point-set-topology

[hidden crag]

no

[odd flame]

that did feel wrong

[hidden crag]

read van kampen again

[odd flame]

ik you said quotient of a normal subgroup but idk how that comes in

[hidden crag]

ignore that

that's irrelevant for the way we're doing it rn

[odd flame]

pi(U cup V) = pi(U) * pi(V)

[hidden crag]

no

[odd flame]

for path connected U cap V

[hidden crag]

plus V is contractible anyway

the intersection is S^1

of U and V

you need to look at how pi_1(U cap V) embeds and take the amalgamation

[odd flame]

maybe im missing something but what does this intersection have to do with van kampen

[hidden crag]

my brother in christ

have you read the theorem

[odd flame]

holy shit i flipped it in my notes

it's not even a result of me rn it literally have cup instead of cap in the notes i took

i'll see myself out

[obtuse meteor]

Note: this is why having intuition for thm statements is important

[odd flame]

ok putting that silliness behind me

how does P#P = K help w this iso

i unspoiled but im assuming it has to do with this

[pseudo coral]

In showing the identity map from Sn to itself is not nullhomotopic so I suppose it is then there exists an extension to the whole closed disk but there is not retract of B^n+1 onto Sn

[unreal stratus]

Well you just need to show S^n isn't contractible right? That can be done by various methods

[bitter smelt]

Yes, including the one they seem to be referencing

Using brouwers

[bitter smelt]

Assuming you have it for generic n, not just n=1

[pseudo coral]

Yes I have it for generic n

[brittle basin]

X, Y being two non empty bounded subsets of a normed vector space. Claim: The union is bounded.
My approach was to just pick max{r,s} then
X U Y c max{r,s} B (B notation for open ball)

I then looked it up and most results are using the triangle inequality. Is my approach not correct? If its correct why would you do it with the triangle inequality?

[gritty widget]

well this is a little vague but if you mean that X is a subset of rB and Y is a subset of sB, then sure

B being the open unit ball centered at zero

[nocturne basalt]

I'm trying to show that if V is simply connected, j_1 must be surjective. I understand the argument which shows that j1 is epic (and thus surjective) but could some one point me in the direction that doesnt use the fact that epic morphisms are surjective homomorphisms?

[brittle basin]

Yeah thats what i meant. But is there any reason to do the proof with the triangle inequality? I feel like its much less "intuitive"/simple.

[gritty widget]

in general you won't have the origin, the balls can be placed pretty arbitrarily

though I wouldn't say complicated as ultimately it should be trivial

[brittle basin]

Okay great that makes sense, thanks.

[coarse night]

do you have any reference to this?

[nocturne basalt]

yeah its from fultons book

[coarse night]

ohh tks, it's a nice proof of VK using G covers

[nocturne basalt]

yep

[odd flame]

i got this one on my own B)

timo if u see this u are very cool and based and thank you for helping

i promise im not a shit student i just got shit coming up all the time

[hidden crag]

You’re welcome

[hidden crag]

I‘m not judging mate don’t worry lol

[pseudo coral]

does anyone have Munkres with them I coulld ask a question about

[gritty widget]

i have a copy somewhere

[pseudo coral]

so question out of the Retractions and Fixed points section, does lemma 55.3 extend to Sn and B^n+1 as apposed to just for S1 to B2

states that if h:S^1 \to X is cont then TFAE:

h is nullhomotopic, h extends to a map k:B^2 \to X, h_* is the trivial homomorphism of fund groups.. im only interested in the first two implications

does it extend to:

if h: S^n \to X is cont then TFAE:

h is nullhomotpic, h extends to some k:B^n+1 \to X

[unreal stratus]

Yes

Proof is p much identical

[pearl holly]

ye more generally X-->Y is nullhomotopic iff it extends to the cone of X

[unreal stratus]

Nerd

Jk yes

[pearl holly]

it's a nice thing to know

[unreal stratus]

It is

!

[bitter smelt]

I've used it fewer than 5 times therefor it isn't nice to know

[pseudo coral]

Ok cool

Thanks you guys

Cause I’m using it in showing the identity map from Sn to itself is NOT nullhomotopic

[pseudo coral]

😂

[pseudo coral]

Thank you 🙏

[unreal stratus]

hm i don't see why you'll need that to show that

How are you gonna use it

[pseudo coral]

Because

I suppose BWOC that it IS nullhomotopic

Then it (the identity on Sn) extends to the full disk

Creating a retraction which is a contradiction

No?

[unreal stratus]

Why is that a contradiction

That is true though

But I feel like you're gonna prove it via a route that already shows the identity is not nullhomotopic

[pseudo coral]

Cause I’m assuming there’s no retraction of Bn+1 onto Sn

[unreal stratus]

Ye sure

Anyway my point was uhhh

Usually one proves that by using the fact that some homology/homotopy groups of S^n non-trivial, agreed

Which already implies your result

[pseudo coral]

🤔

Well I haven’t seen any homology yet so

[unreal stratus]

Lol okay

[pseudo coral]

And I have yet to witness higher homotopy groups..

[unreal stratus]

So you are just like blackboxing that result?

Weird question if it's in munkres then lol

[pseudo coral]

Wdym

[unreal stratus]

Oh I mean like

[pseudo coral]

By black boxing it

[unreal stratus]

where does this come from?

[pseudo coral]

That’s an assumption in the exercise

look :

[unreal stratus]

Yeah that's why i find it a weird questino lol

Hey ho

[unreal stratus]

Ye sure

[pseudo coral]

Yeah right?! It pretty much gives it to you

Using the assumption it’s pretty clear

[unreal stratus]

Ye

Okay then your method is gud lol

[pseudo coral]

Thanks lol

[unreal stratus]

owo

[pseudo coral]

Btw are higher homotpy groups defined for loops whose domain is n copies of the unit interval i.E., π_n

[unreal stratus]

Well, those aren't loops, but yes that is one way to think about it

[pseudo coral]

I was gonna say

[unreal stratus]

So I think the most elegant description is like

[pseudo coral]

Using the term loop is a bit iffy

[unreal stratus]

The set underlying π1(X) is given by like

(pointed) homotopy classes of (pointed) maps S^1 -> X, that fine?

[pseudo coral]

Yes

[unreal stratus]

So uh

change 1 to n

and ur gucci

[pseudo coral]

Ahh

[unreal stratus]

But yeah you can equivalently think of it in terms of maps I^n -> X subject to some conditions, which is more convenient for some proofs and stuff lol

Like maps I^n -> X sending the boundary to the basepoint lol

[pseudo coral]

Ahh I see

[unreal stratus]

Which is again analagous to the n=1 case

However like

Usually a first course in alg top won't deal with these afaik, at least not in the uk ig lol

It's more likekly you'll just do homology and it's much more easy to compute and as powerful for the applications required

[pseudo coral]

Yeah idk if we’re touching on them

I sure hope so !

Sounds p cool

And the homology group is the abelianization of the homotopy group if I’m not mistaken

[pearl holly]

sort of, the first nonvanishing homotopy group of a space is isomorphic to the first nonvanishing homology group

[unreal stratus]

Except for the case where it's the first case in which you take the abelianisation

[pearl holly]

right

was about to add that in lol

[unreal stratus]

So yeah in general what you said was right, just for the first one

This is called like the Hurewicz theorem and I guess like it has two parts

the first being what you said (π1(X)^ab \cong H_1(X;Z)) and uh

that's not too hard to prove

A more general Hurewicz is what Toki said

And then as usual there are relative versions and stuff lol

[pearl holly]

ye so I guess the "catch" is that homology groups are harder to define than homotopy groups, but once they have been defined it's a lot easier to compute them compared to homotopy groups

[next crystal]

i need to show that X* is not Hausdorff. I'm pretty sure the issue is with the points p(0 x 0) and p(0 x 1) (the two origin points). So i need to show that if U, V are open in X*, p(0 x 0) \in U, p(0 x 1) \in V, then U and V intersect. I'm really not sure where to start with this though

[unreal stratus]

Your intuition is correct

What do (smol) neighbourhoods of the origin of X* look like

[next crystal]

thats what im getting stuck on i think. I know that bc X* has the quotient topology, the preimage of these neighborhoods have to be open in X

I was thinking maybe something like this (the union of the two intervals and one of the origin points)

[unreal stratus]

So the key thing is that uh

The preimage isn't just any old open subset of X

What can you say about it

[next crystal]

does the preimage need to contain intervals on the lines y=0 and 1

since the neighborhood around say p(0x0) has to include points other than itself and p(0x1)

not really sure

[unreal stratus]

It's an open set that must contain both preimages of the origin.

In X

[unreal stratus]

But then yes exactly

That's what leads to the issue

[next crystal]

wait but doesnt this show that the preimages of U and V intersect

not U and V themselves

[unreal stratus]

Yes, and that's enough!

because note p(p^{-1}(U)) = U because p is surjective

You typically wanna just lift to X and work there for these kinds of things

[next crystal]

to be clear we know $p^{-1}(U) \cap p^{-1}(V) \neq \emptyset$, so $p(p^{-1}(U) \cap p^{-1}(V)) \neq \emptyset$. But since $p(p^{-1}(U) \cap p^{-1}(V)) \subset p(p^{-1}(U)) \cap p(p^{1}(V)) = U \cap V$, we have $U \cap V \neq \emptyset$?

[gentle osprey]

michαel

[unreal stratus]

Yes, though actually this is easier

Since $p^{-1}(U) \cap p^{-1}(V) = p^{-1}(U \cap V)$ (always the case for any function) and $p(p^{-1}(U \cap V)) = U \cap V$ (since $p$ is surjective)

[gentle osprey]

potato

[unreal stratus]

Or even more clearly, if you pick x in p^{-1}(U) \cap p^{-1}(V) then certainly p(x) is in U \cap V

[next crystal]

got it that is a lot clearer

thanks so much

[silver umbra]

if X, Y are topological spaces, and there exists a cts bijection f: X->Y and another cts bijection g:Y->X, but g isn't necessarily the inverse of f, can we conclude that X and Y are homeomorphic?

i suspect the answer is false but i'm struggling to come up with a counterexample

[gritty widget]

that's gotta be so false

yeah

https://mathoverflow.net/questions/30661/non-homeomorphic-spaces-that-have-continuous-bijections-between-them

[civic verge]

Let $X=]0,1[\subset \mathbb{R}$. Prove that $d(x,y)=|x^{-1}-y^{-1}|$ is an distance in $X$

[gentle osprey]

Awuita Fria

[civic verge]

Guys, I think it's a distance in X.

[gritty widget]

so prove it

[civic verge]

that would be my test

[unreal stratus]

Propiedad del dolor

[civic verge]

:c what

[unreal stratus]

This is all good :)

[unreal stratus]

was joking cause your writing looked like that at the start of the second pic

[unreal stratus]

And this is good except you seem to have omitted <= signs

But yes, otherwise gud

!

[next crystal]

does my reasoning for this proof look right?

[unreal stratus]

Yes

[pseudo coral]

so question

so show the inclusion map $j: S^n \to \Bbb{R}^{n+1} \setminus {0}$ is not nulhomotpic given there is no reetract from $B^{n+1}$ onto $S^n$

[unreal stratus]

Owo

[gentle osprey]

MyMathYourMath

[pseudo coral]

do i deform retract R^n+1 onto S^n

then apply the same argu,ent as showing identity on S^n to itself was not nullhomotpic to j \circ r where r is the deform retrac of R^n+1 onto S^n

[unreal stratus]

That's what I'd do

But you don't need to repeat the arguement actually

:)

So uhhh

[pseudo coral]

i like that argument lol i came up with it on my own 🙂

[unreal stratus]

Okay so yes we know that we have like $S^n \xrightarrow{j} \mathbb R^{n+1}\setminus {0} \xrightarrow{r} S^n$ where $r$ is the retraction

[gentle osprey]

potato

[unreal stratus]

and that composition is identity

[pseudo coral]

yes

[unreal stratus]

If j is null homotopic, what can you say about rj

[pseudo coral]

and if its nullhomotpic

then it rextneds

to a cont function on unit disk

[unreal stratus]

No i mean

You needn't repeat the argument

You can reduce to the other one

[unreal stratus]

ygm

Like if j is homotopic to a constant map, then ...

[pseudo coral]

rj is a map from sn to itsel

th eidentity

[unreal stratus]

Yes

But with that assumption...

My point is that composition respects homotopies

So if j is homotopic to a constant map...

[pseudo coral]

and so is its ocmposition

[unreal stratus]

yup

[pseudo coral]

then*

[unreal stratus]

But that contradicts the other thing

:)

Or yes, you can just repeat the argument

[civic verge]

$d(x,y)=|x-y|^{2}$

[gentle osprey]

Awuita Fria

[civic verge]

guys a question this function is a distance in R?

For the triangular inequality, I don't think so.

therefore it would not be a distance in R

[unreal stratus]

Sure, try to give a counterexample then

[civic verge]

the problem is that I don't know how the triangular inequality would look like.

uh

[gritty widget]

woops my inequality was backwards

i wrote it wrong. let me try again

[civic verge]

I know, but how it would be in this case, if I have the (d(x,z))^2

[gritty widget]

"d(x, y) <= d(x, z) + d(z, y) for all x, y, and z"

if you want to prove this wrong, then you would have to find specific x, y, and z such that d(x, y) > d(x, z) + d(z, y).

can you find x, y, and z with |x - y|^2 > |x - z|^2 + |z - y|^2?

i'd suggest trying a couple of different x, y, and z and seeing what happens

[civic verge]

in this case complies

Therefore, the triangular inequality is not satisfied.

[gritty widget]

can you give a specific example disproving it?

[civic verge]

If we take as
x=5 y=3 and z=7
we have that
|5-3|^2>|5-7|^2+|7-3|^2
4>4+16
4>20

lol

[gritty widget]

i don't think that 4 is greater than 20

but i can tell you that 420 is great

[civic verge]

jsjsjsjsjs

then if it satisfies the triangular inequality then it is a distance function in R

[gritty widget]

if it did, sure, but it doesn't

it's on you to prove that the triangle inequality doesn't hold

[gritty widget]

you're on the right path picking x, y, and z, but you need to make this hold

since you can pick them, one simplification you can make is to pick z = 0 and see if you can find x and y making the resulting thing true

[civic verge]

Shouldn't it be < ?

[gritty widget]

read this carefully

[gritty widget]

you are trying to prove that the first one is NOT true. this means you need to find specific x, y, and z making d(x, z) > d(x, y) + d(y, z)

[civic verge]

I cannot find any number that does not meet this condition.

[gritty widget]

i gave a hint

since you can pick x, y, and z, you might as well try to make your life easier and pick y = 0 here. then you'd have to find x and z such that |x - z|^2 > |x|^2 + |z|^2

[civic verge]

Likewise, if z=0 it is not possible since there will always be an x and y

[gritty widget]

i don't really know what you mean

there will always be an x and y satisfying what?

and why does that come from z being zero?

[civic verge]

I am trying to tell you that the inequality is not being fulfilled even if I do y=0 or z=0, it will always be false.

[civic verge]

no

there is no

then I can conclude that the triangular inequality does not satisfy

Therefore, it is not a distance function in R

[gritty widget]

if y and z are both zero then d(x, z) is |x|^2 and d(x, y) + d(y, z) is |x|^2, so you'd get d(x, z) <= d(x, y) + d(y, z) which verifies the triangle inequality in this case. so this is not a counterexample

you need a different example

if that's what you meant

i'm having a lot of trouble understanding what you're trying to say

[civic verge]

in this case the inequality is satisfied for any value of x y or z, so I can take any value and the inequality will always be satisfied.

[gritty widget]

the inequality? which one? d(x, z) <= d(x, y) + d(y, z) or d(x, z) > d(x, y) + d(y, z)?

[civic verge]

I just want to know if the triangular inequality for the distance function is fulfilled or not ?

[gritty widget]

i've been trying to tell you it's not

and i've been trying to tell you how to find examples

[civic verge]

d(x, z) <= d(x, y) + d(y, z)

[gritty widget]

yeah, that one fails for some x, y, and z

i'm telling you how to search for x, y, and z for which it fails

d(1, -1) > d(1, 0) + d(0, -1) for example

[civic verge]

ooooh

ook!!!

thanks

[next crystal]

why do we need to discard X \ Y in the case where the finite open covering of X contains X \ Y? Since in the covering of a subspace, the union of subsets can properly contain the subspace, isn't this unnecessary?

[gritty widget]

you need a finite subcollection of A

the finite subcollection of B given to you by compactness will indeed cover Y, but it may not be a subcollection of A since X - Y will not be in A

[unreal stratus]

Suppose we have a vector bundle V over a space X, such that the vector bundle has a riemannian metric. How can one rigorously show that V is homeomorphic to the associated open disk bundle (take an open ball in each fibre)?

Of course, in each fibre we can just map the open disk to the whole fibre homeomorphically (say, x -> x/sqrt(1-|x|^2)) but we need to show that glues together appropriately to form a homeomorphism of total spaces

[civic verge]

Guys a question I am asked to find the boundary of a sphere of center a and radius r in R^{n} institutionally I can say that the intersection of the set with the boundary of the set is disjunct so it is empty to what an open sphere, has no element as a boundary.

[civic verge]

Sorry if you don't understand I use translator but this is my idea, I hope it's understood

[gritty widget]

it's the same set

it's closed, has no interior

[bitter smelt]

I'm a little puzzled by the question and the translation. Do you mean the boundary of a ball (i.e., solid ball) or the boundary of a shere (i.e., hollow ball)

[gritty widget]

ah. Yeah that might have been lost in the translation actually 😅

[civic verge]

This means that the set is only border

[civic verge]

I am asked to demonstrate what is the boundary of a sphere in R^{n}

where S(a,r)

I'm looking at some images, and I can see that a sphere is just a border point.

[quiet pilot]

The sphere is closed and nowhere dense so the boundary is just the sphere itself

[gritty widget]

nowhere dense might be adding salt to the wound here

lol

[gritty widget]

The boundary is just closure without the interior

the closure is the sphere itself, and the interior is empty

so its itself without ... the empty set. So its the same set

[civic verge]

Then a sphere in R^{2} is the set formed by the entire boundary, i.e. it has no interior.

[gritty widget]

sure

[dawn oasis]

Hi, in Engelking's general topology book he warns that some authors do not include the requirement that a topological space be T1 (as he does) in their definition of regular, completely regular and normal spaces. Is one approach more common than the other? Is it usually safe to assume one way or the other? Thanks!

[gritty widget]

spaces which don't satisfy T1 assumption are not really important

where by spaces I mean... those definitions

the only reason to not assume T1 in those definitions, that I can think of, is to be the most general possible

I've seen more books assume it than don't, but it's probably an even split

Hausdorffness is probably a little more common...

[unreal stratus]

Doesn't normal + t1 imply hausdorff?

Oh I guess thats why you say that

Since it is equivalent

[gritty widget]

yes

I've saw theorems before of the sort... this holds if the space is T1 or regular

[unreal stratus]

Hm what are the requirements of a map extending to a one point compactification

Ig just have to think of nhds of infty and it'll not have anything nicer

[gritty widget]

but application of weakening the definitions like this...? I never seen it

[unreal stratus]

Yeah

[gritty widget]

I suppose you're mean, f:X to Y and you want to extend it to f:aX to Y where aX is the one-point compactification of X

[unreal stratus]

Yes

[gritty widget]

and X is locally compact Hausdorff

[unreal stratus]

Ye

Or the case of aX -> aY sending infty to infty ig

But that's easier ig?

[gritty widget]

this should boil down in some way to uniform continuity

I have a description of when such map can be extended but I don't know if it's as general... it deals with metrizable spaces

is that okay?

maybe I don't need metrizability anywhere actually

yeah

@unreal stratus so?

[pseudo coral]

Been stuck on these problems for a while if anyone has any insight

[unreal stratus]

Ah okay, how does that work?

[gritty widget]

I'll tell you tomorrow if I don't forget

[unreal stratus]

Cool thankies

[icy zenith]

Maybe I am saying something wrong but this seems to be a particular case of Rouché's Theorem no ? Correct me if I am wrong

[unreal stratus]

I agree

[winged badger]

rouche's is for holomorphic functions, this is general curves

[winged badger]

take epsilon to be the minimal value of |f|. Then any g satisfying the condition will have the same winding number because you can continuously transform it to f without touching 0

just tg+(1-t)f

[void gazelle]

Hi, guys, suppose p: \tilde{X}\to X is a covering space. Then is it possible that for some x\in X, there are open neighbourhood U and V of x, such that p^{-1}(U) and p^{-1}(V) are sheets of U and V respectively, but the number of sheets are different?

[gritty widget]

it looks like @unreal stratus of course some assumptions like LCH should suffice

[gentle osprey]

Blitz

[gritty widget]

The other one will of course be compact in X

So the condition can be simplified to

[gentle osprey]

Blitz

[gritty widget]

I think it's a neat condition

it doesn't adhere to any properties of the one-point compactifications, just X and Y

like I said for separable locally compact metric spaces, definitely holds

for other LCH spaces, I didn't see any obstructions in the proof but I didn't look too intensely

[unreal stratus]

That's very interesting thank you - i'll have a think about it - I like it

I do like how this doesn't rely on saying anything about the point at infinity etc too, kinda cool

[gritty widget]

one direction is easy

the other one is slightly more complicated

[gritty widget]

this definitely doesn't belong in this channel

[willow viper]

Oh xD

Is this multivariable calc?

[gritty widget]

you can just write down a parametrization and compute the area by definition. shouldn't be hard

yes

[willow viper]

Ok let me delete it

[unreal stratus]

Lol i thoguht you meant the one-point0compactification stuff for a sec

Was like

Also Blitz i didn't realise \alpha X was notation for one poitn compactification but it will be useful for smth i'm writing lol thank

[gritty widget]

I've encountered it before but not sure if it's completely standard

a lot of people write a star * or prim '

Engelking writes wX

Dugundji doesn't seem to give any notation

he does give an equivalent condition for extension of maps from a set to one-point compactification though

[unreal stratus]

Kinda makes sense tbh given that like

βX

or was this like a backformation from βX lol

[gritty widget]

$f:X\to Z$ where $Z$ is Hausdorff and $X$ is LCH extends to a map from $\alpha X$ to $Z$ iff ${f(C^c) : C\subseteq X \text{ is compact}}$ converges

[gentle osprey]

Blitz

[gritty widget]

idk I got influenced by van Mill to write alpha for one-point compactification

[gritty widget]

converges in the sense that the filter generated by the set is a convergent filter

[unreal stratus]

Tbh I prefer notation like [...]X to sub/super scripts for stuff like this

[gritty widget]

Dugundji's condition is definitely a different approach idk if it's more or less useful

[unreal stratus]

Ah okay sure

Yeah I think I saw that in Dugundji

Thankies

Lol had to do some pointset for the first time in a while to justify some stuff this week

[gritty widget]

I think Dugundji's condition is wrong because he forgot to write that X is non-compact or something

[unreal stratus]

Oh lol fair enough

[gritty widget]

well I guess this wouldn't even generate a filter in this case

so the statement wouldn't really make sense

if X were compact

[nocturne basalt]

Theres a question in Hatcher which asks to show that the complement of a closed discrete subspace of R^n is simply connected if n >= 3 and I'm wondering why the hypothesis of closed is necessary. My proof idea was to first find disjoint balls around each of the points in the discrete subspace so that R^n - {balls} is a deformation retract of the original space. Then R^n - {balls} is simply connected since adding n-cells to fill in the balls doesnt change pi_1.

[unreal stratus]

I imagine you can have some fairly gnarly sets if you drop being closed - in R^2 you could have points at (1/n, 1/n) in polar coordinate for all n, for example, and I imagine you can do similar things for n >= 3 by basically taking a space filling curve for the sphere, say

[gritty widget]

What's an example of a limit point in a totally-separated space ?

[stark fog]

I'd say endpoints of the Cantor ternary set but I'm not sure

[gritty widget]

every point of the cantor set is a limit point

Yeah. Cantor set is homogeneous so no matter which point you choose. It's even a topological group

[stark fog]

oh nice

[gritty widget]

Should've specified earlier that an example of a limit point in a stone space would be nice

[gritty widget]

Cantor set is a Stone space

[unreal stratus]

Is this like from viewing it as {0,1}^w?

(The group structure)

[gritty widget]

Ye

[unreal stratus]

With addition mod 2 I mean

Nice

[gritty widget]

Ye

[unreal stratus]

Thank

[gritty widget]

No pro bro

[mighty stirrup]

I have a (countable) sequence of inclusions

X_0 subset X_1 subset X_2 subset ...

where:

1. X_i is closed in X_{i+1} for all i
2. every point in X_{i+1} \ X_i is closed in X_{i+1} for all i

For n=1,2,..., I have x_n in X_n \ X_{n-1}. Let X be the union of all the X_i's. Then I want to show that the subset S={x_1,x_2,...} has the discrete topology in X. Equivalently, for each n=1,2,..., I want to find an open subset U_n of X such that U_n intersects S only at x_n. Any ideas?

[gritty widget]

To show that {x_n} is open in S, intersect S with complement of X_n and complements of {x_k} for k < n

since {x_k} are closed in X_k, they are also closed in X

[bright sedge]

i wanna know if i understand the definition of manifold on wikipedia

the surface of a sphere S^2 is a 2 dimensional manifold because it's homeomorphic to an open subset of 2 dimensional euclidean space?

is that correct?

[quiet pilot]

No, the sphere is compact and an open subset of R^2 is not so they can't be homeomorphic

But every point on the sphere has a neighbourhood that is homeomorphic to such a subset

[bright sedge]

damn back to reading 😄

when you think you understand you don't :S

[hollow grail]

Hello
Can anyone suggest a short video about introduction to general Topology
Thanks

[pseudo ocean]

https://youtu.be/VOKgMJEc_ro

You could go to 3b1b's problem on the inscribed rectangle problem :

https://youtu.be/AmgkSdhK4K8

[limpid fern]

you could also read a topology textbook instead

[pseudo ocean]

'Topology without tears' is the best book thus far

But i originally read the one by James Munkres

[civic verge]

How do I prove that the set is closed?

I already proved that an element belongs to the complement and not to the set but I'm missing the (<=)

[unreal stratus]

I don't really understand what you're saying

what is "the set" for starters

[civic verge]

I was thinking that the union of the limit and the set gives me the closed set, but in this case I don't know how to do it, although I don't know if it is correct.

A set A that is a subset of the space in Rn.

Or to be more specific, let A be a subset in Rn, prove that A is closed if and only if A'=A

[unreal stratus]

I'm still confused huh

wdym the limit for example

Oh

Okay sure the set of limit points of the set yes

[unreal stratus]

This is not true.

e.g. {0} is closed but {0}' is empty

[steel glen]

what should ker(d_1) on a simplicial complex represent geometrically/visually?

for example, i think you can visualize im(d_2) as the edges of the 2-simplices, but im not entirely sure what the kernel looks like

and same for im(d_1) as the edges of 1-simplices, so the vertices

[civic verge]

They just tell me to prove it

[unreal stratus]

Bruh

[civic verge]

They don't tell me to prove that the vacuum is open or is it closed lol

I already did that lul

The problem states

Let A be a subset in Rn, any subset DIFFERENT OF EMPTY(to be more specific), then A is closed if and only if A'=A

[unreal stratus]

Yeah that isn't true

Well

What is your definition of A'

It must be nonstandard

[civic verge]

I don't understand why they get so complicated in the context, A is a subset in Rn, it just says that "xde".

See this is the Spanish version

[unreal stratus]

Oh okay

But still yeah must be a nonstandard definition of A'

What is ur definition of it?

[civic verge]

yep

[civic verge]

Of...?

[unreal stratus]

A'

[limpid fern]

are you sure A' is not the closure of A

A is closed iff closure of A = A is a common property

[civic verge]

I got lost in the note

The exercise tells me that A is closed if and only if
A=A'

Now A' what does it mean ?

[limpid fern]

Well we know that

A is closed iff it is equal to its closure

So im guessing A' means the closure of A

[civic verge]

In my notes of R
it has to be a point of Accumulation of A

[limpid fern]

well there you go

A = closure of A

for R specifically, closure of A = A union L where L is the set of limit points of A

[civic verge]

I'm looking at the
lock
adhesion
lock
closure
it's the same term, right?

[civic verge]

Thanks, I got it

❤️

[gritty widget]

A' means the set of points of accumulation of A

which is different from the closure cl(A)

writing A' for closure of A would be cursed

[limpid fern]

woah

but they said A is closed iff A' = A

maybe it is the closure

[gritty widget]

it's more likely to be a typo

[gentle osprey]

Blitz

[limpid fern]

definitely cursed to say A' as cl(A)

[gritty widget]

someone might have even meant to originally the exercise to be that A = A' implies A is closed

but changed it last minute

that would explain why A' is on the right side of the equation

[gritty widget]

or wrote if and only if by accident, whereas they should have wrote "if"

the point of the exercise might be just to show perfect sets (as defined by A = A') are closed

[gritty widget]

I actually want to go back to this. Does X has the weak topology coming from X_i?

[tidal lynx]

I’m confused, I thought that a quotient map on a set X mapped elements of X to subsets of X (its equivalence classes)

is the bottom just an overloaded definition for something that “behaves” like an actual quotient map?

[steel glen]

i think what you described is the projection map onto a quotient set, which happens to be a quotient map. a quotient map is a more general notion

[tidal lynx]

ah, so the more general definition is the standard one

I guess that’s usually how it is

[steel glen]

you might have heard the projection map be called "the quotient map pi:x\mapsto [x]", but im sure they're just stating the projection map is a quotient map

[tidal lynx]

I mean in algebra isn’t “quotient map” reserved for that?

as in for quotient constructions of other objects

[civic verge]

Guys, I have a question, how can I prove that

[tidal lynx]

what’s that

[civic verge]

$X\cap \partial X=\emptyset $

Where X is an open set

[steel glen]

do you know what a boundary point is?

[civic verge]

yep

[tidal lynx]

define it then

[steel glen]

is B_r(x) a ball of radius r at x?

[tidal lynx]

probably

[civic verge]

ye

[steel glen]

and how did you define an open set?

[civic verge]

For the interior

[steel glen]

hablas espanol si?

[civic verge]

jajaja seee

[steel glen]

como definieron un set abierto?

[civic verge]

Que toda bola B(a,r) es un subconjunto de A

[steel glen]

perfecto

[civic verge]

bueno donde a pertence a A tal que existe un r>0

[steel glen]

claro

mira la definicion de los boundary points
fijate que si sos un boundary point de X, cualquier bola B(x,r) no es totalmente un subconjunto de X

entonces si tomas un punto $x \in X\cap\partial X$, $x$ debe de tener una bola completamente dentro de $X$, pero al mismo tiempo debe de satisfacer la definicion de un boundary point

[gentle osprey]

maximo

[civic verge]

pero ese x no puede pertencer a X ya que es un conjunto abierto no?

[gritty widget]

This follows from $\partial X = \overline{X}\setminus \text{int}(X)$

[gentle osprey]

Blitz

[steel glen]

el x por si mismo no es un conjunto

es solo un punto en el conjunto X \cap \partial X

lo importante es que para pertenecer a X\cap\partial X, debes de:

1. (boundary point) satisfacer que todos los conjuntos abiertos a los cuales perteneces deben de cruzarse con X y X^c
2. (conjunto abierto) tener por lo menos 1 conjunto abierto al cual perteneces completamente dentro de X

deberia ser claro en este punto que es una contradiccion

[steel glen]

this feels a little cheap lol

they prob want them to do it with the definition of open balls

[gritty widget]

?

the equation holds either way

whatever your definition is, just prove it

and question is easily resolved

[steel glen]

well it's easier to say ||if you're a point in X\cap bd(X) any open nbhd is inside X, but also intersects X^c||

[civic verge]

No se porque a veces cuando pregunto algo aca salen cosas tan avanzada o simplemente saca cosas que no entiendo xde

ok vale lo intentare

[gritty widget]

the equation should basically follow from definition

and since X is open, int(X) = X

[civic verge]

No lo digo por si no, por los demas xde

[steel glen]

yeah i get that, but i think it's easier to just show by definition without showing bd(X) = cl(X) \ X

[gritty widget]

no

it's basically the same thing

bd(X) = cl(X)\int(X) follows from definitions very directly

[steel glen]

ah alright

i mean either way it's like a 1 line thing

[gritty widget]

yeah

but that there is some equation we can use is easier to grasp

[steel glen]

hm fair. i think i find what i said above a little easier to visualize but i don't expect everyone to agree

[steel glen]

si es q aca la gente es bien genia jajaja y tambien la mayoria tiene mucha mucha experiencia

[gritty widget]

I don't speak Spanish so idk

[steel glen]

talking about this bit

but like

honestly its not even easier to visualize cause again they're kind the same thing

[civic verge]

yo lo hice asi X es un conjunto abierto en Rn, por lo que si a pertence a X entonces existe r>0 tal que la bola abierta de centro a y radio r esta contenida en X o esta completamente contenida en X, por lo tanto cualquier punto en X no puede ser un punto frontera de X' complemento, ya que la bola esta contenida en X, por lo tanto no interseca a X' complemento finalmente que X cap partial X es igual a vacio

[steel glen]

esa es la idea si

[odd flame]

english server moment (i can read spanish but still)

but also

trying to understand the definition of a covering space

is there an intuitive way of saying that condition

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

[unreal stratus]

A map p: E -> X which locally looks like the map U coprod ... coprod U -> U

[odd flame]

https://tenor.com/view/potato-cat-surf-glitch-surreal-gif-14769633

[hidden crag]

you can say that each point has a nbhd where the preimage under p is the disjoint union of open sets in X tilde

[unreal stratus]

Smol pancakes

[hidden crag]

and p restricted to each of these open sets is a homeomorphism

so basically you locally have sets who each look like the neighbourhood around your points "floating" on top of each other

[odd flame]

ok well first of all are arc components and paths interchangeable here

[hidden crag]

to be honest i have no idea what those are

[odd flame]

[hidden crag]

who uses arc stuff

[odd flame]

massey it seems

,w covering space

[gentle osprey]

Results provided by WolframAlpha

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

Requested by stμ₂dying#9923

[odd flame]

and wolfram

[steel glen]

i may be forgetting something but i recall covering spaces as the cont. surjective projection map p:E->X, where each x in X has a local homeomorphism from some open set U of x and it's preimage

[hidden crag]

wtf

[odd flame]

ok thinking about it like this

how does this work? isn't U a nbhd in X?

[hidden crag]

that's where i said comes into action

"looks like" means is homeo to

and all the leaves of your preimage are homeo to your nbhd U

[odd flame]

ohhhhh i think i see it

things in X have as a preimage a disconnected sum of similar things in E

[hidden crag]

yeah

play around with some examples

[odd flame]

projective plane and S^1 make sense

[unreal stratus]

owo

[odd flame]

im looking at this bc i have to show that something isn't a covering actually

I x I -> I given by (s,t) \mapsto s

im not 100% sure i see it though

take a nbhd of I, it's preimage in I x I is a rectangle

i dont see which part of the defn is broken

[hidden crag]

fiber of a point isn't discrete

[odd flame]

fiber?

[hidden crag]

p^-1({x}) for some x

[odd flame]

is that a covering-specific word for preimage?

[hidden crag]

that's a common (naive) set theoretic word

[hidden crag]

not part of the definition but turns out to be true for every covering space

even every local homeo

[odd flame]

ok silly question which exposes my bad point-set basis but

wdym discrete?

hausdorff?

[hidden crag]

we say a space is discrete in another if the subspace topology is the discrete topology on that subspace

think of it as points lying isolated somehow

for example the integers in R

[odd flame]

that's why i guessed hausdorff lol

[hidden crag]

discrete spaces are indeed hausdorff

[unreal stratus]

Prove it

[odd flame]

[unreal stratus]

That does not count as a proof

[gritty widget]

yeah. A comment is that some authors additionally demand that the subspace is closed

but I don't think that's very common

[potent sky]

if I have a genus g_2 surface and genus g_1 surface, where g_1 -1 divides g_2 - 1. how can i show there's a covering of the genius g_1 surface by the genus g_2 surface?

eh i guess i get it

[hollow grail]

Thankss

[hollow grail]

Alright

[pseudo ocean]

Wait what is meant by "leaves" here?

[gritty widget]

i don't think it's topological spaces themselves that we call invariants, but the properties of them

[pseudo ocean]

That doesnt make sense. A Topological space is an object. It has certain properties we call invariants. These invariants dont change under homeomorphisms.

Again: a topology is an object

I dont think theyre considered invariants

Because, obviously, they change when we perform construction such as product, subspace, or quotient topologies

When we have a topological space with a topology and try to construct a new space from it, the topology changes

[pseudo ocean]

^

But that doesnt necessarily mean the topology IS an invariant

Well you can argue that the topology itself doesnt change at all, but ultimately it's equipping with the set will change when we do some construction on the space.

Oh wait

Think of it this way: the topology is like the "geometry" of the space

Or it's underlying structure

When we change the space

The topology changes

[potent sky]

does anyone know how to see that a non-degenerate 2x2 integer matrix is a covering map from T^2 to itself? How do I compute that it's degree is the determinant of the matrix as well?

[plain raven]

hot take: The word invariant is unnecessarily confusing

Nobody knows wtf it refers to

we should just say 'functor'

[little hemlock]

when we say a problem about vector bundles is "local", say in Atiyah's k-theory notes, we just mean that the truth of the statement for each element of an open cover implies the general statement, right?

[woven badge]

I've seen people define 'topological invariant' as 'a functor from Top' 😅

[gaunt linden]

I naively thought "invariant" was any function of topological spaces that's constant on homeomorphism classes. But "functor" sounds like it demands more.

No, wait. If I have a function that is constant on homeomorphism classes, I can make its codomain into a category by declaring its elements to be objects, and having exactly one non-identity morphism between any pair of objects. My function then becomes a functor by mapping every homemorphism to the identity, and every other continuous map to the appropriate non-identity morphism.

No, wait again, that doesn't work because sometimes non-homeomorphic maps can compose to produce homeomorphisms.

[gritty widget]

I think the point was more that we usually have a property of some spaces, not necessarily topological, to which we apply, say, forgetful functor. If the images by this function are homeomorphic, and one space has property P, then so does the other

[gaunt linden]

That doesn't match what Xonram said, where the functor is from Top, though.

[gritty widget]

Sure. I just thought it's a bad definition

But maybe it does cover a lot of useful cases

[gaunt linden]

Which one, mine?

[gritty widget]

Defining topological invariant as a functor from Top

[gaunt linden]

Ah.

[gaunt linden]

In my intuitive understanding, a prototypical example of a topological invariant is dimension (e.g. defined as the supremum of all kappa such that R^kappa embeds into the space in question, if we want a definition that applies to arbitrary spaces). Can that be viewed as a functor?

[gritty widget]

Interesting, I don't think it will coincide with any familiar concepts of dimension like dim, ind, Ind

[gaunt linden]

It's intended to do for manifolds, and I don't really care what it does to non-manifolds.

I'd be happy with an answer for any of the standard concepts too.

[gritty widget]

Can't you just look at it as a functor into N?

With two arrows between different natural numbers n, m, one from n to m and one from m to n

Sounds artificial though

[gaunt linden]

I suppose so, but then all objects in the target category are isomorphic, which doesn't really feel like it's the right spirit -- in that case, being functorial doesn't even force homeomorphic spaces to map to the same number.

[gritty widget]

Maybe Clerk and Xonram had some construction in mind but it's probably the best to just wait and see if there's any explanation, since they didn't provide us with any

[potent sky]

does anyone know how to see that a non-degenerate 2x2 integer matrix is a covering map from T^2 to itself? How do I compute that it's degree is the determinant of the matrix as well?

[plain raven]

I hadn't really thought about dimension or something like that. In this case it is a functor from the underlying isomorphism groupoid on manifolds to the discrete category of natural numbers, I guess

but yeah my problem is more that 'invariant' is a vague term to me

is it just any homeomorphism invariant

[solemn oar]

Hint: Look at the fundamental polygon.

[potent sky]

@solemn oar you mean the identification space?

im just confused because this chapter of the book is talking a lot about like group actions and somehow making covering maps out of them

my bad on the highlight btw

[solemn oar]

I mean this: https://en.m.wikipedia.org/wiki/Fundamental_polygon

Can you see how you can use this to view a parallelogram tiling (with integer vertex coordinates) of the plane as a map T2 -> T2?

Once you see this you're very close to answering your question.

[civic verge]

I did it this way X is an open set in Rn, so if a belongs to X then there exists r>0 such that the open ball of center a and radius r is contained in X or is completely contained in X, therefore any point in X cannot be a boundary point of X' complement, since the ball is contained in X, therefore it does not intersect X' complement finally X cap partial X is equal to empty.

[tidal lynx]

what is the connection between surjections X -> Y and equivalence relations on X ?

sure, if X has an equivalence relation ~, then the canonical map from X to its set of equivalence classes is a surjection

but I don't really see how every surjection induces an equivalence relation

oh

if f: X -> Y is the surjection

then we can define an equivalence relation ~ by x ~ y iff f(x) = f(y)

so did I just establish a bijection between the set of surjections from a set and the set of equivalence relations on a set?

there's probably a categorical way of saying this but idk what it is

[potent sky]

the equivalence relation you defined is correct

and yeah, any surjection can be use to make a quotient space

[gritty widget]

bijection?

[tidal lynx]

ok def not a bijection yea

[potent sky]

he just means the surjections are 1-1 with equivalence relations

like every equivalence relation can be realized as a surjective map or whatever

[tidal lynx]

well maybe if you fix Y as a set with the same cardinality as X

[potent sky]

it doesn't have to be the same cardinality

err

[tidal lynx]

if Y is smaller cardinality then you don't get the equivalence relation of regular equality

not enough guys to hit

[potent sky]

what do you mean

regular equality?

[tidal lynx]

like the equivalence relation where x ~ y iff x = y

and the equivalence classes are one element sets

[potent sky]

oh well that's different

i didnt mean you can do this for a fixed Y

[gritty widget]

what for

[tidal lynx]

just cuz idk

[potent sky]

but im trying to think of a good example where the cardinalitys are different

all of the quotient spaces i work with are usually some manifolds with uncountably many points

[tidal lynx]

{∅, X}

[potent sky]

idk, what are you worried about specifically again?

[tidal lynx]

oh my original question is mostly resolved now I think

I was just confused when u said "surjections are 1-1 with equivalence relations"

but the details don't totally matter for my original question

[potent sky]

well that's what you were asking

if every equivalence relation could be realized as a surjective map into some set

[gritty widget]

that's because it's not really true
every surjection from X induces an equivalence relation, and every equivalence relation induces such surjection
the maps are retraction and section respectively
but it's not a bijection or anything

[potent sky]

you just said it's a bijection and then said it isn't a bijection

although i guess you could have two different surjections come out to the same equivalence relation or something

shrug

[gritty widget]

No. There are two maps, say f and g, with fg = Id but gf actually isn't Id unless X is empty

where by f I don't mean function but a class map

because it's defined on what's usually a proper class

[plain raven]

you have to consider surjections up to commuting isomorphism

so like, equivalence classes of surjections under the equivalence relation that $f : X\to Y$ is identified with $g :X \to Y'$ iff there exists a bijection $h : Y\cong Y'$ such that $h\circ f = g$.

[gentle osprey]

diligentClerk

[potent sky]

ah yes that's a nice solution

[little hemlock]

im confused how they are getting this last part here. Since dim(ker(phi_x)) is locally constant, don't we get that rank(phi_x) is locally constant as well by rank-nullity?

giving equality instead of <=

at the very least, i don't see how the given proof shows that inequality. anyone have an idea?

[cosmic socket]

I don’t understand the following theorem from obstruction theory:

Given a map f:X^n -> Y and a cochain d \in C^n(B, A; pi_n(Y)) there exists a map g:X^n -> Y such that the difference cochain d(f,g) =d

[plain raven]

I found the treatment of obstruction theory by Hu to be quite readable, I would recommend it if you are having difficulty.
"Homotopy Theory" - Hu

[cosmic socket]

Thank you so much that was very helpful

It was kinda obvious in retrospect but it made it incredibly clear thank you

[hidden sentinel]

hello, I'm trying to understand what open cover admitting a subcover means in the context of compactness

[gritty widget]

(finite, in compactness) subset of the cover which is also a cover

[hidden sentinel]

so if i look at (0,1), and the open cover denoted by union from 1 to infinity of (1/n,1), why does this not admit a finite subcover?

[gritty widget]

what do you think?

[hidden sentinel]

oh wait

oh wait

my bad

i get it

so u can't have a finite number of subcovers for this open cover

i get it my bad

[gritty widget]

you can't have a finite number of elements in a subcover

[hidden sentinel]

yes makes sense

thank you nevertheless!

[trail charm]

Let E and F be Banach spaces and f : E -> F a continuous linear map. Then for each x in E, we have f'(x) = f. (Lang Differentiable and Riemannian Manifolds, p. 9)
why is this true?

[unreal stratus]

You can check that f satisfies the correct property by plugging it into the definition

[trail charm]

really sorry but not quite sure what definition to plug it into

definition of the derivative?

[unreal stratus]

Yes

Okay so how is the derivative of f at x defined (if it exists)

[trail charm]

$f$ is differentiable at a point $x_0 \in U$, where U is open in E, if there exists a continuous linear map $\lambda : E \to F$ such that
[ f(x_0 + y) = f(x_0) + \lambda y + \phi(y) ]
for small $y$, then $\phi$ is tangent to $0$. Then we say that $\lambda$ is the derivative of $f$ at $x_0$.

[gentle osprey]

anamono for anamono

[unreal stratus]

Okay that's an even nicer thing lol

Wait "tangent to 0"

[unreal stratus]

Okay sure then we're on the same page

Yes

Okay, so now if f is linear, it is hopefully clear enough that such λ exists

Indeed, what is f(x0 + y)?

[trail charm]

f(x0) + f(y)

[unreal stratus]

Yes

So what can we take λ to be

[trail charm]

f?

[unreal stratus]

Yes

[trail charm]

ohh

[unreal stratus]

In fact like

The derivative at x basically lets us basically a local linear approximation in a neighbourhood of x

Like if the derivative of a function f at 0 is λ and f(0) =0 then f(y) = λy + small error for all small y

And so if we plug in something which is genuinely linear, we just get out the derivative

[obtuse meteor]

*and f(0) = 0

[unreal stratus]

Sorry I emant to say that lol, thanks

[trail charm]

ohh okay that makes sense

yeah that clicks now

i think the "tangent to 0" part is kinda tripping me

[unreal stratus]

Yeah I'd never seen that notation lol

[trail charm]

still not quite clear what that resembles

[obtuse meteor]

it's essentially saying varphi shrinks a neighborhood of 0 by a lot

in the classical normed case this is clear, and the above is just a refinement of that to TVS

[unreal stratus]

Heuristically you can think of φ as quantifying the error of the linear approximation and the fact φ goes to 0 faster than λy shows that we have the best possible linear approximation and φ is some "higher order" error

[trail charm]

ah yeah okay both of u are making sense

thank u guys!

[odd flame]

i asked about this before but

for the first part of this question

the defn i have is that a map p: X -> X is a covering projection if X has an open cover of connected sets U which satisfy the condition that p^-1(U) is a disjoint union of open sets mapping homeomorphically to U by p

timo gave a reason of why the p here isn't a covering projection before but i dont see how it violates that defn

does compactness/finite cover have anything to do?

[tawdry valve]

your definition as stated is correct, you don't need finiteness/compactness

[odd flame]

the preimage of an open set of I in I^2 by this p should be the union of vertical lines starting at x \in (a,b) for each (a,b) in any open cover

but isn't each line disjoint from the next..?

[tawdry valve]

so one thing is that you want to write p^{-1}(U) as a preimage of open sets, and vertical lines aren't open

[odd flame]

ohhhh

union of squares then

in which case i can visualize them not being disjoint

[tawdry valve]

yup, and you can try to formalize this with connectedness

another point too

is that p restricted to the each square should be a homeomorphism, but this projection map will crush things

it'll flatten the squares, so it isn't injective, so not a homeomorphism

[odd flame]

that actually makes more sense than the disjoint thing

in that case then there cannot exist a covering of IxI over I right?

[tawdry valve]

correct

I think it might be a little tricky to prove, but it really doesn't feel true

yeah I'm actually a little stumped. Proving that your specific p(s,t)=s is NOT a covering map isn't too bad, but proving that no p works might need some algebraic topology... hmm

[odd flame]

that's okay i was mostly just curious about if it was a general result

[tawdry valve]

I think it is, I would be really really surprised if it weren't lol

[odd flame]

i thought this would be enough though

though i guess if you use a different topology on I x I...?

[tawdry valve]

so an argument carried out along those lines will help to show p(s,t)=s is not a covering map, since the preimage of some interval of I will be a rectangle

but to show that there is no possible covering map I^2 -> I, I would need to handle when p is really strange

so for a general p, the preimage need not be like a rectangle or something

[odd flame]

ahhh i see

,w path lifting

[unreal stratus]

Lol

[unreal stratus]

Universal covers are unique up to homeomorphism ig

[tawdry valve]

oh true

[unreal stratus]

But yeah that does use AT ofc aha

But yeah I meannn

I reckon it can be done without much tefchnology

Well ig we wanna say no (nonempty) open subset of R^2 can be homeomorphic to an open subset of R

And that sounds doable with just path connectedness

[tawdry valve]

yeah something like removing points disconnects an interval in R but not an open subset of R^2

[unreal stratus]

Indeed

Yeah too sleepy to think through details rn but nah yes that'll work

Because any path connected open subset of R^2 will remain locally path connected after you remove a point cause you can just go round it ig? etc

[tawdry valve]

yup something like that, the details are an exercise for the reader

[odd flame]

does this imply there is no path lifting then?

[tawdry valve]

covering maps will have path lifting, but that doesn't mean that not covering maps don't have path lifting

[odd flame]

im still a bit unclear as to the defn of a lifting though

it's only defined through the theorems (at least in my mostly accurate notes)

is a lifting a particular map or a property?

[tawdry valve]

so a lift is a particular map, and "path lifting" is a property

hmm maybe I don't like using bolds, normally I'd use tilde's for texing it

oh yeah i can tex

given a covering map $p: \tilde{X} \to X$ and a path $\gamma: I \to X$ (so $\gamma$ is "downstairs"), a lift of gamma is a path $\gamma: I \to \tilde{X} (so $\tilde{gamma}$ is "upstairs") such that

$p \circ \tilde{\gamma} = \gamma$

[gentle osprey]

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

[unreal stratus]

french

[hidden crag]

commutative

[tawdry valve]

thank

[unreal stratus]

Ew copied straight from quiver + with sigma

[hidden crag]

yeah my bad i'll just concern myself with tikz atrocities next time

[unreal stratus]

Hug

[odd flame]

https://tikzcd.yichuanshen.de/ clerk showed me this

[hidden crag]

quiver seems simpler

[unreal stratus]

Simplicity bad

[odd flame]

make everything as convoluted as possible

[odd flame]

so the lift is sigma tilde here right

[hidden crag]

idk i didn't read the context

i just thought i'd make josephs life easier

[tawdry valve]

yup sigma tilde would be a lift of sigma

[hidden crag]

you could also show that every covering space has discrete fibers

but that's more of a trick here i guess and not what's supposed to be learned from the exercise

[odd flame]

ok so in english, theorems tell us that given a covering p: X -> X, a path in X can be "brought up" to X when we get certain fundamental group conditions right?

[hidden crag]

no paths can always be lifted

this follows from I being simply connected

generally for a lift of a map to exist you'd need that the subgroup induced by the induced homomorphism is a subgroup of the induced subgroup by the covering projection

it's german but i guess but it's probably readable if you know the context

a) and b) are equivalent

oh Y is path and locally path connected

[odd flame]

\leq means subset i presume?

[hidden crag]

subgroup

i'm not responsible for the notation

[odd flame]

why's this true though?

im probably missing a thm/corollary

[hidden crag]

because the induced subgroup is trivial

[odd flame]

so I is simply connected => pi(I) is trivial => no non trivial subgroups

[hidden crag]

the induced map is a map from the fundamental group of I (trivial) to the fundamental group of B

there is only one map out of the trivial group

the trivial one

[odd flame]

ohhh i see

so i the diagram you sent Y = I and B = I

[hidden crag]

no

Y=I

B=B

https://cdn.discordapp.com/attachments/496785752936546304/1082414893694713937/image.png

[odd flame]

oh i mean for this problem

[hidden crag]

oh lmao

yeah i guess

[odd flame]

since it's asking about I x I covering I

i mostly see it though

thank u timo

[hidden crag]

you're welcome

funnily enough i have an exam about this tomorrow

[odd flame]

mine is in two weeks

this is just hw

good luck tho

[hidden crag]

thank you

[unreal stratus]

gl sir

Pain

Hope ur exam goes well to both of u

[odd flame]

wait im a little confused again

[odd flame]

why is it possible to get liftings in the first place then

that's a werid question uhhhhh

[odd flame]

if this is our diagram

[unreal stratus]

lift

[odd flame]

since pi(I) is trivial then aren't both sigma and sigma bar maps out of the trivial group

[hidden crag]

you're not comparing those

oh wait is this about your example again

[odd flame]

both example and in general i guess

bc when we're talking about paths we're implicitly talking about I

[hidden crag]

in general you're comparing the induced map of sigma and p

[unreal stratus]

why is this using locally arcwise-connected

[hidden crag]

yeah that's it

[pseudo ocean]

ok

this is probably a simple question

but is the complex circle really homeomorphic to the real-valued circle? What is the homeomorphism?

im thinking for every $z\in \mathbb{T}$, $z\mapsto (x,y)$ s.t $x^2 +y^2 = r^2$ for some $x,y,r\in \mathbb{R}$

[gentle osprey]

messyinterval

[gritty widget]

all i am going to write is z = x + iy and let you figure out the rest

[odd flame]

,av tterra

[gentle osprey]

TTerra#5291's Avatar

Click here to view the image.

[odd flame]

im so fucking pUMPed for my homework

i fucking lOVE topology

(i actually do alg top has been fun)

[odd flame]

is this the "natural" covering of a wedge sum of circles

[pseudo ocean]

hmm

[obtuse meteor]

wdym by natural

if you mean the universal cover, then no this is far from it

[odd flame]

yeah that's what i meant

what makes a cover universal

[obtuse meteor]

if it's simply connected

[pseudo ocean]

$z\mapsto (\text{Re}(z),\sqrt{ \vert \text{Im}(z)^2 \vert})$

[obtuse meteor]

you'll see that this is not simply connected, take the standard unit square

there's a hole in the middle of the standard square in this space

[pseudo ocean]

@gritty widget how's this?

[gritty widget]

you will never reach points with negative y coordinate

[odd flame]

what does simply connected imply

it keeps coming up

[obtuse meteor]

Er

[odd flame]

something about loops i presume

[obtuse meteor]

It means the fundamental group is trivial

aka every loop is contractible

[odd flame]

i guess that's a really general question lol

[obtuse meteor]

Try and think about what the universal cover of the wedge of two circles should look like

[odd flame]

in other words no nontrivial loops right

[obtuse meteor]

after you've thought for a while come back and look at this picture

[gaunt linden]

Dang, exactly the image I was just about to link to.

[gentle osprey]

messyinterval

[odd flame]

mom im scared

[gritty widget]

the exact same problem. also you don't need to spam the tex bot

[pseudo ocean]

sorry. wait what's wrong?

[gritty widget]

.

[pseudo ocean]

i kinda fixed it by squaring it and taking absolute values

[gritty widget]

no, not really

[obtuse meteor]

it's still always going to be positive

[odd flame]

that's the opposite of fixing it

[pseudo ocean]

hmm

dang i concende

[obtuse meteor]

:)

[gritty widget]

did you think about this?

[odd flame]

before i look at spoiler

[gritty widget]

i urge you to write out what |z| = 1 means

[odd flame]

what makes a universal cover universal

[pseudo ocean]

hm holdo n

[odd flame]

some commutative diagram i presume?

[obtuse meteor]

if X is a "nice enough" (there's some stupid conditions) space with Xtilde -> X it's universal cover

then for any connected cover Y -> X

there's a covering map Xtilde -> Y

[gritty widget]

semiquasisomewhatlocally simply connected connected path connected

[obtuse meteor]

so that Xtilde -> Y -> X composed is the universal cover (aka so the triangle commutes like you want)

[pseudo ocean]

well for z=x+iy, |z|=\sqrt{x^2 + y^2}

isnt it?

[gritty widget]

ok

[gritty widget]

yes. and (x, y) is in the "real circle" if...

[obtuse meteor]

I mean

this is a very nice condition

[pseudo ocean]

x^2 + y^2 = 1

[gritty widget]

ok

i am not going to give any more hints

i just want you to write out everything so far

[pseudo ocean]

so for any z=x+iy such that |z|=1, map z to a pair (x,y) s.t x^2 + y^2 = 1

-?

[gritty widget]

the details are for you to grind out

[pseudo ocean]

wait, what details? there's more???

[gaunt linden]

What is "the complex circle", anyway? The subset of C×C that solves z²+w²=1?

[odd flame]

there usually is

[gritty widget]

tropo

[pseudo ocean]

the set of all z in C s.t |z|=1

[gaunt linden]

Ah.

[gritty widget]

interpret "details" as "giving a careful definition and argument that this is a homeomorphism"

[pseudo ocean]

i claimed in a paper of mine and in my notes it's homeomorphic to the real unit circle

just wanna know how it is

[gritty widget]

a paper of mine

[pseudo ocean]

yeah...?

[gritty widget]

troposphere can help you with any further questions i need to do my homework

[odd flame]

bye tterra