point-set-topology

[gritty widget]

In this solution.

—-

I’m gonna see here if we’ve any trivial solution of this equation where δab doesn’t need to be the same than the metric tensor

[pearl holly]

Could I post a question here?

[cedar pebble]

yes lol

[fading vale]

Yes post

[pearl holly]

Hmm when does a space $X$ have a simply connected covering? I've heard that the space must be path connected, locally path connected and semilocally path connected, but isn't semilocally path connected enough? Like if you have a covering $p: \tilde{X} \to X$ then every $x \in X$ has a nbh $U$ that is evenly covered by $p$. Each loop in $U$ lifts to a loop in $\tilde{U}$. Now if $pi_1(\tilde{X}) = 0$, then this lifted loop is nullhomotopic in $\tilde{X}$. So now just "project" this loop down to $U$ and this loop is now nullhomotpic in $X$. So if we want that $pi_1(\tilde{X}) = 0$, then the semilocally path connected criterion is only needed, right?

[gentle osprey]

Tokidoki ✓

[gritty widget]

No problem

[cedar pebble]

connected locally path connected semilocally simply connected is enough

[fading vale]

theyre asking about the non locally path connected case

[pearl holly]

yeah I don't see why the path connected and the locally path connected things are needed in this case

[gritty widget]

@torpid geode Look at you private, I’ll send it to you

[cedar pebble]

yea so if you drop any of these assumptions some not great things happen

[fading vale]

its possible but without locally path connected its not guaranteed and without path connected you end up just picking a path component i think

[cedar pebble]

so first of all, note that a path connected locally path connected space has a universal cover precisely if it's semilocally simply connected

[pearl holly]

ah okay, so that's "if and only if", right?

[cedar pebble]

yes

if you remove the path connected assumption then it's not even clear what the definition of a universal cover should be

you can extend the definition to mean like, take universal covers for each of the path components and then take the disjoint union, then it's fine

[pearl holly]

yeah okay I see. Thank you so much!

[cedar pebble]

yea the main issue is just that some theorems kinda break, or at least have to be reformulated in the general situation. But you're right that the semilocally simply connected assumption here is really the important one

[fading vale]

if you want an example to play around with try thinking about the cone of the hawaiian earring

it is semilocally simply connected but not locally path connected

[cedar pebble]

it is locally path connected

it's just not semilocally simply connected

[fading vale]

The cone isnt?

[cedar pebble]

ooh the cone

[pearl holly]

[cedar pebble]

sure, although it's an example that doesn't have an interesting fundamental group no matter how you define it

the Hawaiian earring on its own is kinda fun because there are multiple things you could mean by the fundamental group here

[fading vale]

wait it should not be hawaiian earring it should be the comb

the comb is not locally path connected but its cone is semilocally simply connected

[cedar pebble]

oh right

[fading vale]

Moth vs point set

[cedar pebble]

deletes message about this being a good example

[pearl holly]

oh yeah, there was an exercise about the comb I think in this section

[fading vale]

tokidoki to help you in coming up with examples: cones are locally path connected iff the space you are taking the cone of is

and they are always semilocally simply connected

[pearl holly]

Okay great! I will keep that in mind, thanks!

[cedar pebble]

ahh for the sine circle yea

that's a fun definition

[fading vale]

society if discord loaded images

I havent actually seen an example of a semilocally simply connected space with no universal cover though

[cedar pebble]

well if it's locally path connected but not path connected things are still mostly okay

idk about the locally path connected assumption

[fading vale]

So true

[fading vale]

yeah i have not seen an example of a semilocally simply connected not locally path connected space with no universal cover

[cedar pebble]

unbased

[fading vale]

i cant believe no ones constructed one on MSE

[cedar pebble]

yea I can never remember how all the shape theory stuff works lol

[fading vale]

or at least that i cant find it

maybe i should ask

[cedar pebble]

oh comb space works lol

[fading vale]

Is there a proof it has no universal cover

[cedar pebble]

e.g. https://math.stackexchange.com/questions/20634/comb-space-has-no-simply-connected-cover?rq=1

[fading vale]

I see

Theorem: Let E be any space and F be a metric space. Let K⊂E be compact, f:E→F be continuous, locally injective near every point p∈K, and (globally) injective on K. Then f is also injective on some open neighbourhood N of K.

Interesting

[honest narwhal]

Okay time 2 shine

Gonna just start this chapter from scratch but reference my ramblings from the first n pages

[gentle osprey]

Sloth King Daminark

Sloth King Daminark

[honest narwhal]

Makes it a homogeneous space

Distractions 😦

Anyway

[gentle osprey]

Sloth King Daminark

[honest narwhal]

So the action is transitive

[gentle osprey]

Sloth King Daminark

[honest narwhal]

And we have our Riemannian metric

[gentle osprey]

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

[honest narwhal]

So okay what's the differential of a Mobius element?

[gentle osprey]

Sloth King Daminark

[swift fjord]

dami are you answering anyone or just talking to yourself?

[plain raven]

i think you can talk about all this with homology too

like homologically locally connected

[honest narwhal]

Talking to myself

[swift fjord]

ah ok, I just wanted to ask a question real quick

[gentle osprey]

Sloth King Daminark

[swift fjord]

Dealing with the implication of equal degree meaning homotopic.

So 3.18 just says that 2 paths to S1 are path homotopic iff they have the same degree, 1.3 says that rotations are homotopic to the identity, and 1.6 gives equivalence between a map from S^n being nullhomotopic, being nullhomotopic relative to some basepoint and being able to be extended to the unit disc. Now I can clearly see using 3.18 and 1.3 I can show that the rotated paths are path homotopic, and that the non-rotated paths are (freely) homotopic, but I can't seem to find where 1.6 comes in (Even if I work with paths as maps from S^1), and either way how I can guarantee the homotopy is a closed path at any time t.

cuz even working with the equivalent definition of closed paths, nothing here is nullhomotopic unless the degree is 0

[swift fjord]

Ok I think I figured it out

but I need a sec

[drowsy yew]

eh 😦

idk what that is

[cedar pebble]

I mean you’re computing simplicial homology here

So presumably this comes up somewhere

[drowsy yew]

it does kinda come up

but it never mentions what H_* is

i just dont understand @cedar pebble

[cedar pebble]

H is Z/B

In the notation you had previously

Cycles moduli boundaries

[drowsy yew]

ohh i understand now

so since S_* is how you derivative Z and B

we call H_* = Z _ * /B_ *

ae

u get it

xd

[cedar pebble]

Yea so S is a semi simplicial set here, and you can Abelianize this to get a chain complex usually called C

Among C you have Z and among Z you have B

[drowsy yew]

thanks :p

[cedar pebble]

The process of linearizing a semi simplicial set to get a chain complex is part of the Dold Kan correspondence

[drowsy yew]

:petngroupoid:

[swift fjord]

Ok so here's my solution which doesn't use 1.6 whatsoever which makes me think I did something wrong but whatever:

Equal degree implies homotopic: Suppose $f,g$ have equal degree, then by definition $\text{deg}(R\circ f)=\text{deg}(R' \circ g)$ Where $R,R'$ are the rotations by $-2\pi \alpha, -2\pi\beta$ radians (Where $a=e^{2\pi i \alpha}, b=e^{2\pi i \beta}$) respectively. Then by 3.18 we have some homotopy
$$F:R\circ f \simeq R' \circ g \text{ rel } {0,1}$$
From exercise $1.3$ we know that $R\simeq 1_{S^1}\simeq R'$, therefore there's some homotopy $G:R\simeq R'$. We define the function:
$$H:I\times I \rightarrow S^1, , H(x,t)=\frac{F(x,t)}{G(1,t)}$$
Then we shall show that this is the required homotopy between $f,g$:

First, we note that $R(t)=e^{-2i\pi\alpha}t$, therefore $R(f(x))=e^{-2i\pi\alpha}f(x)$ and $\frac{1}{R(1)}=e^{2i\pi\alpha}$, similarly for $R'$, we have:
$$H(x,0) = \frac{F(x,0)}{G(1,0)} = \frac{R\circ f(x)}{R(1)} = f(x)$$
And similarly $H(x,1)=g(x)$. Finally, we show that for each $t \in I$, the homotopy is a closed loop, and indeed, let $t \in I$, then (recall that $F$ is a homotopy rel ${0,1}$):
$$H(0,t)=\frac{F(0,t)}{G(1,t)}=\frac{1}{G(1,t)}$$
And similarly
$$H(1,t)=\frac{F(1,t)}{G(1,t)}=\frac{1}{G(1,t)}$$
Therefore $H(-,t)$ has the same start and endpoint and is therefore a closed path, so $H$ is indeed our required homotopy.

I did the other implication similarly so if this does work then i'm done

i'd be happy if someone could take a look

[gentle osprey]

ShiN

[swift fjord]

Note that Rof, R'og are in particular paths based at 1 in S^1, therefore we have that F(1,t)=F(0,t)=1 for any t in I

[flint cove]

Stupid question, but in a simply connected space, can we guarantee that there ia a pointed homotopy between a closed loop through a and the constant loop a?

Because trying to contract the „defect“, i.e. the path t \mapsto F(0,t) which the endpoint takes, to a point, doesn't help me since I don't know if this contraction homotopy preserves a

[empty grove]

Yeah, though usually this is taken to be the definition of simply connected (fundamental groups use pointed homotopies)

Use the lemma that any nullhomotopic map from S¹ into X extends to a map from D² to X

Assuming that your definition of simply connected is path connected + every map from S¹ is nullhomotopic

[flint cove]

Okay then I just identify D² as the quotient of a rectangle whose left edge is my path and whose other thre edges are glued together

The map from this rectangle should precisely be my homotopy

*my desired homotopy

[empty grove]

Yeah I think that works too, but I was thinking that you can draw a pointed nullhomotopy of the boundary on D²

The original map is a composition S¹ → D² → X involving the extension

Keep the second map constant

[empty grove]

Contract the first map along this

[flint cove]

Ohhh

[empty grove]

Horizontal composition of homotopies

[flint cove]

I'm not quite sure why the extension lemma holds tho

[empty grove]

Identify D² with the suspension of S¹

[swift fjord]

You can just generally prove that any nullhomotopy from S^n is a pointed nullhomotopy for any point in S^n by going through the equivalence with the extension to the unit disc

[empty grove]

Universal property of the quotient

[flint cove]

So I'm trying hard to view the contracting homotopy as a map from not a rectangle, but a cylinder S¹xI to X (which is constant on the right circle hence the suspension)

But I don't quite see how, since we cannot glue the top and bottom edges together in said homotopy

[empty grove]

Wait in suspension you only glue the top edge of the cylinder right

oof

I should have said cone I think?

my bad lol

[flint cove]

Oh we're talking orthogonal to each other lol

The time part in the homotopy goes to the right for me

[empty grove]

😵‍💫

Cannot comprehend

Imagine thinking of (1,0) as lying on the y axis

[flint cove]

Oh, I just remember that I dont usually do that, I just got confused because I tried gluing a homotopy by 90deg to another homotopy
I guess I need a break to re-sort my mental imagery

Thanks so far 😌

[empty grove]

[marsh forge]

Wait @flint cove you mean come right

Cone

[reef shore]

Ye I said suspension lol

I confused the definitions

[marsh forge]

No lux also said suspension

A nullhomotopy of a map X->Y is witnessed by a map CX->Y

Where CX is XxI/~

And ~ collapses the copy of C at time 1 only

Equivalently this is half of a suspension

[flint cove]

Yeah we both said suspension and meant cone

[marsh forge]

Ah-do you still have a question abt it

[empty grove]

We were vibing max

[flint cove]

Yes but I'll have to come back later as I'm occupied for the next 2h

[pearl holly]

Okay so this is the exercise that I'm working with: Show that if a path connected, locally path connected space X has pi_1(X) finite, then every map X --> S¹ is nullhomotopic. [Use the covering space R --> S¹]

I don't really know what to do here tbh. Lets say that f: X --> S¹ is a map. Then this induces a homomorphism from pi_1(X) to Z and pi_1(X) is finite so this map can't be surjective. And now what? That doesn't really give me much. I also don't know how I'm supposed to use the covering map R --> S¹

[bronze lake]

The only finite subgroup of Z is the trivial group, which must be the image of pi_1(X) on the induced homomorphism. Then we can use the lifting property to go from X --> S^1 to R --> S^1. Then properties of R help finish the proof

[pearl holly]

wait why is the image of pi_1(X) a subgroup? Don't we need injectivity for that?

no wait lmao

[bronze lake]

I'm pretty sure that's just a property of groups under homomorphisms

[pearl holly]

yeah okay I see. Thank you so much!

[bronze lake]

glad to help!

[swift fjord]

I was able to use the hint to show that I can decompose any path (up to htpy) into a finite product of paths entirely contained in the Ujs, but this multiplication occurs in the fundamental groupoid so it doesn't really help me. I can't see how I can use the hint to decompose my path into yet again closed paths each of which is entirely contained in some Uj.

Of course once I have some path in Uj I can deform it into a closed path (All maps from a contractible space to a path connected space are freely homotopic) but I don't know if I can actually construct a homotopy from this to the closed path I started with

Is that actually the way he wants me to do it?

I think this can't be it cuz in that case i'm not using the information that the intersection of every Ui, Uj is path connected at all

Tho rotman sometimes has redundant information in his exercises

[winged badger]

I've asked this already in #help-2 , and someone said that here would be more suitable. The problem is as follows: #help-2 message

[swift fjord]

As I said I can deform the segments into closed paths, but I don't see why I need the bit about the intersection for this

The path segments should agree on intersections anyways

Wait actually it's possible that some of the path segments are contained in an intersection

Yea ok the way I constructed it I forgot about that but it's a possibility

I think

Ok so basically if $x_j$ is the intersection point, and $f_j$ is the path between $x_j$ and $x_0$ I wanna stick a $f_j \ast f_j^{-1}$ between them right? Then
$$f_{j-1}^{-1 } \ast \gamma_j\ast f_j$$ is a closed path and when I multiply them all together the $f_j$'s cancel (in the homotopy class) and i'm left with the original path

Right?

[gentle osprey]

ShiN

[swift fjord]

How did I not see that before

Like I thought about doing that but for some reason didn't think it would work

Thanks!

Yea that's like a standard group theory trick tbh

Yea seems useful

Tyvm

[drowsy yew]

is this correct for the question: "prove that there exists two automorphisms on a 2-sphere such that they have an equivalent homology"

wait wa

wait oops

I thought it meant like

a ball

eee

would this be correct if it was a 3-ball?

t_{S^2} = P(S^2)

(the topology is equal to the power set)

yeas

f

bruh

my wifi decided to have a heart attack

had to restart my pc lol

couldn't u say it is though?

oh wait nvm

wdym

like wdym using the euclidean topology

nah but wdym by Using it

eh

i dont understand

[cedar pebble]

There’s the standard topology on S^2 coming from its embedding into R^3

[drowsy yew]

can u tell me

the prereqs

for it

?

[cedar pebble]

Point set topology

[drowsy yew]

is that it?

[cedar pebble]

Yes, if you know basic point set topology (and standard examples, including things embedded in Euclidean space or manifolds) then yes

But knowing basic stuff about basic examples is kinda important

[swift fjord]

I think this might be a language barrier issue?

[drowsy yew]

iik what u mean by its embedding in R^3 but i dont understand like

what youi mean by

using it

[cedar pebble]

I mean S^2 has a topology induced by this

That is not the discrete topology

[drowsy yew]

ohhhhhh

That's what you meant

sorry

I didnt understand

so R^3 induces the topology of S^2

[cedar pebble]

Yes

[drowsy yew]

how it induce

[cedar pebble]

This is the subspace topology

You should probably learn about this before worrying about AT

[drowsy yew]

oh.. ok thanks

[drowsy yew]

oh wait a minute

sorry I didnt quite understand what you meant

but I think you mean i set the topology of the 2-sphere to it's euclidean topology

which is induced by R^3

right

or I'm mjust wrong

[cedar pebble]

This is the subspace topology of the Euclidean topology on R^3 yes

[drowsy yew]

so you mean that topology of $S^2 = {S^2 \cap R : R \in \tau_{\bR^3}}$

[gentle osprey]

Muffin Man

[drowsy yew]

where t_R^3 is the euclidean topology on R^3

[cedar pebble]

Yes this is the subspace topology

[drowsy yew]

alright

Thank's

[winged badger]

did someone look at this?

[bronze lake]

I'm not familiar with that stuff, but it might be helpful to repost the full question here, so that people don't have to click on the link to read the question, then come back here to answer

[winged badger]

Let $X$ be a metric space, and $\mu$ a borel measure. Assume that $\mu(K)<\infty$ for each compact $K\subseteq X$, and that $X=\bigcup_{n=1}^\infty K_n$ where the $K_n$ are compact. Prove that $\mu$ is a regular measure on $X$.

[gentle osprey]

Phorphyrion

[flint cove]

What are you having trouble with, outer or inner regularity?

(not that I would know the answer to either off the top of my head)

[winged badger]

||Both actually... My idea was to define a sub-algebra consisting of the inner regular sets. I'm having trouble proving it's closed under completion though...||

||You could maybe define a sub-algebra of the sets that are both inner and outer regular, but then how do you show it contains some generating set of the Borel algebra?||

oh actually I got it

i think

[hollow harbor]

I'm just sitting here trying for the life of me to remember how to do this

And pretending as hard as I can that I don't need to retake measure theory in grad school

Oh well

[flint cove]

No joke please respond with spoiler tags @winged badger, I wanna think about this as an exercise because my measure theory is very much lacking 😄

[swift fjord]

Ok I don't understand, when you wanna take the boundary of a singular simplex, do you fix some orientation for all the simplices in $S_n(X)$? Or are we just always looking at the 'canonical' orientation

[gentle osprey]

ShiN

[swift fjord]

Also, there's no nice way to picture this beyond like, a 3-simplex right?

[swift fjord]

It seems like, in singular homology the concept of boundary lives solely in S_n(X), and it doesn't have a concrete representation like the boundary of a simplex?

Or at least, not a useful one

[swift fjord]

Alright yea so that's what I thought

I'm having a bit of a hard time grasping the induced orientation in higher dimensions tho

Like in 2d it's pretty simple but anything more and I don't see why this is the proper generalization

Why does the sign alternate

[fading vale]

I mean honestly "it makes the formulas we want to work work" is a valid answer to that

Im not sure if theres a nice geometric interpretation

[shy moss]

what means that the classifying space classifies G-principal bundles?

[trail tiger]

What are some minimalish regularity conditions we can toss onto a vector field so that it's induced by some global flow?

[hollow harbor]

You probably need locally lipschitz?

I could be wrong about that

[trail tiger]

Not a bad place to start

[hollow harbor]

I mean I think you can find counterexamples for any lower holder regularity

You should be able to get blowup unless I'm crazy

I might be crazy

[trail tiger]

It's not obvious but it feels... plausible.

[remote beacon]

Does this definition seem correct?

A set $A$ \textit{encloses} a region $R$ if there exists another region $R'$ such that $R\subseteq R'$ and $A = \partial R'$.

[gentle osprey]

KingArthur

[remote beacon]

Or are there better definitions?

[trail tiger]

i like this definition. you could appeal to the jordan curve thm or something but this is succinct

or the more general jordan-brouwer

[remote beacon]

I feel like this is a definition most people can understand

It's not that jordan brouwer is hard to understand but the theorem is a bit

complicated?

[empty grove]

Are regions always bounded? Because otherwise you get a weird definition of enclosing

[remote beacon]

I guess I was trying to imply that when saying region

[true robin]

Also, empty set encloses the entire space

[empty grove]

Yeah exactly, if you don't put the boundedness condition

[rancid umbra]

is there a way to show that if a metric space is compact, then it is complete without appealing to the axiom of countable choice via utilizing the notion of sequential compactness?

?? why the delete?

[swift fjord]

I said smth incorrect

I just look at the proof of cantor's intersection theorem and you choose a point from every set in the sequence

so it relies on countable choice

[rancid umbra]

hmm.

[swift fjord]

actually, I think maybe the implication that satisfying cantor's theorem implies complete doesn't rely on it, but i'm not sure

[empty grove]

You should be able to show that a Cauchy sequence viewed as a set is compact iff it contains its limit. Assuming this, take some compact space X, and a Cauchy sequence in it. If the sequence didn't have a limit in the space, it would form a closed subset and hence be compact which contradicts the first statement

Let me see if the first statement is true/uses choice though lol

[rancid umbra]

thanks guys. it’s no big deal anyway. i was just wondering. maybe there is some sort of equivalence between something and aocc here?

[swift fjord]

ok yea I think that's the general idea

[empty grove]

Yeah the lemma holds because you can embed the Cauchy sequence in its completion, and then take an open cover of it as a subset by taking complements of smaller and smaller open balls around the limit, and compactness is independent of the ambient space

So no limit → not compact

And limit → compact because any open set containing the limit will be cofinite

Doesnt seem to use choice anywhere

[rancid umbra]

gonna have to digest some vocab there but i’ll try and work through it that way too

[true robin]

Does compact implies sequentially compact even use countable choice?

[rancid umbra]

not sure. i’d have to run through the proof again. i remember choice was used somewhere in there tho

[empty grove]

Should be avoidable

[fallen whale]

Since X is a metric space so any any compact subset of X is closed hence any sequence in the compact space is convergent in that space so same for any Cauchy sequence also. Hence complete.

[rancid umbra]

compact iff sequentially compact has to use aocc somewhere. it’s not true otherwise i don’t think.

[empty grove]

You take a sequence and then produce a cover which consists of balls at each point such that no ball contains more than one point of the sequence. You can take the radius of this ball to be the inf of all radii that don't have this property, and then you don't have to invoke choice

[rancid umbra]

“any sequence in the space is convergent in that space”

do you mean any sequence in the space has a convergent subsequence in the space?

[empty grove]

Closed means convergent sequences converge within the set

Not every sequence

[true robin]

So i’m pretty sure seq compact implies compact needs aocc but the other way around i’m not so sure

[fallen whale]

Let X is the metric space and C is the compact subset of X. So, any sequence in C is convergent.

[empty grove]

Not true

0,1,0,1,... is not convergent in [0,1]

[rancid umbra]

take X = [-1,1] and the sequence x_n = (-1)^n

on the same page lol

[empty grove]

[fallen whale]

Yes yes!

I was wrong

[rancid umbra]

i’m just going to try to run through those proofs again lol. thanks

if that’s true, it avoids choice then

[fallen whale]

I mean any convergent sequence

Then the proof is OK right?

[true robin]

No

[empty grove]

Cauchy sequences need not be convergent

[rancid umbra]

unfortunately no. you have assumed that cauchy sequences are convergent, which is what we’re trying to show

[empty grove]

We want to prove that they are, in a compact space

[rancid umbra]

just mulling this over, i think because you can “separate” each point in the metric space from the sequence with an open ball (except maybe the point itself, but that’s fine), then the union of each of those open balls is an open cover which contains no finite sub-cover, since there can be at most n terms from the sequence in any finite subcover.

[empty grove]

Right that works

[rancid umbra]

i have to think about why your infimum arg. works. not quite seeing it rn

[empty grove]

It has the discrete topology on denumerably many points

Take sequence (x_n). If it had no convergent subsequences, then for each point y, you can take B(y, r_y) with r_y small enough that it contains no points of the sequence other than possibly y itself. This is an open cover without finite subcover, the only thing we have to change is the choice of r_y

To make the choice of r_y not depend on AC, take the set R_y of all radii r such that B(y, r) contains some point of the sequence different from y. This is non empty because distances are finite. Set r_y = inf R_y

Then r_y > 0 because there is some positive radius which isn't in R_y, and R_y should obviously be a ray in R

So we aren't leaving things upto choice

[rancid umbra]

ahh makes sense now.

thanks

[empty grove]

[fallen whale]

Yes now it is clear, nice proof

[shy moss]

what is an example of non-paracompact space?

[hollow harbor]

long line?

[restive shuttle]

hi i have a question about this page

so here they say that G = A A^T where A is the jacobian matrix

when I try to compute G for polar coordinates it doesn't work it's supposed to be line 1 of the matrix 1 0 line 2 0 r^2

but the jacobian matrix A for polar coordinates is line 1 cos theta -r sin theta line 2 sin theta r cos theta and if you compute A A^T then you won't have the matrix that I just mentioned

also on wikipedia they say that G = A^T A not G= A A^T so I'm confused https://en.wikipedia.org/wiki/Metric_tensor#Examples

[pearl holly]

Okay so this is the exercise that I'm working with: Let $p: \tilde{X} \to X$ be a simply connected covering space of $X$ and let $A \subset X$ be a path connected, locally path connected subspace, with $\tilde{A} \subset \tilde{X}$ a path component of $p^{-1}(A)$. Show that $p: \tilde{A} \to A$ is the covering space corresponding to the kernel of the map $\pi_1(A) \to \pi_1(X)$.

I assume that I should show that $p_*(\pi_1(A))$ is the kernel of the map. I don't know what the map $\pi_1(A) \to \pi_1(X)$ is since it isn't given but I assume that they mean the homomorphism induced by the inclusion $i: A \hookrightarrow X$. This is my work so far:

So let $[f] \in p_*(\pi_1(A))$. By a theorem in my book, this set contains all loops such that when you lift them, they become paths starting at a point in the fiber of the the basepoint in the original loop. I will ignore basepoints tho. This means that $\tilde{if}$ is homotopic to a constant loop since it is a loop in $\tilde{X}$ and this set is simply connected. Now just "project" this bad boy down (by composing the homotopy with $p$) to get that $if$ is nullhomotopic so $f$ is in the kernel.

Now let $f$ be in the kernel instead. We will show that $\tilde{f}$ (the lift) is a loop in $\tilde{A}$ which will end the proof. Let $\tilde{i}: \tilde{A} \hookrightarrow \tilde{X}$ be the inclusion. We know that $if$ is nullhomotopic, but $if$ is homotopic to $\tilde{i} \tilde{f}$ so this guy has to be nullhomotopic. Now $\tilde{if}$ is a loop so $\tilde{i} \tilde{f}$ must be a loop and so $\tilde{f}$ is a loop. I am very skeptical of the very last sentence here, I think that it is very wrong.

[gentle osprey]

Tokidoki ✓

[upper basalt]

How can i check general linear group over C is path connected or not?

[gritty widget]

in the complex case the matrix exponential is surjective, so you can try to use that to get paths

[upper basalt]

Can you please elaborate?

[coral pivot]

how i proved it was like

||consider how to get from a matrix A to the identity||

[upper basalt]

But why A to I , not A to any other matrix?

[coral pivot]

well this is sufficient to prove path connected, and i said that because you know how linear algebriacly you can do this.

[upper basalt]

Hmm ,i have to make a path like

A+(I-A)t

[coral pivot]

so that wont work necessarily

because you dont know if everything on that is invertible

so think back to how you can obtain I from A

like describe to me an algorithm

[upper basalt]

I form A by multiply A with its inverse

[coral pivot]

sure, but lets try to be more elementary here

yes, the word elementary is a hint kek

[upper basalt]

On that path?

[coral pivot]

yeah, cause for instance use that path from -I to I

0 is in it, which isnt invertible

[upper basalt]

Okk let me try to solve this further

[empty grove]

Not gonna read all that, draw a commutative diagram 😌

I mean I read the problem

The solution is

Draw a commutative diagram

[pearl holly]

hmm but how am I supposed to draw a commutative diagram that solves the whole question tho?

[empty grove]

Also yeah if they just say "the" map from pi1A to pi1X, they always mean the induced one

[empty grove]

You're not, draw the only thing you can draw from that information

And it will give you most of the solution immediately

Like don't try to work backwards from the solution

Just draw whatever you know

[pearl holly]

Okay I will give this a think

[empty grove]

😌

[pearl holly]

% https://q.uiver.app/?q=WzAsNCxbMywwLCJcXHRpbGRle1h9Il0sWzMsMywiWCJdLFswLDMsIkEiXSxbMCwwLCJcXHRpbGRle0F9Il0sWzAsMSwicCJdLFsyLDEsImkiLDIseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFszLDAsIlxcdGlsZGV7aX0iLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFszLDIsImYiLDIseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkb3R0ZWQifX19XV0=
[\begin{tikzcd}
{\tilde{A}} &&& {\tilde{X}} \
\
\
A &&& X
\arrow["p", from=1-4, to=4-4]
\arrow["i"', hook, from=4-1, to=4-4]
\arrow["{\tilde{i}}", hook, from=1-1, to=1-4]
\arrow["f"', dotted, from=1-1, to=4-1]
\end{tikzcd}]

[gentle osprey]

Tokidoki ✓

[pearl holly]

so what's step two

is that even right lmao

[empty grove]

YES

[upper basalt]

Can you please give. Me one more hint, i am not getting this

[empty grove]

Step 2 is π_1 😌

[coral pivot]

you get to I by elementary row operations right

[upper basalt]

Yup

[empty grove]

Also first make sure the diagram is commutative

Not just that it's a random square

Then see what it tells you about the fundamental groups and the induced maps

[coral pivot]

show that doing a row op on matrix A keeps it in the same path component. if you have shown it for all of the row ops then you are done

[empty grove]

Also what is f, why is it dotted

[pearl holly]

because that's the map that I'm "focusing" on now I guess

It isn't supposed to mean anything, I just wanted to highlight it

[empty grove]

Ah I see

But you didn't define ir

[pearl holly]

Well I'm trying to show that f: tilde A to A is the covering space corresponding to the kernel map of pi_1(A) -> pi_1(X) so I just highlighted it

[empty grove]

Yeah, but do you know what the map f is

Like can you describe what it does

or maybe just try to prove that the diagram you drew commutes

[upper basalt]

Should use determinent?

[empty grove]

You will need to know what f is to be able to do that

[coral pivot]

sure, use det to show its still invertible everywhere in the path ig

[empty grove]

Toki just read the question carefully you literally typed what f should be

[pearl holly]

it's just the restriction, right?

of p that is

[empty grove]

Yeah lol

[pearl holly]

lmao okay then it's quite easy to see that the diagram commutes right?

[empty grove]

Yep

[pearl holly]

and now we can "transfer" this thing to a diagram involving fundamental groups and that diagram will also commute

[empty grove]

Yes

Because π_1 is a functor

[pearl holly]

maybe I should draw it too

[empty grove]

Also have you shown that the restriction of p is also a covering

That's also part of the problem ig

[pearl holly]

yeah I did, it was an earlier exercise

[empty grove]

Always draw commutative diagrams if you have more than like 2 maps😌

[gentle osprey]

Tokidoki ✓

[pearl holly]

tf are you doing moldi

[empty grove]

😌

[upper basalt]

I can't , not getting it.

[coral pivot]

ok what did you try so far

[upper basalt]

I want to show path connected, so there should be a path ,

But i am not getting that path .

[coral pivot]

so remember i told you to break it up into row operations

find one for each, i.e changing rows, adding em and constant multiplication

[upper basalt]

Transformation

[coral pivot]

yeah find a path from A to the transformed matrix

for each of those

[upper basalt]

Please can you write that path?

I will try this in morning.

[coral pivot]

id rather you figure it out yourself

its a good exercise

[upper basalt]

Okk

Thank you 👍

[gritty widget]

if $A\in GL(n,\bC)$ then $A = e^X$ for some $X\in M_n(n,\bC)$. $t\mapsto e^{tX}$ is a path in $GL(n,\bC)$ joining $I$ to $A$. this suffices for path-connectedness as, given $A,B\in GL(n,\bC)$, concatenate the path from $A$ to $I$ and from $I$ to $B$ to get a path from $A$ to $B$

[gentle osprey]

TTerra

[gritty widget]

oh i just completely ignored the previous conversation

well this is a "not elementary" way to do it since i'm not sure how elementary "every invertible complex matrix is an exponential" is

[coral pivot]

i litteraly meant elementary operations

when i said elementary kek

(to generate matrices that is)

[gritty widget]

lol

[pearl holly]

But now what tho? Doesn't this only prove one inclusion? Because now I have that $i_* f_(\pi_1(\tilde{A}))$ is trivial so every element in $f_(\pi_1(\tilde{A}))$ is in the kernel

[gentle osprey]

Tokidoki ✓

[empty grove]

Ye it proves only one immediately

For the other one start with a loop in the kernel in the bottom left group

Prove that it's in the image of f_* 😎

[fading vale]

i again recommend drawing diagrams and pictures

it will help a lot

[empty grove]

So like start with this loop and track it around the diagram

Via liftings, images, etc

[pastel linden]

I'm a bit confused with whats going on here

when he takes the exterior derivative of fdz, is he doing this formally, or are dz and dz bar actual 1-forms?

are they being treated as 1-forms on C identified with R^2 as a real manifold?

[pearl holly]

(sorry for interrupting but I got it now. You just compose a loop in A, "push" it forward with i to get something nullhomotopic and then lift it up, restrict the range and then you have a loop in tilde A that lifts the original loop. Thank you so much! I'll try to draw a diagram next time!)

[empty grove]

Interrupting what

[pastel linden]

i think i get it now actually

[empty grove]

Did you justify the restricting the range step

Toki

[pearl holly]

Oh crap

[empty grove]

And also why that "lift" really is a lift along f

[pearl holly]

Frick

but why can't I just restrict the range tho?

bro I was chilling in my bed and now this

[empty grove]

Is the lift along p, of the image of the loop under i_*, guaranteed to lie is tilde A?

Like the lift has to be a path in the image of the inclusion of coverings for their to be a valid restriction

[pearl holly]

yeah because it's homotopic to a constant loop right?

no wait, that doesn't say anything, does it?

[empty grove]

[pearl holly]

no it does because of the uniqueness

[empty grove]

How

[pearl holly]

well idk, consider a constant loop lying in tilde A?

[empty grove]

Look at the definition of tilde A

[pearl holly]

well it's a path component of p(A)^(-1), but I don't know what that gives me

man I didn't think this through lmao. I thought that the constant map always is in tilde A or something idk

[empty grove]

Constant map would be

But this isn't a constant map

It's a nullhomotopic map

Not quite the same

So the image of the path under i lies in the subspace A

So its lift lies in p inverse of A

But it's path connected

[pearl holly]

yeah so I can change basepoints right?

So then I choose a basepoint in tilde A?

[empty grove]

No changing basepoints

The first diagram you drew should've been a pointed diagram

All lifts you do are pointed lifts

When you lift along a covering, you choose a basepoint

Each basepoint gives you a lift starting at it

So in tilde A you already have some basepoint and you use that as the starting point of the lift

[pearl holly]

yeah okay I see, but now what? How do I know that the lift is inside tilde A tho?

[empty grove]

This

[empty grove]

And this

Together tell you that the lift lies in the path connected component of p inv A containing the basepoint of tilde X

ie in tilde A

[pearl holly]

but why is path connectedness needed in this case?

[empty grove]

Path connected subsets lie in path components

Otherwise you could jump back and forth between components

[pearl holly]

ooohh lmao okay I see

[empty grove]

Now to prove that this path really is the lift along f 😌

[pearl holly]

man I'm tired as hell I can't think

just restrict this bad boy and boom

[empty grove]

Restrict?

[pearl holly]

restrict the range I mean

[empty grove]

You apply f though

[pearl holly]

no wait

[empty grove]

Ah

Restricting gets you a loop in the top left space yeah

You have to show that this loop in the top left space maps to the original loop under f

So that the original loop is in the image

The idea for this whole inclusion is that informally you should just think of f and p as the same map

[pearl holly]

well f is just p restricted so then this must be the case

[empty grove]

Or that's what I'm doing at least 🤡

Yeah exactly lol

[pearl holly]

okay and now we are done right?

[empty grove]

✓

[pearl holly]

okay great thank you so much! I really appreciate that you wasted all this time on me lmao

[empty grove]

[upper basalt]

A=e^x ? How

[gritty widget]

you can google a proof of this fact

probably something like do a jordan decomposition (we're in C) to reduce to the case of a jordan block

[glossy pine]

Small question for knot theory: if you have a braid word $\sigma_i^{-1}\sigma_i^{-1}$ (for any $i$), does that condense to $\sigma_i^{-2}$ or $(\sigma_i^{-1})^2$? I saw a prof use the former and I just wanted to check that it was appropriate

[gentle osprey]

gmod

[bitter yoke]

it condenses to both of those, they're equivalent

[glossy pine]

oh okay that's convenient

ty

[bitter yoke]

https://en.wikipedia.org/wiki/Presentation_of_a_group maybe look at this for more information

the braid groups are in the examples

[glossy pine]

swag

[mossy ermine]

swag

[bitter yoke]

@mossy ermine ur not swag

[glossy pine]

damn

the poor man

[mossy ermine]

ur stinky

Why r u in this channel

Go back to modular forms

[bitter yoke]

modular forms r topology and geometry

[mossy ermine]

ok analyst

[hollow harbor]

to be clear: we analysts want nothing to do with modular forms even though they're cool.

[upper basalt]

does $f(\overline{X}) \subseteq \overline{f(X)}$ for every kind of $f$

[gritty widget]

you can use \overline to make the bar go over the entire expression f(X)

[gentle osprey]

honey99

[upper basalt]

so if this holds then function is continous?

is it iff?

okk i will try

Does f=x+y maps discrete set to discrete ? Any hint ,

[reef shore]

Look at previous examples 😌

[upper basalt]

Ohhh z*sqrt2 z it is dicrete in R^2 but its map is not?

[reef shore]

Also you can see that all lines of slope -1 have singleton images, so far away points can map to close together elements. Tweak this slightly to get an easier example

[reef shore]

Yep

[upper basalt]

But one thing is that how to show that image of z*sqrt 2Z is dense in R?

[reef shore]

Many ways, main idea is that √2 has arbitrarily close rational approximations

You can also show that you get things arbitrarily close to any integer without equalling it, using pigeonhole, and then translate stuff around to get good approximations of anything

[upper basalt]

I think i got this, by integers a+bsqrt(2) is dense in R ,this is already proved on math stack

[reef shore]

hmm this isn't exactly a+√b

[upper basalt]

Edited

[reef shore]

ah yeah

Wait so your argument is that it's already proved on math stack?

[upper basalt]

The dense part ,

[bleak helm]

[reef shore]

[reef shore]

This argument is really neat btw, my discrete math prof mentions this every year to each batch that he teaches 😌

Very nice application of pigeonhole

[flint cove]

Wait now I'm interested, how does the argument go

[true robin]

For any natural number n, make n boxes by saying box i contains all the elements of the set that are between i-1/n and i/n above a natural number. As there are infinitely many numbers in the set, two elements are in the same box, so subtracting and translating you’ll get a number in the set that is less than 1/n and greater than 0 (the last part from irrationality arguments). Now taking integer multiples of this, we see that all reals are within 1/n of some element of the set

[fallen canopy]

I'm trying to show that in a compact set K, every sequence has a convergent sub sequence. My strategy is to take an open cover and produce a finite subcover, pick an open set V in the subcover that has infinitely many points, and consider the part of V that is disjoint from every other U_i (so V - the union of all the other U_i). This is closed in a compact space thus is compact. I can construct the subsequence by picking a point in this space, and then repeating the construction to get a subsequence and a nested sequence of compact sets descending. But something that I am worried about is if this process is a jerk and just picks a subcover of one open set over and over and over. How do I work around this?

eg if i picked K to be [1,2] and chose the lame open cover {(0,4)} - it seems like I need to do something at the step of picking an open cover

[true robin]

V\the union of Ui’s need not be compact. Also we have the freedom to choose any open cover we like, why use a generic open cover when we could possibly define an open cover using the sequence that will give us interesting info

[fallen canopy]

I thought it would be compact because it's a closed subset of a compact set

[true robin]

It isn’t closed

[fallen canopy]

Oh no I really meant to say V intersect K-unionU_i for all U_i that is not v

my bad

but the part of v that is not in any other U_i is what im focusing on

[true robin]

How is that any different?

[novel acorn]

Can somebody maybe explain how the quotient of a cylinder $S^1\times I$ under the identification $(z, 0) \sim (e^{2\pi i /2}z,0)$ and $(z, 1) \sim (e^{2\pi i /2}z,1)$ gives the Klein bottle
Cuz i ain't seeing it

[gentle osprey]

Irony Incarnate

[fallen canopy]

Now that I think about it that isn't different lol. But I do fail to see how it isn't closed.

[true robin]

Why do you think it is closed?

[marsh forge]

Is there a source this isn’t a presentation of it ive seen

[novel acorn]

Hatcher page 65

[marsh forge]

Can u post it I’m in an airplane

[novel acorn]

Ah OK lol gimme a sec

[fallen canopy]

Okay, so our finite subcover is {U_1,...,U_n} and I'm picking one of these open patches that has infinitely many points and calling it V (so that when I discuss U_i i am not talking about V). The other U_i are open in K and so their union is open in K. Thus, the compliment of their union is closed. Now, we know this closed compliment is a subset of V since V helps to cover K. Thus, V intersect the compliment of the union must be the closed set itself.

The idea is that $(\bigcup U_i)^c \subseteq V$ is clsoed

[gentle osprey]

Michael Harp

[true robin]

I see, V was in the open cover, sorry I thought it was just some random open set

[fallen canopy]

No worries, I should have probably made that clear

[marsh forge]

Oh boy

Irony I think maybe you can prove this with classification of surfaces and van kampen

I’m not sure it’s easy to picture

[true robin]

In any case, I still fail to see how that will help, you kind of showed it yourself by taking the trivial covering, doing this sort of thing kinda gives you no info

[fallen canopy]

I see, and I agree - do I ditch the whole effort?

[marsh forge]

Oh wait no it describes it well if you know how to build a Klein bottle from Möbius bands

Okay maybe threads aren’t a bad idea

[true robin]

You should really try to come up with an open cover using the sequence and show that this has a finite cover implies that the sequence has a convergent subsequence

[fallen canopy]

Pick a basis open set for every term in the sequence?

[marsh forge]

I might be able to convincingly draw it with fundamental polygons if you remind me in like

10 hours

Lol

[novel acorn]

My question at first was actually how he built those mobius bands but then I thought about asking for everything

I can probably imagine the mobius to klein

https://math.stackexchange.com/questions/907176/klein-bottle-as-two-möbius-strips
Thank you math stack exchange for your service to our great cause

[true robin]

Ok so go by contradiction, assume no convergent subsequence for any x in the metric space, find a neighbourhood that has only finitely many elements of the sequence in the neighbourhood. Use this to get an open cover. The problem in your case is that even if we cover all elements of the sequence that most likely won’t cover the whole space.

[marsh forge]

https://math.stackexchange.com/questions/474466/mapping-cylinder-of-z-rightarrow-z2

I tried to describe this myself

And decided

Not to

So that link might help

[quartz edge]

this makes me feel so stupid. why is he able to say dx=dw1e1+dw2e2

[marsh forge]

@novel acorn

[novel acorn]

Thank you and enjoy your flight

[quartz edge]

like yeah, i get that x only has derivatives in those directions, but like, why not have de1 and de2 as the basis for this

[fallen canopy]

@true robin I always forget contradiction exists because I don't like it too much. I'll see how this works out though, thanks for the help fam

[true robin]

No problem

[marsh forge]

Point set topology has a ton of results whose basic proofs tend to be contradiction

I don’t know how many of them actually end up having constructive proofs

[swift fjord]

Is every subspace of a fin-dim normed space closed?

[grave maple]

No. Consider R.

[swift fjord]

R has only 1 proper subspace tho which is trivially closed

And R itself is closed

Wdym

[hollow harbor]

this is true

[swift fjord]

Can you give me a direction for proving it?

[hollow harbor]

try finding a linear homeomorphism with R^n which carries your subspace to something pleasant

there are more direct ways to doing it too

[swift fjord]

Hmm alright

[hollow harbor]

the more direct ways involve literally just showing the complement is open

pick a basis and work in coordinates

[swift fjord]

In an IP space this would just be the This would just be the coordinate vector transfotmation w.r.t. some Orthonormal basis which is an isometry

But is it still an isometry for a general normed space?

[swift fjord]

I would think the complement in general would be more complicated

Since it's not even a subspace

[hollow harbor]

you don't need an isometry

linear maps are continuous

[swift fjord]

Wait right

Right I forgot

[hollow harbor]

i mean why is R^3 minus R^2 x {0} open

[swift fjord]

Cuz you can find an open ball around every point ofc

I get the general idea

I'll play around with it tmrw

Actually wait

I've proven smth stronger than this before

That every fin dim normed space is complete

Therefore as a complete subset of a complete space it's closed

But maybe i'll try doing it directly for sport

[hollow harbor]

that's a good way to do it

so that's like a sequential closedness proof as opposed to a complement is open proof

[flint cove]

For the sake of completeness one should add the „black magic“ proof where you just state that ℝ²×{0} is the preimage of {0} under the projection on the third coordinate π₃:ℝ³→ℝ, which is continuous
But that is only easier if the product topology is defined by the universal property – otherwise the „complexity“ of this argument lies in showing that π₃ is indeed continuous.

I think thhe „nitty-gritty“ approach to verifying open/closedness is really helpful to understand the concept, but the „preimage of continuous map“ approach can provide much more elegant and less error-prone proofs

(Although it is not always easy / possible to find such a continuous map)

[glossy pine]

sorry to interrupt, but I have a question: why is going around a circle twice fundamentally different than the constant loop?

[hollow harbor]

because if i take a string and loop it around a doorknob two times and then tie it tight

i can't take the string off without cutting it

[glossy pine]

ohhhh

[hollow harbor]

well

that's kind of not it

it's one reason

but even if i took a rubber band

i would need to take it off the doorknob

[glossy pine]

so would going around a circle n times be fundamentally different than going around n+1 times?

[hollow harbor]

i couldn't leave it behind the knob

yes

[glossy pine]

true

ok that makes sense

[hollow harbor]

these are all different "homotopy classes" of closed paths on the circle

turns out, they're the only ones

[glossy pine]

I see

ah

[hollow harbor]

every path on the circle can be stretched or shrinked into a path going around n times in one direction with constant speed

[glossy pine]

so what you're saying is if I make a half loop around a circle, I can stretch it to make, say, 10 loops if I wanted, right?

(just making sure I understand what you're saying)

and why does the "constant speed" part matter?

[marsh forge]

No ryc means like

For a given path

There is some n

But only one

For half way around this n=0

[glossy pine]

oh I see

[marsh forge]

The best way to think about this in my opinion

Is you take a string

And a piece of paper

And punch a hole in the paper and like stick a pencil through it

Then you tape on end of the string down and wrap it however you like, double back on yourself, whatever

But the rule is you have to end where you started

And tape it down again

Then the statement is that you can arrange your string to look like something that smoothly goes around the pencil n times for some n

[glossy pine]

where do you tape the string down?

[marsh forge]

Doesnt matter just has to be the same spot for both ends

[glossy pine]

oh I see

[marsh forge]

Oh obviously

The pencil is there to stop you from crossing the “hole” in the circle

Don’t lift the string over the pencil lol

[glossy pine]

yeah lol

okay this makes sense

thank you :)

[marsh forge]

Np

[flint cove]

I feel stupid
If I have injective maps f, g: X→Y, is f(x)↦g(x) between the subspaces Im f → Im g continuous?
My feeling says it should only hold for embeddings (because then the inverse of f is a continuous map from Im f → X) but I fail to find a counterexample

I mean if it is continuous it must always be a homeomorphhism because g(x)↦f(x) would also be continuous

But I guess there are injective continuous maps X→Y whose images have different homeomorphism types

like f = embedding ℝ→ℝ² into first coordinate, g = infinity sign parametrized by ℝ

nvm then :>

So I guess this is yet another case of „hatcher does not mention why stuff is continuous although it isn't obvious at all“

but yeah I guess in this case it's not hard to see that the ψᵢ in question are embeddings.

okay, no, the ψi aren't, but their graphs x↦(x,ψᵢ(x))

[high ember]

I understood the first half of what you said.. then my brain slowly melted. But Id like to learn more; did you solve your query?

[true robin]

I think I have a counter example, let X be a 2 element set {a,b} with any topology (say discrete). Let Y be the four element set {1,2,3,4} which is the disjoint union of the trivial topology on {1,2} (that is, only empty and {1,2} are open) and the discrete topology on {3,4} (everything is open). Let f map a->1 and b->2 and g map a->3 and b->4. The map you describe will not be continuous

[reef shore]

@flint cove did you see this

[flint cove]

Yes, but thank you

It's like 6am over here and I should really stop doing things now.

[reef shore]

[flint cove]

[swift fjord]

generally, if I have $A \subseteq X, B \subseteq Y$, and I know that separately, A,B are homeomorphic (In the resp. subspace topologies), X,Y are homeomorphic, then

1. Can you always find a homeo' between X and Y such that A is sent to B by it?
2. If not, are the relative properties of the subspaces still preserved? (i.e. closedness, compactness etc.)

[gentle osprey]

ShiN

[true robin]

No, here is an example, let X be the topology on two points {a,b} where the open sets are empty set, {a}, and the whole space. We let Y be the same thing as well. Let A be the set {a} and B be the set {b}. Obviously they are homeomorphic. But there is no homeomorphism from X to X that sends b to a, because such a map, would have to send b to a and a to b. The preimage of the open set {a} under this map will be {b} which is not open.

[swift fjord]

Alright I thought as much for the first question but couldn't come up with an example, but what about the 2nd question? In your example the {a},{b} still have the same relative properties

thanks

[true robin]

Relative properties?

Could you clarify what you mean by that exactly

[swift fjord]

Properties relative to X and Y

Just like I said

Topological properties of a subset/subspace

e.g. closedness, compactness

Well compactness is preserved under homeomorphism so that should be true

But properties that rely on the bigger space like being open or closed

[reef shore]

In saketh's counter example, the closed set {b} is mapping to the not closed set {a}

[swift fjord]

Yea I know

I mean even though there is not a homeomorphism that sends A to B

Is it still true that A is open iff B is open

[reef shore]

Wait I was talking about the homeomorphism from A to B not the big space homeomorphism

[swift fjord]

Oh waiy

You're right

[reef shore]

No, {a} was open, {b} wasn't

[swift fjord]

Yeye I misread the example

I thought the topology on Y had {b} oprn

Alright that settles it

[reef shore]

[swift fjord]

I guess it might hold under nice enough spaces

Thanks!

[true robin]

Np

[swift fjord]

specifically the question arose in the context of spaces homeomorphic to (subsets of) eucildean spaces where I realised that closedness would be preserved in this case cause of Heine-Borel

in the case of bounded sets at least

[reef shore]

Yeah only in the case of bounded sets

[true robin]

So closed and bounded is the same as compact, and compact is an “intrinsic” property

[reef shore]

Because (0,1) for example is homeomorphic to R which is closed

[swift fjord]

yea ofc

[reef shore]

I can't think of a counterexample for openness while keeping dimension constant

Open sets in R^k would be k manifolds, and I guess all k submanifolds of R^k would be open?

And being a k manifold should be intrinsic

[swift fjord]

could be, that's beyond me

[reef shore]

The above reduces the problem to "are there non open subsets of R^k, homeomorphic to R^k?"

If there is one you get a counterexample immediately

If there are none then all k submanifolds of R^k should be open

So that argument above would work

[upper basalt]

here f(x)=f(2x) so that means f(x)=f(2x)=f(4x)...., but how this help me to check 1st option?

[marble socket]

what exactly did you mean by "does it work for (0, inf)?"

(0, inf) isn't compact, so you can't say the function would be bounded directly.

another way to think about this is by using that t(x) = 2^x is a homeo from R --> (0, inf)

[upper basalt]

wait let me think about this

[marble socket]

yea sure

[pallid lion]

well maybe this hint is a bit too much but:

||the values f admits on the interval [1,2] already determines all values of f||

[gritty latch]

Let $X,Y$ be two compact hausdorff spaces and $f$ be a function such that the graph of $f$ is closed in $X \times Y$. Show that $f$ is continuous

[gentle osprey]

d/dx

[gritty latch]

I checked that this indeed holds for a function from [0,1] to [0,1] and while that gave an idea it isn't working as well as I expected

I was thinking that I could use a net that approaches the "hole" left by the discontinuity (and thus doesn't converge to anything) and use that to show the graph cannot be closed

[hollow harbor]

Maybe you could find a closed subset in Y whose preimage is not closed in X

[gritty latch]

lemme think about that for a min

[empty grove]

If the graph of f is closed, then f is a composition of a homeomorphism and a projection

nvm that was circular

wait no

it works

[gritty latch]

oh lmao

yeah that works

[empty grove]

[gritty latch]

hmm, but we never assumed f is injective

does it still work?

[empty grove]

The problem doesn't state injectivity though

[gritty latch]

nvm it works

[hollow harbor]

[upper basalt]

how? can you please elaborate

[hollow harbor]

Suppose x is between 2 and 4. What is f(x) in terms of some number between 1 and 2?

In terms of f(some number between 1 and 2) I mean

[upper basalt]

let suppose x=2.4 , then f(2 times 1.2) like this?

[hollow harbor]

Yep!

And we know f(2x) = f(x)

So f(2.4) = f(1.2). Can you prove that we can push any positive number into this interval by doing this over and over?

[upper basalt]

i think we can because f(x)=f(2x)=f(4x)=f(8x)=... so on

[empty grove]

Try to make it more precise

(ignore what I said, you don't necessarily need to do it that way)

[upper basalt]

i was also thinking about that , but still can we do that ?

[empty grove]

Ye it works

[upper basalt]

can you please given any hint for next part, as i know that continous function on compact set is uniformaly continous,

[empty grove]

Try to come up with a counterexample

[upper basalt]

but think about such function which satisfies f(x)=f(2x) not an easy task for me

[empty grove]

Try drawing graphs. You would have to presuppose values, but try doing this: presuppose a value, see what other values you can deduce. Then do this for multiple points instead of a single one. Do this for a bunch of choices and I think you would get pretty good intuition for what these functions look like

[lyric quartz]

any help with this? https://math.stackexchange.com/questions/4217532/the-equivalence-between-cubical-sets-with-connection-and-cubical-omega-groupo

[shy moss]

hi

why the poset of cosets of abelian subgroups of a discrete group is
simply–connected if and only if the group is abelian.?

[pearl holly]

Let me get this straight: a delta complex is basically a way to divide your object into triangles/simplicies, right? So basically it is a family of maps from some standard n-simplex to X and those maps allow us to divide your object/space into triangles. Did I get the intuition right?

and then there's these fancy properties that has to be satisfied and those properties make your "cutting object to triangle" natural in some way

[shy moss]

see https://arxiv.org/pdf/0809.4221.pdf

[pearl holly]

oh wow, this is great! Thank you so much!

[shy moss]

a delta complex X_0, X_1,..., is just a secuence of sets with face maps, so X_n (n-simplexes) is not necesarily formed with the vertices X_0

[pearl holly]

yeah okay got it! Thank you!

[upper basalt]

in option a , here it is given that 2 is not root so (x-2) is not factor of polynomial, and i know the defination of connected set, and knows that continous function maps connected to connected set, how can i use these things to solve this question

[gentle osprey]

Nobody

[upper basalt]

So it will get mapped to non connected set

[abstract pagoda]

@pearl holly for the example of 1.21 right after van kampen on pg52 pdf, whats ur visual intuition for A_\alpha

because Im not sure how the implication X_\alpha is a deformation retract of its open neighborhood A_\alpha

[abstract pagoda]

Is it just me or is Hatchers proof of Van Kampen a little too visual and sort of messy?

Because I remember going over it briefly in a class but it didnt put so much emphasis on this rectangle nonsense

[cedar pebble]

yea it's not a great proof

May's concise has a better proof

[honest narwhal]

You mean May? :P

[cedar pebble]

yes lol

sorry I am tired

[honest narwhal]

Hi tired

I'm daminark

[abstract pagoda]

Wow the linking circles and torus knot parts are cancer

Its either that im tired, my visual iq is low, he is explaining weirdly.

Im honestly feeling like the visualizing explanations are sorta important because they are meant to buikd intuition, but at this point its causing my brain to get overly tired

ESP showing R^3-A deformation retracts to S1 V S1

with A=S1 in R3

HOLY FUCK

I might be dyslexic

It was saying it retracts to S1 V S2

ok ig thats my signal to sleep

[upper basalt]

does we get all the values of R-{0} by this map?

[reef shore]

Do you need all the values?

[upper basalt]

but map is onto so , all the values should be there

[reef shore]

Yes, but see if you actually need to prove it

And what you get from surjectivity

Something much weaker than surjectivity suffices too

But of course you should also prove surjectivity (you probably would end up doing that anyway)

[upper basalt]

does evaluation map work here? like F : Set -----> R such that F(g(x))=g(2)?

[reef shore]

That is the map Nobody was talking about

[upper basalt]

for rest of part i can use path connectivity (1-t)f+tg?

[reef shore]

For which part?

[upper basalt]

for b,c,d

[reef shore]

Yeah that works

[lean marten]

Anyone here know any shape theory?

Looking for a decent quality book to learn it from

[cedar pebble]

Dydak-Segal is the standard textbook account for shape theory

but definitely look through the references in https://ncatlab.org/nlab/show/shape+theory, there's sort of no one size fits all book for this

[novel acorn]

The torus knot thing is pretty hard tbh

He will sometimes have these examples that are just super weird and hard to imagine but they're not too common
Maybe one or two per chapter

[pearl holly]

Oh sorry for not responding, John was explaining some stuff is discussion last night and then I fell asleep lmao. Tbh I don’t think that I have a visual understanding of those sets, I just understand the example intuitively and that’s enough for me

I might re-read that example when I get home tho because I don’t think I truly understand it

[pearl holly]

Yeah I guess you could imagine that those A’s, in the case of S^1 V S^1, is the wedge sun of a circle and a small arc

Might draw a picture once I get home lmao

[reef shore]

I think Hatcher does Delta complexes badly (at least I had a pretty hard time with it)

You might want to use other resources (or skip it) I don't think they're that important for singular homology

[pearl holly]

But it’s really cool

I might try to find something else if I struggle with it more tho, thanks for telling me!

[novel acorn]

Fellow Algtop is cool enthusiast

Covering spaces are pog

Although I'm on the permutation thing and Imma try rereading that to get a better grasp

[empty grove]

"algtop is cool enthusiast"

[pearl holly]

Yeah they are quite nice but tbh I literally find simplicies a lot cooler than covering stuff

Like only the concept of simplicies

Covering stuff is at the bottom of my list tbh

[novel acorn]

I remember reading Crossley and never fully understanding homology so I'm wondering how Hatcher does it

[novel acorn]

[pearl holly]

Hmm idk, I heard people saying that it is bad but I find it nice but I haven’t read a lot so I’m not the one to talk here

[empty grove]

[swift fjord]

What conditions are sufficient for a subgroup of a torus to be itself a torus? Also do I need to understand lie groups to prove that it's indeed sufficient?

[pearl holly]

Yoooo such a nice emote

Rem where

[pearl holly]

Moldi explain clown

Maybe I don’t truly don’t understand it yet idk

[empty grove]

covering good 😌

just draw commutative diagrams bro

Have you seen homology yet

[pearl holly]

Yeah some very basic stuff

Brb

[empty grove]

[pearl holly]

It’s basically ker/ im lmao

[empty grove]

Yes

Pretty amazing right

[pearl holly]

Cycles and then you quotient out the boundary stuff where you glue your cell

[empty grove]

Oh you are on simplicial homology

I skimmed that part 😌

[pearl holly]

Yeah I’m almost there

The cool thing is that you have these nice simplicies and with them you can cut your object into triangles/simplicies and that’s pretty cool imo. And then I guess that with this you can calculate the homology group of some space or something

[empty grove]

😌

[pearl holly]

And then apparently they “measure holes” so that’s pretty cool

Like the Betty number thing I guess

But I’m not there yet, I just remember nG talking about this a while back

[empty grove]

Idk what Betty numbers are

I stopped right after the proof of excision

[swift fjord]

the n-th betti number is the rank of H_n

[empty grove]

Oh That wasn't too scary

[swift fjord]

yea, except calculating ranks when H_n is not free abelian can be a bit annoying sometimes

[pearl holly]

Rank

[empty grove]

Ye

[swift fjord]

the rank of an abelian group can be thought of as the cardinality of the maximally independent set in it (Which exists since zorn), where independence here is being independent as a Z module

so basically the maximal set such that every finite combination with whole coefficients is 0 (the group identity) iff every coefficient is 0 (in Z)

equivalently, it's the cardinality of the maximal free abelian subgroup F contained in G such that G/F is torsion (every element has finite order)

try and see if you understand why these are equivalent

[pearl holly]

Yeah I think I see that, but are these numbers related to holes in some way?

[swift fjord]

yea

[pearl holly]

Or did I misread something a while back?

Oh

[swift fjord]

betti numbers pretty much tell you how many n-dimensional holes your space has

[pearl holly]

Daim that’s really cool, looking forward to reading about that!

And these numbers can be used as a definition of the Euler characteristic I think

From Wikipedia

I think

[swift fjord]

yea

[pearl holly]

So that’s cool too

[swift fjord]

I think yea

[pearl holly]

Yeah this stuff is really nice!

[swift fjord]

Honestly I get the meaning of the homology group, but I don't exactly understand why torsion free elements correspond to holes (yet?)

I might actually

So if anyone can tell me if my intuition is correct that would be nice: Basically, if some element in Hn is torsion, that means that if you 'go around' (In the sense of adding it to itself) enough times you eventually get a boundary, so this isn't a hole, but torsion free elements correspond exactly to those cycles which never become boundaries, and so correspond to holes

I think I might not exactly understand the geometric meaning of addition in the homology group though (SIngular homology specifically but simplical homology too ig)

but what does addition in the singular complex correspond to geometrically

like if it's just 'going around it twice' then why can that make something that's not a boundary suddenly a boundary

like adding a singular simplex to itself

I understand it's formal addition but it must correspond to something geometrically

[analog lark]

Hi this is a topology question:

So we know a straw has 1 hole, but how many does this have?

https://cdn.discordapp.com/attachments/486841106298961930/873206816752959488/unknown.png

If you put a pin prick in a straw then you would say it has 2 holes but saying this has 2 seems silly

and if it look slike this will the answe change

Please @ me

[sweet wing]

a once-punctured S^2 is homeomorphic to the real line

[pearl holly]

ah yes, I love these questions

[west spindle]

@analog lark are we considering this as a surface?

it's homeo to a sphere with three holes, or to a disk with two holes

[analog lark]

im gonna be hnoest with everyone, I know nothing about maths but this is rly bothering me

[west spindle]

ok let me phrase this differently

do these tubes have one "layer" or two?

i.e. is there any thickness to the metal?

[analog lark]

hmmm, well i would say yes

[west spindle]

right

so this is a solid, and you're asking about the hole count of its surface

which means that there's both an inside and an ouside surface to these tubes

[analog lark]

ahh i see

[west spindle]

hmm

[swift fjord]

is it actually homeomorphic or just homotopy equivalent

[analog lark]

so wouldnt that mean a straw has 2 holes

[west spindle]

a straw, if considered as a solid, is homeomorphic to a solid torus with one hole

[west spindle]

i mean, it's both, isn't it? unless i'm missing something

let me think about this for a moment.

[swift fjord]

I don't have a good enough intuition for when something is just homotopy equivalent vs when it's actually homeomorphic yet

[analog lark]

imagine im 3

[west spindle]

okay so like

i think

if we take your shape

it may be possible to deform it, without gluing or tearing anything,

into a torus with 2 holes

that's the topological answer to 'how many holes does this have'

[analog lark]

so a torus has 1 hole

right?

[pearl holly]

I like to think about homotopy equivalent as a deformation retraction. Hatcher says that X and Y are homotopy equivalent if and only if there is a third space Z that deformation retracts onto X and Y. So thinking about this makes the intuition easier in my opinion. So if you have two spaces, you can just look at if they both can be "squished" down from a third space. One example of a deformation retraction is obviously from S^1 X I to S^1, so they are homotopy equivalent. But these spaces are not homeomorphic. Maybe another way to see the difference is to remember the invariance of dimension, which says that if U is open in R^n and V is open in R^m, then these two open sets are homeo iff n = m. Homotopy equivalence doesn't need to satisfy this

someone please correct me if I said something stupid, I am not yet qualified to answering questions here lmao

[swift fjord]

gotcha

that's a good way to think about it ye

[marsh forge]

is there still a question here

[swift fjord]

chazer's question ig

[marsh forge]

depends on your definition of hole i guess

a topologist would probably say a torus has two holes

on through the center and one due to its hollow tube

@analog lark

[swift fjord]

isn't the second one a 2d hole, and the first one a 1d hole tho?

[marsh forge]

no both are 1D

oh wait no i guess so

i mean

the real issue is that the holes metaphor is bad

the real issue is that the holes metaphor is bad

[pearl holly]

Okay so here's another question: I don't really understand how C_2 "looks like". Like what does it mean for a group to be generated by the 2-cell A? What operation is in this group?

[marsh forge]

C2 is Z/2

oh wait

no its not

[swift fjord]

free abelian group is what he means

[marsh forge]

what the fuck hatcher