Mathematics - 2020-10-19 (page 2 of 3)

Does every multifunctor $F\colon\mathbf{Vec}_{\mathbb{R}}^k\rightarrow\mathbf{Vec}$ induce a multifunctor $\tilde{F}\colon\mathbf{VecBun}^k\rightarrow\mathbf{VecBun}$ on the fiberwise level?

Balarka Sen

5:07 PM

lol

the heck's a multifunctor dude

Thorgott

it's just a functor from the product

Balarka Sen

LOL

terrible terminology

Thorgott

as terrible as saying "function of two variables" for a function with domain R^2

Balarka Sen

even nlab doesnt agree with your terminology

multifunctor is some messed up thing

Thorgott

since when do you care about what nlab says

Balarka Sen

5:11 PM

bifunctor is a functor between bicategories, which are 2-categories

this is a garbage terminology

Thorgott

what's a bicategory

a bifunctor is a functor from a product man

Astyx

The best of terminology. Let me tell you. They love it-- everyone loves my terminology.

Hanan N.

@Asty What's $kb$

Balarka Sen

i dont know man

do i look like a category theorist

Astyx

@HananN. $a$ and $b$ are big1 and big0. $k$ is any integer

Thorgott

5:13 PM

whatever, my question has hardly anything to do with cat theory, it's just language

Tobias Kildetoft

a bicategory is a bit like a 2-category, except stricter or maybe more lax

Thorgott

can I beef up the local trivializations, is the question

Balarka Sen

@Thorgott finally the language of gods

i think you just have to check the cocycle conditions

Thorgott

nooooo, not the cocyles again

Balarka Sen

cocycles are gr8

Thorgott

5:15 PM

ok, tell me the order of composition without looking it up then

Balarka Sen

$\phi_{\beta \gamma} \circ \phi_{\alpha \beta} = \phi_{\alpha \gamma}$

easy

take that atheists

Astyx

Why would you put you indices like that ?

And not $\phi_{\alpha\beta}\circ \phi_{\beta\gamma} = \phi_{\alpha\gamma}$ ?

Thorgott

cause function composition is from right to left

Balarka Sen

I fucked up HAHAHA

its not this

Thorgott

I was about to say this makes too much sense

Balarka Sen

5:17 PM

HAHA

yeah

Thorgott

atheism prevails

in this moment, I am euphoric

Balarka Sen

wait lets see

$\phi_{UV} \phi_{VW} \phi_{WU} = 1$

$V = \alpha, W = \beta, U = \gamma$

Thorgott

no wait, that still makes sense

Balarka Sen

Yeah

that's what it is

YOU CONFUSED ME

i got it from WIKIPEDIA

so it must be true

Astyx

Man

Thorgott

5:20 PM

no, I believe you now swap all the indices to make it as nonsensical as possible

Balarka Sen

hahah yeah i have definitely seen this fucked up shit somewhere

Thorgott

let's trust the logs

Balarka Sen

why is that standard

Thorgott

May 15 at 13:30, by Mike Miller

@Thorgott The correct cocycle identity should be I think $I = \varphi_{\gamma \alpha} \varphi_{\beta \gamma} \varphi_{\alpha \beta}$. It is famously irritating to get it correct.

Edward Evans

@BalarkaSen niseeeee

Balarka Sen

5:21 PM

@Thorgott massive

Astyx

That's some elitist notation if you ask me

"Plebs will never understand us" evil laugh

Balarka Sen

No, no, it's actually super easy to remember it. Treat it as a triangle $(1, 2, 3)$ with edges oriented so that $i \to j$ iff $j > i$.

Edward Evans

lel I always get the maps wrong in a projective limit

Balarka Sen

$\varphi_{12} \varphi_{13}^{-1} \varphi_{23} = I$

So you recover Mike's formula $\varphi_{12} \varphi_{31} \varphi_{23} = I$

or a cyclic permutation of it

Astyx

Actually it's just the sensible way but each pair of indice is swapped

Balarka Sen

5:24 PM

Screw this

Thorgott

projective/inductive limit is confusing terminology, just use limit/colimit

Balarka Sen

inverse limit, direct limit

Edward Evans

left arrow limit right arrow limit

Thorgott

still confusing

Balarka Sen

no

only for mutants

Thorgott

5:26 PM

limit/colimit makes it clear where the maps go

Balarka Sen

garbage

Astyx

cogarbage

Thorgott

limit objects come with maps into the diagram, colimit objects come with maps out of the diagram

Edward Evans

projective limit helps me remember by analogy with $\Bbb Z_p$

Balarka Sen

inverse limit patches shit up

Edward Evans

5:27 PM

then inductive is just not projective

Balarka Sen

direct limit zooms into shit

to patch shit up you need to use product

Thorgott

how does it zoom

Balarka Sen

so the universal property is clear

Thorgott

it's like union

Balarka Sen

thats because you think backwards

direct limit of $C^\infty(U)$ for $U$ open nbhd of $p$ is the stalk at $p$

germs of smooth functions

zooms into infinitisimal info

Thorgott

5:29 PM

it's just a downwards union

Balarka Sen

How is germs union of functions mf

you literally zoom

Astyx

(insert meme with guys throwing chairs)

Edward Evans

link.springer.com/chapter/10.1007/978-3-319-45032-2_12

realshit

Thorgott

its a colimit

Balarka Sen

ur mum is a colimit

Thorgott

5:31 PM

its not really a union

but a union is also not a zoom

Balarka Sen

an increasing union is a zoom

anakhro

zoom was a gr8 show

Thorgott

oh

zooms dont always have to be in

its a zoom out ig

hmm, let me quickly come up with another reason for why youre wrong regardless

vzn

@BalarkaSen congratulations! ps wigderson has some interesting ideas, like em. plus theres all the connections to QM. but alas you already said you dont care lol :(

Balarka Sen

yeah you kinda forget which $A_k$ the elements in $\bigcup_{n \geq 1} A_n$ come from

thats a kind of zoom

direct limit forgets, inverse limit remembers

Azmuth

5:34 PM

@Rithaniel Quick Maffs!

Thorgott

hmm, I think a zoom only really works for sequential colimits

Astyx

Pepperidge farm is an inverse limit

Thorgott

or filtered ones, probably

Balarka Sen

where you have a cofinal subsequence yeah

general limit diagrams i dont understand anyway

Thorgott

it still makes sense to think of, like, a module being the colimit of its f.g. submodules as a zoom

but is a coequalizer a zoom?

it's more selective

but it does "forget"

so colimits forgetting is a sensible notion

but I'm not sure in which sense limits "remember"

also, I believe t he answer to my original question is actually yes, though I have to check the technical details

anakhro

5:48 PM

I am unsure if "zoom" in this context is a serious mathematical term or if you guys are throwing some new meme around.

Thorgott

it's intuition

Adam L

ah yes this reminds me of that time as secretary of state, I landed in Bosnia and with only a moments notice we were under heavy sniper fire

the gut instincts served me well there, yes, intuition God bless

ASillyGuy

is promoting one's own question in chat disallowed/frowned upon here?

Make_Stackexchange_Great_Again

Ok, here.

anakhro

@TedShifrin got any more fun ways of relating group actions to topology that (at least geometrically) are explainable to high school students? I am going to be doing orbifold stuff, and I want something more to discuss group actions.

ASillyGuy

6:00 PM

cool cool , well I just wanted to maybe direct people to my question here:
https://math.stackexchange.com/questions/3872478/any-example-of-this-kind-of-algebraic-structure

because it's interesting imo and not so obvious, especially when attempting to attack it with great generality

basically want to find an algebraic structure, in particular some infinite (co)product that has a neutral element only when you look at finite "sub products"

anakhro

What do you mean by "finite reduction"?

ASillyGuy

if we think of the larger structure as an \omega indexed product of (possibly finite) structures, only take n many of those structures as your new product

so if all binary sequences are the larger structure, a "finite reduction" would be all binary sequences of a fixed length

Make_Stackexchange_Great_Again

Someone has posted an answer.

Thorgott

coproduct in which category

ASillyGuy

in the algebraic category of the relevant algebraic structure

ideally, we could talk about magmas that are unital when this "finite reduction" situation happens, and not unital whenever it does not happen

so the answer posted thankfully is a sanity check that there is nothing special about Z_5, but I guess I still would like something general, ideally as general as unital/nonunital magmas

anakhro

6:13 PM

Could you force this to occur? Inductively add two elements: the new identity and an element for which the "old" identity fails? Then kind of expand as needed?

Thorgott

hmm, is the coproduct in Rng the direct sum?

Mike Miller

@Thorgott Certainly not, a map from Z x Z has f(1,0) and f(0,1) multiplying to zero

Which imposes a condition on the map from the "coproduct" it shouldn't have

ASillyGuy

@anakhro this is precisely the kind of thing I would love to achieve

Thorgott

right

Alessandro Codenotti

Coproduct in rings is some kind of weird free product

Thorgott

6:17 PM

just twice the identity Z->Z doesn't factor through that

ASillyGuy

starting somehow from some infinite set of generators (computable so you can just think of them as tags x_0, x_1, ...)

Thorgott

ok, so what actually is the coproduct in Rngs

ASillyGuy

I imagine it is a free product that mods out ring identities, but I am not well versed

Thorgott

it's probably very ugly

anakhro

have you tried building it that way, @ASillyGuy to see what you get after a few steps?

Alessandro Codenotti

6:22 PM

It's just a free product, but with rings. It's somewhat painful to write down, but you know what it should be

Thorgott

@Alessandro Rngs, not Rings

Alessandro Codenotti

Note that coproduct of nonzero rings can be zero tho, if they have different characteristic

@Thorgott bye I'm out

Thorgott

understandable

Mike Miller

I feel like this is an exercise in Hungerford

ASillyGuy

@anakhro I have not, no. I mostly want this for a (turing) reduction to COF (just literally the requirements in my question being satisfied), but I basically just needed this special structure to kind of "aim" towards. I guess a semigroup with local identities works, and, more generally, a magma (since you can split up the associativity by simply having all (possibly not equal) parenthetisations respect the (local) identity)

I will look up Hungerford, thank you

It is really just an idea I had to solve a question I posed myself, more related to computability

oh, or maybe you mean the coproduct of r(i)ngs

robjohn

6:34 PM

@Thorgott I see "rng" and I think of random number generators...

Thorgott

that's better

geocalc33

6:49 PM

Is $x^Tx=0$ the correct expression for $xy=1$ in matrix form

anakhro

@geocalc33 Can you explain more of what $x$ and $y$ are in the equations?

geocalc33

$\mathbf{x}^T\mathbf{x}=0.$

I meant to write the $x$ in bold

$\mathbf{x}=\begin{pmatrix} x \\ y \end{pmatrix}$

Tobias Kildetoft

bolding a letter has no intrinsic meaning

anakhro

Okay so $x^Tx = 0$ says that "the dot product of x with itself is 0", that is, $\|x\|^2 = 0$.

So what did you mean by $xy=1$?

geocalc33

$xy=1$ I meant as a rectangular hyperbola plotted in the x-y plane

Thorgott

6:56 PM

truth is invariant under change of notation

anakhro

You mean the curve $x\mapsto (x,1/x)$? @geocalc33

Thorgott

@geocalc have you ever seen a rectangle

geocalc33

yeah I'm trying to write

Thorgott

did it look like a hyperbola

porridgemathematics

if you have a bunch of propositional formulas $\phi_1,...,\phi_n$ that are mutually non-logically equivalent, is there an easy way to construct a $\phi_{n+1}$ such that $\phi_1,...,\phi_{n+1}$ are all mutually non-logically equivalent?

geocalc33

6:58 PM

@anakhro yes $x \mapsto (x,1/x)$ I need to write it in matrix form

user2103480

@MikeMiller I read through parts of dieudonnés exposition of poincarés ideas but to be honest, I was mostly confused already at the start

porridgemathematics

lets say you aren't working in a language with infinitely many atomic variables, otherwise of course we can just select an atom not in the $\phi_1,...,\phi_n$

anakhro

@geocalc33 I am at a loss for what this has to do with $\|x\|^2 = 0$.

geocalc33

I was thinking it would be like this $z^2-x^2-y^2 = 1 \implies x^TQx = 1$

but never mind that's wrong

Mike Miller

@user2103480 Maybe this is one place where it's helpful to have seen the math ahead of time. I can talk to you about it sometime but not now

user2103480

7:01 PM

Oh no worries! I'll describe the problematic parts anyways and just copy/paste it when I come back to it in the future

geocalc33

@anakrho oh wait I think I know how to write it

user2103480

for example the paragraph on homologies on page 18: there he says that for q-1 dimensional manifolds v_1 to v_k, the sum v_1 + ... + v_k should be homologous to 0 if the boundary of some q-dimensional manifold is (homeomorphic to) the coproduct of v_1 to v_k. And then he goes on to describe what happens when we add homologies, and I just don't know what the "small deformations" are supposed to mean

geocalc33

I think it involves a matrix

user2103480

And how we are supposed to incorporate orientations. My guess would be that the higher-dimensional manifold whose boundary is n_1*v_1,...,n_k*v_k needs to be oriented, and induces an orientation on the boundary manifolds, which determines the homology coefficients (at least intuitively)

Balarka Sen

@user2103480 Oh! From probability to topology?

Alessandro Codenotti

7:10 PM

@BalarkaSen he ventured through the darkness and came back, let's celebrate the right path he has taken now

user2103480

@BalarkaSen yeah topology is the most aesthetic subject imo. So I'll do my best to study some of it in the coming semester

geocalc33

does topology have matrices?

Sophie

Is there a general algorithm for finding out the GCD of two polynomials? Not the polynomial gcd, the gcd of the polynomials evaluated at every integer. For example I just proved that $\gcd(2n^2-4n+3,2n^2-3)=\gcd(n,3)$

user2103480

@geocalc33 the better question is: does matrix have topology?

the answer is yes

@AlessandroCodenotti don't worry, 3 of 4 lectures are probability with an eye on the natural sciences, so every day I stray further from the righteous way

At least I'm not applying the stuff to financial mathematics, y'know

Balarka Sen

@user2103480 Moving a manifolds through space smoothly kind of traces out some path of manifolds, so that you start at $M$ and end at $N$, you somehow end up at a manifold of one dimension more, $W$, with boundary being $M \sqcup N$.

That's what the connection with deformations is.

porridgemathematics

7:14 PM

@Sophie euclidean algorithm?

@Sophie can't you still use $gcd(a,b) = gcd(b,a-bq)$ for any polynomial $q$, and successively lower the degrees that way

Sophie

@porridgemathematics I mean more like... I want to know what all the solutions look like. Is it always a product of the gcd of x and a constant? In particular, while I know how to do this case my approach is not general

user2103480

@BalarkaSen So it's like a "generalization" of homotopies? I mean, I know that homotopy equivalent spaces have identitical homology groups and that there is a natural transformation from homotopy groups to homology groups, but is the intuition also given by "we try something like homotopy, but more relaxed"?

Balarka Sen

In the case of homotopies, $W = M \times I$, essentially. The topology of the manifold doesn't change as you trace it out

porridgemathematics

@Sophie I think the euclidean algorithm answers this question though, it literally tells you that second item in the tuple, as you iterate the algorithm, is decreasing in degree

user2103480

Ah ok nice

@BalarkaSen and that is just a preliminary example, right? So we can also have several disconnected manifolds as being "homologous together"?

Sophie

7:20 PM

@porridgemathematics okay so then what's the next step if you have $\gcd(5n-3,2n^2-3)$?

user2103480

Which constitute the boundary of the higher dimensional manifold

Alessandro Codenotti

Are you talking about cobordism without saying cobordism or?

porridgemathematics

@Sophie divide $2n^2 - 3$ by $5n-3$ and get the remainder , which will be something of degree $0$ necessarily

Balarka Sen

@user2103480 Yeah, for example, $S^1$ and $S^1 \sqcup S^1$ are "homologous" by a pair of pants manifold $W$.

user2103480

And regarding the orientation, does the manifold W have to be orientable or is that just a (useful) analogy

@BalarkaSen Ahh nice reduction to the case for 2

Balarka Sen

7:22 PM

You can shuffle around which boundary terms you want to group togather, because $A = B + C$ implies $A + (-B) = C$, etc. It's an additive theory

Right

@user2103480 Indeed, $W$ has to be orientable for the oriented theory.

Sophie

@porridgemathematics and it will also not be an integer

porridgemathematics

hmm, sorry I guess you need to make sure everything is an integer

yeah

Balarka Sen

And boundary of an oriented manifold has a canonical orientation, of course.

user2103480

Ok cool thank you that helped

@BalarkaSen looked it up before I asked haha

Balarka Sen

@AlessandroCodenotti Yup

Thorgott

7:24 PM

Euclidean division works perfectly fine. I don not understand what the issue is.

porridgemathematics

but if you only do euclidean division in $\mathbb{Q}[n]$ then it isn't guaranteed this will give you the $gcd$ viewing the two arguments as integers, right?

geocalc33

what do you call the operation on a diagonal matrix where you reflect the diagonal elements about the middle?

porridgemathematics

whereas we want the gcd of the two integers for any $n$ in some nice closed form

geocalc33

for example if 1,-1 are the elements, then to reflect them you'd get -1,1

user2103480

haha so yeah I looked it up and so these are bordant. And then it's possible (in some sufficiently nice context) to "translate" this to the identity S_1 + S_1 + S_1 = 0?

Sufficiently nice context should mean something like "all this happens inside some manifold with the right properties", e.g. the pair of pants being part of a genus 2 surface

porridgemathematics

7:35 PM

there are $2^{2^n}$ equivalence classes of logically equivalent propositions , assuming our language has $n$ atomic variables, right?

user2103480

Note that I'm really trying to relate all this to the text by dieudonné

Balarka Sen

@user2103480 Exactly, you want to do this inside some manifold

user2103480

and in effect, to the text by poincaré

@BalarkaSen ok cool

@BalarkaSen btw, I asked a prof whether adding some topology knowledge might be useful for probability, and he basically said "there probably are some interesting questions, but we're not there yet". Sad.

Balarka Sen

:)

I hope we are

user2103480

An intersection between random dynamical systems and topology would be nice

Balarka Sen

7:41 PM

I don't know anything about topology of random simplicial complexes

Those seem to be the first place to start with

user2103480

@BalarkaSen Some day soon, brother, some day soon

The closest thing I've found was this link.springer.com/article/10.1007/BF01845701

But there's also this book arxiv.org/abs/0810.2253 for example

Balarka Sen

looks cool!

I would really like to understand the proof of Gauss-Bonnet using Brownian motions someday

user2103480

Sounds freaky. humboldt university had a lecture about brownian motions on manifolds last semester, but that sounded too scary

Balarka Sen

yeah i definitely dont have background

user2103480

I'm sure it won't take long for you to acquire it

Probability theory on manifolds doesn't seem as focused on awful measurability conditions as the usual stuff, it seems

Or they just swept those under the rug and I didn't read carefully enough

Balarka Sen

7:53 PM

yeah I'll have to try seriously sometime

Mike Miller

8:24 PM

@user2103480 The only thing I can tell you is that $S_1$ is the symmetric group on one letter

$S^1$ is the circle

Balarka Sen

LOL

oh

lol

user2103480

@MikeMiller homeomorphic notations

potato potato

tomato tomato

Mike Miller

Homeomorphic, but not naturally homeomorphic

Thorgott

you can move the 1 from sub- to superscript via ambient isotopy

Mike Miller

Because it's fucking unnatural to put the 1 as a subscript

Balarka Sen

8:26 PM

you know how complex analysts write $\Bbb T$ for 1-torus

and algebros write $\Bbb G_m$ lmao

user2103480

@MikeMiller hahahaha fair enough

Mike Miller

@BalarkaSen Even worse that's $\Bbb C^\times$

Thorgott

I've seen T, but what the fuck is G_m

Mike Miller

And they still call that the torus

The multiplicative group

Balarka Sen

lol yeah

@Thorgott Group_multiplicative

Thorgott

8:27 PM

???

Balarka Sen

it's $k^\times$

the algebraic circle

Thorgott

now I see why you guys hate algebraists

Balarka Sen

fucking G_m

madlad notation

user2103480

that's the discrete torus or the lie group

Thorgott

why would you put a subscript for something that's not an index and not even ambiguous

Mike Miller

8:30 PM

There's also $\Bbb G_a$, which is just $\Bbb A^1$

Thorgott

bruh

Balarka Sen

@MikeMiller proud boys apparently sell T-shirts with "Pinochet did nothing wrong" written on it

Sophie

I'm having a brainfart. That method where you have a curve and a rational point on it and then you trace a line through that point to find other rational points. What's it called?

Mike Miller

Cringe

Sophie

the point and line method? It was something like that but I can't remember it to save my life

Balarka Sen

8:31 PM

im thinking of buying some

user2103480

@Thorgott algebra I was the only wasted lecture I attended

Mike Miller

@Sophie Idk probably something like Euler's parameterization of Pythagorean triples

user2103480

@BalarkaSen nuclear take

Balarka Sen

@user2103480 i love it

user193319

0

Q: Complex Analysis and Bernoulli Numbers from $\frac{z}{2} \cot (\frac{z}{2})$

Define the Bernoulli numbers $B_n$ by $\frac{z}{2} \cot (z/2) = 1 - B_1 \frac{z^2}{2!} - B_2 \frac{z^4}{4!} - B_{3} \frac{z^6}{6!} - ...$ Explain why there are no odd terms in this series. What is the radius of convergence of the series? Find the first five Bernoulli numbers. I understand why t...

Thorgott

8:32 PM

:(

Sophie

@MikeMiller it's a general method not just for that equation

Mike Miller

I know but that's still how I'd refer to it probably lo

l

Check Silverman's book "rational points..."

user2103480

I'll seriously never care about splitting fields

Or... FINITE GROUPS

inser balarka meme here

Thorgott

smh

user2103480

@BalarkaSen blistering take

margaret thatcher take

Edward Evans

8:39 PM

don't say that name

user2103480

@EdwardEvans "dont summon the witch"

She called pinochet a dear friend when he was arrested in london in 2002, and wrote a letter to Blair (probably another name I shall not mention again) that he should be released. But let's stay away from the depressing topics

anakhro

Singmaster's conjecture

Singmaster's conjecture is a conjecture in combinatorial number theory in mathematics, named after the British mathematician David Singmaster who proposed it in 1971. It says that there is a finite upper bound on the multiplicities of entries in Pascal's triangle (other than the number 1, which appears infinitely many times). It is clear that the only number that appears infinitely many times in Pascal's triangle is 1, because any other number x can appear only within the first x + 1 rows of the triangle. == Statement == Let N(a) be the number of times the number a > 1 appears in Pascal's triangle...

user2103480

@Thorgott So we'll also stay away from splitting fields

anakhro

Kind of cute.

Thorgott

splitting fields are a very entertaining topic

Balarka Sen

8:43 PM

splitting fields are great

finite groups, not so much

Sophie

@anakhro that's a nice name

anakhro

@Sophie he was a nice guy, I hear.

user2103480

@BalarkaSen y tho

Thorgott

did you know that $S_4$ is the only group of order $p^3q$ that has no normal Sylow subgroup

Balarka Sen

a lot of galois theory has visual parallels. but instead of repeating myself let me direct you to Alekseev's ""Abel's theorem in problems and solutions"

user2103480

8:48 PM

Also, algebra I destroyed my perfect grade record. I didn't get the A+ in the exam because I didn't name the right Gauss' Lemma in the exam

Mike Miller

Gauss' Lemma (#5)

user2103480

I literally named one Gauss' Lemma from algebra but that wasn't cool enough I guess

@MikeMiller this

Balarka Sen

lmao

i'd have just said exp is a radial isometry

Thorgott

Gauß' Lemma? You mean the fact that R UFD => R[X]UFD?

Balarka Sen

using greek alphabet ohhhhh

smart

Thorgott

8:51 PM

spot the difference: βß

Edward Evans

@user2103480 Blair is somehow the eternal prime minister in my eyes, because he was prime minister when I was born and it felt like he was prime minister forever, given the weirdly dilated time of a child

user2103480

@Thorgott I erased all memory so no clue

Edward Evans

ßpot the differenß

Balarka Sen

GauB-ßonnet

user2103480

Lmao I'm the only german and everyone except me adheres to the ß instead of ss

which, now to think, is probably wise for a german

to do as well

Edward Evans

8:53 PM

so

you're saying

you adhere to the ss

loolol]

lmao

weyyy

ahem so what was i saying

um i mean splitting fields

Edward Evans

hahaha

user2103480

8:57 PM

@EdwardEvans fucc

Edward Evans

wah neyyy

Thorgott

Sanity check: If I have a set $X$, a countable cover $\{U_n\}$ of $X$, bijections $\varphi_n\colon U_n\rightarrow\varphi_n(U_n)\subseteq\mathbb{R}^n$ such that $\varphi_n(U_{nm})$ are open for all $n,m$ and the transition functions are continuous and such that two distinct points of $X$ belong both to a single $U_n$ or each to a distinct $U_n$, then equipping $X$ with the topology generated by $\varphi_n^{-1}(U)$ for $U$ open makes $X$ into a top. manifold and $(U_n,\varphi_n)$ into a chart.

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$Mathematics$ Mathematics