point-set-topology

[marsh forge]

maybe this notation is newer than i thought

C2 normally does mean Z/2 but i see now this is a chain cpx

so this is fine

sorry i havent had coffee yet

[swift fjord]

isn't a chain complex just the free abelian groups generated by n-cells?

[pearl holly]

But like, what is the operation? What does it mean geometrically?

[marsh forge]

infinite cyclic group is isomorphic to Z

[swift fjord]

it's a formal sum, it doesn't really mean anything in itself

[marsh forge]

take Z

take 1

rename it to A

and ur done

so now 2 is A^2

and 3 is A^3

[swift fjord]

You can kind of thing about addition in the chain complex as going around those elements, but that kind of breaks down quickly

the analogy i mean

[marsh forge]

you can also think about it as 2A if you want

and 3A

just notation

[pearl holly]

yeah okay I see. I thought that "A + A" would actually mean something geometrically, like going around the boundary twice but that is not the case right?

[marsh forge]

uhhhhh

uhhhhhhhhh

no

[swift fjord]

it can be thought of like that when the space is nicely behaved

[marsh forge]

sometimes but in general no

[swift fjord]

but not generally

we take the free abelian group cuz we want to give formal structure to the set of n-cells

[marsh forge]

there is geometric meaning to be found in the chain complex but it is hard won

[pearl holly]

yeah okay I see. Thank you so much, both of you!

[swift fjord]

if you keep reading you'll understand the motivation behind this structure better

[marsh forge]

Toki have you done only Z coefficients so far?

if so talk to me when hatcher introduces general coefficients

I think the best way to think about (ordinary) (co)homology

is like

each coefficients group can show you different information about a space

and each has some weird aspect that isn't telling you anything geometric

[pearl holly]

Yes, I have only looked at Z coefficients. I can come back once Hatcher goes over the other stuff!

[swift fjord]

The more I thought about it the more I realised that

[marsh forge]

this is not quite true

For example if you take generators A,B,C then (up to signs) A+B+C should be though of as their union geometrically

I agree things like 2A don't have a ton of geometric info over Z

[plain raven]

you know what cracks me up

[obtuse meteor]

I mean this is what happens in RP^2 no?

[plain raven]

in the book by Gelfand and Manin they spend, like the first 6 sections talking about homology and then section I.7 is titled "Geometry of cohomology"
and it's just a single paragraph that says something like
"Regrettably, cochains are not geometric"

[swift fjord]

Yea, I talked about that later on I think

[obtuse meteor]

Meanwhile Twitter: Geometry is when there’s a contravariant functor

[novel acorn]

mentions RPn
me:

[plain raven]

everybody's got a hot take

[obtuse meteor]

RP^n is good actually?

It has a nice cell complex structure what more could you ask for

[novel acorn]

You missed a not in there

[plain raven]

i like uh, one of the programming language theorists who said that because schemes and domains are both pathological as topological spaces that schemes should be considered a space straight out of PL theory

idk who it was but they were like "specialization!! See? PL"

[obtuse meteor]

This sounds like a sarahz take

[plain raven]

oh yeah i'm looking it up and this is definitely who it was

[obtuse meteor]

This is always so strange and cool to me

[obtuse meteor]

Sarahz has some spicy takes. She’s cool though just extremely PL brained

[marsh forge]

I think up to homotopy ruins this

[plain raven]

i'd love some examples of this if you want to talk about it sometime. I really only have a sense for like, working over Z or Z/2 (because everything's orientable over Z/2 which makes some things easier)

[marsh forge]

I think the correct way to interpret this is maybe through Hurewicz but yes

[plain raven]

but like, working over Z/p feels too much like number theory to me. i wash my hands of it lol

[marsh forge]

like

Knowing which linear combos of chains are "geometric"

is i think nontrivial

and often post-hoc

[obtuse meteor]

Yeah

I should like think through a geometric interpretation of pi_1(RP^2) more

[marsh forge]

i have some very bad news for you about algebraic topology

[obtuse meteor]

I mean actually I know why

It’s just weird

Best way to do this is universal cover RP^2

[plain raven]

I think we should make more of an effort in teaching topology to stress what makes homology qualitatively different than homotopy

[obtuse meteor]

What is the like analogue of universal cover for thinking about homology

Like a universal cover classifies paths

[swift fjord]

What in your mind is the main difference

besides one being a necessarily abelian theory

[marsh forge]

best way to do this is via the fundamental polygon of RP2

[obtuse meteor]

What classifies simplices

I mean that doesn’t feel as geometric to me

But that’s just my brain

[marsh forge]

...

...................

[swift fjord]

and, IG you can have all different kinds of homology theories

[marsh forge]

..........................................................

[obtuse meteor]

Idk why

[marsh forge]

wrong

[obtuse meteor]

Like I get what you mean but

[marsh forge]

............................................................................................

[swift fjord]

max stahp

[obtuse meteor]

It’s just better for my intuition to see the paths between antipodal points on the sphere

[marsh forge]

faye i think u should work harder to embrace intuitions that do not immediatley vibe w you bc that take is jsut bad lol

[obtuse meteor]

I mean it’s the same picture thinking about it now

[marsh forge]

Anyway

Z and Z/2

are probably the most geometric coefficients

for homology

[obtuse meteor]

Max whenever we talk about math you criticize me

[marsh forge]

sometimes Z/p will see something they don't

[swift fjord]

by Z/2 do you mean Z/2Z?

[marsh forge]

its bc u have bad takes

stop making bad takes

[obtuse meteor]

You saying this all the time doesn’t help me get better takes lel

[marsh forge]

nevermind faye thats a worse take

[obtuse meteor]

Lel

[marsh forge]

i mean if ur actually asking my issue is like, this sentence would be better phrased as a question about why this is (in my opinion) the better geometric approachj

[novel acorn]

Whats cohomology then

[marsh forge]

bc sometimes u feel close minded to me about approaches you dont immediatley vibe w

[swift fjord]

just reverse the arrows bro

[marsh forge]

deRham cohomology is lame

[novel acorn]

Dual of the derivative

[marsh forge]

and bad

[obtuse meteor]

Why do you think so then

Often I feel like you just say “that’s a bad take” and then don’t explain it

[marsh forge]

I think that the best way to approach CW complexes and their geometry and its relation to algebra is basically always to think about generators and relations induces by the cellular constructions

[plain raven]

@swift fjord
Don't have a good answer. i would say that the historical origin gives a hint. Homology is more suitable for like, integration theory, you can integrate over simplices. there are operations you can do on simplices (adding them together and subdividing them) which make sense for the purpose of integration (you can integrate over the function over a sum of simplices by taking the sum of integrals over each simplex; the integral of a function over the subdivision of a simplex is the sum of the integrals over each simplex in the subdivision)
Subdivision is really useful on like a manifold because you can refine a simplex until each simplex lives in a coordinate chart. then you can apply the normal integration theory of R^n. A lot of this to me is the basic difference. Homology should be thought of as like, a variant of homotopy which doesn't necessarily think about deformation as a geometric concept so much as "a thing which integrals are invariant with respect to", it's kinda something a complex analyst might have come up with while thinking about the argument principle.

The other big thing to think about is poincare duality, which is why homology was invented.

Theoretically the "correct" answer to this question should lie in the excision axiom or equivalently the Mayer-Vietoris theorem. both of these are satisfied by homology but not by homotopy, and this is the main difference between them. but idk how to derive too much from that starting point

[marsh forge]

this is what the fundamental polygon sees

[obtuse meteor]

I do this for anything other than pi_1 or whenever I can’t see the universal cover immediately

[marsh forge]

Z/2 and Z/(2) are probably the most common notations for this

Z_2 is bad

and C_2 is just hipster

[obtuse meteor]

So I actually agree with you

:vidya:

[swift fjord]

I usually think of C_2 as multiplicative

and Z/2Z as additive

[marsh forge]

C_2 is used most often when we want to like

think about the generator of Z/2 as something other than 1

also C_2 de-emphasizes the ring structure

#ConceptWithAttitude

[swift fjord]

ig

I just write C_2 cuz it's faster

and I prefer multiplicative notation for general cyclic groups

[marsh forge]

Yeah I would only say the one time to def not use it

is when you care about the field/ring structure

in particular I would never write like

$H^*(X,C_2)$

[gentle osprey]

MaxJ #MiuArmy

[marsh forge]

because then the cup product gets weird

[frosty sundial]

its funny you used this example because this group has a unique generator

[marsh forge]

oh no i mean like

the same element up to iso

just renamed

[frosty sundial]

I mean, I think you have a good point in general

Z/3Z has two generators

[marsh forge]

Like if we think of $Z/2=<\gamma\mid \gamma^2>$

[gentle osprey]

MaxJ #MiuArmy

[frosty sundial]

and to me writing Z/3Z sort of means you've chosen a generator (which you aclled 1)

but writing C_3 means you havent necessarily chosen a generator

[swift fjord]

why would you use < > instead of langle rangle

that's a bad take

[marsh forge]

are u actually asking

why i use the one-keypress symbol

[swift fjord]

bruh

My langle rangle is shortened to \< \>

\<

[marsh forge]

$<test>$

[gentle osprey]

MaxJ #MiuArmy
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[novel acorn]

Type \left< \right> if you wanna piss even more people off

[marsh forge]

did not work

[swift fjord]

or even \ip if i'm feeling fancy

well yea cuz you gotta define it

in your preamble

[marsh forge]

Can I make a texit preamble

[novel acorn]

$\left< test \right>$

[swift fjord]

yea

[gentle osprey]

Irony Incarnate

[marsh forge]

oh

well anyway

dont care

$<>$ is fine

[gentle osprey]

MaxJ #MiuArmy

[swift fjord]

you can edit your preamble

You are evil

[novel acorn]

$\fund{no}$

[gentle osprey]

Irony Incarnate

[marsh forge]

Anyway @frosty sundial I was more getting at like, sometimes you want to think of this cyclic thing without thinking about its relation to modular arithmetic and thats the vibe i get from C_n

I agree is also doesn't have a chosen generator

[frosty sundial]

yeah that makes sense

[plain raven]

I wish the standard textbooks on algebraic topology would at least introduce some of these notions lol. Like in retrospect I wish I had known more about analysis on manifolds when I read hatcher. or that hatcher had included some basic facts about smooth manifolds.
Personally, to me, de Rham and Cech cohomology are the easiest to visualize/ most natural to explain in terms of some kind of concrete problem like solving differential equations on the surface of a manifold. Singular cohomology is hard for me to get a read on other than by analogy to de Rham cohomology, i.e. thinking of a cochain as a kind of formal integral operator (or a formal... form?)

[marsh forge]

I think this would be orthogonal to Hatcher's goals and a large digression in material

Not because of his choice in material

his writing could be better

[plain raven]

well it's a 600 page book of which like 300 pages is appendices lol

[marsh forge]

I don't see how that is a bad thing

There's also an unpublished extra chapter 5

[plain raven]

i'm not saying it's a bad thing i'm saying if you call something a large digression, you know, there's 300 pages of digressions already

[marsh forge]

there are entire textbooks on manifolds lol

Stokes Theorem is a good theorem

[plain raven]

you can defend hatcher if you want to i'm still never reading chapter 3 of his book

[marsh forge]

there are many good theorems not in hatcher

ok this is just strange now lol

[novel acorn]

Why not

[marsh forge]

Hatcher's cohomology chapter is quite good iirc

And covers most of the important material for alg topologists

i feel like you just might not like alg top

[plain raven]

alright, agree to disagree then

[marsh forge]

what do you think it is missing?

[plain raven]

I didn't say it was missing anything I'm just complaining that it's hard to read. And it might have been easier to read or understand if notions of cohomology were motivated more in terms of geometry of manifolds. I've probably started his section 3.3 on Poincare duality several times and I can't make heads nor tails of it.

[marsh forge]

Yeah I mean there is nothing wrong with liking geometry on manifolds but I would not say that Singular cohomology should be motivated this way

[plain raven]

Fuck you lmao don't say that shit to me. I've studied Spanier's book on topology for a good 18 months trying to master this shit don't fuck with me and tell me what I do and do'nt have intuition for. I've worked hard as shit at this

[marsh forge]

I'm not saying you aren't good at it

[plain raven]

Really don't fucking tell somebody what kind of mathematician they are or aren't

[marsh forge]

I'm saying its a perspective that you might just not like

[plain raven]

What you do is algebraic topology and the way I prefer to think of things using category theory is what then?

[marsh forge]

huh?

prefering category theory is completely orthogonal to the stuff you just said

[frosty sundial]

preferring category theory is literally max's point

[marsh forge]

Hatcher's approach is much closer to a categorical one that you might see in concise

than geometry on manifolds

which you would never see in concise

(outside of his obligatory chapter on PD)

I really don't mean to imply that you can't understand anything. It just seems like you keep saying that the singular perspective doesn't vibe with you

Why

I mean absolutely don't read May for your first reference imo

unless you really like may's style

in which case I don't see how you could say you prefer the geometric approach

I never implied there was!

I vocally hate dogmatic approaches to the "right" way to understand something

I mean hatcher doesn't try to do either

Hatcher tries to teach """categorical""" algebraic topology to people who know no category theory

All textbooks are functionally free

Concise is free

Concise is just like, not pedagogically great for a first pass

too hard

not enough exercises

BUT it's a fantastic book

[plain raven]

ok let me be more clear, i'll start over. What I said was

1. I find it hard to get geometric intuition for singular cohomology on its own. (However, I have fairly good categorical intuition for it and could derive most of its important properties in a satisfactory way from general nonsense about simplicial Abelian groups.)
2. In retrospect, I wish that I had first been exposed to a perspective on singular cohomology from the get-go in terms of forms on manifolds
3. I got mad because from "I don't have geometric intuition for singular cohomology" you said ** you don't like alg top ** which implies that like, if I don't understand and appreciate geometric intuition for singular cohomology I can't or shouldn't be an algebraic topologist. I also don't like chapter 3 of hatcher (because it's handwavy as shit and I can't follow his bullshit proofs.) which again, I suppose, disqualifies me from algebraic topology because I prefer the book by Spanier, which is not real algebraic topology

[marsh forge]

oh no you completely misinterpretted me

1. I think this is hard because it is very hard

2. there are many places to learn this, I think it has no place in hatchers book

3. I would never say this, because I think being comfortable with not having geometric intuition is the algebraic topology perspective

I feel like you were very uncharitable here

I never said anything about Spanier

[plain raven]

You literally said "You don't like alg top"

[marsh forge]

Spanier's a great book

Yes

You seem to dislike the singular perspective

[plain raven]

The parts of algebraic topology that I like aren't alg top then

What are you talking about Spanier is an encyclopedic reference on singular homology and cohomology

[marsh forge]

I didn't say it wasn't?

[plain raven]

Why would I say I like SPanier then if I didn't like or wasn't interested in the singular perspective

[marsh forge]

I've never made a single claim about Spanier other than saying its a good book

Because you seem to be frustrated with the lack of geometric intuition inherent in singular cohomology's general definition

that's all

like my whole point is that searching for a general geometric intuition for singular cohomology is a genuinely hard and often basically impossible task

[plain raven]

ok

[marsh forge]

I really did not intend to come of condescending with that

I fucking despise most of the geometric perspective

so I don't consider it an insult

to say someone doesn't like a certain perspective

[plain raven]

ok. i wasn't insulted by that, i was insulted by "You don't like alg top" which, idk how to interpret that other than the parts of algebraic topology I do like are not actually alg top for some reason

[marsh forge]

I mean you could come up with a much more charitable interpretation that is not so hostile

But i agree that was poorly phrased

what I meant was, algebraic topologists often don't stress the geometric perspective because in a large large number of contexts topologists work in, those geometric interpretations simply are not possible

[plain raven]

Well, you know, everybody's out here trying to be recognized for the work they put into things, and to say "You don't like this", which can easily be read as "maybe this isn't really the subject for you", you gotta be careful with that.

[marsh forge]

I spend like

no not at all

read the convo before throwing extra vitriol into it ultra

[tight agate]

what

lol

[marsh forge]

it was already not great

you aren't helping

Anyway I was saying, I spend a lot of time in this channel trying to help people learn alg top (that I know) and it is never my goal to keep someone out of the subject because I love it

And I would hope that I could get a more generous intepretation than as a ridiculous gatekeeper

[plain raven]

anyway i think of myself as a category theorist and i'm like, pretty comfortable with simplicial sets and manipulating things formally by appeal to categorical principles so I don't need the geometric approach either

[marsh forge]

I mean even the deRham perspective of geometry basically breaks the second you stop working over vector spaces

[plain raven]

but the way Hatcher does things doesn't really appeal to me either from a geometric point of view or a categorical point of view

[marsh forge]

Yeah I would say like

Hatcher is written with the assumption

that you don't know any category theory

and if you do

then there is probably a better book for sure

I think it would be bad if Hatcher didn't exist though

because not everyone wants to wade through Riehl before approaching alg top

[reef shore]

What's a better book assuming category theory?

For a first reading

[tight agate]

May

[marsh forge]

thats a good question that idk the answer to

May might be fine

[tight agate]

also this might be weird

[marsh forge]

May needs more exercises tho

[reef shore]

Ty

[tight agate]

but just reading a bunch of papers from 1950-80 is not a bad way to learn

[pearl holly]

moldi reading May?

[marsh forge]

I disagree lol

[reef shore]

I might

[tight agate]

idk that helped me learn

[marsh forge]

im not saying it wouldnt

i just feel like its inefficient to do it that way first

reference and textbooks exist to like

crystalize the important stuff

in a bunch of papers

and to brush all the wrong directions and changed defintions under the rug

[tight agate]

I did not do any alg top classes

[marsh forge]

Like I think for alg top specifically, a lot of the way things used to be done is seen as bad and old fashioned

[tight agate]

and jumped straight into a kan seminar

and filled in background when needed

it wasnt bad

[marsh forge]

Brofib i think you are very talented and that you very much could have learned that way

I don't think that makes it a good idea

lol

I don't think there is a good argument for not using a textbook to learn the basics of alg top

[novel acorn]

Same
I mean I'm still in HS so I couldn't take them anyways lmao

[marsh forge]

admittedly there is no good second textbook in alg top lol

[novel acorn]

Hatcher is my saviour

[tight agate]

I think I would not have learned it well if there was no seminar ig

[marsh forge]

Yeah that helps too

[tight agate]

as youre forced to learn it well so that you dont look stupid

[marsh forge]

I mean just like

knowing which papers

and what parts

are important

is hard enough

I think thats the main purpose of even writing a textbook

to distill, reflect, reformulate, and update stuff

Unless you are lurie

and by textbook

you mean

700 pages of novel mathematics

[novel acorn]

Is that a post grad level Algtop textbook?

[tight agate]

it's a book on simplicial stuff

[novel acorn]

Ah

Interesting
So that's where most of Algtop research is currently focused on?

[tight agate]

I also had friends helping me out so in hindsight this is not good advice

[marsh forge]

yeah haha

uh

i dont think most alg top research is focused on anything in particular

[tight agate]

not all of alg top no

[novel acorn]

I wonder if there are OCW lecture notes for algtop

[marsh forge]

I don't think most alg top papers work with the simplicial stuff super directly

and I killed it

(its not dead)

[novel acorn]

The Hague math court accepted to hear this case

[tight agate]

nobody are you an undergrad/grad student?

[trail tiger]

What do low-dimensional topologists do lmao

work on the surgery theory of singleton sets?

[fading vale]

Nobody is kill

[reef shore]

no

[bronze lake]

lots of heegaard splittings

[tight agate]

oh nice

postdoc/prof?

cool

what did you work on

during your phd

[trail tiger]

What R&D do you do now?

I presume you're not using your algtop skills in industry though?

Yeah I might not pursue academia past postdoc, there are some research projects I want to finish but past that...

God I hate teaching so much

I hope you're happy where you are

[tight agate]

teaching looks like fun

grading does not

[trail tiger]

Yeah I like giving workshops and talks. I hate hate hate grading

Did I tell you about the year I graded 1000 pages of PDEs proofs

[novel acorn]

Sounds like pain

[trail tiger]

I was dyin man

[tight agate]

my smooth manifolds TA might have graded more than that in a quarter

[novel acorn]

Hehe I'm going into physics so no 10 page proofs for me

[tight agate]

there were 10 HW sets

and each one was easily over 25-30 pages

(handwritten)

and there were 25 people in the class I think?

[bronze lake]

DiffEq in general is an affront to god

[tight agate]

I think he stopped reading the solutions towards the end of the quarter lmao

[trail tiger]

when i teach i will assign 1-2 problems a week at most

[tight agate]

the prof was the one assigning problems so the TA had no control over it

[bronze lake]

I've never had homework take up more than 8 pages, and even that was just because I had to manually do polynomial division BS

[paper wedge]

ew

[swift fjord]

Once you finish being a TA you're probably qualified to be a handwriting analyst

This tbh

Sadly tests still have to be handwritten

Mostly

[tight agate]

oral exams

you get to do that even for written exams

lol

[swift fjord]

I hate multiple choice tests

Oral exams are p lit tho

[swift fjord]

Once had an exam where it was multiple choice, but you HAD to explain/prove.your choice was the correct one

So it was just mean

Like you couldn't even guess

Might as well have been open ended

[bronze lake]

in my first class on group theory, I gave an "unusual" proof (in the words of my professor) that the image of a normal subgroup under an isomorphism is normal, and they made me defend it during the oral portion of the test

very scary for sophomore undergrad me

[abstract pagoda]

o?

[swift fjord]

Now i'm curious what the unusual proof was

If you rememver

[bronze lake]

If $f:G\to H$ is an isomorphism and $K$ is a normal subgrounp of $G$, then $K$ is the kernel of some homomorphism $g:G\to H$. You can use this to construct a homomorphism $h:H\to G$ with kernel $f(K)$

[swift fjord]

That's pretty cool tbh

[bronze lake]

That's the only definition of normal subgroup I can ever remember

[gentle osprey]

cgodfrey

[bronze lake]

there we go, h is H to G

[swift fjord]

Makes sense

Definitely unusual

[bronze lake]

I don't even remember what the other definition of a normal subgroup is lmao

[swift fjord]

Invariant under conjugation

[swift fjord]

That's the important one to make quotient groups well-defined

Neat

Although the answer follows pretty immediately from the other definition

[bronze lake]

I suppose so yeah

[abstract pagoda]

during a test?

looks wtf to me

[abstract pagoda]

thank you for the explanation

[swift fjord]

^

[obtuse meteor]

Perhaps a simpler way to do this same technique.

Let q: G -> G/K be the quotient map, then f(K) is the kernel of q compose f^{-1}.

@bronze lake note, this technique is how you prove a very similar fact in general Abelian categories

[pearl holly]

First

Second

Looking back at this example, I still don't kind of get it

Why is a-b+c and 2c a basis for the image of dell_2?

It feels like a-b+c and a-b-c should be the basis instead. Where does the 2c come from?

[obtuse meteor]

Those are both basis-s

There’s a Z-linear change of basis map between them

Hint: if x, y is a basis then x, x - y is a basis

[pearl holly]

Oh so I can use a-b+c and a-b-c as basis? Would I get the same homology group then?

[obtuse meteor]

Yes, but you’d probably want to like hrm

Some bases are easier to work with

The 2c one is more natural for this application

[pearl holly]

yeah okay I see. But what is a Z-linear change of basis?

[obtuse meteor]

It’s just a abelian group isomorphism sending certain generators to other generators

The way I phrased it is natural when you realize abelian groups == Z-modules

So if you write down a matrix which has coefficients in Z whose inverse has coefficients in Z (for example if the determinant is 1), then you have a Z-linear transformation

[pearl holly]

Okay and this comes from the fact that x-(x-y) = y and so both x and y generate the same "set", right?

Sorry but I really want to make sure that I understand this

[obtuse meteor]

Yeah basically

They generate the same abelian group

The like matrix that represents this is [[1, 1], [0, -1]]

[pearl holly]

Yeah okay great! And I can also consider a different orientation without changing the homology group, right?

yeah okay because the group is always abelian

[obtuse meteor]

Ye

[pearl holly]

Okay great thank you so much!

[swift fjord]

I wanna say something about closed subgroups of the torus

But when I looked into it it seems to require a lot of lie group stuff

Is there any way to prove smth about, say, closed path connected subgroups of the torus with more elementary tools?

[cedar pebble]

idk that you can avoid much Lie theory here

like what's going on with the closed subgroup theorem in the case of a torus is you're really leveraging the fact that you generate such a subgroup by geodesic flow starting at the identity; this is exactly what the exponential map in Lie theory helps you do

[abstract pagoda]

Ok I have been stuck on this for a while

Ok thats bad. Im just desperate for a clarification since I have been on this for 6+ hours. This one detail has stumped me and I took many breaks.

Specifically the detail that is stumping me can be found in the second image. How does his second "move" doesnt change image of [f] in pi1(A_a) in the quotient group so? g:pi1(A_a)->*_alpha pi1(A_alpha)/N has g([f])=[f]?

Also how does showing equivalent factorizations under his definition imply q:Q->pi1(X) q([f])=q([g]) => [f]=[g]

My understanding is that elements of $Q$ are cosets $([f_1][f_2]\ldots[f_n])N$ where $N$ is the group generated by $i_{\alpha\beta}(w)i_{\beta\alpha}^{-1}(w)$ where $i_{\alpha\beta}:\pi_1(A_\alpha \cap A_\beta)\to \pi_1(A_\alpha)$ and $i_{\beta\alpha}^{-1}:\pi_1(A_\beta) \to \pi_1(A_\alpha \cap A_\beta)$.
So $i_{\alpha\beta}(w)i_{\beta\alpha}^{-1}(w)$ are equivalence classes of loops $[f][g]$ where $[f]\in \pi_1(A_\alpha)$ and $[g]\in \pi_1(A_\alpha \cap A_\beta)$

[gentle osprey]

nine9s

[abstract pagoda]

Also how is Q->pi_1(X) a homomorphism incuded by phi?

Is it because if you have a homomorphism f:G->H and a normal group N then F:G/N->H is also a homomorphism given by F(gN)=f(g)

Any help would be appreciated in the future. Just in case ill ping these great people. @pearl holly @swift fjord

[bronze lake]

Recall that the kernel of $\Phi$ is generated by elements of the form $i_{\alpha\beta}(\omega)i_{\beta\alpha}(\omega)^{-1}$ for $\omega\in\pi_1(A_\alpha\cap A_\beta)$. If we have a $[f_t]\in \pi_1(A_\alpha)$ and $f_t$ is a loop in $A_\alpha\cap A_\beta$, then $[f_t]$ is such an $\omega$ for the kernel of $\Phi$. So this second move that you talk about can be thought of as multiplying $[f_t]$ by something in the kernel of $\Phi$, and so it doesn't change the image in the quotient group. At least that's my understanding of it

[gentle osprey]

cgodfrey

[abstract pagoda]

Doesnt the second move imply that $i_{\alpha\beta}(\omega)i_{\beta\alpha}^{-1}(\omega)$ turn $[f_t] \in \pi_1(A_\alpha)$ into $[f_t] \in \pi_1(A_\beta)$ by the rule. So we take$i_{\alpha\beta}([f_t])i_{\beta\alpha}^{-1}([f_t])$ which is a multiplication of an elements with different domains. One in $A_\alpha\cap A_\beta$ and another in $A_\alpha$.
Also I know he defines the kernel to be generated by elements of the form $i_{\alpha\beta}(\omega)i_{\beta\alpha}^{-1}(\omega)$ but it isnt intuitive for me.

[gentle osprey]

nine9s

[abstract pagoda]

Thank you so much for the reply

I think I see the second move as convenince for defining i_\beta\alpha^-1(w) so that its in the correct domain

[bronze lake]

You're right that it's a multiplication of elements in different domains. But remember, this is regarding the factorization of $[f]$ as a word in $*\alpha \pi_1(A\alpha)$. In that sense, it's fine to do that

[gentle osprey]

cgodfrey

[abstract pagoda]

But I dont see how it doesnt change image of element in quotient map

oh snap

[bronze lake]

As for the quotient part, is it that you don't understand where those generators come from, or how those generators apply to the 2nd move?

[abstract pagoda]

I dont see where the generators come from. And I think I know what the generators do to elements in A_\alpha turning them into A_\beta so that it can be applied to the i_\beta\alpha^-1

I had some ideas. At first I was thinking of a regular old venn diagram. If you have a loop in A then you have a homomorphism that brings it to a loop in A \cap B then you multiply by a loop in A that should keep you with a loop in A

Because from the way I understand it the generators multiply two loops. One loop brought from B into A cap B. And another loop in A cap B into A

So So in essence by multiply these two loops you end up with a product of loops in A

But I dont exactly see how this corresponds to the kernel of phi

I dont want to rush my understanding at any time and this is probably going to take longer to fully understand this proof than it normally takes since im trying to force myself to understand Hatcher's proof

[bronze lake]

I think it might be helpful to try to come up with a situation where you would end up with a nontrivial kernel

[abstract pagoda]

What if I take the classic example of S^1 V S^1

[bronze lake]

sure

Actually, I'm not sure if that would have a nontrivial kernel

Like Hatcher wrote, "most" of the time the map Phi is surjective. And this happens when the A_alpha play nice with each other -- a loop in X can be written as a product of loops in the base space. But this can break for more pathological examples

and if you can find a situation where this breaks and Phi isn't surjective, that should (hopefully) help you see where these generators on the kernel come from

[abstract pagoda]

phi is surjective whenever you cant make a path in X into a product of paths of A_\alpha no?

[bronze lake]

that's right

[abstract pagoda]

This happens in two cases im thinking. One when you dont have the necessary path connectedness

And another weird one if the homotopy classes act strangely

[bronze lake]

I haven't thought about finding an example before, so idk how easy of a task it would be, but I think that'd be the best way to understand the kernel of phi

[abstract pagoda]

So like one loop may be written as a product of loops. But not every loop in its homotopy class

[bronze lake]

oop, to be specific, we're talking about loops, not paths

since this is for the fundamental group

[abstract pagoda]

ye mb i meant loops

wait

oh i wrote path xd

I dont think the case of not every loop in a homotopy class can be written as a product of loops works as an argument though.

Because every loop that is self contained in its own A_\alpha can be if the connectedness properties hold true. So forget I say that part

Anyways the kernel of phi are supposed to be the product of loops in each A_\alpha that when mapped to one loop in X, they become homotopic to the constant loop

So this is where my confusion starts. How does $i_{\alpha\beta}(\omega)i_{\beta\alpha}^{-1}(\omega)$ correspond to this

[gentle osprey]

nine9s

[bronze lake]

Gimme a second, I'm trying to find a document that might be helpful

Okay, so $i_{\alpha\beta}(\omega)$ is $\pi_1(A_\alpha\cap A_\beta)\to A_\alpha$, and similarly, $i_{\beta\alpha}(\omega)^{-1}$ is $\pi_1(A_\alpha\cap A_\beta)\to A_\beta$, and we can consider their product as an element of the free group $*\alpha \pi_1(A\alpha)$. If we have a loop in the intersection of two subsets, then the product of that loop in $A_\alpha$ and that loop in $A_\beta$ is homotopic to the constant loop in $X$

[gentle osprey]

cgodfrey

[bronze lake]

that's the idea for the generators of the kernel

[abstract pagoda]

Me thinks I have it

[bronze lake]

okay I think I have a good example

[abstract pagoda]

im just wondering why the inverse of i_ba

ok

woo

[bronze lake]

imagine you have the sort of wedge sum of three circles like this

[abstract pagoda]

wait

isnt wedge sum with one fixed point?

like a rose

this is linked circles

[bronze lake]

yeah, that's why I said "sort of" wedge sum

idk what else to call it

[abstract pagoda]

o ok

[bronze lake]

the point is, each pair touches at a single point

so if your A_1 is the left two circles, and your A_2 is the right two circles

pi_1(A_1) is the free group on two generators, and pi_1(A_2) is the free group on two generators as well

[abstract pagoda]

ye

[bronze lake]

so if you naively take the free product on this for pi_1(X), you would get the free group of four generators

and this is because we have nontrivial loops in the intersection

which is the middle circle

the inverse thing is just notational stuff

[abstract pagoda]

wait?

shouldnt nontrivial loops in intersection mean pi_1(X) wouldnt be isomorphic to free group of four generators since pi_1(X) isomorphic to pi_1(A_1)*pi_1(A_2)/N where N is nontirival?

[bronze lake]

exactly

this is a case where we have a nontrivial kernel

because the free product alone doesn't give the right fundamental group

[abstract pagoda]

ok

[bronze lake]

and regarding this

[abstract pagoda]

and if we apply our generators of N

so does it need to be inverse lol?

[bronze lake]

the i_ab(w)*i_ba(w)^-1 is also written as i_ab(w)*i_ba(w)^-1=1

so really it's the relation i_ab(w)=i_ba(w)

[abstract pagoda]

oh

but why does it equal 1?

is it that earlier relation with the inclusions i glossed over?

$j_\alpha i_{\alpha\beta}=j_\beta i_{\beta\alpha}$

[bronze lake]

that's sort of just what it means when you talk about modding by something

[gentle osprey]

nine9s

[bronze lake]

it's nothing specific to this

[abstract pagoda]

oh ok

i thought modding meant making set of distinct cosets into a group

[bronze lake]

or if you're being technical, kernel of Phi contains elements of that form (without the equals sign)

[abstract pagoda]

these are compositions though?

oh wait

I see what ur saying

j_b can be abstracted away since i_ba takes it to the intersection

[bronze lake]

but when you actually mod out by the kernel, you could write $\pi_1(X)=\pi_1(1_1)*{i{\alpha\beta}(\omega)=i_{\beta\alpha}(\omega)}\pi_1(A_2)$

[gentle osprey]

cgodfrey

[bronze lake]

where that *_... notation is the coproduct

you could write it either way

[abstract pagoda]

idk what coproduct is

[bronze lake]

it's the same as what we're talking about, the images of elements in the intersection being identified in each base space

[abstract pagoda]

oh

[bronze lake]

so going back to the three circles, if the generators for pi_1(A_1) are a and b, and the generators for pi_1(A_2) are c and d, what we end up with is the free group on a, b, c, d where we have the relation that b*c^-1=1

[abstract pagoda]

so like A intersect B -> A

[bronze lake]

yeah

[abstract pagoda]

and A intersect B -> B

looks scary

[bronze lake]

so does what I said about bc^{-1}=1 make sense?

[abstract pagoda]

Yea I get it

[bronze lake]

awesome

in that case, I gotta head to bed

[abstract pagoda]

because c^-1 corresponds to middle circle

rip

[bronze lake]

exactly

[abstract pagoda]

ok

ill try to come with my own explanation for my next question and share

[bronze lake]

hopefully someone else will be able to fill in any remaining gaps for the proof

[abstract pagoda]

its 3:16 for me

good night

[bronze lake]

same

night

[abstract pagoda]

I guess my next question would be, Why can't doesnt showing factorizations equivalent imply phi is injective. Since factorizations are products of paths

oh wait

ill come back to this tommorow

[swift fjord]

Because you can have multiple factorisations that are not equal, so they are different elements in the free product

But they're the same element in the quotient

If all factorisations are indeed equivalent

[swift fjord]

Oof rip. Guess that means i'm going in the wrong direction tho so that's something

[raw sedge]

I'm wondering how one would define the way a topological space looks like locally near a point

so like, given a point x in a space X, and a point y in a space Y

maybe we can say that if there's a neighborhood of x homeomorphic to a neighborhood of y, they are equivalent

but I think that might be too weak

[raw sedge]

with x and y preserved of course, so the homeomorphism from the neighborhood of x to the neighborhood of y would have to take x to y

I mean, I definitely want it to hold in this direction, but I think I might want to allow more equivalence

I ultimately want to turn the "locality" near a point into its own object, and I want continuous maps X -> Y taking x to y to induce morphisms from the locality of x to the locality of y

I'm thinking of doing something analogous to how you can "turn" vector spaces into affine spaces by weakening the maps to allow translations

I could turn pointed topological spaces into localities by only requiring the the maps be continuous "near" the basepoint

I'm not entirely sure about that though

[reef shore]

Remove all open sets except the ones containing x This would also have a nice universal property

[wanton marsh]

that sounds like infinitesimal neighbourhoods and germs of functions

[flint cove]

Couldn't we define a cat with objects pointed topological spaces and morphisms (X, x) to (Y, y) equivalence classes of continuous maps defined on an open neighborhood of x only (equivalent whenever they locally agree)? Isos would be something like local isos there, right?

[reef shore]

Yeah that'd be exactly the same as what Intel suggested I think

[abstract pagoda]

So the definition of the normal group how they make it means that two factorizations are equal if i_ab=i_ba

but then you can write that the group generated by i_abi_ba^-1

[swift fjord]

Well not exactly

2 factorisations are equal in Q if you can go from one to the other by the 2 moves. The fact that we're looking at quotient by N exactly means that [i_ab(w)]N=[i_ba(w)]N

In N

[abstract pagoda]

er

oh ok

[swift fjord]

This pretty much just means that regarding a loop in A_a as a loop in A_b when it's in the intersection doesn't change the factorisation

[abstract pagoda]

wouldnt it be [f]N=[g]N where f and g are factorizations and the factorizations are equal if you follow those two conditions which translate to reducing the word and taking letters of A_a as A_b when they are in the intersection

[swift fjord]

It would be a product of loops that corresponds to a factorisation yes

That stems from what I said

[abstract pagoda]

and w are the loops in the intersection

[swift fjord]

Yes

Factoring out by a normal subgroup generated by some set of words is basically adding relations to the group (You are forcing some words to become 1, or equivalently forcing some words to be equal to others in the quotient)

[abstract pagoda]

oh ok

i guess that was my main confusion

but i have one more question

‘The second move does not change the image of this element in the quotient group Q=∗απ1(Aα)/N, by the definition of N.‘ Does this mean that taking the factorization that get moved by the second move to Q doesnt change the image of *_api1(A_a)->Q?

[swift fjord]

Yea

[abstract pagoda]

oh wait I think I know what it also means

[swift fjord]

By image he means image under the quotient map

[abstract pagoda]

and the purpose of that move is to?

[swift fjord]

So basically when you identify i_ab(w) woth i_ba(w) (As you do in Q) the move doesn't actually change anythint

I don't know, I haven't read the rest of the proof

I assume it's useful for some algorithm to show that factorisation is unique

[abstract pagoda]

This informal stuff

That is a drag to visualize

I got through half of it

I got lost on the verticies part

[swift fjord]

Imma be honest with u chief

I've had enough of hatcher's bullshit

I don't feel like reading that

[abstract pagoda]

Im starting to punch air with this dude

I remember going through van kampen last year with a pushout diagram but I learned about it in lecture and we weren’t being tested on it

Im going to stick with it for one more week

If I absolutely hate the covering spaces chapter or the exercises im done

[novel acorn]

The van kampen proof is kinda doo doo

[abstract pagoda]

his*

i know it doesnt need to be this convoluted

[gritty widget]

that's too many words

[novel acorn]

Yeah probably
But things are more chill after this

Plus all the problems are fun imo

At least they're chill until homology which I just started so well see how this goes

[swift fjord]

Hatcher in a nutshell

[abstract pagoda]

Lol what

He doesnt draw the hard part

He sats each vertical side can be perturbed so that eaxh point in IxI lies in at most three points

[novel acorn]

I think that mainly because most of the space taken up is examples
And some of them are sooooo long like damn
He comes up with the most complicated shit

[abstract pagoda]

does he mean to draw each box to the top

each box extends all the way from the bottom to the top is only what I can assume

[novel acorn]

I'd have to read the proof again lol gimme a sec

[abstract pagoda]

because it should be at most s-cells if thats the case

because his drawing s has 3 cells but it could be more

[novel acorn]

Look at the drawing again

He uses shitty language

But he basically means leave the top and bottom alone and move the left and right sides by a bit to make it so each point lies in a maximum of 3 Rij

[bronze lake]

I'd also recommend Munkre's proof of van Kampen in his topology book

[novel acorn]

And he basically does this to make the next part easier

The proof is a mess in general and you can find better ones

[abstract pagoda]

So you leave bottom rectangles alone and you stretch the middle row rectangles height to the maximum?

But how does that mean each point lies in three A_ij?

[novel acorn]

No so
Leave the top and bottom rows alone

Now

Look only at the middle rows

In the drawing there's only one

[abstract pagoda]

ya

[novel acorn]

And move the left and right sides of these rectangles by a bit to either make them longer or shorter in the horizontal direction

(in his language he says that the left and right sides of Rij are the vertical sides because they're vertical lines but that's his inability to describe things)

[abstract pagoda]

Ok but how does this mean each point lies in atmost three Ai? Does this mean that corners only contain either two bottom boxes and one middle box or two top boxes and one middle box?

Is it like wood plank floor tiling?

[novel acorn]

At most three Rij's*

[abstract pagoda]

lol

So are those at most three points like th3 points where all three of the planks meet?

[novel acorn]

So what we did is that we made each corner look like this

[abstract pagoda]

oh ok i got it

[novel acorn]

So if you zoom in far enough into any corner on that drawing this is what it'd look like

And as you can see each corner lies in only three Rij's

[abstract pagoda]

oh one more question

what is gamma _r then?

is it the bordering lines of each rectangle?

because idk how else u seperate the recrangles

[novel acorn]

Gimme a sec

This part is weird

[abstract pagoda]

pushing across the rectangle

hmm

i can only imagine that the path is the border

[novel acorn]

Yeah so

They're paths completely different from gamma that follow the borders

[abstract pagoda]

bruh wtf

so each gamma r is different

[novel acorn]

Hmmmmmm

[abstract pagoda]

and say we look at our picture

[novel acorn]

He left out like 2 steps here gimme a sec

[abstract pagoda]

gamma 1 would start from left side corner, move up and across the boundary of rectangle 1, down back to the bottom line, and across to the finish?

but wtf

that doesnt sound close to correct

oh wait maybe

and gamma 2 would start from bottom left, move all the way to right corner of 2nd rectangle, and then go down to bottom line

but idk for gamma 4

oh maybe thats the pattern?

[novel acorn]

I think that the point is that they're all homotopic to each other so it doesn't matter which way it goes as long as it separates that rectangles

[pearl holly]

Honestly just watch this lecture https://youtu.be/0msWLft-lT0

I watched it instead of reading that whole proof

[novel acorn]

Lol yeah
I was trying to find better proofs but it's annoying that I can't figure out what he does

[abstract pagoda]

So these paths are examples of gamma r im thinking

Please come up picture

[pearl holly]

Yeah same here, I didn’t really get Hatcher proof so I just watched the lecture and boom

[abstract pagoda]

So red path is gamma 4

gamma 1 is bluee

gamma 3 is 4

[abstract pagoda]

ok ig its the lecture time

[pearl holly]

I wonder how Rotman and dieck proved it

[novel acorn]

Proof is at 33:00

[abstract pagoda]

i swear if hatcher pulls this shit two more times before i get to and finish covering spaces im permadeleting the pdf

[novel acorn]

It's kind of annoying tho since a lot of his other proofs are a lot nicer

[novel acorn]

What Hatcher does to a person

[abstract pagoda]

really tho

math stack exchsnge kept saying how hatcher is great

or that its best intro book

[pearl holly]

Yeah that is true. So far I only had problems with this proof and the proof that pi1 of S1 is Z. The rest is pretty fine

[abstract pagoda]

i swear this is a lie

yea i got the rest of his proofs

[novel acorn]

No
He can be a bit annoying since he leaves out details like it's a scavenger hunt but he will teach you a lot

[abstract pagoda]

just those two got me rut

[novel acorn]

The first proof is iffy but I like how he has a call back to it near the end of the deck transformations part

[pearl holly]

Yeah exactly that is nice

I wonder how other books look like. ShiN and John are reading different books

ShiN is doing Rotman and John is doing Dieck and ShiN hates Hatcher so I wonder how Rotman is

[abstract pagoda]

hopefully before the 21st I can finish homology 2.1 . After that im moving to concise

[novel acorn]

Author?

[abstract pagoda]

may?

[pearl holly]

Isn’t cat theory a prereq for May?

[abstract pagoda]

https://www.math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf

[pearl holly]

Idk

[abstract pagoda]

he introduces it a little

[novel acorn]

Could be I dunno

[novel acorn]

Imagine being time constrained to read Hatcher

[abstract pagoda]

bruh

lemme do the math

btw what grade ur in irony?

[empty grove]

May defines categories in ch2 (and the basic notions)

[novel acorn]

Going to 4th year of Hs in September

[empty grove]

Haven't read it yet though so idk how much detail

[pearl holly]

Ah okay I see

[empty grove]

But probably intended to remove it from prereqs

[pearl holly]

There was some discussion about it yesterday right?

[empty grove]

Ye

[novel acorn]

But like books on cat theory are generally circa 200 pages so they're relatively short

[empty grove]

Modern AT is all using categories anyway afaik

mac lane 300 pages

[pearl holly]

Yeah I really need to step up my cat game

[novel acorn]

I found even shorter ones

[empty grove]

I have read over 200 pages of mac lane now

do it toki

big language upgrade

[abstract pagoda]

💀

[pearl holly]

I’m kind of noticing it now. When I google stuff up about AT there’s always a fricking diagram

[abstract pagoda]

67 pages excluding appendeicies

[novel acorn]

Gotta love cat theory

[abstract pagoda]

can i read 67 and understand before the 21st

[pearl holly]

And Hatcher has no diagrams at all, until later on

[swift fjord]

That's also a big problem with Hatcher tbh

[novel acorn]

Coverings are good
And relatively easy to understand

[empty grove]

You don't really need cat thy to understand diagrams though

[swift fjord]

Diagrams are so helpful even early on

[empty grove]

They are just saying certain compositions are equal

[swift fjord]

And basic categorical notions make talking about things much easier

[pearl holly]

He defines a commutative diagram on page 100 something

[abstract pagoda]

ok

[novel acorn]

IT WOULD HAVE BEEN VERY NICE TO HAVE A PUSHOUT DIAGRAM FOR VAN KAMPEN

[abstract pagoda]

i got this 🙂

[swift fjord]

Tru

[empty grove]

but you already know what a commutative diagram is

[abstract pagoda]

5pages a day should get me there

[pearl holly]

Yeah I know I’m just saying it because why not

[empty grove]

you only need a proper formal definition when proving theorems about diagrams themselves

[swift fjord]

Like even just saying 'so and so is a functor' or 'so and so is an equivalence is such and such category' helps put things into order imo

[novel acorn]

I remember how daunting the statement was the first time I read it lmao

[empty grove]

]rn you only need to use diagrams as notation

I saw the definition of a Kan extension today

[novel acorn]

He says pretty early on that the fundamental group is a functor but doesn't go too much into it

[empty grove]

So insanely long lol

[abstract pagoda]

he does say that

[empty grove]

But builds on previous stuff

so can at least understand what's going on

broadly

[abstract pagoda]

bruh juggling hatcher with other books is a pain

hatcher has too many words

[novel acorn]

I gave up everything else for Hatcher ngl

[empty grove]

toki how about you switch to May and we read together

saketh might wanna read it too

[abstract pagoda]

o:

[plain raven]

here's a mind bender

[pearl holly]

But like eve on if I learn more cat theory, will it even help me with Hatcher? Like I wonder how much cat theory Hatcher actually talks about later on

[plain raven]

the word "diagram" is syntactic sugar for functor

[empty grove]

ono

[pearl holly]

Wait

[abstract pagoda]

?

[empty grove]

[pearl holly]

Really?

[swift fjord]

Yea a diagram is a kind of functor

[empty grove]

Clerk I said that in a discussion channel once

[plain raven]

like formally in category theory there is no such thing as a diagram. a diagram is just a functor from another category (the shape of the diagram)

[novel acorn]

Ur kidding
How

[abstract pagoda]

but how

to graphs?

[empty grove]

and it became a big argument

[plain raven]

like say you have a commutative square

in a category $C$

[gentle osprey]

diligentClerk

[empty grove]

yes

[novel acorn]

[abstract pagoda]

ok

[novel acorn]

Imagine going to a better book smh

[plain raven]

well then introduce another category $J$ which only has four objects, $a, b,c,d$ and morphisms between them forming a square

[gentle osprey]

diligentClerk

[abstract pagoda]

can u start a week from now 😎

[plain raven]

then the commutative square in $C$ is a functor from $J$ to $C$

[gentle osprey]

diligentClerk

[pearl holly]

But bruh you will probably start on like page 200 while I will be at like 100 so we won’t even read together because of the big gap

[swift fjord]

Toki

[empty grove]

not on pg 200 lol

its not that big a book

[swift fjord]

Move to rotman

[empty grove]

I will skim the early part yeah

[swift fjord]

And let's read together

[abstract pagoda]

thats cool

[pearl holly]

[novel acorn]

Lmao shills

Can't believe this

[empty grove]

sure I can read other stuff till then

[abstract pagoda]

actually

[empty grove]

irony is just too devoted to Hatcher

[swift fjord]

Btw maybe one of you guys can help me

This is the start of the proof of the homotopy axiom for singular homology. I just don't understand where the motivation for the naturality condition came from. Is this a special case of some categorical construction? It seems so out of nowhete and the way it's used in the proof is pretty whack.

[novel acorn]

I just reached homology

[abstract pagoda]

im filtered

[plain raven]

i saw that and i don't have a good explanation sorry

[empty grove]

wew good time to switch

[swift fjord]

Oh oof

[empty grove]

[swift fjord]

Well thanks anyways clerk

[pearl holly]

Hmm but the thing is that I won't be having time reading stuff in a couple of weeks so is there any point in actually switching now?

[swift fjord]

Moldi maybe?

[pearl holly]

in like 2 weeks actually

[swift fjord]

It just comes out of nowhere, the whole construction of the proof is so bizzare

[novel acorn]

👉 👈

[swift fjord]

Like it reduces to an entirely different problem

[novel acorn]

I dunno I'm pretty interested in cohomology does May cover that?

[empty grove]

sorry what

I didn't follow

[swift fjord]

Dis

Clerk said he doesn't have a good explanation

I thought maybe you could enlighten me

[empty grove]

yes, very late though it seems

ok lemme see

[pearl holly]

Rotman be proving axioms

[novel acorn]

[swift fjord]

He's probably building up to the eilenberg steenrod axioms

[empty grove]

[swift fjord]

So he's calling these theorems axioms

[novel acorn]

And I guess that he doesn't touch on homotopy theory
Although for that I can probably find a book

[pearl holly]

How much free time does a undergrad in the first year have?

approximately

[empty grove]

May does homology and cohomology

[swift fjord]

Depends

[empty grove]

After some general homological algebra

[swift fjord]

It can very a lot depending on the default course load and if you take any extra stuff

[novel acorn]

I asked cuz Hatcher has that at the very end so I wondered if other books do that

[abstract pagoda]

a lot

for me

i took one math class

[swift fjord]

Wtf

[empty grove]

May is much shorter than Hatcher

[pearl holly]

yeah okay but do you think that it's enough to actually read a whole book?

[empty grove]

Categories help a lot I suppose

[swift fjord]

Like I was pretty busy in my second semester but that's because I took 5 courses on top of being a TA

[empty grove]

Also May has a lot more stuff it seems

[swift fjord]

First semester was pretty chill but it was also an adjustment period

So felt more stressful than it was

[novel acorn]

Does he calculate the torus knot fundamental group thing tho

[pearl holly]

I hope not

[novel acorn]

That messed with my chromosomes I feel like

[empty grove]

Damn good point

[novel acorn]

There was also that cell complex X in covering spaces that was kinda fucky to imagine

[empty grove]

Is this Rotman?

[novel acorn]

In conclusion Hatcher has hard examples

[empty grove]

I'll just look at the book lol

[swift fjord]

Yes moldi

P76

[empty grove]

thx

[novel acorn]

OK well since my life is empty I'm back to Hatcher for the while
If you two start reading together (😳) ill join you

[paper wedge]

is algebraic topology just this visual and stuff 😦

[novel acorn]

Have you ever seen a hole in a hole in a hole

[paper wedge]

no

[novel acorn]

Well if you read Algtop you probably will

It's very visual most of the time

I just think that cohomology is where is stops being so but I dunno

[swift fjord]

Oooo turns out my library has rotman

I might borrow it, physical books are nice

[novel acorn]

Pog
I'll have to join my university library and start stealing borrowing from there

[empty grove]

I am really getting lost in the notation 😵‍💫 But generally the motivation for how to strengthen an inductive hypothesis usually just comes from trying the induction and seeing what extra hypothesis are required and just adding them and trying to prove them parallely. For the exact intuition for what is happening it might actually be helpful to look at Hatcher Hatcher constructs the prism operators (the P_n) visually and the square will probably be obvious from it (though Hatcher doesn't do induction)

[swift fjord]

Oh yea the notation is kinda rough since it builds up from the last 2 lemmas

[empty grove]

Too sleepy to decipher what the proof is doing rn

[abstract pagoda]

my last sem of 2nd year I took 5 courses, including analysis topology algebra and graded

[empty grove]

Ye well it is mostly standard I think

[abstract pagoda]

but then i got fired from grading

because i realized i can get paid doing no work

[empty grove]

It is just that I haven't looked at the notation in a while

[swift fjord]

Lmao

[empty grove]

so will need to reconstruct from context

😵‍💫

[swift fjord]

It's just that this naturality conditions seems to come out of nowhere

[novel acorn]

Lmao
You have a scholarship I assume?

[abstract pagoda]

lol full scholarship

[swift fjord]

Like why would you think that this commutative square is what you need to strengthen the hypothesis

[abstract pagoda]

i kinda regret getting fired tbh

undergrads cant TA

but we can grade

[swift fjord]

This whole proof looks really out of nowhere tbh. Like it's all clear to me what is happening but I probably couldn't reconstruct it without memorising

[empty grove]

Well the manipulation seems to be going smoothly until you hit the point where this has to be used

[novel acorn]

Damn

How does one attain such things

[abstract pagoda]

minority circumstances

[novel acorn]

[marsh forge]

this is common and not a bad thing

[swift fjord]

I TA maths for the on campus prep school

[empty grove]

But yeah someone who's seen this before might have some insight

[marsh forge]

sometimes proofs be hard and weird

[abstract pagoda]

slightly above piss poor and unrepresented minority in math

[marsh forge]

rotmans notation is not anything I recognized I would have to read the relevant chapter

[abstract pagoda]

• loopholes

[pearl holly]

Tutoring

[swift fjord]

Like for the prism operators I can see the motivation, you wanna construct some relationship between the complexes that makes it so the induced maps differ by a boundary element

[abstract pagoda]

is that the highschool?

[marsh forge]

Oh is this the proof of singular = some other thing

[paper wedge]

read tom dieck

[swift fjord]

No, it's for people who couldn't get in/don't have the requisite knowledge or classes

[swift fjord]

It's the homotopy axiom for singular homolofy

[abstract pagoda]

oh

[swift fjord]

2 homotopic maps induce the same homomorphism on the homology groups

[marsh forge]

oh i see

[pearl holly]

homolofy

[marsh forge]

this proof is not important if you want to just move past it and come back later if you are curious

[swift fjord]

I've already read it, but I'll probably review in a few chapters

[marsh forge]

I guess my point is that things like prism operators or what have you

don't come up again

in any useful fashion

[swift fjord]

I just always like to think how I would have been motivated to come up with the proof i'm reading

And it's a bit frustrating when the motivation is so opaque

[marsh forge]

this is sadly a problem that only gets worse lol

[swift fjord]

But on the other hand the person who proved it might have been using techniques i've never seen before

[marsh forge]

the point of a good lecturer imo is to like

1. explain motivations that aren't written down

2. explain when something is legit just weird

But also, the intuition to prove this kind of stuff

[novel acorn]

I honestly don't know how to accent homotopy

[marsh forge]

can often come from stuff you simply don't know

or intuition you don't have yet

or previous constructions

[swift fjord]

Ye exactly

[marsh forge]

etc etc

[swift fjord]

Maths books are kinda written backwards a lot of the time

[marsh forge]

homo - toe pee

[swift fjord]

HoMOtopy

[marsh forge]

not a joke

thats how you pronounce it

[pearl holly]

lmao

[marsh forge]

homo-tah-pee is for weirdos

[pearl holly]

what

[swift fjord]

Pronounced ha-motahpee

[marsh forge]

no

[swift fjord]

Where the first a is a schwa

But I don't have an ipa keyboard

[marsh forge]

hoe-moe-toe-pee

[abstract pagoda]

toe pe

ye

[swift fjord]

The accent is on the Mo

[marsh forge]

if u literally read hoe-moe-toe-pee

you'll pronounce it right

[swift fjord]

You say homotopy and homology with the same cadence

[abstract pagoda]

quick question

[marsh forge]

what

[abstract pagoda]

say you just finished ur first algtop class

[marsh forge]

hoe-mah-low-gee

[abstract pagoda]

and u have three teachers that study stuff related to it

[marsh forge]

wait no thats bad

[abstract pagoda]

what type of research could you do with them

[swift fjord]

I mean the stress is on the same syllable in both

[marsh forge]

ask them nines

lol

[abstract pagoda]

is it really like that

[marsh forge]

yes

[swift fjord]

You don't say low you say it more like luh

[marsh forge]

just to be clear

its very much not

[pearl holly]

you can do research in homo - toe pee

[reef shore]

hoe maw law gee

[marsh forge]

hoe-moe-logy

[reef shore]

[swift fjord]

Max come to vc

[marsh forge]

okay im just gonna get in voice at this point lmao

[plain raven]

ho m o l o g y

[abstract pagoda]

idk asking prof is scary, i feel like they either going to say i need more prereqs or jus laff in my face

[plain raven]

h o m o t o py

[marsh forge]

they wont

probably