point-set-topology

[gritty widget]

not subtract

why is it called subtract ?

[rancid umbra]

again, it makes sense in the context of sets

[gritty widget]

oh i see if anything it is abuse of notation

cus u should really be writing A x B \ A x {.} if u want to mean what i meant

[rancid umbra]

what’s the abuse of notation?

[rancid umbra]

what did u want it to mean?

[gritty widget]

oh lol

like A x B - A

being removing a fiber from A x B

in A

[rancid umbra]

moldi not “what’s abuse of notation?” lol

[gritty widget]

ig since i emded up here

walk me thru this a bit

say I want to take Tor(Z_3 x Z^2,Z_3)

first I have to find free resolution of Z3 x Z^2

which should be Z^3 -> Z^3 -> Z_3 x Z^2

[empty grove]

The first one should be Z right?

[gritty widget]

is it because Z^2 free or something?

[empty grove]

Kernel of next map is 3Z x 0 x 0

[gritty widget]

oh yeah thats kernel of next map

[empty grove]

Assuming that the next map is (x,y,z) maps to (x mod 3, y, z)

[gritty widget]

oh so it is Z -> Z^3 -> Z3 x Z^2

ecahse you want kernel of Z^3-> Z3 x Z^2 to be equal to image of Z-> Z^3

ok yeah

then i forget what happens at this step lol

i tensor by

Z3

[empty grove]

You take tensor with Z_3

Then compute homologies

[gritty widget]

uh wait

oh

the sequence is exact and composing two maps is 0

so its a chain complex with boundaries being their original map tensor identity? i think

i forgot what boundary maps become after tensoring

[empty grove]

Ye

That's what the tensor functor does

[gritty widget]

yeah wtf

idk what it is explicitly

but if you have T:V to W

T tensor id_H: V tensor H to W tensor H

is explicitly im guessing

T tensor id( v tensor h) = Tv tensor h

so it acts linearly

[empty grove]

Ye, you can prove that

Using uni prop of tensors

[gritty widget]

yeah the thing is tensoring morphisms isnt something that makes sense in my head

im just viewing it as a notation thing

like i can understand tensoring sets

like say you have direct products right

f: X to Y and g: A to B

f x g: X x A to Y x B by f x g((x,a))= (f(x),g(a)) feels like a notational trick

so im guessing for tensors product it looks genuinely identical though

like f tensor g ( x tensor a) = fx tensor ga

[empty grove]

You can also use the universal property to get this, that will feel more natural

[gritty widget]

so i see no difference in what happens to morphisms

[empty grove]

Uni prop of products

[gritty widget]

isnt it like its initial or something

doesnt uni prop of tensor product have tensors as initial objects and isnt direct product have as final?

because for tensors iirc

[tough imp]

Why is this in this channel

😭

[empty grove]

Yes

But it is similar idea for all uni props

[gritty widget]

its X x Y -> A has X x Y to X tensor Y , with X tensor Y to A uniqurly

[gritty widget]

idk tbh

in alg top class

i forget our motivation for doing tensors

wait

i remember

we were talking about total complex

and to understand better we need to know its relationship to crossed product

and all of this involves tensors, but i dont know why tor

something about UCT for homology

but i dont know why we need UCT for homology or what it does

i just know for UCT cohomology it is useful for proving equivalence of cohomology theories

so thats why we went over Ext, but its probably used so much more ways than i can imagine

[swift fjord]

Delta(P) is the simplicial complex obtained by taking {x_1<x_2<...<x_k} as simplices. For (b), I think this space is contractible, but i'm not sure how to go about showing this for n>3

I'd like a hint

[swift fjord]

Another question, I defined the poset $P$ (Doing this for 2 simplices but the general case is the same) on the product $\Delta_{k-1}^2$ by $(x_1,y_2)\leq (x_2,y_2) \iff (x_1 \leq y_1 \land x_2\leq y_2)$. Then $\Delta(P)$ is in bijection with $\Delta_{k-1}^2$, and clearly the simplicial maps of the projections $\Delta(P)\rightarrow \Delta_{k-1}$ induce a continuous bijection $\abs{\Delta(P)}\rightarrow \abs{\Delta_{k-1}} \times \abs{\Delta_{k-1}}$, but i'm not sure how to show this is a homeorphism. I'm not even sure how to exhibit its inverse directly (The inverse of the map of underlying spaces, that is).

[gentle osprey]

ShiN

[swift fjord]

Wait

I can use continuous bijection from compact to hausdorff implies hausdorff right

so it's a homeorphism

[gentle osprey]

ShiN

ShiN

[true garden]

Hello. Can you give a simple intuitive example of a subspace whose (covering) topological dimension is greater than that of the space containing it?

[shy moss]

let $\partial: \pi_1(X,A,x_0) \rightarrow \pi_0(A,x_0)$ defined by $\partial ([f])=[f|\lbrace 0\rbrace]$ and $j_:\pi_1(X,x_0) \rightarrow \pi_1(X,A,x_0)$ the map induced by inclusion, how can i prove that $ker \partial \subset im j_$ ?

[gentle osprey]

Or x1

[vast estuary]

Could I have some intuition for this?

Why consider the "abelianized" fundamental group at all?

This turns out to be isomorphic to the first homology group

Assuming X is path-connected

[coral pivot]

probably the last sentence there tells you why.

so for example, in complex analysis the value of an integral of a closed differential depends on the homotopy class of the loop its integrated over

so we have a homomorphism from pi_1 -> C

by sending the loop gamma to int_gamma d omega

and these are things we like studying, but ofcourse in reality we only need the abelianization of pi_1 (or homology group if you will)

[fallow venture]

Hey hey, just wondered if I could have a bit of help on a subject regarding topology ...
Sometimes, we need to take account of the fact that a set is countable or not. So I wondered if anyone had a general way of getting that proven, for any set. If there is not and it's as creative as convergence/divergence tests, would you mind if we talk about a few examples that would show what are different ways of getting to prove them. I keep thinking about ways to prove it, but unfortunately don't find a way to do it, and we didn't have like a list of ways that could prove it or anything. It's very much like the unknown here, and every idea falls into finding a counterexample in my own thoughts.

[empty grove]

showing a surjection from/ injection into a set you already know to be countable

I don't think you have to show the countability of a set very often in point set topology though, I can only recall a couple theorems which would do that in an intro course

Mostly theorem statements just say stuff like "Let S be a countable set"

[fallow venture]

I have been asked to show that a set is countable with multiple restrictions on a non-countable set already

From memory, there was something like that, but I wouldn't be sure to be able to find it back at all

[fallow venture]

Completely forgot about this one ! Thanks a lot, so basically linking the set to a countable set with a surjective or an injective application

Got it !

[vast estuary]

ohh okay, cool. i'll think more about it but i understand it better ig

[gentle osprey]

Hausdorff

Hausdorff

[vast estuary]

oh yes

this means that x and f(x) are homologous since they differ by a boundary

[vast estuary]

and so the induced map is the identity?

i'm new to this stuff lol

[pearl holly]

wait how do you know that c_x is a singular simplex?

I might be really dumb here

[gentle lark]

You can just make the map explicit by parametrizing the path

As in, parametrize by arclength, then send t in [0,1] to cx(t).

It's a singular simplex cause [0,1] is the 1-simplex

[pearl holly]

aren't we considering a 0-simplex here tho?

[gentle lark]

Yeah but two n-simplexes are homologous if they are the boundary of an (n+1)-simplex

[pearl holly]

so how can c_x be a 0-singular simplex?

[gentle lark]

c_x is a 1-singular simplex

[pearl holly]

oh yeah lol

[gentle lark]

Which has the two 0-singular simplices you're interested in as boundaries

Namely x and f(x)

[pearl holly]

ah yeah right

[swift fjord]

Bump. First message is this but all of them are relevant

[gentle lark]

Gonna go ahead and bump an old question of mine too while I'm here

[gentle lark]

For a), isn't delta P just an n-simplex?

I think b) is an n-simplex x [0,1] as well

Hmmm not sure though

[gentle lark]

I'm about 80% confident of this

[gentle lark]

And 20% confident of this

It's not quite n-simplex x [0,1] but I think it's not far from it

It's like you've got two n-simplices

And you draw attach new k-simplices to connect those two by choosing one vertex in one simplex and creating an n-simplex with it and k vertices from the second simplex

It's possible to prove both are spheres by induction

(homologically)

I'm about 90% sure I've got a valid proof, but you said you just wanted a hint so I'll leave it at that and I can share it later if you want?

[swift fjord]

Yes a is easy

So you're saying that it has the Homotopy type of a sphere or that it's homologically a sphere?

Or a product of sobers

Spheres*

Not exactly sure what you mean

I mean I see what you mean about how it's constructed

Say for n=3 it ends up being like a triangular prism with almost all vertices connected to each other

[gritty widget]

how is tensor 3 x 3 tensor times 3x 1 vector a vector?

so like a tensor b times v a vector

[gritty widget]

ok

so physics weird ig?

they define angular momentum as a vector

but its angular interia times angular velocity

this might be wrong channel mb

[swift fjord]

probability more suited to #diff-geo-diff-top

[tidal cedar]

as a kinda basic answer tensors can be described as elements of a tensor space

and that can include linear functionals on a vector space

Inertia tensor is a rank 2 tensor

and so the inertia tensor here is taking a 3-vector and giving you back a 3-vector

[gritty widget]

rank(0,2)

right?

[tidal cedar]

yeah

(m,n) rank here should be m copies of a vector space and n copies of the dual

if you've seen that definition

[pearl holly]

okay I'm coming back to this now

and now I think that my confusion lies in $\nabla$. How exactly is it defined? Hatcher says "There is an inclusion $\mathbb{R}P^\infty \times X \to \Gamma X$ as the quotient of the diagonal embedding $S^\infty \times X \to S^\infty \times X \wedge X$" and I was overlooking it, thinking that it wasn't important. But now I think think that this plays a huge role but I don't understand what the quotient of the diagonal embedding means

[gentle osprey]

Tokidoki ✓

[pearl holly]

and I just set everything to p=2

and I forgot to say in this post that $\mathbb{Z}2$ acts on $X \wedge X$ by the map $T[x_1, \ldots x_p] = [x_p, x_1, \ldots, x{p-1}]$

[gentle osprey]

Tokidoki ✓

[orchid forge]

if you have two projection maps, and f sends the kernel of one to the kernel of the other, it induces a map on the quotient spaces

[pearl holly]

so I'm thinking that maybe when Hatcher says "quotient of the diagonal embedding" he means some sort of quotient with some relation involving T

wait so Hatcher uses this here?

[orchid forge]

that's just what the quotient of a map is

[pearl holly]

oh lmao

[pearl holly]

wait could you elaborate? How is the induced map defined?

[orchid forge]

if you want a map X --> Y to induce a map X/~ --> Y/~, there's a natural definition

send [x] --> [f(x)]

[pearl holly]

am I doing that with nabla?

[orchid forge]

with the diagonal map (s,x) --> (s,x,x)

i should have said "fiber," not kernel, but it's essentially the same as with groups

if you want [x] --> [f(x)] to make sense, you must show that if you pick a different representative of [x], you get another representative of [f(x)]

i.e., $\pi_1^{-1}(x)$ gets sent into $\pi_2^{-1}(f(x))$

[gentle osprey]

Kogasa

[orchid forge]

that is the necessary and sufficient condition for a map $X \to Y$ to descend to a quotient $X/\pi_1 \to Y/\pi_2$

[gentle osprey]

Kogasa

[gentle lark]

Homotopy type of a sphere I think

||You can prove it by induction by using that an n-sphere is two (n-1)-discs attached at their boundaries||

(I think)

[pearl holly]

I'm really sorry but I don't see how I get nabla from the diagonal map with this quotient thing

[orchid forge]

you have the diagonal map $\Delta : S^\infty \times X \to S^\infty \times X^{\wedge 2}$ given by $(s,x) \mapsto (s,x,x)$. You quotient $S^\infty$ by the action of $\mathbb{Z}_2$ to get $L^\infty = \mathbb{RP}^\infty$. On the domain of $\Delta$, you quotient just on the left, getting $L^\infty \times X$. On the right, $\mathbb{Z}_2$ acts on both $S^\infty$ and $X^p$, so you quotient by the diagonal action to get $\Gamma X$

[gentle osprey]

Kogasa

[gritty widget]

i asked because a physics student was trying to explain angular momentum to me

but from context i shouldve known it was a (3,0) tensor

acting on a 3 dim vector

(a tensor b tensor c )(v)= (av,bv,cv)

[orchid forge]

let $\pi_1 : S^\infty \times X \to L^\infty \times X$ and $\pi_2 : S^\infty \times X^2 \to \Gamma X$ be the two quotient maps. then $\Delta$ sends $\pi_1^{-1}(x) \to \pi_2^{-1}(x)$, since $\pi_1^{-1}([s], x) = (\pi^{-1}([s]), x) \mapsto (\pi^{-1}([s]), x, x) \mapsto ([s], [x, x])$

[gentle osprey]

Kogasa

[orchid forge]

where $\pi$ is the quotient $S^\infty \to L^\infty$

[gentle osprey]

Kogasa

[gritty widget]

lol what in the etale is this

[orchid forge]

so $\Delta$ descends to a map $L^\infty \times X \to \Gamma X$, and then you take the quotient $\Gamma X \to \Lambda X$ to get $\nabla : L^\infty \times X \to \Lambda X$

[gentle osprey]

Kogasa

[orchid forge]

to be clear, the action of $\mathbb{Z}_2$ on $X \wedge X$ is just cyclic permutation, $(x,y) \mapsto (y,x)$. which fixes the diagonal

[gentle osprey]

Kogasa

[pearl holly]

okay wait let me process this a little more

[swift fjord]

Oh I didn't remember this. I think that might work. Not sure I see where the induction helps tho

[pearl holly]

okay so in this case the map here is given by [x] --> [f(x)]

[orchid forge]

yes

[pearl holly]

how about here tho? Here I have products and stuff?

[orchid forge]

since $\pi_1$ is just acting on the first component, $\pi_1(s, x) = (\pi(s), x)$, an equivalence class is written $([s], x)$

[gentle osprey]

Kogasa

[orchid forge]

and $\pi_2(s, x, y) = [(s,x,y)] = ([s], [x, y])$

[gentle osprey]

Kogasa

[orchid forge]

since $\Delta$ sends $([s], x) = {(s, x), (-s, x)} \mapsto ([s], [x,x]) = {(s, x, x), (-s, x, x)}$, it's well defined on the quotient

[gentle osprey]

Kogasa

[pearl holly]

daim okay I see

But thank you so much for typing all of that out! It feels good to break things up in these baby steps sometimes, there was just too much detail Hatcher skipped for a noobie like me to understand lmao

[orchid forge]

np

[gentle lark]

||If you prove it for i = 1,...,n, you can prove that the case i=1,...,n+1 is two copies of a sphere attached at the sphere from the n case||

||Basically just add a point to the one of the two simplices and prove that you end up with a "contraction" of the sphere you used to have, ie a disc. You then add a point to the other simplex and the same thing happens. They are attached at the n-sphere you started with. Thus, it's an (n+1)-sphere.||

[swift fjord]

Haven't opened the 2nd one, but Isn't it 2 balls? Like at each end is gonna be a full n-simplex which is an b-ball

n-ball*

[gentle lark]

Yeah it's two contractions of a sphere, ie a disc, ie a ball

Disc is a 2-dimensional ball and I have a bad habit of using the term to mean ball

[swift fjord]

I think i'll keep thinking about this rmw

Tmrw

Thanks

[gentle lark]

Ping me if you'd like a drawing about this or sth

[swift fjord]

will do thanks

[keen urchin]

I have a question about this example.

Here, the set B is said to have a limit point because the superset [a_0, b] has a limit point. But I don't follow. In general if a set has a limit point, does it imply that every subset has a limit point as well?

Couldn't there exist a limit point in [a_0, b] that is not a limit point of B?

[plain raven]

what is the preceding theorem

[keen urchin]

Compactness implies limit point compactness, but not conversely.

[plain raven]

no it's not true in general that every subset has a limit point etc

it's something about this specific construction

[a_0, b] should be the topological closure of B

i think

[keen urchin]

maybe if b is the LEAST upper bound?

[plain raven]

oh

[keen urchin]

which exists, because S_omega is well-ordered, right?

[plain raven]

yeah i read "least upper bound" even tho as you say it's not there

but

i think if you do least upper bound this proof works

[keen urchin]

Because then b is a limit point of B? what if b has an immediate predecessor though

[coarse night]

saying a function/map is closed and saying that the graph of the function is closed are the same thing right?

[honest terrace]

not really

For any continuous map f: X -> Y where Y is Hausdorff, the graph of f is closed in X x Y

Even if f isn't closed

(and there are continuous map that aren't closed)

Like e^{-x} works I think

[coarse night]

oh yeah right, tx.

[swift fjord]

I'd appreciate a drawing, I have trouble visualising this kind astuff

[true garden]

Hello. Can you give a simple intuitive example of a subspace whose (covering) topological dimension is greater than that of the space containing it?

[pearl holly]

Okay I'm yet again fucking back at this. I'm sorry for rambling about this for days now but I still don't understand one thing. I could've asked this yesterday but it was getting late so I decided to sleep. I know that when I restrict $\nabla$ to $x_0 \times X$ I get a map $x_0 \times X \to x_0 \times (X \wedge X)$, which I could view as a map (1) $X \to X \wedge X$ which sends $x$ to $[x, x]$. Now the domains and codomains match up with $\alpha \otimes \alpha$ which I could identify with $\alpha \times \alpha$. So now $(\alpha \times \alpha) \nabla$ is the cup product on $\tilde{H}^i(X \wedge X)$. We can assume that $i \geq 1$ and so I can see this as the cup product on $H^i(X \times X)$. Now if the degree of $\alpha$ coincides with $i$ then I get that $Sq^i(\alpha) = \alpha \smile \alpha$ and everything is fine. But I don't understand one thing, it's (1). That steps seems really "handwavy" to me. It feels like I just make a completely new map when I do this restriction and before the restriction I can't compose $(\alpha \cross \alpha) \nabla$. But at the same time I don't lose much information when doing this restriction since $\nabla$ is just the identity on that one component that I'm taking away

[gentle osprey]

Tokidoki ✓

[pearl holly]

something is wrong here wait

[pearl holly]

nvm I'm brain dead I get it

[gentle lark]

The left part is the case where n=3

Let's add a_41

For every cell you had before in the sphere before adding a_41, you now have a new cell which is the same cell but with a_41 included

Hopefully with the drawing it becomes clear how this is sort of like shrinking the figure on the left towards a point (in this case, a_41)

Now, we're supposing by induction that the figure on the left is a sphere

And a contraction of a sphere to a point is a ball

So the figure in the diagram is a ball with boundary, where the boundary is the sphere we started with

Now we add a_42

This is again a ball, and it happens to be attached to the previous ball at the sphere we started with

So we in fact have two balls attached at the boundary

So this is actually a sphere of one more dimension

The case n=1 is the 0-sphere (since it's just two disconnected points), so the induction works

[keen urchin]

anybody know about the space S_omega, the "minimal uncountable well-ordered set"?

i'm looking at a proof for why it is limit point compact, and i don't quite get it

[empty grove]

I'm guessing given an infinite set S you'd want to take s = least element such that there are only finitely many elements of S greater than it, and prove that s is a limit point

No that won't work

[keen urchin]

You can assume we have a countable subset of S_omega, since if we find a limit point there, any subset of S_omega has a limit point.

[empty grove]

It can be made to work by taking a countably infinite subset of S

Ye

And then s is well defined

Take an open set containing s. It's of the form (a,b). Then there are infinitely many elements of S larger than a, but only finite many larger than b

So there must be elements of S in between

✓

[keen urchin]

there are only finitely many elements larger than b in S?

[empty grove]

Yes because b > s and by definition of s, there are only finitely many elements of S greater than s

[keen urchin]

s is the least element of S, right?

[empty grove]

No

[empty grove]

This

[keen urchin]

s = least element of S_omega?

[empty grove]

No

Least element of S_Omega such that only finitely many elements of S are greater than it

inf { x ∈ S_Ω | only finitely many elements of S are greater than x }

Can write min instead of inf

[keen urchin]

might this set be empty?

[empty grove]

No because S is assumed to be countably infinite

So it is bounded above

In S_Onega

Limit point compact

[keen urchin]

could it be that s has an immediate predecessor? and then, the open set (s-1, s+1) contains s

[empty grove]

It won't

Because it has only finitely many things larger than it

Its predecessor would have infinitely many things larger than it

But there can only be 1 thing in between

[keen urchin]

pretty sure i get it now. i think. thanks!

[kind cedar]

Can I just invoke the constant rank level set theorem to prove that Ker(F) is an embedded subbundle of E, where F:E->E' is a vector bundle homomorphism over a manifold M and F has constant rank?

[empty grove]

Is this differential topology

[swift fjord]

Actually now that you draw it out, it's pretty clear this.is a suspension. And suspension of an n-sphere is homeo to n+1-sphere. At least it's easier for me to think of it like this And you can also derive the homology like that. Thanks a lot!

[gentle lark]

You're welcome 😄

btw the other delta P is actually a ball

Think I accidentally said it was a sphere

[swift fjord]

Oh yea I knew it was tho since it's just the standard n simplex which is contractible so np

[light rivet]

Can someone tell me what is wrong with this. We know the real projective plane has euler characteristic 1, and can be obtained by gluing the mobius strip with a closed disk. However, the Euler characteristic of that is \chi(Mobius) + \chi(Closed Disk) - \chi(their intersection, which is an interval) = 0 + 1 - 1 = 0

[wanton marsh]

isn't the intersection a circle ?

[obtuse meteor]

^

[light rivet]

How is the intersection a circle? We only glue along 1 edge of the mobius strip, so isnt that a closed interval?

[orchid forge]

How many edges does a mobius strip have

[light rivet]

A mobius strip has 3 edges

nvm, I was looking at the square diagram of the mobius

So I am a bit confused... this is a diagram of a mobius strip. so two edges are idenitfied the same

So in general, if I have a surface that can represented as a 2n-gon with pair edges identified like this, only the "paired" edges are edges of the surface?

[kind cedar]

wrong channel, I guess 😦

[wanton marsh]

I think you should make a mobius band irl

no the paired edges are no longer edges

well

rather, they're not boundaries of the band

and you glue the boundary of the disk to the boundary of the mobius band

not on some inside edge

[light rivet]

yea, i see now

[wanton marsh]

the a and c edges make up the band's boundary

when you walk on a and reach a corner you will seemlessly transition to walking on c

and after you're done walking on c you are back at a

they make a circle

[gritty widget]

Wow geom x top gang is so fractured

Hey. I have a nested sequence of compact nonempty sets with diameter being greater than some c. I should prove that their intersection has the diameter greater than c

What keeps me away from making a case when their intersection is a singleton set

[rancid umbra]

what do you mean “what keeps you away from making a case that their intersection is a singleton set”?

the sets [-1/n,1/n] all have diameter greater than c = 0 and their intersection is the set containing zero.

[gritty widget]

Let's say c is 5 but I still get a singleton set as the intersection

[rancid umbra]

im not going to address this point because it can't happen in a metric space and im going to ignore it (lmao i just don't know how to proceed from here). if the sequence of nested compact sets is (E_n), you should try and pick, for each natural number n, points x_n and y_n in E_n such that d(x_n,y_n) = diam(E_n) (note that this can be done due to compactness).

assume, for now, wlog, that the sequence x_n and y_n converge

then see what you can say

after you have (hopefully) reached the conclusion, see if you can justify why we can assume, wlog, that the sequences x_n and y_n converge

[gritty widget]

Yeah I know haha

I've done that as well

But this example bugs me

Probably a singleton set then is impossible

[rancid umbra]

one way i was trying to go about this was to show that
$$\text{diam}\left(\bigcap_{i\in\mathbb{N}}E_i\right)=\inf{\text{diam}(E_i):i\in\mathbb{N}}$$ and the right hand side is clearly larger that $c$. it would kind of get rid of the "example" you have implicitly

[gentle osprey]

c squared

[rancid umbra]

but showing that was kinda hard for me

[rancid umbra]

yes its definitely impossible if c is strictly positive, just approaching the proof by contradiction from here seems difficult, and also not helpful. you would then only be able to say that the diameter of the intersection is positive, which... just doesnt help you prove the rest of the theorem

[honest terrace]

this is wrong without any hypothesis on the E_i (I don't have the context)

you only have <=

Like for example E_0 = (0,1) and E_i = (1,2) for i >= 1

RHS is 1, LHS is 0

[rancid umbra]

yea of course u have to assume they are nested, this used the assumptions to the problem that Craed0x had

[honest terrace]

I should read the context before answering

sorry

[rancid umbra]

nah ur good

[rancid umbra]

do u know how to get >= ?

because i could only get <=

[honest terrace]

I'm thinking about it, but yeah >= seems to be the hard part. I'm trying to use the fact that the E_i's are compact so that I can extract converging subsequences when I want

[rancid umbra]

if u get anything, ping me. i’d be interested to see it

[honest terrace]

Ok got it I think

@rancid umbra

I should probably latex it

give me a sec

\newcommand{\diam}{\mathrm{diam}}
Here's a proof when they're nested:\

Let $E = \bigcap_{j}{E_j}$.\
For each $i$, you have $\diam(E) \leq \diam(E_i)$, so you can take the inf on that $i$, and we get $\diam(E) <= \inf_i \diam(E_i)$.\
Now let's pick $x_n, y_n \in E_n$ such that $|d(x_n, y_n) - \diam(E_n)| \leq \frac{1}{n}$.\
We get $d(x_k, y_k) \geq \diam(E_n) - \frac{1}{n}$.\

Now two "tricks":\

First, since $x_k$ lies in $E_1$, we can extract a converging subsequence $x_{\varphi(k)}$, say converging to $x$.\
And similarly, we can extract a converging subsequence from $y_{\varphi(n)}$, say $y_{\gamma(\varphi(n))}$, converging to $y$.\
Since $x_{\gamma(\varphi(n))}$ is a subsequence of $x_{\varphi(n)}$, it also converges to $x$.\

Let's say $x'k = x{\gamma(\varphi(k))}$ and $y'k = y{\gamma(\phi(k))}$ to make it readable.\
The inequality we had becomes $$d(x'_k, y'k) \geq \diam(E{\gamma(\varphi(k))} - \frac{1}{\gamma(\varphi(n))}$$.\
By taking the limit, we get $d(x, y) \geq inf_i \diam(E_i)$.\

One last thing remains to be proved, the fact that $x$ and $y$ lies in $E$.\
To see that, the second trick: We just need to observe that, except maybe to the first n terms, each term of the two sequences lies in $E_n$, and since $E_n$ is compact, so closed, the limit must lie in $E_n$, so each limit lie in each $E_n$, so in the intersection.

This isn't really well written but it should be readable

[rancid umbra]

lol, to save some trouble and for readability, just assume the sequences converge

[gentle osprey]

Shika

[honest terrace]

I think it should be readable since it's latexed now

[rancid umbra]

def

[honest terrace]

(there's still a few latex typos like an <= instead of a \leq at the beginning but let's ignore it, I'm too lazy to actually fix everything )

[rancid umbra]

lol

[honest terrace]

Anyway basically the idea is "build sequences x_n, y_n such that d(x_n, y_n) > diam(E_n) - 1/n and since you're working with compacts, the two sequences converge, and better, the limits are in the intersection"

[rancid umbra]

okay, cool

[honest terrace]

And since the limit of the RHS is precisely "inf_j diam(E_j)", we're good

[rancid umbra]

nice

[rancid umbra]

although i think ur argument and this outline accomplish the same thing, and because of the way the inequalities work out, you only need one direction of inequality (>=)

it just happens that u can show their equal

[honest terrace]

oh I didn't read that part

that's the second time

but yeah

this is pretty much the same thing

[rancid umbra]

dude i need more ppl active in this channel so i can procrastinate my other hw

[orchid forge]

i can start knotposting if you want

[rancid umbra]

lol go for it

i just really dont wanna write this psychology paper

[orchid forge]

ok i lied i actually have nothing particularly interesting to say

i have some questions but i'm pretty sure they're too specific to be answered

[rancid umbra]

darn

[gritty widget]

please do knot post

[rancid umbra]

yes

[hollow harbor]

[orchid forge]

if anyone knows anything about the steenrod square on khovanov homology then i will knotpost

[tough imp]

Bruh one of my friends is doing a homological reading course at UNC and I think the guy teaching it either invented khovanov homology or like was co-creating it or some shit and they’re just doing khovanov homology and he’s like wtf is happening

[true garden]

Hello. Can you give a simple intuitive example of a subspace whose (covering) topological dimension is greater than that of the space containing it?

[orchid forge]

wait that's sick

who's the professor

pretty sure khovanov was the guy who invented khovanov homology but there are a few other names that would be pretty epic

[pearl holly]

Wtf is khovanov homology?

[drifting sundial]

any hints for this question? I'm guessing you explicitly construct a homotopy but I'm not sure how

[orchid forge]

a categorification of the jones polynomial

homology theory whose graded euler characteristic is said polynomial

but itself is an invariant of links

strictly stronger

[pearl holly]

Oof daim I see

What’s graded Euler characteristic?

Or maybe you can just refer me to some pdf

[orchid forge]

alternating sum of (graded) dimensions of the homology groups

[pearl holly]

Oh okay

Yo where can I read up on this? Sounds pretty cool with the steenrod squares on them and stuff

[orchid forge]

Bar-Natan's paper probably

https://arxiv.org/abs/math/0201043

[pearl holly]

Okay thanks!

It’s probably going to be too hardcore for me tho

[orchid forge]

if it's too terse, he has another paper that actually goes through more details https://arxiv.org/abs/math/0410495

[pearl holly]

Oh okay that’s nice

Thank you so much!

[orchid forge]

if you want to talk more about it feel free to bug me, i'm supposed to know things about this anyway

[gentle osprey]

Hausdorff

[vast estuary]

(ignore)

[haughty anvil]

Hey so I'm trying to understand mobius transformations and I'm a bit confused form my professors notes. I was wondering if anyone could help explain them.

Also this is the actual problem itself we're doing. if it helps.

[vast estuary]

I think #real-complex-analysis would be more suited for this @haughty anvil

Somewhere in the proof of Hurewicz theorem. \gamma is a 1-cycle and it is homologous to zero ---> \gamma is the boundary of some 2-cycle. How?

[gentle osprey]

Hausdorff

[vast estuary]

Hmm I think I'm making some mistake

what does it mean for an element to be zero in the first homology group?

[gentle lark]

It means it's the boundary of a 2-cycle 😛

The nth Homology group is ker boundary_n / im boundary n+1

[vast estuary]

B_1(X) = Im(boundary_2)

so \gamma in B_1(X) means that it's the boundary of a 2-chain

why is it the boundary of a 2-cycle, necessarily? @gentle lark

[gentle lark]

Because if it's 0 in the homology group, it belongs to what you quotiented by

And you quotiented by Im(Boundary)

Ah wait

Cycle =/= chain

[vast estuary]

yes exactly

that's my question

[gentle lark]

Ok sorry 😅

[vast estuary]

i'm only able to deduce chain

[gentle lark]

What's a cycle?

[vast estuary]

i think we also have to use that gamma is a 1-cycle, somehow

[gentle osprey]

Hausdorff

[gentle lark]

Ah, well that's the other part of the definition of Homology group

[vast estuary]

mhm?

[gentle lark]

This is basically because boundary of boundary is 0

[gentle lark]

The space you're quotienting

Is ker boundary_1

Hmm wait

[vast estuary]

this only gives

[gentle lark]

I'm tripping up here I think

[gentle osprey]

Hausdorff

[vast estuary]

we're using all the other info

[gentle osprey]

Hausdorff

[gentle lark]

Is this relative homology per chance?

Because I don't see how it makes sense for gamma to be the boundary of a 2-cycle

Since it'd have to be... empty?

But with relative homology it could kind of make sense if you allow cycles to have their boundary in the space by which you're relativizing(?)

[keen urchin]

can someone tell me if I understood this problem correctly?

Show that if X is a countable product of spaces having countable dense subsets, then X has a countable dense subset.

My solution:
Write $X = X_1 \times X_2 \times X_3 \times \cdots$. \
Let each $X_i$ have the countable dense subset $A_i$. \
Now $A = A_1 \times X_2 \times X_3 \times \cdots$ is a countable dense subset of $X$.

[gentle osprey]

reking

[vast estuary]

nope

idk what that means so it's definitely not that

it's just singular homology

[vast estuary]

certainly there exists such a gamma' in C_2, we need to show that gamma' is in Z_2 \subset C_2

[gentle lark]

Can you provide the context?

The more I think about it the less sense it's making

A 2-cycle has no boundary by definition

So gamma could never be its boundary

[vast estuary]

Okay sure wait

https://ocw.mit.edu/courses/mathematics/18-904-seminar-in-topology-spring-2011/final-paper/MIT18_904S11_finalHurewicz.pdf

All the context^

I am on Pg. 4, the "kernel" part. The goal is to show that the kernel of h is a subset of the commutator subgroup.

This part

[vast estuary]

@empty grove You might be able to help

[empty grove]

Gaming time

[vast estuary]

haha later is alright

[vast estuary]

exactly!!!

[empty grove]

what's the doubt?

[vast estuary]

i think it's incorrectly stated, but i'm shocked because it's mit?

[vast estuary]

"gamma is a 1-cycle, and it is homologous to zero .... so gamma is the boundary of some 2-cycle"

@empty grove

[vast estuary]

this is certainly wrong lol

2-cycles have zero boundary

did they mean 2-chain instead?

[empty grove]

probably yeah

[vast estuary]

ahh okay. i hope it doesn't affect the argument after this, lemme read

thanks!

umm their proof of kernel \subset commutator subgroup is very convoluted

do you have any other ideas?

[fading vale]

I mean if you understand whats going on geometrically it isnt horrible

[vast estuary]

...what's going on geometrically

i'm afraid i don't understand the commutator subgroup geometrically

[gritty widget]

Hey, I have a topological subspace (0, 1) U [2, 3] of R. I'm trying to get why (0, 1) is closed there. I understand why it's closed with the help of it's complement, but I don't understand how for example sequence 1/n for n>=2, is in this space but it's limit is not

[gentle osprey]

when the yass pills hit

[fading vale]

Makes sense so far?

Since $\partial$ commutes with sums and multiplication by the $n_i$ we have $f = \partial(\sum n_i \sigma_i) = \sum n_i \partial(\sigma_i)$

[gentle osprey]

when the yass pills hit

when the yass pills hit
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[severe frigate]

your ambient space is (0,1) U [2,3]. so the sequence 1/n lives in this space, but it does not converge

[gentle osprey]

when the yass pills hit

[fading vale]

@vast estuary Does this make sense

[gritty widget]

OH I see

Thanks!

I was looking at convergence in R lol, 🤣

[severe frigate]

yup np i was getting tripped up as well

[pearl holly]

Daim that’s kind of trippy, y’all have the same gray pfp

I always just look at the pfp

[quartz edge]

I've got the following problem

[gentle osprey]

grist bundle

[quartz edge]

My reasoning goes

(also how do you write topology T in latex)

[dim radish]

\mathcal{T}

[quartz edge]

Take the union of the family of topologies

This may not itself be a topology but it must be contained in the one we want

It does however form a subbasis for a single topology

That topology is necessarily the smallest

Generate it and it's clearly unique qed

Is this reasoning fine?

Because any must necessarily contain the union

How haven't i said anything

Hmm

I could more explicitly show that

But i feel it's unnecessary

Like i can see another criticism

Like "maybe there's a smaller one than the one generated by completion of the union" in which case you just say "well it's then either not going to contain the union, or an elt of the completion of the union, show it's no longer a topology"

[rancid umbra]

i may be misunderstanding this, but the union of all T_a certainly contains all of the T_a, so shouldn't the smallest topology containing all T_a actually be contained in the union of all of these?

[quartz edge]

The union is exactly all T_a

It's no larger

No smaller

Kk thank you

I agree if i was turning in a hw assignment i would be more explicit about why this works

[rancid umbra]

okay... that doesnt answer my question

[quartz edge]

The smallest topology containing all T_a is at least as large as the union

The reason it might not be exactly the union is because the union might not itself be a topology

[rancid umbra]

oh my b

brain went ooga booga

[quartz edge]

(consider if you had two topologies, T_1 and T_2. The union of them contains both but may not, for example, be closed under intersection so that it's a topology)

[quartz edge]

https://tenor.com/view/ooga-booga-goldberg-mighty-ducks-gif-12294379

[pearl holly]

Oogiboogi :checkmark:

(I can’t copy the checkmark in my name for some reason)

[empty grove]

Imagine not having the checkmark on your keyboard ✓

[quartz edge]

Is there a symbol to denote the completion of a subbasis into a topology

[pearl holly]

WHAT

[quartz edge]

E.g. If S is a subbasis do i write \overline{S}

[pearl holly]

lmao I didn’t even know you could have that as a keyboard thing

[empty grove]

It's there by default on gboard

[pearl holly]

I don’t think apple has gboard

Like iPhone and stuff

[empty grove]

I have never seen any notation for this so I'd say define one and use it

I'd probably use \mathcal T (S)

[quartz edge]

Ah! Nice

[empty grove]

gee

Point set topology exam tomorrow

I'm also upset because I need to wake up early for it

Everyone be online at 4:30 am UTC tomorrow no reason

[quartz edge]

When the quiet kid says 'come to school tomorrow'

[empty grove]

I'm the quiet kid?

[quartz edge]

You're the bubbly kid tbh

[empty grove]

https://tenor.com/view/cat-cat-bubble-cat-cute-cat-surprise-gif-21923476

[vast estuary]

yes it does

[pearl holly]

YOOOO daim man good luck!!! you will do great!

[stone cipher]

Is a compact subset of a locally compact Hausdorff space, a closed set?

[rancid umbra]

compact subsets of hausdorff spaces are closed

[stone cipher]

aaahhh yes lol

[ivory forge]

is x^7+y^7<=30 a compact subset of R^2

[gritty widget]

i dunno

what do you think

[ivory forge]

iw anna say yes bc its closed and bounded is it not,,

i had a classmate tell me who is rly good at this stuff dont overdo it

which tells me to not overthink

topology is so bad man

[gritty widget]

well is it closed?

[ivory forge]

i hate iti

[pearl holly]

Do at then

[gritty widget]

so is this set closed and bounded?

i think one of these goes wrong

have you tried figuring out what the set looks like?

what have you tried?

[ivory forge]

hmm

i just put it in a graphing calc website

.

idk why i didnt do tht earlier 💀

[gritty widget]

a graphing calculator isn't really a proof, but it can help you formulate your answer

so what do you think?

[pearl holly]

Proof by graphical calculator

[ivory forge]

^

well

i feel like im overthinking this rly hard 🤣

[gritty widget]

we cant tell you if you're overthinking it if you're not telling us what you're thinking

[ivory forge]

cuz ifeel like this was taught in high school

but

considering it goes to inifinty, its still bounded right

'_'

i dont think bounded by inifinity but idk how to say it

[gritty widget]

"it goes to infinity" is kinda vague, but that tells me it probably shouldnt be bounded

do you know what it means for a set to be bounded?

[ivory forge]

well it has lower & upper bds

& im p sure this doesnt

[gritty widget]

this is a subset of R^2, so saying it has an upper or lower bound doesn't make much sense

[ivory forge]

it needs upper & lower bounds to be bdd in my txtbook

& yea

thts wat i was thinking too

[gritty widget]

for a set in R^2 to be bounded, you just need to be able to stick it inside a ball

do you think you can fit this set inside a ball?

yknow a set like {x^2 + y^2 <= r^2}

[pearl holly]

How about deez nuts?

[ivory forge]

🤣

[pearl holly]

Sorry lmao

[gritty widget]

deez balls

[ivory forge]

no cuz the set is gonna keep going

'_'

i wish i could use the graph as a proof

🤣

💀

[gritty widget]

yeah so your intuition says this set won't be bounded

so now you have to prove that, somehow

if you can do that, then you'll have shown that the set is not compact

[ivory forge]

deez graphs

okie

i think i can do that

[gritty widget]

one way to do it is to show that, given any ball, you can find points in the set outside of it. another way is to come up with a sequence of points in the set that shoots off to infinity

[ivory forge]

just starting it is always the worst part

the way my topology prof grades, a 50 counts as a C so you'd only need a 50 to pass 😭 😭 💀

[long coyote]

Let $0\to C' \xrightarrow{f} C \xrightarrow{g} C'' \to 0$ be a short exact sequence of chain complexes. Write out the definition of the connecting homomorphisms and verify exactness of the long exact sequence.

[gentle osprey]

亜城木 夢叶

[long coyote]

Let $0\to C' \xrightarrow{f} C \xrightarrow{g} C'' \to 0$ be a short exact sequence of a chain complexes such that $f:C'\to C$ and $g:C\to C''$. Let $n\in\mathbb{Z}$, the group homomorphism $d_n:H_n(C'')\to H_{n-1}(C')$ is called connecting homomorphism is defined by
$$d_{n}(z+\text{im}(\partial_n^{''}))=x+\text{im}(\partial_{n-1}^{'})$$
where $z\in \ker(\partial_n^{''})$ and $x\in \ker(\partial_{n-1}^{'})$, and the corresponding long exact sequence is
$$H_n(C')\to H_{n}(C) \to H_n(C'')\xrightarrow{d_n} H_{n-1}(C')\to H_{n-1}(C)\to H_{n-1}(C'')$$
To verify $H_n(C'')\xrightarrow{d_n} H_{n-1}(C')\xrightarrow{f_{n-1}} H_{n-1}(C)$ is exact, we need to show $\text{im}(d_n)=\ker(f_{n-1})$ and $\text{im}(f_{n-1})=\ker(d_n)$.

[gentle osprey]

亜城木 夢叶

[long coyote]

am i doing it right

[onyx crow]

You have to say how to get x from z

[gritty widget]

[orchid forge]

ah yes, I love induced long exact sequence in homology. Big fan of induced long exact sequence in homology.

[plain raven]

the wise man bowed his head solemnly and spoke: short exact sequences of chain complexes induce long exact sequences in homology.

[fading vale]

1. your defn of the connecting homomorphism is not sufficient (you havent related x to z in any way)
2. exactness at H_{n-1}(C') just means that im(d_N) = ker(f_n-1), the other equality is not even defined

[long coyote]

I need to show im = ker at each C

[vast estuary]

snake lemma?

[empty grove]

Indeed I did

[coral pivot]

[pearl holly]

Lessgoo

But I though you already did point set

[empty grove]

Have to do it again for MSc it is apparently a harder course because we will do van kampen at some point which we didn't do in BSc topology

I got permission from instructor to not attend any class

Have attended nothing since the beginning of sem

[pearl holly]

Yooo wtf lmao

That’s

[coral pivot]

Here i dont need permission to do that

[vast estuary]

but van kampen is alg top lol

[empty grove]

[orchid forge]

it's in munkres

that makes it point set

[empty grove]

I didn't need permission either but don't wanna get on bad side lul he is the model theory prof

[coral pivot]

ah i see

[vast estuary]

ooh do you wanna get a rec from him lol

[empty grove]

Ye first half is point set

Second half AT

[empty grove]

Not rec

But I want to TA model theory

[vast estuary]

ooh cool stuff

[marble socket]

moldi best TA

[viral atlas]

What was your TAing stipend again

[empty grove]

[empty grove]

[coral pivot]

moldi catkinging his students

[empty grove]

6k for entire sem

[coral pivot]

is that rupees??

[viral atlas]

(consider leaking class notes and problem sets for your model theory course)

[viral atlas]

Yes

[coral pivot]

wtf

[empty grove]

At the end of summer I TAed 3 week review session for first years who are just getting into second year and made same amount of money

That is way easier

No grading to do either

Just take 3 tutorials

[viral atlas]

And you can make as much in 3h tutoring JEE weirdchamps

[empty grove]

And I catpilled some of them at the end of one tutorial

[gentle osprey]

Hausdorff

[vast estuary]

lol sorry i kinda interrupted a fun conversation

[empty grove]

Is pi tilde the abelianisation

[vast estuary]

yes

[empty grove]

Isn't it by definition of a boundary 😵‍💫

ohh

For gamma a 2 simplex

[gentle osprey]

Hausdorff

[vast estuary]

is it because they're homeomorphic

[empty grove]

It don't matter, (IxI, ∂I) is homeomorphic to (Δ², ∂Δ²)

[vast estuary]

tau?

[empty grove]

Yes tau

Well it actually depends on how they are triangulating I x I

[vast estuary]

ahh cool! nice, so when they say singular n-simplices, i can take anything homeomorphic to the standard n-simplex in the domain

[empty grove]

Ye, assuming they are triangulating that homeomorphic thing via the homeomorphism

[vast estuary]

they're doing this it seems

[empty grove]

Ouch

Then it's not immediate

[vast estuary]

why not

[empty grove]

Because the boundary is not computed the same way

Like in the first case, you have a map from I x I but really you are just treating it as a map from Δ²

By composing with homeomorphism

But here it's different

So from this case, you can prove the case for any 2 simplex

Take the 2 simplex you wanna prove for, embed that into I x I as one of the 2 triangles, define the map on the other triangle by saying constant along the (bottom left to top right) diagonal

This should work intuitively

[vast estuary]

ooh okay cool! thanks

[vast estuary]

the 2-simplex i wanna prove for is arbitrary here right

like gamma is the boundary of SOME 2-simplex

[empty grove]

More general than that, gamma is boundary of a 2 chain

But if boundaries of 2 simplices map to 0

So do those of 2 chains

Because they're sums of boundaries of 2 simplices

Because boundary is a homomorphism

[empty grove]

Lul there might be something easier

Idk

[vast estuary]

yes agreed so it suffices to prove for a 2-simplex

ok for now i'm just gonna assume that it's ok to prove for I x I

and move on

[true garden]

Hello. Can you give a simple intuitive example of a subspace whose (covering) topological dimension is greater than that of the space containing it?

[gentle osprey]

Hausdorff

[gritty widget]

direct product of Z with itself

n times

[gentle osprey]

Hausdorff

[vast estuary]

Oh yes this makes sense, I know how to prove it! Fundamental groups of finite products of spaces = direct sum of fundamental groups of individual spaces. YAY

[gentle lark]

Neat related fact: the (dimension of the) homology group of tori form Pascal's triangle

Intuitively, this is because you write out T^n as a product of circles, then pick k circles and that's an element of a basis for the k Homology (and, specifically, a T^k)

[loud scarab]

If I have a union that looks like this:

$$\bigcup_{ x \in U} { x , -x }$$

[ivory dragon]

\{ \}

[gentle osprey]

Khaled

[loud scarab]

Where U is open in S^n, visually I know that the union of {-x} is open, but can it be shown more formally?

[gentle lark]

Can you use that x -> -x is a homeomorphism?

[loud scarab]

Yes I guess I can do that

thanks

[feral copper]

(oh I see the channels have been renamed to separate topology and geometry, maybe this shall belong there, idk)

[fading vale]

Yeah i think this goes in #diff-geo-diff-top

[feral copper]

Moved 😉

[fading vale]

Nice

Good luck getting answers, maybe tterra will know

[feral copper]

Thanks!

Regarding the change of channels' names: shouldn't the roles from #get-advanced-access be tweaked accordingly?

[fading vale]

Yeah we are working on that

[obsidian socket]

Hi, how would I express a connected sum of two torus, but glued at two places?

[orchid forge]

i would express it just like that

[quartz edge]

Section 16 Problem 5 in Munkres.

[gentle osprey]

grist bundle

[quartz edge]

All that I can tell we know here is that X is in T' and Y is in U'. I'm not sure what the strategy is supposed to be for the problem

I see now

Diagrams help

[rain shore]

Maybe compare the bases of topology by definition of product topology

[quartz edge]

Yeah

I had to get everything written out to be able to look at it plainly

It makes sense now

[empty grove]

What does single set mean 😵‍💫

If X x Y and X' x Y' are different sets can you even talk about topologies being finer or coarser

[rain shore]

I think "single" means that these are the same sets, but with different topologies, but of course this is the terrible notation

[empty grove]

oh wow ok

[true garden]

Hello. Can you give a simple intuitive example of a subspace whose (covering) topological dimension is greater than that of the space containing it?

[gritty widget]

you didn't get an answer the last four times

have you tried google or MSE?

[true garden]

Yes.

[gritty widget]

were there no answers there?

if you found a partial or unsatisfactory answer online, maybe you could post it here to make your question more specific and therefore accessible

[true garden]

There may be some examples, but neither of them are simple and intuitive.

[gentle osprey]

expectTheUnexpected

[vocal anchor]

oh. no.

[orchid forge]

seems a tough ask for an intuitive example of an intuitively false statement

[obsidian socket]

Does the Tietze extension theorem imply there is a unique extension? Or just that there is some extension?

[empty grove]

no unique

only exist

you can find counterexamples to uniqueness in R

[obsidian socket]

very sad 😦

[gentle lark]

Why the sad face?

Topology would be incredibly boring if we had uniqueness here 😛

[empty grove]

Now replace topology with complex analysis

Cry

[gritty widget]

nope question

is the restriction of a homeomoprhims

always a homeomorphism

[empty grove]

Yes onto its image

[pastel coral]

Hello, can anyone explain to me why any basepoint preserving closed loop in a torus is always homotopic to $aba^{-1}b^{-1}$, where $a$ and $b$ are the toroidal and poloidal loops intersecting the basepoint of the torus.

[gentle osprey]

blackiris

[pearl holly]

look at the fundamental polygon or whatever it's called

the "gluing diagram"

[pastel coral]

I see, but what I'm seeing is that I only seem to need ab without the inverses.

Oh wait

[pearl holly]

------>-----
| |
|^ |^
|----->----|

lmao

[pastel coral]

Yes

I think moving once in the loop using the two generators isnt enough to deform into an arbitrary loop.

[pearl holly]

but if you place a loop inside of that square you can always deform it so that it touches the sides of the square

so any loop will be homotopic to the boundary of the torus

[pastel coral]

Yeah I think I can see it now

[pearl holly]

and the boundary is given by aba^-1b^-1, the inverses are there because you go the in the opposite direction

[pastel coral]

It was hard because I was drawing it with the edges of the inverses coinciding.

[pearl holly]

oh I see

[pastel coral]

So all im seeing is ab lol

Thanks!

[soft rivet]

Hi

I’m not sure if I can ask this here

But I’ll post it

Can anyone please help me by checking them

Ping me too

[empty grove]

tf this gotta do with topology

[ivory dragon]

i checked, none of those are topological homeomorphisms

hope that helps!

[empty grove]

I like the smaller big OH tho 👌

[ivory dragon]

actually idk what topology youre putting on ℕ

maybe they are homeomorphisms

[empty grove]

It's always discrete 🔫

[ivory dragon]

shame

[empty grove]

What's MEUM

[soft rivet]

Where can I ask this ?

Discrete?

[ivory dragon]

the best channel is probably #discrete-math but honestly id recommend a cs/programming server.

[soft rivet]

But can y’all check it tho

[ivory dragon]

no.

[soft rivet]

K

[empty grove]

[soft rivet]

U want me to ask there?

[empty grove]

Bruh moment

[soft rivet]

Nvmmm

[gritty widget]

Hey I'm looking at some proof and it claims something I don't get, Let $K=S \cap A$ where $ S \subseteq T = A \cup B $ and $ A, B $ are clopen sets that partition $T$. The author claims that for some point $ a \in A $ we can find a sequence from $ S $ that converges to $ a $. How?

[gentle osprey]

CRAED0X

[bleak helm]

What is the significance of K in what you said? Did you mean a in K?

[gritty widget]

Ah nvm

I got it

[bleak helm]

Okay

[gritty widget]

The thing was that S \subseteq T \subseteq \overline{S}

So yeah

Thanks for reading anyways !

[bleak helm]

Ah, np

[finite heath]

hey guys

how does my work for:

look

[orchid forge]

i think it should say T2 is finer than T1 right?

[finite heath]

i messed up in that Z+ should be Z because I need the basis elements to span R

but how would T2 be finer than T1 here?

[orchid forge]

it contains more open sets

[gritty widget]

why is orientation cover always oreintable?

the justification given in my class seems silly,

like no matter if the cover is over an unorientable manifold the cover is always orientable

[orchid forge]

what was the justification given in your class?

intuitively I would say the reason is that the orientation cover is created by looking at where orientability fails (so a transition map becomes orientation-reversing), and replacing one of those sets with a disjoint copy in the opposite orientation, "fixing" orientability

[gritty widget]

lots of notation but pretty much local homology of open sets of orientable cover are related by a chain of isomorphisms leading back to original manifold

ill post pic

[orchid forge]

but the precise details depend on how you construct the cover

[gritty widget]

seems super tautological lol

ok looks like i wrote sloppy

,rotate

[gritty widget]

prettt much H_n(~M|u_x) is supposed to be H_n(~M,~M-(x,u_x)) where u_x is generator of local orientation at x

and U(u_B) is defined to be the set of all local orientations that are compatible within an open ball around x in the manifold

we are using hatcher as source

this way of thinking about orientations is weird to me. because we say a manifold has orientation if there exist a function from x in manifold to u_x, a local orientation at x such that they are compatible with local orientations of points in open ball around x

[true garden]

Hello. Can you give a simple intuitive example of a subspace whose (covering) topological dimension is greater than that of the space containing it?

[gritty widget]

[empty grove]

Just 😄 and 👋 terra

[ivory dragon]

insert that lame facebook "definition of insanity" quote

[orchid forge]

lol

[true garden]

I was called "insane"?

[wanton marsh]

which one is the topological dimension again

[empty grove]

lebesgue covering dimension

[marble socket]

is that question answered already? i see it everyday in this channel

[gritty widget]

To answer your question

No, I cannot a simple intuitive example of a subspace whose (covering) topological dimension is greater than that of the space containing it.

[fading vale]

Im not even convinced this is possible, doesnt any cover of ply n restrict to one of ply less than or equal to n on a subspace

Maybe im misremembering the defn ot missing something though

[wanton marsh]

if the subset is closed it's not possible but otherwise I don't know

[fading vale]

Yeah i cant see a proof off the top of my head in the non closed case

Hmmm

This cant happen for finite simplicial complexes

Someone find a paracompact hausdorff space whose cohomological dimension is smaller than that of a paracompact hausdorff subspacr

[gritty widget]

bruh what are plys you mean sheets?

[fading vale]

No

Wikipedia lebesgue covering dimension

[gentle lark]

Hmm

I wonder what the covering dimension of the Vitali set is

I'm sure the answer is either very interesting or very trivial

[gritty widget]

oof

so much vocab

minimal number of open sets of covers?

[gentle lark]

Bit more involved than that

Take all possible open covers

And minimize [minimum n such that p belongs to at most n sets of the open cover, for all p in the space]

[gritty widget]

lol

isnt the circle an example?

no ig not because it is embedded in 2d

[empty grove]

it is 1d in 2d

[bleak path]

Hey, quick question, is the real projective space of n dimensions RP^n homeomorphic to S^n?

[empty grove]

only for n=1

[bleak path]

Right, and that's because of the equivalence relation that identifies antipodal points, right?

[empty grove]

I don't know what you mean by "because of the equivalence relation"

Like that is the definition of RP1

but you can construct a homeomorphism using the universal property of quotients

[bleak path]

Hmmmm, alright

Will ponder it for a while, thanks!

[surreal ocean]

is there a notion of one point being closer than another to some reference point in a topological space? I read that if two points x_0 and y_0 are contained in more open sets than the points x_0 and y_1 then y_0 is closer to x_0. Intuitively this does not seem to hold up when I think about R^n with the standard topology. I would expect the set of all open sets containing x_0 and y_0 to have the same cardinality as the sets containing x_0 and y_1. If this is true then every point would be equally close to every other point in R^n.

[tawdry valve]

You’re definitely right about cardinality not being a good test of comparative closeness.

It turns out that the topology doesn’t really specify this—closeness isn’t preserved by continuous maps. For example, you need your notion of “closeness” to define cauchy sequences and what it means for a metric space (or topological group) to be complete. However, completeness is not a topological property, since R is complete but (0,1) is not.

If you want to talk about this notion of “closeness” though, you can look into
https://en.m.wikipedia.org/wiki/Uniform_space

It’s a little bit technical, and at least for me I’ve only ever had to look at specific cases like metric spaces and topological groups (which are uniform spaces) instead of uniform spaces in general.

[surreal ocean]

to summarize you are saying that we need to add additional structure in order to define comparative closeness? also could you go into more detail about “It turns out that the topology doesn’t really specify this—closeness isn’t preserved by continuous maps.“

why would a property not being preserved by a continuous map imply that the topology doesn’t specify that property

[sweet wing]

topological spaces are really only up to homeomorphism

[surreal ocean]

ohh i see that just helped me piece something together thanks

[finite heath]

facts

[vast estuary]

umm quick question

does this ensure that V is connected too?

in other words, is every open subset of an open connected set also connected?

[gritty widget]

no

definitely not

[fleet trench]

V is a disconnected open subset of the connected open set U

[orchid forge]

[gritty widget]

looks smth like presheaf

what u doing?

[vast estuary]

it's complex analysis lol

[orchid forge]

it's gonna be sheaf

etale space construction of riemann surfaces

$\emptyset \subset \mathcal{T}$ so $\bigcup \emptyset = \emptyset \in \mathcal{T}$ from part 2

[gentle osprey]

Kogasa

[orchid forge]

certainly

[coral pawn]

Can someone verify these calculations?

[gritty widget]

oh my gods

[coral pawn]

I'm sorry

Is it wise to call the helpers for this problem?

Maybe someone is willing to go through this computation

[mossy ermine]

No one is

[empty grove]

do not underestimate the greens

[coral pawn]

<@&286206848099549185>

Hopefully I don't get banned for this

Though I would totally understand if I did

[ivory dragon]

i honestly think it might be faster to write a python script to verify this than try to do it by hand

[coral pawn]

I don't really know how to do this in python

[orchid forge]

surely there is a way to not do this computation

[coral pawn]

My professor said that there isn't

We have to do this explicitly

[honest terrace]

Your prof is a bad person

[fading vale]

Intepreting it literally in terms of cardinality isnt very helpful, but its a good intuitive heuristic for what open sets as an idea are designed to preserve

And this works in R^n

E.g epsilon delta

[gritty widget]

I’m confused

What do you mean verify?

What steps are missing?

[fading vale]

Theyre literally asking someone ti verify that the computation has no errors

[gritty widget]

Oooh

I’m going to pass on that thank you

[fading vale]

[gritty widget]

math server afraid of a little computation as always

just a suggestion

[gentle osprey]

TTerra

[gritty widget]

wait i deleted the source is it a ghost ping now

[empty grove]

Not if you ping them again

[gritty widget]

true!

@coral pawn

[pearl holly]

is this differential topology or what is this?

I assume not because it would go in the other channel so what is this?

[gritty widget]

it's multivariable calculus

[pearl holly]

daim I will never touch multivariable calc then

[barren sedge]

Trying to do this question, but I have no idea where L tensor L comes in exactly, can anyone give any advice?

[barren sedge]

<@&286206848099549185>

[bleak path]

Hi, have a simple question on Hausdorff spaces - I know that in a Hausdorff space, all singletons are closed, but is the reverse true? If all singletons are closed in a space, is that space Hausdorff?

[gritty widget]

such a space is called a T_1 space, and it's not true that every T_1 space is hausdorff. an example is an infinite set with the cofinite topology

[bleak path]

Thank you!

Have not learnt about T_1 spaces yet, though have seen that they exist via wikipedia

[bleak path]

Heya, another question on Hausdorff spaces - If we have a relation ~ on a Hausdorff space X, then X/~ is not necessarily Hausdorff (I believe). But, for any space Y, if Y/~ is Hausdorff, does that mean Y is Hausdorff?

[gritty widget]

No I dont think so, cant you just make Y/~ into a point no matter what Y is?

[bleak path]

I haven't thought about that before, but I guess so? By the relation that everything is related to reach other. But are spaces composed of only singletons Hausdorff?

Or rather is the space composed of only a single point Hausdorff?

[gritty widget]

Yeah

"For any two points..."

trivially true

[bleak path]

Right, and so if there is only 1 point then it's automatically satisfied

[gritty widget]

yep

[bleak path]

And so if the relation is true then the statement is true

Thanks!

[digital wraith]

ok I'm struggling with exercise 20 in Hatcher 2.1

I've tried a number of different things and none of them seem to be working

[plain raven]

Hmm mayer vietoris?

[digital wraith]

don't know that yet

[plain raven]

ok.

is this the reduced suspension?

i don't remember what his convention is in this chapter

[digital wraith]

this is unreduced

which I guess follows from "thinking of SX as the union of two cones CX with their bases identified"

[plain raven]

Well if you delete the two endpoints then what remains should deformation retract onto X.