point-set-topology

[knotty vine]

Though you definitely need subdivision/approximation etc, I guess maybe that plays the role of density in what youre hinting at

[heady skiff]

here, do a and b represent coordinates of the form (a_1, a_2), and (b_1, b_2)? (since for example a is an element of C1 which is by definition a set of two-dimensional coordinates)

[knotty vine]

You can represent it how you want

This text is way harder to understand for me than just saying that S1 x S1 is a donut...

[heady skiff]

nah i love this shit

saying that S1 x S1 is a donut feels very hand wavy to me for some reason

because that's basically saying you have to take for granted that it's a donut in 4 dimensions which we can't visualize

on the other hand showing that they're homeomorphic here is great

now i feel like i can properly think of it as a donut

[gritty widget]

you can't visualize it

[heady skiff]

chess not checkers

[knotty vine]

That's just cause you embedded S1 in R2, but S1 is only 1 dimensional

[heady skiff]

wait wat

[knotty vine]

To visualise, just put two circles next to eachother and track a point on both seperatly

That's precisely S1 x S1: two points in S1

[heady skiff]

sorry i don't follow

[knotty vine]

Draw two circles on a piece of paper, put your right pointer finger on one, and your left pointer finger on the other

Now move your fingers on those circles

Each one of those positions is a point on the donut

[muted arrow]

If you hold one finger still, you are tracing a meridinal curve or longitudinal curve on the torus. That helps me visualize why it is a torus

[knotty vine]

This is how professionals do algebraic topology! I recently graduated to using four fingers!

[queen prism]

so wait, are you trying to use more or less fingers

[knotty vine]

Don't tell me you've used 5 fingers before

[muted arrow]

Sym^g moment

[ebon galleon]

[coral pawn]

Why does inverting weak equivalences imply homotopic maps get mapped to the same maps?

Here C is a closed model category

[prisma garnet]

[hot locust]

can we talk about a sequence in a topological space that isn't metrizable?

[rancid umbra]

yes. sequences are just maps from the natural numbers into your space. you can have a sequence of elements in any set. you need a topology to talk about convergence of sequences in your set

a sequence x_n converges in a top space X to a point x if for each open neighborhood U of x, x_n is eventually always in U

[hot locust]

hi, thanks. Can you give an example of a convergent sequence in a non-metrizable space?

[rancid umbra]

take X to be any non-empty set with more than one element endowed with the indiscrete topology.

this space is not metrizable since it is not hausdorff

every sequence in X is convergent and converges to every element of the set

[hot locust]

ahhhh. Wait, now since every sequence will converge, every subsequence will converge so this space is sequentially compact. But this is not a compact space, so this is in a way a counterexample for the fact that sequentially compact doesn't imply compact. Am I right?

[rancid umbra]

no this space is compact. the only open cover is the whole space

[hot locust]

but it doesn't have a finite subcover?

[rancid umbra]

yes it does. its the same cover

consisting of one element

[hot locust]

okay then

[hot locust]

what would be an appropriate counterexample in this case?

[rancid umbra]

a sequentially compact space that is not compact?

[hot locust]

yea

[rancid umbra]

the long line is the first example that comes to mind

[hot locust]

long line?

alexandroff line?

that one?

[rancid umbra]

yes

[fickle elm]

Say f,g:X-->Y is left homotopic in C. Take f',g':X'-->Y' where X' is cofibrant replacement of X and Y' is fibrant replacement of Y. Since you already invert all weak equivalences in D, f'=f and g'=g in D, so f' and g' are left homotopic in D, again the weak equivalences are already inversed, so they are the same map in D.

[coral pawn]

D is an arbitrary catgeory

So what does it mean to say f' and g' are homotopic in D?

Also, what do you mean by f = f'?

[coral pawn]

Unless X and Y are cofib and fib respectively, this won't be true

I think I got it

If f and g are left homotopic

Take a cyclinderical object for X

It comes equipped with a weak equiv CX --> X

Then the images of f and g are both equal to the composition F(homotopy) F(CX ---> X)^-1 F(codiagonal: A U A --> A) F(inclusion A --> A U A in the first/second slot) = F(homotopy) F(CX --> X)^-1

For reference

[fickle elm]

The idea is to work in the category D, where all weak equivalences are inverted. So we can freely take fibrant and cofibrant replacement as they are now all isomorphisms.

we only care about f and g in D

[coral pawn]

That might show isomorphic but we need literal equality, don't we?

Anyway, I think what I said earlier works. Appreciate the help tho

[fickle elm]

The cyclinderical object CX for X is the cofibrant replacement of X, my understanding is that if all weak equivalences are inverted in a category, then we can assume every object in such category is both fibrant and cofibrant.

[coral pawn]

I am not sure how to make sense of fibrant/cofibrant objects in D

[fickle elm]

Model structure on a category is three collections of morphisms and some diagrams. Functor keeps diagram so I think we can talk the model structure on D.

Not very sure though

BTW I checked the appendix of HTT of Lurie, there the homotopy in a model category is only defined from a cofibrant object to a fibrant object.

[hot locust]

Which sequence in this space doesn't have a converging sub-sequence?

I am trying to prove that this is not sequentially compact

[knotty vine]

https://math.stackexchange.com/a/1558796

[ebon galleon]

Let i_1 and i_2 be the components of i on each respective copy of A. Then since σ is a weak equivalence, it gets sent to an isomorphism. But σi_1 = 1_A = σi_2. So after applying F, we can cancel Fσ to get Fi_1 = Fi_2. So then if h is your homotopy from f to g, we get Ff = FhFi_1 = FhFi_2 = Fg

[ebon galleon]

This is because it's (generally) only an equivalence relation when that holds.

Iirc

[coral pawn]

This seems different than by proof here. Can you verify if mine is correct?

[coral pawn]

This one

[ebon galleon]

Well, F(CX --> X)^{-1} is both i_1 and i_2, so it's kinda the same conclusion. I think you need some justification along the lines of what I said to conclude that though. Since the equality you gave is just noting that (codiagonal) o (inclusion) = (identity)

[hot locust]

Thanks

[ebon galleon]

Oh wait, I suppose what you're using is that (A U A --> CA) = σ^{-1} o (codiagonal)

So then you get that F(f or g) = Fh o Fi o F(inclusion 1 or 2) = the first equality you gave. So then you go from Ff to (Fh o Fσ^{-1}) back to Fg. So that should work

[median sand]

k here is a field and | | is a function with |x|>=0 and |x|=0 <=> x=0 and |xy|=|x|*|y|.

I don't understand why these "balls" should define a topology w/o the assumption of the triangle inequality.

[hot locust]

This makes little to no sense to me 💀

Okay no wait it does

Nvm

[gaunt linden]

It doesn't take much to define a topology. If you declare a set to be "open" if it contains a ball centered on each of its points, all that's needed for that to be a topology is that a) every point is the center of at least one ball, b) the intersection of two balls with the same center is itself a ball with that center.

The topology might not be nice unless you have additional properties, but "topology" itself is a quite lenient concept.

("Balls" is probably not the right terminology unless there's a triangle inequality -- but your quote doesn't say that, it says "neighborhoods")

[heady skiff]

could we have defined $\tilde{f}(s) = (p \mid V_\alpha)^{-1}(f(s))$ for any $V_\alpha$ in the partition of $p^{-1})(U)$ since the restriction $p \mid V_\alpha: V_\alpha \to U$ gives a homeomorphism?

[gentle osprey]

okeyokay

[heady skiff]

<@&286206848099549185>

[fickle elm]

$\tilde{f}(s_i)$ lies in $V_0$ by assumption and you want to define $\tilde{f}(s)$ for s in $[s_i,s_{i+1}]$.

[gentle osprey]

Dong_Valentino

[fickle elm]

So everything should be done in V_0.

[heady skiff]

but $p \mid V_\alpha$ gives a homeomorphism with $U$ doesn't it?

[gentle osprey]

okeyokay

[heady skiff]

so $\tilde{f}(s) = (p \mid V_\alpha)^{-1}f(s)$ would be defined?

[gentle osprey]

okeyokay

[heady skiff]

cuz $f(s) \in U$

[gentle osprey]

okeyokay

[fickle elm]

The preimage of U is a collection of V_\alpha. Each of them is not connected to another as the fiber of the covering map is discrete.

You fix a point in V_0. So when you define the continous map \tilde{f}, it should happen in V_0.

maybe my english is bad, let me try to draw a picture.

[heady skiff]

thanks, i appreciate it

[unreal stratus]

Ur welcome

[fickle elm]

ahh, maybe my drawing is also terrible.

You have the fixed underlying map f and you have a fix point \tilde{f}(s_i) in V_0.

The way to extend the curve is to use the homeomorphism p restricted to V_0

[heady skiff]

i see... i think that makes more sense

thank you

[heady skiff]

what do they mean by "using the preceding lemma to extend..." do we just use the fact that 0 X I and I x 0 are both homeomorphic to I?

[heady skiff]

<@&286206848099549185>

[heady skiff]

The preimage of $U$ is a collection of $V_\alpha$. Each of them is not connected to another as the fiber of the covering map is discrete.
You fix a point in $V_0$. So when you define the continous map $\tilde{f}$, it should happen in $V_0$.

[gentle osprey]

okeyokay

[heady skiff]

ohhh...

so if we defined it in terms of V_\alpha then it wouldn't be defined for si

cuz they're disjoint

thanks @unreal stratus

[heady skiff]

so just to be clear, $\tilde{f}$ is continuous by the pasting lemma since it agrees on each of the endpoints $s_i$; for example, if we consider the two intervals $[s_{i - 1}, s_i]$ and $[s_i, s_{i + 1}]$ with $f[s_{i - 1}, s_i] \subseteq U$ and $f[s_i, s_{i + 1}] \subseteq U'$ and $\cup V_\alpha = p^{-1}(U)$, $\cup V_{\alpha}' = p^{-1}(U')$, then we would define $\tilde{f}(s) = (p \mid V_0)^{-1}(f(s))$ for $s \in {s_{i - 1}, s_i]$ and $\tilde{f}(s) = (p \mid V_0')^{-1}(f(s))$ for $s \in [s_i, s_{i + 1}]$, but we could have $s_i \in V_0 \cap V_0'$?

[gentle osprey]

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

[heady skiff]

i guess how we would prove that they're in the intersection is my main question

i give up...

<@&286206848099549185>

[muted arrow]

Helpers ping is meant for the help channels, not the advanced channels

[heady skiff]

oh really

shit my bad

[muted arrow]

np!

[gentle osprey]

wizard

[fickle elm]

I don't understand the question. To get a lift $\tilde{f}$, you start with a given point $\tilde{f}(0)=e_0$. You do it for small interval and you successfully define $\tilde{f}(t)$ from 0 to a small positive number s, repeat this process with the starting point $\tilde{f}(s)$, and you ca finish in finite step since the unit interval is compact.

[gentle osprey]

Dong_Valentino

wizard

[median sand]

But if you want these neighbourhoods to be a basis for a topology, then the intersection of 2 of them should contain a 3rd and that's what's not obvious to me. Or is is not required here that these neighbourhoods be a basis?

[tribal palm]

nooo i slept right through an important topology lecture

[gaunt linden]

It won't necessarily be a basis, no. In fact I'm not even sure they will necessarily be neighborhoods of the centers according to the resulting topology. All I'm saying is that the sets you end up with do satisfy the conditions for being a "topology".

[pseudo ocean]

do yall know about Heegard splittings?

there's limited resources about it and the wikipedia shows jackshit

I wanna know how it affects the homologies of a space I apply it on

[hidden crag]

What’s the problem

[languid patrol]

This is very vague

[silver umbra]

is it true that if H^(X; R) =/= H^(Y; R) for some R, then H^(X) =/= H^(Y)?

H^ here is the cohomology ring

[pseudo ocean]

let's say Hk(X) is the k-th homology group of X. how does a heegard splitting on X affect Hk(X)? does it change? does it stay the same?

the space becomes disconnected (I think) so surely H0(X) changes

also where can I find resources on this? precise definitions?

cuz all wikipedia goes is "oh it's a thing you do with 3-spheres"

[dry jolt]

I'm not qute sure I understand your question. A Heegaard splitting is a way of describing a 3-manifold as a union of two handlebodies of genus g glued along their boundary, it isn't a construction of a new space. In particular, if you're given a Heegaard diagram for a 3-manifold X, you can explicitly describe a presentation for pi_1(X), from which you can obtain a description of H_1(X)

[hidden crag]

I don’t think you understood this correctly

[pseudo ocean]

hm I probably did...

[pseudo ocean]

ohhhh

where can I find resources on this?

[dry jolt]

Saveliev's lectures on the topology of 3-manifolds is a pretty good resource for some of this stuff

[last marlin]

Can anyone help? I am reading Hatcher when I encountered this

"RP^n is the quotient space S^n/(v~-v), the sphere with antipodal points identified. This is equivalent to saying that RP^n is the quotient space of a hemisphere D^n with the antipodal points of boundary of D^n identified."

I fail to imagine that for RP^2 and so on. Can anyone help me with this?

[fathom steeple]

Let A,B be subsets of X (X has a topology) such that A=B/~ fo
r an equivalence relation ~. Then are the quotient topology on A and the induced topology on A the same?

[last marlin]

Induced from where? As a subspace of X?

[fathom steeple]

yes

specifically the case im thinking of is induced toology on the unit circle

which can also be thought of as Z/~ where a~b<->a=b+2pi*k

(as a set)

[last marlin]

I assume B=A/~, and ? means I'm not sure, but if this holds, then I've shown that any set open in B in subspace topology is open in B under quotient topology

[heady skiff]

i'm trying to verify that the hypotheses of Lemma 54.1 are verified (in extending $\tilde{F}$ to $0 \times I$ and $I \times 0$). is it because $F(0, t)$ for $t \in [0, 1]$ is a path from $f(0)$ to $f'(0)$ where $f$ and $f'$ are both paths in $B$, where $f(0) = b$? and because $F(t, 0)$ for $t \in [0, 1]$ is a path from $f(0)$ to $f(1)$ in $B$?

[gentle osprey]

okeyokay

[fathom steeple]

This seems to answer it: https://math.stackexchange.com/questions/705840/quotient-topology-vs-subspace-topology

not sure what you did, seems kinda circular?

[last marlin]

I just proved what OP could prove here, that is, the quotient topology is finer than subspace topology.

But yeah, I get it

[muted arrow]

Imagine what

Like you want to visualize RP^2

This isn't generally something that people do

[last marlin]

To some extent, yes

[muted arrow]

You can get intuition for it though

I find describing it as a quotient of R^3-(0,0,0) as more insightful

[last marlin]

I understand that

[muted arrow]

Then the construction from S^n should have the same intuition

S^2 is just a deform retract of R^3-0, then the quotient map is the same

[last marlin]

My main issue is the motivation behind S^n/(v~-v) and D^n/(v~-v in boundary of D^n) being the same

[muted arrow]

Ahh

[last marlin]

I know that S^n=D^n/del(D^n)

del being boundary, ofcourse

[muted arrow]

So yeah the point is that you can just cut S^n in half (obtaining D^n) and identify antipodal points of its boundary

[knotty vine]

Why do you need f and f'? Isn't it enough to just say that F(0,t) is a path?

Have you ever used a proof assistant? I think it would appeal to your sense of formality

[heady skiff]

proof assistant?

[last marlin]

Okay, I see it now

[heady skiff]

yeah but doesn't the hypothesis require that it start at b0

also true if i can't see something i get stuck on it

[last marlin]

Visually, yes, it makes sense (only till R^3 ofcourse), but why can I claim it in general?

[knotty vine]

But F(0,0) does start at b0, no?

[muted arrow]

It's worthwhile constructing a homeo from the S^n construction and the D^n construction

[muted arrow]

In general you can write down the homeo

In arbitrary dimension

[last marlin]

Yes, I should try that

[knotty vine]

A programming language that helps you write formal proofs checked by the computer. Examples are Coq, Lean, Agda, Isabelle, HOLLight etc

[last marlin]

With the visual intuition, I don't think it should be that hard

[heady skiff]

ye i was just tryna visualize it as a homotopy

[heady skiff]

i've heard of them

i might look into it

[knotty vine]

I don't really know what you're asking but it seems right

Guess that doesnt help you tho

[heady skiff]

i'm confused again lmfao

how does F(0, t) = b_0 for all t in [0, 1]?

like okay

what are we even assuming F is a path homotopy between

because say F is a path homotopy between f and f'

then by the definition of a path homotopy we must have F(0, 0) = f(0) and F(0, 1) = f'(0)

no?

[heady skiff]

its aight i moved past that part lol

i'll just read it again at some point

[knotty vine]

How is path homotopy defined?

[heady skiff]

well first of all you need to have two paths in order for a path homotopy to exist

so i'm trying to figure out what the paths here are lol

[knotty vine]

Is there a definition of the word path homotopy in the book?

[heady skiff]

yeah let me screenshot it

one sec

[knotty vine]

So, path homotopies are identified with the F part. The two paths f, f' are extracted from it by setting the right argument to 0, resp. 1

So, really, any F : I x I -> X with F(0,t) = constant and F(1,t) = constant is a path homotopy

What it is a path homotopy between can be extracted from that info

[heady skiff]

so are we just considering the two paths to be constant paths at b0?

[knotty vine]

No, but the two paths must start at b0

They must have the same initial point, as in the definition

[heady skiff]

i'm just taking s = 0 and letting t range in the first line of the requirements of a path homotopy

[knotty vine]

Yes, all those are b0

[heady skiff]

ohh

OHHHHHH

i see now

yea because by definition

both paths have to start at b0

like you said

so f(0) = b0 = f'(0)

[knotty vine]

Imho it's a bit of a weird definition. For some reason a path homotopy isnt just a homotopy between paths. The paths must also have the same endpoints

I think it's more usual to call those "based" path homotopies or something

[heady skiff]

ok wait now i'm a little bit more confused about the points in between 0 and 1 LOL

[heady skiff]

yae it's confusing

because we know that F(0, 0) = f(0) = b0 = f'(0) = F(0, 1) sure

but how do we know that that holds for all intermediate values of t

[ebon galleon]

I think I've seen the term path homotopies before. It's just a special case of relative homotopies

[knotty vine]

So says the definition

Have you made a drawing of these things yet?

[heady skiff]

i've attempted

[knotty vine]

Does it look like the cross section of an american football?

[heady skiff]

uh no lmfao

[knotty vine]

Draw two points x0 and x1 in your favorite simply connected space (mine is a piece of paper). Now draw two different curved paths from one to the other (keep it simple). This could look like an american football or a "Vesica piscis". Now lets draw the paths F(1/4, t), F(1/2, t), F(3/4, t)

[heady skiff]

i'll try that out on my second reading

thanks

why exactly does this hold? I wrote p^{-1}(f) = tildef and used the fact that f(1) = g(1) but p is not necessarily invertible

wiat yea

cuz i know we have to have p o ftilda(1) = f(1)

similarly p o gtilda(1) = g(1)

but hwo does that imply that gtilda(1) = ftilda(1)

cuz p doesn't have to be injective

[knotty vine]

Use phi(f) = phi(g)

Write out the definition of phi

[haughty yew]

I'm stuck on part c, and haven't been able to come up with a counterexample so far, but haven't been able to prove it either

I can do it if they don't have to be closed though

[muted arrow]

Try X = {a,b,c} with the discrete topology

[haughty yew]

A U B is only connected if one of them is empty

In this space

[muted arrow]

or if A U B = X

[haughty yew]

X is disconnected

[muted arrow]

Wait yeah you're right

I suppose we aren't happy with empty set?

Okay I've proven it

There's not a good way to give a hint without telling you the answer.

But I'll try! Hint 1: ||Suppose one of A or B is disconnected. Write it as union of disjoint sets. Use the fact that everything is closed to prove that one of the intersections with B must be empty||

Ping me if that isn't enough I can't really come up with a hint 2 that doesnt complete the problem

@haughty yew (didn't see your bio)

[haughty yew]

Thanks!

I'll try it out

[tame magnet]

Hey! There was a math major entering another Engineering server I am in questioning about basic hollow/beam sections (things they never heard of before) for a purpose. I suddenly asked them to name every axis of a particular matlab graphical output. They answered with "What's an axis?" and for a moment we thought they were just joking or probably very bad at math. Then they surprisingly gave responses such these:

So. For this I want to know what's an axis for topology? and how an axis doesn't exist in it. Thank you :))

[pallid delta]

I'm no topologist (I've had 4 weeks of topo classes), but axes make the most sense when you consider a canonical basis of a finite dimensional vector space. Though it can be extended to an infinite dimensional vector space with an infinite basis, saying each span(b_i) is an axis.
In a topological space though, you'd struggle to make it always exist meaningfully. You don't have a metric, you don't have much structure at all really.
Consider X = {1, 2, 3}. With one topology, it has 2 open sets, with another, it has 8 open sets, you'll hardly make axes out of that, unless they're the singletons, but you still don't have the "span" part.

Imo, axes fail to make sense if you're not in a vector space. It's related to the idea that each axis is 1D and the vector space is the direct sum of all these axes. This is pure linalg and I don't believe you can translate that meaningfully at all.

Now ofc if a real topologist comes in and says actually they can have meaning, then they might. But I don't think so

[tame magnet]

I see it makes much more sense in LineAlg. Great to see the drop! But I'd still wait if a "topologist" as you said would mention any hidden relation of this. Over that I really appreciate your answer it's very helpful and simple to grasp :)) Thank you.

[pallid delta]

can't make it hard to grasp when you yourself know nothing complicated

[tame magnet]

Yes! I have the bare minimum or less than that to build any complicated understanding of Topology. But this question pushes me more into it :)) I found some applications of it in Fluids so.. it might be very helpful too later

But since we rely a bit on LinAlg on some of stuff that are computationally related to Fluids I don't know the limits yet

[pallid delta]

wait, it's the math major who didn't know what an axis was ?

and you're the engineer ?

[pallid delta]

cause he managed to give the orthogonal complement of an axis instead

[tame magnet]

I hope what I wrote was clear! The answers are coming from the math major. I am the person in the Engineering server :)) I cut some answers they probably had more to say?

[pallid delta]

he just failed linalg, that's all

[tame magnet]

What they said

[pallid delta]

that I agree with

[tame magnet]

@pallid delta Does this mean that a metric space doesn't have an axis significance?

Oh wait..

I mean "Metric Space" in general without vector space being under it

[pallid delta]

a circle is a metric space

a disk too

the big cities that are connected by railroads are a metric space

find axes in that one I dare you

in general, a weighted graph is a metric space

[tame magnet]

I gotcha. I need to have a good read on the different spaces

Thanks a bunch!

[pallid delta]

A subset of a metric space is a metric space

[tame magnet]

I see

[solar juniper]

any idea for the <== direction

[queen prism]

#discrete-math or #combinatorial-structures idk

[prime sigil]

Hello how can I increase my knowledge in topology in normed I am new in this field

[haughty yew]

In normed what

[prime sigil]

Sorry topology in normed space

[pallid delta]

It's quite common to start with this, to apply it to analysis, rather than start in general topologie and see the specifics of the metric case afterwards

[unreal stratus]

Bad that when I see something circled in like this i immediately assume it's one of those dumb memes with everything circled etc

[knotty vine]

Did you see the ghost in the picture?

[bronze wind]

Hello ! I am trying to undersand what is the pushout of f:Z->Z (x|->2x) and g:Z->{0} (in the category of groups). I know that it will be the coproduct of Z and {0} (namely the free product generated by Z and{0}) modding by the normal closure of {f(x)0 where x in Z} in the free product . But how does it really looks like ?

[ebon galleon]

If it helps, {0} is the initial and terminal object in Grp. Hence it acts as a unit for both product and coproduct, so Z coprod {0} ~= Z

So now consider what the normal closure of that is.

(hint: it's an abelian group)

[unreal stratus]

Tbf this would normally go in #groups-rings-fields or smth but fair

[ebon galleon]

True, but it's also probably an application of SVK

[unreal stratus]

Ye but still lol

[bronze wind]

Yes sorry it's indeed because this exercice is from a course in algebraic topology, I will be more careful next time !

[bronze wind]

I am sorry, of which set am I supposed to take the normal closure ?

Oooh I think I am starting to get it, I have to show that every "formal" sequence in the pushout which contains a 0 is basically 0 ? So then it will be obvious that Z coproduct {0} ~= Z

[prime sigil]

Alright but I need recommendations please !

[knotty vine]

Which channel is for functional analysis? Don't tell me it's this channel

[ebon galleon]

#advanced-analysis

[knotty vine]

I propose you ask in #advanced-analysis !

[ebon galleon]

You have the set {2x | x in Z} in your coproduct Z. Take the normal closure of this.
For the second comment, in the free product you can get rid of any identity in a string to reduce it [i.e., (g_1, h_2, e) and (g_1, h_2) represent the same element of the free product]. So in G coprod {0}, you can reduce your strings to just strings in G, modulo the multiplication within G What you end up getting is isomorphic to G.

[bronze wind]

The normal closure of this set is itself in the coproduct since as you said it's an abelian subgroup. If you mod by this group in the coproduct (which is normal obviously) then you get something isomorphic to Z/2Z,but I am supposed to get Z 😦 I think I am tired lol

[ebon galleon]

Uhh well I'm pretty sure this pushout should be Z/2. So if you got this pushout through Van Kampen, it might be worth it to check that you set up the pushout correct if it was supposed to be Z

[bronze wind]

No it's just I thought that you previously said that it was Z so I believed that I was wrong. Thank you for the help !!

[ebon galleon]

Ah sorry, that was just for the coproduct, not the pushout. So if that's correct now then

[chrome ridge]

Just for reference; what are the homology groups of U(2)?

[wind moat]

Z when k is 0,1,3 or 4 and zero otherwise.

[heady skiff]

so i'm trying to rigorously verify why W is evenly covered by p. I understand that Ryx mentioned we're supposed to look at the restriction of p to W_\alpha (where the W_\alpha form a disjoint union of the preimage of p^{-1}(W)) (which is contained in each V\alpha) which should be a homeomorphism but i'm not entirely sure how surjectivity follows instantly. i know that injectivity is inherited from p, but i'm unsure how to prove surjectivity. similarly the inverse being continuous follows from p restricted to V_\alpha

[knotty vine]

$p|{W\alpha}$ can be seen either (1) as a map $W_\alpha \to B$, or (2) as $W_\alpha \to U$, or (3) as $W_\alpha \to W$. We need to show that (3) is a homeomorphism.

We know from the even covering of $U$ that (2) is injective. We also know that the codomain of (3), $W$, is precisely the image of $p|{W\alpha}$ (by definition of $W_\alpha$). Any injective function whose codomain is its image is a bijection.

There are much more general/easy ways to say this, but this is at least rigorous.

[gentle osprey]

Semer

[heady skiff]

thanks, i think that makes sense

here are we assuming that F and F tilda are homotopies?

[ebon galleon]

Well a homotopy is just any map with X x I as its domain for some space X. Here, the extra assumption is that F is a path homotopy, i.e. it's constant along endpoints

[heady skiff]

ah right

thanks!

what would be the induced automorphism here? would it be $[f] \mapsto [\overline{\overline{\sigma}* \gamma}] * [f] * [\overline{\sigma} * \gamma]$?

[gentle osprey]

okeyokay

[heady skiff]

nvm

i'm guessing it's left multiplication

[ebon galleon]

That's correct (or maybe the other way around with the inverses, idk)

inner automorphism = conjugation

[heady skiff]

oh huh okay cool

oh wait yea the inverse would be on the right

analagous to $x \mapsto gxg^{-1}$ i'm guessing

[gentle osprey]

okeyokay

[heady skiff]

well i guess it doesn't matter lmao

well

also it it true that $\overline{\alpha *\beta} = \overline{\beta} * \overline{\alpha}$?

[gentle osprey]

okeyokay

[ebon galleon]

Yeah one of those, not sure what the autho's convention is for which side the inverse is on. Doesn't really matter though, since you can replace g with g^{-1}

[ebon galleon]

And yes. It's a group.

[heady skiff]

Yea but here it wouldn’t necessarily apply to sigma and gamma right since they’re not elements of the group

[ebon galleon]

Oh uhhh well that still applies since that forms a groupoid (category with only isomorphisms; here the objects are points of X and morphisms x --> y are homotopy classes of paths from x to y)

And that formula holds for inverses in any category

[heady skiff]

fancy stuff

okay cool because checking that was gonna be a pain in the ass

ok wait conjugation by inverse on the right wouldn't even be defined

cuz say we define $f \mapsto (\sigma^{-1} * \gamma) * f * (\sigma^{-1} * \gamma)^{-1}$

[gentle osprey]

okeyokay

[heady skiff]

then we get $\sigma^{-1} * \gamma * \bigl((f * \gamma^{-1}) * \sigma^{-1}\bigl)$

[gentle osprey]

okeyokay

[heady skiff]

but then $(f * \gamma^{-1})$ is a path from q to p

[gentle osprey]

okeyokay

[heady skiff]

and sigma inverse is a path from q to p

so their product doesn't make sense

similarly for left multiplication of the inverse

nvm

ignore all that

forgot to take the inverse of sigma inverse LOL

[obtuse meteor]

This is overkill. Those maps are equal on the nose as loops

[ebon galleon]

Well I didn't read carefully so it wasn't a group anyways

But my point was just "yeah that's how inverses pretty much always behave". You can also see it directly by just writing out the formulas though, as you note.

[obtuse meteor]

Alpha and alpha bar aren’t inverses !

Not as straight up paths

[ebon galleon]

And 1 and -1 aren't inverses as straight up objects, but rather wrt to addition. The question had an explicit reference to the operation of concatenating the paths, so I don't see the comparison/problem.

I suppose maybe the point is that they could be between different points as in the problem, which is fair.

[obtuse meteor]

1 + (-1) = 0 as objects not as a homotopy.

But bar(alpha * beta) = bar(beta) * bar(alpha)

My point is this is not a thing abt equivalence classes up to homotopy

And I think it’s important to distinguish when you have an equality and when you only have a homotopy

(E.g. what if these are geodesics. You don’t want to argue this as inverses up to htpy anymore bc you might care you are 1 geodesic then another)

[ebon galleon]

Okay, that's a fair point. It is a stronger statement than what I suggested. Thank you for clarifying

[urban zinc]

For alpha ≥ 1, every C^alpha structure contains a C^beta structure for each beta > alpha

Anyone know if there's a simple proof of this?

Spivak mentions it but doesn't prove it

[gritty widget]

hirsch's differential topology book, theorem 2.9

this book's got all of the stuff on smooth structures and function/mapping space topologies

it's basically the place to go for it

a lot of books avoid going into detail. hirsch IS the details

[urban zinc]

thank you!

[winter falcon]

Im just starting to learn point-free topology and currently trying to di some exercises on frame homomophisms.
I have no idea on how to approach the second question? Any hints ir sources would be appreciated

[marsh fulcrum]

Perhaps what I asked in #help-7｜zen1thxyz might also belong here? (It's about fractals)

[hidden crag]

No

[marsh fulcrum]

I was told fractals belong to topology

[hidden crag]

The question has nothing to do with topology

[marsh fulcrum]

Into what area would you categorize fractals then?

[hidden crag]

Never concerned myself with them, combinatorics perhaps

[knotty vine]

Maybe dynamical systems?

[marsh fulcrum]

Oh, alright.
Hm, it seems Wikipedia always gives topological interpretations of fractals, so it's possible that they belong to Topology, but with their topological interpretation, and the "usual" fractal stuff, just looking at their dimension, ... is something else

[knotty vine]

Topology is everywhere so that's not too surprising

[ebon galleon]

Just note fact 1 to get that it preserves all joins, no?

Maybe the first point helps with some boundary cases

But uhh to prove fact 1, it's a little silly but I think you can just do this: ||directed means that every pair of elements in S have an upper bound in S.... So just attach the join of S to the set to make T. Should trivially be directed with the same join||

[winter falcon]

I was more interested in proposition 3. How to prove that the right adjoint is expressed as a join in that manner??

[ebon galleon]

Well i think this might be one of the easier conditions to check for this one, using that h preserves joins. (and this itself is not terrible to prove from the definition of adjoint)
Usually, this is expressed as the join over {x | f(x) <= y}, not =, but I guess surjectivity makes it the same in this case.

[winter falcon]

Sure thanks let me write down the proofs, do you have any texts/references on pointfree topology besides Picardo and Pultr ?

[ebon galleon]

I like Picado's book generally from what I've read. The other main reference I've seen is Johnstone's "Stone Spaces" (which really is largely about locales, despite the name), but that's a bit more terse.

Joyal has some notes as well, "An Extension of the Galois Theory of Grothendieck", which is nice as a reference if you need a specific result. But this one is even more terse (and takes a quite high level approach, so it's less accessible)

[clever sable]

Let $(X,\mathcal{T})$ be a topological space and let $(Y,d_Y)$ be a metric space. Let $f_n:(X,\mathcal{T})\to(Y,\mathcal{T}_{d_Y})$ be a sequence of continuous functions. Let $x_n$ be a sequence of points of $X$ converging to $x$. Show that if the sequence $(f_n)$ converges uniformly to $f$, then $(f_n(x_n))$ converges to $f(x)$.

[gentle osprey]

jsidind810

[clever sable]

Does it suffice to show that $f$ is continuous?

[gentle osprey]

jsidind810

[clever sable]

Since this thm in munkres:

Showing $f$ is continuous would mean showing that given any $x\in X$ and $\varepsilon>0$, there is a $\delta_\varepsilon >0$ such that $d_X(x,y)\Rightarrow d_Y(f(x),f(y))<\varepsilon$. But what is the metric of $(X,\mathcal{T})$?

[gentle osprey]

jsidind810

[clever sable]

Also since $(f_n)\to f$ uniformly we know that $d_Y(f_n(x),f(x))<\varepsilon$ for any $\varepsilon>0$, for any $x\in X$, and if $n$ is greater than some $N_\varepsilon\in\mathbb{Z}$.

[gentle osprey]

jsidind810

[clever sable]

Help pls !!

[unreal stratus]

There needn't be one

[clever sable]

thats why im confused

what am i supposed to do instead?

[unreal stratus]

Do you know what it means for a function between topological spaces to be continuous?

[clever sable]

i cant find the definition

[clever sable]

?

The question states $f_n$ is continuous from $(X,\mathcal{T})\to(Y,\mathcal{T}_{d_Y})$

[unreal stratus]

Well that's doesn't work as X needn't have a metric in general

[gentle osprey]

jsidind810

[clever sable]

but what the

yeah

theyre implying some metric

[unreal stratus]

No they aren't

[clever sable]

damn

then what

[unreal stratus]

A function $f: X \to Y$ between topological spaces is called continuous if for all open $V \subseteq Y$, $f^{-1}(V) \subseteq X$ is open

[gentle osprey]

potato

[unreal stratus]

I suggest you go back and review these notions since they must've been covered before this

[clever sable]

oh yeah

ik that but

ye

idk what to do next

[queen prism]

you don't need a metric to appropriately define continuous functions

all you need is some conception of open sets
metric spaces can give you this, and it's nice, but you don't need them

[clever sable]

well idk how to proceed on this question man

can u help

Please

someone

anyone

[tidal lynx]

Here's a hint:

to show that the sequence f_n(x_n) converges to f(x), you need to be able make the LHS above arbitrarily small for sufficiently large n

now the inequality above, which is an application of the triangle inequality, bounds the LHS, so if we can just make the RHS arbitrarily small...

[clever sable]

can u give another hint?

[tidal lynx]

Bound the first term on the RHS using the uniform convergence of the f_n, and bound the second term using the continuity of f and this fact

[clever sable]

does |f_n(x_n) - f(x)| < epsilon imply uniform convergence?

[clever sable]

.

is taht in munkres

[tidal lynx]

In your question, the functions you are given that f_n uniformly converges to f. I don't know why you are trying to reprove that.

[tidal lynx]

Yes, it is the one you posted a screenshot of earlier

[somber phoenix]

https://puu.sh/JTjef/c7a39b5bc5.png

doesn't the start of this proof take both sides of the iff statement, to then prove the right side of the statement that it's already assumed to be true?

[tidal lynx]

It just says Y is compact and takes an arbitrary cover of Y in X

And then it extracts a finite subcover, proving the forward direction

[somber phoenix]

but by definition, Y being compact means that the cover comes from Y.

https://puu.sh/JTjeS/5687ac9c57.png

[tidal lynx]

Yeah, that is correct

But in the forward direction of the lemma, you are trying to prove that covers in X also have finite subcovers

[somber phoenix]

so not sets in X, and we'd have to match these sets with ones that could've been made by interesections of something in X

[tidal lynx]

Unpacking definitions, the forward direction of the lemma states:
Suppose every open cover of Y in Y has a finite subcover. Then every open cover of Y in X also has a finite subcover.

[somber phoenix]

if we use the topology of X, then this should be easy. Since each finite open cover in Y, will have been created by some union of Basis elements in X, in intersection with Y. And then you'd pick the corresponding unions of those basis elements, without the intersection with Y, to find this finite cover in X

but I don't know if that's allowed

[tidal lynx]

I don't think I understand your explanation, but do note that you aren't given any bases in the hypothesis of the lemma

[somber phoenix]

well, since Y is a subspace of X, then all open sets in Y are a consequence of topology of X (not basis elements I guess). Then each element in the finite open subcollection would be constructed by some subcollection of T elements in the topology of X, and then each of those union intersection with Y, would correspond to one of the elements in the finite open succover of Y. And of course, each element in the open cover has to be constructed by some union of elements in the topology.

ah, X doesn't have a topology? but then how would Y be a subspace?

[clever sable]

is this showing continuity?

which then implies UC?

[somber phoenix]

I mean, I guess the book is kinda right. just taking an arbitrary covering of sets in X... but I feel like it's cheating? Because from knowing Y is compact, we can't use the elements in the backward direction sorta. I don't know, not happy with the proof, but mmm....

[tidal lynx]

Sorry but I can't really follow this; to prove the forward direction you need a way of extracting a subcover from an arbitrary open cover of Y in X. So I recommend you begin your proof with "Let A be an open cover of Y in X, then..."

[tidal lynx]

It seems you ignored this message @clever sable

[somber phoenix]

But imo you can't start like that, because A is an open cover of sets in Y, not in X. Because of it was sets in X, it would not be a compact space, since the definition requires the open sets to come strictly from the same space (Y in this case).

[clever sable]

So |f_n(x_n) - f(x)|< epsilon is just the definition of convergence to f(x)

[knotty vine]

You must start like that

Then the proof fixes it so that we find opens in Y

by taking the intersection with Y

[tidal lynx]

When we say "A is an open cover of Y in X", we aren't using the compactness of Y.

It's just a step in the proof of the forward direction of the lemma

[somber phoenix]

oh, I see, I jumped the gun

[tidal lynx]

That doesn't seem like a definition to me

[clever sable]

are you using that definition

[tidal lynx]

There is no "definition of convergence to f(x)". You have to specify a sequence.

[clever sable]

Are you using Thm 21.3

[tidal lynx]

In the definition? No

[clever sable]

in |f_n(x_n) - f(x)| < epsilon

being sufficient to prove the question

that is just a definition of f_n(x_n) -> f(x), correct?

[tidal lynx]

Ok

Can you write out what it would mean for f_n(x_n) to converge to f(x)?

[clever sable]

$n\geq N_\varepsilon\Rightarrow |f_n(x_n)-f(x)|<\varepsilon$

[gentle osprey]

jsidind810

[tidal lynx]

That's just an inequality

[clever sable]

im saying if that is true then we have convergence

[tidal lynx]

(I'm being intentionally pedantic, I want you to write everything out clearly. It will be easier to find the proof then."

[tidal lynx]

you haven't defined what N_eps is

I want you to write out the "for all eps..."

[clever sable]

No

[tidal lynx]

Ok

[clever sable]

for all epsilon >0 Ne is an integer

[tidal lynx]

Right, for all eps > 0, N_eps is an integer so that this inequality holds.

[clever sable]

yes

[tidal lynx]

So then for f_n(x_n) to converge to f(x), for every eps > 0 we would need to find an N so that
n > N implies that |f_n(x_n) - f(x)| < eps

Now since f is continuous and x_n converges to x, we have by Theorem 21.3 that f(x_n) converges to f(x), meaning that for every eps > 0, there is an N so that
n > N implies that |f(x_n) - f(x)| < eps

Moreover, since the f_n converge uniformly to f, we know that for every eps > 0, there is an N so that
n > N implies that |f_n(x) - f(x)| < eps for all x.

Does this make sense @clever sable ? I am unpacking the definitions of convergence and uniform convergence.

[clever sable]

ye

oh

so

$|f_n(x_n)-f(x)|=|f_n(x_n)-f(x_n)+f(x_n)-f(x)|$

[gentle osprey]

jsidind810

[clever sable]

$\leq |f_n(x_n)-f(x_n)|+|f(x_n)-f(x)|$

[gentle osprey]

jsidind810

[clever sable]

$\leq \varepsilon/2+\varepsilon/2$?

[gentle osprey]

jsidind810

[tidal lynx]

Yeah.

That's it

Hope you understand why I was being anal about the definitions lol. As you can see, after writing everything out explicitly and one application of the triangle inequality, everything falls into place.

[clever sable]

how do we know $|f_n(x_n)-f(x_n)|<e/2$

[gentle osprey]

jsidind810

[tidal lynx]

Oh

You kinda jumped the gun

So now we are actually going to prove that f_n(x_n) converges to f(x). Start with fixing an arbitrary eps > 0.

then there is a N_1 so that
n > N_1 implies that |f_n(x) - f(x)| < eps/2 for all x

and there is also an N_2 so that
n > N_2 implies that |f(x_n) - f(x)| < eps/2

Now take N = max(N_1, N_2)

Then
n > N implies that |f_n(x_n) - f(x)| < |f_n(x_n) - f(x_n)| + |f(x_n) - f(x)| < eps/2 + eps/2 = eps,
by the above two inequalities

[clever sable]

uh

but

why do those use f(x_n)

but

the above ones use f(x)

lol

yeah thats simple

[tidal lynx]

You can also see why we need the f_n to uniformly converge to f for this proof to work?

[clever sable]

mhm

[tidal lynx]

Uniform convergence was what allowed us to produce this eps/delta statement to work for all x

hence in the proof we could choose x = x_n

[clever sable]

mhm

Now how about this simpler one

Let $d$ be a metric on a set $X$. Show that the collection $$\mathcal{B}_d={B_d(\mathbf{x},\varepsilon):\mathbf{x}\in X\text{ and }\varepsilon>0}$$ of balls forms a basis for a topology. In other words, check that $\mathcal{B}_d$ satisfies the two conditions of being a basis.

[gentle osprey]

jsidind810

[clever sable]

For each $\mathbf{x}\in X$, is there at least one ball $B=B_d(\mathbf{x},\varepsilon)$ in the basis with $\mathbf{x}\in B$? Yes -- for any $\mathbf{x}\in X$ and any $\varepsilon>0$, the ball $B_d(\mathbf{x},\varepsilon)\ni\mathbf{x}$ is in the basis.

[gentle osprey]

jsidind810

[clever sable]

of course we have the first condition? or do i need to say more

[tidal lynx]

Yeah that works

Every point is the center of some ball, and is of course contained in it as well

[clever sable]

For the second condition

it seems i need to construct an epsilon for D?

[tidal lynx]

Yea, you need a ball centered at x that fits inside the intersection

[clever sable]

so given epB and epC

how would i do that

[tidal lynx]

Guess

[clever sable]

dude i have no idea even though its so simple

lol

i have no idea why

[tidal lynx]

Surely you can give a guess

[clever sable]

min epB with something idk

sorry i mean

min(epB,epC)

idk

[tidal lynx]

Let b = d(x_B, x) and c = d(x_C, x)

try to work with these

[clever sable]

dude idk

[tidal lynx]

epB - b

epC - c

[clever sable]

that would be it

im an idiot again

[tidal lynx]

I mean that's not fully it

[clever sable]

min of them

right?

[tidal lynx]

yeah

[clever sable]

but then dont i need to further show that eD<max(eB,eC)

[clever sable]

?

[tidal lynx]

Let eps = min(epB - b, epC - c). Then for any y in B_d(x, eps), we have that
d(x_B, y) <= d(x_B, x) + d(x, y) < d(x_B, x) + (epB - b) = epB

[clever sable]

and B was taken wlog?

[tidal lynx]

Yeah, I'm using your diagram

[clever sable]

o

diagram is not exact

[tidal lynx]

x_B and x_C are arbitrary points in X, eps_B and eps_C are arbitrary positive reals, and x is an arbitrary point in the intersection of B_d(x_B, eps_B) and B_d(x_C, eps_C)

At least, that's how I interpreted your diagram

[clever sable]

then why did you take B

[clever sable]

here

you did that wlog?

you could have taken C?

[tidal lynx]

oh

yeah the same argument works for C

[clever sable]

are you sure

[tidal lynx]

like if you change every occurrence of b with c, and every occurrence of epB with epC, it's a true statement

And of course the same eps works

[clever sable]

and thats necessary to remark, no?

[tidal lynx]

Yeah

[clever sable]

i have one more hard question for you

[tidal lynx]

Ask away

[clever sable]

If $X,Y$ are metric spaces and $f:X\to Y$ has the property that $d_Y(f(x_1),f(x_2))=d_X(x_1,x_2)$, show $f$ is an imbedding

[gentle osprey]

jsidind810

[clever sable]

where $X,Y$ have the topology induced by the respective metrics dX dY

[gentle osprey]

jsidind810

[clever sable]

and x1,x2 in X

Ive shown f is injective

[clever sable]

.

anyoen else?

[shadow charm]

Note how the inverse of f (from its image) also preserves distances

Which makes it clear that it is also continuous

[clever sable]

ok

what about hte rest of the question

[clever sable]

.

[fathom steeple]

Why is the fundamental group of R^2 not trivial? Why can't every curve be homotopic to constant.

[knotty vine]

Who told you that it wasnt?

[tidal cedar]

the fundamental group is trivial

[clever sable]

Can anyone help with my question

[shadow charm]

What rest

[clever sable]

this

[shadow charm]

You already showed injectivity

[clever sable]

and continuity

[shadow charm]

And I completed by telling you how it’s a homeomorphism on its image

So there’s nothing left

[clever sable]

what

how can you just conclude that

is that a theorem

[queen prism]

what's an embedding

[shadow charm]

.

So inverse is continuous, ie the corestriction of f is a homeomorphism

[fathom steeple]

I see.. I saw online that the fundamental group or R^2-{0,0} is Z so thought the same thing for this

I guess I can see the distiniction here

[tidal cedar]

R^2 - {0,0} is not trivial because of this obstruction

In general removing a point is going to change homology in most spaces

[pulsar lagoon]

is the set of all convergent sequences open in Romega with the uniform topology?

[half vortex]

How do i know what basis to write for $d_{\infty}$ where $(X_1, d_1)$ and $(X_2, d2)$ two metric spaces endowed with metric topology. The function $d{\infty} \colon (X_1, X_2)^2 \to \mathbb{R}$ via $d((x_1, x_2),(y_1, y2))=max_{i=1,2}d_i(x_i,y_i)$

the main questions asks "Show that the metric and product topologies on $X_1 \times X_2$ coincide."

like i'm struggling to notate $\beta_{d_{\infty}}$

[gentle osprey]

張嘉棋

[limber wren]

I think no, if you have a convergent sequence x(n) in R omega, the sequence x_r(n) = x(n) + r (-1)^n is within r > 0 of x(n), but doesn't converge (eventually fluctuates by r(-1)^n around the limit of x(n)). So you can't have any r-neighborhood around x(n) containing only convergent sequences

In fact it seems very non-open, like the set of non-convergent sequences is dense

[pulsar lagoon]

Oh I see I see

[normal umbra]

if i have two exact sequences $A_i$ and $B_i$, how do I proof that if $A_{n} \cong B_{n} \implies A_{n+1}/A_{n} \cong B_{n+1}/B_{n}$?

[gentle osprey]

habuki (dms open)

[somber phoenix]

https://puu.sh/JTkHJ/d18f774122.png

this feels like something like non convergent sequences, where a subsequence converges.

for every B we can find a U that does not contain B, but for every U, we can find a B that fits inside it. Which one came first?

so I guess U = B, but U should be able to have ratius just slightly smaller than B, or slightly larger

[unreal stratus]

well every metric space is first countable which seems relevant to this

[somber phoenix]

https://puu.sh/JTlTB/514df64660.png

does induce mean "the same"?

[muted arrow]

Sort of. Given a metric d on a set X, you obtain a topology called the induced topology. Given a topological space (X, tau), we call it metrizable if there exists a metric d such that the induced metric (X,d) is equivalent to the to topology (X, tau)

The induced topology from a metric has basis given by open balls B(x,r)

[swift fjord]

Maybe equivalent isn't the best word here. The topologies should be equal, which is probably what you meant

[merry geode]

Was trivial topology discrete topology or indiscrete topology?

[opaque cloud]

isnt trivial topology just {\phi, X}? and discrete topology is the powerset?

[tribal palm]

why assume fancy F contains X - A if we’re just going to discard X - A straight away afterwards?

in the proof of the theorem here

[pallid delta]

It's to say that no matter whether X-A is in the cover, the cover of A won't contain it

[merry geode]

Ahh

[pallid delta]

\phi is indeed disgusting

[merry geode]

Huh wdym

[pallid delta]

nobody should write the emptyset that way

[merry geode]

How would you write it

[pallid delta]

empty

[merry geode]

If not for $\phi$

[gentle osprey]

Absta

[pallid delta]

$\emptyset$

[gentle osprey]

Bezier

[merry geode]

Oh right

How could I forget this

[opaque scroll]

The symbol for the empty set is actually derived from the Danish-Norwegian letter Ø, which is completely unrelated to the Greek letter phi. The more you know 🌈

[merry geode]

Ohh interesting

Time to use $\varphi$ instead

[gentle osprey]

Absta

[merry geode]

To avoid confusion

[tribal palm]

sze-tsen hu uses an empty box symbol for the empty set, and i love it, ngl

[hidden crag]

[pallid delta]

proofs finishing 3 times per proof be like

[merry geode]

Honestly {} would be more canonical.

[pallid delta]

yes
Python has sets, denoted as {1, 2, 3}
Hence {} denotes the empty dictionary

[hidden crag]

I don’t see anything wrong with the standard way

[tribal palm]

rebel

[unreal stratus]

Lol

In topology the box often means some sort of cube

I find it cute when people do that

[ebon galleon]

I hate this

[tribal palm]

god

proving urysohn’s lemma and metrization theorem in one lecture is a bit too much

oh p is contained in Bi is contained in V is contained in Cl V is contained in Bj is contained in U is contained in X you say? yes i’m totally following

at least it was kind of the lecturer to draw this enlightening picture

[civic verge]

wow name of book

[tribal palm]

lecture notes

https://www.uio.no/studier/emner/matnat/math/MAT4500/h23/topology.pdf

[fathom steeple]

Intuitively why is S1 X S1 the torus?

I know that S1 is the same as R/~ where a~b iff a=b+2pi*k and is thus [0,2pi)

[tribal palm]

S1 is the circle, so S1 x S1 is a circle at every point in a circle

[trail charm]

start from a circle, then draw a circle perpendicular to that circle

[heady skiff]

i'm confused as to what midstrong wants me to show. does he want me to find a path from 1 to f(1) or something?

[fickle elm]

If you already know the fundamental group of S^1 is integer group, then f_* is a group homomorphism from Z to Z. You need to describe this group homomorphism.

Is it only the identity? or it may sends 1 to n

Well, may have spoiled the fun

[heady skiff]

i don't understand what they mean by "describe"

moreover how is this a path

$f(e^{i \cdot 0}) = f(1)$

[gentle osprey]

okeyokay

[fickle elm]

Point out the group homomorphism

[heady skiff]

doesn't a path by definition have domain [0, 1]

wait

well

i guess f(e^i2pi) = f(0) lol

what's the purpose of having f(e^i\theta) in the input

[fickle elm]

recall what 1 in Z=\pi_1(S^1, 1) represents and imagine how it changes by f_*

[heady skiff]

so i think this is just going to be the path that winds around the circle in a clockwise direction

wait nvm

so f(0) = (-1, 0)

on the unit circle

right?

[fickle elm]

1 represents the path that winds around the circle "once" in a clockwise direction base at 1.

Now imagine what the path looks like after you apply the map f

where it starts, how fast does it go and where it ends

[heady skiff]

i see

yeah i'm a bit confused because when they talk about the induced homomorphism $f_*$

[gentle osprey]

okeyokay

[heady skiff]

they mean the homomorphism given by $[g] \mapsto [f^{-1}] * [g] * [f]$ right?

[gentle osprey]

okeyokay

[fickle elm]

No. Suppose $\alpha: [0,1]\rightarrow S^1$ is a path on $S^1$, $f_*(\alpha)=f\circ \alpha$.

[gentle osprey]

Dong_Valentino

[fickle elm]

wait a min. I guess you may confuse the time interval [0,1] and the point on S^1 as complex number exp(i theta).

The 1 in your exercise refers to (1,0) in the coordinate plane, or as a complex number 1.

so $f_$ sends a path $\alpha$ based at (1,0) to a path $f_(\alpha)$ based at $f(\alpha(0))=f(\alpha(1))=f((1,0))$.

[gentle osprey]

Dong_Valentino

[heady skiff]

ah i think i see now

thanks

[heady skiff]

so it's not this map then?

i'm so confused

what's the purpose of e^i\theta inside of f

is that supposed to be the map alpha?

then why do they not denote f as f*?

like

how would you even compose $f(e^{i\theta}) \circ \alpha$ lmao

[gentle osprey]

okeyokay

[fickle elm]

I guess you are confused.

[heady skiff]

do you think you could do the first example for me?

i have no clue where to start

or at least part way

[fickle elm]

The theorem in your picture is about change of base point for a connected space.

Though the notation is similar, gamma_* and f_* are completely different things

[gritty widget]

you can backslash your asterisks to stop them from italicizing text

[heady skiff]

oh so it'd be this one then

[fickle elm]

here it should be $f(\alpha(e^{i\theta}))$

[gentle osprey]

Dong_Valentino

[heady skiff]

ohh ok

god i fucking hate this book

ok thanks

[fickle elm]

BTW what book is this?

[heady skiff]

armstrong topology

it's horrible

[heady skiff]

this one is munkres

so much better

[fickle elm]

Thanks for the tip.

[gritty widget]

same goes for underscores and other formatting stuff

\||v\||
lets you type out norm brackets without getting spoilers, for example

[fickle elm]

||v||

Oh, thank you for that.

[heady skiff]

oh so here does exp(i\theta) take the place of [0, 1]

[fickle elm]

exp(i\theta) is the circle, f is defined on the circle.

The image of the path \alpha is on the circle.

[heady skiff]

yeah i really don't get this. i'll probably skip it. for example $f(\alpha(e^{i \cdot 0})) = f(\alpha(1))$, and how am i supposed to know what that maps to

[gentle osprey]

okeyokay

[heady skiff]

yes

ok i'm just going to guess that this takes a loop and flips it across the x-axis or some shit

i have no idea how to rigorously show this but then again armstrong doesn't really rigorously show anything so it's probably what he expects

thanks for the help

[fickle elm]

You purpose is to determine the homomorphism between fundamental groups. Namely from Z to Z. Recall that a group homomorphism from Z to Z as additive group is determined by where you send the element 1 in Z.

The rest is just by using the law of group homomorphism.

So what you need to do is find out what kind of path the element 1 represents in the fundamental group Z, then look at how this path changes after applying f_*, what kind of element it is in the target fundamental group.

[heady skiff]

i was able to figure it out

thanks

why couldn't we say by the uniqueness of path liftings, $\tilde{F}(s, 1) = \tilde{f}(s)$?

[gentle osprey]

okeyokay

[heady skiff]

because they both start at e_0

oh

F(s, 0) = f(s)

what's the intuition behind the lifting correspondence?

[gritty widget]

https://tenor.com/view/hang-in-there-arm-day-cat-working-out-lifts-gif-14487393

[heady skiff]

cat?

category theory?

[queen prism]

cat is lifting

[heady skiff]

nah i know lol

how is this a loop at b_0?

f(1) = p o ftilde(1) = p(e_1)

oh wait

[knotty vine]

Do you have some favourite examples of covering maps? Work out some simple liftings for them

[heady skiff]

e_1 in preimage

got it

[heady skiff]

actually munkres gave an example which was nice

munkres the goat

[knotty vine]

which one is that?

[heady skiff]

love munkres

[knotty vine]

Ya thats classic

Some variations are where R is replaced by R/nZ in this story

That looks like a sort of spiral that goes around n times and then connect back to itself

[heady skiff]

ooh interesting

[knotty vine]

It looks like the edge of one of those foldable greenscreens. Ever seen one of those?

Note that those arent simply connected, so they provide some counterexamples

[heady skiff]

mm i think?

[knotty vine]

these annoying little things

This kind of stuff boggles the mind https://youtu.be/eHq9hT6l8vo?feature=shared&t=130

Anyways, I think the black edge of that thing is basically R/3Z --> S1

Actually, is it 3?

I tried it with a piece of rope and I think it's 3. I'm still amazed that this actually works with a disk as well

[unreal stratus]

Is anyone familiar with profinite sets ( = totally disconnected compact hausdorff spaces?) i'm interested in the following lemma: if $X$ is locally compact Hausdorff, there's a collection $f_i: X_i \to X$ of maps ($X_i$ disjoint union of profinite sets) such that for all compact $K \subseteq X$, there's finite $J \subseteq I$ and compact $K_j \subseteq X_j$ s.t. $K = \bigcup_j f_j(K_j)$

Appears in some work of Scholze

[gentle osprey]

potato

[unreal stratus]

Basically what this involves is a efinition of a grothendieck topology on the category of locally compact hausdorff spaces, and then saying each X admits a cover by disjoint unions of profinite sets in that topology lol

[gritty widget]

In what order would you recommend someone to study algebraic topology to come to an advanced level? I have studied basic algebraic topology (as in Hatcher) and vector bundles (sth like Milnor, Stasheff). I also studied some homotopy theory. Where to go next? What would be a logical progression?

[fickle elm]

You have already covered the basics. There are various branches of algebraic topology you may want to have a look? e.g. knots ( or more generally low dimensional topology), topological K-theory, stable homotopy theory, certain (co)homology theory in symplectic geometry or algebraic geometry. They are quite different and it really depends on your interests.

BTW if you really want to learn about homotopy theory, I suggest reading a bit about simplicial homotopy theory if you haven't. They are quite useful.

[gritty widget]

Ok nice thank you a lot! I will definitely look into simplicial homotopy theory, i've encountered simplicial srts on occssion but haven't studied them extensively. And I will also likely look more into K theory, it seems to be pretty active currently

when you entered your home today you were doing K-theory. we're always doing K-theory

[fickle elm]

Not an expert on homotopy theory, but nowadays homotopy categories are viewed as infty-categories, at least in some new papers. You may want to read some introductory books on infty-category with some discussion on simplicial sets. This seems to be the modern way to do things. (I am now reading Markus Land's introduction to infty-category).

[sonic crane]

Am i able to learn munkres’ topology without having taken a “formal” real analysis course

[gritty widget]

yes

you might miss out on some motivation or examples here or there, but strictly speaking it's not necessary to know beforehand

[sonic crane]

Cool, i have the book and it looks understandable to me but im only concerned that maybe without the examples shown in analysis i will lack some …

U said what i was gonna say

[gritty widget]

so just learn the analysis through the examples

[sonic crane]

How “deep” do these motivating examples go?

I would still be able to visualize the topic and questions properly u think?

[viral atlas]

My experience informs me that intuition from metric spaces in particular goes a long way to understand and get comfortable with topological spaces in their generality

Of course, they eventually only serve heuristic value as you realise that a space carrying a metric has way much more structure (lots of conditions collapse to being equivalent for these, while there are several layers of nuance to general topological spaces)

[sonic crane]

I see thank you

In terms of working out the problems etc, what kind of flavour is it like? Im doing group theory right now and of course it feels much different than when i did epsilon delta type inequality stuff in some previous courses

Do topology problems feel more like abstract algebra , analysis, or is it its own feel?

[viral atlas]

A different feel, at least that's how I found it to be

[sonic crane]

Interesting

I way prefer abstract algebra than doing stuff with inequalities and estimating

I always felt like i lacked good intuition with estimating

I like the structure of algebra

[viral atlas]

Honestly pointset topology is a lot about keeping track of lots of sets and definitions, at least when dealing with the basic constructions

[sonic crane]

I see

[ebon galleon]

I mean a fair amount of stuff in point-set is motivated by analysis

But it does have a different feel when you're working with it

[sonic crane]

I see what u mean

[sonic crane]

Yea thats what i was wondering

[high hill]

i keep feeling its analysis smh

with this counterexample and that

[fathom steeple]

Why do we need locally trivial for sections to be a map from base space to total space

like, isnt this definition of a section? for any bundle

ignoring that line, my understanding is that for every p in the physical space, its fiber is isomorphic to C and a section is just a function from physical space -> C

also sidenote question: sections need not be cotinuous correct? just a map?

[knotty vine]

sections are (usually) continuous

[fathom steeple]

is that part of the definition though?

my book & wikipedia don't mention it but idk

[knotty vine]

The definition of the word "section" depends on which category youre working in I guess

[fathom steeple]

bundles

[hexed steppe]

not a map to C unless the bundle is trivial.

the fibers are C though, yes.

[fathom steeple]

well, sections are maps from the physical space to the fiber of the point right? so if the fiber is C then its a map to C?

[knotty vine]

The bundle may not be trivial

[hexed steppe]

what

just because f(p) is a complex number

[knotty vine]

Imagine the line segment bundle over the circle that looks like a mobius strip

for example

[hexed steppe]

does not mean that you can glue this data into a continuous map [physical space] -> C

[fathom steeple]

base space being S1?

[knotty vine]

yes

[fathom steeple]

right that was my next question, are sections defined to be continuous?

[hexed steppe]

certainly.

[fathom steeple]

i see

[hexed steppe]

otherwise you would be right, but then the notion of bundle would be devoid of meaning

i mean it may not be continuous…it may be some other structure depending on your context. but sections definitely aren’t arbitrary functions

i strongly suspect that they are continuous though.

[fathom steeple]

👍

[hexed steppe]

if not something stronger

[knotty vine]

(bundles and section are useful outside of topological spaces, so in the most general sense they dont have to be continuous (or smooth or holomorphic or w/e) and they arent devoid of meaning, but that's not under consideration here)

[hexed steppe]

surely there has to be some structure preserved though?

even measurability is something

[ebon galleon]

all functions/maps are often assumed to preserve whatever structure is needed to be in your category

[hexed steppe]

otherwise it is just a fancy way of saying e.g. “function valued in C”

[knotty vine]

Of course structure has to be preserved, but structure isn't always there in the first place

[hexed steppe]

do you have an example?

[ebon galleon]

Like some terms like that (in particular, you see section fairly often) can broadly be used in category theory

[fathom steeple]

does this mean that if the bundle is not locally trivial then there does not exist a continuous function from physical space to C?

[hexed steppe]

uhh not quite

[knotty vine]

It may, but not necessarily

[hexed steppe]

to be clear

[knotty vine]

Certainly the trivial section works

[hexed steppe]

the question is about existence of a map B -> E

[fathom steeple]

B->C

[hexed steppe]

no

sections are maps into the total space.

the fibers are each C

but the point is that the total space may not be of the form B x C

[fathom steeple]

ok so every output is in C but the function is continuous only with codomain M?

[hexed steppe]

what is M

[fathom steeple]

sorry, E

[hexed steppe]

then yes that is one way to say it

[fathom steeple]

lol

[hexed steppe]

i think it is easier to think just purely in terms of the total space

thinking of it as copies of C attached to each point

[fathom steeple]

yes

weird, if f:B->E is continuous, and f(B) \subset C then f:B-C is not continuous?

[hexed steppe]

well the entire point is that f(B) subset C is meaningless unless you specify a trivialization

you need to identify each fiber in f(B) with the same copy of C

and unless you do so continuously, the result need not have any reasonable topological structure.

[knotty vine]

A proof of the statement \forall (x : X), P(x) is a section of the map P : X -> Propositions

[hexed steppe]

💀

ok fine

[hexed steppe]

a trivialization is precisely a “continuous way to identify all the fibers with one copy of C”

a local trivialization is that but local.

[fathom steeple]

oh im braindead

kekw

it just dawned on me....

what i was saying was a map to some infinite product of C

[hexed steppe]

yes exactly

[fathom steeple]

something isomorphic to that

ol

lol

thanks

[hexed steppe]

idk it took me a while to figure this out.

how bundles work etc.

[fathom steeple]

ya

[hexed steppe]

actually i remember being confused about this exact point in undergrad

[fathom steeple]

it has come full circle

[fathom steeple]

Can someone give me intuition for the name "pull back" of a bundle E -> B with respect to f:M -> B

why do we care about M

or will i see this later when i learn tensor fields.... and should just accept this definition right now

[fickle elm]

If you forget about the bundle structures and view the square in the category of sets, it is exactly the pullback square in the category of sets.

[gentle girder]

hi, the pullback can really be generalized to any structure that you want. If the pullback bundle is the first time that you are seeing pullbacks, then it won't make much sense why we are calling this extra structure on M a "pullback" of the one on N. Here is how to view it from an abstract viewpoint: Suppose you have a "structure" on N (by this, it can really mean any structure, such as a topology on N, a connection, a bundle with a projection pi: E-> N, or if N is a manifold, a 1-form, which is a map from TN -> R, or if there is no additional structure on N, even a function g: N-> W), and you have some map f: M -> N a suitably nice map (for a bundle, f is continuous, for a 1-form, f is differentiable), then you can define the pullback by saying "what would M look like if the structure on M was just as though it was the image f(M)?" The simplest example of a map g:N ->W yields the pullback map of g - it'll be a function from M to W satisfying f*(g)(x) = g(f(x)). If you have a 1-form omega: TN -> R, then you get the pullback form: f*(omega)_p(v) = omega_f(p)(df_p(v)). And for the bundle pi: E -> N then we get the pullback bundle, f*E, where if f(p) = q, then we get that the fiber at p of f*E is just the fiber at q of E

lmk if i made a typo - I always get nervous to write responses like this because they are likely to have typos/errors

[heady skiff]

To show that the mapping $f: S^1 \times \mathbb{R}+ \to \mathbb{R}^2 - \textbf{0}$ given by $(x, t) \mapsto tx$ is onto, for ${x} \in \mathbb{R}^2 - \textbf{0}$, would it suffice to take $(\frac{x}{\norm{x}}, \norm{x}) \in S^1 \times \mathbb{R}+$?

[gentle osprey]

okeyBOOkay

[ebon galleon]

Yes.

[heady skiff]

thanks

bruh moment

what's the easiest way to show injectivity

i wrote $f(x_1, t_1) = f(x_2, t_2)$, so $t_1x_1 = t_2x_2$, but at least to me, there' sno obvious way to go about showing $(x_1, t_1) = (x_2, t_2)$ - i tried doing some stuff with letting $x_1 = (a_1, a_2)$ and $x_2 = (b_1, b_2)$ but that got messy

[gentle osprey]

okeyBOOkay

[ebon galleon]

look at the norms

if t_1 is not = t_2 this should be clear. And from that, if t_1 = t_2 = t, when x_1 is not = x_2 show tx_1 and tx_2 are not equal using norms

hint ||look at the difference: tx_1 - tx_2||

[heady skiff]

ah i see so ur going (x1, y1) \neq (x2, y2) => f(x1, y1) \neq f(x2, y2)

ok i'll try that

[lusty trench]

Are there any interesting fiber bundles whose total spaces can be embedded in R^3? Besides the Möbius strip as a bundle over S^1, and the cylinder as the orientation bundle over the Möbius strip.

[heady skiff]

i mean, is it just that t(x_1 - x_2) \neq 0 since t > 0 and x_1 - x_2 \neq 0? is it just that?

[ebon galleon]

Yes

[heady skiff]

ah ok

[gentle girder]

here is a neat object: for any curve, you can make the tangent developable of that curve, which is the tangent bundle embedded in R^3. For a curve c: R -> R3 take s: TR (= R^2) -> R^3 by s(x, v) = c(x) + dc(v)

[heady skiff]

wait this is just cancellation law

bruh

[ebon galleon]

Oh or that

[lusty trench]

That would be an embedding in TR^3 ~= R^6, rather than R^3, right?

Also, the tangent bundle of a real curve is trivial.

[knotty vine]

Is it interesting if the fibre is discrete? The classic covering spaces give examples

[lusty trench]

Oh. That would be interesting.

The orientation bundle of the Möbius strip would be an example, but I had already mentioned it.

[gentle girder]

hmm, I was thinking it would give us a surface with a parameterization (x,v) -> c(x) + vc'(x), (i think this is what happens if you identify TpR3 with R3?) Yes it's the trivial bundle, but it gives us a way to make some ruled surfaces, which I find cool - I don't think I actually meant dc when I said it the first time sorry

[heady skiff]

could i get a hint for constructing the inverse? i feel like this is very elementary but for some reason my brain is not working

also quick question which i suspect to be true; if h is a homeomorphism and f is a covering map, is h o f a covering map?

(dw i will attempt to prove this)

[knotty vine]

ya

[heady skiff]

just came up with something that i think works: suppose we have $f: X \to Y$ a covering map and $h: Y \to Z$ a homeomorphism. Let any $z \in Z$. we must find a neighborhood $U$ of $z$ such that $(h \circ f)^{-1}(U) = f^{-1}\bigl(h^{-1}(U)\bigl)$ is evenly covered. since $h$ is a homeomorphism, $h(y) = z$ for some $y \in Y$. Since $f$ is a covering map, we can find a neighborhood $U$ of $y$ which is evenly covered; since $h$ is a homeomorphism, we see that $h(U)$ is a neighborhood of $z$. But [f^{-1}\bigl(h^{-1}(h(U))\bigl) = f^{-1}(U)] which is evenly covered, as desired.

[gentle osprey]

okeyBOOkay

[knotty vine]

Pretty much whenever you have some interesting/useful property of topological spaces or continuous maps, then it's gonna be invariant under homeomorphism

[ebon galleon]

The property of "contains \emptyset as an element" is not invariant under homeo

[knotty vine]

quiz: name a non-topological property that doesn't come from other categories

[gentle girder]

good thing it's not useful or interesting as a property of topological spaces

[knotty vine]

unfortunately, it is a useful property of sets, but not invariant under iso

thats why sets SUCK! Boooo!

[quaint silo]

@knotty vine Is that a Halloween scare?

[ebon galleon]

"name something that doesn't come from topological spaces or any other category"

[knotty vine]

You ever go to a standup comedy bar on halloween? Very scary

[knotty vine]

You know what I mean

[quaint silo]

Ask me a question

[ebon galleon]

I actually don't

[knotty vine]

What's a good question?

[knotty vine]

To make it more rigorous, I mean properties that are invariant under homeomorphism 🙂

[quaint silo]

Ask the helpers

[gentle girder]

how about homeomorphism class

👍

[ebon galleon]

Anything interesting in topology is witnessed by functors, so it is impossible to name something not from another category

[knotty vine]

What I meant to exclude with requiring that the property doesnt "come from another category" are properties like "does it contain some specific element" which really arises from the forgetful functor from top. spaces to some weird cat of sets where you can talk about specific elements somehow

That's not a good example...

[heady skiff]

someone :(

[ebon galleon]

bruh

[gentle girder]

then the fundamental group is the low-hanging fruit if we are just allowed to take a functor

[ebon galleon]

you already gave the inverse

[knotty vine]

polar coordinates?

[ebon galleon]

in your original question

[heady skiff]

LMAO

oops