point-set-topology

[odd flame]

chmoki went p2w

is there an intuitive way to think about orientability of a manifold

and when it's not orientable

the mobius example makes sense but beyond that idk

[pearl holly]

My nitro was free, so this was a win win moment . Anyway, if you know homology and stuff I recommend reading https://en.m.wikipedia.org/wiki/Orientability under “homology and the orientability of general manifolds”. The idea is that you first look at local orientations at each point p, which is a generator of H_n(M, M\ p), which turns out to be the homology, in appropriate dimension, of a sphere, which is Z. The local orientation is just a choice of a generator here, so either +1 or -1, which basically says that you choose whether or not “a sphere around that point goes positively or negatively”. Now in some sense M is orientable if all these local orientations agree somehow. This is all made precise in the wikipedia article I sent

I guess it’s still worth reading the wiki if you don’t know homology and stuff tho

[feral copper]

I think it makes morese sense when talking about orientability along a curve.
Consider a point in your manifold. Choose an orthonormal basis of the corresponding tangent space. For instance, you're the point and the manifold is the universe, and your orthonormal basis is your first three fingers (thumb,index,middle).
Now, start moving the point around, along a closed loop. There is a way to carry the orthonomal basis along with that point into the different tangent spaces (out of context here, but you could use a Riemannian metric and use parallel transport for this).
If the curve does not exit a very small neighborhood of the point, well, because you're on a manifold locally modelled around R^n, this will not do much, and after you're back at the initial position, you'll have applied a rotation to the basis. That is, you have used an element of SO(n).
Now, for any loop, you always end up applying an O(n) map to the orthonormal basis. Turns out: sometimes, this element is not in SO(n)! This means that the determinant of the change of basis is -1. Now, for our analogy, this means that if you looped around the entire universe, you would still be you, but everyone else would have their heart on the right, and the US would be driving on the left (note: there is no such thing as left or right in a non-orientable manifold).
Hopefully, the universe is orientable, so this doesn't really happen. But on the Mobius band, it does happen: your basis is only your first two fingers (thumb,index). Now, if you start moving along a loop that turns around once, you'll end up back at the original position, but with the orientation reversed!
You call a manifold orientable if along any closed loop, you end up with an SO(n) change of basis. The manifold is non-orientable if a disorienting loop exists: one that has O(n) but not SO(n) change of basis.

Tbh, orientability is more of a geometric and concrete notion, you don't need homology to 'understand' it. Of course, there's homological interpretations of it, but in the end what matters is change of basis along closed curves.

Like, I've never really managed to represent the homology group H_n(M,M\p) in my head x')

[pearl holly]

yeah I agree, this is a nice explanation, but I just thought that I would mention a description of it terms of homology and stuff because I personally like it

[feral copper]

Oh I don't dislike that explanation, it has great use in orientability of vector bundles over non-orientable things!

[pearl holly]

yeah true, but I think your explanation does more justice to this lol. I was actually about to edit my post to say "if you don't know homology and stuff, wait for someone else to answer this" lol

[civic verge]

I made a mistake typing this last night sorry

Do you know where I can find this demo?

[gentle osprey]

Awuita Fria

[gritty widget]

Still wrong though lol

Last one should be R

[civic verge]

A set of $X\subset \mathbb{R}^{n}$ is bounded if, and only if, its projections are $X_{1}=\pi_{1}(X),...,X_{n}=\pi_{n}(X)$ are bounded sets in $\mathbb{R} $

[gentle osprey]

Awuita Fria

[gritty widget]

What's a demo

[bitter smelt]

Proof?

Formal statement?

It's unlikely you'll find a proof of this in any textbook. It's clear enough that it's probably left as exercise

[unreal stratus]

What is ur question

This seems to be an unrelated theorem

[civic verge]

I'm really sorry I'm tired and I haven't slept well, I just woke up and I'm a little sleepy.

I would like to do it in a different way than the one in the book so I was wondering if there is another way to solve it.

[bitter smelt]

Can you translate? What different way are you seeking?

[nocturne basalt]

you can just say if X \in rB^n then Xi \in rB^1 right

i guess for one direction

[unreal stratus]

I mean I don't really know how else you could rephrase it

In the 2D case, the proof is saying (informally) "you can fit a subset of R^2 in a box iff of finite length and of finite width"

i'm not really sure how one can give a nicer proof

[civic verge]

I am reading about sequences and Bolzano's theorem so, first I have to say that the sequence is bounded which I have proved effectively, now as the sequence is bounded I have to say that the projections are bounded, now by Bolzano's theorem I know that all the projections have a convergent subsession then

1. I have proved that the sequence is bounded.
2. I have not proved that the projections are bounded.
3. if the sequence in Rn is bounded and its projections are also bounded then the projections have subsucesion by Bolzano's theorem.
4. there is a part of the proof that I do not understand about subsessions and nested intervals of naturals (N),

[bitter smelt]

The proof.......

You're asking for a "different proof," but not translating the original

Unless you're asking for another way to state it?

What do you want??

[next crystal]

Does this work to show a topological space X with the discrete topology is locally compact?
Suppose $x \in X$. We show ${x}$ is compact. If ${U_\alpha}$ is an open cover of ${x}$, then some $U_{\alpha_i}$ contains $x$. So ${U_{\alpha_i}}$ is a finite subcover that also covers ${x}$. So $x \in {x} \subseteq {x}$, and since ${x}$ is open, X is locally compact

[gentle osprey]

michαel

[civic verge]

Sorry, I use the translator as I don't speak English.

the proof tells me that if I have a bounded subset in Rn then the projections are bounded.

I would like to make the proof different from the book

Seriously sorry sorry sorry for the English, sometimes I have complications to be able to make myself understood.

[bitter smelt]

I'm sorry, but I cannot help you construct a proof different from the one in the book if I am unable to read the proof in the book. Hopefully a Portuguese (?) speaker comes along that can help you.

[grizzled ibex]

I can help you, i'm from the same country. (sou do msm país, me manda dm)

dm me

im going to sleep rn, but when i wake up tomorrow hopefully we can talk

[next crystal]

what does "f extends to a homeomorphism of their one-point compactifications" mean?

specifically the f extends part

[stone temple]

i think it means that there exists a unique homeomorphism g on the compactifications such that g|_{X_1} = f, so show existence first, then show any two homeomorphisms on the compatifications, which if they both restrict to f, must be the same

[long hornet]

Any hint on why the subscript of the last x^2 in the bottom line isn't a + b - 2s? Or, more generally, does anyone know where I can find a similar proof?
(this is from Fuchs and Fomenko's proof of Adem's relations)

[unreal stratus]

Hm the statement doesn't say it's unique tbf

(Though uniqueness is obvious just set-theoretically)

[civic verge]

Can I write to you privately?

[grizzled ibex]

done bro

[marble kraken]

what is the best way outside a standard group theory/algebra course to learn the group theory required for the second half of munkres?

[pearl holly]

I don’t think you need a bunch of group theory to read it, a standard group theory/algebra course would do it

[marble kraken]

i dont have any group theory knowledge yet

[pearl holly]

Oh sorry, I misread your question

How about this then? https://people.bath.ac.uk/gt223/MA30237/lnotes.pdf

Up to page 34 should be enough I think

I haven’t actually read the thing so maybe I’m in the wrong by recommending it, but it looks very friendly with pictures and stuff. Otherwise every basic algebra book should cover the prerequisites

[marble kraken]

thanks sm

[odd flame]

any pointers on this please

this isnt really clear on what a double cover means lol

i asked about this before but forgor to reply sorry matplotlib

[bitter smelt]

it says what it means in the parenthesis

it means the preimage of any point consists of two points

more generally, an n-fold cover is a covering map for which the preimage of a point consists of n points

[nocturne basalt]

just a 2 sheeted cover

[long hornet]

Lazy answer:
https://math.stackexchange.com/questions/1073425/prove-that-there-is-a-two-sheeted-covering-of-the-klein-bottle-by-the-torus

Make sure to read all the answers

[novel talon]

Is there an error in the claim pi^{-1}({pi(p)}) = [p]? I think that [p] is not an element of S, but an element of S/~ so how can this be closed in S?

I mean we have defined that pi(x) = [x] so pi(p) = [p] and no we are saying that pi^{-1}([p]) = { x \in S | pi(x) \in [p] } = { x \in S | [x] \in [p] } = [p], but I don't think that [x] \in [p] means anything?

[gritty widget]

[p] as an equivalence class on S, is a subset of S

It's not pi(x) in [p] but pi(x) in {[p]}

That is [x] = [p]

When you write f^-1(a) here its meant to be as a fiber - like f^-1({a})

Sorry didn't sleep well

[tepid vale]

idrk where to post this, i know cartan decomposition is probably used somewhere but other than that i have no idea where to begin

[odd flame]

ooooooooohhhhhhh i thought that was a z lmao

[hidden crag]

[tepid vale]

i solved it myself

[broken nacelle]

let q be a quotient map from P to M, since P is second countable there's a countable basis B for the topology of P

I fail to see why {q(U) for U in B} doesn't satisfy this definition of basis

I know that's wrong since I'm not using the locally euclidean assumption

but I don't see why

oh, q^-1(q(U) might not be open coz q might not be injective

nvm then

[gritty widget]

yeah q(U) might not be open

But {q(U) : U in B} still form a network so you do get that M is second countable e.g. when M is compact

As to see where this fails, R/Z (topological quotient identifying Z with a point) is not second countable

In fact not even first countable

Interesting thing I found
https://dantopology.wordpress.com/2009/11/16/network-weight-of-topological-spaces-i/
It's also second countable if we assume M is metrizable or locally compact it seems

So M locally Euclidean is a special case of M being locally compact

I mean it makes sense that if nw(X) = w(X) holds for compact spaces then it should hold more generally for locally compact ones

So it does follow from {q(U) : U in B} being a countable network

[tender halo]

in metrizable everything is equal, weight, network weight, cardinality of dense subsets, etc etc

[gritty widget]

Good point. I actually knew that because I read it in handbook of set-theoretical topology

[tender halo]

I am at the midpoint of engelking, really enjoying it

some of the results are really beautiful, its a shame they are not that useful outside general topology

Cech-complete spaces are a very neat abstraction for unifying baire category theorem for locally compact spaces and for complete spaces

[gritty widget]

I never read Engelking, mostly picking up theory as I go when reading other stuff

[tender halo]

its a tough read id say, very compressed and not much in terms of exposition

[gritty widget]

I have plans to read it, just not a priority

[tender halo]

but the results are very cool

for example, a metrizable space is complete iff it is a countable intersection of open sets of any space its embedded in

[gritty widget]

What I find interesting is dimension theory for Tychonoff spaces

[tender halo]

and all separable completely metrizable spaces are embeddedable in R^\aleph_0 as closed subspaces

[tender halo]

why do you find it interesting?

[gritty widget]

It's a setting where dimension theory still somewhat works

It's interesting to see how the different dimensions behave for those spaces

I've been zero-set pilled

[tender halo]

from what ive read so far i found Stone-Cech compactification to be the most interesting part

apparently its an important object in logic and set theory, so im looking forward to that perspective on it

[gritty widget]

It also appears in functional analysis

[tender halo]

oh yeah? how?

[gritty widget]

Stone-Cech compactification has some interesting properties but it's usually very far from simple as an object

βN for example

[tender halo]

yeah, engelking discussed some properties of it, seems like a very...peculiar object, to say the least

[gritty widget]

The dual of bounded continuous functions on X is the space of finite Radon measures on βX

[gritty widget]

ad this

[gritty widget]

Yeah my suggestion is to not try to understand it intuitively, but in terms of its properties.

[tender halo]

thats how i took it to be

well you inspired some confidence into me, while i really liked the material i was a little afraid that im spending time on something disconnected from the rest of math

thanks

[gritty widget]

You're welcome. It's nice to see someone interested in topology!

[tender halo]

what about realcompact spaces and perfect mappings, are they useful tools?

they seemed a little disconnected from the rest of the material

[gritty widget]

Perfect/proper maps are used in some AT ig

If its used in AT then I'd consider that it has applications

Realcompact spaces I think are used in some topological group stuff but not certain

[tender halo]

icic

[gritty widget]

I don't think this is true tbh idk where realcompact spaces are applied

I guess to characterize compactness

A lot of the time you see something like that and don't think about it much other than "it's just another way to generalize or modify this and that concept"

It's definitely something that conceptualizes a chunk of topological spaces so ig that's good enough

[tender halo]

I see, makes sense

[alpine bolt]

Is the trefoil knot and the unknot diffeomorphic as submanifolds of R^3?

[gaunt linden]

Not in the sense that there's an automorphism of R^3 that takes one to the other. The complement of the unknot has fundamental group Z, the complement of the trefoil knot has a more complex (not abelian) fundamental group.

[cedar pebble]

yes, they just aren't ambient isotopic

their knot complements are also different as troposphere mentions, but this isn't really relevant to the question you're asking

[gaunt linden]

I'm mostly confused about whether "diffeomorphic as submanifolds of R^3" means something different from "diffeomorphic as differentiable manifolds in their own right".

[umbral panther]

It’s not a standard term

[cedar pebble]

I think most people will read "diffeomorphic as manifolds" unless ambient isotopy is specified

[odd flame]

asked about this before but im having trouble seeing that this is a double covering

https://math.stackexchange.com/questions/1073425/prove-that-there-is-a-two-sheeted-covering-of-the-klein-bottle-by-the-torus

what's with the mod 1

[odd flame]

better question but what's the map here

[gaunt linden]

Cut the torus into two halves, each a tube with a 180° bend.
Then each of those two halves maps to an entire trip around the Klein bottle, but with the cross-sections of the tube mapped to circular cross sections of the Klein bottle in different directions for the two halves.

[odd flame]

ok i was already half way there

i just cut a torus in half

but wdym a 180 degree bend

like just an arbitrary path wrapping around it?

[gaunt linden]

I just mean, suppose your torus is a bicycle tube. The each of the halves of the torus covers 180° of the wheel perimeter.

[odd flame]

quoii

the bicycle tube didnt help

ok i kinda see it nvm

[feral copper]

The covering T²->Kl is geometric, I usually think of it as a quotient of the torus by a free action of Z/2 on it. Here, I found the fixed-point-free involution on the torus that works: https://math.stackexchange.com/questions/3395391/proving-quotient-space-of-torus-homeomorphic-to-klein-bottle

I mean, I think of it the same as I think of S²->RP²: your involution identifies pairs of 'antipodal' points on the torus

[odd flame]

what's our equivalent of antipodal though

[feral copper]

The involution (z,w)->(z*,-w)

So you reflect the torus along a plane, and you rotate by 180°

Try to pick different points on the torus, and see where they are mapped to

Actually, what I drew also works for the 2-sphere and for any higher genus orientable surface (which happens to be the oriented cover of the corresponding non-orientable surface)

[odd flame]

oh i see

i find it easier to imagine it kinda inside a sphere

and then it's literally just "normal" antipodal-ness

[odd flame]

the the fiber of any point in this composition of maps is two points right

[feral copper]

Yes!

I mean

[odd flame]

the RP2 analogy helped

[feral copper]

If you consider the map T²->Kl

[odd flame]

it's like mapping to the upper hemisphere of the torus in a way

[feral copper]

The involution itself is a map T²->T²

[odd flame]

ok you keep saying involution but we havent used that term in my class

[feral copper]

Yes, although the fundamental domain of the Z/2 action is going to be the 'upper torus' here

[feral copper]

A map f:X->X is an involution if the composite with itself is the identity: f²=id

This corresponds to a group action of Z/2, where 0 acts as id and 1 acts by the involution

But basically, here, if I denote the involution as f:T²->T², then the Klein bottle is the topological space T²/f, where the equivalence relation is x~f(x)
Endowing it with the quotient topology still makes it a smooth 2-manifold, and it happens to be diffeo to Kl (however you defined it)

[odd flame]

my question is to prove that there is a double covering of T -> K, is this enough?

[feral copper]

I mean, it depends on what stuff you have available in your lectures

[odd flame]

it's such a guess tbh

[feral copper]

If you have the classification of surfaces, then it's done yeah

[odd flame]

he's very handwavey in lectures

but then expects varying levels of rigor in pfs

[odd flame]

we did that early on yeah

[feral copper]

Do you have group actions on manifolds?

[odd flame]

briefly but yes

actually my first idea was to look at their fundamental groups

is that related

since you mentioned group actions

[feral copper]

The proof could go as follows: the quotient is a non-orientable closed connected surface (the involution is orientation-reversing, and you have a free action of Z/2). Basically, you can check that this induces a covering, and the degree of the covering is the cardinality of a fiber (two). You compute Euler characteristics, and use classification of surfaces to show that below is a Klein bottle

[odd flame]

i dont see the classification, isnt the klein bottle P # P

[feral copper]

But again, it depends on if all those words I use are something you did in your lectures. your professor could also be a meanie and be expecting you to work with a parametrization of the torus and the Klein bottle (if he's a diff geometer for instance)

[odd flame]

he is not thankfully

[feral copper]

The classification tells you that the only non-orientable closed connected surface with Euler characteristic zero is the Klein bottle

[odd flame]

ahhh i forgot about that

okay im gonna switch gears a moment

because i mostly understand the covering so i'll come back to that later

this is begging for SVT right?

[hidden crag]

wait this smells like you can deform it into an easy space

[odd flame]

sniff sniff

[feral copper]

Like a wedge of four circles?

[hidden crag]

yes

that's what i was thinking

[feral copper]

Yeah I agree

[hidden crag]

but my brain is tired so i wasn't sure if i'm making things up

[odd flame]

ohhh we can push those points in

like those four little dashes on the circle are all identified under this quotient right

that makes it easier to see the wedge at least

well

im not sure actually but i'll draw it anyways

it'd look like a four leafed clover

[hidden crag]

yeah

[odd flame]

so it's an immediate result of SVT

Z/4Z

ok for b)

[feral copper]

No, the pi1 is not Z/4

It's the free goup on 4 generators

Wait, no

The space is not a wedge of 4 circles

It's for all theta that you identify it with theta+pi/2

So yeah the Z/4 thing makes more sense

[odd flame]

hm

ok so i was right to doubt originally

[feral copper]

Yeah, SVK gives you that

[hidden crag]

ohh i missed that

[odd flame]

yeah so ever point is identified

not just the marked ones

[feral copper]

For b), you need to know how many subgroups there are of Z/4

If you have a generator in the subgroup, then you're screwed you have the whole group, so the only non-useless subgroup (ie not the whole or the trivial) is the Z/2 spanned by the class of 2

[odd flame]

so just 1

i get that

[feral copper]

No, you have two

The universal cover, and the one corresponding to the subgroup Z/2

(if you exclude the trivial cover id:X->X)

[odd flame]

ok understood

i'll come back to c) in a bit

[odd flame]

to show that a covering of X onto Y does NOT exist is it enough to shown that the fundamental group of X is a subset of that of Y

[nocturne basalt]

do you mean subgroup?

for connected covers this comes from galois correspondence

bc fundamental groups of connected covers correspond w subgroups of fundamental groups of the base space

[feral copper]

Do you have a specific example? A direct computation of the Euler characteristics could be enough

Otherwise, yes, you should prove that the fundamental group of the thing above is not a subgroup of the fundamental group of the thing below

[long hornet]

Suppose (E, f, B, F) is a fibration. Then, under suitable compactness hypotheses, I think we have
p(E) = p(B)p(F),
where p is the Poincaré polynomial.
This doesn't work for Poincaré series (when we drop compactness). Is there anyway to save this?

[odd flame]

it was actually the converse of the klein bottle thing i was talking about earlier

showing that there is no double covering of K onto T

[lunar yoke]

If your definition of poincare polynomial is that x^n has coefficient the n-th betti number, in particular you're not ignoring the 0th one, then the hopf fibration S^1 -> S^3 -> S^2 should be a counterexample, since x^3+1 != (x^2+1)(x+1)

[odd flame]

what's a covering trasformation

it's in a friend's notes but not defined in my textbook

[nocturne basalt]

a covering space map?

[odd flame]

ooh is that it

i just be overthinking shit

[unreal stratus]

Hm

[unreal stratus]

I would understand that as a deck transformation

Which is not a covering map/projection

[odd flame]

what's a deck transformation

ive also never seen that lol

[nocturne basalt]

its an automorphism of the covering space that preserves the covering map

[long hornet]

Sad
It is true that chi(E) = chi(B)chi(F), though. So I'd still like something similar for infinite-dim spaces

[rough cedar]

lol is this someone's name or did someone do the funnies

[shadow charm]

Deck as in card deck I think

Since you get an induced permutation on the fiber

[hidden crag]

You had a question yesterday that asked for isomorphism classes of coverings

That usually refers to two coverings s.t. there is a homeomorphism between the two that respects fibers

A covering transformation is (for me) a continuous map that does that

[opaque cloud]

@limpid fern wdym?

isnt the discrete topology just the powerset of the space?

oh do you mean, for a particular point x \in X the smallest open neighborhood you can choose is {x}?

[limpid fern]

yep

perhaps your question could be

"given some open nbhd, can we always find a smaller one?"

and the answer is no

[opaque cloud]

huh then you need a different definition for convergence in topological spaces?

[limpid fern]

no?

the topological definition of convergence is given an arbitrary open set around the convergence point

we can always find the tail of the sequence there

it goes wonky in the discrete topology btw

[opaque cloud]

I see okay my intuitions of R^n have lead me astray

[limpid fern]

like say we have some limit point x

we can literally take {x} (which would be an open set)

I mean you can see how weird that gets

[opaque cloud]

right you just say "for every open nbhd around the convergence point" instead of for every open ball in the metric space case

ok im gonna do electrodynamics now

[rain ether]

the convergence point?

[hidden crag]

lol

[limpid fern]

*a convergence point

it's only "the" if its hausdorff

(but you don't have that wonkiness in hausdorff spaces anyway)

[opaque cloud]

oh

[limpid fern]

(you can converge to more than 1 limit if you have weird spaces)

[opaque cloud]

oh goddamn

so a given sequence can have any number of limit points? (on a general top space)

that's totally not insane

[odd flame]

last thing about this discussion hopefully but how does this act (choosing my words carefully here bc there's a notion of a group action here i think) on the fundamental group of T and K

[odd flame]

[odd flame]

any hints on this please

in general, im havinig trouble really seeing the relationship between covering spaces and properties of the fundamental group

[bitter smelt]

Are you reading a textbook sebb or just notes?

[odd flame]

prof is loosely following Massey

though he gives us problems from hatcher because yes

[bitter smelt]

Right, but are you reading a textbook, or just his notes

[wanton marsh]

do you have a concrete path on RP² representing a generator of pi1(RP²)

[odd flame]

oh i use the textbook as reference

mostly his notes

[odd flame]

is concrete a formal term?

[wanton marsh]

it's something you can draw or you can define with honest formulas

because it's only asking to move it in pi1 of the connected sum, and then lift it to your covering

and once you have lifted it you know the action on all the (2n,0,0)

and hopefully you can deduce the action on everything else from that

[opaque cloud]

what's an example of a net that's not a sequence?

[gritty widget]

Riemann sums

[odd flame]

i just realized my midterm is on thursday and not tomorrow

last semester i thought my midterm was on thursday when it was on a tuesday

this is a much better feeling than that

[gaunt linden]

In a space with the diffuse topology, every point is a limit of every sequence.

[opaque cloud]

[gaunt linden]

Non Hausdorff spaces are weird.

[rain ether]

Is diffuse another term for indiscrete (the trivial topology)? Or something different? I haven't heard about it.

[opaque cloud]

Oh that would make sense actually

[gritty widget]

What's a diffuse topology?

[gaunt linden]

Yes, I meant indiscrete topology. It was called "diffuse" when I learned point-set many years ago (in a non-English language), but Google now tells me that it an exceedingly rare name.

[opaque cloud]

smh tropo trying to fool me with a trivial example

[gaunt linden]

Like, the only Google hits are course notes from the one professor who taught that course.

[gritty widget]

With respect to convergence at least

And extensions of continuous functions from dense sets
Or compact sets being not necessarily closed

It's just that assumption of Hausdorffness leads to beneficial properties which may often fail in another setting

[bitter smelt]

huh?

A topology is something you put on a space. Any space can be endowed with the indiscrete topology

To ask for an indiscrete topology doesn't really make sense

There is only one

[rough cedar]

lol

[bitter smelt]

Anyway, what does it mean to "splice by a sphere?" For example in the construction of pi_2 using whitney disks in OS's HF homology, what does this mean

Are all of these being used in the same way?

If so, what do any of them mean

The vibe from the last one is just, you stick the two disks together and get a new one from x to z

But this seems somewhat sketch just for existence on RHS versus LHS if we are just naively sticking them together

[gritty widget]

I'd say 'set' instead of space but yeah

Indiscrete topology on a set X is the smallest topology you can give on a set, it consists only of the empty set and X, that is {empty set, X} is the indiscrete topology on X

Intuitively it's a topology which doesn't distinguish between any two points of X

[tidal lynx]

dox

[feral copper]

Okay so basically, there are only three index 2 subgroups of Z², which are:

1. {(x,y) | x=0 (mod 2)}
2. {(x,y) | y=0 (mod 2)}
3. {(x,y) | x+y=0 (mod 2)}
   I know how to draw a 2-fold covering for the first two (they are a torus wrapped around twice, just like the 2-fold cover of the circle). What is a covering corresponding to the third? Been trying to see this all afternoon, and I'm starting to think the space above might just not be a torus...

[bitter smelt]

I would like to announce I have figured it out

When in doubt, approximate

[odd flame]

[abstract wigeon]

quick question: if two topological spaces have different fundamental groups then they are not homeomorphic, but if two topological spaces have the same fundamental group, that does not imply that they are homeomorphic right?

[dry jolt]

Correct. For example, S^n has trivial fundamental group for n > 1, but S^n is not homeomorphic to S^m if m is not equal to n

[abstract wigeon]

right okay thanks

im not sure how useful this is, but does homeomorphism imply identical fundamental groups?

[dry jolt]

Homeomorphic spaces have isomorphic fundamental groups. In fact, homotopy equivalent spaces have isomorphic fundamental groups, the isomorphism being induced by a homotopy equivalence

A broad theme of algebraic topology is to understand spaces up to homotopy equivalence (because this is a more tractable problem than understanding spaces up to homeomorphism). Knowing that the fundamental group is preserved under homotopy equivalence is good and useful because it tells us that the algebraic invariants we assign to our spaces will help us distinguish between spaces which aren't homotopy equivalent

[abstract wigeon]

very interesting, thank you very much

[pearl holly]

ye exactly, now you might ask for a converse: do homotopy groups (which are just maps from S^n to X up to homotopy with a "natural" group structure, e.g. the fundamental group is the first homotopy group) characterize your space up to homotopy equivalence (as Walter said above, the fundamental group does not suffice for this, but maybe looking at the higher homotopy groups will)? The answer turns out to be no, but you can get a nice "algebraic" partial converse to this if you restrict to CW complexes, where Whiteheads theorem holds, which says that two CW complexes are homotopy equivalent iff there is a map inducing an isomorphism on all the homotopy groups

this is irrelevant to your questions but I thought it would be nice to mention since it's a cool thing

[abstract wigeon]

woah that's pretty intense but sounds really cool. i think i need more background to really understand it. do you prefer hatcher's book?

[plain raven]

👀

[pearl holly]

Lmao clerk ready with those eyes

Idk there seems to be a lot of controversy around it, but I personally like it. There are other books too, e.g. Dieck, Spanier, May, Rotman, Switzer maybe if you want to go deeper I guess

Maybe clerk reacted with the eyes because of something different?

[unreal stratus]

Hm maybe I'm being dense, but okay so one form of Bott periodicity says that $\Omega^2 (BU \times \mathbb Z) \simeq BU \times \mathbb Z$ and since $K^0(X) \simeq [X,BU \times \mathbb Z]$ whilst $K^{-2}(X) = K(\Sigma^2 X) \simeq [\Sigma^2 X, BU \times \mathbb Z]$ we have $K^0(X) \simeq K^{-2}(X)$ as sets via $\Sigma-\Omega$ adjunction. But why does this homotopy equivalence lead to an isomorphism of \emph{groups} $K(X) \simeq K^{-2}(X)$?

[gentle osprey]

potato

[gritty widget]

Iirc Clerk is not a fan of Hatcher but maybe I remember wrong

[pearl holly]

Ye iirc they like Spanier, don’t quote me on this tho

I just like Hatcher because it’s very geometrical and it honestly covers a lot

[bitter smelt]

Hatcher bad.

dC eyes emoji because someone said the word complex and their simplex OCR got triggered

[pearl holly]

where is Max and brofib

[plain raven]

the eyes were more like "Can't wait to see who's going to start a fight over Hatcher this time"

I do not like Hatcher.
I like Spanier.

[unreal stratus]

I am in the same boat

might buy a copy of Spanier tbh, keep loaning it

[pearl holly]

Whatever floats your boat

[next crystal]

I want to prove that $[a,b] \cap \mathbb{Q}$ is not compact in $\mathbb{Q}$ for all $a,b \in \mathbb{R}^+$ with $a<b$.\ Since $[a,b] \cap \mathbb{Q}$ is countable, we can enumerate it as ${q_1, q_2,...}$. I want to say ${[a, q_i)}{i \in \mathbb{Z}+}$ is an open cover of $[a,b] \cap \mathbb{Q}$ with no finite subcover, but if $b \in \mathbb{Q}$, I think this may not actually cover the set. Is there a way to fix this?

[gentle osprey]

michαel

[next crystal]

would including a set like (q_1, b] in the open cover do it?

[unreal stratus]

You can play around with the left end points too

But you can just adapt the cover a bit and use a similar idea

So like, if b is irrational, what you've done works nicely.

So try adapting it so that you use a number in [a,b] you know is irrational

[next crystal]

this might be more complicated than it has to be but how about this
Pick $\alpha \in (a,b)$ that is irrational. Say $[a, \alpha] \cap \mathbb{Q} = ${p_1, p_2, ...}$, and $[\alpha,b] \cap \mathbb{Q} = {q_1, q_2,...}$. Then ${[a, p_i)} \cup {(q_i, b]}$ is an open cover of $[a,b] \cap \mathbb{Q}$ with no finite subcover

[gentle osprey]

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

[next crystal]

and this should work for rational or irrational a and b

[unreal stratus]

That works yes but can be made simpler

You don't need to explicitly enumerate stuff

[next crystal]

{[a, n) | n < alpha} U {(n, b] | n > alpha}?

[unreal stratus]

That works, nice

!

What I had in mind was like

[next crystal]

wait would i have to intersect it with Q though

[unreal stratus]

yeah

but i assumed that's what you meant lol

But yeah uh what I had in mind was like

$\mathscr U = {U_n}_{n=1}^{\infty}$ where $U_n = { x \in [a,b] \cap \mathbb Q : |x-\alpha| \ge 1/n}$

[gentle osprey]

potato

[next crystal]

ah so the cover elements here are just single points

[unreal stratus]

No

[next crystal]

oh wait nvm

[unreal stratus]

Yeah so like

stuff of the form [a, alpha - 1/n) cup (alpha + 1/n,b ], basically

But yeah

Also, like

[a,b] is compact Hausdorff, so a subspace of it is compact iff it is a closed subspace

Well okay that is overkill but let's simplify that

One thing you can use is that for metric spaces X, X is compact iff it is sequentially compact

And it's easy enough to find a sequence in [a,b] \cap Q which doesn't have a convergent subsequence - just approximate an irrational by a sequence of rationals

That corresponds to a cover without a finite subcover

[gritty widget]

https://www.academia.edu/96262306/δ_small_Subobjects_and_Universal_Categories

[next crystal]

we actually just convered compact iff sequentially compact in metric spaces so that makes sense

thanks so much!

[unreal stratus]

Np

[odd flame]

what is this saying in english

[odd flame]

might i ask why

my prof uses Massey and takes problems from hatcher

[odd flame]

but also here i think im just getting lost in the symbols

[gritty nest]

what's the conclusion of theorem 2.1?

[odd flame]

oh right

kek

[plain raven]

hatcher is a bit verbose and handwavy with his arguments. A lot of intuition and it can be hard to track down the rigor. He leaves a lot of work to the reader of filling in the rigorous content behind what he's saying, too, and this gives the overall impression that rigor is just not very important in topology and you can just vibe with the playdoh

[odd flame]

that makes sense with the way my prof has taught this so far lol

his proof of pi(S1) was basically "look at it"

i cant deny that the relaxation in rigor has made it more fun though

but i guess i should go through spanier when i inevitably redo this

[thorny agate]

Anyone know anything about "Topology and Groupoids" by Ronald Brown?

[plain raven]

I'm sure it's a good book but just hearing the name of it makes me laugh because Ronnie Brown shills his book on stackexchange every single time he answers a question about topology

Q: "How do I compute the fundamental group of this polyhedron?"

A: "In my book, Topology and Groupoids,..."

[thorny agate]

lmao

[odd flame]

i like to imagine professors sitting patiently in front of their monitors waiting to be like "ackchually"

[odd flame]

bump to this though

it's like a slightly different universal property...?

[plain raven]

What is Proposition 2.1

oh i see

[odd flame]

.

[plain raven]

you posted it later

Oh

I see.

Basically like

In category theory, a universal project defines an object up to isomorphism

This is like a very general theorem/metatheorem that reoccurs in lots of places in category theory

so like

Once you prove that there is at least one object X satisfying the universal property

then you know it's unique

Uniqueness basically always holds for universal properties

even if existence doesn't

[odd flame]

ok so it's not just me, this feels strangely abstract

[plain raven]

So you can read proposition 2.1 as being in two parts:

1. There is a certain universal property P
2. the weak product of Abelian groups has this universal property, i.e., there exists an object having the universal property

then prop 2.2 is uniqueness
3. There is a unique object with property P up to isomorphism

[odd flame]

i see

[plain raven]

I can spell this out for you if you like lol

Let ${ G_i }_i$ be a family of Abelian groups.

[gentle osprey]

diligentClerk

[plain raven]

We say a group $A$ has the universal property $P$ if, for any Abelian group $B$, there is a one to one correspondence between\
\begin{itemize}
\item families of homomorphisms $\phi_i : G_i \to B$
\item homomorphisms $A \to B$
\end{itemize}

[gentle osprey]

diligentClerk

[plain raven]

Moreover, this correspondence should be "natural" in B. Don't worry about what this means rn

But note that if you take B = A and the homomorphism A -> B to be the identity map id_A, then this corresponds to some family of maps \psi_i : G_i -> A.

Now proposition 2.1 says:

The weak product G of the Abelian groups has the universal property P.

Proposition 2.2 says:

The object satisfying the universal property P is unique, up to isomorphism; any object satisfying property P is isomorphic to the weak product.

[odd flame]

ok that def helped

thank you clerk

[steel glen]

is H_n always the abelianization pi_n/[pi_n,pi_n] or is that only for n=1?

ok i should have googled it first

[solemn oar]

Even for n=1 it is only true is your space is path-connected.

[patent quarry]

cf. hurewicz theorem.

[next crystal]

If X is limit point compact and A is a closed subset of X, then A is limit point compact.
To prove this I need to show every infinite subset B of A has a limit point b. Should I be showing this in the smaller space A? So every open set in A containing b intersects B\{b}?

vs every open set in X containing b intersects B\{b}

[unreal stratus]

lol well if anyone saw this, Bott answers this question in another paper lol

[supple sable]

Hey im trying to proof $C:=[-1,0]^2\cup[0,1]^2$ is contractible. Visually it makes sense that you can collape any point to $(0,0)$. To proof it is contractible I think it is easiest to proof $Id_C$ is nullhomotopic. For this i set up a homotopy: $H(a,t)=(1-t)a$. It is easy to verify $H(a,0)=a (Id_C)$. $H(a,1)=(0,0)$. But i struggle what the steps are to proof it is continuous. It is just the product of two continuous functions, is it that easy? I can also check that H maps each point to another point in C, are these two enough?

[gentle osprey]

AlexSchopbarteld

[unreal stratus]

Yes

:)

[supple sable]

Well

[unreal stratus]

Tbh usually one would just say this is obviously continuous lol, but also it is clear from the metric on R^2 (if you want another way)

[supple sable]

Fair enough haha

[hidden crag]

This is so funny

Reading some answer and as soon as I read that books name „oh it’s Ronnie again“

[unreal stratus]

Yup

[odd flame]

okay then

[gritty widget]

True lol. He has to gain recognition somehow

[odd flame]

how is knowledge of free groups/products relevant to fundamental groups

obv example is SVT ig

[odd flame]

is the free product of groups the same as their direct sum?

[dry jolt]

NO

[odd flame]

[dry jolt]

The free product of groups is a kinda weird and messy thing, but it's characterized by the same universal property as the coproduct, of say modules. Namely, if you have group homomorphisms G -> K and H -> K, then there is a unique group homomorphism G * H -> K such that the canonical inclusions G -> G * H and H -> G * H make the relevant diagram commute

I'm saying that because I know you've seen categorical coproducts before

[odd flame]

yeah i saw the universal property thing and figured they might be at least similar

[dry jolt]

As for the actual construction of the free product of groups, it's a little messy to deal with in a hands on manner as you have to start talking about reduced words and such

[odd flame]

i have this example

[dry jolt]

Right, given groups G and H, a word is some element of the form w1 w2 ... wn where each w_i is in G or H. You can reduce this word by removing copies of the identity or multiplying adjacent words if they lie in the same group

Yes, precisely. Every word has a unique reduced form. The free product of G and H is the group whose underlying set is the set of reduced words where the group operation is concatenation

There are some technical issues to settle, like concatenating reduced words need not yield a reduced word so you have to rewrite it as a reduced word

[odd flame]

ig the word stuff at least makes it kinda clear how it relates to polygons n stuff

[dry jolt]

Free products are annoying and messy in practice, they become a bit more tangible if you opt to work with group presentations since then you can explicitly write presentations for free products and free products with amalgamation

[odd flame]

that's the next section actually

considering skipping ahead then

[dry jolt]

Perfect

[odd flame]

4. Free Products of Groups
5. Free Groups
6. Presentation of Groups by Generators and Relations

[dry jolt]

Fun

[odd flame]

lot's of free stuff

[tidal lynx]

learning free object stuff is satisfying after only seeing their constructions in an algebra class

[hidden crag]

The abelinization of the free product is the direct sum of the abelinization of the components

Useful fact for the abelinization connection between pi1 and H1

Because of left adjoints preserving colimits etc

[marble socket]

what is this sorcery >.<

[odd flame]

ok hi walter, if you're not busy can we talk about what you meant by this please

[dry jolt]

sure, but let's move to #groups-rings-fields

[odd flame]

specifically looking towards SVT tho

[dry jolt]

ah ok

never mind then lol

Right, so first of all do you feel like you understand group presentations?

[odd flame]

i mean im trying not to overthink it

just a choice of generating set and the relations it needs

[dry jolt]

Or at least why they might be useful? Like we can describe every element in the group as a product of generators, and we know which words are trivial based on relations

Right, exactly

[odd flame]

the word stuff is weird

[dry jolt]

One more useful property worth pointing out is that it's very easy to describe maps out of a group given a presentation

All you need is to specify where you map the generators and make sure that the images of the generators satisfy the same relations that the generators in the domain satisfy

[odd flame]

ive seen that come up in hw's yeah

[dry jolt]

great

ok, so now how do free products interact with group presentations?

[odd flame]

im gonna assume nicely

[dry jolt]

Recall that the universal property of the the free product G * H tells us that given maps G -> K and H -> K, there is a unique map from G * H -> K making stuff commute

In particular, if we have presentations <S_1 | R_1> for G and <S_2 | R_2> for H, then we know how to specify maps G -> K and H -> K

Then I claim < S_1 union S_2 | R_1 union R_2> satisfies the universal property of the free product, hence is isomorphic to G * H

To see this, the natural way to define a map from this presentation to K is to map the generators to what they are sent to by G -> K and H -> K. Then these satisfy the same relations, hence we have a homomorphism

So yes, free products interact with group presentations very nicely

A similar thing is true for free products with amalgamation, but I'll let you mull over that for a little while

Or maybe not, idk, it's more relevant for van kampen stuff

[odd flame]

ok a little bit behind on what you're sending but gonna back up real quick to the actual defn of a free product

[dry jolt]

Sure

[odd flame]

i can see that it has the same universal property as a coproduct of modules, like you said before

should i concern myself with anything beyond that property though?

like my textbook section just has the universal property, proof of uniqueness and existence and examples

[dry jolt]

Hm, maybe think a little bit about why the direct product of groups doesn't satisfy the universal property

But other than that, I mostly just use the universal property and then use presentations for any actual computations

[odd flame]

i mean it's the same reason why it doesnt for modules right, the infinite case

we can treat it the same as a coproduct of modules then is what im asking

sorry if im fixating on something unrelated

[dry jolt]

Ah, yeah it's not really the same reason

[odd flame]

my other big boy class this semester is fields and modules so i have that to go off of

[dry jolt]

Like you can form a direct sum of groups, it just doesn't satisfy the universal property of the coproduct

[odd flame]

ohhhhhhhhh

[dry jolt]

The issue is that in a direct sum of groups, elements corresponding to different groups automatically commute, so that's a relation any image will have to satisfy. In particular, if you have maps G -> K and H -> K whose images don't commute, then there won't be a map G x H -> K making the relevant diagram commute

[odd flame]

ok im taking the relationship between general groups and modules too far then

they have the same universal property but the objects of the category satisfying them are of a different form

[dry jolt]

yeah, the module/abelian case is very nice

that's right, it's probably worth taking a closer look at the construction of the free product because it's very different from the construction of the coproduct of modules

[odd flame]

ok that makes more sense and is probably part of my confusion

[dry jolt]

Right, so as a quick example, let's take G to be the cyclic group of order 2 generated by g, and H to be the cyclic group of order 3 generated by h. Then the free product consists of reduced words in g and h, so an arbitrary element will look something like g h^(k_1) ... g h^(k_n) where k_i = 1 or 2. Also a word need not start with g or end with h, that's just the way I've written it

[odd flame]

maybe a silly question but how many elements are in the free product then

cuz we can just make the word as long as we want

within the relations of the presentation right

[odd flame]

"unique up to unique isomorphism"

[dry jolt]

That's right, you can show that the free product of non-trivial groups is always infinite

[odd flame]

that's why the free product we see in SVT is funky right

[dry jolt]

Right, so Van Kampen actually uses the free product with amalgamation, which is a pushout in the category of groups. This is a little different from the free product

[odd flame]

https://tenor.com/view/timi-gif-20112884

[hidden crag]

[odd flame]

you mentioned pi1 and H

what was H

[hidden crag]

The first homology group

H1

H_1 to be clear I guess

[odd flame]

oh we havent done homology yet

well we have

but it wont be on the exam im studyinig for

[hidden crag]

Oh okay

[odd flame]

so how is abelianization related to pi1 tho

[nocturne basalt]

H1 is the abelianization of pi 1 for path connected spaces

[hidden crag]

Right I forgot the path connected assumption earlier

[odd flame]

all rings are commutative and have identity

right

[blazing oxide]

does anyone have a good resource of motivating the definition of a topology?

im writing a paper about the fractional integral operator, been looking at metricizable spaces, but want to better understand the topologies of the R->R functions

[gritty widget]

Metrizable

Wdym by topologies of R -> R functions

[nocturne basalt]

both operations need not be commutative

[odd flame]

i was kidding lol

[nocturne basalt]

also some people require a multiplicative identity, others do not so it depends on context

[blazing oxide]

like idk we have L^p spaces and C^n spaces, what other interesting things can we do if we have topological non-metric function spaces

[gritty widget]

Those are all metrizable

You can have topology of pointwise convergence ig

[little hemlock]

a sometimes useful topology in functional analysis is the weak* topology on the dual of a banach space, which sort of captures the notion of pointwise convergence. It's usually not metrizable iirc

[gritty widget]

It's weak and weak* topologies on infinite-dimensional Banach space will never be metrizable

[odd flame]

im trying to understsand SVT and using pi(T # T) as an example

i know pi1(T) = Z x Z

and T # T has an obvious was of being broken down into open sets A U B

[hidden crag]

what's T

[odd flame]

torus

basically how do you find pi1(T # T)

[hidden crag]

uh

ah nvm i thought you said pi1(T # T)= Z x Z

[odd flame]

or is there a more enlightening example

no no

[hidden crag]

hmm i guess you get the two torus, stretch it a bit in the middle and choose U and V s.t. they are one torus plus the stretched part each

the intersection is basically a square then which deforms to S^1

[odd flame]

that's what i was thinking yeah

[hidden crag]

U and V should be punctured Tori then, homotopy equivalent to S^1 v S^1

this should give the result expressed as amalgamation

[hearty jacinth]

the topology of pointwise convergence is Hausdorff but not metrisable

[hidden crag]

ok

[odd flame]

ok

ok i need to review the amalgamation stuff

i understand the idea behind SVT but not how to arrive at the word presentation

[supple sable]

Can someone explain how pushouts are used to define gluing in CW complexes?

Or in gluing in general

[hidden crag]

Do you know how a pushout in top can be constructed

[hidden crag]

yeah i'm not exactly good at this either

[odd flame]

ish ok

sometimes i forget the math server is just a bunch of people and not an all knowing wizard that understands exactly what's in my head

i still appreciate you timo

[hidden crag]

i'm just a random undergrad as well

[odd flame]

what year are you

[hidden crag]

what is that

what is a super reaction

[odd flame]

it does a little dance when you hover

[hidden crag]

third

[odd flame]

crazy

[hidden crag]

[fading vale]

you just take the pushout of all the attaching maps into the n skeleton and the inclusions into the n+1 disks and define that to be the n+1-skeleton

like if you wanted to do this with a single cell you have an attaching map S^n+1 -> X_n and an inclusion map S^n+1 -> D^n+1 and then you take the pushout

[supple sable]

I think i finally get it

I never encountered category theory so I got kindof lost at what was geometrically happening with all the mumbo jumbo

[hidden crag]

@odd flame another SVK task is to compute pi1 of S^1 v S^1 ... n-fold

generalizing the S^1 v S^1 case

basic free product stuff:
Let $g \in G_1 \ast G_2$ prove that if g has finite order then it is either in $G_1, G_2$ or conjugate to an element of one of the two.

[gentle osprey]

Timo

[hidden crag]

And prove that the center of G_1 * G_2 is trivial if G_1 and G_2 are non-trivial

the klein bottle is another neat example but you already did that one if i remember correctly

[woeful coral]

Given a compact set A in R^n such that it is the closure of it's interior, can it be partitioned into finitely many compact convex sets?

Where a partition is defined in the usual sense as a set B of subsets of A where the union of all elements in B is A and all elements have NULL intersection.

[tender halo]

take segments [1, 1/2], [1/4, 1/8],... and {0}

its a closure of its interior but can't be partitioned

[coarse night]

I mean saying compact already means it's closure of it's interior

[gritty widget]

Take a set shaped like an arrow

In the plane

If it had a partition of more than 1 compact set, it'd not be connected

Thus its partition needs to be itself, but its not convex

[coarse night]

not true for n>1 nvm

[tender halo]

not true in R either

[coarse night]

lol

[tender halo]

a point is compact

[coarse night]

yeah dum moment

[gritty widget]

You probably want union of compact convex sets whose interiors are pairwise disjoint

[gritty widget]

This is a counter-example to that ig

Additionally you probably want to assume that A is compact and connected

[woeful coral]

How would one describe a set that has “volume” topologically then. Im tryna get rid of cases like S^1 where we have a hollow interior

[gritty widget]

Wdym

[woeful coral]

Where the red denotes that its got a volume to it

[gritty widget]

Compact connected set can have positive volume but no interior

Usually volume & topology aren't very compatible

[woeful coral]

Yeah

Okay

[tender halo]

like what

fat cantor set?

wait that one is extremely disconnected right

[gritty widget]

You can take [0, 1] x (fat cator set sum {0, 1})

[tender halo]

i see

[charred sundial]

Isn't that still disconnected?

Sierpinski's carpet works though

[gritty widget]

Yes because I wrote it wrong

[0, 1] x (fat cantor set) sum {0, 1} x [0, 1]

Any point can be reached by moving around the border of the square

[potent sky]

I know if 2 spaces are homotopy equivalent, they have isomorphic fundamental groups. But if they have isomorphic fundamental groups you can't say that they're homotopy equivalent can you? Is there anything you can say?

[charred sundial]

The sphere and the plane have the same fundamental groups for example

[potent sky]

ty also:

If I'm trying to show that there is no covering map from T^2 to RP^2 can I just say that RP^2 has a universal cover from S^2 which has trivial fundamental group. T^2 has fundamental group Z^2 so T^2 and S^2 are not homotopy equivalent and so T^2 is not a cover of RP^2?

[dry jolt]

I'm not sure how that shows T^2 is not a cover of RP^2

[charred sundial]

Covers can be homotopically different

[odd flame]

what is 2 saying here

[potent sky]

i see

[dry jolt]

You might want to try the the result saying that if you have a covering map, then the induced map on fundamental groups is injective

[odd flame]

actually what's a deck transformation in the first place - this is a random set of notes i found from a different uni but my prof never defined this

[odd flame]

this is a result of the lifting thm right...?

[potent sky]

ah yes that's more immediate

[charred sundial]

A deck transformation of ~X is a transformation f: ~X -> ~X such that the projection doesn't see that anything changed, i.e. pf = p

[dry jolt]

yes, this follows from lifting

[charred sundial]

For example, the classic spiral R -> S1 you can translate floor up

[odd flame]

is this the same as the set of automorphisms of a covering space then?

[charred sundial]

Sure thing

[odd flame]

:O

the general statement is that a homeomorphism of covering spaces induces an injective map of the fundamental groups right

[potent sky]

no i think any covering map induces an injection on fundamental groups

[potent sky]

what are you asking here

what i just said is correct though, i just checked

$$p: \widetilde{X} \to X$$ induces an injective homomorphism on the fundamental groups if p is a covering map

[gentle osprey]

*-algebra

[potent sky]

are you asking if a homeomorphism induces an isomorphism or something?

[unreal stratus]

Hm

Is there a nice description - modelling BU(n) by Gr_n - of what the maps BU(m) -> BU(n) induced by maps U(m) -> U(n) look like?

I suspect the maps may only be well defined up to homotopy but yeah

[potent sky]

Awhile ago I wrote a proof that any map from S^2 to T^2 is homotopic to a constant map.

I start by saying T^2 has R^2 as a universal cover, and then I say because S^2 is simply connected that any map f: S^2 --> T^2 lifts to a map S^2 --> R^2.

Does anyone know why I pointed out that S^2 is simply connected and what the relevant theorem is here?

[odd flame]

simply connected = path connected and trivial fundamental group if im remembering correctly

R^2 being the universal cover of T^2 means that it is simply connected

so im assuming the combination of those things can give you your homotopy

[potent sky]

im not actually constructing any homotopy here

i just use the fact that R^2 is contractible so any map out of R^2 is homotopic to a constant map

the simply connected part plays a role in the lifting I think?

@odd flameI found the theorem in hatcher!

i actually had no idea about that iff condition. But it's true here because the fundamental group of S^2 is trivial

[unreal stratus]

Yeah it's nice, you can think of this as saying that "the obvious obstruction is the only one"

I think it's known as the lifting criterion

Also, I wanted to point out (if you've not seen it already) that this sort of problem is super important - it allows you to show that if a space X has a contractible covering space, then all the higher homotopy groups vanish

[odd flame]

does * as used in this hint always just refer to homomorphism of fundamental groups

[dry jolt]

yes, in this context it refers to the induced map on fundamental groups

[snow fulcrum]

in general i think it's used to denote a push forward by a functor

in this case it's the homology functor

[odd flame]

how does the proof here go anyways

ive seen the f(x) - f(-x)/| f(x) - f(-x) | map

but how does that give the contradiction with projective planes

[plain raven]

Are you asking for a solution to the exercise?

[odd flame]

i'll rephrase

[odd flame]

https://math.stackexchange.com/questions/2417185/proof-of-borsuk-ulam-theorem first of all what is the double covering self map in this answer

second why is such a nontrivial homomorphism a contradiction

[dry jolt]

the double cover sends theta to theta^2

[dry jolt]

what's the fundamental group of RP^2 and what's the fundamental group of RP^1?

[odd flame]

rp1 is just s1 right

[dry jolt]

Right

[odd flame]

so Z

oh a hom from Z/2 to Z

[dry jolt]

Right

[odd flame]

kek

so we assume for contradiction that an f with no antipodal preservation exists

define a map g: S2 -> S1 in terms of this f

see that this map g implies a bad homomorphism and arrive at contradiction

[little hemlock]

Let $X$ be a connected CW complex. How come $\widetilde H_{n+1}(SX) \cong \widetilde H_n(X)$?

[gentle osprey]

kxrider

[little hemlock]

it doesn't seem to follow directly from the LES for (SX, X). im guessing its some kind of cellular homology argument?

[dry jolt]

You could do a Mayer-Vietoris

[little hemlock]

oh, like take SX to be a union of cones?

[dry jolt]

That's right

[little hemlock]

aight lemme try that

okay yea i see

ty walter

[dry jolt]

happy to help

[odd flame]

is it accurate to state the subgroups of the fundamental group of a space X fully determine the covering spaces X can have

so for example S1, with fundamental group = Z

there are infinitely many possible coverings...?

also i know this theorem is really strong but im not quite sure what it's saying

[odd flame]

tho if im understanding, this is a result of that thm

[hidden crag]

We’ve talked about this exact theorem before here

[odd flame]

last min review session

[hidden crag]

there is one cover with n-sheets up to isomorphism for all n

[odd flame]

hmr

[hidden crag]

Z has one subgroup of index n for all n

[odd flame]

yeah each of those n has a cover associated to it right

[hidden crag]

yes

[odd flame]

this is relatively simple just double checking

kk

[nocturne basalt]

the connected covers yeah

if your space is "nice"

[hidden crag]

yeah i'm always assuming that your space is nice enough for the classification

[grizzled ibex]

Wikipedia says that a subbase of (X, \tau_X) is not required to always cover X, but if you want to prove alexander lemma you should assume so.
Isn't this kinda misleading? idk

it might make people kinda confused idk

[flint hamlet]

hello guys, does anybody have any resources for learning about other models of quantum computing than gate-based

especially topological qis

qc

[coarse night]

not sure if that's topology

[flint hamlet]

its computing based on braid groups which act as logic gates

braids formed by anyon world lines

[grizzled ibex]

#modeling #computing-software

[honest thicket]

Hey everyone,
I am an undergrad student and I want and have to learn basic topology. Therefore I am looking for a good video series for learning topology of Euclidean space (open, closed, limit point, compact, connected) specifically and topology in general.
I have already found these 3 series and was wondering which one you already know/can recommend. I can't watch them all due to limited time.

https://youtube.com/playlist?list=PLbMVogVj5nJRR7zYZifYopb52zjoScx1d

https://youtube.com/playlist?list=PL6763F57A61FE6FE8

https://youtube.com/playlist?list=PLd8NbPjkXPliJunBhtDNMuFsnZPeHpm-0

If you would recommend a different course, feel free to suggest it.

[bitter smelt]

Sounds like you're looking for point set, not algebraic?

Honestly I'd recommend just reading some point set book over a video series. But if you insist, the second one is not the same subject.

I'm not going to watch these, but the first one is indeed point-set, so at least it's the subject you're looking for.

And lastly the third one also appears to be point-set, but the order of lectures does seem a bit strange, maybe.

[odd flame]

just read munkres

learning to read textbooks is an important skill

[stark fog]

I also think reading is most important but I watched the videos from the third playlist and it is very helpful when you have some idea on what's he talking about.

[odd fjord]

I'm trying to solve this problem:

Construct a compact set of real numbers whose limit points form a countable set

I made the set $S = \left{0\right} \cup \left{\frac{1}{n}: n \in \mathbb{N}\right} \cup \left{\frac{1}{n} + \frac{1}{m}: n \in \mathbb{N}; m = n, n + 1, ...\right}$. It's clear that $0$ and all the $\frac{1}{n}$ are limit points of $S$, so now I'm trying to show that these are the only limit points (then the set of limit points is countable because they are all rational). The smallest element of this set is $0$ and the biggest one is $2$, so if there is a limit point that is not the ones I mentioned it must be in this interval

Also, since it is in the bounds of $S$ it must be between $\frac{1}{p + 1}$ and $\frac{1}{p}$ for natural $p$. So I considered all the points of $S$ between those numbers to be of the form $\frac{1}{p + 1} + \frac{1}{y}$ for $y > p(p + 1)$. Now we can choose a $y$ that is so large that $\frac{1}{p + 1} + \frac{1}{y} < x$. It's clear that there is a finite number of points between that number and $\frac{1}{p}$, so $x$ (which is between them) cannot be a limit point of $S$, which in turn shows that $S$ is closed and hence it is compact

Is there anything wrong with this proof?

[gentle osprey]

Aiya

[gritty widget]

By countable you mean infinite countable?

[odd fjord]

Yes

The book I'm studying seems to define countable as always infinite. But now that you mention it I'm not sure

It's probably infinite

[gritty widget]

"Since it is in the bounds of S..." wdym?

[odd fjord]

It's between its least upper bound and greatest lower bound

I meant to write "within the bounds" of S. Maybe that would've made it clear

[gritty widget]

There's two logical skips here

1. numbers > 1 in S form discrete set
2. 0 is irrelevant

I guess you're assuming its not of tge form 1/n or 0 though so whatever

[odd fjord]

0 is irrelevant in what part? If the set didn't have 0 then it wouldn't be closed

I need it to be closed for it to be compact

[odd fjord]

Also you're right about the first one. I didn't have to consider numbers greater than 1. That makes sense

But the things I wrote still work if we consider only numbers between 0 and 1

[gritty widget]

So is x a limit point? You didn't write it

You write "it's clear that ...". How is it clear

I'd elaborate on that part

Other than that it looks good, good job

A_n = {1/n+1/m : m = n, ...} and notice that sets A_n are bounded by 1/2n etc is what I want to see, like a more formal justification/better sketch of the argument

[odd fjord]

Learned that from Rudin
It's clear that there is a finite number of points between those numbers where x lies between because all the members of S between 1/(p + 1) and 1/p are of the form 1/(p + 1) + 1/y with y > p(p + 1)

And by choosing a y big enough to make 1/(p + 1) + 1/y less than x you'll only have finite many numbers of that form that are bigger than it and less than 1/p

I'm saying that there are finitely many points of S between 1/(p + 1) + 1/y and 1/p

Obviously there are infinitely many points between 1/(p + 1) and 1/p, by how S is defined

[gritty widget]

They don't have to be of the form 1/(p+1) +1/y for natural number y if thats what you mean

[odd fjord]

They do

Because that's how S is defined

[gritty widget]

No, look at n = 3 for example.

We have 2/3

Thats between 1/2 and 1

The condition that m = n, n+1, ... is too weak to guarantee they're all in disjoint intervals

Still, only finite amount of terms will be there

If you did powers of 2 instead

It'd be easier to write down this set but each interval would only have desired elements

[odd fjord]

You're right

That's what I was doubtful about, if those are the only numbers between 1/(p + 1) and 1/p

[gritty widget]

No but it doesn't matter much. But if you chose powers of 2 the argument would be easier

[odd fjord]

When can 1/n + 1/m be between 1/(p + 1) and 1/p for n > p + 1?

[gritty widget]

Because then between 2^(-(p+1)) and 2^(-p) you'd have what you want

1/n+1/m <= 1/2n so for large enough n it's never between 1/(p+1) and 1/p

[odd fjord]

So I can just rule out those edge cases and the argument will work, right?

[gritty widget]

So 1/n+1/m is between those only for finitely many n

[odd fjord]

Right

[gritty widget]

Since n > p+1, also for finitely many m

And by your argument there is finite amount of them of the form 1/(p+1)+1/m

Which contradicts x being a limit point

(points of S between 1/(p+1)+1/y and 1/p)

Using powers of two (or letting m start from some value so that 1/(n+1) +1/m < 1/n like m = (n+1)n+1) would made the argument easier

Though it's not that relevant

[odd fjord]

I understand everything now and I fixed my proof

[flint hamlet]

why such strict time constraints

[thorny agate]

Is taking algebraic topology with no topology background a bad idea?

cause supposedly the topology prof next semester isn't that good so I may not take it

[cedar pebble]

kinda yea

you can fill in the gaps on your own but you shouldn't go in without being reasonably familiar with basic point set topology

[thorny agate]

I'm familiar with some from analysis

Ehhhhhh I'll power through a shit prof

I've done it before I'll do it again

[solemn oar]

Ideally you should know a little point-set topology and a little differential topology/geometry before doing algtop.

[little hemlock]

if by "familiar with some from analysis", you mean the topology of metric spaces, i think you could get up to speed on general spaces pretty easily tbh

[shadow charm]

Just get familiar with basic concepts of separability, compactness, and connectedness, and get familiar with the basic operations of products, disjoint unions and quotients

(By separability i really just mean hausdorff and non Hausdorff spaces)

[paper wedge]

@thorny agate read bredon's geometry and topology and do not skip chap 1

[supple sable]

I am struggling with how to come up with CW complexes of glued structures. An exercise I had was to glue two solid tori ($D^2xS^1$) along their boundary($S^1xS^1). I can come up with CW complexes for the tori themselves, first a point, connect two edges to form an 8, and then connect a plane to form a torus. This can be made solid by adding a 2-cell to one of the circles to form a cap within the torus and then fill it in with a 3-cell.

But now i struggle how to glue this to some other torus.

The map that identifies the gluing is the picture where X,Y are the solid tori and Z their boundary. I also fail to visualize exactly what this map is doing, my current intuition is that it wraps one tori around the other helically (or something along those lines) to create a knot.

The excercise states this defines a (p,1)-lens space (pushout(X<-Z->Y), but i have never seen any lens spaces before

sos

[thorny agate]

Ya that's what I mean

[thorny agate]

Sounds good

[supple sable]

As someone currently going through it I can definitely say that you really want to know your point set topology, as well as some category theory if the course uses it.

I have never encountered the latter and I'm struggling

[thorny agate]

I'm learning some category theory rn but it's light

I may work through Awodey's text a little more in my own time

[supple sable]

My algebraic topology course just introduces the concepts that are relevant and I've yet to take a course on it so I think you'll be fine then if this is your first algebraic topology course too

[gritty widget]

Are p-adic spaces homeomorphic to the Baire space ω^ω ?

[civic verge]

Guys a question a plane in R3, is it a closed plane?

Could someone explain me what it is exactly?

I know that the plane in R3 is conformed by limits is not an open set since the ball or sphere that I make overflows me, but then in the case that it is closed I know that I take an element of the set of the plane and I show that indeed the set of the complement is open, now if I draw a sphere or "ball" this ball is not completely contained in the set then as such it would not be closed, right?

In my exam, I showed that it is not open, but I also said that it is closed..... but I got to thinking and I say no, it is not closed (after my exam).

[unreal stratus]

Yes, it is closed, for example because you can define a plane as the vanishing set of a continuous function

Or, more concisely, by translation we can assume the plane is given by the points (x,y,z) with x = 0, and then this is clearly closed

[feral copper]

A (linear, not affine) plane is closed, as the kernel of a linear form (continuous linear map from R^3 to R). It is not open. For instance, the origin (0,0,0) is always a point on the plane, but any ball centered at the origin will contain a vector normal to the plane, and so points not in the plane (i.e. there is no ball centered at the origin, no matter how small, that will be entirely included in the plane)

For affine planes, it's the same, because it's the translation of a linear plane

if you want to show that the complement of the plane is open directly, you can.

[rain ether]

Every topological space is open and closed in itself.

[feral copper]

Take a vector not in the plane, and take its orthogonal projection onto the plane. Call R the distance between those points. Then the ball of radius R/2 is entirely included in the complement of the plane

[feral copper]

Uh right x')

Got a brain fart there x')

Quick question: if I let f:S^4->S^4 be the map f(a,b,c,d,e)=(a,b,c,-d,-e), then Fix(f)=S^2. What is the quotient S^4/f? Is it also a 4-sphere?

The 4-manifold below at least has the same homology as the 4-sphere xD

[frank bolt]

How can I think about S^2 x S^2?

I read that you can think about S^2 x S^1 as a thickened sphere with the inner and outer spherical boundaries identified

So for S^2 x S^2, I think you could do the same, but imagine the thickness as changing over time?

Yes ok I am more convinced of this now. The thickness would follow a sine curve over time

[unreal stratus]

Not sure how visualising it would necessarily help, given this is most naturally embedded in like lol R^6

[frank bolt]

Apparently Thurston was good at thinking about 3- and 4-manifolds, so I think it's worth trying

[umbral panther]

No, p-adic spaces are locally compact, unlike the Baire space. The p-adic integers are compact. The p-adic rationals contain the p-adic integers as an open subset

[gritty widget]

Already got it, they're homeomorphic to countable disjoint union of Cantor sets.
Thanks though

[untold dew]

i was reading some math overflow and wiki pages whats this whole closure thing? closure, closed, etc?
(does it even belong in this channel idk)

how would u explain it to a highschooler (totally not cuz im in hs)

[untold dew]

looks like an amazing emoji btw

[plain raven]

no

oh

you got it

my bad

but the 2-adics are homeomorphic to the cantor set which is fun

[tepid vale]

let E->X be a topological complex vector bundle where X is compact and E is hausdorff. then the space of sections is a module over the ring C(X;\bC)

my question is how does C(X;\bC) act on this set

[gritty widget]

2-adic integers you mean?

[hidden crag]

pointwise multiplication

wait all the assumptions that you wrote down make me question this since this works for any (real) manifold like that

[gritty widget]

Basically: I never worked with p-adic integers or p-adic numbers, and I could find info about p-adic integers being homeomorphic to the Cantor set but no info about p-adic numbers, hence the question.

[hidden crag]

maybe the complex case breaks something here?

[tepid vale]

oh right, ok

[tepid vale]

no this still works there, i was asking in the context of the serre-swan theorem

[hidden crag]

ah okay i see

[shadow charm]

aight maybe im just being dumb but ive got an exercise where im asked to prove a bunch of spaces are homotopy equivalent, but it seems to me like they're just homeo and I dont really get why my prof is asking to prove homotopy equivalences on exercises where homeomorphisms are true. Two examples of this are the suspension $\Sigma A / (A\times {1/2})\simeq \Sigma A \vee \Sigma A$ or $S^2/(\cup_{i=1}^nm_i)\simeq \bigvee_{i=1}^n S^2$. It's not like showing homotopy equivalences is any easier than showing homeomorphisms here with the knowledge we have (for the second one, fair enough we have contractible subspaces of a CW complex so it would follow easily from that but we havent seen that result in the course so it seems odd to me he'd make the distinction)

[gentle osprey]

𝓛ittle ℕarwhal ✓

[shadow charm]

(here mi are distinct meridians)

[hidden crag]

tbh i'm not sure what the problem is

[shadow charm]

i just want to make sure im not being stupid and these are indeed homeo and not homotopy equivalent

[hidden crag]

I think it's reasonable to just always say htpy equivalence instead of homeo at some point because it's the thing you care about

[shadow charm]

because the formulation of the exercise would suggest they are just homotopy equivalent

[shadow charm]

ig

[hidden crag]

even if they are homeo i'd still say htpy equivalence depending on the context

[shadow charm]

hmm okay

guess i need to stop reading into the exercises

[hidden crag]

I do think that htpy equivalence is easier to show here though

oh i didn't read the "with the knowledge we have" part

[shadow charm]

yeah we havent really done much to help with homotopy equivalences (CW complexes and homotopy extension property are done in another course)

otherwise id get it

[hidden crag]

do you have any assumptions on A

[shadow charm]

nah it's just some space

but i mean the homeomorphism is so explicit

[hidden crag]

if you can just write that down then sure why not

i'm just kinda bad at that so i try to avoid it :D

[shadow charm]

ill be honest for the other ones i just did drawings

[hidden crag]

that's what i'd do

or well the argument that you hinted at at the end of your msg

[shadow charm]

that's what id like to do but the criteria for rigor are oddly inconsistent

i mean for only doing drawings

[hidden crag]

yeah Alg top is sometimes a bit weird in that aspect i agree

[shadow charm]

like the teacher will sometimes ask us to show that some very obvious maps factor through to a quotient and other times ask us to just draw

[shadow charm]

at least the alg top course is more consistent in that drawings are almost always accepted

[hidden crag]

lol fair enough

[gaunt linden]

A more charitable reading might be that you really need to be able to do both.

[shadow charm]

good point

[hidden crag]

i remember spending like two hours with a friend on finding explicit homeos

[shadow charm]

but parameterisations can be so disgusting sometimes

[hidden crag]

just for the TA to draw a pic and move on

[gaunt linden]

Drawings and geometric intuition are indispensable for having a feel for what you're doing, but in order to be a reliable intuition you also need to be confident that the drawings you make could in principle become a symbolic, formalizable argument.

[shadow charm]

yeah that's fair

[hidden crag]

hmm

[quasi steppe]

Is this reduced suspension or regular suspension?

[steel glen]

if Z = XvY, are X and Y open in the wedge sum topology?

[plain raven]

No, not in general.

For example, take X and Y to be circles.

[steel glen]

i see, thank you

im curious if there's any way to visualize the interior of X and Y then

[plain raven]

I would think of it something like this.
First you always have X - cl( x0 ) as an open subspace of X v Y.

The question is when the interior is larger than that.

If y0 is open in Y (i.e., isolated) then X is open in X v Y.

[steel glen]

hm

i think this is an xy problem, i wasn't really looking for the interior

[plain raven]

Oh, ok.

[steel glen]

ok this might be another xy
but if a set Z is the disjoint union of X and Y, then X and Y are open right?

yeah ok the preimage of each cannonical injection is either the whole space or the empty set

[plain raven]

Yes, that's correct. Presuming that we have the disjoint union equipped with the standard topology, if this doesn't go without saying.

[steel glen]

yes that was assumed. thank you

we actually figured out the question

[snow fulcrum]

is there an example of a continuous bijection from a space X to X that isn't a homeomorphism

i.e. both X have the same topology

all the canonical examples have different topology on the two spaces

[limpid fern]

continuous just forward?

[hidden crag]

such examples do exist but i don't know the construction off the top of my head

i remember reading a MSE post about this where several ones are given

[snow fulcrum]

ty

i'll see if i can find it

[hidden crag]

https://math.stackexchange.com/questions/20913/are-continuous-self-bijections-of-connected-spaces-homeomorphisms/20932

i found it

this is restricted to connected ones though, i didn't remember that

[snow fulcrum]

the first answer is really nice actually

[hidden crag]

yeah

[supple sable]

I am struggling with how to come up with CW complexes of glued structures. An exercise I had was to glue two solid tori ($D^2xS^1$) along their boundary($S^1xS^1) and find its CW complex. I can come up with CW complexes for the tori themselves, first a point, connect two edges to form an 8, and then connect a plane to form a torus. This can be made solid by adding a 2-cell to one of the circles to form a cap within the torus and then fill it in with a 3-cell.

But now i struggle how to glue this to some other torus.

The map that identifies the gluing is the picture where X,Y are the solid tori and Z their boundary. I also fail to visualize exactly what this map is doing, my current intuition is that it wraps one tori around the other helically (or something along those lines) to create a knot. Could someone give an example of a CW complex of a structure glued to another? Perhaps this excercise or a reference to another. The excercise states this defines a (p,1)-lens space (pushout(X<-Z->Y), but i have never seen any lens spaces before

[gentle osprey]

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

[obtuse meteor]

A trick that might help is that when gluing two things together up to homotopy you can attach a contractible thing to both

You can also just start with a common boundary and never build up two things

[quasi steppe]

I don’t know if I’m correct but my first thought is to think of how the map behaves locally

And not to think of it globally

Since it’s hard to visualize

And then it’s easier to think about since their boundaries are homeomorphic

[supple sable]

Then I would have to construct a second torus from the boundary?

Well since the boundaries are assumed to be equal i guess that makes sense

So I should just construct the hollow torus and then glue to fill on one hand as identity and on the other differently?

[obtuse meteor]

The gluing maps would be the same you’d just have two 3-cells making up the inside of each one

[supple sable]

Yes but would in making the CW complex the gluing map be how i attach the 3-cells?

I also need to attach a 2cell right? Cuz the hole in the torus is not homeomorphic to a ball

[obtuse meteor]

Yes to both :)

[trail charm]

can anyone give me a hint on this? my first thought is to take $p_1, p_2 \in X$. then there exist functions $f_1, f_2$ such that $f_1^{-1}(0) = {p_1}$ and $f_2^{-1}(0) = {p_2}$

[gentle osprey]

anamono for anamono

[abstract furnace]

just use that \R is hausdorff

note $f_{1}(p_1)\neq f_{1}(p_2)$, find open disjoint open neighbourhoods and then take preimages

[gentle osprey]

delirated

[next crystal]

when we use the notation [f] to mean the class of all functions homotopic to f, what do we mean by class?

is it the same as set

[abstract furnace]

equivalence class

[tacit coral]

I absolutely love this alternative definition of convergence of a sequence: a sequence $a_n$ in a topological space $X$ converges to $L$ iff the map $n \to a_n$, $\infty \to L$ from the one-point compactification of $\mathbb{N}$ to $X$ is continuous.

[gentle osprey]

Morphy

[tacit coral]

uniqueness of limits in a Hausdorff space is because a continuous map is determined by its values on any dense subset

and then, for example, it follows that if $f : X \to Y$ is continuous and $a_n$ converges to $L$ in $X$ then $f(a_n)$ converges to $f(L)$ in $Y$ simply because the composition of continuous maps is continuous.