point-set-topology

[grizzled ibex]

not the mapping

[tiny ridge]

i disagree

[grizzled ibex]

you find it easy to visualize the embedding of rp2?

oh god

[tiny ridge]

i dont visualize the embedding

[grizzled ibex]

thats my point, i am still at the stage of trying to visualize embedding(s)

[bitter smelt]

How do you visualize it ibsen

Other than just parametrizations of lines for example, which isn't really a visualization

[grizzled ibex]

yes

[grizzled ibex]

agreed ffs

[tiny ridge]

first glue a disk to the boundary of the möbius strip. i claim this is rp^2. why? because throw away the center circle of the mobius strip, thats just a disk with an annulus attached to the boundary which is still a disk. the boundary of this bigger disk is glued to the center circle of the mobius strip by z -> z^2 because the boundary of the mobius strip spins around twice around the center circle

rp^2 is nothing but disk glued to the circle by z -> z^2

[grizzled ibex]

obs: it wasn't a exercise or something

i was just wondering myself

btw nice sketches

[white oxide]

Is there some fact about possible deck transformations of S^2 for any space X that S^2 covers.
If not, how does that argument work?

[tiny ridge]

theres not many group actions on S^2 which are free

[coarse night]

try showing that only group that does is ||Z/2||

[white oxide]

Right, properly discontinuous and covering imply the action being free?

[tiny ridge]

iff for hausdorff spaces

throw in compact maybe

in fact every effective finite group action on S^2 is conjugate to a linear action

[white oxide]

Messes with associativity or sth?

[tiny ridge]

for the free S2 problem?

think degree

hm i dont know how to come up with Boy’s immersion of RP^2 in R^3

i know an unnatural cut and paste construction

but its magic that it works

[unreal stratus]

Oh boy

[gritty widget]

wrong channel

[tiny ridge]

ok i have a big idea

draw the “not seifert surface” of the trefoil knot

this is, topologically, a mobius strip

we have to cleverly attach an immersed disk to get the boys surface

probably

at least this should be the source of the Z/3 symmetry of the boys surface

[white oxide]

Okay yeah. Haven't gotten there yet but it seems really handy. Thx

[tiny ridge]

think i see it kekw

uh

no

[tiny ridge]

also we absolutely need a triple point for an immersion of RP^2 in R^3. proof: suppose f : RP^2 -> R^3 immersion, no triple points. let C = {x in RP^2 : exists y s.t f(x) = f(y)} be the double point locii in RP^2. this is a bunch of non intersecting circles in RP^2

for each pair of circles, consider the following modification of the immersion

this changes the surface by adding a tube S^1 x [0, 1] and the effect it has on the double point locus is that two components of it get connect summed

this doesnt change the Euler characteristic of the surface modulo 2. keep repeating until theres a single connected component, a circle of double points. a regular neighborhood of this can look like S^1 x (the letter X) in which case we can desingularize it to S^1 x )(

in which case Euler characteristic mod 2 has remained the same as that of RP^2 ie 1 but it is embeddable in R^3 so it is orientable, which forces Euler characteristic 0. bogus

theres one more case in the last desingularization process, it can look like mapping torus of S^1 x X under the map X -> X which rotates X by 180 degrees. but either way we tube by replacing the X fibers by )( ie xy = 0 becoming xy = eps

Euler chi mod 2 is invariant under this process

[abstract saffron]

Are we doing knot theory?

😄

[tiny ridge]

Still at it, this is so hard lol

I think we need six more immersed “thickened trefoil” like the grey region union the appropriate half-planes that has the blue trefoil as one boundary component

Then one exhausts all the white edges

But then one ends up with a complicated identification problem

Boy was a lunatic

[balmy field]

Very active!!!

[urban zinc]

where does the name "deck transformation" come from?

[gritty widget]

https://tenor.com/view/faro-shuffle-cardistry-world-xm-card-shuffle-gif-16263605

[tiny ridge]

exactly this

[tiny ridge]

its an honor

[urban zinc]

honorable next

[urban zinc]

are the cards in this like the individual sheets and then you shuffle them around

ohhh

[gritty widget]

pretty much exactly that

[urban zinc]

okay that actually makes a lot of sense

ty!

[tiny ridge]

I think I have constructed the Boy’s surface

This is an immersed annulus in R^3. It can be constructed piecewise linearly from the union of three coordinate squares by attaching strips to the edges, as indicated, or envisioned as follows. Take a trefoil knot, then consider a planar projection. Imagine a homotopy of this which modifies it to a circle with a triple point, and then makes it a circle with three double points.

This is an “illegal move” in that it is not a knot isotopy, but the movie of the homotopy gives the surface above.

This immersed annulus cobounds the trefoil knot and the unknot. Cap it off on the unknot end by an embedded disk, and on the trefoil end by the “illegitimate Seifert surface”, namely this image:

[tiny ridge]

.

This should be the final immersed RP^2, if I am not mistaken.

This seems cleaner than any existing exposition I have come across if this works lol

And I don’t see a problem for now. The whole thing boils down to constructing an immersed disk bounding the trefoil knot (which always exists for any smooth knot), and then capping it off by the illegitimate Seifert surface.

[violet summit]

I showed that X is normal, how can I use the given sequence to show that no metric induces the topology O?

[coarse night]

do you have a candidate for the sequence mentioned in the hint?

[violet summit]

Since (0,0) is a limit point for all $L_n$, there exists $x_n\in L_n\cap B((0,0),\frac{1}{n})$ so that $x_n \neq (0,0)$

[gentle osprey]

pramana

[pliant hornet]

It seems to me that, by normality, on each L_n you could choose a neighbourhood of (0,0) which does not contain x_n, no?

[coarse night]

yes

[violet summit]

Do I do that for all L_n? I’m not sure how that proves no metric exists

[pliant hornet]

Think about the topological definition of convergence vs metric spaces.

You want an open subset of X which is disjoint from your sequence, to prove convergence fails topologically.

[coarse night]

if a metric did exist, your space would have to be first countable, what the hint says is assume your space is metrisible then from first countability, you can find a sequence converging to (0,0) but show that does not happen

consider picking a point on L_n that eventually goes closer to (0,0)

[violet summit]

Do you have an exact definition for topological convergence? It wasn't covered in the chapter

[copper raven]

I want to find a fault in my reasoning:
I want to show that every T_3 (regular) space with a countable basis is T_4 (normal).
So if we have two closed disjoint subsets A, B of a space X
I would start by taking points of B and taking union of their disjoint neighbourhoods we're given by T_3 (and that's how we obtain two disjoint neighbourhoods of A and B)

But that doesn't use the countable basis argument we're given so I'm probably wrong somewhere

[coarse night]

your conclusion is infact right but not the proof

[copper raven]

Perhaps for this to work all neighbourhoods of A and points of B should me disjoint pairwise

[coarse night]

look at theorem 32.1 in Munkres

[copper raven]

Yeah I'm looking at it right now, wanted to attempt to prove it before but it seems like I've caught my error anyways, thanks

[eager vigil]

Why does it suffice to check only the case when A is a closed interval?

[copper raven]

There is a theorem that states that g^{-1} is continuous if and only if g is a closed map (maps closed sets to closed sets)

Also, here you could have an intesection of closed intervals, not neccessarily a single closed interval

[eager vigil]

Well, yes, but we don't have g^{-1} continuous given.

[copper raven]

Yes, that's what we want to prove

[eager vigil]

How about arbitrary closed sets?

[copper raven]

Okay so

To start with open sets in R, every open set can be written as the union of open intervals (a, b) (since they form a basis for standard topology on R)

That means that a complement of a closed subset of [0,1] can be written as a union of open intervals of (a, b)

Which means that a closed subset of [0,1] can be written as an intersection of complements of open intervals

[nimble portal]

Is S^k homeomorphic to C^k for k odd natural?

[coarse night]

what's C

[eager vigil]

Agreed, but doesn't seem to show that it suffices to consider the closed interval case

[nimble portal]

Complex numbers

[unreal stratus]

No

[coarse night]

lol

[unreal stratus]

I'll try to think if there's some like particularly elementary way to argue this lol

As in, w/o appealing to more alg top lol

[coarse night]

proof: trust me

[unreal stratus]

lol

[tiny ridge]

try k = 1

[unreal stratus]

Lol it holds for k=1 therefore must hold for all

[tiny ridge]

experimyent, as we say here in russia

[nimble portal]

💀

That’s what made me ask

[unreal stratus]

If you want to appeal to stuff you've probably seen though feather

S^k is a k-dimensional real manifold

C^k is 2k-dimensional

[nimble portal]

Oh so it’s just because (k+1) = 2k is satisfied by k = 1

[unreal stratus]

But yeah I mean this obscures the more important fact that S^a is never homeomorphic to R^b for any a or b

[copper raven]

Okay, so complements of open intervals can have the form: (-infty, a] U [b, +infty)

[white oxide]

One is contratible, the other isn't

[unreal stratus]

The easiest way to see that is using alg top but idk how much you've done lol

Indeed

[nimble portal]

Negative

[white oxide]

Ignoring the dimensionality problem as well

[tiny ridge]

do you know compactness

[unreal stratus]

Oh yeah lmfao

I forgot about that

[nimble portal]

Yes?

[tiny ridge]

prove that image of a compact space under a continuous map is compact

[nimble portal]

wat

[tiny ridge]

X compact, f : X -> Y continuous, prove f(X) is compact

with the subspace topology

then prove that S^n is never homeomorphic to R^m

[unreal stratus]

Lol

Maybe Kerr and I are too alg top brained

[tiny ridge]

also i do not see the relevance of this numerical observation

[coarse night]

point set is pointless

[unreal stratus]

Oh

I think they are saying S^k is k+1 dimensionl or smth lol

[tiny ridge]

my point was S^1 is not homeomorphic to C (can you see that?)

[unreal stratus]

But I said it is k dimensional

[white oxide]

Idk, the fact that C^k has dimension 2k seems like a decent argument against S^k and C^k being homoeomorphic too

[unreal stratus]

I only said that cause I know Lee blackboxes invariance of dimension or smth

[eager vigil]

Yes...?

[copper raven]

Just a second writing more

[white oxide]

Then again, depends on how careful you wanna be with dimensional arguments before getting the homological machinery to proof things like R^n and R^m not being homeomorphic for m unequal n

[eager vigil]

Ah, ok, sry

[unreal stratus]

Oh lol this is funny Ibsen

So I was gonna write out a meme argument for this and it turns out that it is just a more complicated way to appeal to compactness lel

If S^k were homeomorphic to R^b then their one point compactifications would be homeomorphic too

But the one point compactification of the former is like disconnected right lol whilst the other is S^b

💀

[white oxide]

You can compactify R^n by adding a point to make it into S^n tho

So in that sense they do have homeomorphic one-point compactifications

[unreal stratus]

Isn't that what I said lol

[tiny ridge]

no one point c

compactification of S^n is S^n disjoint union a point (as potato pointed out)

[unreal stratus]

yes exactly ye

But this is just what ibsen said in a dumb way lmfao

[copper raven]

And by taking intersection of all those intervals in that form ((-infty, a] U [b, +infty) ), you can obtain two cases either (-infty, sup(a)] U [inf(b), +infty) or just and interval [sup(b), inf(a)] here (a, b) is an open interval we're considering from the union

@eager vigil

But the catch is, that intersection has to contain [0, 1], so it has to be the other case

Which is a closed interval

[unreal stratus]

I also feel obligated to say like

[nimble portal]

No

[nimble portal]

I was saying it was, I thought it was

[unreal stratus]

Usually one would prove this using compactness and things so ye

[nimble portal]

Because S^1 is just the unit circle right?

Oh I guess that’s not all of C

LMAO

But you can normalize all of C to be on the unit circle

So why not?

[tiny ridge]

except 0. but i claim even S^1 and C \ 0 are not homeomorphic

prove all of this by using the compactness exercise i gave you

and figure out why the normalization map is NOT a homeomorphism

[nimble portal]

Once I’m home

:sadge:

[eager vigil]

I see. Thank you

[coarse night]

gimme example of a manifold whose boundary is non orientable

[tiny ridge]

take the Klein bottle, this is an S^1 bundle over S^1. Fill in the fibers by disks

This is a “solid Klein bottle”, a D^2-bundle over S^1 which bounds the Klein bottle

[coarse night]

hmm okay

[tiny ridge]

more precisely its the mapping torus of (x, y) -> (x, -y), D^2 -> D^2

[coarse night]

wasn't there a result saying a mani is a boundary of a compact space iff all the streenrod powers are 0

[tiny ridge]

i think you mean pontryagin numbers

[nimble portal]

wtf did I just read

[coarse night]

it's the "stiefel-whittney numbers" not streenrod powers lol

[coarse night]

are they same?

[tiny ridge]

stiefel whitney, yeah, youre right

pontryagin is for orientable cobordisms

stiefel whitney is when you allow nonorientable cobordisms also

[coarse night]

yeah

[tiny ridge]

this is a huge theorem tho

its by Thom

[coarse night]

yeah lol

I like how milnor drew it as a headless guy

[tiny ridge]

its a nice exercise to prove in an elementary way that RP^2 is not nullcobordant

so while people say that the Klein bottle has no “interior”, thats actually wrong, it is nullcob — but RP^2 isnt

[coarse night]

nullcorb meaning cobordant to empty set?

[tiny ridge]

yeah and more simply that theres a manifold one dimension up with this as boundary

[coarse night]

I know the fancy way of SW numbers

[tiny ridge]

theres a sw number which is not so fancy

[fading vale]

Lol

[coarse night]

w_1 you mean

[tiny ridge]

is that an sw number

what are the sw numbers for a 2-manifold

[coarse night]

yeah it's w²_1

which is not zero

what's the elementary way

[tiny ridge]

just take the other one

w_2

what is it called

[coarse night]

euler class mod2? idk what you mean

[tiny ridge]

yes

prove that euler class mod 2 is a cobordism invariant

euler characteristic, really

[unreal stratus]

hot

wanna learn some cobordism

where is good 2 go for it lol

[fading vale]

this stuff is in milnor stasheff

[coarse night]

kinda remember doing something similar, M is nullcobordant then EC is even

is this correct?

[unreal stratus]

oh i mean i recognise this stuff lol

[tiny ridge]

ye

[fading vale]

Ya

[unreal stratus]

I mean better stuff

/ more advanced stuff

[tiny ridge]

learn the Thom cobordism theorem from somewhere

dunno where

[fading vale]

I think it depends what direction you want to go in

[coarse night]

lol 'ye' for my question?

[tiny ridge]

ye

[coarse night]

cool

[fading vale]

read the green book

🙂

[coarse night]

which one?

[tiny ridge]

complex cobordism

[fading vale]

the green book is complex cobordism and stable homotopy groups of spheres by ravenel

[tiny ridge]

?

[fading vale]

Yeah

[tiny ridge]

ok

lol

[coarse night]

I'll have a look in the summer

[fading vale]

this was not the most sincere suggestion

[tiny ridge]

cobordism theory puts me to sleep

Thom’s proof is just algebra

[fading vale]

i think this stuff is neat though i am certainly not very well read on it

its not something i am interested in getting actually good at for the time being but its fun to read about

[nimble portal]

I like how math has so many names for books

[tiny ridge]

i know nothing beyond the basics

dont plan to either

[fading vale]

chromatic moment

[nimble portal]

Baby Rudin, Papa Rudin, Blue Book, Green Book

[coarse night]

mappa jk

[tiny ridge]

every oriented 3-manifold is oriented nullcobordant is the only theorem worth knowing

which has an elementary proof also

every 3-manifold is a surgery on a link in S^3

surgery is a cobordism operation

S^3 is nullcob

i guess milnor uses the corresponding 7 dim result for exotic lulz

theres a really stupid proof of hirzebruch signature theorem via thom cobordism theorem by showing both sides are cobordism invariant then calculating it out in dimensions 2^k

i think this proof is the original one by hirzebruch

ag brain

[fading vale]

AG brain can be fun

imo

[tiny ridge]

fun fact

CP^2 # CP^2 is not homeomorphic to CP^2 # CP^2bar

[fading vale]

this is an excellent fact

have you heard of the 11/8ths conjecture ibsen

[tiny ridge]

yeah

[fading vale]

we talked about it in class last semester once it was so delightful

[tiny ridge]

i get bored by the quantiative things because i don’t understand them

[abstract saffron]

We have 1/3rd conjecture, and also 11/8th?

Lmao, what's next?

[tiny ridge]

theres a conjecture in freed-uhlenbeck which says that any simply connected closed smooth 4-manifold is a connect sum of complex algebraic surfaces (possibly with reversed orientation). do you know the status of this moth?

[fading vale]

oh wow

No i did not know about this

[tiny ridge]

its like the uniformization theorem in 4D

should be wide open

[fading vale]

i dont know anything about 4-manifolds i need to read gompf and stipsicz or kronheimer-donaldson some day

[tiny ridge]

gompf and stipsicz is very nice but huge

[fading vale]

Yeah

[tiny ridge]

theres a book by akbulut which you can try if youre inclined to do a lot of kirby calculus

[fading vale]

See the thing is

[tiny ridge]

i powered through a good chunk of that last year

[fading vale]

I feel like GS is probably the better book but KD might honestly be closer to my interests

[tiny ridge]

KD seemed super terse

[fading vale]

cut and pasty topology is cute but probably not my calling lol

[tiny ridge]

i dont understand this instanton stuff lol

so much analysis

[fading vale]

Yeah i am scared of the analysis

[tiny ridge]

but thats all it is

[unreal stratus]

Just read his thesis

[fading vale]

if i bash my head against it for a few years surely by the time i go to grad school i will be able to do one (1) computation 🙂

Lmao

[tiny ridge]

french lol

but i think thats a good suggestion

thom writes incredibly well if you can read french

there might be translations available, dunno

[tiny ridge]

my sincere belief is there must be a flexible way of understanding the technical garbage

im years behind on a working knowledge of the field though

[fading vale]

I no longer allow myself to hope

[tiny ridge]

youre still in undergrad, losing hope is for grad students

keep trying to understand and have irreverent dreams as long as you can

[bitter smelt]

Can confirm, hopeless grad student here

By the way, ibsen. The reason I asked the real line with two origins findamebtla group question was that hatcher answered it on math overflow

He really wanted to use SvK but could not, so he took a mapping cone over it and then took SvK anyway

Really Chad move tbh

[tiny ridge]

lmao

[bitter smelt]

Mapping cylinder I mean

[fading vale]

My plan is to switch it up and develop an insane ego in grad school in an attempt to counterbalance the inevitable grad student psychosis

[tiny ridge]

well it looks like a circle to me and i know pi_1 of S^1 cannot be computed by svk so i didn’t even go there

[tiny ridge]

lol

its a thought

[fading vale]

im rationing the happiness

[tiny ridge]

i think the universal cover picture is the cleanest tbh

its like Z many real lines with every other pair glued along the positive or negative half line

tire tracks of a car in a tight parking lot

LMAO: https://sketchesoftopology.wordpress.com/2022/10/03/an-illustrated-counterexample-to-the-smooth-four-dimensional-poincare-conjecture/

https://www.flickr.com/photos/sketchesoftopology/52402424843/in/dateposted-public/

this one killed me

[coarse night]

is it correct? M_Z → M be a orientation cover when M is non orientable, then there's no section?

[tiny ridge]

what is a non section

yes this is correct

if there was a section thatd give you an orientation

[coarse night]

yeah exactly right

In part a why should a nontrivial one exist unless M is R orientable

Are we assuming R orientability? No right?

[tiny ridge]

guess in the nonorientable case lemma is just vacuously true

[coarse night]

Or is it a local section?

[tiny ridge]

around an open nbhd of A? possible

its still true

just replace M by Op(A)

[coarse night]

Yeah it’s not a wrong statement but feels wrong that’s all

[tiny ridge]

i dont like how hatcher spends a million years on M_R

non è importante!

his exposition of poincare duality is quite poor actually

[coarse night]

Yeah

[white oxide]

What does exactness mean for the zeroth homotopy groups?

[coarse night]

Does not make sense to call it exact for pi_0

[sturdy notch]

you think of them as pointed sets, an exact sequence of pointed sets is what you'd "expect it to be" (the base point of pi_0 is the component of the base point of the space)

Ie the kernel of eg p_* is the subset of pi_0(E,x) getting sent to the connected component of b \in B

[white oxide]

Maybe, but whats a meaningful notion of a "kernel" for maps between pointed sets?

[tiny ridge]

preimage of the basepoint

[white oxide]

That's pretty convenient, I'll take it.

[hidden crag]

Hatcher mentions that

I remember being stuck on that as well

[unreal stratus]

As an example: say $p:E \to B$ is a covering map, so it's a fibre bundle w discrete fibre, and let's suppose $E$ is path connected. Then this gives us teh standard bijection $\pi_1(B,b)/p_* \pi_1(E,x) \simeq \pi_0(F_b,x) \simeq F_b$ which is cool, and the point b corresponds to the coset of 1

[gentle osprey]

potato

[white oxide]

Based short exact sequence moment. But what type of "isomorphism" is that there?

Nvm, the left hand side is a group proper. So probably interpreting the factor group as a set of cosets

[unreal stratus]

Yeah

[white oxide]

Can you give this more structure than just being notation that behaves like one expects in specific instances or is there a natural way to make this into an exact sequence in some nicely behaved category?

[tiny ridge]

What will you gain from it

Other than aesthetic pleasure, maybe

[white oxide]

I don't know, that depends on if such a "nicely behaved"/useful category exists. Or framed differently: is there more to this than making the slight abuse of notation work out smoothly?

[gentle osprey]

crazy morally russian topologist

[unreal stratus]

Nice

[fickle hamlet]

Why do we need to use the restriction of Y to Z that is surjective? Isn't it sufficient to say O,V open in Y means f-1(O), f-1(V) open in X by continuity, then f-1(O) intersected with f-1(V) is empty since if there was a shared element then there is a particular x from X that f(x) is in O and f(x) is in V, O and V is empty, so this is a problem with uniqueness of f(x) for a particular x.

[unreal stratus]

Are you claiming that the codomain is always connected?

[fickle hamlet]

if f isnt continuous then f-1(O) and f-1(V) may not be open in X

[unreal stratus]

f is by hypothesis

I'm saying like do you think Y is necessarily connected here

I mean a key point is that you need the preimages of A and B to be non-empty

[fickle hamlet]

oh ok so like one of f-1(O) or f-1(V) could be empty is what you are sayin

[unreal stratus]

Yes

So for example consider the inclusion {0} -> {0,1}

[fickle hamlet]

ya thanks makes sense

[unreal stratus]

np

[fickle hamlet]

would it also be ok to argue for case 2 that if c is in A_o then supA_o=maxA_o and open sets in L intersected with an [a,b] interval cannot contain the max (and we know from case 1 that c is not b)

[hollow grail]

Hello
Any one can give me a simple defenition about "limit point"
Thanks

[stark fog]

if X is a set, a limit point of X is a point for which every open interval around it contains a point other than it on the set

mathematically, if $U$ is an open set containing the limit point $y$ of $X$, then

$$U\cap (X\setminus{y})\neq \emptyset$$

[gentle osprey]

SubGui

[pulsar lagoon]

intuitively we want a limit point of a set to the limit of some sequence(non constant) of member of that set. In metric spaces this is easy to do, in more abstract topological spaces we have to be a little more careful when doing so. We then look back at epsilon-delta for what it means for a sequence to be convergent and come up with the fact that there is an open neighbourhood of your point that intersects your set other than at the point you chose

and this is the definition that @stark fog gave

[stark fog]

they are equivalent in fact

a point x is a limit point of X iff it is the limit of sequence x_n of elements of X, such that x_n are not equal to x

mainly because convergence of a sequence is related to the elements being in an neighborhood of the point for large n

[pulsar lagoon]

the way I interpret is that the sequence definition is the intuition behind the idea

[stark fog]

[hollow grail]

So, if X is a set and have a limit point {a} then every open interval around {a} contains a point other than {a}
@stark fog
Am i right

[pulsar lagoon]

yes

so in R with standard topology the limit points of the set (a, b) U {c} is [a, b]

[hollow grail]

Okay

Thankss
@stark fog @pulsar lagoon

[urban zinc]

construction of the universal cover is so cool :o

but also why do the first three sections of hatcher cover the same material as this 43 video playlist of half hour videos

I think hatcher overestimated my reading comprehension abilities

noooo section 1.3 is like 23 pages long

[tiny ridge]

very important section

[urban zinc]

hm how do I pace this so that I finish it by friday

basic arithmetic is failing me

time to make a grocery list instead

[white oxide]

Like 6 pages a day? which might be a bit steep with hatcher. Idk how much time you are investing

[urban zinc]

my current state of mind is "however much time it takes"

because I have a reading group that meets every friday lol

I think I have the intuition for this section now after watching a bunch of youtube videos, but I need to actually read the details

[white oxide]

Well, good luck with finite Galois th- I mean covering space theory then

[urban zinc]

Thank you!!!

[urban zinc]

probably a dumb question, but how do we know p^-1(x) is locally constant? I see that x has to have a neighborhood (namely the evenly covered neighborhood) where p^-1(y) ≥ p^-1(x) for any y in the neighborhood, but not sure how to prove the reverse inequality

[white oxide]

Part of the definition of a covering space is usually that it has locally the structure U x F, so correspondingly you define the sheet counting function on that U to be the cardinality of F

[coarse night]

it's constant for some nbd U of x, which is obvious if you take U to be evenly covered

[white oxide]

Usually also have the discrete topology on F, but I am not entirely sure about that one.

[tiny ridge]

both p^-1y and p^-x has cardinality of the number of connected components of p^-1U

because you’re picking out one point from each of the slices over U

in either case

[urban zinc]

oh wait

nvm yeah this makes sense lol

ty

[eager vigil]

Hey, I know it ain't pretty, but could anyone check whether it's even correct?

[tiny ridge]

The Mobius strip with a full twist (thats still topologically an annulus! The usual Mobius strip has one half twist) is not isotopic through immersions to the standard annulus in R^3, but the Mobius strip with two full twists (so 4 half twists) is isotopic through immersions to the standard annulus in R^3

[urban zinc]

you might want to ask in #real-complex-analysis instead

[eager vigil]

Ok, thx

[tiny ridge]

Isotopy of the Mobius strip with 2 full twists into a standard untwisted annulus through immersions

[urban zinc]

hype, I understand the proof of the lifting criterion now

[urban zinc]

Me: I don't understand what he's saying here... Maybe it would help if I looked at his drawings?
Hatcher's drawings:

pain

[nimble portal]

ITT: Eric doesn’t know what a snowflake is

:p

It’s clearly a snowflake and some sort of fucked up ladder ;)

[urban zinc]

oh shoot you're right feather

thanks you're a lifesaver

[nimble portal]

Gotchu bb

[urban zinc]

Wow we should just teach algebraic topology instead of group theory, things are making sense

On the other hand this construction of the universal cover is hurting my head

[coarse night]

just skip it, there's another proof using fibre bundles

[nimble portal]

LOL What made you say this

[lime sable]

why study G when you can study K(G,1)

[coarse night]

as an exercise, can you find a covering transformation that is not a deck transformation

By covering transformation I mean map f: X → X that with pf=p. p: X → Y being covering projection

hatcher uses different notion of covering transformation

[urban zinc]

Writing these down to return to later

[bitter smelt]

pf=p?

f continuous, bijective?

[tiny ridge]

just continuous i think

but yeah pf = p should be the case

Puzzle: Is there a word on two letters (a, b, a^-1, b^-1) which is equal to the identity in every finite group generated by two elements a, b (and with some relations)?

[coarse night]

yes

[tiny ridge]

Oops sorry I meant a reduced word ie one where there are no consecutive inverses.

No appearance of aa^-1 or bb^-1 strings in the word

[coarse night]

@tiny ridge we have a lie group G and let S={g ∈ G | gⁿ=e for some n} then is S necessarily countable?

[tiny ridge]

no

there are uncountably many order 5 elements in SO(3)

[unreal stratus]

No because F({a,b})

[tiny ridge]

one for every axis

[coarse night]

oh lol

[tiny ridge]

Oh sorry mistake again, let me edit the question I meant in every finite group generated by two elements

[unreal stratus]

Lol interested why you picked 5

[unreal stratus]

Ah

Lol

Lemme think then

I have an exam w groups stuffs tomorrow lol

[tiny ridge]

just randomly :p

i like 5 order symmetries

[unreal stratus]

Ye fair lol

UwU

[bitter smelt]

Does something simple like this not work? ||Let your word be a^nb^m (after commuting, I'll take ab=ba as one of the relators).

G = <a,b | a^{n+1} = b^{m} = aba^{-1}b^{-1}>||

[tiny ridge]

It should vanish in every finite 2-generated group, Migil.

[bitter smelt]

Which is why my answer was no

[unreal stratus]

Hm I'm confusd what you mean like after commuting lol like

[tiny ridge]

Oh, but a word need not be of the form a^nb^m

[unreal stratus]

How do you know the word isn't smth which always vanishes in abelian

[bitter smelt]

It does after commuting

And my group is commutative

[tiny ridge]

Oh ok

[bitter smelt]

So whatever your word is, it'll look like that to this group

[tiny ridge]

Interesting

One second

[unreal stratus]

But how do you know the word isn't 1 in that group I mean (like it could be n = m = 0)

Like aba^-1 b^-1

Like say there's a word which evaluates to 1 in every abelian group but not in every group

[bitter smelt]

Oh hm

[tranquil brook]

Hi everyone can someone help me for test

[tiny ridge]

Yeah OK you have written down some word which doesnt vanish in some group

Thats not good enough

Logic is hard

[bitter smelt]

I thought I got every word

I missed when n and m are 0

[unreal stratus]

I'll have a think about this lol hmm

[tiny ridge]

So the only remaining case is when its in [F_2, F_2]?

Now the problem becomes about trying to show if there’s a word in x(m, n) = [a^n, b^m] which vanishes in every finite group generated by x(n,m) and some relations lol

Coz [F_2, F_2] is freely generated by those

[coarse night]

is there such a word?

[tiny ridge]

Nope

That would be crazy right lol

[unreal stratus]

I have an idea but i am think

[coarse night]

yeah ofcourse

so what are we trying to show? there is no word? lol

[tiny ridge]

Yes, we are trying to show that in fact all words are trivial

F_2 is the trivial group

[bitter smelt]

Ibsen what math is it you do?

[tiny ridge]

I think I have mentioned something along the lines of h-principles before

[bitter smelt]

Ah right

[unreal stratus]

Okay yeah I have an idea which is a bit hacky

So

Oh wait are you done lol

Rip

[tiny ridge]

Nope

Go on

[unreal stratus]

Oh nvm

Okay cool

[bitter smelt]

Done?

[unreal stratus]

Well basically sketchy so far but it seems to work so like

Say I've got a word $w = a^{\alpha_1} b^{\beta_1} \dots a^{\alpha_n} b^{\beta_n}$ and pick some big natural $N$ to be chosen later. If we impose the relation $ab = ba^N$ on $F({a,b})$ then in this group $G$ we have $a^k b^\ell = b^\ell a^{k N^{\ell}}$. In $G$ we can then rearrange $w$ to some word of the form (I think) of the form $b^{a_0 + a_1 N + \dots + a_k N^k}$ and I need to check but p sure not all the $a_i$ can vanish. So we can pick $N$ large enough that this means $b^M = 1$ for some $M \ne \pm 1$. Then consider a group with $ab = ba^N$ and $b$ forced to be of some other order like $M-1$

[gentle osprey]

potato

[unreal stratus]

Obviously what I'd need to make rigorous is that bit with not all a_i vanishing but it seems to work for examples but actually may be doubtful, and I'd need the check the a_i don't depend on N but that seems true - actually yeah it's a polynomial in the N

But the idea of using some weird relation to mess around with the stuff in the word seems like it should work

Maybe I am a crank though

eh there must be a nicer proof even if this is mendable lol

[tiny ridge]

I like this idea. Hm

[unreal stratus]

Actually so like the fact I'm assuming not all a_i vanish means like

I need to show that w doesn't vanish in G ig lol

But G is infinite so that may be easier

[tiny ridge]

I suppose a_i = 0 will give some equations in alpha and betas

[unreal stratus]

Actually

Wait a second

Okay odn't wait but I will think

[bitter smelt]

How about something simpler, like G = <a,b | a^2=a,b^2=b>. Then your generic word will be (ab)^n (ish). You just need to finitize it. Morally you just take (ab)^{n+1} a relator but you need to deal with the ish

And also fix that this group is trivial

I wonder if instead of these constructive proofs there is a topological proofs

[tiny ridge]

[bitter smelt]

[urban zinc]

Yknow in most books the examples are there to elucidate the definitions, but I think in Hatcher, the examples are just there to knock you down a notch in case you're getting confident about the material...

What does he mean here by "These permutations obviously determine the covering spaces uniquely, up to isomorphism"?

Like can someone give a more precise statement of this fact? I'm having trouble wording it

[tiny ridge]

if you take a fiber of a covering space and then loop de loop the basepoint the fiber maps to itself after going around but the points in the fiber gets scrambled up

if you know how this scrambles for every loop in the base you pretty much know the covering space yeah

[urban zinc]

oh wait so isomorphic covering spaces <=> isomorphic actions of pi_1 on the fibers

[tiny ridge]

yes

[urban zinc]

so if we have
p_0 : (Y_0, y_0) → (X, x)
and
p_1 : (Y_1, y_1) → (X, x)
then the former means that there exists a homeomorphism F : Y_0 → Y_1 such that
p_0 = p_1 F and p_0 F^-1 = p_1
and the latter means that there exists f : p_0^-1(x) → p_1^-1(x) such that for any loop [gamma] in pi_1(X, x),
[gamma] · f(y_0) = f([gamma] · y_0)

[tiny ridge]

yes

[urban zinc]

ayyy okay I see

ty

[urban zinc]

Can you just take two copies of Y as a cover of Y and then the covering transformation is just sending both copies to one copy

Which is clearly not invertible

[coarse night]

Oh I require connected

[urban zinc]

Oh

[coarse night]

Should’ve have mentioned

Hint: ||try some cover of S1vS1||

[hidden crag]

It’s always that one

[coarse night]

Yea

[urban zinc]

okay to put it in group theoretic terms we're looking for a subgroup H of G and an element g of G so that gHg^-1 is strictly contained in H? not sure if this is right

[coarse night]

Correct

[urban zinc]

hmmm

[coarse night]

Can you elaborate for those who might be reading the answer

[urban zinc]

If our base is Y and our covering space is p : X→Y, the lifting criterion tells us that a covering transformation : (X, x0) → (X, x0') exists iff p_*(pi_1(X, x0)) is contained in p_*(pi_1(X, x0'))

and changing a basepoint in the covering space is just conjugation in pi_1(Y, y0)

[coarse night]

Try solving it purely group theoreicly and translate to covering space

[nimble portal]

group theoreicly

[quiet thorn]

Theoweicwy

[nimble portal]

Hey Catto, make sure to stay hydrated :) eat some food too bestie

[quiet thorn]

Thanks will do

[cosmic socket]

What is the functor $\Omega^\infty: \text{Spectra} \rightarrow \text{Spaces}$

[gentle osprey]

bigradedSphere

[cosmic socket]

I know this is basic question but I can't find the answer

This old book im reading defines it as $\Omega^\infty E = E_O$ (but I can't tell from the typesetting if this is just a $0$

[gentle osprey]

bigradedSphere

[cosmic socket]

or if its something else

I think it should be some colimit

nvm I think it is just E_0

[tiny ridge]

@cosmic socket Bizarre. Is the point that if X is a space, there is a spectrum {\Sigma^n X} you can cook up from there, and X -> \Omega^n \Sigma^n X is a homotopy equivalence by loopspace suspension adjunction?

So \Omega^infty of the suspension spectrum of X is like X, ie the 0th graded guy in the spectrum

[cosmic socket]

I think I'm still trying to understand it

its not quite what you said

There is a role played by a functor $Q X := \text{colim } \Omega^n \Sigma^n X$

[gentle osprey]

bigradedSphere

[tiny ridge]

Oh yeah \Omega \Sigma X is definitely not homotopy equivalent to X

Take X = S^1

Even stranger then

[cosmic socket]

but you define $\Sigma^\infty$ not naively as a the spectra with $E_i = \Sigma^i X$ but rather as $\Sigma^\infty X = {Q \Sigma^i X}$

and then you get that $\Omega^\infty \Sigma^\infty X = QX$

[gentle osprey]

bigradedSphere

[cosmic socket]

maybe this is the point of the definition

[tiny ridge]

im totally confused by your description lmfao

[cosmic socket]

me too

I need to think more about it

[gentle osprey]

bigradedSphere

[sly ruin]

What kind of space is X = {(x, y, z) in R³ | at least one of x, y or z is an integer }?

[tiny ridge]

Is this the Quillens Q construction

[unreal stratus]

Not really sure what kind of answer you want lol

[sly ruin]

I mean geometrically

[unreal stratus]

Ig one way to describe it is "a bit weird"

[cosmic socket]

I guess the point is you want $\Sigma^\infty$ and $\Omega^\infty$ to be adjoint and also there to be a weak equivalence $\Omega^\infty \Sigma^\infty X \rightarrow \text{ colim} \Omega^n \Sigma^n X$ (these are two of the `desirable' properties of spectra

[gentle osprey]

bigradedSphere

[cosmic socket]

do you mean cofibrant replacement? I don't know if it is

[tiny ridge]

😵😵‍💫

[cosmic socket]

I think it depends on what model of spectra you are using too

in this book it's just using that the structure maps are homeomorphisms which is a lot stronger than what I have usually seen

but I think apparently the homotopy category will be the same

[tiny ridge]

it ought to be

[cosmic socket]

to be honest there are too many models of spectra and I don't even know what is the correct model to have in my head

but someone told me the details are usually not too important (unless you care about them) and you can work with them formally

[tiny ridge]

my knowledge of stable homotopy theory is very close to nil i should admit

[cosmic socket]

me too :/

[tiny ridge]

there must be a topological way to get this. Omega Sigma X <- X is not a homotopy equivalence, but maybe Omega^n Sigma^n X <- X becomes one as n -> infty?

[marsh forge]

I’ve been having an Eric Andre let me in moment as I waited for my 10 minutes to end and allow me to speak here haha

The functors Omega^n Sigma^n X can be thought of as the free n-fold loop space on the space X

And in the colimit they give the free infinite loop space on X, and this is canonically equivalent to the suspension spectrum of X after the identification of infinite loop spaces w connective spectra

So one way to think about QX is that it’s a space whose normal homotopy groups are the stable homotopy groups of X

[tiny ridge]

Ah

OK, that makes sense!!

[tiny ridge]

Ok, so this is as simple as saying, pi_n+k(E_k) = pi_n(Omega^k E_k), taking colimits as k -> infty gives pi_n^s E = pi_n(Omega^infty E)

Are your spectra Omega-spectra?

Then Omega^infty E = E_0 makes sense

[marsh forge]

Yeah making the suspension spectrum Omega is exactly the point here

[tiny ridge]

Cool I see

this feels a bit like group completion lol

[marsh forge]

It is!

[fading vale]

omg

return of maxj

[marsh forge]

the prodigal son

[tiny ridge]

Ok, I don’t understand this 100% yet but this is obviously morally russian topology, not homotopy theory

good math, approve

[marsh forge]

Well

This is pretty importent in homotopy theory

[tiny ridge]

but homotopy theory is not important

so

just joking

[fading vale]

Watching grad students fight is like watching monkeys in the same enclosure brawl for food in the zoo

[marsh forge]

homotopy theory is unbothered

[tiny ridge]

thats because its too stable

[cosmic socket]

Wdym but free n fold loop space

[marsh forge]

Well

[cosmic socket]

Isn’t it more like 2n fold loop space

[marsh forge]

No, you are only looping n times

It might help to think about the role of the suspension

If we just looped, we’d be chopping off homotopy groups that would otherwise live in negative degrees

The suspension provides the breathing room needed to loop w/o losing info

I mean you do lose info but that’s the intuition

[cosmic socket]

Ok I see what you mean by this

[cosmic socket]

But I don’t see how this fixes it

[marsh forge]

N-fold suspensions are n connective

So you don’t lose info

An important theorem is that n-fold loop spaces are equivalent to n-connected spaces (homotopically)

And the functor is exactly just looping n times

[tiny ridge]

(pi_n+k Sigma^k X = pi_n Omega^k Sigma^k X, and the left hand side stabilizes as k -> infty)

[cosmic socket]

Ok so the point is as k gets large

Homotopy groups of RHS are stable homotopy groups of X

[cosmic socket]

Which explains this comment

So why can’t you do the naive definition and define \Sigma^i X = \Sigma^\infty X_i

[marsh forge]

Many authors do

But it’s not cofibrant

[cosmic socket]

Oh ok

[marsh forge]

So you have to do a cofibrant replacement

[cosmic socket]

In which model structure

[marsh forge]

The sequential ones w the omega ones cofibrant

Anyway the defn using the Q functor is basically just a manual cofibrant replacement

[cosmic socket]

Ok makes sense

Why is it desirable to have your suspension functor land in cofibrant spectra

[marsh forge]

Well you want to really only ever work w the cofibrant ones

[cosmic socket]

I thought sphere spectra is no longer cofibrant

In modern category of spectra

[marsh forge]

Not sure what modern means here

[cosmic socket]

Like EKMM

[marsh forge]

The actually modern approach is to eschew model categories entirely

EKMM is pretty old

[cosmic socket]

I see

[tiny ridge]

funny field

[cosmic socket]

Modern approach is to use infinity categories?

[marsh forge]

Yeah

[cosmic socket]

Can you do everything you can in model categories in infinity category techniques

And is everything nicer?

[marsh forge]

This type of question can get one in controversial waters

But yes

Well, the thing about infinity categories is that like

Most of the technical complexity is in the foundations

It’s way easier to define a model category and get the theory off the ground than it is to introduce eg quasicstegories

[cosmic socket]

In practice do people blackbox foundations

[marsh forge]

Yeah mostly

[cosmic socket]

I’ve been told I can blackbox stuff like how smash product of spectra work. But in the infinity category approach is it nicer?

[marsh forge]

I mean most homotopy theorists have a working understanding ofc but it’s something you gain in tandem w working w them

[marsh forge]

Much nicer

I mean the theory of infinity operads shows up

But if u blackbox that then yes

[cosmic socket]

So I should just keep learning actually stuff and eventually figure out how infinity category picture works

[marsh forge]

Basically yeah

Basically all modern papers use the language of HTT and HA, and eventually you’ll read stuff that uses them, and you can back-fill as needed

[cosmic socket]

I tried learning some infinity category stuff but I found it a bit dry without motivation

Also I didn’t know if it was actually worthwhile

[marsh forge]

It’s absolutely worthwhile, but it’s very frustrating to learn

[cosmic socket]

Ok makes sense

Thank you

[cosmic socket]

Why do you say so

[marsh forge]

Np, feel free to @ if you have questions

[tiny ridge]

if i speak i am in big trouble

[marsh forge]

Lmao

[tiny ridge]

no just that at some point it starts becoming like formal logic at least for those of us who care about topology of manifolds and is not a working homotopy theorist

[marsh forge]

It’s formal but I doubt the logicians would agree w that

[cosmic socket]

I guess there is the categorical side but I think there is also the very topological side right?

[marsh forge]

But there’s tons of cool very-homotopical work that has direct consequences for manifolds

[tiny ridge]

i don’t disagree, but curious what you have in mind

[marsh forge]

A great recent example is the Burklund Hahn Senger paper which is titled “on the boundaries of highly connected almost closed manifolds” or something

I think HHR is in this realm as well

[tiny ridge]

interesting result

whats hhr

[marsh forge]

HHR is a little closer to the Russian spirit

Hill Hopkins Ravenel

Something something kervaire invariant 1

[tiny ridge]

ah ok

good theorem

ive always thought of that as an algebraic topology theory though i will admit

theorem*

[marsh forge]

The line is blurry but the extensive uses of stable equivariant homotopy theory I think makes it a bit more modern than say Adams paper on hopf invariant 1

[tiny ridge]

i should read the Adams paper someday

[cosmic socket]

I don’t know too much about this but I think there are also applications to (classical) problems in number theory via the motivic side.

[marsh forge]

Oh homotopy theory is super influential in number theory these days

Mostly via connections to arithmetic geometry

[tiny ridge]

no wonder

similar kind of subjects

[cosmic socket]

Actually could you elaborate more on this?

Just key terms I guess?

[marsh forge]

So there’s a few directions. The connections to K-Theory start with Nikolaus-Scholze and Blumberg-Gepner-Tabuada. The connections to arithmetic geo can be seen in the recent explosion of work on prismatic cohomology and related invariants

Basically Bhatt-Morrow-Scholze show in their first paper that there is some big cohomology theory Ainf that specializes to a bunch of other known ones, and then in their second paper they show that this is deeply connected to THH

But honestly most “derived” work at all these days uses infty cats

Recalling how to compute things on resolutions where helpful

Since I’m on a soap box anyway I believe the correct take on all of this is that infinity categories are what we were trying to get at all along and model categories are akin to group-presentations

Ie model categories are just one of many ways to present an infinity category

[cosmic socket]

Yea I’ve heard of this analogy before

[abstract saffron]

I just have one question: all of this stuff, has anyone used them to prove any result that someone outsider can understand?

[marsh forge]

How outsider

[abstract saffron]

Say, a first-year grad student

[marsh forge]

With what background

[tiny ridge]

thats quite broad

[marsh forge]

(I’m a second year grad student to be clear)

[abstract saffron]

I guess "grad student" means different things then 😄

I have heard about this infinite category before, just not sure where to go from there

[marsh forge]

Well, it depends on how and whether they relate to things you care about

[abstract saffron]

Category theory to me is still only a convenient tool to state stuff in Alg Topo

[marsh forge]

I think you don’t need to force the issue w infinity cats

If they are relevant to your interests they will come up naturally in other peoples papers and you can learn as you need

[tiny ridge]

i learnt about them from Lurie’s cobordism hypothesis paper right after getting done with Hatcher. It should be an accessible introduction to a place where they crop up somewhat naturally, for first year graduate students in topology.

[marsh forge]

Luries On Higher Topoi also has good intuition

[tiny ridge]

but if you know what a simplicial set is, infty cats are a small walk from there, definition wise. and simplicial sets are useful everywhere, even in more geometric subfields of topology

[marsh forge]

Yeah. The issue is more so that like the definitions and the intuitions have a big gap

Like essentially 0 of my inf cat intuition comes from thinking of them as simplicial sets satisfying horn filling conditions

[tiny ridge]

makes sense

i guess im telling him to learn what a simplicial set is instead of an infinity category immediately lol

more probability of using them irregardless of specialization

[abstract saffron]

simplicial set...?

I'm pretty sure I know abstract simplicial complex, but I'm pretty sure they are not the same thing

[tiny ridge]

theyre simplicial complexes but, like, kinda fat

[abstract saffron]

.... right

makes total sense

so... a polyhedron?

[tiny ridge]

you know how [0, 1, 2] is a 2-simplex? three vertices, three edges, one face?

[abstract saffron]

Yeah?

[tiny ridge]

in then world of simplicial sets, you have lots of and lots of degenerate vertices, edges, faces and higher cells even in the 2-simplex.

for instance, [0, 1, 2, 2] is a 3-cell for the 2-simplex

[abstract saffron]

ahhhh

[tiny ridge]

its homotopy-fat

[abstract saffron]

Ok, I saw this one I think

Yes, def saw this one, just didn't remember this name

[solemn oar]

Ye olde analogy
infty-cat <---> model category
manifold <---> local charts

[abstract saffron]

Ok, now I know neither

But the analogy at least makes sense now

[marsh forge]

Hm

I like that one less

[thorny agate]

I have a quick terminology question. Some places seem to use the term "analytic basis" but as far as I can tell this just looks like the standard definition of a basis? Is this correct or am I missing something?

[marsh forge]

What’s the term analytic basis defined as

[thorny agate]

this is one place: https://proofwiki.org/wiki/Definition:Basis_(Topology)/Analytic_Basis

this looks to me to be the definition of a basis

just more asking for a yes/no sanity check

[solemn oar]

Just looks like a basis

[tiny ridge]

bizarre terminology

[marsh forge]

I think this is mostly a distinction between a basis for a given topology vs the combinatorial data needed to generate a topology from a prescribed basis

It’s entirely psychological tho

[tiny ridge]

i tell you homotopy theorists can never say “equality”

[marsh forge]

Huh

[abstract saffron]

That's the def of basis I know and love in topo, that's weird

never knew ppl call it analytic basis

[marsh forge]

no one calls it that

Ive never heard this term in my life haha

[thorny agate]

ok I'll keep this term in the back of my head and never use it then lol

[solemn oar]

No such thing

[tiny ridge]

i need to explain to geometrically-minded topologists, very briefly, what a derived category is. does anything at all get lost by saying “objects are chain complexes, morphisms are roofs”?

[marsh forge]

Roofs?

I’d at the very least say that you quotient by quasi-isos

[tiny ridge]

A <- B -> C are the morphisms A -> C, where the left one is a quasi iso

[marsh forge]

Omg what

[tiny ridge]

?

this is what localization is

[marsh forge]

That’s a funny name for a span

[solemn oar]

That's the classical construction of the derived category

[marsh forge]

I’ve never heard the word roof used before

[tiny ridge]

oh i have always called it roofs

[marsh forge]

Anyway honestly I would just say that you take the category of chain complexes and force quasi-isos to be isos

I don’t think you need to mention the construction at all if you’re not gonna use it

[tiny ridge]

i will use it :p

[solemn oar]

Yeah maybe stick to the universal property if possible

[marsh forge]

Ah

Then I’d say what I said and then briefly remark on why your thing is a model for it

[tiny ridge]

cool

good idea

[cosmic socket]

You can have much bigger zig zags tho right

[tiny ridge]

they compose

[unreal stratus]

Can I ask a boring π1y thing lol just to sanity check smth lol

[tiny ridge]

quasi isos are multiplicative

[solemn oar]

You use the Ore condition to compress them into that shape

[tiny ridge]

think of a pyramid of cards

[unreal stratus]

So like I was thinking uh take S^2 and identify exactly one pair of (wlog i guess up to homeo) antipodal points, what's the universal cover

[tiny ridge]

the bottom is zigzags, the big edges form the big roof

[unreal stratus]

I imagine what we do is we just take a long line of S^2s glued at one point between each

And then just identify all of the points which are used to connect them?

[solemn oar]

So just S^2 wedge a circle?

[tiny ridge]

not upto homeo

homotopy theorists…

the universal cover is a string of S^2s

[unreal stratus]

Yeah S^2 wedge a circle has a cover which is fairly easy to write down

[tiny ridge]

i dont understand the second line

[unreal stratus]

Oh I wondered if you had to do some identifications to this I mean

So uh

[tiny ridge]

nope

[unreal stratus]

No you don't

Nice sure

In the wild are you meant to find these sorts of thing mostly by geometric reasoning lol

[marsh forge]

Note that a nice feature of the theory is that if you have a guess for the universal cover it’s pretty straightforward to check if ur right

[marsh forge]

Yeah

[unreal stratus]

Thankies

Yeah idk why I was basically concerned about it being a local homeo about the points you identify

But no, a nbhd of that is easy to visualise and matches up with the string of spheres

[tiny ridge]

take spaces X, Y glued along a subspace Z. what is the universal cover in terms of Xtilde, Ytilde, Ztilde?

as a corollary compute universal cover of torus with a mobius strip glued in along the meridian

[novel acorn]

I calculated this and I must comment that I definitely did not enjoy it at the time lmao

[tiny ridge]

this is good stuff

in this part of the world we call it bass serre theory

[novel acorn]

iirc it's like R^2 x {0} U R^2 x {1} and you glue strips of height 1 along Z x {R} x {0} and Z x {R} x {1}

[tiny ridge]

is that simply connected

[novel acorn]

tbh we weren't sure it's the only thing we got after like trying for 3 days straight lmao

plus we found some paper supporting us

[tiny ridge]

its much worse

at least i think so, i just cooked up a random example and didnt give much thought

[novel acorn]

I think that's a problem from Hatcher

[tiny ridge]

your thing doesnt look simply connected

hatcher asks T^2 v RP^2

iirc

[tiny ridge]

oh lol

i guess i am hatcher

[novel acorn]

ah yeah we made a mistake lol that isn't correct it's something else much worse

[tiny ridge]

just have to unfurl more

nbd

[unreal stratus]

Oop

Pain

I am doing some interesting thing with an application to groups

So like if G is a finitely presented group with a presentation <x1,...,x_n | r1,...,r_k> it's standard that we can form a 2-dimensional complex X with pi1(X) = G by taking a 0-cell, adding n 1-cells and then attaching 2-cells via the relations

But this is saying like "subdivide the cells" to show that we can give G a presentations with all relations of length <= 3

Inch resting

[tiny ridge]

sounds useful

yup, this is good

havent seen this before

[unreal stratus]

Hm any ideas on how to subdivide the cells as they say? Feels slightly odd to me hm

[tiny ridge]

my first thought was to triangulate stuff

because 3 = triangle

the cell complex can be made into a polyhedral one easily and then one can triangulate each polyehdral cell

now continue

[solemn oar]

So for every relation with length 4 or more (abc...) you just add another generator z and a relation z^{-1}ab and iterate?

[tiny ridge]

yes but good to see that geometrically in terms of a triangulation

i might fall asleep so ill add more hints now that cloudberry revealed one way to do it. think of a square cell attaching to wedge of four circles, then subdivide the square in half. interpret the diagonal as z and sliding part of the relator loop (given by roaming around the four circles) into the diagonal as setting z = ab

in general triangulate and consider the 1-skeleton of the triangulation as the new wedge of circles (after contracting the maximal tree inside) and the triangles as the new cells

side note, its good to think of presentations in terms of the cayley complex, it remembers a lot about the group. for example, surface groups (genus >= 2) have the property that they have a presentation in which any two 2-cells in the universal cover of the cayley complex will share at most 1/6th of the length of their respective boundaries. a group is said to have the C’(1/6) cancellation property if they satisfy this and is a strong imposition on the structure of the group, eg the universal cover of the cayley complex becomes contractible

this stuff falls under the rubric of “small cancellation theory”, used to great effect in both geometric group theory and low dim topology

[violet summit]

Is there a clean explicit construction to this?

[steel glen]

i assume p and q are in the interior of D^2?

[lime sable]

1. rotate them to be aligned horizontally
2. expand the top and contract the bottom (like rotating a sphere and looking at it from the side, but you keep some region near the edge to expand or contract) or vice versa so they lie on a diameter
3. do the same thing horizontally to bring p to q
4. undo the first two steps

there's also probably something you can do with a bump function

also there's the thing where D^2 (modulo boundary) is a model of the hyperbolic plane

which might be cleaner

actually you might not be able to keep the boundary fixed there

[steel glen]

i like the idea of using a bump function for this

[lime sable]

yeah you can't

[steel glen]

maybe this
$$f:D^2 \to D^2, , f(x) = x + \psi(x)(q - x)$$ where $\psi:D^2\to D^2$ is a bump function on some open ball $\beta\subset D^2$ containing $p$,$q$, and such that $\psi(p) = 1$

[gentle osprey]

maximo

[steel glen]

i may have oversimplified but this looks ok to me

[lime sable]

i think you need psi(p) = psi(q) = 1 at least

because here f(p) = f(q) = q

but not really sure what else you need

[steel glen]

ah yeah

[lime sable]

i think blowing it up to a plane, translating, and shrinking it back down might be the easiest way

[steel glen]

i think it's an easier visual, but i think your idea of a bump function will be a cleaner construction

[lime sable]

but it is a little tricky since you have to show that doesn't mess with the boundary

[steel glen]

ah ok maybe it's simpler than what i wrote above. $$f(x) = x + \psi(x)(q-p)$$

that should be a plus

[gentle osprey]

maximo

[lime sable]

i think you need to choose a good bump function so you don't go beyond the boundary

[steel glen]

i don't think there's bijection issues, but i can't tell

[lime sable]

actually it's not so bad, you scale the interior by 1/(1-|r|) to go to the plane and 1-1/|r| to go back, and then you can show that translated lines in the plane still end at the same point on the boundary of the disk

or you can do this with a sequence approaching the boundary of the disk

or whatever

[violet summit]

so the bump function would be 0 outside that ball B?

[violet summit]

Also thank you, this construction makes sense

[coarse night]

Good exercise

[sly ruin]

Is X = {(x, y, z) | at least one coordinate is in Z } homotopy equivalent to the 3-torus?

[steel glen]

yes, though i think there’s injection issues with my definition

[coarse night]

Don’t think so

[sly ruin]

What would be pi_1 of this space?

[coarse night]

You get something like hollow boxes stacked together

[coarse night]

0 I believe

[sly ruin]

Is this a contractible space?

[coarse night]

Is it like an exam ? Are you asking the parts of a problem

[sly ruin]

No I'm trying to calculate the fundamental groups of some non-trivial spaces. I think that each hollow box here would be at least homotopy equivalent to S²?

[coarse night]

Yeah

[sly ruin]

Not sure if this is even a thing but would this result in the wedge sum of S²'s?

[coarse night]

Feels like it, sure

Feels like a 2D version of a graph where you can contract your maximal tree

[sly ruin]

Cool. What if instead of Z we use Q? We'll just get more spheres in the sum?

Like a denser grid

[coarse night]

Tbh idk

[white oxide]

Seems like it. It's a lattice with faces filled in everywhere, so any loop restricted to some face would look like a path between boundary points, then you have a fixed end-point homotopy to the boundary and the process repeats

[coarse night]

yeah

[white oxide]

Image you inflate each sphere to a hollow cube with rounded edges, this doesn't change the homotopy but the edges wouldn't actually touch at any point, but in the hollow box lattice you do have "diagonal" hollow boxes touching.

So it's like a wedge sum with some added relations

[lofty rain]

how would i approach this?

[stark fog]

you can use openness of both discs

to find open balls around your point entirely contained within each

use that to find the third disc within their intersection

[lofty rain]

right, but i couldnt think anything explicit in terms of this arbitrary point (a,b) that would be surely contained within both

[stark fog]

start writing something using the open condition

what does this give you?

[lofty rain]

if any <x,y> in D_<a,b> is a point such that (x-a)^2 + (y-b)^2 < r^2 for some r>0 then D_<a,b> is open 🤔
so we want some disc such that we can have this for an arbitrary point in this intersection
... but im not sure where this is going unfortunately

im being really cautious since the notion of metric spaces hasnt been introduced yet

could i instead somehow show that instead we could form... some open rectange instead within this intersection and therefore we can form some open disc within this open rectangle which would transitively also be within this interesction?

[stark fog]

the idea is that these discs forms a basis for the topology on R^2, so by definition of a basis, if there is a point in the intersection of two basis elements, there is a third basis element containing the point, and contained in the intersection

[lofty rain]

yeah thats what the exercise is heading towards haha

[stark fog]

yeah, the topologies are also the same

euclidean, product and the basis given by open discs

[stark fog]

but yeah, your idea was close enough

you just need to manage to find this r

but the thing is not to show D_(a, b) is open, but contained in the intersection

[lofty rain]

right yeah of course

alright ill give it some more thought

[arctic relic]

Do symmetric monoidal categories have that for a pair of morphisms (f,g), f composed with g equals g composed with f (I’m assuming the functor equipped with the category basically acts as composition/precomposition)

This is in the context of topological field theories

The definition of symmetric monoidal category didnt distinguish between the action of the functor on objects and morphisms

[solemn oar]

morphism composition is directional. Composability of (f,g) does not imply composability of (g,f).

[arctic relic]

In the paper I’m reading, there’s a criteria that there’s an isomorphism between functor( C,D) and functor(D,C) which seems to be equivalence of compositions

But yeah i was confused about it

[umbral panther]

Look at an example. Sets form a symmetric monoidal category under disjoint union. The composition of maps of sets is not commutative

[solemn oar]

Even if the case they do, it's not true. Consider the following in Cob_2:

$\tau : 2 \rightarrow 2$ twist map

$T : 1\rightarrow 1$ the torus with two holes.

Now
$$\tau \circ (1\sqcup T) \neq (1\sqcup T)\circ \tau$$

[gentle osprey]

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

[arctic relic]

Nice. So know I know the functor associated with a sm category does not act like composition. But what does it act like on morphisms?

[lofty rain]

would it suffice to say something like
suppose we are in the first 'half' of this intersection, where half is just modelled by a side of a line that joins up the two points of intersections
then picture a line perpendicular to one of our D's (depending on the half we chose but including the "half line") that goes through a,b
clearly, the distance between length of the line between our intersection of the line and circle is our shortest distance between the point and the circle
so we can set our 'r' to this distance and so we can have an open disc D_<a,b> radius r which must be contained in our intersection by symmetry

[arctic relic]

In the category of cob, morphisms are associated with bordisms (or are bordisms not sure which) has the functor which acts like a tensor product. So what’s a tensor product between morphisms supposed to suggest? That you just take the disjoint union of the bordism as you normally do in cob?

[solemn oar]

In Cob, the symmetric monoidal product is disjoint union, yes.

[arctic relic]

Thanks

[fickle hamlet]

since basis is an open covering for X isnt 2 the same thing as part a of the theorem?

oh maybe there is some issue with the intersection not containing a basis element if i use the theorem

[patent quarry]

not every cover is a basis. i don’t think either of these statements implies the other.

it’s true that exercise 2 implies part a when the open cover happens to be a basis.

[fickle hamlet]

ya I see

thanks

[urban zinc]

Btw part a is known as Lindelöf's theorem (or lemma)

[livid tundra]

If I have an infinite compact Hausdorff space, and two different points, can I always separate them "twice"? Find two disjoint neigbourhoods of them and then again a proper smaller closed neighbourhood of each of the points?

It's also fine if the bigger ones are open neighbourhoods, they just have to be disjoint.

[solemn oar]

No. Take any infinite compact Hausdorff space and add two disjoint clopen points.
In this discrete subspace you get a counter-example.

[livid tundra]

But the second should be possible? Because it is T4 and T3?

[solemn oar]

It's sufficient to add just one point but two is easier to visualize

[livid tundra]

I can take two disjoint closed neighbourhoods because T_2.5 and then separate them by T4?

[solemn oar]

It reduces to the discrete space on two points.

No, you don't get proper inclusions.

[livid tundra]

Ah yeah, I see, I actually didn't need proper inclusions when the second one is open, so that's fine.

[unreal stratus]

Nice example cloudberry

[livid tundra]

Thanks for the help!

[mighty frigate]

Hello can u guys suggest books or references for countable and denumerable sets

[limpid fern]

that aint topology. just read some set theory book lol

[mighty frigate]

Our prof said that this is a need for our topology subject. I guess I chatted the wrong channel hahaha.

But do u guys have book suggestions for topology?

[solemn oar]

Do you mean in terms of induction proofs? Or can you be a little more specific about what you need?

[mighty frigate]

like this

[mighty frigate]

our activity is like this.
and i would like to study it further.

[solemn oar]

I see. I don't know of any text that treat this specific notion in great detail, but there should be lecture notes floating around the web.
If you google "Countable sets lecture notes" you should be able to find a few to choose from.

[neat stag]

a book in intro to proofs should work, I heard the one by rosen is good

[mighty frigate]

Thank you @neat stag and @solemn oar

[lofty rain]

is hyouka a worthwhile watch?

[stark fog]

[coarse night]

yes

very much recommended

[lofty rain]

ok ill watch thx

[trail charm]

"Suppose $X$ is locally Euclidian of dimension $n$, and $f : X \to Y$ is a surjective local homeomorphism. Show that $Y$ is locally Euclidian of dimension $n$."

Let $U_x \subseteq X$ be a neighborhood of $x$. Since $f$ is a local homeomorphism, $f(U_x) \subseteq Y$ is an open subset and $f|_{U_x} : U_x \to f(U_x)$ is a homeomorphism. Then each $f(U_x) \approx U_x \approx$ an open ball in $\bR^n$, where the last equivalence is from definition of $X$ being locally Euclidian. Since $f$ is surjective, each $y \in Y$ has a neighborhood $f(U_x)$ which is homeomorphic to an open ball in $\bR^n$, and is thus locally Euclidian with dimension $n$.

does that proof work? it feels kinda odd

\approx denotes homeomorphic

[gentle osprey]

anamono for anamono 🍓

[urban zinc]

more or less the right idea but be careful bc not every neighborhood of x is homeomorphic to its image, and not every neighborhood of x is homeomorphic to an open ball in R^n

[trail charm]

every nbhd of x is homeo to its image is defn of local homeo i thought