point-set-topology

[gritty widget]

@cold vine I’m testing, k?

[cold vine]

yeah

[gritty widget]

do you know what gf means?

i can give more explicit hints, but I think if you write out really explicitly what f, gf, [gf] all mean and then use the fact that g is null homoptopic you will get the answer

[cold vine]

Is it not the map g composed with the map f from S^n to X right

and then I think gf is null homotopic and I would have a nullhomotopic map from S^n to Y?

Is there something here

[gritty widget]

yeah this the idea

just fill in all the details

sorry for being so not directly helpful but this is such a good exercise

[cold vine]

ok nice! thanks a lot!

No this was very good ^^

Good to think through it

that was a good direction

[gritty widget]

welcome

[cold vine]

So I'm trying to show that p:SxS -> S#S (# is the smash prod ) induces trivial maps in homotopy and nontrivial in homology. I guess since we know S#S is S^2 helps us in the homology case but I'm completely stuck in the homotopy case

Anyone have a tip on where to start?

[marsh forge]

is S just S^1

[cold vine]

yeah

[marsh forge]

compute the homotopy groups

you can compute the map in homology explicitly i think but visually it makes sense

[cold vine]

Ah I got the homotopy now I believe. I first calculated the homotopies incorrectly so I though it wouldn't come explicitly, but I think I'm getting it to work

[marsh forge]

well i think the problem is pretty trivial if you just write down pi_n of both spaces

which you know well enough

(obviously pi_n S^2 is hard but it turns out it doesnt matter)

[cold vine]

Yeah the groups turned out trivial on one side

[marsh forge]

yeah

then for homology if you think about the actual quotient for T^2-> S^1 smash S^1

the cellular/simplicial approx is ‘obvious’

[cold vine]

Yeah I've only done singular homology but I'll think about the quotient

thanks!

[marsh forge]

lmk if u want help computing it but yeah just get a grip on why its true visually first

then you can make it work w any formalism

[cold vine]

ok thanks ill try!

[tepid quarry]

This is a basic question so idk if it belongs here. Is a Topological space obtained by $\mathbb{R}\times [0,1)$ paracompact? I've seen $\omega_1 \times [0,1)$ being used as a counterexample because $\omega_1$ is an uncountable set, so it seems like replacing that with $\mathbb{R}$ should also work if that's the case.

[gentle osprey]

Exynouz

[cold vine]

So I guess this map turns out to be a map from reduced homology of T^2 to S^2 but how do I make sure it's not trivial

[marsh forge]

take a generator of H^2T^2 and think about visually what happens to it

under the quotient map

one such singular generating chain consists of the entire surface of the torus

what is the image of the surface in S^2 thought of as T^2/~

[cold vine]

okay thanks!

[tepid quarry]

Well... Nvm

[prisma seal]

ok I might be stupid but I'm having a ton of trouble parsing this definition

could someone give me an intuition for what this is saying?

[gritty widget]

you're not stupid for having trouble parsing this definition because this is garbage

seems like an extremely complicated way to just say "equals the graph of a function up to a permutation of coordinates"

who writes like this

Overall I'm not sure what is the point of the condition that the elements of K are increasing

for the sake of digesting this definition it's good to keep an example in mind

e.g. the circle

[prisma seal]

this prof has no lectures and only notes that look like this :^)

[gritty widget]

near (1, 0) it's locally the graph of the function sqrt(1 - x^2), in the form (x, sqrt(1 - x^2)), but near (0, 1) it's locally the graph of the function sqrt(1 - y^2), in the form (sqrt(1 - y^2), y)

[lost estuary]

This is the shittiest definitions I've ever read I think, the idea is drowned in an ocean of futilities

[prisma seal]

these are literally all the notes

with like 2 examples

[gritty widget]

my condolences

[prisma seal]

no lectures

[lost estuary]

F

[prisma seal]

and minimal pictures

[gritty widget]

F

On the other hand there I was recently reading evans, who seems a bit too handwavy about "locally C1", defining this as equal to the graph of a C1 function up to rotation and relabeling coordinates

would you rather have this?

I might say yes

[prisma seal]

anyways with the circle, what is it saying?

[gritty widget]

i outlined it here

but i can go into some more detail if you'd like

[prisma seal]

oh lmao sorry I'm blind

[prisma seal]

right, that's only local because that equality stops somewhere

or uh

it would be not local if that equality worked everywhere

[gritty widget]

i think you're on the right track

the circle isn't globally the graph of a function because then you'd run into issues (if it were then the function would be multivalued)

[shut moat]

wth

[prisma seal]

I guess the part I'm having the most trouble parsing is what K and L actually represent

[gritty widget]

but you can patch together the graphs of functions to get the circle!

[prisma seal]

I think that's the part that isn't clicking

[shut moat]

the idea is basically "locally a graph but we're being looser about which coordinates are functions of which"

[prisma seal]

So is it just that around each point, you can find a sufficiently small open ball, the intersection of M and the ball looks like a graph?

[gritty widget]

yes

[prisma seal]

what the fuck

why was that not written

[gritty widget]

[prisma seal]

what the fuck that's so clear

[lost estuary]

Man

It's an emergency situation

[gritty widget]

this was clearly written with that one type of extremely pedantic student in mind

u know the one im talking about

[lost estuary]

You have two steps in front of you

[prisma seal]

ok now I need to connect that to this definition to this

[prisma seal]

the prof is writing these notes as we go, maybe that's just how his brain works lol

[lost estuary]

I disagree, a pedantic student would prefer a short def

[prisma seal]

he said extremely pedantic

this one

[lost estuary]

Step 1: find a good reference for your course

Step 2: pump libgen

[gritty widget]

time to exercise my title of "lee simp"

read lee

:petthecat:

[prisma seal]

Lee - the book is called like Smooth Manifolds or something right?

[gritty widget]

Introduction to Smooth Manifolds

i can dm a copy

[prisma seal]

I can probably find a copy

if I can't, I'll take you up on that

already found a pdf lol

[gritty widget]

popular book huh

[prisma seal]

anyways, what are K and L here

is it just a way to like pick different restrictions on different coordinates?

[shut moat]

the list of indices labeling the function coordinates and the ones labeling the variables, I think

[prisma seal]

errr sorry

what's the meaning behind even taking those indexing sets

[shut moat]

like terra's example for the circle, in some regions, the graph looks like $(x, y(x))$, while in others $(x(y), y)$

[lost estuary]

I think he really wanted to point out that they could be in any configuration or something

[gentle osprey]

~S^1

[shut moat]

you wan to account for different cases, because the default definition of a graph would be $(\bm{x}, f(\bm{x}))$ where the variables are neatly separated

[gentle osprey]

~S^1

[prisma seal]

OH I THINK I SEE

[lost estuary]

Like it's not necessarily (x1, ...,xK) = f(xK+1,..., xn)

[prisma seal]

Right you're just picking out some of the coordinates

and then delta is making them be very close to the corresponding coordinates of the point p

and you're sending those to the other variables

[lost estuary]

Yes

[prisma seal]

which are also not too far from the corresponding coordinates of the point p

and you take the graph of that function

and it looks like your M

or if you intersect it with the small epsilon/delta rectangle, it equals M

[lost estuary]

I think we can "improve" his definition by introducing a permutation $\sigma \in S_n$ ....

[gentle osprey]

Jean-Jérôme

[shut moat]

this was a nice illustration of it in hubbard

[gritty widget]

anything u could do to the given definition would be an improvement

[prisma seal]

did you see this profs statement of the gauss-bonnet formula

[gritty widget]

i might have before

show me again

[gritty widget]

inb4 lee has a perfect and flawless statement of it

is this like

at least accompanied by a picture

[prisma seal]

no

[gritty widget]

this might be okay if it were

[prisma seal]

scroll up a lil to see my profs definition of "locally equal"

hahahahahhaa

I'm saving all of your comments for the course eval

[lost estuary]

Hahaha

[gritty widget]

no differential geometer mathematician writes like t his

[prisma seal]

"or a book that actually got edited" holy shit haha

[shut moat]

find literally any book on diff geo, I doubt it can be worse than the notes

[prisma seal]

definitely going in the course eval

[gritty widget]

pummel him

[prisma seal]

"Lectures would be helpful"
"I'm sure they would be"

[shut moat]

wait, the class doesn't have any lectures??

I didn't know that was a thing lmao

[gritty widget]

this dude does not deserve even a shred of decency in the course evaluation

[prisma seal]

he's widely regarded as one of our best profs, and apparently he's a great lecturer. Even his non diffgeo notes are usually positively regarded

But he's just... not doing online school well

[gritty widget]

you're paying for the notes

[prisma seal]

and the fact that this is a degree requirement

and the GPA boost

I don't know why diffgeo is a degree requirement

but

[gritty widget]

just convince ur uni to let u take 367 at uoft instead

or something

[prisma seal]

I'm like 80% done this course though

[gritty widget]

oh

well

[prisma seal]

well, 60%

but in a week I'll be 80% done

there are 5 assignments worth 100% of the grade

[gritty widget]

on the bright side it's almost over

[prisma seal]

and I'm done 3 of them

on the last one I wrote "I'm turning dL into -4dt here because it gives me the right answer. I don't know why it works" (paraphrased) and still got full points on that question lol

[gritty widget]

lmao

nice

[prisma seal]

ok he isn't even proving inverse function theorem. this was proven in calc 3 lol

[gritty widget]

inverse function theorem

[rotund thicket]

if f is injective it has an inverse

done

[sleek thicket]

@prisma seal you should take this up with a person in your department

complain

this is not an acceptable amount of instruction

Like it won't help you now, class is almost over. But it's worth reporting it in case this person is teaching online in the future

[prisma seal]

this person has been doing this for 2 terms

[sleek thicket]

have students complained to people who aren't the prof?

[marsh forge]

that prof should lose tenure for that level of bs lol

thats below the bare minimum by a lot

[prisma seal]

he's widely regarded as a great prof in non covid terms

[rotund thicket]

yeah but the dept could try giving him a slap on the wrist

[prisma seal]

course evaluations aren't just read by the prof? but I don't know if anyone has gone to the dean or anything

[rotund thicket]

probably wouldnt do much

[marsh forge]

course evals are often read by others

for non tenured profs they are incredibly important in fact

[sleek thicket]

I agree it probably wouldn't do anything, but I would recommend contacting the dean

Like this is literally not a course

[prisma seal]

this person is not a prof, they're a lecturer, but they are on an indefinite contract

[marsh forge]

oh

then yes i assume someone reads their evals

[rotund thicket]

i mean it's best to be direct about it

rather than an eval

you could get maybe a couple of students to sign some google form

[prisma seal]

maybe yeah

what would that accomplish though?

he's normally a great prof, I don't think the department can compel him to record lectures

and I definitely don't want him fired

no, no lectures at all @gritty widget

there are office hours once a week that last 40 mins

[shut moat]

what's his objection to virtual lectures??

[gritty widget]

2 much effort

[prisma seal]

#point-set-topology message this is all I know lol

[shut moat]

they make it sound like they have no choice lol

[gritty widget]

this looks horrible

[gritty widget]

is this a correct solution? and is there a better method

[gritty widget]

where is this from

[tough imp]

This is more succinctly stated as saying a map Spec B -> Spec A has the scheme theoretic fiber over a prime p in Spec A as
Spec (B (x)_A kappa(p))

In this case you have A = Z, B = Z[x1,...,xn], and kappa(p) = F_p

[delicate hollow]

ok

quick question

[uncut surge]

that was quick

[thin nova]

https://math.stackexchange.com/questions/4075940/methods-of-calculating-great-circle-distances

could anyone please answer this

basically my question is,

1. is it possible to calculate the great circle distance using calculus of variations?
2. Will it assume a spherical or an oblate spheroid Earth?
3. Is this any more of less accurate than the Haversine or the Vincenty's formulae?

[fading vale]

im not sure this belongs here lol

[thin nova]

yea but no idea where to put it

i just saw geometry in the channel heading so

and within "early university" there was no geometry so i had no idea where to put it

[gritty widget]

Yeah, this sounds quite geometric, if you ask me

[gritty widget]

the contraction part of this proof makes me feel uncomfortable

is it even right

this is in lawson’s spin geo btw ty

[proud ridge]

Is covering the fundamental group normal for a first topology course?

[gritty widget]

mine did

not in much detail but we at least saw it and like

computed pi_1 S^1 = Z

[gritty widget]

cool, thanks

mine did 1-5 and 9 2-5, 9

[shut moat]

my class is trying to speedrun chapters 1 through 9

[gritty widget]

we skipped some of the smaller and less relevant stuff throughout the chapters

also

we did not do 1

i forgot that 1 is set theory

my copy calls it 1

[shut moat]

I'm low key annoyed by some of the sections we're skipping, like why skip quotient topology >:(

and I haven't found the time to independently sit down and go through that yet

I only know some embedded manifold stuff from self study

[sleek thicket]

Thst sucks sphere

[shut moat]

¯_(ツ)_/¯

it's a weird class

the space filling curve section seems cool tho

[sleek thicket]

Space filling curves aren't that like, deep imo. You can pick it up

They're really cool

Don't get me wrong

but you can sort of just read the section

(or not read it)

[wide kayak]

@ripe moss it seems to say that the surface parametrized by (x, y, f(x,y)) with f being C^1 is contained in a plane iff the isothermal coordinates (u,v) of f are obtained from a nonsingular linear transformation of (x,y)

it's a well known result that any sufficiently smooth surface (dim = 2) locally has isothermal coordinates exist

so this seems to say if you can get those "easily", i e. just via a linear transformation, then the surface is (part of) a plane

what book is it from?

[desert bloom]

I am rather stuck in ii.)

I'm not quite sure what the hint is trying to hint at

what does the 1 point space have to do with this?

[gritty widget]

in the definition of proper map, what Z can you plug in so that you get a statement about the closedness of f: X → Y

[fathom cave]

how do i write an open set as a countable union of closed sets?

I need a hint

[gritty widget]

i'm pretty sure this is false

endow the cofinite topology on an uncountable set, then consider any non-empty open set

@fathom cave

[fathom cave]

no the topology im trying to prove this is on euclidean topology @gritty widget

[gritty widget]

oh

lol

[fathom cave]

i should have mentioned it

also i have a confusion with a definition. so i wana make sure i have understood it
a defintion of saturated set says "A subset B of X is saturated if B is an intersection of open sets" so does this intersection has to be countable? i dont thnk so because the defintion does not mention it.

[bleak helm]

No, it doesn't have to be countable

[fathom cave]

cool

ty

[gritty widget]

my first idea is to take balls of radius 1/n around every element of the complement

[fathom cave]

fuck me 😂 its open interval, not open set

damn i cant even type correctly

[bleak helm]

You have to write (a, b) as an intersection of closed intervals?

[fathom cave]

ye

[bleak helm]

[a+1/n, b-1/n]

[gritty widget]

lol

prove that a closed set is F_sigma

[bleak helm]

I hope you can do the closed interval one on your own

[fathom cave]

that one is trivial

[bleak helm]

Everything in life is trivial

[fathom cave]

ty

[bleak helm]

Just like two boxing is trivially the right answer

Np

[gritty widget]

you don't get the million

[bleak helm]

That was never an option then

[gritty widget]

lol no

[bleak helm]

I know we joke around, but it's genuinely astounding to me that one boxers aren't just trolling me. Every now and then I have to remind myself yall are being serious 😂

[gritty widget]

it is also astounding to me that two-boxers are so petty that they would give up a million for a <10% chance of getting $1,001,000

whatever man

[bleak helm]

😂

[fathom cave]

I feel so lost rn

[gritty widget]

about the exercise?

[fathom cave]

no this boxing thing

[gritty widget]

oh lol

it's an inside joke

[fathom cave]

lol

[marsh forge]

wait whats the box thing

i wanna know

[viral atlas]

Newcomb's Paradox

https://en.m.wikipedia.org/wiki/Newcomb's_paradox

[marsh forge]

oh thats easy to answer

both me and the predictor know

[gritty widget]

box = space homeo to [0, 1]^n
two box = box \sqcup box

[marsh forge]

i dont need 1000 really

[viral atlas]

So, you pick one box or both boxes?

[gritty widget]

the "paradox" is to explain how two-boxers can be so silly

anyway, i will not talk about it further in this channel

[marsh forge]

one box

dominant strategy is a bad model here

util maximization is the correct model

its not a question of anything but what game theory to use imo

the wikipedia analysis is more or less correct that the only hard part of this problem is the under specification

actually i change my mind this is trickier than i thought bc of how underspecified it is

idk are questions that just dont have enough information interesting

[uncut surge]

i don't think the interesting part about this question is finding the correct answer, and rather thinking about why both answers feel weird

[wide kayak]

I saw somebody say they would counter the predictor by just flipping a coin to decide whether to pick 1 or 2 boxes

but if the thing is truly an omniscient predictor, it necessarily knows the result of your coin flip as well

in a deterministic universe, perfect knowledge of the future of any subset of the universe should extent to perfect knowledge of the future of the entire universe, or at least everything in the same causal bubble, no?

[marsh forge]

they only feel weird because the question itself is insufficiently phrased though

[uncut surge]

I mean I don't think there's a better phrasing

It basically just comes down to "do you believe a perfect predictor like that can exist"

[marsh forge]

huh

how is that relevant its a hypothetical

if you assume the predictor is perfect the answer is trivial is it not

there is barely a decision tree at that point

the issue is mostly about how reliable the predictor is

[uncut surge]

yeah well then i guess you're questioning if the predictor is really perfect or not

hence "do you believe a perfect predictor like that can exist"

[marsh forge]

again not relevant

its not about whether i believe it, it is a hypothetical

the question does not specify whether the predictor is perfect

[gritty widget]

in the situation, your know the predictor is highly reliable

[marsh forge]

that is meaningless

[gritty widget]

why

[marsh forge]

if he has a .00000000000001% chance of being wrong

its very different

from a 10% chance

[wide kayak]

I think 2-boxers don't really take seriously the proposition that if you converge on taking 2 boxes, then the predictor will have successfully predicted you do just that

[marsh forge]

both are “very reliable”

“very reliable” is the entire problem

[wide kayak]

they think of the predictor's decision as "set in stone" and that they can then cheat it

[marsh forge]

it doesnt tell you anything

[gritty widget]

even with a 10% error probability you should take one box

#point-set-topology

[marsh forge]

again my entire point is

it depends on the % chance

of the predictor being wrong

or even like

whether its a percent chance

it could depend on the choice

or the player

or anything

the predictor is so underdefined that the question is meaningless

[wide kayak]

it's not though

[marsh forge]

it is

[gritty widget]

let us say 99% accuracy

[wide kayak]

the details of it don't matter, you're supposed to assume it has the advertised property

[marsh forge]

thats not the original question oxide

[gritty widget]

what would you do them

what, why

[marsh forge]

it says “very reliable”

this is like being asked to show that x is greater than 5

and choosing x to be 6

[gritty widget]

lol

[marsh forge]

anyway

if its a pure % chance

[gritty widget]

well in most discussions i've seen, they always take the accuracy to be >=90% at least

[marsh forge]

it just boils down to basic arithmetic i dont wanna do

[marsh forge]

it probably still matters as 1mil>>1 thou

[wide kayak]

I thought it does just that

[bleak helm]

You could tell me the predictor has 100% accuracy and I would still take both boxes.

[wide kayak]

now that doesn't make sense

[gritty widget]

if the predictor is omniscient, then you can either get 1000 or 1 million, no other choice

[bleak helm]

The accuracy is irrelevant because he has already made his prediction when I make my decision.

[gritty widget]

lol

[wide kayak]

correct

[marsh forge]

this is nonesense to me

[gritty widget]

if you take both boxes you'll only get 1000 then, no chance of 1mil+1000

[marsh forge]

draw out the decision tree

and think to yourself

about your mistakes

[bleak helm]

😂

[marsh forge]

oh ok

[bleak helm]

I'm legit not

[wide kayak]

I think 2-boxers don't believe the predictor has the property "If you end up doing scenario X, the predictor correctly predicted your doing scenario X"

[marsh forge]

the problem is trivial if the predictor is perfect

[wide kayak]

but that should be implied by it being 100% accurate

[bleak helm]

I thought oxide was at first. One boxing makes no sense to me.

[sleek thicket]

you are all children, I would simply be the predictor

I know which box contains the 1 mil and so can take it myself

[marsh forge]

do you mean 1-boxer ultra

[uncut surge]

i hope you know you're all falling into the trap

"To almost everyone, it is perfectly clear and obvious what should be done. The difficulty is that these people seem to divide almost evenly on the problem, with large numbers thinking that the opposing half is just being silly."

[bleak helm]

Anyway, watching tv, I will argue again another time.

[marsh forge]

Again

thats not my take at all larto

Oh i was misremembering

yes

[sleek thicket]

oh lol sorry ultra

[marsh forge]

this is just pretentious and annoying lol

[bleak helm]

I'll ask one more question to max just in case he interprets it differently

[sleek thicket]

also, montero is very very good

[marsh forge]

it is

its so fucking good

[sleek thicket]

Yeah

[marsh forge]

if the album is this good

i will cry

lil nas x will bring the hot boy summer we so desperatley need

[bleak helm]

You wake up, you've never heard of the problem before. You are told the predictor has already made his decision. So what you do now does not affect the prediction. Do you take one or both boxes?

[marsh forge]

imagine montero going off the first night i go to a US club

ill lose my mind

[sleek thicket]

I simply cannot

[marsh forge]

no thats not a sadcat

[sleek thicket]

hope is illegal

Imagining a future where clubs are a thing is sadcat

[wide kayak]

I don't see how this changes anything. In the problem it's your first time hearing about it

[bleak helm]

I don't think it changes anything either. But I think it makes 2 boxing more obviously correct.

[wide kayak]

but if you take 2 boxes, the predictor will have predicted your taking 2 boxes

[marsh forge]

Luna I feel like you are making a classic error

of trying to use like

weird real-world logic

for a hypothetical problem

[bleak helm]

😂

[marsh forge]

the predictor is perfect

you can make any choice

and the predictor already predicted it

thats the assumption

[wide kayak]

that's what I'm saying also

I think people are just split on this point

[marsh forge]

I think the issue is that like

hyptheticals really aren't that interesting

if you don't have faith in the hypothetical

if you start saying "what if" you realize how limited hypotheticals are

[bleak helm]

Well, yes, I am interpreting it as my choice cannot change the prediction because the prediction happened first. If you just remove that, then sure you can one box

[uncut surge]

i think the most interesting thing is really what people take away from this

[bleak helm]

But if you remove that assumption, there's literally no paradox so that's obviously not what is intended

[marsh forge]

this is equivalent to a perfect predictor

[wide kayak]

if you assume the predictor is genuinely accurate as advertised, whatever you end up doing will be predicted by it already

[marsh forge]

what is intended is to ask a question that is malformed to produce faux-intrigue

its just silly

but under an additional assumption of a perfect predictor the answer is easy

[bleak helm]

No comment. Back to my TV.

[wide kayak]

I just realized this whole scenario is reminiscent of Minority Report

[gritty widget]

this problem is always a nerdsnipe

that is amusing

[marsh forge]

is it a true nerdsnipe

i feel like a nerdsnipe involves solving a problem

this is just inflammatory hahaha

[gritty widget]

well maybe not an actual one, but in the spirit of it

[marsh forge]

me and ultra came to the same conclusion basically instantly after I re-read the problem statement

[gritty widget]

same

[wide kayak]

ahh! wow, that takes me back. I did an undergrad research project on minimal surfaces, that was one of my sources

[gritty widget]

minimal surfaces

[desert bloom]

I have no idea how to do iii.)

I think they are talking about this in the hint

[bitter yoke]

Is the fibre here just f^{-1}({y})?

[desert bloom]

yeah

he uses upper star for pre image

instead of -1 because it is not strictly an inverse

[bitter yoke]

ok sure

[desert bloom]

just some sort of direction to what I'm supposed to do is good

cuz I don't even know how to start for some reasons

[bitter yoke]

then I think the idea is that for any space Z, f*({y}) x Z is a subspace of X x Z and you can use the proper property to show (ii) of compact

[wide kayak]

suppose you have an open cover of f*({y})..... reasons reasons QED

[desert bloom]

Yeah, I did start with that at some point

but I didn't know what to do with that fact

how would I be looking for a finite subcover is seriously beyond me

[wide kayak]

what's meant by "the product characterization of compact spaces"?

[desert bloom]

proposition 4.19

I posted it underneath

[wide kayak]

ah

[desert bloom]

its the ii.) statement

[wide kayak]

woah, never heard this

[desert bloom]

well, I am equally as confused

apparently the proof requires zorn's lemma

and I'm using apparently because I have no idea what the proof is

since its non-examinable so its not given

so I don't have the slightest bit of intuition why this is even remotely true

[marsh forge]

wtf textbook is this lol

[desert bloom]

and I am in that stage of doing my problem sheets where my single braincell is just playing bohemian rhapsody non-stop

its my notes

[marsh forge]

yeah its just an odd approach to this stuff

okay let me think for a second sorry was distracted

[desert bloom]

no problem man

at this point, any help is appreciated

[marsh forge]

Okay I am trying to look up one lemma to make sure something is true haha but I know how to do it i think

[desert bloom]

take your time

[marsh forge]

Yeah okay i am correct

So i am not sure how much of a hint you want, but the strategy here is indeed to use 4.19.ii and the original definition of proper together

In particular you want to notice that

Let F be the fiber

FxZ -> {y}xZ -> Z is the same as the projection FxZ -> Z

So the steps I would take are

1. Show (or cite, hopefully) that the restriction of a closed map f is closed, so that FxZ->{y}x Z is closed

2. Show the projection is closed

3. Hence F is compact

[desert bloom]

oh, so you apply f x idz to the F x Z?

[marsh forge]

yeah

i should be more careful and say that the restriction of fxid to FxZ is closed

rather than just f by itself

[desert bloom]

oh right, F doesn't have to be closed...

[marsh forge]

It will be for nice spaces but in general no

[desert bloom]

damn I'm dumb

thanks dude

[marsh forge]

np

[quasi forum]

Question, how does the question of convergence translate from metric to topological spaces?

Better yet, does it?

[tough imp]

Yeah

You take sequences and just say for any nbd of a point, eventually the sequence lands inside that nbd

If this is satisfied for every neighborhood of a ppint L, then L is a limit point of the sequence

[quasi forum]

So does the neighborhood have to be an element of the topology?

[tough imp]

By definition

But limits need not be unique in general

[quasi forum]

Why is that by the way?

[tough imp]

Consider a topology X,empty set

[quasi forum]

That is definitely a new concept

[tough imp]

Every single point is a limit point of ever sequence

[quasi forum]

Oh because the only neighborhood of each point is X

Okay, so if we considered an element $x_n\in \R^+$ where $\tau :={\R^+,\emptyset,(0,n) for each n\in \N}$ then we say x converges to any point in the set $(0,\ceil{x})$

There we go

There we go. Does that logic check out?

Well, technically it is a discrete space anyways

[sleek thicket]

Are you talking about the constant sequence with value x?

[tough imp]

x isn’t a sequence so it can’t converge to anything unless you do what sham says

[quasi forum]

One second, I want to rephrase

[sleek thicket]

also, is x in the set N? If so, why are you taking the ceiling? If not, it's not in the set on which τ is a topology

So the issue is that x is an element, not a sequence. Are you thinking of the constant sequence with value x?

[quasi forum]

Technically I am trying to think of a sequence that would 'normally' converge to x in the normal euclidean metric

And I guess another notable restriction is that (x_n) has to be increasing here

Man, this did not pan out how I wanted it to 😆

[tough imp]

The issue is that you’re still saying x converges to...

You need the sequence of x_n to converge to points

[quasi forum]

Yea, I'm scrapping that. I really don't think my thought process represented at all what I wrote

Okay, let's say $(x_n)\to x\in \R^+$ with the normal euclidean metric

[gentle osprey]

dackid

[quasi forum]

Another requirement is $(x_n)\leq x$ for all $n\in \N$

[gentle osprey]

dackid

[sleek thicket]

sure

[quasi forum]

Okay, now consider the topology I had earlier

[sleek thicket]

considering

[quasi forum]

With respect to that topology, would we say that $(x_n)$ converges to any point in $(0,\ceil{x})$?

[gentle osprey]

dackid

[sleek thicket]

no

[quasi forum]

I definitely agree with that

[tough imp]

Pretty sure it converges to any point in [floor{x}, ceiling{x})

[quasi forum]

So what exactly would we say this converges to if we consider that topology

[quasi forum]

But that's not a set in the topology

[tough imp]

Who cares

It’s points in your set

Actually I think it converges to any point in [floor{x},infinity)

[quasi forum]

So then why don't we just say it converges to x?

[tough imp]

It does

[sleek thicket]

it does converge to x, but also lots of other things

Convergence isn't unique outside of metric spaces (and certain classes of topological spaces)

[quasi forum]

The 2nd part is what I am trying to understand

[tough imp]

Your topology doesn’t distinguish points very well so you can converge to lots of things

Take a point in the set I described

[quasi forum]

Sure 🤔

Umm, there needs to be an x_n in any open neighborhood of 3?

Yes

That I'm not as sure as I thought I was

Oh wait....

We see in the open sets (0,3),(0,4),(0,...) That every x_n is in each set

Well, because we explicitly said $x_n\leq 2$ for all $n\in\N$

[gentle osprey]

dackid

[quasi forum]

Hmm, okay

Ah okay, that is true

[sleek thicket]

well, you might not converge to x anymore

right?

[quasi forum]

Yea, that is true

[sleek thicket]

err I think I'm wrong

Ignore me

Yeah sorry slim, I'm being silly

Yeah

[quasi forum]

So what exactly is an open neighborhood? I think that is where the confusion is

I know it w. r. t metric spaces

Oh, that's vague af

Well, I mean that can be any set containing x

I thought it needed to be restricted by the open sets in the topology

[sleek thicket]

open neighborhood always means that slim, the terminology concern would be if you just said "neighborhood"

[quasi forum]

Well, for our topology, (1,2) is not an open set in the topology

So since it wasn't, I thought that we couldn't use it as a neighborhood

I understand that, I was just trying to tell you why I was confused

Okay, anyways, so then I am actually not too sure I understand why this converges to 3 anymore

I'll give you the metric variant, and you can translate over to topological spaces

A sequence $(x_n)$ is convergent to a point x if for any $\epsilon>0, \exists N>0$ so that: $\forall n\geq N: d(x_n,x)<\epsilon$

[gentle osprey]

dackid

[quasi forum]

So for sufficiently large n, we can always find an open set, in this case $B_{\epsilon}(x)$ so that $x_n\in B_\epsilon(x)$ for all $n\geq N$

[gentle osprey]

dackid

[quasi forum]

How so?

Ah, the first part

So for any open set, we can find a sufficiently large N so that...

Okay sure. Let me restart that part

So for any open set containing {x}, we can find a sufficiently large N so that $x_n$ is in the open set for $n\geq N$

[gentle osprey]

dackid

[quasi forum]

How's that?

Okay sure

Let's let x be 2 like last time

This certainly still means that a limit point is 2.

But I am not sure sure how this would suggest 3 is a limit point

Because can't we choose a sufficiently small open set, say $(3-\epsilon,3+\epsilon) in which it is not true?

So it does have to be an open set in the topology

Ohhh okay, cool! In that case any open set that contains 3 also contains the necessary x_n

Ahhh, okay

Open sets of X helps me get that down

So then the limit points of this sequence would be in $[2,\infty)$

[gentle osprey]

dackid

[quasi forum]

Got it! I understand now.

So is that why we only really considered Balls in a metric space, since those are the only open sets?

Ah okay. So it was sufficient enough to show convergence

[gentle osprey]

slimvesus

[quasi forum]

Ahh, okay. That makes a lot of sense now.

Ohhh, interesting

This was very helpful. Thank you very much

[west spindle]

so here's something i'm wondering about:

how would i go about computing the complement of a trefoil knot?

for context, i know some basic topology and very very basic knot theory, so like. ELIU i guess?

[cedar pebble]

what do you mean by computing this?

Do you mean computing various invariants of this complement?

[west spindle]

i mean the complement is a 3-manifold, right

what i'd like is some kind of intuitive representation of that manifold

like... "start with a polyhedron and glue these sides in this way"

that kinda representation

if it makes any sense

the most i know about knot complements is the Not Knot video if that helps any

[cedar pebble]

Oh I see

[feral copper]

Hi ! Would anyone have a reference in mind (other than Cerf's paper) about Cerf theory please ? :)

[sweet wing]

really when looking at hyperbolic 3-manifolds you care a lot about the fundamental group (see Mostow's rigidity theorem), so really what you can do is compute the fundamental group then if you want something explicit just find an explicit representation G in PSL(2,C) and then your quotient is H^3/G

[west spindle]

hm

[sweet wing]

"The polyhedron and glue these sides" is specifically for 2-manifolds because they can be (all?) constructed like that

[west spindle]

for 2-manifolds you would start with a polygon no?

okay, how do i compute the fundamental group of the complement of a trefoil

[sweet wing]

oh yesh polygon mbmb

unfortunately 3 manifolds are quite finicky to handle but the rough summary on how to deal with them is usually from geometricization conjecture it reduces to looking at discrete subgroups of one of 8 groups

so the fundamental group is somewhat tricky to compute but a rough idea is

• select a bunch of generators by kinda cutting the complement up into simply connected components
• figure how to glue them by van karpen

intuitively van Kampen is kinda like

observing what relations are there between possible loops around the knot

lemme try to draw how we can do it for trefoil

[west spindle]

man

[sweet wing]

so you kinda like draw a bunch of loops then i think we can move g1g3 and g2g1 to each other

[marsh forge]

@west spindle they keyword to look up it called the wirtinger presentation

it provides a nice algorithm for computing fundamental groups of knot completments

(under the hood the wirtinger presentation is just repeated van Kampen but its much simpler than trying to work this out yourself)

[sweet wing]

if you imagine hard enuf you can see that g2g1 = x = g1g3

[marsh forge]

I once knew of a great book on this

but i no longer do

[sweet wing]

ngl that image i sent is garbage i should learnt how to draw these better

but yea look up what max said there's prob a book with way nicer images

for more general treatment of hyperbolic 3-manifolds i quite liked Albert Marden's book Outer Circles (i heard he has a more updated book may want to see that instead)

[wide kayak]

Help! I got tangled up with an Alexander horned sphere!!

[sweet wing]

[uncut surge]

"Help! I got tangled up with an Alexander horned sphere!!" this sounds like one of these bad mobile game visual novel screenshots

this kinda stuff

[stray skiff]

Hey, maybe a trivial question but how would I prove that the composition of smooth maps is smooth? It feels like I can just say "trivial!" so I'd appreciate a pedant chiming in 🙂

[uncut surge]

Smooth maps between open subsets of R^n or between arbitrary smooth manifolds?

[stray skiff]

smooth manifolds

[uncut surge]

Then I guess it just follows from the correpsonding statement for open subsets of $\mathbb{R}^n$, since a map $f : M \to N$ is smooth if it is smooth if composed with charts, i.e. $\phi \circ f |_{\psi^{-1}(U)} \circ \psi^{-1} : U \to V$ is smooth, for charts $\phi,\psi$ of $M$

[gentle osprey]

Lartomato

[stray skiff]

yep ok so I needed to mention that I am composing with charts

[uncut surge]

Yah, that's the manifold-philosophy

A function between two manifolds has property P if it has the property in every chart

[stray skiff]

nice okay, seems like I will be saying that a lot

[uncut surge]

And then the important trick is that, if you have a composition $f \circ g$ of smooth functions, you can write the composition in charts $\phi \circ f \circ g \circ \psi^{-1} = \phi \circ f \circ \tau^{-1} \circ \tau \circ g \circ \psi^{-1}$ for appropriate charts $\phi, \psi,\tau$, and then it's simply a composition of smooth maps on $\mathbb{R}^n$

[gentle osprey]

Lartomato

[uncut surge]

Some details to carve out, making sure that all domains and codomains make sense, but that's the general idea

[stray skiff]

yeah good enough for me, thank you

[uncut surge]

np!

[gritty widget]

so

i'm trying to think about ramification in ℤ[i] geometrically

since ℤ[i] = ℤ[x]/(x² + 1), it suffices to consider prime ideals of ℤ[x] containing (x² + 1)

wait, let me find that picture

it would be the [(x² + 1)] curve right

but then how would you know that (2) is ramified and (4n+3) is inert purely through algebro-geometric techniques

[jaunty bear]

Hey guys, i'm doing a paper in the link of symplectic geometry and analytical mechanics (with extension toward Weyl/Clifford algebra), have you got any goooood book/paper to read about the link between symplectic and mechanics ? because all i have read so far isn't that great, i mean, doesn't emphasis the link

[frigid patrol]

@gritty widget probably the better channel for this is #groups-rings-fields or #advanced-number-theory

[gritty widget]

oh thanks, but i'm now in the process of understanding the answer on my own

[thin bramble]

Would anyone be willing to help out?

[thin bramble]

<@&286206848099549185>

[marsh forge]

This should be in #geometry-and-trigonometry

[lucid dragon]

Can anyone prove that the perimiter of Minkowski sum of convex figures is the sum of the figures perimiters(basically perimiter is linear to Minkowski sum)? I see why this is intuitively true, it's trivial for polygons, but i don't know how to provide a rigorous proof.

<@&286206848099549185>

[pale sky]

does anyone have an intuitive explaination for why the first betti number of the projective plane is 0? basically this thing in the picture. Why don't we have gaps or holes? Is it obvious that any loop on this thing is homotopic to the identity?

[flint cove]

The first betti number being zero doesn't necessarily have to mean that the space is simply connected, iirc it only means that the abelianization of the fundamental group is trivial

good question tho, never thought about it

Wiki says the first fundamental group is ℤ/2 which makes sense when you think about the loop goingfrom the center up and from the bottom back to the center; when you move the lower part to the right, the upper part is going to go to the left, so you won't be able to contract it

But if you take double that loop then you can leave the first peek-through as is and move the other peek-through around

Oh wait I forgot that ℝP² is literally SO(3, ℝ), that's why that reminded me of the SO(3)-Argument

[sleek thicket]

isn't SO(3) = RP^3?

[flint cove]

Ah yes, sorry.
But it also has 2-torsion for similar reasons, doesn't it?

[sleek thicket]

yeah

universal cover is degree 2 over it

kind of by definition

RP^3 I mean

[flint cove]

Yeah, our good boy SU(2)

[sleek thicket]

lol

nice

[flint cove]

I remember that because years ago I claimed that SU(2) and SO(3) were isomorphic, and haven't been corrected. This was very embarrassing in retrospect.
(of course their lie algebras are isomorphic, which is what I meant)

[sleek thicket]

I always mix up which one covers the other

(if I think about it for a second as like, unit quaternions acting by rotation I remember, but formally I mix them up)

[flint cove]

idk you can try remembering a phrase like „real rotations have torsion!“

so SO(3) is not the one being simply connected

[sleek thicket]

yeah I mean as I said, if I think about it I can rederive which is which

But SO(3) and SU(2) are very similar symbols

so my monkey brain mixes them up

[flint cove]

Anyway, my intuition for that would be that ℝP² is double-covered by S², which is simply connected. Now this fact almost transports over to ℝP² – we just introduce a little torsion in the form of π₁(ℝP²)=H₁(ℝP²) = ℤ/2ℤ. But the betti number doesn't see that, so it's not a „one-dimensional hole“.

[pale sky]

is there anyway to translate that so a monkey like me can understand? xD

[flint cove]

For that I think I'd have to understand the matter better, lol

But also like what's your background, are you comfortable with notions like „simply connected“?

[pale sky]

no xD im not sorry

does that mean if theres holes?

[native raptor]

if you can draw a loop in the space and continuously deform it so that it becomes a single point, it's called simply connected

so like if you have a surface, it would be something without holes

or something with genus 0 like a sphere

so a disc, a sphere

whereas an annulus or a torus are not simply connected

[flint cove]

So now the set of homotopy classes of loops in X (given a chosen basepoint, let's ignore that atm) is called the fundamental group π₁(X)
Group law is given by [γ] [δ] ↦ the loop we get from traversing first γ then δ

This is one of the many constructs that allows you to talk about 1D holes. Other ones are the first homology group H₁(X) and the first Betti number b₁(X)

H1 is, although constructed entirely differently, actually the abelianization of π1, i.e. the group you get when you add the equation „gh=1“ to π1.
So in some sense π1 is a bit more general, but that difference doesn't matter in our case, since ℤ/2ℤ is abelian.

[sleek thicket]

@pale sky how have you defined the Betti numbers?

I think lux is thinking of Betti numbers in terms of the homology groups

(correct me if I'm wrong lux)

But since you don't seem to be super comfortable with the term simply connected I think you might be thinking of these things in terms of like, the simplicial complex structure

if not, sorry for assuming

[flint cove]

I'm largely making stuff up as I go, my algtop knowledge is all over the place. But yeah, Betti numbers are the free rank of the homology groups for me.

[pale sky]

yeah thats how betti numbers were defined in my course as well

lux's definition

i was wondering about the intuitive understanding of betti numbers, i believe for betti 1, this is the number of non trivial loops, but im not sure if this is the case for non orientable spaces like the projective plane

[pale sky]

we have learned it rigorously in this way but i want to develop then intuition too

[sleek thicket]

Yeah so number of loops is mostly okay but it gets weird in the presence of "torsion"

In the case of RP^2 we have a nontrivial loop which if you do it twice is trivial

The Betti numbers sort of don't see elements like this

(but the homology groups do)

[flint cove]

Yep, so essentially the betti numbers only measure such types of loops that cannot be „undone“ by just walking along them a few more times. Which is not a phenomenon that would make sense with our visual interpretation of a „hole“.

[sleek thicket]

the thing that's weird here is that finite order elements in an abelian group are linearly dependent just on their own

So the rank can't detect them

thank u for the endorsement gristle

[quartz edge]

this has been a certified gristle moment

learning about parallel transport rn

[gritty widget]

parallel transport

[sleek thicket]

parallel transport is when you pick up a lil vector and push it along a curve

[bitter yoke]

im doing the exercise in do carmo that says that parallel transport not depending on curve implies that the manifold is flat

[sacred gust]

[quartz edge]

Does anybody know of a simple example of a smooth surface with principal curvatures k1 > k2 > 0 with a point P at which k1 is minimized and k2 is maximized?

I had to construct one of these this morning and it was really nasty to build up. Nobody else seemed to be able to come up with any simple example.

[sleek thicket]

sorry so by k1 > k2 do you mean you want this to happen everywhere?

Like pointwise I mean

this seems annoying so I want to make sure I've got the question right first :P

[quartz edge]

Nope, just at the origin of some nice parametrized chart

[sleek thicket]

Wait so just at some point?

[quartz edge]

Yep.

[sleek thicket]

the one which minmaxes?

[quartz edge]

Yes

I tried constructing this extruded catenary (which clearly has k1 > k2 = 0 at the "trough" of catenaries), "pinching" the extruded surface while leaving the catenary arclengths unaltered (getting k2 maximized at the pinched catenary), then holding the "trough" of catenaries constant while squeezing the extrusion ends together (varying arclengths of the catenaries)

for sufficiently short extrusions (given a particular "pinch" magnitude) this makes k1 a local minimum

hard to explain

[sleek thicket]

I'll think about it

[quartz edge]

it does work but it's not simple by any means

[sleek thicket]

Unfortunately I am bad at geometry 😔

[quartz edge]

me 2

this was my shitty illustration of the deformations

first is the extruded catenary

second is after pinching

after that you just squeeze the ends together to get a diamond shape while holding the red line at k2 in the second image constant

with catenaries hanging from the perimeter

[past brook]

hi, im working through munkres' introduction to topology, and im baffled by this sentence: why do we need X to be compact for the sup metric on C(X, Y) to be defined?

[gritty widget]

because the metric rho(f, g) can be infinity if X is not compact

[past brook]

sorry im not really seeing why

oh nevermind i figured it out

[brittle widget]

Hello, for the red part, isn't $\overline Z_0\subsetneq\ldots\subsetneq\overline Z_n$ only locally maximal so idk. For the blue part, isn't it $n+1=\dim A\left(\overline Y\right)$?

[gentle osprey]

@brittle widget

[brittle widget]

This is from Hartshorne chapter I

[bitter yoke]

the dim of k is 0 since the transcendence degree is 0 over k, so the dim of A(\overline{Y}) is only the height of m which is n

@brittle widget idk if this answers your question, and idk what you're asking with the first part

[brittle widget]

Ah yes, we are considering Krull dimension right, not topological dimension

[bitter yoke]

yeah

[brittle widget]

For the first part I don’t get how the correspondence implies $\operatorname{height}\mathfrak m=n$

[gentle osprey]

@brittle widget

[brittle widget]

I only saw how $\overline Z_0\subsetneq\ldots\subsetneq\overline Z_n$ cannot be extended to a longer chain

[gentle osprey]

@brittle widget

[bitter yoke]

since each of the \overline{Z_i} are closed and irreducible, you know that I(Z_i) are prime ideals and this gives you a chain of length n contained in m @brittle widget

[tough imp]

Epicgay are you using Hartshorne to learn varieties?

Unless it’s a course text only use it for exercises, it has such a shit treatment of varieties

[brittle widget]

Yeah I heard, but I thought learning AG with a single book would be more convenient

[long gorge]

read shafarevich to start

you can't really learn about varieties in that few pages and you can go decently far without seeing all of that algebra right at the beginning

[brittle widget]

I’ll have a look. About my question it suffices to show if $I$ is irreducible in $\bar Y$ then $Y\cap I$ is irreducible in $Y$.

[gentle osprey]

@brittle widget

[brittle widget]

@bitter yoke but how do you know that that chain is maximal over all prime ideal chains?

[brittle widget]

Ah maybe not, die texit

[tough imp]

why do all your messages have strikethrough lol

[brittle widget]

so people look at TeXit's output not my code

[sleek thicket]

you can have texit delete your orignal post

[gentle osprey]

Shamrock (chazzed up)

[brittle widget]

it doesn't work after 5 mins

[sleek thicket]

oh

[brittle widget]

oh, I like editing my original post

[sharp yoke]

@bitter yoke you came back

[obsidian socket]

Hi does anyone who first proved that a general quartic has 28 bitangents? I've searched for a while and think it might be Plucker, but am not sure. Thanks!

[native raptor]

hey this is a really stupid question, but for hatcher 2.3.2, what is $\prod_i \tilde{H}_i(X)$ if not $\bigoplus_i \tilde{H}_i(X)$?

[gentle osprey]

seastaralgebras

[wanton marsh]

is the indexing set finite or infinite ?

[gritty widget]

i believe hatcher does mention that it is 0 if X is a finite-dimensional complex

[marsh forge]

the indexing is N

So the two do not need to agree in general

and you have to think about what universal properties you want

at a guess, Hatcher wants to talk about projections or the universal property related to them

(although for most spaces we care about we have homology bounded above in which case the two agree even though the indexing is infinite)

[native raptor]

wait i thought the indexing was finite, it doesn’t actually specify in the problem

[marsh forge]

I would very much assume that he is taking the product of all the homology groups

if it is written exactly as you have it

[native raptor]

ahhhh okay that makes much more sense

[marsh forge]

this is very common notation for topologists

Because most of our invariants have obvious indexing

so if you omit the indexing

you mean all of them

[native raptor]

i just took a break from a lot of riemannian geometry so i got too used to finite indexing

thank you!

[frigid river]

Can someone give me a hint with the this? Prove that every Cauchy Sequence has a subsequence that is a Fast Cauchy Sequence.

[gritty widget]

If I recall the definition of fast cauchy sequence correctly, it should be trivial

[marsh forge]

Here's a slightly strong hint, that I will put in a spoiler

||if you want to find a subsequence that converges faster, the naive idea is to just pick a term in the sequence, and then wait until you see another term that satisfies the "fastness" bound. Your goal is to prove this naive approach works||

[marsh forge]

this is unhelpful and pointless

[gritty widget]

Well, it distinguishes it from a problem that involves eg using heavy theorems

I generally prefer to give fewer hints rather than many. (I will admit that your hint is probably more helpful than mine, and is not too "giving it away")

But eg slims is too helpful

Well, it is up to vhp, how strong of a hint he wants 🙂

[frigid river]

Honestly, I am having such a hard time learning Topology that I really don't care if you give me too much of a hint xD Thanks, guys!

[sharp yoke]

does anybody use Massey for algebraic topology lmao

there isn't a single rigorous proof in the first chapter i skimmed through

in fact there aren't even any proof sketches

[marsh forge]

Yes

I used it in a class once

[sharp yoke]

he just vaguely describes everything

[marsh forge]

It was a good companion to Hatcher but it is not recommended often

I think you are wrong about a lack of proofs, you are probably reading an intro chapter

which many math books have

[sharp yoke]

perhaps

is this supposed to be a rigorous definition of connected sum

feels like just a vague description to me

[marsh forge]

thats perfectly rigorous

[sharp yoke]

is it really lmao

is this what all of algebraic topology is like

[marsh forge]

not really

theres nothing algebraic about that

[elder yew]

just fold and glue

[marsh forge]

so i wouldnt even really call it algebraic topology

I am not even sure what you think is lacking rigor in that description

its honestly more rigorous than I'd write it

[elder yew]

Low Dim'l Topology frequently has descriptions like this

[marsh forge]

well you can use the same description for a generalized connect sum in higher dimn

but yes

low dim topology is often visual where there is no risk of ambiguity (or, provably no ambiguity, as in this case)

[sharp yoke]

"it is clear that S1#S2 is a surface"

[marsh forge]

Yes

[elder yew]

What's the definition of a surface?

Go through the properties

And check they're all there

[marsh forge]

Omitting details is not the same thing as a lack of rigor

[sharp yoke]

a connected 2-manifold i think

[marsh forge]

if your defn of "rigorous" is spelling everything out

then every subject is "like this"

[sharp yoke]

ok im bad at math so i prefer to have everything spelled out for me

[elder yew]

This is how you get better

You ask yourself a simple question "Wait - this is not clear why this is a surface"

Then you start thinking

[marsh forge]

(in higher level textbooks, there are no exercises, and filling in details like this is how you make sure you understand what you are reading)

[elder yew]

When you read papers, they expect you to have these things down

Or be able to get up to speed fast

[sharp yoke]

im gonna try hatcher and see if i feel more comfortable

does hatcher have manifold stuff

[elder yew]

Hatcher doesn't really do things all that rigorously in the way you're expecting it

Are you trying to learn manifolds? Alg. Top? Diff top?

What is your background?

[sharp yoke]

im taking algebraic topology

[marsh forge]

Hatcher has manifold stuff to answer your question but he doesn't focus on manifolds

(nor do most algebraic topologists)

he certainly doesn't have what most people in this channel would call manifold stuff

[sharp yoke]

does the Euler Poincare characteristic have something to do with algebra lol

[marsh forge]

e.g. metrics, smooth structures, etc

[sharp yoke]

why are manifolds in an algebraic topology book

[marsh forge]

$\chi(M)=\sum_i (-1)^i b_i$

[gentle osprey]

bored and salty

[elder yew]

because alg. topology concepts are very well illuminated by manifolds

[marsh forge]

where $b_i=Rank H^i(M,\matbb{R})$

[gentle osprey]

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

[marsh forge]

i might have gotten the sign wrong there i didnt google it

but the point is that euler char is determined by (co)homology

[sharp yoke]

yea i saw in decarmo that is was a topological invariant that lets you classify compact 2 surfaces

[marsh forge]

There are manifold-specific results in AT

its just not the central focus

[sharp yoke]

guess i'll see the cohomology stuff later this quarter

[marsh forge]

Yes dim 2 is odd because its compact orientable manifolds are completely determined by euler characteristic

its very rare to get such a nice classification even of a rather tame group of spaces

[sharp yoke]

i see

can't wait to finally see what tf a cohomology is lol

i've heard so much about them, hope im not disappointed

[marsh forge]

cohomology is a central tool in basically every algebraic-y field of mathematics

not just topology

[sharp yoke]

is there quick and dirty explanation of it

[marsh forge]

The classic is that rational cohomology counts holes in a space

More honestly, cohomology measures obstructions to important phenomena

and its hard to get less vague than that bc it is really general

[sharp yoke]

so ur saying it can be used to calculate topological invariants like genus and the euler characteristic basically

[marsh forge]

it is a topological invariant

that is far more important than either genus or euler characteristic

(both can be thought of as summaries of cohomological information, but cohomology has far more information than either in general)

So like, usually if you know cohomology of a space X you don't really care about its euler characteristic anymore

because euler char is just less useful

[strong heron]

In differential topology, the deRham cohomology (which turns out to be equivalent to the cohomology you'll learn about in your algebraic topology course) helps us glean global information about a manifold using some local information.

[cedar pebble]

Euler characteristic is still sometimes more useful than individual cohomology groups

That being said, Euler characteristic is like an alternating sum of (dimensions of) cohomology groups

So it’s a looser invariant, you can have a lot of cancellation in this alternating sum that is separated at the level of individual cohomology groups

[marsh forge]

i said this above haha

my point is if you ask whether I'd rather have euler char or cohomology

the answer is obviously cohomology

[marsh forge]

this seems false unless you mean literally just one cohomology group

[cedar pebble]

?

[marsh forge]

oh sorry i somehow quoted the wrong msg

[marsh forge]

this one

[cedar pebble]

Ah

yea so there are examples where the cohomology groups behave very irregularly but their alternating sums behave very regularly

[marsh forge]

Are you worried about like, cases where the alternating sum is somehow hard to compute

from the individual summands

[cedar pebble]

no generally the alternating sum is easier to compute once you know what it is, but the individual terms that end up cancelling in the end are a mess

[marsh forge]

i mean even if the sum is all you get from coho its certainly no less useful

and its often much more info

i just dont really get what you're saying about euler char ever being more useful

[cedar pebble]

I'm thinking of a few examples where we only know conjecturally what the individual cohomology groups are but we know what the Euler characteristics are just fine

[marsh forge]

oh sure

weaker invariants are often easier to compute

[cedar pebble]

and you can verify this conjecture in particular examples but it involves a lot of analysis

[marsh forge]

my point is just, the user I was responding to said

or guessed

[cedar pebble]

yea fair enough I'm just splitting hairs here

[marsh forge]

that the importance of cohomology was to compute things like euler

my point is if you already know cohomology

this is kinda pointless

[cedar pebble]

it depends on what you're trying to compute, in general cohomology is a finer invariant and it shows up all over the place

[marsh forge]

or, if not pointless, an afterthought

[cold vine]

I have shown that Mf is the pullback of f:X->Y and ev1:Y^I->Y, where ev1 is the evaluation function at the endpoint. And I got that the universal map from Z ->Mf is (a(z),b(z)) when a:Z->X and b:Z->Y^I. And now if I have the pointed nullhomotopy H I can make a map phi:Z->Y^I by z -> Hz that is the path of the homotopy at z. Now by the universal property (since this given diagram can be filled to commute by g and the map phi) g~ is the unique map taking z to (g(z), Hz). And now I'm a bit stuck as to how this would be in bijection with the homotopy. Any hints? I do think this would work.

Ah, I think by this construction I have shown that indeed a pointed nullhomotopy gives us a unique lift g~. Now I need to somehow show that such a lift gives rise to a nullhomotopy but I can't tell how to approach this. Could this have something to do with homotopy lifting property?

And furthermore since Hz at every z defines H completely I think g~(z) = (g(x),Hz) and H being in injective correspondence is pretty clear but surjectivity is unclear

So I think the only part I'm missing is that given such a lift I would need to have a nullhomotopy of fg, no?

[sweet wing]

what exactly is $\mathcal M$ here

Based on the descriptions it looks a lot like the pullback of $X\overset f\to Y\overset{ev1}\leftarrow FY$ where $FY$ is the subspace of $Y^I$ consisting of pointed maps seems to somewhat match your description

[gentle osprey]

a cute cat ٩(˃̶͈̀௰˂̶͈́)و

[sweet wing]

with this you can draw your pullback square. The commutativity condition tells us that given $g:Z\to X$ and $h:Z\to FY$, we require $fg=ev1\circ h$

[gentle osprey]

a cute cat ٩(˃̶͈̀௰˂̶͈́)و

[sweet wing]

you can get some info from this and using the some form of the tensor-hom/smash product/something adjuncstion

you should obtain that h is a nullhomotopy of fg

[sweet wing]

(for sufficiently nice spaces we have the smash product is a tensor product)

[quasi forum]

I need a bit of help. I'm not too sure how to prove the function in (ii) is well-defined 🤔

[sweet wing]

show that if [x1]=[x2] then p(x1,y)=p(x2,y)

by assumption you have p(x1,x2)=0

and throw into the integral

or jus throw into the axioms

works as well

cuz if you have a<=b and b<=a, then a=b

[quasi forum]

The integral was just an example, we don't care about that one

[sweet wing]

then have to use this

[quasi forum]

Yea, I figured, and I can see exactly where that is going since x_1 relates to x_2

[sweet wing]

yesh

[quasi forum]

Is that the only thing we need to consider for it being well defined?

[sweet wing]

also poke the y

and thats really it

[quasi forum]

Poke as in change?

Well, the arguments the same

[sweet wing]

cuz it shows that for any a,b in [x],[y] p(a,b) a constant, hence independent of the representative

yeee

in general if you see quotients to show map is well defined jus show it is constant on each "coset"

[quasi forum]

Out of curiosity, how would we go about it if we changed both x and y at once

[sweet wing]

it is equivalent to changing x first then changing y

equality is transitive

[quasi forum]

Actually, it's a direct consequence

Sweet. That was super easy!

And (iii) is even easier

[cold vine]

Yes correct, this Mf is the mapping fiber/homotopy fiber exactly as you described. Can you expand a bit on how to show that h is the nullhomotopy? And isn't it a problem we need to assume the existance of the map h since it wasn't given, or can I assume it?

[sweet wing]

it comes from the smash product adjuncation thingy if you've seen it before

if not it isnt too hard to proof

define $F^0(X,Y)$ as the subspace of $Y^X$ consisting of pointed maps and let the constant map be its basepoint

then we have for locally compact something

$$Hom(X\wedge Y,Z)\cong Hom(X,F^0(Y,Z))$$
where hom consists of pointed maps

this plays well with homotopy so can quotient by homotopy

[gentle osprey]

a cute cat ٩(˃̶͈̀௰˂̶͈́)و

[cold vine]

Hmm alright I might have seen that one, thanks! Do we get the map h straight from the diagram given? How do we know we can complete the xommutative square for Z

[sweet wing]

mm ye so here we let Y=I