point-set-topology

[knotty vine]

Did I beat some record for longest message? damn

[unreal stratus]

||semisimplicial spaces: ||

[knotty vine]

||In my Ms thesis I used the name semisimplicial set instead of Delta complex, but theyre the same thing||

[unreal stratus]

I did actually wonder what the correspondence between the two is, since I've not really seen that written up

[knotty vine]

For some more context:

• Any simplicial complex whose vertices are ordered is a Delta complex
• Any simplicial complex (without order) can be made into a homotopy equivalent Delta complex
• Any Delta complex can be made into a homeomorphic simplicial complex using subdivision (I think...)
• Any Delta complex can be made into a homeomorphic simplicial set by freely adding degeneracies
• Any simplicial set can be made into a homotopy equivalent Delta complex by making degenaracies non-degenerate (you can do slightly better so that compact simplicial sets stay compact Delta complexes but idc)
• All are CW complexes

[unreal stratus]

Is it just like a delta complex is the realisation of a semisimplicial set, and the data used in constructing a delta complex lifts to that of a ssset

[knotty vine]

Yep

[unreal stratus]

and by realisation i guess i mean like

left adjoint of the inclusion sset -> ssset followed by the normal realisation?

or is there some other realisation lol

[knotty vine]

We dont need that, just use the usualy realisation story for the obvious functor Delta_inj -> Top

[unreal stratus]

Okay sure

[knotty vine]

Some Kan extension or other

[unreal stratus]

Yeah

kan extension of Δ_inj -> Top along yoneda embedding ig

nice

[knotty vine]

Keeps the realization compact

[unreal stratus]

i imagine that amounts to teh same thing right

hm

uh

idk

[knotty vine]

The problem with using the inclusion functor Delta_inj -> Delta and realizing that way is that already the point simplicial set becomes the infinity-sphere...

Since all of the degenerate simplices suddenly become real

Luckily, the infinity-sphere is contractible (??? !!!)

Let me @grim knot, see #point-set-topology message

[grim knot]

may I gift you something for the effort? @knotty vine I'm kind of speechless

[knotty vine]

Feel free to ask if you have any questions!

[unreal stratus]

real

[grim knot]

I need to work through your explanation, but it is surely clearer now. Thank you again for taking your time by giving me such a wonderful answer, I am really grateful for that!

[median sand]

I don't know if you guys know this, but apparently this guy https://friedl.app.uni-regensburg.de/teachingindex.html has a 4k-page topology book on his website, covering just about anything. The bookmarks in the PDF were a little wonky and I spent half an hour fixing them, so if anyone wants it, here you go.

[knotty vine]

Nice, but how is it possible that people still compile their latex looking like this

Looking like the Harry Potter logo

[median sand]

Weird, it might be from my editing the document, let me check

Tbh I'm not sure what this is, the document is displayed fine on my computer.

Ofc when I re-ordered the bookmarks and saved, the file shrunk from ~65mb to ~50mb, so maybe some compression happened and caused this.

[knotty vine]

It's already present in the original file. It's just the old way of compiling Tex files using bitmap fonts or something

If you zoom in you will see the pixels and if you zoom out you will see the absence of hinting

[median sand]

This is present for me as well, yeah.

[median sand]

But it's not as wonky as in your case.

[knotty vine]

Zoom out a little further

[knotty vine]

It's perfectly nice if you are at precisely the right zoom level

[median sand]

I'm as zoomed out as can reasonably be, looks about right.

In any case, small price to pay for so much goodness in 1 pdf!

[lofty stag]

bit of a strange question but is it possible to use variational calculus (Euler's equation) in some way to find the shortest distance between two points in hyperbolic space?

[umbral panther]

Yes, but it’s tricky.
Reparameterizing a path gives many degrees of freedom
Have you tried?

[lofty stag]

nope, but i was thinking of trying it

[spice basalt]

we learned the universal cover is the most general cover in the sense that it “covers all other covers”.

a simply connected space is its own universal cover, e.g. [0, 1] is the universal cover of [0, 1]. but then consider the space [0, 1] x {a, b}, two copies of the unit interval. isn’t this also a cover of [0, 1] by a projection? if so, it doesn’t seem to covered by the unit interval alone

[white oxide]

I thought the universal cover was universal over those that are connected?

[red yoke]

And covers other connected covers

[white oxide]

yeah

[red yoke]

Or "simply-connected cover of a connected space"

[spice basalt]

is it connected or simply connected

wikipedia says simply connected

[hexed steppe]

connected

[red yoke]

The universal cover is a simply-connected cover of a connected space and covers all other connected covering spaces

[hexed steppe]

what does wikipedia say exactly

[white oxide]

You want to characterize any connected cover using the fundamental group and how this determines them up to homeomorphism

[spice basalt]

beta is simply connected

[hexed steppe]

that is not a statement about covering

[spice basalt]

wait nvm

youre right my bad

[hexed steppe]

just uniqueness of simply connected covers (for connected base spaces)

[white oxide]

you want a certain "galois correspondance" between the subgroups of your fundamental groups and possible covers.

[umbral panther]

The universal cover is not universal because it has isomorphisms. You can rigidify it by choosing a base point. This chooses a base point in the space below. And thus chooses a component. The based universal cover of the base point component of the space is universal. It lifts uniquely to any based covering space. No need to mention connected. Based implies non empty

[hexed steppe]

isnt this just saying “covering space theory for connected spaces also applies to an arbitrarily-chosen connected component”

[umbral panther]

No, it’s (also) giving a universal property

[plain raven]

Simplicial complexes are a fully faithful and reflective subcategory of "symmetric" simplicial sets. I would imagine they're a fully faithful and reflective subcategory of symmetric semisimplicial sets as well.

@knotty vine

[umbral panther]

What are these symmetric categories?

[plain raven]

I don't think the terminology is standard, but I don't know standard terminology. I mean take the usual base category of finite ordinals and add all the permutations [n] -> [n] as well.

So "symmetric simplicial set" = presheaves on finite cardinals with arbitrary functions between them
"symmetric semisimplicial set" = presheaves on finite cardinals with arbitrary injections between them

[umbral panther]

Reflective? Really?

[knotty vine]

I can only imagine the name of the category of symmetric semisimplicial sets...

Thats one way to do it. Either impose an ordering, or just use all orderings simultaneously

[unreal stratus]

Oh yeah that's the first time in a while i've seen that kinda remark like

well to avoid having to make choices in mathematics either do all choices at once, or include a choice as part fo teh data lol

[plain raven]

Imma be honest I haven't written down a pen and paper proof I'm just vibing here. Upon further reflection (ha) this probably fails for semisimplicial sets. But for symmetric simplicial sets here's my informal argument. There is an obvious nerve functor from the category of abstract simplicial complexes into symmetric (semi) simplicial sets which sends each simplicial complex to the simplicial set of "singular" simplices in the complex. Simplicial complexes are cocomplete so this has a left adjoint given by "realization". I'm claiming the counit of this is probably an isomorphism which would imply that the right adjoint is fully faithful. Do you buy that? Like I should be able to recover the original simplicial complex given all maps from the n-simplex into it, all permutations and such considered.

[plain raven]

Yeah that's why I'm bringing this up because we like functoriality

We can't impose orderings uniformly although we can work in a category of ordered simplicial complexes

[umbral panther]

What is your definition of simplicial complexes as a category? I’m not convinced it’s cocomplete

[tidal cedar]

presheaves from the simplex category I'd assume

[umbral panther]

But forget that and just write down a full subcategory and argue that it’s reflective

[plain raven]

I started to write out a detailed argument and I will finish it and send it to you if you really want to see the details worked out, but I found a reference so I won't bother unless you want me to.

https://ncatlab.org/nlab/show/simplicial+complex
This page gives one definition of simplicial complex that is reasonable, and it asserts that

From this point of view, it is immediate that simplicial complexes are the separated objects for the Lawvere-Tierney topology on the category of [symmetric simplicial sets] whose sheaves are sets.

So, simplicial complexes are the category of separated presheaves in this topology.

Lemma 5.7.7 of Categorical Logic and Type Theory by Bart Jacobs establishes that the category of separated objects for a Lawvere-Tierney topology is reflective.

[plain raven]

nah fiona i meant like, traditional abstract simplicial complexes

[sterile prawn]

Idk if this is top, but is there a concept of like "degree of concavity"
Like V has 1 concavity, N has 2, W has 3, etc.

[feral copper]

For the graph of a function, that's not concavity, that's just local extrema

Unless you also want to say the graph of the cubic y=x³ has 'degree 1', in which case you'll need the number of zeroes of some (the first and/or the second) derivative of the function

Additionally, that's not even a topological/smooth invariant: a circle is diffeomorphic to a blobby circle which looks like the contour of a U, and you'd like to say they have differing 'degrees'

[knotty vine]

I guess it's relevant for Morse theory? idk about morse theory

[red yoke]

Or minimum size of open cover such that concave / convex on each set?

[coarse night]

looks like you're just counting critical points of height function.

[sterile prawn]

Its not really about critical points

Its like if you mafe a convex hull of a shape, how many regions would be formed

Like if you were to have a convex hull of H, you would create two new regions

[coarse night]

what about a shape like let's say M. It makes 2 regions not 3.

[red yoke]

You're just counting the number of components of the intersection of the boundary of the convex hull with the (simple) curve

Or number of gaps, actually

[feral copper]

Thet very much depends on the embedding; my circle example still works

Also, what's the deal with shapes like Y or X?

[knotty vine]

And what about O ?

[opaque scroll]

Do you take the convex hull minus the initial set, then count components?

[knotty vine]

I feel like M should have three holes, not two

3 has 3 holes

[opaque scroll]

Depends on the font perhaps

[knotty vine]

Take the space of all pairs of points such that the straight line segment between them intersects the ambient space

Then count path components?

[opaque scroll]

You mean that it only intersects the ambient space, or just that it intersects?

[knotty vine]

points in X so that the line between them goes outside

[opaque scroll]

Won't that make M have just 1?

[knotty vine]

Then a set is convex if this space is empty

Why?

[merry geode]

Oh, is there e.g. homology which counts this

[knotty vine]

Take one point in the left leg of M and one point in the left middle line

o wait

[opaque scroll]

Yeah, so that's one component

[knotty vine]

youre right

[red yoke]

Yes, H0(convex hull - curve)

[opaque scroll]

Though if you change it to, only intersecting the ambient space, then it might get more interesting

[knotty vine]

You mean, lies entirely inside the ambient space?

[merry geode]

Ah, quite simple

[knotty vine]

But that misses out on 3 having 3 holes 😭

[opaque scroll]

Yeah, except the end points of course

[merry geode]

Font-dependence sucks a bit tho

[opaque scroll]

Not necessarily, depends on the font again

[red yoke]

What if you make the bend less sharp

[red yoke]

Whether there are 2 or 3 would depend

[knotty vine]

How many holes does capital xi have?

[merry geode]

Can you remedy the issue to make this a meaningful invariant

[red yoke]

This has 4 for Ξ?

[knotty vine]

I think so yeah

doesnt make much sense intuitively

[merry geode]

Like the rough shape shall be similar for M in each font.

[red yoke]

What if we require that the curve be simple

[knotty vine]

simply connected?

[red yoke]

And we only count components intersecting the diagonal

Like, an arc

[knotty vine]

booooring

[red yoke]

So we consider components in I²

Then M has 3 holes

[merry geode]

Wdym simple curve?

[red yoke]

Injective continuous function I → R²

[merry geode]

Hm

[red yoke]

But holes of 3 would still depend on bend ig

[merry geode]

Yea, I’d like independence to the e.g. location of the middle.

[knotty vine]

idk how you guys draw your threes smh

[red yoke]

Like depending on whether the tangents at the bend pass above or below the endpoints of the 3

[merry geode]

Like the direction of curvature should be same

Maybe too tall of an order..

[knotty vine]

??? You draw your 3 like a sideways gamma

[red yoke]

???

[knotty vine]

Oh wait

does 3 have 4 holes

[red yoke]

3 or 4 depending on bend by the original defn

2 or 3 by my defn

[knotty vine]

I dont think I understand yours

intersecting the diagonal?

[red yoke]

The line segment has to be able to shrink to a point without intersecting the curve

[knotty vine]

Which line segment?

[red yoke]

So the line connecting top and bottom endpoints of 3 doesn't count

[merry geode]

Ah I thought I was just too dumb

[red yoke]

Diagonal of [0, 1]²

[merry geode]

It counts, wdym

Oh, it does not intersect the diagonal ig

[knotty vine]

Can you make a drawing or write out a more formal definition?

[red yoke]

Let S be the subset of I² s.t. the line connecting f(x) and f(y) is a line segment whose interior does not intersect f(I)

Count the components of S whose boundary intersect the diagonal

Oops should be boundary

[red yoke]

i.e. A line segment is counted if the endpoints can be moved together without the line intersecting the curve

[red yoke]

So one of those would be excluded

[knotty vine]

You mean something like this?

[red yoke]

Ye so left has 2 holes while right has 1

Wait were you asking about my 3 or my defn

[merry geode]

Why did you add the condition of intersecting the diagonal?

[red yoke]

Yes the right is the 3 I was mentioning

[knotty vine]

Yeah now I understand what you meant

[red yoke]

Boundary intersects diagonal = can be shrunk to a point

[knotty vine]

I though you drew your 3's like

[red yoke]

Lmao

[merry geode]

I draw like this

[knotty vine]

[red yoke]

Wait number of components / 2 probably

[knotty vine]

But the diagonal is when x = y, i.e. f(x) = f(y)

[merry geode]

Btw how does 3 have 4 holes by original definition?

[red yoke]

Ye boundary of component intersects diagonal

Not the component itself

[knotty vine]

idk if I like that or not

[merry geode]

Ah, like how M has 4 holes

[red yoke]

I like how the number of holes reduces if you draw the endpoints of the 3 closer

[knotty vine]

Ye not so good

[merry geode]

I was thinking of in terms of convex hull - curve

[merry geode]

I am dumb, so what is f here, and what is S

[red yoke]

A spiral has one hole which is good

[knotty vine]

I dont like it cause it makes weird instabilities like this

Not topology anymore imho

[red yoke]

Yea

[knotty vine]

but idk how to fix it

[red yoke]

How do you do "3 has 3 holes" without something like this tho

[knotty vine]

idk

but it does feel intuitive no?

[merry geode]

Actually why does 3 have 3 holes

[knotty vine]

It has one tiny hole on the right, thats not debatable. Then it certainly has one big hole on the left

But then it also has 2 smaller holes inside that big hole

Should we count the two smaller holes in addition to the big hole, or count only the smaller holes, or only the big hole?

Should we assign a degree to each hole?

[red yoke]

How about minimum number of line segments such that each bounded (by lines and the curve) component is convex

[knotty vine]

Then 3 has 2 holes right?

[red yoke]

3 has 3?

By bounded component I mean components of R² - curve - lines

[merry geode]

Ah, nice

Now dialects of 3 has the same number of holes

[red yoke]

*and each point on the curve is a boundary of some such component

Otherwise every curve has 0 holes

[merry geode]

Vacuous truth moment

[red yoke]

Lines have 2 holes now noo

[knotty vine]

Yeah, convexity of the complement is not such a nice property it seems

Wait

[red yoke]

I can't think of any bad cases other than lines tho

[merry geode]

How about minimum number of convex component after adding a few line segments, so that its boundary covers the curve

[red yoke]

Wait M is bad, it has 2 holes now

3 has 2 holes too

[knotty vine]

Count the components of the convex hull - shape?

[merry geode]

Huh

[merry geode]

Not exactly the convex hull, because each conn component may not be convex

[red yoke]

No I failed to properly define what configuration of lines are "sufficiently many lines"

Minimum number of line segments such that components are convex and any path from a point in (convex hull-curve) to the curve must pass through the closure of a bounded component

There, 3 and M have 3 holes, line has no holes, spiral has one hole

2 has 2 holes

[merry geode]

Why does 3 have 3 holes then

[red yoke]

The 3 components are convex

[red yoke]

That too ig

[merry geode]

Is the right side component convex tho

[red yoke]

Fuck

[knotty vine]

oof

scrap it

its done for

[red yoke]

H0(f(I)) = Z

👍

[hidden crag]

Fuck is going on

[knotty vine]

I want to count the holes in a spoon

[red yoke]

I love algebraic topology

[knotty vine]

no algebra in sight yet

[red yoke]

Homology is best hole

[merry geode]

Does it have 2 holes

[knotty vine]

the cross section of a spoon I think so yes

my spoon is too big

[red yoke]

The spoon as a surface has 1 hole though

The hole under the spoon is the same hole

[knotty vine]

huh?

[red yoke]

I think the line segment on top can be moved to the line segment below for the spoon manifold in R³

[knotty vine]

How many holes do these vases have?

[red yoke]

The last one has 4 by your original defn and 3 by the diagonal defn

I think we can all agree spirals have 1 hole

[knotty vine]

idk anymore

[red yoke]

Wait

What if 1 + number of turning points of θ(t)

For curves with differentiable θ

Or actually

The size of partition of I such that θ' is nonpositive / nonnegative on each

Back to turning points lol

[knotty vine]

Then these are the same

[red yoke]

Both 3 holes

[knotty vine]

ya

[red yoke]

I mean the right is a 3

[knotty vine]

does make sense

[merry geode]

Finally

[woven trail]

so in R^n are open balls the usual basis of the topology?

or well ig in metric spaces in general

and that would make the concept of open as it is in metric spaces

?

[unreal stratus]

All are contractible so 0

[unreal stratus]

Yes

And yes

[woven trail]

:0

[unreal stratus]

Works for all metric spaces

[woven trail]

so I think I'm understanding the idea of a topology

cus when I read the definition of a basis I started drawing what a basis would look like

and it clicked that it seemed like a generalization of an open ball

for other kinds of sets

[alpine nest]

Yeah, ultimately it's mostly about generalizing the arguments that start with "there's a ball centered around x on which we have this nice property"

[woven trail]

idk if that's how it has gone historically but probably

[unreal stratus]

Yeah they're like "small" open sets you build others out of so balls r a good picture

[woven trail]

hehe balls

[woven trail]

:0

[alpine nest]

Yeah, one disadvantage of general topology vs metric topology is that you don't get to talk about balls so much

[woven trail]

sad

[knotty vine]

This is what algebraic topology orthodoxy would have you believe, but a vase obviously has a hole!

[unreal stratus]

Lol

[unreal stratus]

Concept of a hole is somewhat nebulous

So bleh

[alpine nest]

Oh no, the "how many holes does a straw have" argument draws near

[unreal stratus]

It is not a well defined thing but well understood as natural language

Gotem

[high hill]

no its just non topological

in the usual sense

[alpine nest]

It's an open issue

[knotty vine]

Topology is what we topologists make it

[alpine nest]

Subject to some constraints of decency, I hope

[knotty vine]

[woven trail]

amazing

[alpine nest]

I draw the line at non-Hausdorff spaces

[high hill]

,w hole

[gentle osprey]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by shuri2060

[high hill]

[alpine nest]

Finally a definition we can get behind

[high hill]

something made by a shovel

[alpine nest]

I wonder if there's a cantor shovel.

to go with the Cantor brush and the Cantor tent

[merry geode]

Opinion: Basis is just small open set that is enough to describe the neighborhood behavior

On a metric space, balls are small enough (not so many balls) yet it describes the behavior when "close".

On an Affine Zariski topology, D(f) forms a basis; this means {f \neq 0} is enough to describes the topological behavior in algebraic geometry.

[hexed steppe]

opinion: bases generate topologies

[white oxide]

Seems pretty objective to me, the topological behaviour is the topology

[hexed steppe]

topological behavior is in the eye of the beholder

[white oxide]

Mfw when topological behaviour isn't invariant under homeomorphism

[alpine nest]

In what way is that an "opinion" rather than a fact?

[hexed steppe]

see here

[alpine nest]

It's this kind of stuff that gives moral relativism a bad name

[high hill]

did someone mention spanning sets

[alpine nest]

You, just now

[empty grove]

{}

[alpine nest]

I mean, you can't discuss convergence of sequences/series without some kind of topology, so fair enough

I mean, you can, but you really shouldn't

[empty grove]

A series is the span of a sequence of you view the set of numbers as a vector space over F_1.

if*

[unreal stratus]

lol yh

[ebon galleon]

Opinion: bases are siwwy

[vagrant sand]

Time to develop baseless topology

The long lost cousin of pointless topology

[ebon galleon]

No bases in pointless topology

[alpine nest]

The opposite of based topology is cringe topology

[high hill]

bases... can we calls them spanning sets...

spannings

[alpine nest]

NO

[high hill]

spannings it is

[alpine nest]

Stop trying to make algebraic topology happen, it will never happen

[high hill]

so the half open intervals of the reals is a spanning

wait thats bullshit

[alpine nest]

I thought that was the idea

[high hill]

in some appropriate topology*

i feel like ive mixed measure theory up here but its ite

[white oxide]

if we use basis in the sense of a minimal spanning set, then they are

[high hill]

exact wait

minimal

[white oxide]

if you want the standard topolog you kinda to get rid of the half opens

[alpine nest]

Spanning in what way?

[white oxide]

producing the topology or finer

[high hill]

yeah so im sure i thinking measure theory things

i just mean the open intervals

is a spanning

[white oxide]

oh

les go

[alpine nest]

I mean, they generate the topology and they generate the Borel sigma-algebra

I'm vaguely uncomfortable with the term "spanning" in this context, since I think that's strongly associated with linear combinations

[white oxide]

thats why the minimality as a sorta analogue

[alpine nest]

Open intervals aren't a minimal generating family though

[white oxide]

under the order relation $A \geq B$ if $\tau_A \subseteq \tau_B$

[gentle osprey]

Kerr

[alpine nest]

You'd need a subset of the open intervals if you want to have minimality though

Zorn Lemma probably tells you that you can always find a minimal generating set, but "all open intervals" ain't it

Actually Zorn Lemma might not work

But either way, you can remove quite a lot of intervals and still have a generating set

[high hill]

thats exactly why i think spanning

[white oxide]

Yeah I am sucking at expressing it

[high hill]

linear independence where

a basis with more open sets is still a basis

[white oxide]

but ideally A < B only if B generates a strictly finer topology

[alpine nest]

What are the vector space operations on subsets of the Reals?

[high hill]

asking too many questions!

but nah like

idk

basis doesnt suit to me

basis is meant to be like the smallest

in a sense

for LA

but that's not it for topo?

[alpine nest]

If you're identifying a set with its indicator function, then sets are identified with real-valued functions of the real variable, but only 0-1 valued functions

[alpine nest]

Yeah, topologically you don't have anything like linear independence/minimality in the definition of basis

It's the same word but there isn't much in the way of analogies

[white oxide]

if we mean by "smallest" the actual cardinality of the set and not that of the topology it generated, then we are looking at aleph null

[alpine nest]

Maybe a bit in that in algebra every vector is a linear combination of basis vector, and in topology every open set is an union of basis sets

[high hill]

ok ok think like this

[alpine nest]

But there's no uniqueness or anything like that

[high hill]

groups u have generating sets

thats exactly what bases are in topology

but calling them basis for groups is...

[alpine nest]

I'm not sure if exactly, but that's closer than trying to equate it with the vector basis

[white oxide]

but with groups the generated group can only ever be a subset

[high hill]

?

[white oxide]

if you add more opens to your topology than you can just getting finer topologies

[high hill]

disagree

you can add more elements to your group

[alpine nest]

They're all subtopologies of the discrete though

[high hill]

but if u start with the same bunch of opens

[white oxide]

what?

[high hill]

or the same bunch of elements in group

ok ok so ur saying

[white oxide]

by generating set you mean the generators then?

[high hill]

{1, 2, 3} generates Z group
open intervals generates R top

ok?

no matter what I add to 1, 2, 3 from Z

i end up with Z or a subgroup of it at most

[white oxide]

yeah

[high hill]

but the same applies from R top I argue

[white oxide]

Okay sure

[high hill]

As long as you are only picking open sets

already in the top

If u add more sets that werent open

that's like adding more elements to the group not in it

i would say

[white oxide]

makes sense

take Z as embedded in Q

to extend the analogy

[high hill]

I compare generating sets for groups and topological basis because you do the same things.
Take a collection. Take the minimal algebraic structure containing that collection

and thats your thing generated

(minimal is intersection of all structures containing it)

and this also applies to rings and fields

and spanning sets for vector spaces

===
but tbh ive brought this up before in here and I think the general "why" for the term is more historical than anything else iirc

some ppl use it and it stuck

[empty grove]

I recently realized that this "subthing generated by S is the intersection of all subthings containing S" is an instance of the general statement that a complete meet-semilattice is also a complete join-semilattice and hence just a complete lattice. You can prove it the same way, but it is more fun to nuke it with Freyd's adjoint functor theorem (if you unpack that proof it's basically the same, just phrased in categories by viewing posets as cats)

It's so wacky

[vagrant sand]

this is epic

[plain raven]

That's pretty cool.

[merry geode]

Did I start this stream of convo about "opinion:", was what I said that bad

[white oxide]

your opinion was so uncontroversial that it caused integer overflow

[knotty vine]

By Freyd's theorem you mean this one? https://ncatlab.org/nlab/show/complete+small+category#CompleteSmallCategoriesArePosets

[empty grove]

Nope

If D has all limits and R: D → C preserves them and a technical set theoretic condition, then R has a left adjoint

You can find it in Mac Lane with the name "Freyd's adjoint functor theorem"

[knotty vine]

Right but how does that imply cocompleteness?

[empty grove]

The Wikipedia page on the Stone-Cech compactification also carries out essentially its entire proof so you can look at that to see it in action

[empty grove]

Existence of all colimits of shape J is equivalent to the diagonal functor C → C^J having a left adjoint

So you show that this diagonal functor satisfies the conditions of the theorem for any small J

[ebon galleon]

I love nuking things

[empty grove]

You'll love deep rock galactic

[ebon galleon]

I gave a presentation on this!

I think it's a fun application of the adjoint functor theorems

[knotty vine]

ultrafilters my beloved

[empty grove]

One of my favorite useless things

Idk if SC compactification ever comes up

[ebon galleon]

Adjoint functor theorems can be useful

[knotty vine]

So the only reason complete categories arent already cocomplete is the solution set condition?

[empty grove]

I have used it to cook up counterexamples

[ebon galleon]

Reflective subcategory

[empty grove]

Ye

[ebon galleon]

That fact itself is more important than the construction itself lol

[knotty vine]

funny

[empty grove]

But how do you use it?

I guess it tells you about colimits

[ebon galleon]

Cocompleteness is a useful one

[empty grove]

You use completeness of CompHaus if you wanna use Freyd lol

[ebon galleon]

I don't have examples offhand but generally (co)reflective subcategories are very nice.

[empty grove]

That comes from Tychonoff anyway so not a big deal

[ebon galleon]

Missing a co, sorry

[empty grove]

Ah ok

[knotty vine]

coco

you mean mpleteness

[empty grove]

I think it's used in some obscure analysis proof

[ebon galleon]

One other thing I've used with them is the fact that the (co)unit is a natural isomorphism

[knotty vine]

As a side hussle I am also a median: hire me for parties and weddings

[empty grove]

Where you show some property for discrete spaces and that stone cech compactification preserves the property and that proves it for a large enough class of spaces that you are done

[ebon galleon]

Oh, because you're the dual of funny?

[empty grove]

Yeah but you only have a counit after you've constructed the adjunction

[ebon galleon]

But yeah SC is a fun application of SAFT, but I can't point to a particular usage of it offhand. It gets thrown around here and there though! But again I think from what I've seen it's often just about the reflective subcat part

[plain raven]

if everything's finite dimensional

[empty grove]

Idk what the counit would encode a priori

[knotty vine]

True, I have to betray my constructive soul to make that joke

[empty grove]

Ok fair enough

I've just been looking for a concrete use of it lol because it's one of my favorite theorems

Kinda like p hacking here

Find a theorem then find a use for it

[ebon galleon]

If you didn't know about the compact open topology, you can use SAFT to derive the existence of exponential spaces

[knotty vine]

Actually, is the fact that the initial object is the colimit of the empty diagram, but also the limit of the identity functor an instance of this?

[ebon galleon]

By showing product preserves colimits. I almost had a use for that in another category but I fucked up the proof of preserving colimits

[empty grove]

Yeah that is good

But I meant use of Stone Cech specifically lol

I already have uses for the AFTs

[ebon galleon]

Oh I'm confusing myself by this point >.<

[empty grove]

[woven trail]

is the definition of a topology basically defining what the open sets are?

I guess that'd make sense since all the so called topological properties of for exemple R^n are about open sets and how they interact and stuff

[viral atlas]

Yes, and we insist that the collection of sets has the you-know-what properties

[woven trail]

cool

I think I'm getting it all I'm absorbing it

[viral atlas]

[woven trail]

feeling what topology does

I'm genuinelly getting very excited cus that seems like such a smart way of generalizing these kind of stuff

[red yoke]

How does that work

Say, for a topology generated by a subbasis B, do you take the poset of all topologies on S and show the join of topologies contained in B is the meet of all topologies containing B

Where "contained" is defined by taking this poset to be a subset of P(P(S))

[empty grove]

You gotta prove that "principal subthings" exist first (generated by singletons - in this case I'd take a subbasis rather than a basis because non trivial bases can't be singletons). This is usually very easy to describe explicitly.

The topology generated by a (sub)basis B is by definition the join of the principal topologies generated by the elements of B (definition being "smallest topology containing B" which is exactly the condition for being this join).

The AFT argument (or the direct argument that takes the meet of everything above) gives you a non constructive construction 🤡 of this join.

By non constructive I just mean that you don't describe its elements.

[arctic relic]

Where did you get stuck?

[gentle osprey]

Leno

[arctic relic]

Is this the correct problem statement?

Why are you taking an open partition? f^-1(C) U V is not necessarily clopen in f^-1(C).

Right, I think you want to assume that V is a clopen subset of f^-1(C) not that V union f^-1(C) is clopen

Where V is not the empty set or the whole fiber

It’s definitely true the image of V will be closed

Just to make sure: you’re following this argument

For openess in Y: We know that V is clopen and thus open in X. We also know f^-(C) is closed in X. What can we say about the quotient of these two sets and the image of the quotient of these sets under f?

Edited

[gentle osprey]

Leno

[arctic relic]

Clopen means its closed and open right?

So its certainly open

I mean i guess I should’ve said clopen in f inverse C

Dont think open wrt to X matters thats on me

But like its the same thing i think

[red yoke]

How does one visualize 1-surgeries on a 3-manifold

Say, for surgery on 3-sphere along borromean rings, each 0

I'm getting the impression that surgery along a knot may somehow be equivalent to subtracting a solid torus and then quotienting each longitude (+n meridians) into a point

But that be complicated quickly

[woven trail]

there's a real term called surgery in topology???

[high hill]

yh...

someone once chucked me this

idk why kek

[hexed steppe]

https://arxiv.org/abs/math/0303109

[red yoke]

Thanks, they look pog

[rancid umbra]

give it one more shot in the simplified case when C = Y

if you get stuck, ill help you out

[prisma garnet]

topology basically determines which maps between two spaces are continuous

the easiest way to motivate that definition is by considering euclidean spaces

where you already have an intuitive notion of continuity (by means of analysis)

that is, a map $f\colon \bR\to\bR$ is continuous if for all $x_0\in\bR$ and given some $\epsilon >0$ there exists a $\delta >0$ such that
[\abs{x_0-x}<\delta \implies \abs{f(x_0)-f(x)}<\epsilon]

[gentle osprey]

DarQ

[prisma garnet]

turns out that there exists a set of subset of $\bR$ denoted $\mathcal{O}\subset P(\bR)$ that detects exactly which of the functions $f\colon \bR\to\bR$ are continuous. i.e. $f$ is continuous if and only if
[f^{-1}(O) \in \mathcal{O}]
for all $O\in\mathcal{O}$

[gentle osprey]

DarQ

[prisma garnet]

this set of subset satisfies the usual definition of a topology

[prisma garnet]

this is useful for generalizing the notion of continuity coz you don't need to be able to add or substract the elements of your space as in this defenition

your space doesn't have to be imbued with a metric

you need only determin which subsets of your space you care most about

hope this helps

anywho

does anyone know what the fuck does T do?

that geometric motivation makes 0 sense

also, tf are the simplices in \partial \Delta^n \times I?

[fading vale]

I have no idea what this is referring to without context, what is LC_n

[prisma garnet]

linear n-chains on Y

Y is a convex subset of a euclidean space

[prisma garnet]

the context is the algebraic barycentric subdivision of linear chains

$b_\lambda$ is the image of the barycenter of $\Delta^n$ under $\lambda$. $b_\lambda$ also denotes a map $LC_n(Y)\to LC_{n+1}(Y)$ defined by
[[w_0, \dots, w_n] \mapsto [b_\lambda, w_0, \dots, w_n]]

[gentle osprey]

DarQ

[prisma garnet]

I'm sorry, thix construct is like 5 pages long, I realize now how much context is missing

here's the definition of S

[fading vale]

T takes a simplex Delta, thickens it, divides this thickened simplex up into simplicies using barycentric subdivision like in the image, projects this subdivision onto Delta, and then uses it to define a new simplex of rank one higher

look at the image they have for n = 2

theyre thickening the triangle and taking a barycenter there, and subdividing. if you project that subdivision down you are left with the usual barycentric subdivision of Delta^2

[fading vale]

which defines a 3 simplex via this map

I would assume that for n > 2 the subdivision becomes more non trivial and probably stops agreeing with the standard one

is this the proof for excision

[prisma garnet]

what does the barycentric subdivision of a thikened simplex mean tho?

hatcher here is sayin that the subdivision is obtained by joining all the simplices in $\Delta^n\times {0} \cup \partial \Delta^n \times I$ but what're the simplices in $\partial \Delta^n \times I$?

[gentle osprey]

DarQ

[prisma garnet]

oh, wait, I think I get it

ah yes, the good old "stare at it for long enough" technique coming in clutch

[broken nacelle]

thx moth!

[forest prawn]

If I have a manifold with non trivial first cohomology, and take a non-trivial $\omega$ of it, can I embed a circle times some interval in it, with some “angle function” $\theta$ such that $\omega =d\theta$?

[gentle osprey]

Ben

[fading vale]

Im not sure what exactly you mean by an angle function. are you asking for a cylinder embedded your manifold X such that the pullback of omega is exact?

if dim of the manifold is at least 2 then you can just take any contractible (embedded, by taking a chart about some point and embedding a circle in it) loop, take a cylinder about it that extends to a disk in the manifold, and then the restriction of omega to the cylinder factors through the restriction to the disk, where it is trivial, and thus it will be exact

Or do you mean something more analogous to the case of the circle in R^2, where d theta is not actually an exact form

[forest prawn]

Let me formalize it more carefully

Let $0\neq[\omega]\in H^1_{dr}(M)$ is there necessarily some non contactable embedded cylinder, and some representative $\eta \in [\omega]$ such that the pullback of $\eta$ is the angle map of the cylinder, maybe up to some constant

[gentle osprey]

Ben

Let $0\neq[\omega]\in H^1_{dr}(M)$ is there necessarily some non contactable embedded cylinder, and some representative $\eta \in [\omega]$ such that the pullback of $\eta$ is the angle map of the cylinder, maybe up to some constant

[woven trail]

I remember something like that from a real analysis book

like, topological version of continuity

[woven trail]

:0

not using a metric was what most surprised me for defining those stuff

I'm completely surprised

[forest prawn]

If your looking for a cool not-so-easy challenge, try finding a property a topology induced for a metric has, that not all topologies have, and use it to find a topology which necessarily isn’t induced by any metric

[fading vale]

so off the top of my head i think you need some assumptions on pi_1 or the dimension of the target space (but not both) but otherwise yes

Since omega is non-trivial there exists some non-trivial first homology class on which it does not vanish. since H_1 is the abelianization of pi_1 this is represented by some (smooth, up to homotopy) map S^1 -> X

If we can homotope this smooth map into an embedding, then since dim X > 1 you can take the normal bundle to S^1 in X and then pick some copy of S^1 x I in it (im not totally sure why you want an embedded cylinder since I don't think it particularly changes anything from just getting an angle form on a copy of S^1 itself though). picking eta representing this and restricting it to omega, we have that eta = c d theta for some constant c, so their difference is given by df, where f is some function on S^1 x I whose integral along the circle is 0. extending f by 0 using a partition of unity to all of X, eta - df is in the class of omega and should give you the form that you want

So the question is whether we can resolve the map S^1 -> X

since dim X > 2(1) - 1 = 1 we can always resolve it to an immersion. if dim X is at least 3 we can further resolve it to an embedding, as desired

[forest prawn]

Thank you!

[fading vale]

Hmmm

No we dont need any pi_1 assumptions

They do nothing for us here

I was thinking in analogy with the case of 2nd homology classes

[forest prawn]

Does not vanish means integrating the form on the homology class isn’t zero?

[fading vale]

Right

[forest prawn]

Okay cool, so assuming compactness it must be a circle

And now it’s just analysis of functions on a circle

[fading vale]

off the top of my head im not sure about the dim X = 2 case. if X is actually closed and orientable I think you can probably do something slick by using that omega corresponds to a Z2 bundle (you can use a similar argument to show that 2nd homology classes in a closed orientable 4 manifold are represented by embedded spheres)

this could also just probably be done concretely and directly

But i am a bit too lazy to think about that right now so i hope this suffices for your purposes

[forest prawn]

It does in fact

The goal was to feed curiosity - thanks!

[fading vale]

np

[woven trail]

hmmm

by not induced by a metric you mean

as basis being open balls in the metric?

I don't know any properties about topologies rn sooo I think I'll think about it later

[heady skiff]

i'm not really sure if i've got the hang of this yet, but could we just use the fact that X x Y is a (co? always forget the arrows )product in the category of topological spaces, so we get a factorization X x Y -pi-> X --> Z = F, and since pi is continuous h: X --> Z must be continuous

i guess the only problem would be that we don't know that h is the map that pops out from X to Z

so maybe not

lol

[hexed steppe]

what

that is very circular lol

[heady skiff]

oh wait nvm this is just the composition x |-> (x, y0) |-> F(x, y0)

[hexed steppe]

“assume h is a morphism in Top, then …”

why is the first map continuous

[heady skiff]

nvm

i was viewing X as a subspace of X x Y but that doesn't make sense

oh wait

the first component could just be the identity map and the second the constant map at y0

that's continuous since each are continuous as maps from X to X and X to Y

[hexed steppe]

sure

actually your definition of categorical product is not even right

you should probably focus on understanding the basics of topology before trying to understand that approach

it is a powerful language but it doesnt allow you to avoid proving things

[prisma garnet]

oh, I thought you already knew which sets I was talking about

they're the sets where every point is an interior point

[lean knot]

how do you use the Gram–Schmidt process to prove there is a direct product decomposition of GL(n,R) to the direct product of O(n) and positive definite symmetric matrices

[high hill]

#linear-algebra bruh

[golden gust]

is there a name for a collection of subsets of a topological space, whose interiors form an open cover?

[prisma garnet]

I was wandering the same thing yesterday actually!

bonus question, does $C_n^\mathcal{U}(X)$ and $H_n^\mathcal{U}(X)$ have any special names?

[gentle osprey]

DarQ

[woven trail]

open sets?

[prisma garnet]

yes indeed, these are the open sets of the real line.

but that's what I'm trying to explain, we call them open sets because they detect exactly which functions $f\colon \bR\to\bR$ are continuous. The same way the open sets are the sets that detect exactly which functions $f\colon X \to Y$ between two topological spaces are continuous

[gentle osprey]

DarQ

[prisma garnet]

that's how I think about the definition of a topology, it generalizes the notion of continuity.

[prisma garnet]

hopefully this explains exactly what topology does

[woven trail]

:0

[naive trench]

Gives you a certain type of isomorphism called homeomorphism just adding continuity to isomorphism. Some properties of an space are invariant by homeo and thats pretty useful

[unreal stratus]

any idea what lambda_-1 means here?

Ah nvm, it's a case of λ_t from the λ-ring structure

[heady skiff]

can somebody explain to me why we need this display? if $y \in V$, then $\frac{\bar{d}(x_i, y_i)}{i} < \bar{d}(x_i, y_i) < d(x_i, y_i) < \epsilon$ for $i = 1, \dots, N$, so shouldn't that be enough to show that $y \in B_D(x, \epsilon)$?

[gentle osprey]

okeyokay

[merry geode]

You mean which display? The former part is rather obvious part, the important part is about the i > N imo.

[heady skiff]

the highlighted part

i don't understand why it's needed

[merry geode]

I believe it is to highlight the i > N case.

The i <= N case is likely included here to briefly show how this part is not a problem

[heady skiff]

i see, okay

thanks

[woven trail]

does an intersection of a set with itself count?

[gleaming warren]

Well $A\cap A=A$ and $A$ is a finite intersection of $\mathcal S$ as $A=\bigcap_{n=1}^1A$.

[woven trail]

good

indeed

[gentle osprey]

Max

Absta

[woven trail]

🙏

[tribal palm]

i had a slight existential crisis when i first tried evaluating the intersection of all the sets in the empty set

[red yoke]

You mean, universal crisis

[gaunt linden]

The crisis is about whether the result exists, though.

[ebon galleon]

It exists by convention/definition

[thorny agate]

it exists by convenience

[unreal stratus]

Replace union and intersection by lower and upper bound in the poset of subsets

[merry geode]

Empty set is the identity in the monoid of set unions

[high hill]

[unreal stratus]

If $X$ is a spectrum, to show that $H\mathbb Q_(X) \simeq \pi_(X) \otimes \mathbb Q$, is it enough to say that 1) both sides preserve filtered colimits and shifts and 2) they agree on the sphere spectrum (by classical work of Serre)?

[gentle osprey]

potato

[empty grove]

Is every spectrum a filtered colimit of the sphere spectrum? I feel like you should also need colimits at the very least

I mean coproducts

[white oxide]

What is a filtered coproduct

[woven trail]

by contained in T1 and T2 does this mean that it is a subset of both at the same time or that it's a little bit of the subset of one and a little of the other

like a subset of the union

[plain raven]

it should be a subset of T1 and T2.

[woven trail]

that is very convenient

since the intersection of two topologies is also a topology

[empty grove]

Filtered colimit is when you union things in a larger thing but possibly many times

Formal definition is colimit of a filtered diagram but I'm not gonna define that

[ebon galleon]

The "[moldilocks] means coproducts" part was for the second sentence

There is no "filtered coproduct"

[unreal stratus]

Okay sure but both sides preserve finite coproducts too ig

How would you argue it?

I think it's pretty cute that agreeing on the spectrum is work of Serre, like it's a really pretty rephrasing of a weak version of his computations

[woven trail]

why is the Rk topology that seems so random

[unreal stratus]

wdym

i cannot parse that question

[naive trench]

Rk topology? Which one is that?

[woven trail]

why was it invented what is its purpose

[woven trail]

K = {1/n for all n in N}
basis are open intervals -K

[woven trail]

maybe I'm asking questions too fast

[white oxide]

Oh the K-topology

[ebon galleon]

Various counterexamples.

[naive trench]

Never heard of it before

[woven trail]

hum

[white oxide]

It's hausdorff but not regular otherwise I don't see the point either

[woven trail]

that was one of my hypothesis, that it was made to be a counter example of something

[naive trench]

Like sorgenfey topology

[white oxide]

Point-set topologist are too occupied with if they could make another pathological space, but never even bothered thinking if they should smh

[alpine nest]

True

[merry geode]

Are the points the problem of point set topology

Or is the "point set" part misnomer?

[unreal stratus]

I believe the point is that things like algebraic topology and homotopy theory care about spaces more categorically/algebraically, whilst point set topology is more occupied with the precise space itself (qua space rather than through the lense of those invariants etc)

roughly

[merry geode]

Ah, space itself. Hmm

Does many topologists mainly deal with manifolds?

[novel talon]

I'm new to topological groups, but what does it mean formally when we say that rotating the unit sphere S^2in R^3 about the z-axis is an example of an action of the circle S^1 on the sphere?

[red yoke]

The circle is the group of unit length complex numbers under multiplication

[white oxide]

The rotation of S^2 can be uniquely described where it sends points on its equator

Which is homeo to S^1

[red yoke]

S¹ acts on the unit sphere by rotating the xy coordinates

[novel talon]

We should have a map S^1 x S^2 -> S^2 right?

[red yoke]

That is, e^iθ sends (x, y, z) to (x', y', z), where (x', y') is (x, y) rotated by θ

Or if you take S² ⊂ R³ = C × R, then f: S¹ × S² → S² is given by f(z, (w, r)) = (zw, r)

(actually (z^n w, r) is also rotation)

[white oxide]

consider the action of the group of rotation matrices of R^2 G and consider the action G x R^3 -> R^3. Convince yourself that when restricting R^3 to S^2 you get image in S^2

the action works like arki described

then you can convince yourself that, going back to the complex of unit length, that those form a topological group which is homeomorphic to S^1

Unless that was already your definition of S^1 as a top group

[empty grove]

Once you have this agreement on S, shifts, coproducts and filtered colimits, you should have agreement on all CW spectra. If you know that these respect weak equivalences, then this should be enough.

[unreal stratus]

Yup cool beans

So yeah the same thing aha

Though preserving coproducts + filtered colimits implies preserving all colimits right? at least for 1-cats (since ig these are homotopy colimits)

but i imagine the result is still true more generally

[empty grove]

oh does it?

That seems cool

[unreal stratus]

Well 1-categorically this is like

all colimits can be constructed via coproducts and coequalisers

But then if you have finite coproducts and filtered colimits then you can construct coequalisers as a filtered colimit of like finite ones

and then preserving them is a relative version of that

[empty grove]

Yeah thats what I am struggling to see

Intuitively I see that it should be true

[unreal stratus]

which bit?

Or just all of it lol

[empty grove]

This

[ebon galleon]

I don't believe that's true

[empty grove]

Doesn't homology preserve filtered colimits and coproducts? But it doesn't preserve all colimits

[ebon galleon]

You need to finite colimits and filtered colimits iirc

[empty grove]

That would make sense

[unreal stratus]

Hm if you have finite coproducts and filtered colimits you immediately have all coproducts and coequalisers right?

Like arbitrary coproducts are filtered colimits of finite ones

[ebon galleon]

You construct arbitrary coproducts in this fashion.

[unreal stratus]

And coequalisers are filtered right

lol

[ebon galleon]

Coequalizers are not filtered

[unreal stratus]

Wait yeah lol that was dumb

[empty grove]

In this particular case though I think it works by essentially coyoneda

[unreal stratus]

Oh yes sorry you are right

[empty grove]

Take all the coproducts of shifts of S mapping to your given spectrum X

[unreal stratus]

my thing only servers to show you have all coproducts

[empty grove]

Right

[unreal stratus]

but not all coequalisers

sadge!

Mb

[empty grove]

This, along with maps of the slice category, should be a filtered diagram

This would express any spectrum as a filtered colimit of coproducts of shifts of S

Ok something might be wrong here

The topology needs to work too

And for that I might need X to be CW

[unreal stratus]

But then it's fine due to wweak equivalence right

[empty grove]

Set theoretically this works I am just not sure if we get the same topology. There should be a way to reduce this to coyoneda if it works generally because the shifts of S are representable

[empty grove]

Assuming that we do get the CW stuff even

I will need to think about this properly which I can't really do rn

[unreal stratus]

To be fair, here's a funnier way to do it

It's true for suspension spectra by normal topology things

Then use homotopy colimits and shifts lol

[empty grove]

lol

Cool

Again seems intuitive that every spectrum should be a homotopy colimit of shifts of suspension spectra but idk how to prove it

[unreal stratus]

Oh i mean just like

[empty grove]

Do we need homotopy colimits? What about plain colimits

I think plain colimits work too

[unreal stratus]

Actually

I may have messed up

[empty grove]

Ah set theoretically they do but I am not sure about the topology again

Damn it

I keep applying the unenriched coyoneda

But the coyoneda to be applied here is the Top_*-enriched one

[unreal stratus]

No it's oay

*okay, I checked what i had in mind lol

as in if we view a spectrum $X$ as a sequence of spaces $X_n$ then $X \simeq \mathrm{hocolim}_{n} \Sigma^{\infty-n} X_n$

[gentle osprey]

potato

[empty grove]

Oh cool

Cant be isomorphism very sad

Because I can't do an abstract nonsense proof of that

[unreal stratus]

Eh just do it infinity categorically

[empty grove]

Imagine doing topology

[unreal stratus]

:)

[empty grove]

🤮

[empty grove]

oh

nvm this good

lol

I need to get there I have been doing so much other stuff but now infty cats are very high priority I feel

[unreal stratus]

same for me oopies

[empty grove]

Missing out on a lot of important tools

lol

[unreal stratus]

though i wonder if i'm gonna get an advisor who doesnt use infinity cats lol

well relatively low chance

[empty grove]

I have choices

So many homotopy theorists here

[unreal stratus]

Hot

[empty grove]

so true

[unreal stratus]

You're already doing a phd though right?

[empty grove]

yep

Are you doing a masters?

[unreal stratus]

Yeah just applied for PhDs

Actually got one more to do lol

[empty grove]

epic atb come where I am with toki

[unreal stratus]

But yeah basically homotopy theory within uk

o

[empty grove]

ah fuck im not in the uk

[unreal stratus]

sorry

[empty grove]

F

[unreal stratus]

or rather i shoudl say

lucky you

lmao

[unreal stratus]

for not being in the uk

[empty grove]

oh fair

USA numbe 1

[unreal stratus]

lol

Reading some CHT notes by Lurie and it's funny when he uses AG words i don't know

[empty grove]

He is such a show off

[unreal stratus]

Well we have sets X,Y and a group G and instead of saying like X/G iso to Y he says X is a principal homogeneous space

and stuff

Oh okay there is a more specific meaning

[empty grove]

Yeah that's what happens when the french study your subject

[unreal stratus]

but it's still funny like words i've never heard

[empty grove]

Weird terminology

[unreal stratus]

mixed in with stuff that's fine

yeah fair

étale

[empty grove]

espace

[umbral panther]

étalé

[unreal stratus]

real

[strange talon]

réel

[plain raven]

this is not a term of algebraic geometry it is a term of topology

you can look at fiber bundles by husemoller to read about this stuff

[unreal stratus]

Ye ok

I just found it funny when they are only sets

[tribal palm]

what the hell is this exercise asking??: “prove that any base for the canonical topology on R can be decreased”

particularily it’s the word decreased i’ve no idea what they mean with

[red yoke]

Every basis has a proper subset that is also a basis, probably

[tribal palm]

waw that’s an interesting property

[unreal stratus]

Never heard that but I would say "there is no minimal base"

[tribal palm]

can you do something like pick any d > 0, then take as a new basis all the old basis elements with diameter less than d ?

then you can get an infinite descending sequence of bases by taking smaller and smaller d

if this even is a basis which i admit i haven’t checked

oh actually and i suppose for the new basis to be a proper subset you need to take a d smaller than the diameter of some basis element in the old basis

[red yoke]

👍

[knotty vine]

In R, any (nonempty) open can be written as the union of two proper sub-opens. So any single basic open can be removed

[red yoke]

Ig this works for any metric space without isolated points

[tribal palm]

can you elaborate on this, i’m not quite following

i’ve been embarassingly struggling a bit with working with open sets in R

i’m not sure what you mean by basic open

[knotty vine]

A basic open is just an element in the basis

[tribal palm]

oh

[knotty vine]

If a basic open $U$ is nonempty, we can write it as the union of two proper sub opens $V_1, V_2 \subset U$. These $V$ can consequently be written as a union of basic opens that do not include our original $U$. Hence we can remove $U$ from the basis and still generate the same topology.

[gentle osprey]

lax semer

[tribal palm]

ohhh right

[unreal stratus]

A way to think about it intuitively is like

Topology is sorta local stuff so you can always chuck away all "big" things and shrink

[tribal palm]

am i mixing this up or does the property “at any point x there is a infinite descending sequence of nbhds of x” have a name

[unreal stratus]

Hm i mean it's not always true

But idk lol

There is a relevant notion of a first-countable space ig nd stuff about neighbourhood bases

[alpine nest]

Any non-isolated point in a Hausdorff space will have this property, probably

So in Hausdorff it's going to be equivalent to being non-isolated

Or globally, for the space to have no isolated points

And I refuse to consider non-Hausdorff topologies as a matter of principle

[knotty vine]

This also works in T0 I guess

But something similar can even be said about non T0 spaces, as long as the Kolmogorov quotient (making topologically indistinguishable points equal) has the property

[tribal palm]

on second thought, this is a different, but equivalent statement to first-countability

[alpine nest]

I wouldn't say that, a discrete two-point space is first-countable

And doesn't have your property

[tribal palm]

yes? so both conditions fail

to refute the equivalence one’d have to come up with an example for which one but not the other holds

[alpine nest]

That space is first-countable, and not every point has an infinite descending sequence of neighborhoods

[tribal palm]

oh, right finite is also countable

at least if a space is first countable, let (Bn) ba a seq of nbhds at x, then the seq of intersections B1, B1 n B2, B1 n B2 n B3, … is a (i suppose not strictly) descending seq of nbhds of x

[limpid fern]

You could call it the nested neighborhood basis property?

[unreal stratus]

if one talks about the "order" of a (real) vector bundle V over a space X, is that the order of it as an element of the reduced KO-group of X?

i.e. least n such that the n-fold sumV (+) ... (+) V is a trivial vector bundle

[broken nacelle]

why save you from homology?

homology is love

homology is life

[alpine nest]

Homology isn't real and can't hurt me

[limpid fern]

No need

I finished the homology presentation a month ago

[umbral panther]

If I had to guess, that’s how I’d interpret it, but something has gone wrong if it’s not spelled out. What’s the context?

[broken nacelle]

you did a homology presentation??

whoa

what did you cover

[unreal stratus]

Just vector bundles, like

[limpid fern]

only simplicial homology

[unreal stratus]

To be fair I can check the reference there to be 100% sure but wasn't sure if this is common terminology

[limpid fern]

i only had 10mins to talk about it

[umbral panther]

Do they at least mention KO?

[unreal stratus]

Not afaik

Yeah I checked the reference and it is just least n such that the n-fold sum is stably trivial

i.e. 0 in reduced KO

I think here is it just a remark which isn't used anywhere in the paper, unless I'm mistaken

[gritty widget]

Let X and Y be two metric topological spaces. Let $A \subset X$ and $B \subset Y$ . Show that $\overline{A × B }= \overline{A} × \overline{B}$.

solution : let $x \in \overline{A×B}$ then this is equivalent to say that there exists $x_n\in\overline{A×B}$ such as $x_n\to x $ let $x_n = (a_n,b_n)$ and $x = (a,b)$ then we have $(a_n,b_n)\to(a,b)\iff a_n\to a \mbox{ and }b_n \to b \iff a\in\overline{A} \mbox{ and }b\in\overline{B}\iff (a,b)\in\overline{A}×\overline{B}$ then we have $\overline{A × B }= \overline{A} × \overline{B}$.

[gentle osprey]

john.

[gritty widget]

the Ö is ×

didn't know latex will interpreted like that

is it correct if i reason like that?

[red yoke]

*exists xn ∈ A × B

I'm not sure if you've proven cl(A×B) ⊂ cl(A)×cl(B) or cl(A×B) = cl(A)×cl(B)

You might need to fix your presentation

[gritty widget]

isn't my argument by equivalence ?

[red yoke]

I'm not sure if you mean:
Let (a, b) ∈ cl(A×B), then there are sequences (an, bn) → (a, b), which is equivalent to an → a and bn → b, thus (a, b) ∈ cl(A) × cl(B)

Or:
(a, b) ∈ cl(A×B)
↔ Exists (an, bn) → (a, b)
↔ Exists an → a and bn → b
↔ (a, b) ∈ cl(A) × cl(B)

[gritty widget]

the second

[red yoke]

Ye then that's correct

Mod presentation

[lean knot]

If open sets U ⊂ R^n and V⊂R^m are homeomorphic, then why is n = m?

[red yoke]

If f: U → V is a homeo and n > m, then we have a continuous injection U → R^n by considering R^m as a subset of R^n. By invariance of domain, the image of this map is open in R^n, which is clearly false.

Unless U is empty

Invariance of domain can be proven using algebraic topology I believe

[woven trail]

in metric spaces, are products of any open balls also open balls?

not open balls

just open

are they open

just realised this question doesn't necesserely makes sense

cus you'd need to define the product metric

[alpine nest]

If A is open in X and B is open in Y, then A x B is open in X x Y

[woven trail]

I know

[alpine nest]

Which generalizes to finite products of more spaces, but not to infinite products

[woven trail]

but I was thinking of the open balls thing

[alpine nest]

The products of two balls might not be a ball

[woven trail]

yes I know

but like

[alpine nest]

Although that might depend on how you define the product metric

[woven trail]

I was thinking of metric spaces

yeah

[feral copper]

Why ask a question if 'you know'?
Ask a more precise question if you want more precise answers...
A product of balls may or may not be a ball, it depends if you take the product metric or something different (that can or cannot induce the product topology)

[woven trail]

I asked and immediately realised my question didn't make any sense

[naive trench]

Product of open sets is open but there are open sets that cant be expressed as product of open sets

[unreal stratus]

Puzzled about smth - how should I think about the Borel construction as a homotopy colimit?

I want to say, for example, that if X is a G-space and we use one of the standard model structures on G-spaces then X//G should be (-)/G applied to some sort of resolution of X

But it seems that X could already be fibrant-cofibrant and it still isn't good enough to apply (-)/G

I guess the point is you use a simplicial resolution which just spits out X x EG essentially

Actually no i messed up a bit lol

[umbral panther]

Free is cofibrant in an appropriate sense. Cofibrant in the category of objects with G action. But that’s kind of circular. First you should convince yourself that freeness is enough that the derived functor is well defined. And then use that to motivate a theory of cofibrancy

[unreal stratus]

Sure okay

In what sense do you mean "well-defined"? As in G-equivalences are sent to (standard) weak equivalences?

[umbral panther]

Yes

[unreal stratus]

Hm I'm not sure how to show that it preserves weak equivalences other than using like

Well the LES in homotopy groups fibre sequence X -> X//G -> BG and naturality?

As in we should get this uh

... -> π_n(X) -> π_n(X//G) -> π_n(BG) -> ...
| | |
... -> π_n(Y) -> π_n(Y//G) -> π_n(BG) -> ...

And then apply 5-lemma

Or is there some lower-tech way to see this

[umbral panther]

I think you should be able to do this without any algebraic invariants

[unreal stratus]

Well this was about weak equivalences so some reference to π_n is necessary right

Well ig you could use some CW replacement etc

[umbral panther]

Yeah, you have to use whitehead

[unreal stratus]

Sure

Wait in what way do you mean?

I was interested in weak equivalences so I'm unsure how thatd help

(thank you for your help btw)

[umbral panther]

You have to use algebraic invariants or you have to use cw complexes. Whitehead is to get between then

But if you can use whitehead for ordinary spaces with no mention of G, that’s better

[unreal stratus]

Sure

[unreal stratus]

Further research seems to imply this is the standard way other than some formal nonsense with model cats

i'm not entirely sure how you'd use model cats in a way that isn't almost assuming this result though

[umbral panther]

Yeah, I guess what you need is that if X, Y are free and cw and X -> Y is an equivariant that there is a map back

I think you can do this by hand, but it might not be worth it on a first pass

[unreal stratus]

Sure

Hm

Anyway thanks

I guess one thing I'm finding hard is thinking about homotopy colimits invariantly rather like

"Here's a recipe which gives something good"

I think ideally it can be phrased in terms of homotopy commutative cones though right

[umbral panther]

How do you think about derived functor in homological algebra? The recipe is for a resolution. But what’s important at the resolution is that it has a lifting property so you can get maps both ways between different resolutions, so any one resolution is approximately universal

[tidal lynx]

First ever ascii commutative diagram

[thorny agate]

read any book old enough and all commutative diagrams are ASCII for all intents and purposes

[broken nacelle]

the blue book

it was so bad even people that don't study algebraic topology agreed it was bleak

[empty grove]

Not quite (if it could, then homotopy colimits would just be colimits in the homotopy category). The universal property is infinity categorical, so it is about cones which commute up to all higher homotopiss

Maybe that's what you meant too but just making sure

[unreal stratus]

that's what i mean ye

uh homotopy coherent cones i suppose

[empty grove]

[unreal stratus]

An online resource for homotopy-coherent mathematicians.

[empty grove]

Me

[unreal stratus]

Yes

[empty grove]

jk idk infty cat theory

[unreal stratus]

just use vibes

[prisma garnet]

I don't get the part about the second isomorphism

[gentle osprey]

DarQ