point-set-topology

[gentle osprey]

Morphy

[tacit coral]

so, let's say $a_n$ converges to $A$ and $b_n$ converges to $B$ in $X$, and $* : X \cross X \to X$ is some continuous binary operation (say $X = \mathbb{R}$ and $$ is $+$ or $\cross$). then there's a continuous map $n \to (a_n, b_n)$, $\infty \to (A, B)$ by the universal property of $X \cross X$, and when you compose it with $$ you can conclude that $a_n * b_n$ converges to $A * B$ in X.

[gentle osprey]

Morphy

[unreal stratus]

Yeah this is p dope

[grizzled ibex]

I think i saw something like this but with X=Q as a alternative for dedekind cut

[tacit coral]

maybe you mean the construction of R as equivalence classes of Cauchy sequences? I think it's only related in the sense that it involves sequences and convergence

[grizzled ibex]

i don't think so ;c

[tacit coral]

interesting

[grizzled ibex]

it does not involves sequences directly if im not mistaken, it's more the idea of topological closure

[tacit coral]

if you can find it I'd love to see it

[grizzled ibex]

of the (Q,+,1)

(as a topological group)

[tacit coral]

meaning Q as a pointed topological group with addition and 1 as the basepoint?

[grizzled ibex]

prob, i didn't finished this chapter yet, it was some kind of "spoiler" from the author

it's in my native language

[tacit coral]

do you remember which book?

[grizzled ibex]

so i don't think it helps to send you

[tacit coral]

cause R does end up precisely adding the points that make it topologically connected

[grizzled ibex]

i can look on the references

w8 a sec

but

i think the main reference is bourbaki

not pretty sure

the book itself is bourbaki being digested

there are over 100 references

i'll send you pm

[rancid umbra]

u good?

[honest narwhal]

Not good

In fact ded

[rancid umbra]

lul

[hidden crag]

[honest narwhal]

Dam Timo beat me to the punch

[hidden crag]

easy

[trail charm]

ty!

[trail charm]

if we define the 2-torus to be the product manifold S^1 times S^1, Lee considers it to be a subspace of R^4. someone on mathSE said that there is also an embedding of T^2 to R^3, where we consider the "surface of the bagel"

what does that embedding look like?

also in the general case, if we regard T^n as the product of n copies of S^1, is it a subspace of R^2n? and if we generalize the "surface of the bagel" analogy, would that embedding be T^n to R^2n-1?

[stoic eagle]

T^n is usually thought of as subset of R^n+1

[feral copper]

So I did something today 😇

[umbral panther]

Which subset? Is it so easy to write down?
||I guess you could define it inductively as the boundary of a small neighborhood of T^n-1. But do you really want that?||

[feral copper]

Which one? The embedding of T(n) into R(2n) is called the flat torus. If you type that, you'll see a representation of the flat T(2), although I'm not sure what's being drawn.
Flat comes from the fact that the Riemannian metric induced by the embedding is flat.
However, if you mean the embedding of T(n) into R(n+1), then this is the standard embedding (non-standard terminology here). The embedding of T(2) inside R(3) is, well, the outside of the doughnut.

[feral copper]

https://math.stackexchange.com/questions/307476/embedding-of-n-torus-in-mathbbrn1-via-implicit-function-tn-f-10

[trail charm]

oh didn’t know they had names

i’ll look into this, ty!

[umbral panther]

Like I said, do you really want that?

[feral copper]

What I linked is just technology saying that T(n+1) is a surface of revolution of T(n), and that's how visual it can get

Unless you have a mental image of euclidean 4-space 😛

[feral copper]

Which, in turn, is saying that T(n) is the product of n circles, by induction

[tulip bridge]

Do you guys have a good alg top book recommendations?

preferably pretty beginner since my course is more of an intro to topology class and at this point we've barely even covered quotient spaces

[winged viper]

Lee's intro to topological manifolds is good as a beginner book if you aren't yet comfortable with point-set

the algebraic topology starts halfway through when he goes over the fundamental group

before that it's mostly the point-set topology that is useful for algebraic and differential topology (so he doesn't go into things like tychonoff's theorem and the details of countability/separation axioms). i think he also covers CW complexes and proves some results with them using only point-set topology

but you could probably start reading from when he starts talking about the fundamental group, and go back if you come across terms that are unfamiliar

it might be useful to read the sections on quotient spaces since those are pretty important

but imo it is a lot more beginner friendly than hatcher, especially when he covers things like van kampen and covering spaces

[trail charm]

also, lee’s intro to smooth manifolds is a sequel to intro to top manifolds

[zinc siren]

May's book, A Concise Introduction to Algebraic Topology. You should be warned, however, that it is difficult. It is, in some sense, "beginner level" in that little knowledge is assumed

[hidden crag]

I would definitely stay away from Mays book if you’re a beginner

[slow marlin]

it's full of category theory, i had a prof put it on reading list, gave me PTSD

[chrome plinth]

Why is euclidean metric of R^2 on the unit circle S^1 not a path metric space but measuring the angle in radians is? To my understanding the induced intrinsic metric should find the infimum of the lengths of all paths. Does the straight line between points not count as a path?

[stoic eagle]

It’s not contained in S^1 though

[long grail]

Is there a one size fits all strategy for showing an arbitrary open cover has a finite subcover (assume compact)

I'm trying to prove in a separable Hilbert space, the closed unit ball is compact in the weak topology

[chrome plinth]

What does it mean to be contained in S^1?

[stoic eagle]

As a subset

[unreal stratus]

Basically, π is not 1

[chrome plinth]

I'm afraid I don't understand what you're trying to say, how can we define what curves are contained in the set

[stoic eagle]

Taking the induced metic on S1 (from R^2) would have that the distance between points is the Euclidean one, but no path between them contained in S^1 has such length

So there’s no length minimizing path

[chrome plinth]

What constitutes a path being contained in S^1? Why is the direct path between the two points not in S^1?

[stoic eagle]

S^1 is the unit circle right?

[chrome plinth]

Obviously I understand intuitively that the chord is not in the circle

[unreal stratus]

That's it though

[chrome plinth]

But formally how do we define when a path is in the set

But thats intuition. Its hardly a formal definition of an intrinsic metric

[stoic eagle]

It’s the image of a cont function from the unit interval

So it’s just a subset

You can show by calculation that these straight lines don’t have only points of norm 1 (which is what it means to be in the circle)

So that would prove it

[chrome plinth]

Does an intrinsic metric only work on sets with the same cardinality as the reals then?

[stoic eagle]

I’m not sure what you’re asking

[chrome plinth]

Well intuitively an intrinsic metric on a circle must be a path that follows what we intuitively see as a solid curve. If you have a finite set then how do you define an intrinsic metric on it? What would be your continuous map?

[stoic eagle]

It’s more about path connectedness than about cardinality but sure

I don’t really know why that’s relevant to your original question tho

[chrome plinth]

Sorry I will limit myself to one question

Wrt my original question I just wanted to try and understand what really constitutes a path being inside the set. Where can I read a formal definition of what they mean by path?

[tawdry valve]

a path $\gamma$ in a metric space $(X,d)$ is a continuous function $\gamma: [0,1] \to X$

[gentle osprey]

Joseph

[tawdry valve]

and when we say a path $\gamma$ is inside a set, for example $S^1 \subseteq \mathbb R^2$, we mean that for all $t\in [0,1]$, $\gamma(t) \in S^1$

[gentle osprey]

Joseph

[tawdry valve]

maybe you'd appreciate a set of notes. I had taken a course following these notes, page 13 is relevant to this discussion https://jypan10.github.io/113notes.pdf

[long grail]

What are some big implications when a space is seperable? Anything to do with showing compactness?

[unreal stratus]

For metric spaces, separability is equivalent to every cover having a countable subcover I believe

[untold lily]

yeh, for metric spaces lindelof iff separable iff second countable

[tawdry valve]

also for studying hilbert spaces, separable is useful because it is equivalent to a having a countable orthonormal basis

[untold lily]

coincidentally this is what I'm going through right now

[chrome plinth]

So if I have two points in S^1 (1, 0) and (0, 1) as you'd see it on the euclidean plane and with the chordic metric the distance between them is sqrt(2). If I define a map y: [0, 1] -> S^1

Such that [0, 0.5] yields (1,0) and (0.5, 1] yields (0, 1) can we say this is discontinuous because at some point there is a discontinuity in the metric applied over this function y? That is, for [0, 0.5] the path length is 0 and then suddenly a little bit further into the interval it is sqrt(2)

[tawdry valve]

yeah, that's a way to see that your example function is discontinuous, since given a continuous function $\gamma: [0,1] \to S^1$, this should induce a continuous function $L: [0,1] \to \mathbb R$ which gives you the length up to time $t$, but as you said, this length function has a discontinuity at $0.5$.

[gentle osprey]

Joseph

[chrome plinth]

Okay that's far more understandable. Thanks a lot

[tawdry valve]

the more traditional way to see that your example is not continuous would be using the definition of continuity. Here's the definition of continuity: a function $\gamma: [0,1]\to S^1$ is continuous if for all $t_0\in [0,1]$, for all $\varepsilon>0$ (think of this as your approximation error), there exists $\delta>0$ such that for all $t\in (t_0-\delta, t_0+\delta)$, you have $d(\gamma(t_0), \gamma(t)) < \varepsilon$.

Then, your example $\gamma$ is not continuous for $t_0=\frac{1}{2}$, since you cannot find $\delta$ for $\varepsilon=0.1$ for example. For any $\delta>0$, you will have $\gamma(\frac{1}{2}+\frac{\delta}{2}) = (0,1)$ while $\gamma(\frac{1}{2}-\frac{\delta}{2})=(1,0)$, and these points have distance greater than $\varepsilon$.

[chrome plinth]

Are you sure vareps is a token?

[gentle osprey]

Joseph

[chrome plinth]

Yay

[tawdry valve]

thanks lol yeah I need to setup my texit bot config or something

in my latex I have a command "\newcommand{\eps}{\varepsilon}"

[chrome plinth]

I made an entire command that would render a perspective view cube and then my config got wiped 😦

[tawdry valve]

aw sadge

[chrome plinth]

Sorry to be pedantic but should t_0 belong to the closed or open interval of 0 to 1?

I suppose it doesn't matter does it

[tawdry valve]

it should belong to the closed interval, but I should have $t\in (t_0 -\delta, t_0+\delta) \cap [0,1]$ for this to make sense

[gentle osprey]

Joseph

[tawdry valve]

so there was an issue

[chrome plinth]

Skill issue, only kidding thank you so much

Thats really helpful

[tawdry valve]

lol no problem

[unreal stratus]

Or just use intervals within [0,1] with the induced ordering and it's already correct 😎

[pastel coral]

I need some sanity check. Given a metric space $M$, is the collection of all open balls of the form $B_{\varepsilon}(y)$ for $\varepsilon>0$ and $y\in M$ still a basis for the topology if we uniformly bound $\varepsilon<r$ by some $r>$? I.e., if we have a uniform upper bound on the radii of the balls, do we still get a basis for the metric space topology?

[gentle osprey]

blackiris

[unreal stratus]

Yes

[long prairie]

Yes

[pastel coral]

Okay that makes sense

The topology generated must be the same. This is the same reason why in epsilon-delta proofs, we can always take epsilon smaller than some number.

[unreal stratus]

Basically if O is open, then you have a ball of positive radius about each point in O, and then you can just shrink each of those balls until they're small enough

because the union will still be the whole of O

[pastel coral]

Yeah, I just wanted to ask it here just in case I'm being stupid, and what I have in mind is not true.

Thanks

[unreal stratus]

Ye nws

W metric spaces ig you can also replace the set of radii with just like some (positive) sequence going to 0

by the same logic

[pastel coral]

This probably proves second countability of metric spaces iff separable

[unreal stratus]

Yeah ig this is an ingredient in it

It shows that metric spaces are first countable anyway

[little hemlock]

P(V) is the projective space of V, and P(E) is the projective bundle of E. Why does having natural isomorphisms P(V) -> P(V otimes W) imply that there is an isomorphism P(E) -> P(E otimes L)? Namely, how do we know the map between bundles we get is continuous?

[long prairie]

This is an old book

[little hemlock]

if there is a nowhere-vanishing section $\sigma : X \to L$ then there is an isomorphism $\varphi : P(E) \to P(E \otimes L)$ defined by $\varphi(e) = e \otimes \sigma(p(e))$ where $p : L \to X$ is the projection. i.e. we just need to make some continuous choice of nonzero elements in each fiber of $L$. On the other hand, a line bundle with a nowhere-vanishing section is trivial, so my map $\varphi$ is basically too rigid of a construction

[gentle osprey]

kxrider

[tidal lynx]

If S^3 is the unit sphere and is written as S^3 = {(x, y) \in \mathbb{C}^2 | |x|^2 + |y|^2} = 1}, and M = {(x, 0) | |x|^2 = 1} and N = {(0, y) | |y|^2 = 1} are the intersections of S^3 with the x/y axes in \mathbb{C}^2, why is the fundamental group of S^3 \ M equal to Z (the integers) ?

[unreal stratus]

why no latex lol

[tidal lynx]

If $S^3$ is the unit sphere and is written as $S^3 = {(x, y) \in \mathbb{C}^2 ~|~ |x|^2 + |y|^2 = 1}$, and $M = {(x, 0) ~|~ |x|^2 = 1}$ and $N = {(0, y) ~|~ |y|^2 = 1}$ are the intersections of $S^3$ with the $x/y$ axes in $\mathbb{C}^2$, why is the fundamental group of $S^3 \setminus M$ equal to $\mathbb{Z}$ ?

[gentle osprey]

mrean

[unreal stratus]

I imagine S^3 \ M deformation retracts onto N

[tidal lynx]

you're saying there exists a (onto) continuous map r: S^3 \ M -> A that restricts to the identity on A ? In that case the there's a homomorphism between the fundamental groups of S^3 \ M and A induced by the injection j: A -> X.

how would that even help tho because that's only a homomorphism not an isomorphism

[unreal stratus]

A deformation retraction induces an isomorphism on fundamental groups

What you described is just a retraction

[tidal lynx]

Ok I forgot what that was so time to review

also chatgpt's solution lol:

[unreal stratus]

Lol

The homomorphism Z -> Z is multiplication by 2pi

Oh ig makes sense but badly written

Anyway

The second paragraph claims A is homeo to D^2 x S^1 which is already enough

Lmao

[tidal lynx]

this one actually looks right

well hmm did it get the complement wrong

yeah wtf chatgpt you suck

[tidal lynx]

Ok time for actual solution

$r_t: S^3 \setminus M \to N$ defined by
$$ r_t(x, y, z, w) = (\sqrt{1 - t^2}x, \sqrt{1 - t^2}y, z, w) $$
for each $t \in [0, 1]$ should be a deformation retract. Moreover, $N = S^1$ by the map $(0, 0, z, w) \mapsto (z, w)$ so $\pi_1(S^3 \setminus M) = \pi_1(N) = \mathbb{Z}$

[gentle osprey]

mrean

[tidal lynx]

is this right @unreal stratus

[unreal stratus]

The image isn't contained in S^3

But it is fixable I think

Maybe

[tidal lynx]

image of which one

[unreal stratus]

Like

The norm of most elements changes through time

[tidal lynx]

you're saying the first condition fails?

[unreal stratus]

Well

This isn't even a map S^3\ M x I -> S^3 \ M, I mean

[tidal lynx]

oh damn it

so you're saying sometimes points (x, y, z, w) in the image are in M, i.e., x^2 + y^2 = 1 holds

also yeah it should be r_t: S^3 \ M -> S^3 \ M not r_t: S^3 \ M -> N

For a point (sqrt(1 - t^2) x, sqrt(1 - t^2) y, z, w) in the image
(sqrt(1 - t^2) x)^2 + (sqrt(1 - t^2) y)^2 = (1 - t^2) (x^2 + y^2) = 1 can never happen, because (1 - t)^2 < 1 so we would need x^2 + y^2 > 1, right?

@unreal stratus

I'm lowkey not sure why we can't just use like 1 - t tho

[unreal stratus]

No i'm not saying that lol

I meant that for example like

if x^2 + y^2 + z^2 + w^2 = 1

then (1-t^2)x^2 + y^2 + z^2 + w^2 < 1 typically

[tidal lynx]

ohh

ok from now on I write (z, w) (two complex numbers) instead of (x, y, z, w)

damn this is hard uhh

ok are u even sure S^3 \ M deformation retracts onto N lol @unreal stratus

[unreal stratus]

No but I bet it does, honestly not thought about it much tho

But identifying the complement w smth more tractable may be a good start idk

[tidal lynx]

Cry

[tidal lynx]

still can't come up with the deformation retract 😭

Wait @unreal stratus I just realized

why do we need the homotopy to be a deformation retract

doesn't homotopy equivalence between path-connected spaces in general mean that their fundamental group are isomorphic?

[little hemlock]

I think it’s just that the most obvious homotopy equivalence is a deformation retraction

[tidal lynx]

hmm

[tidal lynx]

wait

there's this observation

Every $p$ not on $M$ or $N$ is uniquely expressible as $p=\cos(\theta)m+\sin(\theta)n$ for points $m\in M$, $n\in N$ and convex angle $\theta\in[0,\frac{\pi}{2}]$, i.e., all $p$ lie on an arc between points of $M$ and $N$.

[gentle osprey]

mrean

[tidal lynx]

so close

[little hemlock]

as a hint i think you might be overthinking

either that or im underthinking

not sure how to share my idea w/o just giving away the answer

basically, the trick with constructing maps into spheres is to try to construct a map f into R^n - {0} (or C^n - {0} in our case) having roughly the properties you want, then define g(x) = f(x)/||f(x)||

[maiden pilot]

anybody have a clue as to prove the following?

Suppose we are given two basepoint-preserving maps from S^n to a space Z that agree on a disk A containing the basepoint p. Then if the two maps define the same element of pi_n(Z), they are homotopic by a homotopy that stays fixed on A.

Ive tried viewing the homotopy induced by equality in pi_n(Z) as a cylinder by viewing S^n as D^n with del D^n corresponding to the base point and A corresponding to an annulus in D^n. But i dont think that gives an obvious homotopy

[tidal lynx]

uh if I have the map go into a bigger space it's no longer a def. retract

[little hemlock]

if you have a map into R^n - {0}, then you have a map into a sphere by dividing out by the norm like i did to define g

[tidal lynx]

is it a def. retract tho

hm probably

ok so you're saying to find a homotopy between S^3 \ M and C^2 - {0} ?

and then divide out the norm

[little hemlock]

yeah, the goal is the same: find a deformation retract of S^3 \ M onto N, but instead of "trying to get things right on the nose", you find a homotopy into C^2 - {0} and then divide out by the norm to get the answer

[tidal lynx]

I've only dealt with homotopy equivalence between maps/paths

finding maps f and g between S^3 \ M and N such that f \circ g and g \circ f are homotopic to the identity map seems like a way more daunting task

[little hemlock]

you don't have to do that

[little hemlock]

you just need this

[tidal lynx]

minus the F(a, 1) = a part?

[little hemlock]

depending on your definition, you might only need F(a,1) \in A, and for our purposes, this suffices

the map i have in mind does satisfy F(a,1) = a

[tidal lynx]

I just thought that def. retracts had to go from a space to a subspace

and not a subspace to a space

[little hemlock]

not quite sure what u mean by that

[tidal lynx]

I asked this because the addition of the F(a, 1) = a condition turns F from just a homotopy into a def. retract no?

[little hemlock]

okay, yes, i agree

the whole idea with those conditions is that it is equivalent to there being a map
r : X -> A such that ri = id_A and ir homotopic to \id_X

where i : A -> X is the inclusion

[tidal lynx]

right ok

so really the homotopy (that I construct, of course it is invertible) should be from F(x, t): S^3 x [0, 1] -> (S^3 \ M)

and this you're saying is easy to construct to be a def. retract

[little hemlock]

should be from (S^3 \ M) x [0,1]

[tidal lynx]

how's that a def. retract then

[tidal lynx]

.

oh

was this what you had in mind

[little hemlock]

okay i get your confusion i think. We have a retraction r : X -> A. This does map onto a smaller subspace, but by "deformation retraction", I am referring to the homotopy we have to construct to show that r is a homotopy equivalence. namely, the homotopy of ir with id_X

[little hemlock]

yes, this was my answer to your MSE question

[tidal lynx]

You saw that too lol

are you EM

[little hemlock]

yes

[tidal lynx]

coolio

now let me think why this works

wait uhh

@little hemlock

shouldn't f_0 be the identity

[little hemlock]

f_1 is the identity in my construction. homotopy is a symmetric relation, i.e. you can replace the "t"s with "1-t"'s if you prefer

[tidal lynx]

Ok I see

so those are the family of maps if you want the homotopy from N into S^3 \ M

and replacing "t" with "1-t" is the same as reversing the maps, i.e., the resulting family is a construction in the other direction

[little hemlock]

hm no that's not quite what I mean. I'm just saying that if a map f is homotopic to g via a homotopy F such that F_0 = f and F_1 = g, then there is also a homotopy G such that G_0 = g and G_1 = f given by G(t) = F(1-t)

[tidal lynx]

ok I'm mixing up definitions sorry

[little hemlock]

so it doesn't matter that f_0 is the identity or f_1 is the identity as long as one of them is the identity, and the other is a retraction

[tidal lynx]

do you know what the "class" of a subspace means in a fundamental group

the fundamental group is an equivalence class of loops at some basepoint

so the "class" of a subspace doesn't really make sense to me

[little hemlock]

yea i don't think i've heard that before. i would need more context

[tidal lynx]

There's another question in this problem that asks "What is the class of M in π_1(S^3 \ M) = Z ?

[little hemlock]

strange, im not sure what that could mean

[tidal lynx]

Oh hmm looks like that's a term to be defined next lecture

[little hemlock]

bumping my bundle question

[fading vale]

The point here is that you're trying to get an isomorphism between the projective bundles

[little hemlock]

yes

[fading vale]

so your map is basically correct in a certain sense

in that over any subset on which L is trivial it works

Or okay like here is the point: cover X by subsets U_i where U is locally trivial and take some section s of L

then you can define, on the level of ur usual bundles, isomorphisms E -> E (x) L on the preimages of the U_i by doing x |-> x (x) s

this map is given by w_i for some w_i in the fibers of L

and when we pass to the projective bundles its an isomorphism P(w_i) : E -> E (x) L over U_i, right?

[little hemlock]

yes

[fading vale]

so if L is non-trivial it is impossible to choose all the w_i such that their induced isomorphisms agree on intersections and hence patch together to an isomorphism E -> E (x) L

But! On U_i cap U_j, P(w_i) = P(w_j)

hence on the projective bundles the P(w_i) are a bunch of isomorphisms that agree on intersections, and hence patch to an isomorphism P(E) -> P(E (x) L)

this is what they were trying to express by the existence of a "natural" isomorphism between the projective spaces

[little hemlock]

ohhh

[fading vale]

like u can think of the non-triviality of ur line bundle as an expression of the non-naturality of the isomorphism on the fibers and what this is saying is that when you pass to the projective bundles everything is identified so it works out

[little hemlock]

dam, that makes sense now

i appreciate it, moth

[fading vale]

np

i hope it was coherent i woke up like 10 minutes ago and didnt get out of bed to get my glasses so i can barely read what either of us are saying

[little hemlock]

well, i can't 100% confirm because its 6am and i still haven't slept

[coral pawn]

Let X and Y be smooth manifolds. Is there a (natural, whatever this means) smooth structure on $C^\infty(X,Y)$ such that $C^\infty(point,Y)$ is naturally isomorphic to $Y$ as a smooth manifold?

[gentle osprey]

Finitely Many Bananas

[coral pawn]

If we were working with topological spaces, then the topological space structure would be given by putting the compact open topology on the set of cts maps X --> Y

I'm wondering if doing so is possible in the category of smooth manifolds

[sturdy notch]

I think as soon as X and Y are infinite you get spaces which cannot be homeomorphic locally to Rn (like X=Y=R) so you might have to look towards infinite dimensional manifolds

[pseudo coral]

i like the new names for the channels

question about this

[gentle osprey]

MyMathYourMath

[pseudo coral]

then abelianize it to find that $H_1(K)= \Bbb{Z} \times \Bbb{Z}_2$?

[gentle osprey]

MyMathYourMath

[unreal stratus]

lol i assume that is a typo but ye i think that is the presentation for it

[pseudo coral]

so im GIVEN that the Klein bottle is generated by $\langle a,b : aba^{-1}b=1 \rangle$

[gentle osprey]

MyMathYourMath

[pseudo coral]

and using this i need to determine the group in question

[grizzled ibex]

Is this the so called right hand rule? 😮

[pseudo coral]

does this make sense @unreal stratus

[unreal stratus]

well lol

The question seems just to be a ytpo

But uh

[pseudo coral]

since $aba^{-1}b=1 \Rightarrow b^2a=a \Rightarrow b^2=1$

[gentle osprey]

MyMathYourMath

[unreal stratus]

I assume what you mean is like

pi1(klein bottle) is isomorphic to <a,b | aba^-1 b>

[pseudo coral]

yes

exactly

and so i have to determine that group based on the relation

but if $b^2=1$ then $b$ is an element of $\Bbb{Z}_2$ no?

[gentle osprey]

MyMathYourMath

[unreal stratus]

Uhhh lol

That doesn't really make sense

[pseudo coral]

well the only group where all elements have the property that

[gentle osprey]

MyMathYourMath

[pseudo coral]

is z mod 2

[gritty nest]

no it isn't, the trivial group also has that

[pseudo coral]

ah shux

[coral pivot]

such groups are called 2-groups

there are many such groups

obvious examples are (Z/2Z)^n

[pseudo coral]

well that was my try at it lol

ahh ur right

[unreal stratus]

But also like lol

asking if b "is an element of Z/2Z" is just kinda nonsensical to me idk

[pseudo coral]

idk how else to attack this exercise 😦

should i get a condition on a as well to determine

the group it is

[coral pivot]

well you just work out the presentation as you suspected

like you have b^2 =1, then figure out how a and b interact

and then conclude

[pseudo coral]

ahh ok

then i need another contition on a

and see how they interact as youbve mentioned

[gritty nest]

i'm probably just missing something but i don't get how you concluded that b^2*a = a here

[pseudo coral]

i did this

wait i forgot what i did but it checked out

i do have this

$ba^{-1}=(ba)^{-1}$

[gentle osprey]

MyMathYourMath

[pseudo coral]

@gritty nest

thats not very helpful : (

[gritty nest]

well that looks true but i'm not sure it implies anything useful

[gritty nest]

this is wrong @pseudo coral

let $a = (12), b = (123)$ in $S_3$ \
then $aba^{-1}b = 1$, but $b^2 \neq 1$

[gentle osprey]

bee

[gritty nest]

...although wait hang on a second

our goal is to find the abelianisation of this group right? (i only vaguely know what homology is but you said at some point that that's what you're going to do)

[zenith basin]

Guys

What s going on with the server

[coral pivot]

Right yeah they are looking for homology, so they are abelianizing pi1 which should immediately give ‘em what they want

[gritty nest]

can't we just use this presentation to find a presentation of the abelianisation by adding that everything commutes with everything else

[zenith basin]

(Sorry for interrupting)

[gritty nest]

and then it all becomes trivial because then aba^{-1}b really is b^2

[unreal stratus]

algebra in topology channel lol

[balmy field]

got hacked

[pseudo coral]

our goal is to find pi1 of K then abelianize it

[gritty nest]

is "pi1 of K" the thing with a presentation of $\langle a,b : aba^{-1}b = 1\rangle$ or is that something else?

[gentle osprey]

bee

[pseudo coral]

yes

[gritty nest]

...what exactly would "finding" that mean then?
like, we've described it up to isomorphism in a relatively simple way, and we have enough information to abelianise it pretty easily

[pseudo coral]

then i think thats what he wants us to do

abelianize that group

so mod out by its commutator

[maiden pilot]

bumping this

[unreal stratus]

hm i feel like you need more conditions

Wdym exactly by agreeing on a disk here

[maiden pilot]

i mean that their restrictions to that disk are the same

[unreal stratus]

No i mean like when you say "a disk"

Do you mean like a small closed ball about that point (in the metric on S^n)

[maiden pilot]

oh, D^n viewed as disk on the surface of S^n

[unreal stratus]

Ofc you need some restrictions like being small (i.e. not the entire sphere)

But sure

[maiden pilot]

so like an upper hemisphere for example

[unreal stratus]

Sure

[maiden pilot]

In particular im looking at this stackexchange post: https://math.stackexchange.com/questions/1837247/x-y-is-finite-where-x-is-finite-connected-cw-complex-and-y-has-finite-h

One of the comments quotes the "elementary fact": Lemma: Suppose we are given two basepoint-preserving maps from S^n to a space Z that agree on a disk D^n containing the basepoint. Then if the two maps define the same element of pi_n(Z), they are homotopic by a homotopy that stays fixed on D^n.

but i dont really see why its true lol

[umbral panther]

Yes, the space of smooth maps from X to Y turns an infinite dimensional manifold.

The most basic structure called a smooth structure is just a notion of tangent vectors. The tangent space to a particular map from X to Y is the space of sections over X of the pullback of the tangent bundle of Y. In particular, if you have a smooth family of maps, its derivative is such a thing. There are more refined notions of an infinite dimensional smooth manifold, but this is the important part

Oh, I guess I should specify a topology first. The space of smooth functions on X is a Frechet space, where convergence is given by convergence on all derivatives. There is an obvious generalization to when the target is Y. This space is a manifold locally iso to a frechet space
But the tangent vectors are more important

[coral pawn]

I'm not really familiar with infinite dimensional manifolds

Is there a way to realize the set of smooth functions from X to Y as a smooth manifold?

When I said smooth structure, I was referring to compatible charts to Euclidean space

[umbral panther]

Yes, the topological space of smooth maps is locally isomorphic to a topological vector space, namely the tangent space (which I described above). In general, if you know the tangent space, you should exponentiate tangent vectors to get everything. If X is compact, you can do this. If not, it’s still true that it’s locally iso to a topological vector space, but it’s harder to prove and not as useful as just playing around with vectors

[coral pawn]

Sorry I'm not really sure I understand what you mean

What charts do you use on the space of smooth functions from X to Y?

And do you still use the compact open topology?

[coral pawn]

What are the open sets in this topology?

[umbral panther]

No, not the compact open topology

What is topology on smooth functions with values in R? Look up frechet spaces

[cedar pebble]

pretty much never, no. If you enlarge the category you're working in to include Frechet manifolds then the set of smooth maps from X to Y is a Frechet manifold, when X is compact

if you want to remove the compactness assumption on X then you need to enlarge your category again to something like diffeological spaces

[golden flume]

where is the dontus???!?!?!?!?!

[rain ether]

You can take a look at Kriegl's and Michor's The Convenient Setting of Global Analysis if you're interested in smooth mappings between manifolds.

[maiden pilot]

Can anyone give me a hint on this problem?
Let $X \subseteq R^{n+1}$ be the union of the infinite sequence of spheres $S_k^n$ of radius $1/k$ and center $( 1 / k , 0 , \dots, 0 )$ . Show that $\pi_i(X) = 0$ for $i < n$.

[gentle osprey]

OmnipresentCoffee

[unreal stratus]

i think this is a compactness moment

[gaunt linden]

Wait, wouldn't i=0, n=1 of this claim be saying the hawaiian earring is simply connected?

[unreal stratus]

Hm not quite right (like showing it's path connected)

[gaunt linden]

Argh. The numbering of the homotopy groups always confuses me.

[odd flame]

ermmmm... wtf homology?

but like actually im looking at my notes and im so confused

we've been using massey but maybe i'll independently switch to hatcher

i have that given some chain complex, H(C) depends on the kernel of the boundary map into it some C_n and the image of a boundary map out of it

is that a correct statement?

[pseudo coral]

This is what my prof said

[hidden crag]

look at the definition

[hidden crag]

i mean the definition is a quotient group of kernel and image

[odd flame]

ok perhaps a random question

but how do chains come up in the wild, like what's the motivation

this sorta helps

[coarse night]

what's the question

[bitter smelt]

Chains give you homology, which has many applications. you can get topological invariants, knot invariants, etc out of a homology theory

[hidden crag]

maybe give some context on what homology theory you're currently learning about

then we might be able to provide examples and context

[odd flame]

well this is the question i was looking at but then i realized i havent understood as much as id thought from lectures

so i was just going back to remind myself of definitions and general notions

[bitter smelt]

Do you want help with definitions or motivation

this question is definition unpacking and diagram chasing

[odd flame]

both pls

yeah the diagrams for this already got a bit crazy in lecture

[coarse night]

https://www.3blue1brown.com/blog/exact-sequence-picturebook
try reading it, gives a little visual for homology

[bitter smelt]

You essentially just draw the diagram and "chase" an element around to try to prove ker \subset im and im \subset ker

[coarse night]

so you can think of homology as the "defect" of something being "nice"

[bitter smelt]

So pick an element in ker f_*, try to show there is something which maps to it by \partial by using the exactness of the SES, which will appear in your diagram somewhere

[pseudo coral]

question in computing the fund group of klein bottle given the presentation aba^-1b=1

I saw somwhere they consider the subgroup A generated by a and B the subgroup generated by B then A is clearly normal if you shuffle the equation up just a tad

and they got that the fund group is the semi direct product of Z with itself? i dont see this

any help

[coarse night]

ok so try cutting and pasting on aba'b to get the form x²y²

the reduction should follow

[pseudo coral]

whato do you mean by cut and past aba^-1b to get x^2y^2

and how does x^2y^2 become the semi direct product of z with itself

[coarse night]

have you done the exercises where you change the presentation of a surface form one to another

[pseudo coral]

nope this is the first hw in this section

[coarse night]

ex Klein bottle can be represented by aba'b as well as x²y²

[pseudo coral]

ok so somehow i need to get from aba^-1b=1 to x^2y^2

any reason why youre using x and y and not a and b

[coarse night]

not really just not to get a clash of notation

[pseudo coral]

ahh ok

[coarse night]

wait a bit

reference Top manifolds by Lee

[unreal stratus]

OH lol

When they said aba^-1 b they meant like the fundamental polygon

[coarse night]

using this representation, G=<a,b | a²b²>
now change the base to a, ab and see what do you get

[unreal stratus]

I was so confused 😭

[pseudo coral]

woah mind blown

not intro to smooth manfiolds but top mani? by lee?

[coarse night]

this follows from the classification of compact surfaces, they are either T²#...#T² or P²#...#P²

[pseudo coral]

true big theorem

or S^2

[coarse night]

sounds big but actually not that hard to prove

[coarse night]

lol yeah

[pseudo coral]

on our exam we just had to state that theorem not prove it

so c is a point not on the line that is a nor the line that is b right? like in the picture of things

or a line i should say

[coarse night]

anyway, x=a, y= ab then a²b²=x² (x'y)²=y² so the relation reduces to
G=<x,y | y²> which is the fund group

now to get H1, abelianise which gives Z×Z2

[nocturne basalt]

c is the line you cut across

[coarse night]

I'm too lazy to explain this now, refer to John M lee

[pseudo coral]

the topological mani or smooth mani text

[coarse night]

you just cut along one diagonal and attach either on a or b

[coarse night]

top mani

or just look at what shd has sent

[nocturne basalt]

the xi are all the same--this was from an exercise in hatcher on how identifying edges of \Delta^3 deformation retracts onto a Klein bottle

[pseudo coral]

did you mean G = < x,y : x^2y^2 >

[coarse night]

chane of generator

read carefully

[pseudo coral]

oh i see what u did

[rough cedar]

what's an intuition behind the profinite topology

[odd flame]

what did you mean by this

[hidden crag]

homology usually measures something

in a topological setting we can think of that as holes

H_n being non trivial means we have an "n-dimensional hole"

we can think of this as measuring the failure of our space to be contractible

If your chain complex is exact all homology groups are trivial

[rough cedar]

what is an n dimensional hole

[hidden crag]

i put it into "" for a reason it's not something that's rigorously defined

[rough cedar]

yeah but what is it intuitively

[hidden crag]

a hole

[odd flame]

is there an easy example with like

a torus or something

[hidden crag]

for example S^1 and S^2 have different kinds of "holes"

[coarse night]

one way you can think of chain complexes in top is that a "chain having 0 boundary then it encloses a region", which is not always true like S¹, the S¹ itself is a cycle which do not enclose a solid region. The homology measures such defect, when a region with 0 boundary fails to enclose a region.

[hidden crag]

this works out nicely with visual intuition when looking at simplicial homology

[coarse night]

that's exactly the quotient
(chains which are cycles)/(chains which encloses a region)

[hidden crag]

i recommend working through some examples

[odd flame]

and this is the ker/im quotient

[odd flame]

any suggestions

[hidden crag]

yes you want to kill stuff that does enclose a region

[coarse night]

yes

[hidden crag]

some simple simplicial complexes helped me with this

[odd flame]

i havent seen the word simplical before what's that

also unrelated

[hidden crag]

it might not be the best idea for me to just dump an entire homology theory on you

[odd flame]

isn't 2 here just the transitive part of 3?

[hidden crag]

your course will probably cover a certain one as soon as the foundations are laid

[odd flame]

yeah this homework doesnt seem too bad, at least compared to how confused i felt in lectures

[rough cedar]

🏃

[lunar yoke]

note though that there are (even nice) spaces with trivial homology that are not contractible

[odd flame]

such as...?

[hidden crag]

indeed i should have added that

[lunar yoke]

moore space for A_5

[odd flame]

is contractibility gonna keep coming up?

[lunar yoke]

basically 1-dim connected cw complex with fundamental group A_5

by pontryagin the 1st homology will be the abelianization, which is then 0

all higher homology groups are 0 for dimension reasons

and 0-th one is Z for unreduced and 0 for reduced homology

[coarse night]

not an intuition but a motivation would be in Galois theory where infinite galois extensions are equipped with the profinite top of all its finite extensions called the Krull topology which gives a nice association with subgroups with subfields which holds for infinite extensions

[rough cedar]

yeah that's where my question comes from actually lmao

I want to show that Gal(L/K) is a topological group

equipped with the krull topology

so I have to show that multiplication and inversion are continuous

[odd flame]

ok last silly question and forgive the terrible wording, but where do chains come from? like do we take a given space X and consider a chain of subspaces?

[rough cedar]

but I'm not sure what open sets there are

[rough cedar]

also just for my own saneness, when we say that G x G -> G is continuous, we mean that the fiber of an open set in G is an open set in the product topology of G x G?

[coarse night]

yes but fiber is not a correct terminology, use preimage

[odd flame]

when is fiber more appropriate?

[nocturne basalt]

singular chains?

[coarse night]

inverse image of a single point

[odd flame]

im asking what kind of object A,B,C might be in 0 -> A -> B -> C -> 0 or something like that

[nocturne basalt]

oh chain complexes?

[bitter smelt]

abelian groups, R-modules, vector spaces, etc

[odd flame]

im gonna assume that in the context of alg. top theyre abelian groups?

[rough cedar]

fiber sounds cooler tho

[bitter smelt]

Sure, abelian groups is common

[nocturne basalt]

for homology yes they are free abelian groups on n chains

[bitter smelt]

they can be defined in any abelian category

[coarse night]

definitely

[nocturne basalt]

but you can do the connecting homomorphism for ses -> les w just homological algebra

[coarse night]

you mean R-mod

[lunar yoke]

no, of course they are objects in a stable infinity category

[coarse night]

abstract nonsense go brr...

[lunar yoke]

like D(Z)

= mod_HZ

[hidden crag]

or D(N)

[coarse night]

don't know what these are

[lunar yoke]

they're the thing triangulated categories should have been

[cedar pebble]

[coarse night]

[lunar yoke]

since we're on this topic

quotients in higher algebra are weird

or i guess just homotopy theory in general, you already get this with just cones as homotopy quotients

if you have f : X -> X, then the induced map X/f -> X/f by f need not be zero

but it turns out that the square of the induced map is always 0

so like in spectra, which are the infty-cat analogue of abelian groups where you replace Z with the sphere spectrum S, if you look at S/2, then multiplication by 2 is nonzero

the proof of this is surprisingly easy

if the map were 0, you would get a decomposition of S/2 ⊗ S/2 as S/2 + suspension of S/2 from the cofiber sequence, which would imply that the F_2-homology of S/2 ⊗ S/2 decomposes as module over the steenrod algebra, which you can just directly compute cannot happen

anyways this stuff is fun

does anyone have this meme of jacob lurie with a gun that says something like "Read Higher Algebra. I'm not asking anymore"

[hidden crag]

I don't think i have that

[pearl holly]

maybe this

[lunar yoke]

found it

[gentle osprey]

MyMathYourMath

[nocturne basalt]

uhh idts

i think you just get the free abelian group on 2 generators

i.e. Z x Z

[bitter smelt]

What obstructions other than dimensionality is there to embedding a manifold with a (finite) CW structure into another manifold with a (finite) CW structure

[pseudo coral]

then how come the first homology group of kelin bottle is Z x Z_2 :/

[bright acorn]

If $G$ is a topological group admiting a transitive action on some compact space $X$, is it true that $G$ has to be compact? (is it true if we impose further conditions on $G$ or $X$?)

[gentle osprey]

MisterSystem

[nocturne basalt]

the fundamental group of the klein bottle is not the free group on two generators (thats the fundamental group of S1 v S1)

[pseudo coral]

is it Z x Z??

[unreal stratus]

(I guess any group acts transitvely on the one point space and you can similarly get stuff for other finite spaces like GL_n(R) acting on {1,-1} where A flips the sign if det A < 0 and fixes it otherwise? Or am I being silly?)

Like I imagine there are many examples which are less trivial than that but idk

[bright acorn]

hmmm

[unreal stratus]

(And then the first can be extended to nontrivial actions if you take products with smth noncompact and let that act trivially and stuff like that)

[bright acorn]

I was trying to think of a way to prove that O(n) is compact by exploiting its action on S^n.

[unreal stratus]

Oh lol well that is easier

[bright acorn]

Yeah, there's a way to do this by showing it is bounded and closed in R^n²

using e.g

[unreal stratus]

Okay I guess you don't want an alternative argument and want it in terms of actions

Yeah

[bright acorn]

the fact that orthogonal matrices have operator norm equal to 1

[unreal stratus]

Ye

[bright acorn]

and a limit of orthogonal matrices being orthogonal

[unreal stratus]

Ye

Anyway so uh hm in terms of actions is interesting

O(n)/O(n-1) is homeomorphic to a sphere and I imagine that should let you do it

Like take a net in O(n), it gives you a subnet net in O(n)/O(n-1) convergent to some point, then lift it to a net in O(n) and use compactness of O(n-1) to get yet another subnet

[bright acorn]

Oh, you get O(n-1) as the stabilizer at some point x of S^n of the action of O(n), right?

okie

I think this should do it

[unreal stratus]

Ye

[pseudo coral]

So again what’s the fund group of the Klein bottle lol it has to be a group whose abelianization is Z x Z2

[nocturne basalt]

<a, b | aba^{-1}b>

[pseudo coral]

is that associated with a known group

[nocturne basalt]

wdym

i think you just compute it by looking at the klein bottle

[pseudo coral]

like Z x Z orrrr

[nocturne basalt]

oh

no

[pseudo coral]

ahh so its just the group generated by a b such that aba^-1b = 1 and thats that?

[nocturne basalt]

well Z x Z is <a, b | aba^{-1}b^{-1}>

yeah

well do you see why

[pseudo coral]

why Z x Z is what you wrote

[nocturne basalt]

why the fundamental group of the klein bottle is what i wrote

[pseudo coral]

i can see it pictorially

[nocturne basalt]

ok

so now to abelianize

[pseudo coral]

you mod out by

<a,b:aba^-1b^-1=1>

?

[nocturne basalt]

well

you can mod out by commutator

or just look at the relation

i think

[pseudo coral]

does modding out give b^2=1

[nocturne basalt]

like aba^{-1}b = e, but ab=ba so b^2 = e

it should give you the same thing

[pseudo coral]

and the b^2=1 implies z2?

and no relation on a implies Z?

so we get Z x Z2?

[nocturne basalt]

yeah i think so

[pseudo coral]

thx! i may be back when im typsetting the solution if i get stuck again!

god bless this discord server

[warm quiver]

you can also think of this as a semidirect product: https://en.wikipedia.org/wiki/Semidirect_product#Fundamental_group_of_the_Klein_bottle

[tepid vale]

after all to the well organised mind

[noble osprey]

Hi, im slightly confused what the question is trying to say. Is it right that Vn+1 is another closed subset in Vn and Vinf is the intersection of a collection of closed subset Vn of X ? Thank you.

[grave solstice]

the statement is quite precise. It's just a nested sequence of closed sets $V_1\supseteq V_2\supseteq V_3\supseteq V_4\supseteq\cdots$ and you want to show that $\bigcap_{n=1}^\infty V_n$ is non-empty

[gentle osprey]

Croqueta

[grave solstice]

I'm not sure what your question is tbh

if you just have a sequence of closed sets in X, then the statement is obviously false (generally speaking)

[grave solstice]

ie you want to show that there exists an x such that x in V_n for all n

[noble osprey]

Ah i see, I just dont see the picture at first . Thanks

[mossy night]

Hello I need your help guys

If E is a real vector space, C is a convex subset of E, and x is a point in C, we say that x is an extreme point of C if C \ {x} is convex.

Here, we consider the space E of continuous functions from [0, 1] to R. Determine the extreme points of the closed unit ball in E for the norm of convergence in the mean, for the norm of convergence in the mean square, and for the norm of uniform convergence.

[supple sable]

I struggle with product and coproduct proofs, an example i am working through is in the category of open sets on a space x.

The product would then be defined as the intersection and the coproduct as the union. Then if I have two continuous functions f and g going from c to U and V respectively i need to come up with a (unique) function h from c to U\capV.

To do this i think i have to set h(w)=(f(w),g(w)) to make the diagram commute. But i fail to understand how the projection map works in this example since we only have a single set so the product notation confuses me. But if we need a single function I dont know how to guarantee that f(w)=g(w) on the intersection to make h(w) well defined (cus from the definition they dont seem to require this, or do they?)

[plain raven]

you can't consider arbitrary functions. you have to consider morphisms in the category

you're mixing up two different categories i think

when people say "the category of open sets on a space" it's implied that Hom(U, V) has a single element (the inclusion map) if U is a subset of V, else Hom(U,V) is empty if U is not contained in V

When you talk about h, f, g in this scenario, these variables are supposed to range across these inclusions.

If you were talking about the category of all topological spaces and continuous functions then h, f, g could range across arbitrary continuous functions but then as a result the product would be different, it wouldn't be the intersection

[supple sable]

Yes that makes sense

And that also messes up the well defined issue i had

Since the inclusions must agree on the intersection

Thanks!

[silver umbra]

how can one show that a particular sequence of points in a topological space X will converge in every possible metric of X?

(it's unknown whether X is metrizable)

[coarse night]

If X is not metrizable then what do you mean by “converge in every possible metric”

[silver umbra]

well its not known whether it is

[unreal stratus]

Wait lol I'm confused

[silver umbra]

well the statement "every possible metric" is a bit unprecise

what i more so mean is

if we were to assign an arbitrary metric on X

[unreal stratus]

Converging in a metric which induces the topology is equivalent to converging in the space (as a topological space) anyway

Like

[silver umbra]

but we're not assuming the metric here induces the topology

we're just ascribing a metric on X treated as a set

[unreal stratus]

Well then saying "metrisable" isn't precise

[coarse night]

What’s your question exactly! Rephrase it

[silver umbra]

sure

we have a topological space X, and it's unknown whether it's metrizable

my guess is that it isn't

in order to show this, I want to demonstrate a particular sequence of points that will always converge (trivially to the same point) given any metric ascribed to X

however, i will simultaneously show that this point cannot be the limit of this sequence in the topology on X that is already given

thereby showing its not metrizable

[unreal stratus]

Hm that doesn't seem a very effective strategy

Like with the discrete metric, only the (eventually) constant sequences converge

And you can do perturbations of that metric too

[coarse night]

If you are free to choose the metric the giving it the discrete one just forces the sequence to be eventually constant

[unreal stratus]

Usually you wanna check stuff like normality / Hausdorffness etc

[silver umbra]

hm this is true

[silver umbra]

yeah this is generally the strategy for showing that a space isnt metrizable

the one that im working with here is normal but not compact

[unreal stratus]

what is teh space if i may ask

Like we may be able to give better hints or whatever if you'd like

[silver umbra]

hm i wanted to phrase my question more generally so that i can think of the specifics myself

but sure, the space in question is the infinite broom

https://en.wikipedia.org/wiki/Infinite_broom

[unreal stratus]

Lol

isn't the topology on the infinite broom like

defined by a metric

[coarse night]

Lol

[silver umbra]

no

[unreal stratus]

I mean it's a subspace of R^2

[silver umbra]

it's not

that would make it a compact space

[unreal stratus]

Why

[silver umbra]

the topology that's ascribed is such that the open sets are unions of sets that are open in each of the lines of the broom

[coarse night]

It’s literally written a subspace of Euclidean space

[silver umbra]

its a subset, but not a subspace

with the topology that it's defined by

[unreal stratus]

So what you're talking about isn't the standard infinite broom then

[silver umbra]

i thought that it was?

[silver umbra]

but this^ is the definition im working with

[unreal stratus]

Sure

[silver umbra]

an explicit example of a set thats not open in the subspace topology but is in our definiton

[coarse night]

Think that gives you the Euclidean metric?

[silver umbra]

would be like

segments that are decreasing

u take each line in the infinite broom and take a segment of it

that gets smaller and smaller

(open segments)

[silver umbra]

if we're treating it as a subspace of R2, yes but it's not given the topology we've defined on it

ah hm

i realized ive been approaching it wrong

i was taking the sequence a priori, and then trying to show that it converges in every metric

but what i should do instead is take a metric a priori, and demonstrate a sequence that doesnt converge in it

[trail charm]

can i have a hint for 4-18 (a)? my instinct is that i only have to find charts for points within the “adjunction” part; points in the interior of M1 and M2 already have charts since M1 and M2 are manifolds

[unreal stratus]

Your intuition is correct

Consider what ("small") eighbourhoods of the boundary points look like

[narrow harbor]

/theorem of engel

[trail charm]

can we use f as our chart homeomorphism?

[uneven ridge]

$\mathcal{U} \in \mathcal{O}{\text{standard}} \Leftrightarrow \forall p\in \mathcal{U} :\exists r\in \mathbb{R}^{+} :B{r}( p) \subseteq \mathcal{U}$

[gentle osprey]

berketozlu

[uneven ridge]

about this statement

the points at the boundaries of U are not included, right?

B is an open ball

since, the balls would be out of the U

how do we write down the points at the boundaries of the set U?

[balmy field]

Like $\partial U$?

[gentle osprey]

UwUChad

[balmy field]

Or am I not understanding the question

[uneven ridge]

yeah, i didn't understand when we we write p \in U, it is the elements in the set U, but how the elements of partialU is related to the U in the same notation?

[balmy field]

Well \partial U is defined as the set of all boundary points

And a boundary point $x$ is defined as $\forall r>0 B(x,r) \cap U \neq \emptyset$ and $B(x,r) \cap U^c \neq \emptyset$

[gentle osprey]

UwUChad

[uneven ridge]

ah thanks, this is what I was asking basically

[balmy field]

I see

There is also the concept of limit point if you want to look into that

[wispy veldt]

the boundary set can be written as the closure of U minus the interior of U if you're familiar with these concepts

so you cant really pick points in the boundary when testing if its open

[uneven ridge]

i see

[uneven ridge]

i've seen the defns but this is my first time studying it..

thank you both, i'll check them more in detail

[balmy field]

For this the closure is the set union its limit points

This is where limit point comes in…

[odd flame]

what does $H_{*}(C)$ usually denote in the context oh homology, where C is a chain complex

[gentle osprey]

not sebbb not stμ₂dying

[unreal stratus]

Well the homology of a chain complex forms another chain complex which may be denoted thusly

[odd flame]

https://tenor.com/view/nerd-nerdy-nerds-nerd-emoji-gif-25380417

"which may be denoted thusly"

sorry potat id never heard that phrase lol

all love

i need to review my definitions though, i dont know enough to ask proper questions yet

[balmy field]

Thusly is a dope word

[unreal stratus]

It's also like grammatically incorrect ig lol

Well

[unreal stratus]

yes i am aware

i mean apparently it was coined like by people trying to seem smart since thus is already an adverb

[hard pumice]

Ah lol

Now that i have caught your attention

I have to prove that, given a manifold M, if U is an open subset of M, with its induced topology U is also a manifold

Let me outline my proof

Idk if it's legal or not lol

Given that M is a manifold there exists a set of charts {(U_a, f_a)} that cover U. Denoting t as the induced topology on U, taking the intersections of all V_a in t with the U_a's should give me a set {(U'_a, f_a)} with the U'_a open. Thus, this set would define an atlas on U which turns it into a manifold.

[hearty jacinth]

https://tenor.com/view/my-reaction-to-that-information-meme-my-reaction-to-that-information-mike-o-hearn-gif-13257779959215558930

[odd flame]

what does it mean to add things here

like whats addition of those maps doing

[hard pumice]

Does this work?

[coarse night]

It's an abelian category and by def all Hom sets are equipped with an addition structure

i.e. you can add maps. Think of it in R-mod, f,g ∈ Hom_R(M,N) then (f+g)(x)=f(x)+g(x). It's basically that but in a abelian category

you can just replace abelian category by R-mod

[coarse night]

you need to fill the details why those charts should be compatible

[gaunt linden]

It's a slightly more roundbout that it needs to be. You can just intersect each U_a with U itself.
You many need to do a bit more footwork if you find that U_a intersect U = U_b intersect U for some a,b, but the trouble is mostly notational -- e.g. you can just drop one of them.

[marble socket]

also i found this cute thing recently...
instead of defining your abelian cat to have hom-sets as abelian groups you can actually recover this by assuming certain morphisms in your category are isos.

if you assume your category has a finite products and coproducts and that the map A + B --> A x B given by the identity matrix is an isomorphism, i.e. you have biproducts. then you can define an operation on Hom(A, B). given f,g: A --> B, look at the composite
A --> A⊕A--> B⊕B --> B
where first map is the diagonal, second is f⊕g and third is codiagonal/sum map.
so this way your hom-sets are (commutative) additive semi-groups.

if you assume your cat has a 0-object then hom-sets are automatically pointed sets and that the 0 morphism is 0 in the semi-gropu making it an abelian monoid.

finally if you assume that the map A⊕A --> A⊕A given by the matrix (1, 1\0, 1) is this also an iso for all A. then you get that these abelian monoids are actually abelian groups.

[coarse night]

interesting

[gaunt linden]

noice

[marble socket]

i find this nicer because it makes me feel like the definition of abelian category is less artificial lol

[coarse night]

assuming you need (1, 1 \ 0, 1) to define the inverse

[gaunt linden]

That step still smells slightly of artifice.

[marble socket]

yea that's just a weird way to ask for a map i : A --> A such that 1+i = 0

but the point was that, we can describe these axioms just by asking some weird class of morphisms are isos

and the last axiom is that if further you have kernels and cokernels, then natural map coimage --> image is an iso

[coarse night]

I see

[marble socket]

inverse of this would be (1, i \ 0, 1) as you can expect

[gaunt linden]

And if you want an inverse for a general f: A --> B, you simply compose it with an inverse of one of the identities for A or B?

[marble socket]

yep

i should've mentioned that the way you define the operation is automatically then compatible with composition

[coarse night]

so treating a map like Matrix just refers back to the elements of A

so aren't we going back to underlying elements?

[marble socket]

nah, we didn't mention elements at all

a map from a coproudct of A_i to a product of B_j is just a collection of maps f_ij : A_i --> B_j

i'm just putting them nicely in a matrix

because if you define the addition like above, then composition of maps like you can expect is just matrix multiplication

there are lots of tiny details to check

but it was quite surprising for me that how little your axioms are asking and how much structure is automatically induced

also another advantage of such a definition is that, you don't need to ask a question like "does the same category admit different abelian category structure by possibly changing the abelian group structure of the hom-sets"

cause the structure can be defined intrinsically, its not some extra data

[coarse night]

yeah this feels like a more natural way to define Ab Cats ngl

where did you read it from?

[marble socket]

from my prof's head

(i mean he told it in a lecture, i didn't invade his mind)

[coarse night]

we were just told it's R-mod

[marble socket]

i've come to think of the freyd mitchell theorem in reverse now

earlier i used to think that if you wanna prove something in an abelian category, you can pretend you're in some R-mod and do stuff

but now i feel like, it should be said as, if you're proving something general enough for R-mods, then you might as well work in an abelian category and use more categorical tools

[coarse night]

well I like element chasing

[marble socket]

chase generalized elements

[coarse night]

[marble socket]

[marble socket]

like for example if you develop homological algebra by doing everything for abelian cats, you only have to do it once for projectives/injectives, and then appeal to the opposite category

and another cool thing is that the functor categories A^J are also abelian

so for example arrow category of A, Ar(A) is abelian and if A has enough injectives, so so does Ar(A)

and clearly Ar(R-mods) isn't naturally a R'-mod for some R'

so abstraction is uwu

[coarse night]

true

[marble socket]

so the same theory lets do somethign stronger, if A --> B is a morphism, then you can find compatible projective/injective resolutions directly!

[odd flame]

can i get a hint on this

i tried the obvious thing but it didnt work

it's hard to explain without a diagram though...

[odd flame]

do you have a more concrete example by any chance

[odd flame]

nvm got it

[unreal stratus]

Lol ye where is homological algebra meant to go

[odd flame]

im just hoping things distribute as im assuming they dont

"as shown in lecture"

[umbral panther]

Actually, it is for R’ upper triangular 2x2 matrices

[marble socket]

oooh really?

so like if M is a module over that matrix ring, do we recover the stuff by looking at E11 * M and E22 * M?

[odd flame]

quick question

oh no

well if that's visible enough

are the d_A and d_B "the same"

[hidden crag]

No

[odd flame]

right ok

https://math.stackexchange.com/questions/1024567/how-to-show-that-homotopy-of-chain-maps-respects-composition

in the answer here how come they are kinda treated the same

for ease

[hard pumice]

Thanks!
What does it mean for two charts to be compatible?

[odd flame]

when rewriting the f1 - f2 the answer uses the same d in ds + sd

shouldnt it be like d's + sd

[hidden crag]

Wait I’m at a bar rn so I can’t look into details properly

[odd flame]

on discord

at a bar

[hidden crag]

On my phone for a second

[odd flame]

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

i appreciate you still looking at my silliness while at da club timo

[hard pumice]

Thanks!
Would i really need to drop one if all i wanted was to prove the existence of an atlas for U?

[gaunt linden]

Depends on exactly how your atlas formalism has been set up.

[marble socket]

oh, i should think of it like a quiver then the triangular matrix thing would just be the path algebra right

[gaunt linden]

Sometimes an atlas is a set of U's and exactly one map into E^n for each.
Sometimes an atlas is juset a set of maps, each from some open subset into E^n.

[umbral panther]

Looking at E11 and E22 gets you a module over diagonal matrices, which is a module over RxR, which is two modules, the source and target. The action of E12 gives you the arrow

[umbral panther]

Right

[hard pumice]

oh ok, depends on the definition i would be working with

[gaunt linden]

Yes.

[odd flame]

quick bump asking why there arent different boundary maps and instead only d

[hard pumice]

i'm following nakara and "gauge fields, knots and gravity" by john baez. that was exercise 4

[odd flame]

homology is weird man

[odd flame]

ok stupid question because i cant make sense of these definitions

what am i being asked here with respect to H_*(C)

i know what chains are and what a free abelian group is

[nocturne basalt]

well its asking you to compute H_n(C)

[bitter smelt]

it says to compute it, what do you mean what are you being asked?

[odd flame]

i think im just not seeing what a homology group is

or what the point of one is

assigning abelian groups to a topological space just feels kinda random i guess

im reading my textbook again though

[nocturne basalt]

this should be defined in any textbook, it's the ker / im of boundary maps

I think the intro to hatcher 2.1 motivates it quite well

[odd flame]

i have massey

it starts with some stuff on singular n-cubes

yeah these books take different approaches i think

hatcher is using simplexes and i have cubes I^n

[nocturne basalt]

you should still be able to compute the homology just from the bases and boundary maps

like for H_0, ker = C_0 and im = 3C_0 so you get Z/3Z

[odd flame]

how come it's 3C_0 though, b isn't in the image

[nocturne basalt]

look at what the boundary operator does on the basis elements of C_1, it sends a_1 to 0 and b_1 to 3 a_0

[odd flame]

ohhhh my b my b i looked over the defn of C_i for i = 0

ok then for H_1(C)

ker = (a_1)

[nocturne basalt]

yes

[odd flame]

im = (2a_1)

so Z/2Z...?

[nocturne basalt]

yeah

[odd flame]

:O

[untold lily]

^wtf is this <@&268886789983436800>

[odd flame]

is this right then? or would it be more accurate to write like (a,b | a_2 - 2b_2 = id)

im also not sure what happens with H_3 if H_2 is trivial

more general concept/understanding questions: what information about the chain do we get from these groups?

[coarse night]

[bitter smelt]

What do you mean? The groups are literally built from the maps of the chain.

They give you information about the kernels and images of the maps in the chain

If the homologies are 0 the chain is exact

You can think then of homology as measuring how far your chain is from being exact

[little hemlock]

im trying to parse what they mean here. I assume by "union of all these", they mean that $H^* = {(a, e) \in P(E) \times E : e \in a}$ and that $H^*$ should be interpreted as some kind of generalization of the tautological line bundle over some projective space?

[gentle osprey]

kxrider

[next crystal]

Suppose, for contradiction, that there is a $b_1 \in B$ such that $|p^{-1}({b_1})| = j \neq k$. For each $i \in \mathbb{N}$, let $A_i = { b \in B : |p^{-1}({b})| = i}$. Then $A_i$ is nonempty for $i=j,k$. I then show that $A_j$ and $B \setminus A_j$ form a separation of $B$, contradicting connectedness. What happens if $p^{-1}({b})$ has infinitely many elements for some $b \in B$ though? My argument doesnt consider this and im not sure how to account for it

[gentle osprey]

michαel

[quiet thorn]

If I have that for $G$ a topological group with identity $e$, and $\gamma_1, \gamma_2 \in \pi_1(G,e)$; then $\gamma_1 \cdot \gamma_2$ is homotopic to the pointwise product $t \mapsto \gamma_1(t) \gamma_2(t)$. Is it a one-liner that $\pi_1(G,e)$ is abelian or am I severely overlooking something?

[gentle osprey]

Tubular Cat

[quiet thorn]

because there is a hint on this problem and yet it doesn't feel like it needed one

[odd flame]

this helped!

so we have almost exactness here

[odd flame]

this is a little obvious in retrospect actually

[prisma geyser]

probably a trivial question, but I’m working through Lee’s top mfds and I fail to see why $a^{-1}$ which I’ve found to be $$a^{-1}: , z \mapsto \frac{ln(z)}{2\pi i}$$ is not continuous.

[gentle osprey]

humefanboy

[hidden crag]

you could also argue by compactness

[odd flame]

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

[hidden crag]

like you can rule out the possibility that the two are homeo so no cont. bijective map can have a cont. inverse

[odd flame]

i still have trouble visualizing these, cuz each of A B C are their own chains right

[prisma geyser]

This is however before compactness is introduced in Lee, so is there a sort of analytic way to see why that specific map is not continuous?

[unreal stratus]

You can show it isn't an open map

[hidden crag]

this is the visualization

do you need a translation or is it clear

[odd flame]

a translation would probably make me fixate on unnecessary details lol that helps

[unreal stratus]

What do Sequenz and Diagramm mean

[prisma geyser]

which open subset should I focus on?

[odd flame]

im more confused by R-Moduln

could it be a field?

[hidden crag]

[unreal stratus]

Uh have a try looking at certain basis elements

Like very simple open sets

[prisma geyser]

Also I’ve been able to come up with an example where f is a continuous bijection but not an inverse, though I’m not 100% sure if it works. You take id: X —> X, equip the first with the discrete topology and the second with the trivial topology. For any reasonably big X this should work?

[unreal stratus]

Yes that is good

[gritty widget]

can you please help me to understand an exercise 😔

[unreal stratus]

It works for any X with more than 1 element

[odd flame]

be specific about your questioon

dont ask to ask

[odd flame]

[rancid umbra]

try drawing out the map. its essentially stretching out the half open interval into a circle, but the circle isnt "all the way closed"

is there an explicit form of (u_n)? or is it asking to give a counterexample

[unreal stratus]

Wait how are you defining adherent point

Oh okay sure

And then a cluster point is the same but you exclude x ig

That difference should help suggest what kind of sequence to consider

[odd flame]

are the delta's in these two diagrams the same?

[unreal stratus]

There are boring examples essentially

[unreal stratus]

No

The deltas in the second are the induced boundary maps

[odd flame]

oh so they're just named the same to fuck with me

[unreal stratus]

This is standard basically

Name all differentials the same thing

[prisma geyser]

I’m still not too sure. Like I suspect there’s an issue here, but I cannot pin it down exactly. The image of [0, 1) is just S^1 since the map is surjective if I’m not glossing over something huge

[unreal stratus]

It is fine in the long run

[odd flame]

pain

then in the last paragraph, i(a) = db

that delta is from the above map right

[unreal stratus]

Ye

From context you can deduce which delta it is

[odd flame]

busy potat

[rancid umbra]

focus on neighborhoods of 1 in S^1

[unreal stratus]

Lol I would say focus on neighbourhoods of 0 in [0,1)

Wdym are the same

[odd flame]

is that first diagram in my img the same as the one in hatcher?

or a slice of it

[unreal stratus]

Ye it's a slice of it

[prisma geyser]

Oh yeah I see it now, [0, a) is open in [0, 1] but its image isn’t open

[unreal stratus]

Perfect

Another way which doesn't explicitly identify an open set is like

Wair hm

If you consider a path in the circle which loops right twice, say

Then in [0,1) that corresponds to like

Going from 0 up to 1 back to 0 etc

Very discontinuous

[odd flame]

does mine not have the same kind of delta then?

or is each A,B, C in my picture also a chain

[prisma geyser]

Oh I see, thanks for the help guys

[unreal stratus]

It says they are chain complexes

I misread ig but it is still a slice of it kinda

[odd flame]

it said SES so i thought A -> B -> C was that SES

OHHHHH

im so dumb

"ses of chain complexes"

[coarse night]

the spectral sequence proof of this is very nice, did it recently

[coarse night]

@odd flame try reading the 3b1b article I had sent recently about exact sequences

[odd flame]

i read part of it but need some more time to really sit through it

modules exam tomorrow

[coarse night]

Then go back to abstract alg, what are you doing here

[odd flame]

hw due

[odd flame]

i'll be there after my class lol

a bleakified version of wanwan would be nice timo :)

[coarse night]

Btw is this wanwan from mobile legends?

[odd flame]

idk i just like how naive and gullible she looks (she's just like me)