point-set-topology

[empty grove]

Manifold here might not include second countability

[gritty widget]

is pi1 ever uncountable?

and why not?

i never thought about this

why cant pi1 be Real

[empty grove]

Uncountable wedge of circles

[empty grove]

It can be any group

All groups can be realised as fundamental groups

[gritty widget]

ok so im not crazy

i probably read his statement without context

also isnt that the eilenberg maclane proof

[empty grove]

I think they're talking about second countable hausdorff manifolds in particular

[gritty widget]

that there exists a cw complex

with pi1 being G for any G

[empty grove]

Yes you can always get a 2 dimensional CW complex

[gritty widget]

is the construction simple?

[empty grove]

There are many

[gritty widget]

is there a canonical one?

[empty grove]

Should be many natural ones idk

But there is one I like

[gritty widget]

like say D12

is the group

[empty grove]

You write G = F/N for free groups F and N

And then take a graph that realises F

We can do this because we know how to realise free groups

[gritty widget]

when you mean realized?

like cayley graph?

[empty grove]

Pi 1 = F

Can take wedge of as many circles as rank of F too

[gritty widget]

so like wedging a circle for each generator of F

ok

[empty grove]

Ye

[gritty widget]

oh

and relations hmm

[empty grove]

Then there's a cover of this corresponding to the subgroup N

By the Galois correspondence

[gritty widget]

yeah

there is a 1-1 correspondence between covering spaces and subgroups of G

[empty grove]

So you have f: E → X a covering map from pi 1 = N to pi 1 = F

[gritty widget]

normal subgroups*

[empty grove]

Then the mapping cylinder of f has fundamental group G

Can be checked easily from van kampen

[gritty widget]

fuck

is mapping cylinder a map from E to X I

i never learned what is was useful for

[empty grove]

Lol

It's a space

Just see definition

That is enough for this proof

[gritty widget]

ok ty

[empty grove]

But there are more elementary (harder to prove) constructions in Hatcher

Hatcher gives 2

[gritty widget]

i had suspicions that it was like suspension

[empty grove]

[gritty widget]

it isnt at all

but it that was my suspicion which was wrong

[empty grove]

Lol

[gritty widget]

also moldi i need advice

are you into algebraic Nt by any chance

[empty grove]

No lol

Det is

[gritty widget]

what are you in grad school for?

[empty grove]

I'm doing an MSc, still haven't decided

[gritty widget]

very cool

[empty grove]

🙈

[gritty widget]

does this mean you are doing topic other than math?

or is msc just the name of degree

[empty grove]

Nope

Ye

[gritty widget]

o nice

what are some topics you are learning about rn?

or books

[empty grove]

I was reading model theory and AT before my semester started but then I was hit with my busiest semester ever

[gritty widget]

o

why model theory

[empty grove]

I plan to continue those now that vacations are almost here 😌

[gritty widget]

ive heard it talked about so often here

im wondering why

[empty grove]

it cool

[gritty widget]

gimmie an example

[empty grove]

[gritty widget]

like what motivated you to learn it

[empty grove]

You can prove that every first order theory which has ∀ ∃ axiomatization has direct limits, which proves this for groups, rings, modules and a lot of other things at once

I find stuff like this cool

[empty grove]

It might be ∃ ∀ axiomatization tho

[gritty widget]

lol order matters🤷‍♀️

[empty grove]

Ye I don't remember I just remember it being very cool

That includes fields

So must be the first order

[gritty widget]

i can see how existence of direct limits is cool

ive always found limits to be super cool when used

[empty grove]

And Ax Grothendieck and Nullstellensatz are commonly cited applications

Ax Grothendieck was first proven using model theory

[gritty widget]

so is it just logic applied to bigger algebraic objects?

ax grothendick

[empty grove]

You can apply it to any class of objects that is first order axiomatised

[empty grove]

Yes

[gritty widget]

o

it applies to AG

fuck

[empty grove]

Yes

[gritty widget]

i still dont know what a variety or scheme are in general

this sucks

[empty grove]

Variety is like manifold but instead of smooth functions we care about algebraic functions

And then you can throw away R and C and use any field

[gritty widget]

i just have understanding that varietys are solution sets to things but idk exactly what, i only know definition in terms of solution sets to system of polynomials for affine algebraic variety

but then i see shit like abelian variety

[empty grove]

😵‍💫

[gritty widget]

and it makes me question wtf

[empty grove]

Sad

(also same)

[gritty widget]

many such cases of internet learning

[empty grove]

many such cases

[gritty widget]

we gotta stick to the books

[empty grove]

[gritty widget]

to be off topic

what is motivation for higher homotopy groups because I always percieved homotopy invariance as only tool cared about

why in particular is pi2 important

or pin

i thought i knew

because of this homomorphism

huerwochz

or something

[empty grove]

I have no clue 🙈

[gritty widget]

but idk why it is even true

i just know there is this thing called heurswixz homomorphism between homology groups and homotopy groups

but idk why anyone cares about it even

i suspect it makes calculations for homology groups easier?

but it probably doesn’t

[empty grove]

Probably because homology groups lose some of the information and you can use that to quantify amount of information lost

😵‍💫

[gritty widget]

o mayb

[long hornet]

I thought covers were always connected

[empty grove]

No

I haven't seen that included in definitions

Might be in some books idk

[shy moss]

Count the holes of a space

[shy moss]

Under some connected conditions hurewickz homomorfism gives you an isomorphism, so it help you calculate homotopy groups

[gritty widget]

yes

but why do we care about higher homotopy groups

doesnt homology count holes with betti numbers

[pearl holly]

higher homotopy groups counting holes

I guess AT want's to "classify" maps up to homotopy [X, Y] but this is hard so we can at least try to do it when X = S^n

[shy moss]

also homotopy groups classify spaces up to homotopy equivalence better than homology

Whitehead theorem tells you that if a map induces isomorphism in homotopy groups then this map is a homotopy equivalence

[shy moss]

Assuming the space are CW complex

[gritty widget]

makes sense ig?

i always did find that weird

my intuition for what holes meant in homotopy was better than in homologu

oh wtf ty

oh great

i can only understand vocabulary in homotopy theory section

im lost on the other two

its 😎

[still harness]

what's a really accessible introductory topology online course?

[stone cipher]

In book that I follow, it includes second countability

How do I prove the transitivity of the action?

[stone cipher]

my attempt:

Let $x\in X$. By surjectivity of $q$ there is $e\in E$ such that $q(e)=x$, so $e\in q^{-1}(x)$. For any loop $f$, there is a unique lift $\tilde{f}$ such that $\tilde{f}(0)=e$. So we define the function $\beta:\pi_1(X,x)\to q^{-1}(x)$ to be $\beta([f])=\tilde{f}(1)$.

[gentle osprey]

RaD0N

[stone cipher]

Now, I want to prove that, for any $e'\in q^{-1}(x)$ there is a loop $f$ and a lift $\tilde{f}$ such that $\beta([f])=\tilde{f}(1)=e'$.

[gentle osprey]

RaD0N

[stone cipher]

I note that, by construction of $\beta$, we always have $\tilde{f}(0)=e$.

[gentle osprey]

RaD0N

[stone cipher]

What theorem can provide me a lift of $f$ such that $\tilde{f}(0)=e$ and $\tilde{f}(1)=e'$?

[gentle osprey]

RaD0N

[stone cipher]

OK...................................................

https://math.stackexchange.com/questions/1074341/covering-map-is-proper-iff-it-is-finite-sheeted

If $q:E\to X$ a covering map is proper, then it is finite-sheeted. I don't understand why the fibers are discrete spaces?

[gentle osprey]

RaD0N

[fading vale]

The fibers of any covering map are discrete in general

because the fiber is a disjoint union of separated points

if you want to show it explicitly take a point b in B, and a U around b such that p^{-1}(U) is the disjoint union of open V_i, each containing some point x_i with p(x_i) = b

since p restricts to a homeomorphism on each V_i, if V_i contains two points x, y in the fiber of b, then x and y are equal

so basically each point of the fiber of b is contained in one and only one of the V_i

so p^{-1}(b) cap V_i = x_i, and thus x_i is open in p^{-1}(b) with the subspace topology

hence its discrete

(side note: this is kind of a defining property of a covering map, if you drop the requirement that the fibers of discrete and just say that the total space is locally a product, you get something called a fiber bundle which is a really important construction in its own right)

slim sotruing me... seethe

[stone cipher]

Ok. I feel kinda dumb, because I knew that and I didn't identify the fibers being discrete space itself

thanks anyways

I knew that there weren't 2 elements in each open set..

[clear storm]

Question : Suppose that $(X,A)$ is a relative CW-complex (meaning that the zeroth skelton is $A$ for some Haussdorf top space $A$) and $A\subset Y \subset$ subspace such that $Y\cap Xn\setminus X_{n-1}$ is the homoeomorphic image of characteristics maps for all $n$. (So this set is homeomorphic to open n cells). I need to show that the filtration $Y\cap X_n$ defines a CW complex structure on $Y$.

[gentle osprey]

Sei

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

[clear storm]

the commutative diagram didnt render properly so here it is

[stone cipher]

is a covering map closed?

[empty grove]

I think any local homeomorphism should be closed

because you can check whether a set is closed locally

like if you have an open cover {U_i} of X, then V is closed in X iff each V intersection U_i is closed in U_i

Then cover the domain of the covering map with neighbourhoods that it maps homeomoprhically onto open sets of the covered space

And then use the fact that homeomorphisms are closed

[stone cipher]

thx. I will see it later

[stone cipher]

https://math.stackexchange.com/questions/1074341/covering-map-is-proper-iff-it-is-finite-sheeted?rq=1

If q is finite sheeted, then q is covering map. I can't understand the proof. I've been torturing myself for hours to understand. I feel so dumb these days.. never felt that way

[long hornet]

It's okay! But you probably shouldn't do that - it's much more useful to take a break.

[surreal ocean]

in munkres topology a linear order is defined as being 1) comparable/complete 2) nonreflexive and 3) transitive. This differs from every other definition of a linear order i have seen. Wikipedia defines a linear order as being 1) comparable/complete 2) reflexive 3) transitive and 4) antisymmetric. Why does munkres use the alternate definition?

[surreal ocean]

ok thanks

[stone cipher]

Thx for advice, but I can't do that yet. The semester isn't over :|

[still harness]

bump

[spiral stratus]

Quick dumb question: why is, for a real VB, the pullback a ring iso on cohomology?

(p:E->B the VB ofc)

[tight agate]

contract all the fibers

[spiral stratus]

locally thats clear to me, but how can i see it works globally?

[tight agate]

try writing it down

[spiral stratus]

is there a homotopy equivalence in general?

[tight agate]

wdym in general

[spiral stratus]

nvm i see what you mean

yeah theres no issue "globally"

thanks

[tight agate]

[pearl holly]

Suppose I have $H^*(X; \mathbb{Z}/p) = \mathbb{Z}/p[\alpha]$ with $\alpha$ a generator of degree $n$ such that $\alpha^2 \neq 0$. If $p$ is a prime, why must $n$ be even?

[gentle osprey]

Kanga Gang Swagger Tokidoki ✓

[pearl holly]

I forgot so say that \alpha^2 is not 0 as well

[dusk heron]

Let's say that we have a complex smooth projective variety $V\subseteq\mathbb{C}P^3$, defined by a polynomial of degree $k$. The second integral cohomology $H^2(\mathbb{C}P^3)$ is generated by the Poincaré dual of the fundamental class of a copy of $\mathbb{C}P^1$ in $\mathbb{C}P^3$ (right?); let us denote this cohomology class by $a\in H^2(\mathbb{C}P^3)$ If $\iota:V\hookrightarrow\mathbb{C}P^3$ denotes the inclusion, then how does the pullback $\iota^*a$ relate to a generator of $H^2(V)$? My guess is that $\iota^*a$ is equal to $k$ times such a generator, but I'm not sure.

[gentle osprey]

gustavn64

[pearl holly]

isn't it the case that when p is 2, alpha has a degree that is a power of 2?

I'm sorry but I still don't get it lmao, I'm extremely tired

oh lmao I see

Thank you so much!

ye I think that if alpha is not a power of 2, then the Steenrod square Sq^n(\alpha) = \alpha^2 would decompose and that would yield a contradiction since all Sq^t(\alpha) =0 for 0<t<n are zero, so \alpha^2 would also be zero

the exact same argument can be made to show that a map S^(2n-1) --> S^n with Hopf invariant 1 can only exist when n is a power of 2 which is really cool

If i is not a power of two, you can look at the Adem relations for i = a + 2^k. Then you can use a theorem that says that \binom(m)(n) is congruent to the product of all \binom(m_i, n_i) mod p where m_i and n_i are the coefficients in the p-adic expansions of m and n. So when p = 2 you get that the coefficient in the first term in the Adem relation is 1

this is iff, so Sq^i is indecomposable iff i is a power of 2. For the other direction you have that i is a power of 2. Now you can use the fact that the total Steenrod square is a homomorphism and by Freshmans dream you get that Sq( \alpha^i) = \alpha^i + \alpha^(2i), so Sq^t(\alpha^i) must be zero unless t = 0 or i. So if it were to decompose, we would get that \alpha^(2i) is zero

[obtuse meteor]

steenrod squares are ridiculous

[gritty widget]

whats the motivation?

for steenrod algebras

[glossy pine]

probably to make your life hell

[gritty widget]

I was reading through a proof that said that since a certain subset of a lie group was zariski closed, it must have measure 0 unless it is the whole space

that wasn't the claim of the proof, it was just a detail that was briefly mentioned as a step

is that a trivial result

to be clear, I can understand this for zariski closed subsets of Rⁿ but measure 0 in Rⁿ doesn't necessarily mean much for subsets of lower dimension

like SO3 is measure 0 in R⁹ but it's not measure 0 in itself hahah

[honest narwhal]

So to state it carefully

Let G be a Lie group... I guess actually a linear algebraic group? And let X be a Zariski-closed subset of G

Then unless X is all of G, it is measure zero... probably with respect to the Haar measure on G

The heuristic should be that if you're Zariski-closed, you're "at least a dimension lower"

[gritty widget]

yeah that's about it

I'm solving polynomial equations in SO3² btw

I'm accepting nowhere dense as well as zariski closed (nowhere dense with the usual toplogy, not zariski)

[honest narwhal]

What's the particular situation?

[gritty widget]

I have a word w(a,b) in the free group on a and b

I want the solutions to w(A,B)=I in SO3×SO3 to be measure 0

unless w is the empty word of course

w(A,B) is given as polynomials in entires of A and B

I just wish I could say "this set feels pretty measure 0 to me" 😔😔

[honest narwhal]

To be precise, ${(A,B)\in SO(3)\times SO(3) \mid w(A,B) = I}$?

[gentle osprey]

several people

[gritty widget]

yes

that set, for a particular word w, is measure 0

I have a loose understanding for why I want it to be true

but the details are mismatched

like the set of solutions is obviously closed in SO3

[honest narwhal]

Are you trying to prove Banach-Tarski?

Like oh countably many words

So there's a copy of F_2

In SO(3)

[gritty widget]

😳

[honest narwhal]

Not sure if that face means I called you tf out or if you're like huh

[gritty widget]

you called me tf out

[honest narwhal]

:)

[gritty widget]

but i want to prove the cool version where most pairs of rotations generate a free group

[honest narwhal]

Yeah tru I didn't even know this proof was a thing tbh

I just see that and I'm like wait up.... :0

My undergrad MT class on our first pset had us show that SO(3) contains a copy of F_2 and ooooo boi was that something

I'd like this proof a lot more

I wonder if we can Sard theorem this 🤔

[gritty widget]

"something" meaning gross?

[honest narwhal]

None of us could figure out anything reasonable so we kept looking through Tao writings until we found something that kinda did the trick

[gritty widget]

hahaha

[honest narwhal]

Basically you choose specific candidate matrices and somehow passed to like, F_5^3

\mathbb{F}_5^3

Finite field not 5 element free group that'd be scuffed

[gritty widget]

i don't even see how that makes things free 😭

[digital peak]

wouldn't be surprised if it contained F_n for all n

wait

[honest narwhal]

F_2 contains all F_n

[gritty widget]

it does

[digital peak]

thta's trivial

yeah

[gritty widget]

we just need the zariski proof 🙏🙏

[honest narwhal]

Actually now that I think about it

To apply your proof you still need to know SO(3) contains one copy of F_2

Since you wanna make sure it's not all of SO(3)

[gritty widget]

also yes this sounds like sort of related to sards theorem but I can't see how it applies

[gritty widget]

I just need one pair for every word that doesn't satisfy it

the hard part is words where the sum of powers of a and b are both 0

[honest narwhal]

Oh oh tru

[gritty widget]

if non commuting rotations satisfy this then I'm fucked

but I hope just like commutators there's sort of a "if and only if they commute" sort of deal with those words

this sort of feels like trying to prove most reals are transcendental over Q

except that proof is easy

but like imagine if you needed to find one transcendental number in order to prove that

like proving a particular number is transcendental is hard

[honest narwhal]

Hmm

So right now we want a measure theory argument

What about a Baire category argument?

[gritty widget]

I tried seeing if using the double cover SU2->SO3 could give me anything (it didn't)

[honest narwhal]

Maybe that's easier

[gritty widget]

wouldn't that just apply to when I'm taking the union over all words

[honest narwhal]

I mean the idea is we say solutions to w(A,B) = I have dense complement rather than being measure zero

[gritty widget]

yeah that's good enough for me

[honest narwhal]

Honestly I'm fairly convinced that proper + Zariski-closed does imply measure zero anyway

[gritty widget]

you're right, my brain just went to measure 0 implies dense complement but only dense complement is still enough

[gritty widget]

but how to prove 😭😭

[gritty widget]

if there's a neiguborhood of points all satisfying this equation, then could we use some wacky analytic extension thing

like SO3² is a manifold

so we could treat the polynomial as an analytic equation in R⁶

and then if it's 0 on a subset with nonenpty interior, it's zero everywhere

[gritty widget]

locally

[honest narwhal]

Hmmmmmmmm

So Zariski closed doesn't mean smooth submanifold

But we can sorta "induct"

I guess

[gritty widget]

I can't show that analytic functions in R¹⁸ necessarily give analyticity when you restrict to a 6 dimensional sub manifold 😔

[gritty widget]

wait are you kidding

how could it be zariski closed and not smootj

[honest narwhal]

You need a condition on the rank of the jacobian

xy = 0 in R^2

[gritty widget]

oh omg

you're right

[honest narwhal]

I don't think this will be a problem for us

[gritty widget]

yeah I say who cares

[honest narwhal]

Because you can say oh take the smooth points

And then the non-smooth points are a subvariety

So that's even smaller

etc

[gritty widget]

but also shouldn't I just care that SO3² itself is smooth??

[honest narwhal]

Well what I'm saying is, I want to reduce the statement about Zariski-closed => measure zero to a statement "submanifold of strictly smaller dimension => measure zero"

[gritty widget]

yeah the hard part is the first half of that

why it would be strictly smaller dimension

[gritty widget]

hey we can actually use sard for this. the problem points are negligible

[honest narwhal]

Tru tru

So yeah basically you're saying something like

Also tbh we should go to #diff-geo-diff-top

Also holds for difftop 😛

[gritty widget]

yeah let's move

[honest narwhal]

Good opportunity to clear our heads a bit

[digital peak]

so I'm a little confused

so if we have a pair of chains, we can talk about a chain map

(e.g. realizing a chain is a functor from a particular indexing category, and then chain maps are natural transformations)

then we have homotopies between chain maps

but say e.g. the "uniqueness of projective resolutions" mentions a "chain homotopy"

what's that about?

chain homotopy equivalence? or what

[empty grove]

Yes it is uniqueness up to chain homotopy equivalence

Because the identity map of modules gets resolved to identity up to chain homotopy

something like that

[digital peak]

ok that makes sense to me

one day I'll prove that proper but black boxing works for now

[crude vector]

Is it true that a subset of a topology must be equal to one of either its interior or its closure?

[empty grove]

That would be equivalent to saying that all subsets are either open or closed

[crude vector]

I see, so false then, right?

[empty grove]

ye

[crude vector]

Many thanks, gentleman

[empty grove]

[rigid gorge]

This is one of the first qs in my topology book so not sure if it should be asked here, but it's Prove that the set of all finite subsets of a countable set is countable.

So it suffices to consider the positive integers. My idea was that to every finite subset, we'll put the elements in increasing order, and then concatenate them to associate to it a natural number. However, this fails for repeat digits e.g {9, 99} and {999} will be the same. So to fix it (after a hint) what I did was instead of immediately concatenating I replaced all the commas with 0's e.g {9, 99} becomes 9099 instead. But I'm not sure whether this works, and if it does, how to prove injectivity?

[gentle osprey]

Nobody

[rigid gorge]

nice, that's clever

i presume there's a way to do this without using \mathbb{N} at all

[finite heath]

[empty grove]

[finite heath]

my man's on demon time folks!

🐴 🤝 🐻‍❄️

[empty grove]

what

[jolly garden]

Dumb question: Is $\langle e \mid -\rangle$ a presentation of the trivial group?

[gentle osprey]

Anton.

[empty grove]

I suppose, but why in the topology channel

Well it should not have e in the generators

No generators

[jolly garden]

Sorry, it has to do with algebraic topology though. I'm trying to figure out the Seifert-van-Kampen theorem.

Hmm? "e" refers to the identity element.

Yeah, I forgot to mention that. It is standard notation, however.

[empty grove]

You can write < e | e > if you want

You have to say that e is the identity in the relations if you want it to be the identity

[jolly garden]

A presentation of Z is given by <s|-> where s is in {1,-1}.

[empty grove]

No

Remove the last part

s can be anything

[jolly garden]

😳

[jolly garden]

Alright, thanks

Okay, thank you

[velvet finch]

P4 from Rotman AT chap 1 asks you to show Brouwer's fixed point thm for spaces homeomorphic to D^n

That is, if X = D^n, show any cont f: X -> X has a fixed point

The proof of Brouwer's given in the book makes use of a lemma that says there is no retraction from D^n to S^n-1, and then assuming f: D^n -> D^n has no fixed points, you can define a retraction from D^n to S^n-1 by mapping x to the intersection of S^n-1 and the ray from x to f(x)

Now, if there is a retraction r from X to dX then I think you can give a retraction from D^n to S^n-1 by taking g o r o g^-1 where g is the homeomorphism between X and D^n

But my question is how do you construct such a retraction from X to dX

[shy moss]

dX is the boundary of X?

[velvet finch]

Yeah

If you have a retraction r from X to its boundary then using the homeomorphism g: X -> D^n you can obtain a retraction from D^n to its boundary

In the form of g o r o g^-1

I think

And this is a contradiction ofc

[severe frigate]

you can construct the retract from X to dX by mapping over to D^n, doing the ray thing there, and mapping back

[velvet finch]

Oh lol

Thanks

[severe frigate]

but there's a shorter way to argue tho

a map X -> X gives you a map D^n -> D^n and extract the fixed point from there

[velvet finch]

Yeah true lol

[supple sable]

any help here?

[rancid umbra]

is it true that if i have a continuous surjection f : X —> Y, then X and Y have the same number of connected components?

[shy moss]

no

[rancid umbra]

ce?

oh wait

it was literally staring me in the face

the problem i’m working on rn is a ce

lmao

[shy moss]

The function that collapses two o circles to a point

[rancid umbra]

what weaker conditions do you need to preserve the number of components other than homeomorphism?

[shy moss]

Homotopy equivalence

[rancid umbra]

oh

going to look up what that is now

[shy moss]

but also homotopy equivalence is so much stronger than have the same number of connected components

Homotopy equivalence implies weak homotopy equivalence and this implies have the same pi_0

[rancid umbra]

hmm. well, i wanted to show that there is no continuous injection from A = (R x 0) U (0 x R) to R. i was going to argue by using connected components.

but i don’t think that works now

like. removing the origin from A produces a set with four connected components and its image then has two connected components

wait. maybe it does work. how does this sound. if i have a continuous bijection f : X —> Y, then f^-1(f(U)) = U and f(f^-1(V)) = V for all U subset X and V subset Y. this gives a bijection from the set of connected components of X to the set of connected components of Y

[rancid umbra]

nvm : (

[velvet finch]

Problem 5 from Rotman Chap 1 gives f, g: I -> I^2 with f starting/ending at y=0/y=1 resp and g starting/ending at x=0/x=1 resp. I'm asked to show the paths intersect, and the idea is to use Brouwer's

The question is, what should the map from I^2 to I^2 be

A hint would be appreciated, I honestly have no clue

[tight agate]

take some sort of difference but normalize it so that it lands back in I^2

I remember doing a similar problem but I dont remember the exact function

hmmm

can we assume f(t)_x > g(s)_x always

no

ignore that

okay

we can assume one of f(t)_x - g(s)_x and f(t)_y - g(s)_y is nonzero

okay if they dont intersect

then f(s) is never equal to g(t)

taking rays from f(s) to g(t) defines something from I^2 to the boundary of I^2

but this isnt a retraction

[tight agate]

hmm

we can define a function

[gentle osprey]

Kanga Gang Consigliere Brtogi

Kanga Gang Consigliere Brtogi

Kanga Gang Consigliere Brtogi

Kanga Gang Consigliere Brtogi

[tight agate]

we have a surjection to {L,R} x {Top, Bottom} iff the curves intersect

blech

this doesnt do anything

@velvet finch did you figure it out

oh fuck this works

yeah you can show that it's always surjective by arguing with endpoints/points where the curves touch y = 0,1 x = 0,1

nvm

[tight agate]

alright define \varphi and \phi as above

claim: I^2 ---> {L,R} x {T,B} is surjective

proof:

pick t s.t. g(t)_x = 0 and g(t)_y is max (among things with g(s)_x = 0)

then \varphi(g(t)) = left as 0 is less than everything

\phi(g(t)) = top due to the max condition

so (left, top) is in the image

now pick t s.t. g(t)_x = 1 and g(t)_y is max (among things with g(s)_x = 1)

If \varphi(g(t)) = left, then 1 \leq min{f(s)_x : f(s)_y = g(t)_y}, which implies that the curves intersect (we're assuming that they dont)

therefore \varphi(g(t)) = right

we have \phi(g(t)) = top

so (right, top) is in the image

using a similar argument for f, we can show that the other two things are in the image

so the map is surjective

the logic so far is: Curves do not intersect implies map is surjecive

but if the curves do not intersect, they bound some region (you probably need the jordan curve theorem here), and you can show that the function (\varphi x \phi) is constant on this region

not really the region bound by the curves, but you should be able to define two new curves which only keep track of "forward" or "upward" progress using max and min as above

these probably still dont intersect

this should give you three regions, and the function defined above is constant on each region

which should give you a contradiction

as the image of the function has 4 things

[empty grove]

There's also an easy way to do this using Jordan curve theorem

You can connect the endpoints of one of the paths from outside the square and then prove that for the other path, one of the endpoints lies in the unbounded component and the other doesn't

But you have to first reduce the problem to one where you can apply the theorem because that needs an injective curve

And I think you can do that by zorn's lemma

Cutting out loops of f until no more can be cut out

[tight agate]

I dont think you need to cut out loops

you can just ignore backward progress

oh nvm

I was thinking of doing something like replace g with t mapsto (max_{s \leq t}{g(s)_x}, g(t)_y )

and do a similar thing for f (but upward instead of forward)

but yeah nvm this doesnt work

unless we can show that f and g don't intersect implies the replacements don't intersect

[empty grove]

But they may not go back along themselves, and need not be monotonic

[tight agate]

I dont mean back along themselves

I mean back in the sense that the x coord decreases

[empty grove]

Oh then you'd have to change the path entirely

Rather than taking a sub path

[tight agate]

.

[empty grove]

I see

[tight agate]

this isnt even continuous lol

no

it is

it is

I think

I am go eat

brb

ping me if you figure it out

[empty grove]

Ye it is

[empty grove]

You can do this though, right? If a value is repeated, then take the first and the last time that it is hit by the path, and make the path constant between them. Call this operation O(s) where s is the repeated value. Then we can take the poset of all repeated values, with s > u if the inverse image of u is in the convex hull of the inverse image of s. Then given a chain, you can argue that the supremum of that chain, S, is also in the poset by continuity, so that's an upper bound. This means that's there's some repeated value v such that when you do O(v) the function no longer has any non constant loops. And we can easily then eliminate this one constant subpath

[tight agate]

this doesnt eliminate all the loops

right?

what about something like this

[empty grove]

Ok I think this works, for the Brouwer approach, not sure if this is what brofib did
Take I² → I² defined by
(s,t) ↦ (s,t) + f(s) - g(t)
But this may not land in I², so just compose with the standard retraction ℝ² → I² after this. Continuity is obvious, and then Brouwer's theorem says this has a fixed point. If the fixed point is on the interior then we are done because then it must have come from the formula itself. On the boundary you'd have to argue some more because the retraction messes with things, but in my head it works out

[tight agate]

[\begin{tikzcd}
& \bullet & \bullet \
\bullet && \bullet && \bullet & \bullet \
& \bullet &&&& \bullet \
&&&& \bullet
\arrow[from=2-1, to=2-3]
\arrow[from=2-3, to=1-3]
\arrow[from=1-3, to=1-2]
\arrow[from=1-2, to=3-2]
\arrow[from=3-2, to=3-6]
\arrow[from=3-6, to=2-6]
\arrow[from=2-6, to=2-5]
\arrow[from=2-5, to=4-5]
\end{tikzcd}]

[gentle osprey]

Kanga Gang Consigliere Brtogi

[empty grove]

Ah right

Then maybe apply Zorn's lemma again to get a maximal set of maximal repeating values

[empty grove]

The retraction I'm thinking of here is the one that projects the area outside orthogonally onto a side, and everything that can't be projected like this onto the corresponding corner

No that doesn't work

Wait maybe it does

[tight agate]

f-g/|f-g| gives you something that lands in the box

|f-g| is never zero by assumption

[empty grove]

Yeah but that won't have the property of fixed point → intersection

[empty grove]

This does

[tight agate]

yeah I think im done for now lol

[empty grove]

Suppose the fixed point is on the boundary. Then let's suppose it's because the first coordinate is 0, so the point is (0,y). Then g(0) has x coordinate 0, so the image of this is R((0, y) + f(y)), so this being the same as (0,y) means that f(y) has x coordinate negative or 0, so must be 0, so this must be the intersection

And then argue similarly for all 4 edges and corners

There should be a more convenient normalisation

But I'm pretty sure this will get the result

Actually I'm not sure how this argument would work on the opposite edge

[tight agate]

what if you do your function but divide by 3

move I^2 s.t. it's [-1,1] x [-1, 1]

[empty grove]

Fixed points wouldn't correspond to intersections though

[tight agate]

oh yeah

[empty grove]

Wait can we say that fixed points also get divided by 3 or something

Nah doesn't seem right

[tight agate]

what about the sup norm

[empty grove]

What about it

[tight agate]

nvm

[empty grove]

[tight agate]

yeah I dont see how to do this with brouwer

I think I can make my up/down/left/right thing work with some more work

but it still uses jordan curve

[empty grove]

[dim meadow]

You can use $h(x,y)=(1-||f(x)-g(y)||/\sqrt 2)(x,y)$

[gentle osprey]

'quid

[dim meadow]

The fixed points should be exactly the points where f(x)=g(y)

[empty grove]

Aren't you dividing by √2 if f(x) = g(y)

Oh division is only on the norm

[dim meadow]

Yeah

[empty grove]

Nice makes sense

@tight agate @velvet finch lol

[limpid leaf]

Liquid is so good

[tight agate]

🗿

[empty grove]

🚱

[finite heath]

yo what

how did u type this!

[\begin{tikzcd}
& \bullet & \bullet
\bullet && \bullet && \bullet & \bullet
& \bullet &&&& \bullet
&&&& \bullet
\arrow[from=2-1, to=2-3]
\arrow[from=2-3, to=1-3]
\arrow[from=1-3, to=1-2]
\arrow[from=1-2, to=3-2]
\arrow[from=3-2, to=3-6]
\arrow[from=3-6, to=2-6]
\arrow[from=2-6, to=2-5]
\arrow[from=2-5, to=4-5]
\end{tikzcd}]

[gentle osprey]

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

[finite heath]

[empty grove]

Lol discord formatting hides some of those symbols

\\ becomes \

[finite heath]

\

\\

lol

(╯°□°）╯︵ ┻━┻

┬─┬ ノ( ゜-゜ノ)

[still harness]

\

huh

[finite heath]

i dont care for your discord bio.

[velvet finch]

Ah I see, yeah I figured it was gonna be some norm related expression in f and g

I just couldn't figure out what

[velvet finch]

I did something like this initially

Because I was thinking of the proof given for Brouwer's when n=1

Or no what am I thinking of

No that's right

And then you argue that if the graph of f doesn't pass through the line, it's disconnected

Oops I didn't send my message

So you draw I^2 with a diagonal y=x line

[velvet finch]

No I went to sleep at 11

Went to bed

Fell asleep

Fell to bed

[velvet finch]

Rotman Ch1 P6: A in Mat_nxn(R+), show it has an eigenvector in D^2

Does it not suffice to just define g: D^2 -> D^2 given by g = cA for a sufficiently small c

Or well not D^2 but that's the case I had in my head

Then g(x) = x by Brouwer's gives Ax = 1/c x

Explicitly, you can take c = 1/max{||Ax|| : x in D^2}

The hint given defines a linear functional giving a suitable scalar for every x in D^2 but I don't get why just taking a single sufficiently maximal scalar wouldn't work

[crude vector]

Is every metric space a topological space?

[glossy pine]

yes

but the converse is not true

if a topology has a metric, it is a hausdorff topological space

[crude vector]

But see, a definition like this is confusing me, since there's no mention of topologies

Is it just implied?

[glossy pine]

eh

there doesn't really need to be a mention of topologies until later perhaps

an introduction to metric spaces uses the motivation of literally distance, doesn't really need to talk about topologies right off the bat

[crude vector]

Ah, I see

[glossy pine]

a connection can be made last

later*

[crude vector]

Yeah I suppose for like an analysis book or something, it doesn't need to dive that deep

[glossy pine]

doesn't that book discuss topology a bit

[dim meadow]

There is a natural topology for metric spaces

[glossy pine]

yes

[dim meadow]

But saying that metric spaces are topological spaces is a bit off I think

Like morally it's true

[glossy pine]

mathematical morality

[crude vector]

They define some natural topology with open balls or something, was that it

I read about that

[dim meadow]

Yeah

[crude vector]

I see, and sets with that topology are what we call metric spaces?

[lunar yoke]

no metric spaces come together with a certain metric

different metrics could induce the same topological space

like in R^n you can look at the euclidean norm or the 1-norm if you know these

and they both induce metrics that are also different

but if you consider the topology generated by open balls then they yield the same

[crude vector]

Ohhh I see

[lunar yoke]

this is nice to visualize since you can place the balls in one norm into balls of the other norm

[crude vector]

So how do we distinguish between them, we say they generate they different metric spaces?

Or well, not generate

I don't know if we use that word here

[lunar yoke]

like this

[orchid forge]

a metric space is just a set with a metric. there is a natural topology associated to any metric space

[lunar yoke]

well metric spaces are just sets together with a metric

[slender burrow]

if you use the open set definition of a topological space you can use that to prove every metric space is a topological space. it goes something like defining open sets in your metric space as neighborhoods i believe. I think usually books have a proof on this somewhere?

[lunar yoke]

and if the metrics are different then the metric spaces are different i guess

[crude vector]

Oh, so we don't actually use the topological space part in the definition, that comes afterwards?

[lunar yoke]

yes

[crude vector]

Okay okay, now I get it

[lunar yoke]

if this is a book on topology, then its probably just there to give an example later. since metric spaces yield rather tame topological spaces and you usually have a good intuition for them

unlike many other wild topological spaces

[crude vector]

Yeah, I'm treasuring this feeling of being able to visualize examples, R^2 is just lovely

In any case, thank you all a bunch, I get it now

[empty grove]

I would say it's not

Completeness is not preserved under homeomorphisms and that's an important property of metric spaces

Actually idk what you mean by morally it's true so nvm

[orchid forge]

there's a natural topology on metric spaecs

which means it's fine to think of metric spaces as being topological spaces

although technically the topology is not part of the definition

[empty grove]

I interpreted that as there is no more structure

[orchid forge]

they're not just topological spaces yeah

[empty grove]

Because existence of a natural topology was mentioned before

[pearl holly]

I don't know if this question makes sense, but say that I have something "nice" in a cohomology theory (say the cup product on singular cohomology). How would one generalize this "nice thing" to an arbitrary cohomology theory?

is there a standard way of just generalizing something or does it depend on what you want to generalize?

[late iron]

This semester I took an undergraduate intro to topology course, the course covered both point-set and algebraic topology, I am fairly confident with the point-set part but have messed up my education in algebraic topology.

Is there any recommendation on how to self study the topic over the winter? Thank you!

[spiral stratus]

I am currently working with Fumenko-Fuchs' "homotopical algebraic topology"
it's got basically all you'd want to know on the basics i think, from homotopy groups, knots, surfaces, fibrations, CW structures, (co) homology, UCT, spectral sequences and also some K-theory etc further back
i think it's quite nice and the presentation is fairly elementary
But there are some typos thrown in so you got to be aware

Also there are really beautiful illustrations in there

they're unrelated but inspired by the content of the book and i found them to be a good source of motivation while reading :)

[late iron]

Oh thank you so much! Is it beginner friendly to someone like me who’s only background of alg top is basically just fundamental groups + van kampen?

And some covering maps etc.

[spiral stratus]

yeah i think it def is

it starts only assuming point set and not even group theory, really

and it goes fairly in depth on fund groups, vK, covering theory too

so you'll find anything you're missing on the way

it seems very self contained if nothing else

it doesn't stress categorical language too much
that can be a good or a bad thing depending on your level but it's usually fairly apparent what's a twisted product etc when you see it in the text

one example for the typos:
in this proof that every map has a h-equiv strong serre fibration, the last line says phi = s when it should say = x
But from context it's pretty much always clear whats meant, i felt

[swift fjord]

My prof called this 'block.decomposition' but I can't find anything about it. The idea is for a simplicial.complex K and a subcomplex L, you take a sequence of increasing subcomplexes
L=M0\subseteq M1\subseteq.... Mn=K, such that Hr(Mk,Mk-1)=0 for r \neq k. You then construct maps from d:Hk(Mk,Mk-1)->Hk-1(Mk-1,Mk-2) which satisfy d^2=0, and apparently the homology of the resulting chain complex is isomorphic to Hk(K,L)

We haven't proven this uet but this was the general idea he gave

[late iron]

Thank you so much for your recommendation!

[slender burrow]

@spiral stratus I was actually looking for a similar recommendation this book looks exactly what i was looking for so thanks too

[late iron]

Just making sure:

Is this the right link?

https://link.springer.com/book/10.1007/978-3-319-23488-5

[spiral stratus]

yes

thats the one

i quite literally found this on the floor
no idea how i would have discovered it otherwise

[gritty widget]

does the fundamental group functor have a right adjoint

I'm guessing no but I don't remember the exact details. I thought there were sum spaces whose wedge sum had a fundamental group other than the free product

[empty grove]

No, it does not preserve all colimits

svk tells you one kind of colimit that it preserves

if it had a right adjoint svk would be trivial

[gritty widget]

yeah that's what gave me the idea

[empty grove]

There are counterexamples when you remove any one of the hypotheses in svk

[gritty widget]

I mean it could have a right adjoint that's much more hard to describe than van kampen is to prove.....

[gritty widget]

this was my idea

[empty grove]

Like S^1 when you remove connectedness of the intersections

[gritty widget]

oh yeah

[empty grove]

would be kinda insane 😵‍💫

[gritty widget]

I mean we know it's false

[empty grove]

ye

I mean if something like that were true

svk has so many hypotheses

cover by open sets, path connected triple intersections and even then it only talks about pushouts

[gritty widget]

yeah 😔😔

you're right

ok well like

that's all still reasoble in a universe where is t had a right adjoint

[empty grove]

ye lol

[gritty widget]

let's pretend it has the most complicated right adjoint ever

then it's fine to give a weak svk

and then the bug fancy people can say "we actually know a stronger version of this but it's too hard"

it's not like people always give you the most general version of every theorem

[empty grove]

Ye but that seems very far fetched though we'd at least have a version for preserving coproducts at least

But coproducts aren't preserved

Wait are they preserved by the fundamental groupoid functor

It looks like it

[gritty widget]

yeah that's what my doubt was but I couldn't remember a specific pair of spaces where it fails (I think something like the hawaiian earing or something)

[gritty widget]

[empty grove]

You need existence of a neighborhood of the basepoint that deformation retracts to the basepoint

Then coproduct is preserved

So ye Hawaiian earring should give a counterexample

[gritty widget]

nice

[empty grove]

But I don't see it immediately at least

Is earring ∨ earring a counterexample?

[gritty widget]

yeah I think

[empty grove]

All I know about the fundamental group of the earring is that it's uncountable

I would be willing to believe that earring ∨ earring is homeomorphic to earring lol

[gritty widget]

me too

[empty grove]

😌

[gritty widget]

but the question is whether it's fubdamental group is it's own free product with itself

[empty grove]

Ye

[gritty widget]

is that even possible for groups

I wouldn't be surprised

[empty grove]

Ye

Free group on infinitely many generators

[gritty widget]

oh yeah

I was just gonna say 😬

[empty grove]

😩

[gritty widget]

anyways on a slightly related note does C2*C3 contain a free for up

free group*

and no I don't count Z 🙄🙄

[empty grove]

The trivial group is free

[gritty widget]

💀

[gritty widget]

yeah I think so

like it just has to

like come on

[empty grove]

Seems too small ngl

Can probably use covering space theory for this

Want to prove that figure 8 or something with that fundamental group can't cover the wedge of some wacky quotients of a sphere

[gritty widget]

ok take 1•1 and 1•2 where the first number is in C2 and the second is in C3

[empty grove]

😵‍💫

[gritty widget]

those have to be independent don't tbey

[gritty widget]

does this sound easier to you?

[empty grove]

Isn't their sum 0

Wait

[gritty widget]

no

[empty grove]

Free product 😵‍💫

[empty grove]

Nah

[gritty widget]

in the free abelian product 🙄🙄

[empty grove]

Ye lol

[gritty widget]

throught so ..

anyways there's this long chain I'm working on

I want a free subgroup of SO3

so I take the double cover SU2

then the complexification SL2(C)

then the subgrouo SL2(Z)

then the quotient PSL2(Z)

then that's the free product C2*C3

then I find a free subgroup

and somehow that gives me a free subgroup of SO3 😬😬

[empty grove]

I am sorry

[gritty widget]

sorry for what HAHA

[empty grove]

For what you are having to do

[gritty widget]

hahahah I am "choosing" to do this

but honestly I don't see how free subgroups of SL2(C) actually give me free subgroups of SU2

something about how if a word in matrices A and B evaluates to 1 uniformly on SU2 then the same must hold in SL2(C) by analytic something???

sounds right to me I just cant prove it

I was trying to figure this out by christmas 😳😳

⏱️⏱️

[empty grove]

Why

[gritty widget]

because I feel like that's a satisfying deadline

well actually the reason is Im making this writing collection of something interesting every day from Dec 1 to 25

and I want this to make it in 😭😭😭

[gritty widget]

do you understand this?

this is actually the step I'm most stuck on

[empty grove]

No clue what that is

[gritty widget]

😔😔😔

I actually don't need to do all of tbus

for every word in the free group I need to show some pair of rotations doesn't satisfy it

I feel like a good way to do that is to show there's a free group

but there may be some other way to force a non-solution to a given word..

like who knows maybe every single pair of rotations has A²B^-7A³B⁷A^-5=1

[empty grove]

Guess every equation and check 😌

[gritty widget]

yesssss

[neon swift]

hi, can someone tell me what's the closure of this {(x,y)∈R2;x>0} ?
also why

{(x,y)∈R2∣x2+y2<2}∩{(x,y)∈R2∣(x−1)2+y2>1}      is equal
= ({(x;y)∈R2;x2+y2⩽2}∩{(x;y)∈R2|(x−1)2+y2=1})∪({(x;y)∈R2|x2+y2=2}∩{(x;y)∈R2|(x−1)2+y2≥1}.)```

[empty grove]

closure of that in R^2 is the same set but with > replaced with >=

[neon swift]

how did u know this

[empty grove]

The y axis is exactly the set of limit points

[west spindle]

okay so i have this problem and i want to ensure my thought process for it is correct

oh

im interrupting

aren't i

[neon swift]

oh so you drew the image in your head, so if this was a bit more complex set how would you find its closure ?

or is it not possible ?

[empty grove]

Drawing the image is usually the first step in any point set topology problem unless you are really comfortable with some other methods

If the set if too complicated you try to understand it locally

in the sense that you look at a small region around that point and see if you can say anything

and maybe use some theorems

but idk any general methods

[neon swift]

alright alright thanks

[empty grove]

@west spindle

[west spindle]

oh i can post mine now?

aight so
i have this quotient space of the unit disk obtained by identifying each point x with -x. and i need to find the fundamental group

i think the space is homeo to the plain unit disk and thus the fund group is trivial

the idea is that it's "naturally" homeo to the lateral surface of a finite cone

cause you can take half of the disk as a fundamental domain and then glue it together into a cone, which you can then flatten

does this hold any water?

or did i do a big dum dum somewhere

[empty grove]

I think this is homeomorphic to RP^2 so that seems incorrect

[west spindle]

how tho

[empty grove]

wait we aren't just identifying boundary points

[west spindle]

no we aren't

[empty grove]

😵‍💫

[west spindle]

we're identifying opposite points within the whole disk

[empty grove]

Yeah then homeomorphic to the disk itself seems legit, I think the complex squaring map should produce the homeomorphism

Like embed the disk into C

(I didn't get your reasoning)

[empty grove]

then D^2 to D^2 you have a squaring map

That factors through the quotient

[west spindle]

the disk is alr embedded in C

[empty grove]

and the factored map is a homeomorphism because all is compact hausdorff

oh ok

ok I think I got what you are doing

same homeomorphism as the one I gave

Assuming all the stretching that you do in yours is sort of uniform

[west spindle]

im not stretching anything rly

but ok

[empty grove]

ye true

I mean it could be like the squaring map but rotated by something or whatever

[vague brook]

Apparently if you order the vertices

that gives an ordering of the faces as well

this is supposed to be a tetrahedron drawing btw

but like what would the ordering of the faces be here, given the ordering of the vertices

[lunar yoke]

if you have a k-dimensional simplex with an ordering on the vertices, you get a canonical ordering of the faces by saying the n-th face is the one opposite to the n-th vertex

not sure if thats what you want

[vague brook]

oh cool

[lunar yoke]

"opposite of nth vertex" is formalized as the unique face not including the nth-vertex as any of its vertices

[vague brook]

So would the ordering say of the edges on the {0,1,3} face be [{1,3},{0,3},{0,1}]

[lunar yoke]

yes

[vague brook]

ok thanks

[hexed holly]

Has anyone here read and understood May's book Concise Course in Algebraic Topology?

I have a question from p117 of the PDF

This part

I don't understand how to prove the corollary
I want to show that a certain diagram commutes, where there are two maps d
Clearly I am supposed to use the fact that d: E_q(X,A) -> E_{q-1}(A) is a natural transformation
For that I would want to find a map (X,A) -> (CA, A) that induces the composition of the topological differential d_star and Sigma^-1 when applied homology functor E
And I want the map to be identity on A
But I can't find any such map, and I doubt if one even exists

Or maybe I'm looking at it the wrong way.

Since this is generalized homology, all we know about the boundary map is that if f: (X,A) -> (Y,B) is a map of pairs of spaces, the following diagram commutes

Any help is appreciated

@wooden falcon@cedar pebble@pearl holly@quartz edge@flint cove

I pinged the above people because they claimed to have read the book or recommended it to others

@tight agate

Yes we have a sequence A -> X -> Ci -> SA -> SX -> SCi -> ...

I don't think that's relevant here

We only care about the boundary map X/A -> Ci -> SA here

[flint cove]

I have been absent from this server for months I'm amazed that anybody remembers me

reads backlog

[hexed holly]

I just searched for people who talked about the book and found you

[flint cove]

it's alright lol was a pleasant surprise
can't promise to be helpful tho because I didn't even manage to get a good intuition for what a cofibration is

[flint cove]

@hexed holly I'm afraid you will have to help me. What's the „topological boundary map“ X/A→ΣX? The reduced suspension is only defined for pointed spaces, right? so what's the base point of X here

Ah, given at the end of p108

Ohhh I cannot read, we map into the suspension of A

[hexed holly]

@wooden falconDo you actually know how to prove it or was it just a guess? If the former, could you be more specific about what I need to use from that chapter?

[pearl holly]

I think this is described here https://ncatlab.org/nlab/show/mapping+cone#HomologyExactSequencesAndFiberSequences but I barely know what’s going on lmao sorry

[hexed holly]

Yeah this is not helpful at all

The proof is supposed to be a "trivial diagram chase"

I just want to see it

[unborn lotus]

hi friends

i am going to ask very basic questions about pointset in here

pls forgive me

wait

no oops

wrong channle

where is the normal one

wtf

oh wait

it says point set is ok here

ok

ok here we go

Let $X$ be a topological space. Suppose that $A$ and $B$ are open subsets of $X$ with $\overline{A} = X = \overline{B}$. Prove that $\overline{A\cap B} = X$.

[gentle osprey]

Kanga Gang Yakuza Yung cofe

[unborn lotus]

idk how to think about this tbh

i know that X is closed so that the closure of A cap B is going to be contained in X

that's fine

but the other direction

brain not working...

hmm a point in X is going to be a point in A closure and also in B closure

means that every open set containing x intersects nontrivially with A and also intersects nontrivially with B

oh

ok

thats it

wait where am i supposed to use the hypothesis that A and B are open

🤨

im doing something wrong

[tawdry valve]

you don't quite get it from that

just because N cap A and N cap B are nonempty doesn't mean that N cap A cap B is nonempty

[unborn lotus]

oh ic

ok

[tawdry valve]

if you drop the open requirement you can actually get a counterexample

[unborn lotus]

oh

so intersection of the neighborhood containing x with A need not actually contain x

same with B

and so it's possible for the N cap A cap B to be empty

but i need to find out how A and B being open make it nonempty

[tawdry valve]

mhm

[hexed holly]

if U is a nonempty open set, then it intersects A, thus U \cap A is a nonempty open set, thus it intersects B, so U \cap A \cap B is nonempty

[unborn lotus]

hmm sure it might be nonempty but can i choose U so that the resulting intersection contains an arbitrary element x of X

i see how you are using the openness to get nonempty

wait

let me think

oh

yeah i see

ok

hmm

but is x in U cap A

aghh im very confused

[tawdry valve]

so our goal is just to get U cap A cap B nonempty, we don't need it to have x

[unborn lotus]

oh

hmm

[tawdry valve]

icosahedron referred to a little lemma that if B is dense, then B cap V is nonempty for any nonempty open set V

[unborn lotus]

right

im not totally convinced on this but we don't need to worry about arbitary elements x in X because U is arbitrary?

[tawdry valve]

take U to be an open neighborhood of x, then U cap A is a non empty open set, but it need not contain x

[unborn lotus]

right

oh

ohhh

i see

it doesn't matter

what icosahedron said is indeed enough

it just has to intersect

because then U cap A is another nonempty open set, which will then intersect nontrivially with B since B is dense

so then overall any U containing x will intersect nontrivially with A cap B

so x is in the closure of A cap B

is this right

[tawdry valve]

yep!

[unborn lotus]

thank you so much both of you

❤️

[unborn lotus]

so im thinking for the closure of {1} it is Z_+ because there are no other closed sets containing 1, right? Similarly also for the closure of {2} the only closed sets containing 2 are those where its of the form Z_+ - {odd numbers and their divisors} and of course also Z_+

so for {2} its closure is the set of even numbers?

[hexed holly]

yes

[unborn lotus]

ok ty

[gritty widget]

does anyone know of a nice problem/ theorem whos proof uses induction on the CW complex

I forget what kind of properties carryover in these pushout squares

[shy moss]

The cellular approximation theorem

[gritty widget]

and another question

what do we actually mean when we write something like $H^{*+1}(X)$

[gentle osprey]

lime_soup

[gritty widget]

what I am particular confused about is, the cohomology ring $H^(X)$ always has a module structure over $H^(\text{pt})$

[gentle osprey]

lime_soup

[shy moss]

what is pt?

[gritty widget]

a point

[tight agate]

X ---> pt induces a map of rings H(pt) ----> H(X)

[gritty widget]

is there any place where i can get a lot of practice and exposure to different types of topological groups/rings

[gritty widget]

a book on lie theory will tell you something about some topological groups

[gritty widget]

sorry i meant

i am confused because

what happens to this module structure when we hvae

H^*+1(X)

i assume H^+1(pt) to H^+1(X) is the module structure

[gentle osprey]

Nobody

[gritty widget]

this is correct in general

My question is, is $H^{+1}(X)$ a module over $H^{+1}(pt)$ or $H^{*}(pt)$

[gentle osprey]

lime_soup

[gritty widget]

I think it should be $H^{*+1}(pt)$

[gentle osprey]

lime_soup

[gritty widget]

i think however that this is just spelling

[gritty widget]

this out

[hexed holly]

@wooden falconhi can you answer my question?

or @tight agate, you said you've read the book

If you don't know the answer that's ok too

To the person who reacted with "just ask", I already asked the question

Nobody has given me a hint to look at cofiber sequences, but I don't know if it's just a guess or if he knows how to do it

If the latter, I would really like him to show me

It's supposed to be an "easy diagram chase". If they've read the book and understood it, it shouldn't be hard to answer

I'm a noob so I can't see it

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

p117 of the PDF

I would really appreciate it if you showed me the proof

[hexed holly]

@wooden falconare you there?

[empty grove]

Nobody is there

[pearl holly]

yooo I'm so confused lmfao. D^n is just an n-cell, right? So the only chain group that is nontrivial is C_n(D^n),= Z, but that would imply that H_n(D^n) = Z

I literally don't see the mistake

[frosty sundial]

The n-cell has to be glued to something.

[pearl holly]

oh daim okay

[frosty sundial]

for example, the most common way to represent D^2 is

two points (north and south pole), two line segments connecting them (left side and right side)

and then one 2-cell in the middle

you can also do it with one point (north pole), one 1-cell making a loop, and then one 2-cell filling it in.

[pearl holly]

right right I see

so if I instead have like, say, S^n and I attach D^2n to it someway or another, woudl the cohomology in degree 2n be trivial?

[frosty sundial]

i'm not quite sure how you're doing the attaching

and the answer would be no

I think what you really want to do is attach D^n to S^(n-1)

[pearl holly]

how would I show that it's nontrivial tho?

[frosty sundial]

waht do you mean

[pearl holly]

when I say attach I mean that I have a map from the boundary of D^2n to S^n and I identify a point with its image btw

[frosty sundial]

i guess it doesnt really matter what the attaching map is

[pearl holly]

so after I glued my disk to my sphere in this way, how would I know that the homology in degree 2n is nontrivial?

[frosty sundial]

for n > 1 look at the chainc omplex

use your exact argument from before

you have a Z in degree 2n

and nothing in degrees 2n+1 or 2n-1

[pearl holly]

oh so NOW I can use that argument

right right I see

[frosty sundial]

the issue with your first argument

[pearl holly]

this completely screwed me over lmfao

[frosty sundial]

was just that your thing wasnt a valid chain complex

err