimage of) p in X_i which don't intersect and then extend its image in the wedge into an open set about p

That's informal but meant to be more a hint - I think it's instructive to, say, consider how to prove it for a wedge of two circles

That gives the geometric motivatiom

[rancid umbra]

how do i know that the nbhds will be open when i take them back through q?

is q an open map?

[rancid umbra]

geometrically i can see this for the wedge of two circles (i think). i’ll try writing it out more explicitly tho

[gritty widget]

Note that $\bigcup {p_i}$ is closed

[gentle osprey]

Blitz

[unreal stratus]

q isn't necessarily open no, but in this case the neighbourhoods can be completely contained within (the copy of) one of the spaces and are saturated

[rancid umbra]

okay, thanks guys, will try something with this

[gentle osprey]

Shaw

[unreal stratus]

If you replace open with closed everywhere that works aha

A isn't necessarily open in X

[grave maple]

It might be easier to argue using filters since you're dealing with closed spaces.

[paper wedge]

is open in A which is open in X is wrong

if u picture it

u got a space X decopmosed into two parts and then you have Y which just assume its one part

u want the inverse image so like a set of Y sent to X

X is two parts so like either A or B

so take a set E subset of Y

f^-1(E) must be either in A or in B

[paper wedge]

is that true?

@gritty widget

yup

now

f|A^-1(U) = f^-1(U) intersect A

now it should work ?

try

using

the fact that A and B are closed

and continuous pre of closed is also closed

but yea ur right

so to finish this up

open subsets of A is open in X

f^-1|A(U) is open in A so..

do the same for the other part

and then just union both of them

do u get it

yea

now rewrite the proof

like fully

A is closed in X given

actually yea ig those are clopen sets

cuz like X is decomposed

ye

open subsets of A are open in X

[unreal stratus]

I mean just swapping open for closed works too

[paper wedge]

closed actually*

yea

[unreal stratus]

Idk where you got clopen from tho

[paper wedge]

just thought about connectedness idk

[unreal stratus]

A and B needn't be complements

[paper wedge]

yea

yea yea my bad

rusty

[gritty widget]

[empty grove]

A and B could overlap

[unreal stratus]

It wouldn't contradict anything tho

[surreal lantern]

You're gluing the disk together along its boundary

hmm

could be a tedious homeo to define explicitly

but then again i think you get exactly S^n and not just something homeo to it when doing that

maybe thats of help?

but i guess jumping between dimensions like that doesn't work formally

[cerulean oriole]

I think a projection to a closed hemisphere (boundary only) followed by stretching it to bring the boundary to a single point would give you a map D^n -> S^nwhich is a homeomorphism of D^n/S^{n-1} with S^n

Something like (x,y) -> (x,y,sqrt(1-x^2-y^2)) composed with (x,y,z) -> (k(z)x,k(z)y,(z-1)/2) where k(z) is chosen to keep the result on the sphere

[swift fjord]

It's not 'exactly' S^n if you're defining S^n as the set of points of norm 1 in R^n

[surreal lantern]

yes thats what i meant with the msg after that

[gritty widget]

with any kind of norm on R^n

continuity

your function is not continuous, and a retraction must be

thinking about connectedness

"map" should be read as "continuous map"

sometimes people have a convention that they don't write that maps are continuous

maybe it's one of such authors

[grave maple]

Anyone familiar with limit spaces?

[gritty widget]

Like convergence spaces?

Or like inverse limit maybe?

[grave maple]

Like convergence spaces.

I'm trying to nail down what the sup of a set of limit structures is, but every definition I try leads to problems.

I'm consulting Foundations of Topology by Preuss, but he doesn't define the sup explicitly.

[plain raven]

Fix a topological group G
i'm interested in the category of spaces equipped with a G-action, continuous maps respecting the G-action, G-equivariant homotopy
I googled this a bit ago and google seemed to suggest that the right definition of 'contractible' in this setting is that
a G-space is contractible if the identity map admits a G-equivariant homotopy into a single orbit of the G action
So sure i'll just accept that this is a good definition.
My question is, is there an analogue of the path space fibration in this setting like
if (X,x) is an ordinary pointed space there's a based path fibration P(X,x) over X
the space of continuous maps (I,0)_->(X,x)
and this fibration admits a section iff X is contractible, more generally a map (Y,y) -> (X,x) is homotopic to a point iff it lifts along this fibration.
My question is like
is there a G-equivariant analogue to the path fibration with the same properties?
I guess like, precisely, take (X, A) to be a pair of a G-space X and a subspace A consisting of a single G-orbit.
Is there a G-fibration over (X,A) such that for any (Y,B) and f : (Y,B) -> (X,A), f lifts along the fibration iff f is contractible to a map (A,A) by a G-equivariant homotopy

[vague brook]

f:[0,1]->S^n t-> sqrt(1-t^2)*x+ty where x and y are points on S^n shows its path connected right

[gritty widget]

sqrt(1-t^2)x+ty doesn't need to be a point of S^n

[vague brook]

oh yeah

guess im assuming they are orthogonal

<ax+by,ax+by> = 1 a^2+b^2 + 2ab<x,y> = 1 , b^2+2ab<x,y>+a^2 -1= b =[-2a<x,y>+-sqrt(4a^2<x,y>^2-4a^2+4)]/[2], which is a continuous function of a as square root is positive, and ax+by will be a member of S^n. Taking a as parameter yields result? ???

basically do quadratic formula to find suitable value of b

given a between 0 and 1

[gritty widget]

Yeah, I think it works

the denominator should be different

[vague brook]

thanks

[cursive flume]

is being homotopic the same as being homotopic relative to the empty subset?

[gritty widget]

yeah

[cursive flume]

So its an iff?

f_1 homotopic f_2 iff f_1 homotopic f_2 relative empty

[coarse night]

yes

[regal mirage]

can someone explain lifting the path with example?

p:X->B, f:A->B maps, there is g:A->X covering f or lifting if p*g=f

is it like composition of functions in math analysis?

[gritty widget]

p*g = g \circ p other way around

it is literally composition

[swift fjord]

p circ g

But yea

It's not like, it just is

[gritty widget]

oops

i saw the star and thought pullback

bad notation!

[swift fjord]

Fair enough lmao

I think they say someone write pg

And thought multiplication

[gritty widget]

p(o)g

[regal mirage]

yes last one

so task is to find g somehow?

[gritty widget]

yes

[rancid umbra]

https://tenor.com/view/do-something-meme-gif-8007840

[cursive flume]

any hint for c first point? i.e. [X,Y] bijective with pi_0(Y)

I did this so far

[gentle osprey]

ProphetX

[cursive flume]

[pastel linden]

Think about what it means to have a homotopy class of maps from a point into a space

[coarse night]

hint: X is contractible means it's path connected. what can you say about f(X) for cont func f

[pastel linden]

Point x I is homeomorphic to I

So a continuous map from the interval is...

[coarse night]

try to show constant maps are homotopic iff their image lie on the same path component

[tardy onyx]

I apologize for the double post, but perhaps this is the more appropriate channel: Is the topological data analysis technique of persistence homology invertible?

[stoic eagle]

By van kampen, the fundamental group of the wedge of two spaces is the free product of their fundamental groups, right?

sry for the wordiness i cant use tex

[coarse night]

yes

[cursive flume]

like this?

[coarse night]

didn't check the details but the idea looks alright

[chrome ridge]

Is there any nice reference for constructing a manifold with a prescribed group presentation ?

This construction specifically

[cursive flume]

this is so much data to keep track of. is f_0, f_1 fixed or arbitrary? is g_1 an extension of f_1? is G extension of F?

[gritty widget]

f_0, f_1 are fixed, yes, yes

[plain raven]

I wouldn't say f0 and f1 are fixed. f0 and f1 are implicitly determined by the choice of F because

f0 can be defined in terms of F as the restriction of F to A x {0} and similarly f1 is the restriction of F to A x {1}

so when it says f0 -> f1 you can just view this as new notation to designate these functions being introduced on the fly

similarly with g1, g1 is determined in terms of G by restricting it to time t=1

[cursive flume]

I am not sure I get what you mean. a homotopy needs be from f_0 to f_1. that s how a homotopy is defined

I can have several homotopies frm f_0 to f_1 though

if I don t fix f_0 and f_1, then what is the homotopy between?

[plain raven]

Sure but you can simply speak of maps F: A x I -> Y

and then after F is introduced you can introduce the notation f0, f1 for the restriction of F to A x {0} and A x {1} respectively

Then F is a homotopy from f0 to f1.

Therefore any continuous map A x I -> Y can be considered a homotopy retrospectively

But if you like you can rearrange the quantifiers so that f0 and f1 are introduced before F

"For all Y, for all f0, f1 : A -> Y, for all F : f0 => f1"

Similarly if you want you can rephrase the last line as
"there exists a map g1 : X -> Y and a homotopy G : g0 => g1, G extending F"

[cursive flume]

sorry I am really confused. what is your definition of homotopy?

For me it looks liek this: Let X,Y be topological spaces and f_0,f_1:X->Y continuous maps. A homotopy F from f_0 to f_1 is a continuous map F:X times 0,1 closed to Y, such that F(x,0)=f_0(x) and F(x,1)=f_1(x)

how you define homotopy?

[plain raven]

yeah i agree with your definition

what's the problem?

[cursive flume]

a homotopy is by defintion from f_0 to f_1

[plain raven]

Ok

[cursive flume]

I need know what f_0 and f_1 is to speak about F here

no?

[plain raven]

I'll repeat what I said a minute ago

Let X and Y be topological spaces.

Let F : X x I -> Y be a continuous function.

[cursive flume]

F is not a homotopy yet! right?

this is what I am missing I think

[plain raven]

Sure, fine. Let me continue

Let f0 be defined as the composite F \circ i0, where i0 : X -> X x I is the inclusion map x |-> (x,0)
similarly let f1 be defined as the composite F \circ i1, where i1 : X -> X x I is the inclusion x |-> (x,1)

Now, f0 and f1 being given

it makes sense to say that F is a homotopy from f0 to f1

even though we've given F beforehand, we can now say in retrospect that it is a homotopy from the newly defined functions f0 and f1

[cursive flume]

yes, in your definition A homotopy is a quintuple (F,i0,i1,f_0,f_1)right?

[plain raven]

I don't consider i0 and i1 part of the data

I don't know if I understand what you mean

[cursive flume]

my point is you have a random continuous map F. Then you compose this to get f_0 and f_1

and after composition you get the homotopy

so from any continuous map,you can construct a homotopy

but I am confused as what it means to be for any homotopy?

any homotopy F means for any continuous map F, I can construct the homotopy via your procedure?

[plain raven]

yeah i see

[cursive flume]

in my def,if I fix f_0,f_1, then I have one homotopy from f_0 to f_1 as one cntinuous map so that

[plain raven]

by "any homotopy F" they mean "any triple (f0, f1, F)"

[cursive flume]

and then simply all homotopy from f_0 to f_1 is the collection of continous maps, which satisfy the property given

[plain raven]

yeah.

[plain raven]

this is an abuse of notation which is acceptable because f0 and f1 can always be reconstructed uniquely from F if only F is given

[cursive flume]

but f_0 and f_1 is constructed from F

[cursive flume]

yes

so they simply mean for all F, and then f_0,f_1 follow (i.e. they are fixed by F)?

[plain raven]

Yeah.

[cursive flume]

thanks for help!

[plain raven]

something similar is going on with the g1

like

g0 is introduced in the assumptions "for all g0" and then we introduce a homotopy G : g0 -> ???

[cursive flume]

yes, if there exists a homotopy G, then g_0 and g_1 are determined

[plain raven]

Yeah. here it's slightly different because g0 is introduced prior to G

but G determines g1

[cursive flume]

yes,true. but as G is extension of F,that means G restricts to F, so both g_0 and g_1 need restrict to f_0 and f_1, right?

[plain raven]

Yes, that follows.

[cursive flume]

apparently this guy satisfies the homotopy extension property

but I have troubles understanding who M_f is in the first place

I am confused by what it means (x,0) sim f(x)

they never define it

(x,0) sim f(x) if what

[plain raven]

yeah

so

you understand that given any topological space we can form the quotient space of this by an equivalence relation

[cursive flume]

yes

[plain raven]

right?

[cursive flume]

I do not know what the equivalence relation here is

[plain raven]

so they're defining it on the fly here by saying, the relation on (X x I) \amalg Y is the smallest equivalence relation \sim such that for all x in X, (x,0)\sim f(x)

[cursive flume]

what does this mean? I never saw this

an equivalence relation is a subset of X x I amalg Y cartesian product with itself.

[plain raven]

Yeah.

so

Let R be a relation on some set X, so R \subset X x X.

Do you understand what i mean by 'the smallest equivalence relation containing R'

[cursive flume]

no. but my guess is this is some univ prop

I only say smallest/largest things when defining things via univ property

[plain raven]

arguably it is but category theory isn't going to be a helpful perspective here.

i mean smallest in the sense of the ordering imposed by \subset

like

there is an equivalence relation R' \subset X x X

such that R\subset R'

and for any other equivalence relation R'' \subset X x X with R \subset R''

[cursive flume]

there could be several equivalence relations on R. how can I know how many?

how do I determine them?

[plain raven]

shhh hold on

lol

you didn't let me finish

[cursive flume]

sry

[plain raven]

Let $R \subset X \times X$ be a relation on $X$.
Then there exists an equivalence relation $R' \subset X \times X$
such that $R\subset R'$,
and having the property that for any other equivalence relation $R'' \subset X \times X$ with $R \subset R''$,
$R'\subset R''$

[gentle osprey]

diligentClerk

[cursive flume]

this is a claim or assumption?

[plain raven]

I claim it

i can prove it

First of all do you agree that if such an R' exists

it's necessarily unique?

[cursive flume]

Suppose $R'''$ is an equivalence relation with those properties. Then $R \subset R'''$ and it has the property that for any. equivalence relation $R'' \subset X \times X$ with $R \subset R''$ it holds that $R''' \subset R''$

[gentle osprey]

ProphetX

[cursive flume]

we apply this to $R''=R'$

[gentle osprey]

ProphetX

[cursive flume]

Then, $R''' \subset R'$

[gentle osprey]

ProphetX

[plain raven]

yeah.

and conversely R' \subset R''' by the same logic.

[cursive flume]

now we apply this for $R''=R'''$

[gentle osprey]

ProphetX

[cursive flume]

and we get the converse

[plain raven]

So they are both subsets of each other

and thus have the same elements

[cursive flume]

yes

[plain raven]

and are equal.

[cursive flume]

this is like universal property proof

similar

so yes,if exists,is unique

[plain raven]

yeah. i guess it technically is a universal property proof except we're in a category where all isomorphisms are equalities. hahaha.

anyway

yeah so it suffices to construct any guy satisfying this property

there are two methods of constructing this guy R'

First of all let $R_0$ be the union of $R$ with the set $ { (x, x) \mid x \in X}$, i.e. the 'diagonal' of $X\times X$

[gentle osprey]

diligentClerk

[plain raven]

It's easy to see that $R_0$ is reflexive.

[gentle osprey]

diligentClerk

[plain raven]

Moreover any relation containing R0 will also be reflexive. We'll construct R' so that it contains R0.

[cursive flume]

is R_0 an equivalence relation?

or just a set

[plain raven]

R_0 is a relation on X, but it doesn't yet satisfy all the axioms of an equivalence relation.

If it did we would be done.

But there's no reason to believe that R0 is symmetric or transitive.

[cursive flume]

R_0 is reflexive,because R is, and we added a set, which is 'reflexive'?

[plain raven]

No, I didn't assume R was reflexive, I think.

R_0 should be reflexive even if R satisfies no interesting properties at all

[cursive flume]

yes,fair

[plain raven]

(R could be empty!)

Ok let's take care of symmetry.

Let $R_0^{\rm op} = { (y, x)\in X\times X \mid (x,y )\in R_0$

[gentle osprey]

diligentClerk

[plain raven]

let $R_1 = R_0\cup R_0^{\rm op}$

[gentle osprey]

diligentClerk

[plain raven]

I claim R1 is both symmetric and reflexive.

Do you buy this?

(Not necessarily transitive.)

[cursive flume]

intuitively makes sense,but I will have to work out the details, I am not convinced by reflexive either yet

[cursive flume]

this makes sense intuitively

[plain raven]

Good, cool

Let $R'$ be defined in terms of R1 as follows : $(x,y)$ is in $R'$ if there exists a sequence $x_1,x_2,x_3,\dots x_n$ of elements of $X$, (some $n\geq 1$), such that $x = x_1$, and $y =x_n$, and $(x_k, x_{k+1})\in R_1$ for each $1\leq k < n$. (In the case $n=1$ we will take this condition to be vacuously satisfied, so $(x,x)\in R'$ automatically because it comes from the singleton sequence with $x_1 = x$.

The point of this construction is to guarantee that R' is transitive.
The parenthetical observation at the end about the vacuous case n=1 guarantees that R' is reflexive as well.

Moreover, this should be symmetric if R_1 is.

[cursive flume]

this is kind of a big fish to me how does this have anything to do with transitivity?

[plain raven]

What is the definition of transitivity?

[cursive flume]

Suppose $x \sim y$ and $y \sim z$. Then $x \sim z$

[gentle osprey]

ProphetX

[plain raven]

ok. you understand what a graph is right

like a directed graph

[cursive flume]

what do you mean by graph?

I only know graph of equivalence relation

[plain raven]

A set of nodes with edges between them

[cursive flume]

I'm really bad with pictures but yes, I know what you mean by graph

[plain raven]

aight nvm

but yeah like

I could tell you why R' is transitive but think about it for a second and unfold the definition of R'

and see if you can figure it out

i won't torture you too long

[gentle osprey]

diligentClerk

[plain raven]

small edit

[cursive flume]

ok,sec,sorry for disturbing this,but I think I managed to get this idea: R_0 is reflexive, because a relation on X x X is reflexive iff it contains all points {(x,x)}

[plain raven]

Yeah.

[cursive flume]

so no matter what R contained,we added those,so the new guy is reflexive

[plain raven]

Right

[cursive flume]

if (x_k,x_k+1) in R_1, then (x_k+1,x_k) in R_1 by symmetry. now my intuition is that from singleton we should somehow get that x=x_1=x_2=x_3=...

but i don't see the last part

is just vague idea

[plain raven]

what are you trying to argue here?

transitivity?

[cursive flume]

no,not trying that yet

just trying to see if i understand what you wrote down

[plain raven]

i'm allowing sequences here to be of varying lengths

n isn't fixed, different sequences can have different lengths

The case of reflexivity comes from considering sequences of length 1

[cursive flume]

transitivity would look like this: if $(x,y) \in R'$ and $(y,z) \in R'$, then $(x,z) \in R'$. Suppose $(x,y)$ in $R'$. then there exists a sequence $x_1,\dots,x_n$, such that $x=x_1$ and $y=x_n$ and $(x_k,x_{k+1}) \in R_1$ for all $1 \leq k \leq n$. Moreover, suppose $(y,z) \in R'$. Then there exists a sequence $y_1,\dots,y_m$, such that $y=y_1$ and $z=y_m$ and $(y_l,y_{l+1}) \in R_1$ for all $1 \leq m$

[gentle osprey]

ProphetX

[plain raven]

Yeah

[cursive flume]

we now need to find a sequence $z_1,\dots,z_p$, so that $z_1=x$ and $z_p=z$ and $(x_m,x_{m+1}) \in R_1$ for all $1 \leq m \leq p$

[gentle osprey]

ProphetX

[cursive flume]

but this is impossible

[plain raven]

lmao

[cursive flume]

I think the sequences must have same length

all length n

[plain raven]

no, sorry

i should have clarified that

they're permitted to have different lengths

||Just concatenate the sequences you have||

[cursive flume]

wait, is the sequence necessarily finite?

[plain raven]

yes

[cursive flume]

I define the sequenze $z_p:=x_1,\dots,x_{n-1},y_1,\dots,y_m$. Clearly, $z_1=x_1=x$ and $z_p=y_m=z$. Left to see: $(z_m,z_{m+1}) \in $R_1$ for all $1 \leq m \leq p$. However, this is clear, as all $x_n$ and all $y_m$ are.

[gentle osprey]

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

[cursive flume]

like this?

[plain raven]

yeah

Does this make sense?

[cursive flume]

yes,except the indexing part

what length do I allow the sequences to have?

any finite length?

[plain raven]

Yes, any length >= 1

[cursive flume]

ok so if R_0 was reflexive, R_1 is and in turn R' is, because I just added elements

R' is transitive,as I just proved

left to see: R' is symmetric

Suppose $(x,y) \in R'$. Then there exists a sequence $x_1,\dots,x_n$, such that $x=x_1$ and $y=x_n$ and $(x_k,x_k+1) \in R_1$ for all $1 \leq k \leq n$. Then there exists a sequence $x_n,\dots,x_1$, such that $x_n=y$ and $x_1=x$ and $(x_{k+1},x_{k}) \in R_1$ by symmetry of $R_1$, so $(y,x) \in R'$

[gentle osprey]

ProphetX

[plain raven]

yeah. does that make sense to you?

[cursive flume]

yes

[plain raven]

so yeah this construction works

[cursive flume]

left to see: R_1 is symmetric.

[plain raven]

oh

Well

hm

[cursive flume]

are you sure it need not be (x,y) in X x X?

[plain raven]

I think i got it right. why do you think this wouldn't work

[cursive flume]

So if (x,y) in R_0, and I add (y,x) to it, where (x,y) in R_0, then R_1 seems to be symmetric

R_1 is symmetric if for all (x,y) in R_1 it is true that (y,x) in R_1, yes?

[plain raven]

yes

[cursive flume]

suppose (x,y) in R_1. Then either (x,y) in R_0 or R_0^{op}. If (x,y) in R_0, then (y,x) in R_0^{op}. So (y,x) in R_1.
If (x,y) in R^{op}, then (y,x) in R_0. So (y,x) in R_1.
Altogether: (y,x) in R_1 in either case

like this?

[plain raven]

Yeah. This makes snese to me

Does it make sense to you?

[cursive flume]

yep,makes sense

just for curiosity, in what books is this treated?

because our topology prof never mentioned this

[plain raven]

Hmm.

one sec

[cursive flume]

also without going through the construction we did for this specific example right now,which would take really much work, I will do it tomorrow,is it fair to say the equivalene relation ~is a subset of X x I amalg Y cartesian product with itself, such that x~x' iff x=x', y~y' iff y=y' , (x,a) ~(x',a') iff x=x',a=a' and (x,a) ~y iff f(x)=y, a=0?

[plain raven]

Idk. lol. Sorry. Equivalence relations are treated in the book "Elements of Set theory" by Enderton and in the first chapter of "Topology" by Munkres. I don't see this specific construction in either one though

[cursive flume]

I got it,I will type it up tomorrow in detail,but I was curious where you learned it

it didn't seem completely trivial to me, without your hints,I would never come up with such construction

[plain raven]

I'd be careful about saying iff here

By the way there's a very slick proof of existence of R'.

The proof is as follows.

1. There is at least one equivalence relation containing R; namely the equivalence relation X x X \subset X x X

in which all elements are identified.

2. If ${R_i}_{i\in I}$ is a family of relations, $R_i \subset X \times X$, then their intersection $\bigcap_i R_i$ has the following properties:

[gentle osprey]

diligentClerk

[plain raven]

• if each R_i is reflexive, then \bigcap R_i is reflexive

• if each R_i is symmetric, then \bigcap R_i is symmetric

• if each R_i is transitive, then \bigcap R_i is transitive

• if each R_i is an equivalence relation, then R_i is an equivalence relation

Moreover if all R_i contain R, then \bigcap R_i contains R.

Therefore we can define R' to be the intersection of all equivalence relations containing R. This is an equivalence relation, it contains R, and by construction it is a subset of every other equivalence relation on X containing R.

[cursive flume]

how is this R' the same as the one we had before?

this is really counter intuitive

[plain raven]

It has to be the same because it satisfies the same universal property hahaha

[cursive flume]

oh

lol,right

[plain raven]

i guess i didn't actually prove that R' as i constructed it before has the universal property

[cursive flume]

yes,we just showed is an equivalence relation

[plain raven]

yeah

well

i don't want to do that rn but it's doable

there's something else i want to point out

namely like

remember that you learned the universal property of the quotient of a set by an equivalence relation

[cursive flume]

yes

[plain raven]

so what's interesting is

if you go back to the definition

then in order to phrase and state the universal property

you don't need that R is an equivalence relation.

You can just say

Let X be a set and R be a relation on X.

[cursive flume]

I need it

otherwise the projection isn't surjective anymore

[plain raven]

Well hold on.

The quotient of X by R is a set X/R, together with a function eta : X -> X/R

such that

1. if R(x,y), then eta(x)= eta(y)

2. if Y is any set, and f : X -> Y is a function satisfying R(x,y) -> f(x)=f(y), then f factors uniquely through eta, i.e. there exists a unique function f' : X/R -> Y such that f' \circ eta = f

This makes sense right

like i don't use the fact that R is an equivalence relation anywhere

ok let me ask now what you mean that 'the projection isn't surjective'

[cursive flume]

in order to prove eta:X->X/R is surjective,we use that R is equiv rel

no?

[plain raven]

Hmmm ... but when you constructed X / R you were working under a specific construction of X/R, namely the set of points of X/R in that case was the set of equivalence classes of X/R

[cursive flume]

but each equivalence class was non-empty

guaranteed by reflexivity

[plain raven]

equivalence relations

[maiden pilot]

How would I use Van Kampen for this? I thought to use $U = R^n\setminus K$ and $V = R^n\setminus{p}$, so $R^n = U \cup V$ and $U \cap V = R^n\setminus (K\cup{p})$. By Van Kampen, we have $\pi(U) {\pi(U \cap V)} \pi(V)$ is trivial. Note that $\pi(V) = {1}$. With $i \colon \pi(U \cap V) \to \pi(U)$ the homomorphism induced by the inclusion map, it follows that $\pi(U)$ with the relations $i (c) = 1$ for $c \in \pi(U \cap V)$ is trivial. Can we conclude that $\pi(U \cap V) \cong \pi(U)$ from this fact?

[gentle osprey]

OmnipresentCoffee

[plain raven]

Interesting expository paper on the historical task of computing the homology of the K(pi, n) spaces

and the methods developed to solve this problem

[gritty widget]

djvu

djvu

[plain raven]

PDF link here
http://www.numdam.org/item/AST_1976__32-33__173_0/

[plain raven]

What is the basic cofibrant replacement notion in the category of G-spaces, G a fixed topological group

I'll be more precise

Conjecture:

Let G be a topological group, X a G-space.
There exists a G-space Y such that the projection Y-> Y/G is a principal G-bundle, and a G-equivariant map Y -> X which is (forgetting the G-action) a homotopy equivalence.

What does the construction of Y look like?

[odd flame]

is a half open interval open in R under the standard topology? no right?

in general my question is: if you can express a subset of a topological space X as a union of basis elements then it is by definition open in X right?

the half open interval came to mind bc [0,1) = {0} U (0,1) which are both open in R - but then that same reasoning would work on closed intervals and closed intervals are… closed

actually i guess a point set isn’t open in R …?

yeup nvm, point set is closed

are these all open in Y? im pretty the first four are but idk about the last one

[cursive flume]

isn't this the Milnor construction?

of the classifying space

[cerulean oriole]

{0} is not open in (the standard topology on) R.

[odd flame]

yeah i realized that

[odd flame]

of these i think the only open one in R is the first

[ocean narwhal]

idk any topological arguments for this but if you take the standard metric topology on R you can show that there is no neighborhood around 0

so it is not open

[odd flame]

What is this example trying to show

Like what does “closed as subsets of Y mean “ if it’s saying they’re open in Y

[grave maple]

It is demonstrating that things can be open and closed at the same time.

[gritty widget]

only the first and last one should be

[odd flame]

Why isn’t the second?

[gritty widget]

interior of B is A

1 is not in the interior for example

oh wait

what's Y?

[odd flame]

This appears before he talks about closure and interior tho

[-1,1]

As a subspace of R

[gritty widget]

Then A, B and E are open

in Y

int(A) is largest open set contained in A

[odd flame]

Then why isn’t C?wouldnt the same logic as B apply to C

[gritty widget]

no, because [-1, 1] has those two points -1 and 1

which act differently from points of (-1, 1)

[odd flame]

An open set in Y is the intersection of Y with an open set in R

Ohhhh wait one sec

[gritty widget]

basic neighbourhoods around any point of (-1, 1) are just open intervals (given they are small enough)

[odd flame]

I see it

[gritty widget]

but for 1 they are of the form (a, 1]

so here everything is fine

[odd flame]

Yeah but 1/2 is on the inside so to speak

[gritty widget]

yeah, that's why you can't include 1/2 there

because there's no points on the left of it

[swift fjord]

Notice that being closed does not imply not being open. In this case it does, but only because R is connected and {0} is a proper.nonempty subset. You can just show directly that [0,1) is not open by showing 0 doesn't have a nbhd which is a basjs element and container in [0,1) e.g. An interval.

[odd flame]

wdym connected

Only on ch2 of munkres lol

But I think that comes up soon

[swift fjord]

You don't have to worry about it now

But it's one condition that precludes clopen sets from existing

In one Equivalent formulation it is the existence of a clopen nonempty proper subset

[odd flame]

I’ll keep it in mind then

[gritty widget]

connected means it can't be decomposed into two disjoint non-empty closed (equivalently open) subsets

or, there is no non-trivial clopen sets, the only clopen sets are empty set and the whole space

clopen meaning closed and open

[plain raven]

The milnor construction of the classifying space is very closely related to what i'm asking here but it's not quite it... the milnor construction solves this problem in the case X = a singleton endowed with the trivial group action

i'm looking for a generalization or adaptation of the milnor construction where the total space of the fiber bundle would not be contractible necessarily but rather homotopy equivalent to a given space X

[cursive flume]

Ah ok i see

Seems tedious

[chrome ridge]

Any nice references that explain regular neighborhoods of CW-complexes ?

[paper wedge]

https://math.stackexchange.com/questions/1448789/understanding-construction-of-open-nbds-in-cw-complexes

@chrome ridge

https://cpb-us-w2.wpmucdn.com/sites.uwm.edu/dist/3/337/files/2016/10/TOP-regular-neighborhoods-25n85yn.pdf

[chrome ridge]

Yeah I've seen the construction in Hatcher's, are there any references that maybe explain some intuition. I don't think I can visualize it the way Hatcher defines it.

[plush sluice]

How are the boundary maps being calculated from the triangulation? I've got zero grasp of this module as a whole, but I cannot for the life of me figure out the method for calculating them

[lunar yoke]

I went through a computation of them a while ago

[lunar yoke]

see if that helps

its a different example though i think

[plush sluice]

Aye, I think so. I've got an example with just a standard triangle that's giving me a larger headache

[lunar yoke]

maybe this picture already does it

you basically use the boundary formula of oriented simplices

[plush sluice]

I could honestly kiss you, that clicked straight away

First year the modules being taught at my uni and the examples given skip any kind of computation. Just jumps straight to the answers for each step

[lunar yoke]

oof

examples are very important

the more explicit the better

[plush sluice]

Exactly. Especially when you're getting to this level of maths where you start losing the abundance of examples online and are heavily relying on the teaching

[pastel linden]

Are there universal properties for vector bundle constructions? I've been thinking about how to prescribe uniqueness to the constructions (sums, tensor products, pullbacks, quotients, etc.) but it feels like there's a general noncanonicity that makes this stuff a bit trickier

Like, if f*E -> N is the pullback bundle of E -> M with f : N -> M, I'm thinking if E' -> N is another bundle with bundle map g* : E' -> E commuting with f, do we get some kind of bundle map f*E -> E'?

[cedar pebble]

I mean you have adjointness

if you regard a vector bundle E as a (sheaf of) O_X-modules

we have a natural isomorphism $\mathrm{Hom}_{\mathcal{O}X}(f^*\mathcal{G},\mathcal{F})=\mathrm{Hom}{\mathcal{O}Y}(\mathcal{G},f*\mathcal{F})$

[gentle osprey]

nGroupoid

[cedar pebble]

so you can characterize maps f*E -> N as you say in terms of their adjoints

also, the pullback f*E is itself a colimit, so there's a universal property there you can exploit

[cursive flume]

isnt the pullback a limit?

limit over cospan

[cedar pebble]

Yes but we’re talking about two different things

[cursive flume]

oh sorry

[cedar pebble]

The pullback f^{-1} of a sheaf is a colimit

[cursive flume]

I see-I only know the definition of pullback in categorical sense

didn't work with sheaf yet

[peak crystal]

A is subset of X, let x in closure of A is there always exist a sequence (x_n) in A such that (x_n) convergence to x.

[unreal stratus]

Depends on the type of space

But for first countable spaces, in particular metric spaces, yes

[peak crystal]

In Hausdorff?

[gritty widget]

no

[gritty widget]

no

proper concept of a "sequence" in a topological space is that of a net

some people use filters instead which are more set-theory vibe

Hausdorff spaces are those for which limits of nets, if they exist, are unique

I don't think this has anything to do with separation properties, you can have pretty strong ones and still can't use sequences in general

first countability is the culprit

[peak crystal]

Okay thanks alot

[cursive flume]

If I only have as toolkit seifert van kampen and the fundamental group of CW complexes, how do I start computing fundamental groups?

This is my toolkit

[lunar yoke]

Fundamental groups of what? Arbitrary spaces?

In theory every space has a map into it from a cw complex which induces an isomorphism on the fundamental group (even on all higher homotopy groups), so in some sense it does suffice to „know“ the htpy groups of all cw complexes

I dont think this is of much help in practice though, as non-cw spaces can be pretty wild, and these maps pretty complicated

[obtuse meteor]

it really depends on what data you have...spaces in mathematics can be described in dozens of ways, so you need to know what description you have in your hands to start with

generally though, giving a cell structure, van kampen to break your space up into smaller spaces, and covering space theory (not listed in your toolkit--but later essential) are all the ingredients you need to compute something like this

[odd flame]

Why is the closure of a set important in general

[golden gust]

important in what sense

[odd flame]

Nothing in particular I’m just reading about it in munkres and I’m wondering if/when it’ll come up again

[surreal lantern]

It will

[golden gust]

it's quite an important idea (so is the interior)

[gritty widget]

small closed sets are good

[swift fjord]

Sometimes elements you want to consider wander outside your set, but close enough to be in the closure

[odd flame]

Why tho

Should I just keep reading for now

Next section is continuity

[odd flame]

I meant the why at this

[ornate berry]

It's a bit hard to explain why they're useful without seeing it

But it really does turn out that these concepts come up often

Maybe here's an example

The boundary of a set is defined as its closure without its interior.

I hope at least you can see why this is sort of an interesting idea, even if you can't see its uses yet

You can use the definition to prove some interesting facts about the boundary. For instance, you can find a description of the boundary of A u B in terms of the boundary of A and that of B

[surreal lantern]

dense sets also use the concept

[ornate berry]

Perhaps that gives you an example of why the closure and interior are interesting ideas

[surreal lantern]

just keep reading for now

[odd flame]

[gritty widget]

it's a central concept to all of topology

it describes the topology of your topological spaces, and it behaves in a particularly nice way on subspaces too

many things can be described using the closure of a set

[tawdry valve]

One thing that often happens is that if some property P holds for all points in S, then P holds for all points in the closure of S, so the closure is useful for proving a property for many points.

This is also where density comes into play. If you want to prove some property about some closed set E, sometimes it’s good to find a subset S inside E, such that the closure of S is E. Then, it might be easier to prove the property first for S, and then prove that the property is true in the closure

[gritty widget]

I can't describe in words how important the concept of closure is in topology, but you'll have to trust me that it really is extremely important without no doubt.
Questioning the importance of closure feels a bit like questioning the importance of topology itself

[stiff light]

The closure of a set is essentially the optimal solution to the problem of finding a closed set containing that set. So anytime you want to find a closed set containing some set (which is often), the closure will probably be useful.

[odd flame]

Maybe I haven’t gotten to this in the book, but I don’t see why this is the case yet

[grave maple]

There are many ways of describing a topology. The usual way involves open sets, but you can also do it with closed sets.

In fact, you can define "closure spaces", which are sets with a closure operator. These are more general than topological spaces.

[stiff light]

One common example where the closure is important are dense sets: a set is dense if its closure is the entire space. This effectively means that any point of the space can be approximated to any degree of accuracy by points in the dense set.