point-set-topology

[hexed steppe]

it is nonetheless fairly common

at least in the US

[white oxide]

... tychonoff is not important?

[hexed steppe]

not the proof

the statement takes 2 seconds to understand

[white oxide]

i am getting culture shock

[hexed steppe]

given the previous content

look i agree

im just saying

when do you actually use that stuff

in math

like the statement of tychonoff is important

[white oxide]

sure?

not everyone can be linear algebra I and II

[hexed steppe]

lol

ok chapter 4 is probably very important though

2-4 would be a good semester course

also you should do the quotient thing

idk why munkres marks it as optional

that is like the single most important construction

[white oxide]

quotients as optional 😭

[hexed steppe]

if you want to learn any further topology

[white oxide]

thats like top 5 most important things from set topology

[coarse mist]

what are the other four

[white oxide]

Connectedness (basic and path) and compactness for sure up there

[hexed steppe]

nagata-smirnov metrization theorem

[white oxide]

ofc ofc

[hexed steppe]

product topology

[white oxide]

like the vodka?

[hexed steppe]

idk i didnt read chapter 6

[white oxide]

i mean, solid strategy. Make students read the proof and then sell them a bottle to cope

[white oxide]

oh that one

iff metrizability theorem

[hexed steppe]

ok the top 5 are probably compactness, various notations of connectedness, urysohn lemma, quotients, products

or something

not necessarily in that order

[coarse mist]

i can do that in winter break ez

[hexed steppe]

that is like 200 pages

[white oxide]

you planned on doing munkres all within winter break initially?

[coarse mist]

uhh as much as i could

[white oxide]

thats like

a good pace for a regular book

[coarse mist]

it makes sense that i need to spend much more time on the problems though

[white oxide]

for a math textbook thats absolutely unhinged 😭

or is your winter break 2 months

[coarse mist]

nah one month

200 pages in a month is 10 pages a day with a bunch of free days just for problems

[hexed steppe]

10 pages a day is a lot

[coarse mist]

yeah true

anyways imma start

[hexed steppe]

unless you know it already

[coarse mist]

expect me in this channel frequently

thanks for the help

[hexed steppe]

good luck

[white oxide]

or they work on it for like 8h a day

sometimes it feels like half the discord is half a year away from burnout

[ebon galleon]

True!

[last topaz]

Do people really learn topology by reading Munkres cover to cover? I tried that as a wee lad and gave up after like 5 seconds. If I recall correctly, I was frustrated by the fact that there are so many notions and theorems proven about them and it was unclear which ones are crucially important and which one you'll never ever meet in real life.

@coarse mist
I think it would 100% be a better use of one's time to pick up some free lecture notes online from a professor that you like and go through that. If they are a competent teacher, they will have done the job of cherry picking the most important topics and discarded the crap. Also they will have curated a finite list of useful exercices for you to look at.

If you don't know where to start, look at Zuoquin Wang's website. All of his lecture notes are fantastic imho.

http://staff.ustc.edu.cn/~wangzuoq/Courses/22S-Topology/index.html

[rancid umbra]

i actually really like munkres. the first couple chapters are worth skipping and coming back to though.

[last topaz]

Maybe my mistake was starting on page 1 🤷‍♂️

[coarse mist]

Yes, that part is what's giving me trouble

I'll look into that. That's extremely helpful

[thorny agate]

My self study of topology was chapter 1 of Bredon

Plus occasionally looking at notes from my uni's topology course

Lecture notes in question: http://lerman.web.illinois.edu/535/f21/535f21hw.html

[umbral panther]

Munkres is thorough. It seems like a bad idea to try to tackle it if you might not get through. Better to study something lighter to get the make sure you get the big picture

[pseudo coral]

Silly question

What’s the topological boundary of rn\0

Is it his

Jus

{0}

Or $\emptyset$

[gentle osprey]

MyMathYourMath

[pseudo coral]

I wanna say it’s just singleton zero

Cause it’s closure minus

Interior

And rn\0 has interior itself and closure rn

[feral copper]

What does rn\0 mean? What is your topolofical space?

[pseudo coral]

Set minus the point zero

[feral copper]

Which set?

[pseudo coral]

Rn

Say R^2

For@simlocity

[feral copper]

Wow, do you actually mean $$\RR^n\smallsetminus{0}?$$

[gentle osprey]

Matplotlib

[pseudo coral]

Yes $\Bbb{R}^n\setminus {0}$

[feral copper]

Woooow, rn\0 really does not look like that...

[gentle osprey]

MyMathYourMath

[pseudo coral]

Sorry

[feral copper]

SO what is the closure of that? What is its interior (note it is open)?

[tribal palm]

[pseudo coral]

I just wrote those down

Interior is itself

Closure is all of R^n

[gentle osprey]

Matplotlib

[pseudo coral]

I’ve never thought about this problem and a friend and I had the argument

That’s what I thought ok

[unreal stratus]

0

Yh

[radiant cedar]

What does it mean to say that "if X is a compact Hausdorff space and C(X) the ring of continuous functions on X, you can recover X from C(X)"?

[knotty vine]

I think Gelfand duality is meant?

[radiant cedar]

What does it mean to recover X from C(X)? You get the points of X and all the open sets from this ring?

[feral copper]

Is it not an application of Urysohn's lemma? You can recover open sets from C(X)

[tidal lynx]

I don’t know about this particular result but my guess is this is saying that if X and Y are CH spaces with C(X) = C(Y) then X = Y

[feral copper]

Because you recover the topology yeah

That's what is meant by the claim. Not just that X=Y as sets, that's kind of stupid, but really as topological spaces!

[tidal lynx]

Right

[umbral panther]

A slightly stronger statement is that all homomorphisms C(X) to C(Y) come from continuous maps from Y to X

[knotty vine]

Is it Urysohn? How do you recover the opens from a general ring? Then again Gelfand duality works with C* algebras

[umbral panther]

The Zariski topology on C(X;R) is homeomorphic to X

[tidal lynx]

Worlds largest dictionary

[hexed steppe]

yes its urysohn

this is an exercise in atiyah-macdonald chapter 1

it explains it in some detail in the text

[merry geode]

I heard one can have a cycle on a space of single point

In a specific homology

Or is it cohomology?

[steel glen]

what are you referring to

[hexed steppe]

homology has cycles

and sure H_0({pt}, Z) = Z

[merry geode]

I mean when one has nonempty H^1({pt}, G)

[gritty widget]

Why does the last statement hold

[steel glen]

which part?

composition with the unknot does not change the knot type

[hexed steppe]

look up extraordinary cohomology theories

[steel glen]

and if the unknot K_0 is some composition of knots K_1,K_2,...,K_n, then any other knot J would satisfy
J = J,K_0 = J,K_1,K_2,...,K_n
it's analogous to how if 1 were composite, every integer would be composite

[feral copper]

Wait, what? Isn't the Zariski topology never Hausdorff?

Like, you mean the Zariski topology on a ring as in Spec(A)?

[red yoke]

https://math.stackexchange.com/questions/106043/rings-whose-spectrum-is-hausdorff

[merry geode]

That is quite an interesting piece

Makes sense, it's all about the zero set

[oak pumice]

Hey

[heady skiff]

this is probably a silly pedantic question, but is there any difference between saying that X is a set on a topology T vs. X is a topological space with topology T?

i mean it's just the same thing right

oh wait i see

they're both the same set on different topologies

nvm then

[gaunt linden]

"set on a topology" sounds like a typo.

[gleaming warren]

This is just confusion overload just say let T, T' be topologies on a set X

[heady skiff]

Ye Ikr

[radiant cedar]

Anyone know a good informal introduction to characteristic classes?

Hatcher is not doing it for me

[winged viper]

What about milnor stasheff

[fading vale]

milnor stasheff is the classic

[stuck sorrel]

Guys I have this dump question about proving an open interval is an open set

The proof on proofwiki use epsilon is min of (b-c,c-a)

And in the middle part it stated that epsilon >= c-a and epsilon <= b-c

Is it really possible to do so?

I thought that min(b-c,c-a) only implies epsilon is greater or equal to either element

[rancid umbra]

epsilon is less or equal to both b - c and c - a

[stuck sorrel]

Oh I think I messed up with the signs, thank you for your input

[last topaz]

The book by Chern Complex Manifolds without potential theory is adorable. It's message is "look how much fun there is to have in here". It's hard to get less formal than that. But the book is tiny and of course it only talks about Chern classes. And I struggle to remember if he talks about just c_1 or not. But anyhow, it's a fun read and a good first contact with characteristic classes.

That said, if you don't have all day to durdle around and need something that isn't as informal, Milnor-Stasheff...

[fickle elm]

Bott and Tu's differential forms in algebraic topology, Chapter IV gives a similar treatment of characteristic classes with Hatcher (projective bundles, splitting principals, etc.). Maybe it is more readable for you.

Alternatively, you can look up Chern-Weil theory and use connections to define characteristic classes, it is in Milnor-Stasheff's appendix.

I found a student paper by googling. From the abstract, it looks like a good introduction.

https://people.math.ethz.ch/~acannas/Student_Papers/Semester_Papers/2022_joeran_schloemer_sp_chern_weil_theory.pdf

[radiant cedar]

Is there a version of this that isn't written with a typewriter?

[last topaz]

I dont know.

[scarlet turtle]

i know how to prove this, but is there a counterexample for the <- direction if we remove "K is closed"?

[red yoke]

{n×n×… : n ∈ N}

[tiny obsidian]

what does that mean?

[hexed steppe]

i dont think this works

how do you find the B_n

[red yoke]

Oh oops

[hexed steppe]

that is compact no?

[red yoke]

1×0×0×…, 0×1×0×…, 0×0×1×…, …

[hexed steppe]

agreed

wait

yes

agreed

[red yoke]

[hexed steppe]

btw @scarlet turtle this topology should be metrizable in a fairly simple way

which ime makes it easier to understand whats going on

and to produce counterexamples to things

[scarlet turtle]

i'll go give it a try, thanks

[scarlet turtle]

ahh i see, i was struggling to see why this was not closed until i put the product metric on it. thanks a lot!

[hexed steppe]

idk if its easy to come up with but it should be easy to google

[coarse mist]

In regards to a metric, what does d|AxA mean

[gaunt linden]

Probably d (which I assume is a metric on some larger set) restricted to inputs from A.

[radiant cedar]

What is meant by saying that if X is a vector field on a manifold, then the # zeros of X on M (counted correctly) equals the euler characteristic of M?

[hexed steppe]

context?

[knotty vine]

I guess thats the Poincare-Hopf theorem

[winged viper]

Yeah I think the precise statement is the oriented intersection number of a vector field (viewed as a map M -> TM) with the zero section of TM is the Euler characteristic

So you need the vector field to be transverse and then you count it with orientations (which turns out to be the index of the zero)

So it’s just Poincaré hopf

[knotty vine]

That reminded me of my favourite cartoon in high school https://archive.org/details/TopoTheWorld-English-JeanPierrePetit/mode/2up

Didnt realize that's what it was about until now...

[merry geode]

A bit of a shitpost but

Can I consider something close to discrete topology Metric topology,
and something close to cofinite topology like Zariski topology?

[fading fern]

wait the singular chain of Q is the same as that of Z right

[lime sable]

like, they're both countably many points in R?

you cannot have an infinitely large chain

[fading fern]

i mean like

you know chain complexes right

if you know homology you gotta know those

[lime sable]

are you asking if their singular chain complexes are isomorphic

yeah should be

[fading fern]

alright thanks

[merry geode]

So shitpost gets ignored

What did I expect..

[opaque scroll]

The Zariski topology is kinda close to being cofinite I guess.

Though I don't feel like metric topologys are usually very close to being discrete.

I don't really get what you're even asking though

[merry geode]

I think one might be able to identify characteristics of topologies this way

[opaque scroll]

Which way?

@merry geode

[merry geode]

Like giving finer topologies approximation to manifolds

While coarser topologies approximating to zariski topology

[tiny obsidian]

Do you unknowingly mean separation axioms though

[merry geode]

What kind of separation does zariski topology have?

[opaque scroll]

It's T0

[merry geode]

Yea, but that lacks characterization of zariski-like topology

[opaque scroll]

What do you mean by Zariski-like?

Like being T0, but not T1 is a somewhat restrictive property.

Of course the spectrum of a field (or product of fields) will have discrete spectrum. So in those cases you can't distinguish discrete from Zariski topology.

[merry geode]

Hmmmm

[opaque scroll]

Having generic points is sort of a big part of Zariski though. Something like "every closed set is the closure of a finite set", might be good

Only works with Noetherian rings though, I think

[merry geode]

Yea that sounds great

Are common rings in AG Noetherian? I vaguely recall that hartshorne largely assumes noetherian scheme

[opaque scroll]

Yeah, stuff is almost always Noetherian

[white oxide]

Does it ever satisfy T3? It being T1 and T2 are both equivalent to A having an oddly specific ring property

[opaque scroll]

Well, like I said a (product of) fields is just the discrete topology, so that satisfies all the Ts I guess

[white oxide]

Okay but like when we have more than one point 😭

[merry geode]

Doesn’t product of fields have more than one point

[white oxide]

Yeah it does. Fair enough then

[white oxide]

For finite products it has as many points as it has factors

[knotty vine]

Fields dont have products...

[unreal stratus]

Spectrum of product of comm rings is iso to disjoint union of product (at least in finite case lol)

[knotty vine]

Oh product of fields as a ring ig

[unreal stratus]

Yeah that's what I assume was meant

[white oxide]

Quick edit...

[white oxide]

Spec always treats whatever you slot in as rings

[knotty vine]

Yeah sorry my field hatred bubbled up. Forgot you guys were talking about algebraic geometry (🤢) in the algebraic topology channel!

[unreal stratus]

Lol

Good point

[knotty vine]

Zariski topology is topology

[white oxide]

On life support

"Spec A is Hausdorff? Ah, of course. The quotient ring R of A with it's nilradical has the property that any R-module is flat"
Statements thought of by the deranged

[red yoke]

Ok quick sanity check:

1. A knot K has Seifert surface M and exterior X.
2. p: Y→X is the cover generated by a meridian of K.
3. The Seifert form α on M is the map H1(M) -pushoff→ H1(M^c) -Alexander→ H¹(M) -UCT→ Hom(H1(M),Z).
4. α* is the extension of scalars to Z[deck(p)].
5. Δ = det(tα*-α*^T) is the determinant of the Z[deck(p)]-module map H1(p^-1(M)) → H1(p^-1(M^c)) induced by both pushouts.

Is this the best way to think of Δ?

How does this compare to, say, treating it as a special case of Homfly

[broken nacelle]

you hate your field?

[alpine nest]

if you don't hate your field, you're not sufficiently advanced yet

[red yoke]

[fading fern]

guys H_0(R,Q) is coprod of an uncountable amount of Z's right?

[red yoke]

No

[fading fern]

hi fox guy

[red yoke]

Hello hsf

[fading fern]

please explain how would you do it

[red yoke]

Wait is it

[unreal stratus]

For example

[fading fern]

how do you guys do it with exact sequence

[unreal stratus]

Uhhh

[fading fern]

i do it by the usual definition

[unreal stratus]

Maybe i wrong tho

Wait no

[red yoke]

• Relative chains: ||Chains in R modulo chains in Q||
• Relative cycles: ||All chains in R modulo chains in Q||
• Relative boundaries: ||All chains in R modulo chains in Q||

[unreal stratus]

But I'd still use the exact sequence lol

[fading fern]

relative cycles and relative boundaries are the same?

[red yoke]

(||All chains in R can be turned into a cycle or boundary by adding some chain in Q||)

Here yes

Relative cycles of (A, B) are chains in A whose boundary is in B

[fading fern]

wait so like

[red yoke]

Relative boundaries are the modulo-chains-in-B classes that contain boundaries of A

[fading fern]

if you wanted to say it more fomally

relative cycles of (A,B) are n-chains in A with boundary on S_n-1(B)?

[red yoke]

Boundary is a chain in B yes

[fading fern]

alright let me write it down

[white oxide]

no way

hsf has left discussy

[fading fern]

yes truly unbelievable

[white oxide]

someone pinch me

[red yoke]

Hsf has graduated discussy

Road to server owner

[fading fern]

so true.....

[white oxide]

lemme dig out my shitty screenshot once more

you have an exact sequence like this one

[fading fern]

wait

isnt that wrong

it goes from n-1 to n to n+1

[red yoke]

Ye the number decreases for homology

[white oxide]

oh whoopsie

[fading fern]

haha i know more AT than you now B)

[white oxide]

anyways i just wanted to test your observations, flip the signs

You know H_n(Q) and H_n+1(R) for n>1?

[fading fern]

yeah i think so

let me re-do it

[white oxide]

namely?

[fading fern]

(i forgot to write it out)

[white oxide]

oh well

It helps an exercise in singular homology boundary maps ig

[fading fern]

i just got H_n+1(R)

[white oxide]

yeah that one is easy once you know whats going on

[fading fern]

H_n(Q) is a little weirder but let me calculate

[white oxide]

ye

[fading fern]

so H_0(Q) is coprod countable many Z's

and the rest are 0s

[white oxide]

some observation for it:
since the only simplex maps are constant the boundary of those maps is also the same constant map

can you see why then the homology groups must be all zero?

(idc about H_0 shut up)

[fading fern]

wait let me process

yeah i get why it is true for H0(Q)

WAIT I GOT AN IDEA

WHAT IF I USE THE LONG EXACT SEQUENCE

[white oxide]

huh

[red yoke]

That's a brilliant idea

[white oxide]

uh, of what?

or do you wanna continue assuming H_n(Q)=0 for now?

[fading fern]

first part of LES (for homologies) should look something like this right

wait let me send photo

[white oxide]

yeah

[fading fern]

wait i think im figuring it out

[red yoke]

Perhaps Z → H0(R, Q)?

[white oxide]

oh yeah thats mixed up

[fading fern]

wait really

[white oxide]

H_0(R,Q) should be last

[fading fern]

isnt the exact sequence
SQ->S(R,Q)->SR

[red yoke]

H0(Q) → H0(R) → H0(R, Q) → 0

[white oxide]

... -> H_n+1 (R,Q) -> H_n Q -> H_n R -> H_n(R,Q) -> H_n-1(Q) -> ...

H_-1(X) is definitionally 0

[red yoke]

Or H0(Q) → H0(R) → H0(R, Q) → Z → Z for reduced I believe

[fading fern]

wait my book says something else

let me google it

wait

guys what are your original exact sequences

like the short ones

[red yoke]

0 → C(Q) → C(R) → C(R, Q) → 0

Definition of relative homology

[fading fern]

wait really

im dumb

thank you

[red yoke]

0 → A → B → C → 0 means C = B/A (roughly)

[white oxide]

roughly?

[fading fern]

yeah because of weird isomorphism not equality stuff

[white oxide]

at this stage I'd just supress my isos

[fading fern]

wait this is kind of basic but like

how do homologies from Q move into homologies of R

since like

[red yoke]

Via inclusion

[fading fern]

countable coprod of Z ---------> Z -----> H0(R,Q)
is pretty much the important part of the sequence

i get that but like

wait let me think it more

[white oxide]

cycles are just singular maps

the elements of $C_n(X)$ are just cts maps from the n-simplex into X. So if $X \subseteq Y$ just compose with the inclusion

[gentle osprey]

Kerr

[fading fern]

yeah i get the inclusion part

[red yoke]

Then you just have to figure out where each summand of ⊕Z is sent

[fading fern]

it's the sum of the elements?

like if you got (1,2,0,0,0,0,0,0,.....) then would it go to [3]?

alright wait let me think my answer

yes 3

[red yoke]

Yes

[fading fern]

so im(inclusion) = ker(projection) right?

because its an exact sequence

[red yoke]

Ye

[fading fern]

im(inclusion)=Z=H_0(R)

kernel is the entirety of H_0(R) then

and since

Z->H0(R,Q)->0 is also exact

H0(R,Q)=im(projection)

H0(R,Q) is trivial then?

[red yoke]

💯

[fading fern]

LETS GO

I DID BIG BRAIN

it took me like 1 month but i did it

thank you guys

[white oxide]

0 -> Z -> H0(R,Q) -> 0?

[fading fern]

no

[white oxide]

or what comes before in the sequence

[fading fern]

wait let me copy-paste the fox guy symbol

⊕contZ -> Z -> H0(R,Q)->0

[white oxide]

we should have a final sequence of the form 0 -> H1(R,Q) -> ⊕Z -> Z -> H0(R,Q)->0,

[fading fern]

so

H1(R,Q) is isomorphic to the sequences that sum to 0 right?

[white oxide]

not sure how you got there but what I mean is

[fading fern]

wait

[white oxide]

you determined correctly that H0(R,Q) is the image of the map from Z, but then you need show that this map is trivial to conclude H0(R,Q) is too

I dont think you can argue purely with the sequence on that final strip. Without some knowledge on the maps you can get some different results

[fading fern]

yeah but it's exact sequence

[white oxide]

Often times you have sequences that look like 0 -> A -> B -> 0 which imply that A and B are isomorphic

[fading fern]

ker(proj)=im(inclusion)=H_0(R) right?

[white oxide]

yeah

[fading fern]

so if kernel is the entirety of domain it must be trivial

[white oxide]

since proj H0(R,Q) -> 0 is trivial you can conclude that H0(R,Q) is exactly the image of Z

[fading fern]

yes

so H0(R,Q) is trivial

[white oxide]

why is the map H0(R) -> H0(R,Q) trivial

you need that to conclude what you said

[fading fern]

because kernel is H0(R)

[white oxide]

no the kernel is the image of ⊕Z

[fading fern]

and the image of ⊕Z isn't H0(R)????

[white oxide]

only if the map from H0(R) to H0(R,Q) is trivial

but you might notice that we are getting a bit circular here

[fading fern]

alright let me re-do it

[white oxide]

i am not sure this is even part of your exercise tbh

[fading fern]

alright let me send photo of my work

[novel acorn]

H_0(R) should just be Z right

[fading fern]

yes i think so

[novel acorn]

Well since the last map is surjective, then H0(R,Q) is something Z can surject onto

[white oxide]

One bit of helpful information is that you can determine one map purely algebraically

[fading fern]

trivial group?

[novel acorn]

So it can't be \oplus Z

[novel acorn]

It can be a lot of things

[fading fern]

yeah but what i got originally was trivial group

and i trust past self

yeah it's trivial

[white oxide]

or nvm

I assumed we already had H_1(R,Q)=0

[red yoke]

From this you got H0(Q) → H0(R) surjective right

[white oxide]

actually that is impossible

[novel acorn]

H_0(Q) is a countable number of Z's

[fading fern]

yes

[red yoke]

Thus H0(R) → H0(R,Q) trivial

[fading fern]

yes that's what i was saying

[red yoke]

Then since last object is 0 this is surjective

And done

I don't see the objection

[fading fern]

yeah me neither

i trust kerr though

[white oxide]

didnt see that one

the objection is that the reasoning presented only used the data in the sequence, which isnt enough.

But if you know some things about the specific maps you can quickly fill out the blanks

[fading fern]

yeah but we have data of inclusion map

it's sum of numbers

[red yoke]

[fading fern]

wait afk

[white oxide]

anyways have you calc'd H1(R,Q) yet?

you have everything you need

[fading fern]

back

took girlnap B)

[lament vortex]

what is Conv??

[fading fern]

best guess ive got is convex hull

[white oxide]

sensors in D? What are you doing there? Context helps

[fading fern]

hi kerr

[lament vortex]

here's the setup, i wasn't sure how much to include my b

[fading fern]

i got that H1(R,Q)isoH0(Q)/Z

[white oxide]

please elaborate your steps bestie

[fading fern]

alright will write steps

so let's say you've got the LES and part of it is:
0->0->H1(R,Q)->⊕Z->Z->0

so we get that H1(R,Q) iso H0(Q)/Z

Q.E.D

[white oxide]

Right so the step you did was:
Observe ⊕Z->Z is surjective. Then consider the subsequence

0->H1(R,Q)->⊕Z->Z
since the last map is surj. you can augment it with 0 to get into a more familiar form:

0->H1(R,Q)->⊕Z->Z - > 0
This is just the short exact sequence that defines Z as ⊕Z/f(H1(R,Q)). Although for describing H1(R,Q) you'd probably need to go over to the map H1(R,Q) -> ⊕Z and figure what elements get send to zero there

namely those elements whose sum of coefficients is 0

[fading fern]

yeah thats what i said earlier

i thought you said it was wrong though (maybe i misunderstood)

[fading fern]

here

[white oxide]

yeah, is that actually isomorphic to ⊕Z/Z?

[white oxide]

thats correct

[fading fern]

idk i guess you could use first iso

not sure though

[white oxide]

that feels like something you can show with the structure theorem but i cant think of it rn

No matter, you described all the groups. Yippee

[fading fern]

yay!!!!!!!!

thank you for the help

[quiet thorn]

Yippee

[fading fern]

also imma search up structure theorem

oh its cool

[white oxide]

yeah, feel free to ask stuff on here in the future

[fading fern]

thanks

[knotty vine]

Touching grass?

Field theory?

[alpine nest]

Pictured: applied field theory

[gritty widget]

https://zenodo.org/records/10429581

[tidal cedar]

[knotty vine]

snarXiv moment

please tell me this is a typo

Also, handwritten references

[empty grove]

These groups are not finitely generated so you can't use the structure theorem. The identification with the quotient here is unnatural since the group we are concerned with is a kernel (and hence really a subgroup rather than a quotient) but it does happen to be isomorphic to the quotient because the SES splits.

[gentle osprey]

Leno

[red yoke]

Every open ball is a union of equivalence classes

If x is in the open ball of radius r at y, and x' is equivalent to x, then x' is in the ball

[gentle osprey]

Leno

[red yoke]

You're basically taking a pseudometric space and constructing a metric space from it such that some points are "identified as the same point"

This "identification" is described by the equivalence relation

In the new metric space, points with distance 0 are treated as 1 point

One way to formalize this is to take the equivalence classes in the old space as the "points" in the new metric space

And the map π is the "quotient map", the map that tells you what each point in the old space corresponds to in the new space

[red yoke]

Can you see why this is equivalent to the statement that needs to be proven?

And can you see why this is true?

If you can answer these two questions then you've solved the problem

The triangle inequality may be helpful

The image of the open ball of radius r around x is the open ball of radius r around π(x)

[gentle osprey]

Leno

Leno

[prisma garnet]

does this theorem have a name?

sounds super important

@bitter shell you should use #latex-testing

you've already spammed this twice in #real-complex-analysis

also, you can just edit your message and texit will rerender your latex

[hidden crag]

The important theorem that carries this is excision

You use excision for the relationship between reduced homology of the quotient and relative homology

(For good pairs obv)

And then you replace terms in the exact sequence of relative homology right

I don’t have Hatcher on me rn but I think that’s how it should go?

[prisma garnet]

dunno, haven't reached excision yet

[hidden crag]

Oh lol I thought that comes before this

Ignore it then

But that’s why this is morally true

Hope that makes some sense at least

The big thing is the relationship between reduced and relative homology

[prisma garnet]

I see, aight

[prisma garnet]

I'm just reading about relative homology so it doesn't yet lmao

[hidden crag]

Accept it as axiom and work in the generalized setting

[prisma garnet]

but I'll keep it in mind

[hidden crag]

[hidden crag]

It will soon then

[red yoke]

More generally you have zig-zag lemma

Whenever you have three chain complexes 0 → A → B → C → 0 you have a long exact sequence of their homologies

This specific instance is the zig-zig lemma applied to relative chains 0 → C(A) → C(X) → C(X,A) → 0

And by "excision" mentioned earlier, you have H(X,A) = H(X/A) for sufficiently nice inclusions A ⊂ X.

(this is partly why cones / relative homology are nicer than quotients in alg top)

[merry geode]

Isn’t relative homology literally from taking quotient of cycles tho

[red yoke]

I mean quotient space X/A

[hidden crag]

This isn’t the same as taking quotient spaces

Quotients in algebra are far better behaved than quotients in topology

[red yoke]

Cuz H(X,A) is always equal to (reduced) H(X∪CA) but is only (reduced) H(X/A) for nice A → X

[merry geode]

Hmm, I never understood quotients by subspace

[prisma garnet]

whoa

wicked

[hidden crag]

Subspace gets glued together

There’s not a lot to just taking quotients but it just changes a bunch of shit in the space

[merry geode]

Yea but I don’t see the motivation, I mean

[hidden crag]

The motivation is exactly that

Gluing stuff together

[merry geode]

Hmmm, can’t one use X / ~

Where ~ is a suitable relation.

[red yoke]

Quotient by subspace is when ~ is generated by a~b for a,b ∈ A

The motivation is for when you haven't learned about cones yet

[hidden crag]

That’s what you do

[gaunt linden]

Can a vector bundle be recovered from its sheaf of sections (viewed as a sheaf of abstract vector spaces)?

[ashen kraken]

I want to try and find a concrete counter-example for this, since I think it is false. Does anyone have a hint?

[gaunt linden]

Um, if the answer is "no", then you cannot prove that by an example.

[white oxide]

In either case you can start reasoning about it through "Suppose there is such a metric, then it must fulfill the metric axioms"

see where that gets you

[high hill]

topological quotients... where my 1st iso huh?????

[thorny agate]

imagine having quotients without iso theorems

[novel acorn]

Welcome to the world of fibrations/cofibrations hehehehe

[red yoke]

What if you do this:
Choose a smooth section f of [Möbius strip → S¹] that intersects 0 only once, transversely.
For any section of S¹ × R, multiply by f to obtain a section of Möbius strip → S¹

Does this give an isomorphism between sheaf of sections of cylinder and Möbius strip

[obtuse meteor]

Yes, if you take it as a C(base)-Module

This is in Milnor Stasheef

[red yoke]

This for R-vector space

[gaunt linden]

Hmm, it doesn't feel like an isomorphism -- it doesn't hit sections that are nonzero at the crossing point of f.

[gaunt linden]

Then I need to recover a Cbase-module structure first, right? Hmm, perhaps with enough handwaving I could do that by approximations and continuity.

[red yoke]

How do you get that from an abstract vector space structure

[gaunt linden]

That's where handwaving would come in. Perhaps with enough assumptions (e.g. base space is a compact manifold or something) I can approximate a general continuous function uniformly as a sequence of locally constant functions on dense open sets, and then appeal to continuity to get a limit.

[red yoke]

You mean defining multiplication by a function as a limit?

[gaunt linden]

Yeah. Not sure it will work, though.

[red yoke]

How do limits work in an abstract vector space

[gaunt linden]

Hmm, good point.

My vague notion was to just assume things are finite-dimensional so the limit would make sense, but the C(U) spaces are definitely infinite-dimensional even if the fibers are not.

[obtuse meteor]

Also this definitely works

[gaunt linden]

How is it an isomorphism if it's not surjective?

[obtuse meteor]

Oh sorry I am not thinking abt an isomorphism how do I want to say this hrm

This is an iso onto things that are 0 at one pt

But yes you’re right not sure if there is one in general

Idk how dumb you expect

0 dim over difft bases for example

[gaunt linden]

Hmm, I was mostly thinking: Suppose we have a fixed space X and two non-isomorphic real vector bundles on X, is it then possible that their sheaves of sections are isomorphic as X-sheaves of real vector spaces?

[umbral panther]

You can extract the subsheaf of sections that vanish at a point, can’t you? From this you can recover the vector bundle as the fibers. An isomorphism of sheaves of vector spaces induces an isomorphism of fibers, so an isomorphism of vector bundles. But it’s not clear to me that an isomorphism of sheaves is a C-module map

[red yoke]

You can extract the subsheaf of sections that vanish at a point, can’t you?
Don't you need multiplication by smooth functions for that

[umbral panther]

I think you detect vanishing from extension properties

[red yoke]

How does that work

[tribal palm]

what do they (Viro) mean by (3) and (4) being “nonsymmetric” ?

[tribal palm]

do they perhaps just mean replacing X with resp a union and intersection of some arbitrary collection of sets?

[tidal cedar]

can't swap X and A maybe?

that wouldn't make much sense but it's the only thing I can really think of?

Oh, wait

Maybe yeah making it symmetric in the sense of union and intersection on left and right

[tribal palm]

hmm

thanks i’ll try that tmrw

[fringe cypress]

https://media.discordapp.net/attachments/666448069751209984/1188982432406454392/image.png?ex=659c812e&is=658a0c2e&hm=01a3e7e33e62f0b7e8bd4ad87e41267b3fd654dc1145c052d881efd4afee6cf0&

Very fun exercise from Willard

[knotty vine]

The topology is defined via closure?

[unreal stratus]

ye

[knotty vine]

Seems to me that ||its T0 if B is a singleton, and T1/T2 if B is empty (ie the topology is discrete)||

[unreal stratus]

ig ||it is T_1 iff B is empty right? since we need {a} = {a} u B for all a in B. okay, ig if X is a singleton then B can be the whole space too but ye lol||

and ye agreed with the T_0 thing

[white oxide]

||T2: the only open set containing points in B is the space itself, so you aint gonna separate nun. B must be empty
T1: Take any point x outside B. The smallest closed set containing it is just {x} \cup B, so B must be empty here.
T0: For the reason mentioned for T2 pairs of points inside B are indistinguishable, so B contains at most one point||

[unreal stratus]

Ofc doing T2 separately is unnecessary lol

[white oxide]

True but this is just for fun anyways lol

[unreal stratus]

Yee

[gritty widget]

can someone explain to me why the open ball metric is swapped?

[white oxide]

Swapped?

The topology induced by d contains any set that is open according to d'. You can cover the open ball of d' by balls that lie completely inside and then you take the union over all x inside that ball

[gritty widget]

like we have Bd in Bd'

[white oxide]

Yeah

Are asking why the topology of d' (curly fancy T's) lies inside the topology of d?

[gritty widget]

i thought that if Td' in Td , that means every open set in Td' is also open in Td

[white oxide]

It is

Think of some more extreme situation. The coarsest topology has only X and the empty set

So it is a subset of any topology on X

Bd'(x,r) is still lies in Td. It just takes a bit of work to show it

[gritty widget]

is it like for every element of x i can recover Bd' by balls of Bd?

[white oxide]

Yeah

For any x in Bd' use the definition. Then union

[gritty widget]

oke

[prisma garnet]

call the homomorphism from $H_n(X, A)$ to $H_{n-1}(A)$ the map $\partial$. To prove this chain is exact we need to prove $\ker \partial = H_n(X, A)$, right?

[gentle osprey]

DarQ

[prisma garnet]

since the homomorphism right before it is surjective

or am I missing something

[red yoke]

Why

[prisma garnet]

this is what hatcher says (replace A, B and C with A, X and (X, A)) respectively

[prisma garnet]

it's the homomorphism induced by the canonical projection $j\colon C_n(X) \to C_n(X)/C_n(A)$

[gentle osprey]

DarQ

[red yoke]

j is surjective

[prisma garnet]

yea

[red yoke]

But this be j*

[prisma garnet]

wait, the induced homomorphism need not be surjective?

[red yoke]

It need not be

Neither do inclusions / surjections of topological spaces induce injective / surjective maps on homotopy or homology

[hidden crag]

If all these were necessarily surjective the sequence would look very odd

Think about it like this DarQ you can easily surject/inject from or into spaces that are „bigger“ or „smaller“ in a certain sense but are less simple

Think about S^1 and D^2 for example

Exercise is to find counterexamples

Don’t spoil it smay

[prisma garnet]

I knew about injections, I didn't know about surjects tho

[prisma garnet]

a counterexample to what?

[hidden crag]

Injection/surjection induces injection/surjection

[prisma garnet]

oh

wait, there's a quotient map from D^2 to S^1 too

ig those are the counterexamples

the injection S^1 to D^2 and the quotient map

[hidden crag]

Quotient map from D^2 to S^1?

[gentle girder]

where does 0 get sent here

[hidden crag]

Actually the counterexample for surjectivity is pretty tough

[prisma garnet]

yea, just colapse the disk to a line and then line to circle

quotient map

[gentle girder]

doesn’t work i don’t think

how are you collapsing this disk

[prisma garnet]

why not?

[prisma garnet]

like, think of it as the unit square instead, and then use the relation (x, y) ~ (x', y') iff x=x'

[gentle girder]

oh yeah that is a counterexample innit

i was confused by you calling it a quotient map lol

[prisma garnet]

another way to think about it is to consider the projection onto the x-axis, which is surjective and continuous. Hence by the closed map lemma it's a quotient map

or, well, not another way but like, I included more details lmfao

[hidden crag]

Yes that works

[hidden crag]

I was thinking of something different here

I had a MSE answer in mind that had additional assumptions

What was even the original question

I lost track

[gentle girder]

it was about the LES in homology

[hidden crag]

Oh

[gentle girder]

about proving exactness at Hn(X,A)

[red yoke]

Long snek go brrrr

[hidden crag]

It’s done by „homological lemma that was invented for this exact thing“

[gentle girder]

you can just do the diagram chase

[hidden crag]

Bit of a classic in AT

[prisma garnet]

best kind of homological lemma

[white oxide]

Back to relative homology huh

Does Hatcher prove the zigzag lemma?

[red yoke]

I think so

[hidden crag]

Now that topology people are assembled I should take that opportunity as well

[gentle girder]

to prove the zigzag lemma?

[hidden crag]

[red yoke]

Zig-zag be spectral

[hidden crag]

i've been struggling with the proof that the Serre specseq has the properties that make it useful

[gentle girder]

i am going to be honest idk anything about spectral sequences, what properties are you trying to prove specifically

[hidden crag]

i'll send a specific question in a bit

the only hard part is to prove that the second page looks the way it does

[white oxide]

Oh.. I think I left my stove on my kids football game, bye

[hidden crag]

LMFAOO

[unreal stratus]

I thought you legally had to blackbox this theorem

Semi-joking

hm

[hidden crag]

if i don't get it by today i'm doing that

[unreal stratus]

fair lol

it's some cubical homology stuff right

[hidden crag]

tf is cubical homology

[unreal stratus]

lol

[gentle girder]

is there a proof in hatcher spectral sequences (hatcher mentioned 🤮)

[hidden crag]

that's what i'm reading yeah

[unreal stratus]

Okay I just googled to check, yeah basically the original thing uses cubical chains

rather than singular chains

Lol

Pretty sure it is just defined in the analogous way with cubes instad of simplicies

[hidden crag]

hatchers proof doesn't really involve explicit chains

but there's some weird pullback fibration thing with details swept under the rug

or maybe i'm just tripping

[unreal stratus]

Oh sure lol

a friend's advisor recommended they read the original serre aha

[hidden crag]

first thing when googling pullback fibration is a question about this exact part of hatcher lmao

[unreal stratus]

best thesis of all time possibly

[hidden crag]

so did mine

[unreal stratus]

Oop

[gentle girder]

god i know i’m gonna have to read this shit at some point

[hidden crag]

why though

[merry geode]

Cubical homology sounds so cute

[hidden crag]

what do you do smay

[gentle girder]

i want to be able to think about/gain an appreciation for stable homotopy theory

[white oxide]

i skipped the chapter about spectral seq in one book, i wasnt feeling like migraine inducing sudoku just yet. Worst case you can do the same and blackbox it

[hidden crag]

you don't really need specseqs for that

for computations in stable htpy however...

[gentle girder]

i would like to be able to do basic computations

[hidden crag]

how much homotopy theory have you done smay

[gentle girder]

idk why

not too much actually

i know what a spectrum is

[hidden crag]

oh i meant like

basic theory of unstable ones like in hatcher

but you're probably done with that then

[gentle girder]

no :)

[hidden crag]

oh okay

[gentle girder]

for some reason i simultaneously think homotopy is really cool and still don’t know anything about it

[hidden crag]

that's probably the first step to appreciation of stable phenomena then

unstable ones suck and stable ones suck less

[hidden crag]

i was like this as well lol

i didn't do some of the basic theory as thorough as i should have to get to stable things faster and i def don't recommend doing it like this

[unreal stratus]

homotopy

[gentle girder]

don’t call me topy!!

i am a bit down on my AT self-esteem because i had a bad semester tbqh

[unreal stratus]

;-;

i am stressed about AT/htpy theory because of grad apps

[hidden crag]

I suck at it and keep doing it anyway

[unreal stratus]

mood

[white oxide]

why?

[unreal stratus]

Wdym lol

Grad apps are scary

[hidden crag]

Yeah they show you how warm it is

[unreal stratus]

wdym lol

Oh okay lol

german joke nice

[merry geode]

You guys are not even graduate student yet so knowledgeable, wow

[hidden crag]

[unreal stratus]

bruder muss los

[white oxide]

a bit outdated but nice

[unreal stratus]

yeah

haven't seen the meme in like 7 years

lol

and i assume was already v old then

[white oxide]

7?

i thought that was like 3 years ago 😭

[hidden crag]

kerr are you german

[white oxide]

like the saying is common as well I thought you meant the pic of the man equipped in fierce german military equipment (pedalos)

yeah

[gentle girder]

yeah i basically struggled through chapters 1 and 2 and I’m going to have to review them again, i did not get a good grade in it at the end because i just sort of didn’t submit homeworks, some of them that i’ve done and some that i didn’t, so moving on from it is going to just be incredibly difficult

[unreal stratus]

Maybe im wrong

Timo back me up

[gentle girder]

kartoffel

[unreal stratus]

Maybe it is only like 5

[hidden crag]

it's not THAT old yeah

[unreal stratus]

But I remember it before I started uni and I'm in 4th year

Lol

[hidden crag]

wikipedia says 2018

[white oxide]

WIKIPEDIA?

[hidden crag]

no fucking way the german wikipedia page for pedalo has a reference to this meme

[gentle girder]

lol

[hidden crag]

HAHAHAHAH

[unreal stratus]

I found it from 2017 in fact

On ich_iel

[white oxide]

why is it formulated to casually 😭

[unreal stratus]

Anyway this is off topic so I'm banning Kerr and Timo

[white oxide]

:..(

wait you brought it up

mods disemvowel them

[hidden crag]

Turning them into Püree as we speak

[white oxide]

I was feeling kartoffel puffer actually

[hidden crag]

Doable

[unreal stratus]

Bruder muss los...

[hidden crag]

We should do german hours sometimes here

[white oxide]

what does that entail lmao

[hidden crag]

German

👍

[white oxide]

and hours? damn

[hidden crag]

How did you know

[white oxide]

i have my ways

[hidden crag]

Kerr are you in bachelor master or phd

[unreal stratus]

high school

[white oxide]

bachelor

[white oxide]

if only 😔

wait nvm hs sucks ass

[unreal stratus]

masters in uk is basically a bachelor by continental europe standards and a phd by american standard

[white oxide]

lol

[hidden crag]

Is the grade very important

[white oxide]

i shocked some US people a few days ago by telling that the currently running 3rd sem algebra I course ironically puts the five lemma in as homework

[gentle girder]

i am an undergrad so the grade matters a little bit

[hidden crag]

You seemed to understand it well when I saw you talking about it

[gentle girder]

but i do feel as though the understanding matters

much more

[white oxide]

like its not super hard, but why tf do they need to learn that

[hidden crag]

Based

[gentle girder]

so i feel like i will end up moving on

[hidden crag]

My 3rd semester alg course had that as hw as well lol

[hidden crag]

Definitely

You’ll be good to go AT wise

[white oxide]

yeah mine didnt but that one was probably much more of a mess

[hidden crag]

But there really is no reason to do it whatsoever

[white oxide]

what reason is there for stuff outside the mandatory courses

if they like it they like it

[hidden crag]

My LA 2 course did TPs of modules, universal properties of kernels, cokernels and even units and counits etc for algebras

The prof was quite something

[white oxide]

i was always happy to see some universal uproperties

did you just say units and counits

the adjoint functor ones

[hidden crag]

No

Actually unit and counit might not be the right translation

[white oxide]

actually why did you cover the tensor product of modules in LA

why would you spare the time even introducing modules

[hidden crag]

We were done with all the standard LA stuff after 3 weeks of LA2

[white oxide]

So it was LA 1.5/2 and Alg 0.8

[hidden crag]

Lmao kind of

[white oxide]

Least busted german math course

[hidden crag]

Exam ended up being very easy though

Prof was really just interested in teaching some more interesting stuff

[white oxide]

Oh that kind of prof

[prisma garnet]

[white oxide]

We have one of those, actually really nice

[gentle girder]

i think the snake lemma is actually really nice

ngl

[hidden crag]

Is it

[white oxide]

A bit rude honestly

[hidden crag]

Bit rude innit mate

[unreal stratus]

chewsday innit

[umbral panther]

The snake lemma is a Hollywood star

[heady skiff]

in arbitrary topological spaces is it true that algebraic combinations of continuous functions are necessarily continuous

oh wait this doesnt make sense since the codomain doesn't have to be equipped with a binary operation

[opaque scroll]

Composition of continuous functions is continuous, so if you're in a setting like a topological group or topological ring were the operations are continuous, then it's true.

[heady skiff]

oh yeah that makes sense

i was just wondering if it was true for arbitrary topological spaces but now i see why that doesn't hold

thank you

[heady skiff]

could i get a hint for this question pls?

[hexed steppe]

try showing that {(y,z) : y <= z} is a closed subset of Y x Y

[heady skiff]

okay i'll try that thanks

[tribal palm]

what is an algebraic combination of functions?

[heady skiff]

just like the sum, product, quotient etc. of two continuous functions

[white oxide]

functions into R?

[heady skiff]

oh nah i was considering them into arbitrary topological spaces but then i realized that that notion doesn't make sense

[white oxide]

good luck with that

[heady skiff]

lol

[frank thistle]

I'll try this one (I know 0 about topology)

[heady skiff]

crazy

[keen marten]

Every ordered set with the order topology is a hausdorff space

This would help you i guess

[heady skiff]

i think i got a solution, i'm writing it down rn and i'll type it up

i'm not really sure if it works/their might be some gaps but at least it's something

[frank thistle]

I read the page for "order typology" making great progress

oh right I totally forgot closed means something else for sets than functions

[heady skiff]

I hope this works: Let $V \coloneqq {(a, b) \in Y \times Y \mid a \leq b}$. We show that $\overline{V}$ is open. Let any $c \in \overline{V}$, so that $c = (c_1, c_2)$ and $c_1 > c_2$. If we let $U = (c_2, \infty)$, $V = (-\infty, c_1)$, $c = (c_1, c_2) \in U \times V \subseteq \overline{V}$ since $c_1 < c_2$. Hence $\overline{V}$ is open, so $V$ is closed. Now the function $h: X \to Y \times Y$ given by $h(x) = (f(x), g(x))$ is continuous since both $f$ and $g$ are continuous. Since $V$ is closed, $h^{-1}(V) = {x \mid f(x) \leq g(x)}$ is closed in $X$, as desired.

[gentle osprey]

okeyokay

[keen marten]

This is correct i guess

[obtuse meteor]

note for other old ppl like me: please don't use Vbar to denote complement it will make ppl cry :)

[quiet thorn]

closure is open

[plain raven]

i denote closure by $V^c$, where $c$ stands for "closure"

[gentle osprey]

diligentClerk

[gentle girder]

based

[unreal stratus]

how is this even old lol

[obtuse meteor]

I think young ppl are too computer science brained

[unreal stratus]

oh bruh

[obtuse meteor]

so they use Vbar

[unreal stratus]

Lol

Gross

[ebon galleon]

I do not like bar for complement

[thorny agate]

^

[unreal stratus]

Honestly I just always write X \ U or whatever

[ebon galleon]

Yeah

[thorny agate]

^c is good

[unreal stratus]

Otherwise you have to remember what it's inside or it can be unclear etc idk

[thorny agate]

but I also just write the complement explicitly

[unreal stratus]

lol

[ebon galleon]

I don't particularly like ^c either, but better than bar imo

[unreal stratus]

One thing i find funny is limit points like

There was a time when X' for me meant limit points but it feels more useful to just reserve it as useful notation to talk about spaces X, X', X'', etc

lol

[thorny agate]

X' for limit points is nuts

sorry

[ebon galleon]

I've seen that

[unreal stratus]

seems p standard right?

[thorny agate]

I've never seen that

[unreal stratus]

Idk what the standard notation would be outside that

[ebon galleon]

Not very uncommon, at least in a first course

[unreal stratus]

Oh lol

[thorny agate]

then again I never took a first course

[unreal stratus]

But yeah honestly idk when the last time i cared about limit points was...

[thorny agate]

I self studied

[unreal stratus]

I guess in pointset it is more important lol

[ebon galleon]

It's funny

[unreal stratus]

or analysis

[thorny agate]

and Bredon I don't think used a specific notation for limit points

[ebon galleon]

In my topology course a lot of people usually went first to limit points when talking about closed sets

I usually avoided using limit points

[thorny agate]

oh huh

[unreal stratus]

cursed

[thorny agate]

I've only seen "closed is complement of open"

and the limit points characterization was a proposition

[unreal stratus]

also limit point is sort of terrible naming

but yh

[ebon galleon]

Tbf, we did the converse I think. Has all limit points = closed

[unreal stratus]

rudin chap 2 has limit points as part of the def

which is annoying

lol

[white oxide]

the proof looks very similar in the end

[ebon galleon]

But this is how I think of them, so maybe that's part of it

[white oxide]

show that any point in the complement has a nbhd not intersecting your set

[thorny agate]

ofc

it's an iff after all lol

[white oxide]

fair lol

I think i like it a little bit more over limit points. You usually do the same argument when it comes to showing some map is cts

[ebon galleon]

It was also interesting to see how frequently people would use contradiction or try to use induction when it wasn't necessary

[unreal stratus]

Induct on the cardinality fo the topological space

[knotty vine]

induction is always necessary

that was my inner computer scientist coming out

[unreal stratus]

Lol

[knotty vine]

funny name

[white oxide]

unnecessary contradiction as in "scratch first and final line to get constructive proof" ?

[ebon galleon]

That type of stuff yeah

[unreal stratus]

lol

[gentle girder]

y'all are so afraid of setminus

\

[merry geode]

Yea it's quite an easy shortcut for proofs

While verbose, it shows you a clear way ahead to do the proof.

[heady skiff]

it looks as if i've started a civil war

anyways yeah writing \ is a lot better the five day break i took from math fucked with my head

i apologize.

[obtuse meteor]

There’s a technique called “induction on the real numbers” that somehow ppl’s are so much more likely to use bc it has a name

I think names are important for students. As mathematicians we have a sort of toolbox of ideas and ways to prove things. But sometimes these have bad names

Proof by induction and proof by contradiction are attempted so much bc they have good names :)

Granted, the proof technique in ug point set topology is more often than not “write down the definitions and do a search on the proof tree”

It just so happens that your intuition can prune the proof tree really fast if you’re familiar with the topology on R and some counter examples :3c

[merry geode]

Do you compute the possibilities on the proof tree

[obtuse meteor]

No but my brain does!

[gentle osprey]

Leno

[merry geode]

Oh wow, this sounds similar with a problem I had regarding local maximum

[gentle osprey]

Leno

[gentle osprey]

Leno

[merry geode]

An easy starting point of connectedness is to assume disconnected

And show by contradiction

Where did you fail?

Could you show what you have tried

Ahh, I see

Oh, you did not use compactness in exam?

Hm yea that would not work

What do you mean by "individual element inverse set", fibers are connected by the problem assumption

[rancid umbra]

they meant fibers there

[merry geode]

I mean that does not make sense, we already know fibers are connected in this settings because it is literally part of the settings

[rancid umbra]

i don’t think this claim is true

[merry geode]

Which claim?

Ah

I mean you would be able to form such a cover, but yea wouldn't be easy

It could be easier to try the case C = Y first.

[rancid umbra]

f is a quotient map (since it is closed - this is crucial), and the fibers are the equivalence classes. if you take C in Y connected, then f^{-1}(C) is a disjoint union of connected fibers.

Suppose for contradiction that there is a non-empty proper subset C’ of f^{-1}(C) that is both open and closed in f^{-1}(C).

There are two cases:
there is a y in C such that f^{-1}(y) meets C’ and the complement of C’ in f^{-1}(C)
or
every fiber is contained in C’ or the complement of C’ in f^{-1}(C)

show that neither case can occur

[merry geode]

Eh that's like mostly giving the answer imo

[rancid umbra]

eh, i think the first two paragraphs were mostly set up.
the cases kind of fall out on their own but that’s all i really gave them

[merry geode]

I think figuring out that

1. The map is closed, and
2. It is relevant to the problem
   is the hard part

For a person who is learning topology in intro analysis

After what you gave, basically what is left is just filling in the details.

[grim knot]

Can somebody give me some intuition about what a $\Delta-complex$ is?

[gentle osprey]

damn_guuurl

[knotty vine]

Thats Hatcher, right?

[grim knot]

it might be, cause my professor uses it as a reference

[hidden crag]

Read some of hatchers chap 2

I think there’s a paragraph for the intuition

[knotty vine]

Actually, is there a difference between a delta complex and a triangulation?

[hidden crag]

No idea, I never bothered too much with delta complexes

Don’t like those

I think Hatcher mentions that delta ones are more efficient for computations sometimes?

Since there’s Computer assisted machinery for simplicial stuff that might be relevant

[knotty vine]

Oh wait I think a Delta complex is a simplicial set without degeneracies

So for example, the circle constructed as a single point with a single line attached is a Delta complex, but not a simplicial complex

Otoh, a sphere as a single point and a single disk attached is a simplicial set, but not a Delta complex

[grim knot]

we constructed our whole alg top on delta complexes, so I was trying to undersand those

[red yoke]

Delta complexes are just simplicial complexes with orientation preserving boundary maps

[knotty vine]

Not quite

See my previous message

[red yoke]

Oh right

[grim knot]

My big problem is that we begin defining delta complex, then simplicial complex and as a last step CW-complexes

And I don't get delta complexes

[red yoke]

But why do we need delta complexes when we have CW complexes

[hidden crag]

Computations

[red yoke]

Are there delta complexes that are not CW complexes

[knotty vine]

No

[red yoke]

Simplicial and cellular homology are computed in the same way I believe?

Oh maybe it's for the exercises in 2.1 since celullar is in 2.2

[knotty vine]

Delta complexes provide a nicer represenation than CW complexes because each n-simplex must attach precisely to (n-1)-simplexes

While a CW complex can attach willy nilly

[grim knot]

(i still can't visualize what a delta complex is)

[knotty vine]

Thats kind of odd because delta complexes are more general (in some sense) than simplicial complexes, while CW complexes are more general than both

[knotty vine]

Ill write you an intuition when I get on a real keyboard if you still need it then

[grim knot]

I will probably still need it hahaha, thank you very much and sorry for bothering

[hidden crag]

Im not sure how the machinery works for cellular things

Especially with computer assistance

Like I’m not talking about MV, LES etc

[red yoke]

Ig the point of preserving orientation is that you don't have to care about signs during computation

[knotty vine]

There are four notions of "spaces that are built up recursively by dimension from basic building blocks" that are popular and under consideration here (there are many more but that's another story):

• simplicial complexes (family of sets closed under subsets),
• Delta complexes (Hatcher, i,e, presheaves on Delta_inj),
• simplicial sets (presheaves on Delta), and
• CW complexes.
  You don't need to worry about simplicial sets, I'm just fixing the names cause they're confusing and people called (and call) them different things in rather unfortunate ways. Another thing: these (except for CW complexes) are usually abstract: they aren't actually topological spaces yet, but we can make them so using "geometric realization". Sometimes people use these names for the abstract things, sometimes for topological spaces that are homeomorphic to the geometric realization of one. We don't really need to worry about this (except for simplicial sets, the geometric realization is obvious), as long as you're aware.

First, the reason why we want to work with these things and not just topological spaces, is that we want some kind of "combinatorial" representation of space that we can analyze, "algebraicize", and compute with. Plain topological spaces are intractable in this regard: they are huge (their set of points is usually uncountable, let alone their set of opens) and they provide relatively little structure (imo: the least amount of structure any notion of space could have, then again that's also their strength).

I assume you have some intuition for simplicial complexes. Their concepting idea is as follows: (n+1) (linearly) independent points in Euclidean space define an n-simplex (using the convex hull or barycentric coordinates or something) and they have the nice property that any subset of those points again defines a simplex: a face of the original simplex. We abstract away from concrete points in Euclidean space and just say that an n-simplex is a set of size (n+1). Then we can find its subsimplices (i.e. faces) by just taking subsets. Now a simplicial complex is just a family of sets (i.e. a subset of some powerset) which is closed under taking subsets.

For the purposes of algebraic topology, there are two problems with this notion:

• We might like to construct the circle as consisting of only a single point (i.e. 0-simplex), with a single line segment (i.e. 1-simplex) that's attached from both ends to the single point. All of our notions of algebraic invariant work for this space (Euler characteristic, homology, with a little more work (the same amount of work we would need for any simplicial complex) homotopy too) so our notion of space should be able to represent it. But a simplicial complex can't represent it (of course it can represent a circle, but it needs at least 3 points and 3 lines for it). The problem is that representing faces as subsets is too restricting: we want a more general way to represent the face relation.
• For homology, we need some kind of ordering on the vertices or orientation on the simplices: there is no ordering on the faces of a simplex but we need that to construct the boundary operator on the chain complex. (We could impose an ordering on a simplicial complex to fix this, but that may seem a bit arbitrary)

Delta complexes kill both birds with one stone: instead of using subsets for the face relation, just list the faces of each simplex explicitly. That is: for every n-simplex we have an ordered list of length (n+1) consisting of (n-1)-simplices which form the faces of the original n-simplex. We no longer need a simplex to be some subset of vertices; a simplex is now just some kind of identifier together with its dimension. We now need an extra requirement: the face relation must be "transitive" in the right way (that's technically not the right word but w/e). For simplicial complexes this is baked in: A \subset B \subset C implies A \subset C. For Delta complexes we need to enforce this: the face list of an n-simplex only lists the (n-1)-dimensional faces so we need some equation that tells us about faces of faces. Since this is only for motivation purposes, I will not spell out this equation... (Hatcher uses a trick so he doesn't have to spell it out).

This face list idea fixes both problems:

• Since we have a list of faces, one face may occur multiple times. This way we can form the circle from our original problem: a 0-simplex v and a 1-simplex e whose face list is [v, v]. (0-Simplices "usually" don't have faces so they don't have a face list. We could allow (-1)-simplices and this would arrive at an interesting notion but that's another story)
• Since we have a list of faces, the faces are ordered.

I'm not quite sure which specific definition of Delta complex you have seen, though it is equivalent to the notion I described here. If you have a question how your definition lines up with mine: please provide your definition and I will tell you.
There is another problem with Delta complexes in the same vein as the first problem with simplicial complexes which leads to simplicial sets. So, if anyone is interesting in this, I can provide a similar motivation, but I will leave it here for now.