point-set-topology

[steel glen]

haha yeah fair enough

[tidal lynx]

Oh yeah also to my original problem, I just realized my solution produces a construction for such a convergent sequence

Though I’m pretty sure the proof of B-W does a construction anyway

[cursive tendon]

Let $S$ be a set, $x\in S$, when can we say that there exists an open set $U \subset S$, $U \ni x$? A sufficient condition seems to be $x \in int S$

[gentle osprey]

Mattuwu

[ebon galleon]

Necessary as well, by definition

[tidal lynx]

This is called U being a neighborhood of x. If x has no neighborhoods inside S then x certainly is not in the interior of S.

[ebon galleon]

x \in int S iff there is an open set U with x \in U \subseteq S. Or, this is also given by the fact that int S is the largest open set contained in S

[cursive tendon]

thanks

[runic sun]

This is not quite right, an increasing sequence might not exist. However an increasing or a decreasing has to. Then the argument works.

[tidal lynx]

@cursive tendon If you change to U \subsetneq S, then the statement only fails when S carries the indiscrete topology. Because the statement is then saying that “x has a neighborhood that isn’t the entire space”, and for any non-indiscrete topology this holds (otherwise the union of every open set that isn’t S would not be S)

[ebon galleon]

Also fails when taking discrete topology and S={x}

Or at least, is empty

[runic sun]

The discrete and indiscrete topologies on a singleton are the same.

[ebon galleon]

Of course

[tidal lynx]

I was still processing where I messed up until you said that entropy lmao

[runic sun]

This isn’t true, such spaces will suffice however for example in the Sierpiński space one point is in the interior of the whole underlying set but has no neighborhoods besides the whole set,

[tidal lynx]

I have no idea why I thought that all topologies should satisfy the union property I mentioned, wtf

If x is not an isolated point of S, then the statement holds.

I think this is right finally

Wait wait

[runic sun]

No, I don’t see how being isolated plays a role here. The Sierpiński example gives a counterexample to this too.

[ebon galleon]

You want essentially the opposite of isolated point

[tidal lynx]

What do we call points x that have no neighborhoods that aren’t the entire space

[ebon galleon]

Isolate means {x} open, here you want the only open set containing x to be the entire space

[tidal lynx]

I tried to intuit that the right word was isolated, but that’s wrong yea

[ebon galleon]

Bad

Worthless points

[tidal lynx]

NPCs

[ebon galleon]

Yeah idk what they're called lol

[runic sun]

I‘ve never heard of a term for them either.

[tidal lynx]

I’m calling them NPCs

[ebon galleon]

Topological indistinguishable, in some sense I guess

[tidal lynx]

Oh right, two points that satisfy the property are topologically indistinguishable

[cursive tendon]

Let S be an open set, is it true that \forall x \in S, \exists open U \subsetneq S, x \in U?

[ebon galleon]

Right, so this is essentially a "global" kind of indistinguishability

[ebon galleon]

Take U = S

[cursive tendon]

subsetneq though

[ebon galleon]

Ah, my b

[cursive tendon]

This seems true in Euclidean spaces?

when is it not true?

[ebon galleon]

Certainly for R^n

[tidal lynx]

Wait lol

This again only fails when the only neighborhood of x is S

[ebon galleon]

Not quite

[tidal lynx]

When S has NPCs

oh uhh

[cursive tendon]

NPCs

[ebon galleon]

Or when you consider subspace topology, yes

S has to be a minimal neighborhood of x for no such U to exist

[tidal lynx]

Oh I assumed S was the entire space

[ebon galleon]

In particular, this should happen iff $\bigcap_{U \in \mathcal{N}(x)} U \in \mathcal{N}(U)$

[gentle osprey]

Ryxiann

[cursive tendon]

like this topology of 3 points

[ebon galleon]

I.e. the intersection of all neighborhoods is a neighborhoods, and S is that neighborhood

[tidal lynx]

Wait what’s the notation N(U) supposed to mean

Don’t you just want = S instead of \in N(U), like you wrote in words in your last message

[ebon galleon]

*that's only possible if x has that property

Is what I meant, sorry

[tidal lynx]

Huh I’m lost

[ebon galleon]

x can only have a minimal neighborhood if that intersection is itself a neighborhood

[tidal lynx]

Oh, I see

right

[tidal lynx]

so was this a typo for \in N(x) then?

because I don’t see why the converse to your iff holds then; the minimal neighborhood could just be inside S

I think you want = S for there to be an iff

[ebon galleon]

It should = S if you want to interpret it in the context of the original problem.
What I meant to write was that x has a minimal neighborhood iff that intersection is a neighborhood

[tidal lynx]

Oh, I should’ve just read your messages in order

Why do we rarely talk about convergence in a topology?

What I mean is like, the generalization of continuity from the reals to topologies is super important, but I don’t really see that much attention given to the generalization for convergent sequences

[ebon galleon]

Sequences are not enough for general topological spaces

Spaces which we can characterize properties by sequences (at least, some properties like continuity, limit points, closures...) Are called first countable (see the current convo in #discussion lmao)

Any metric space (and in particular, R) is first countable, so that works

[tidal lynx]

Oh so is talking about convergence of filters pretty popular then

[ebon galleon]

But one can take convergence of other structures to be fundamental, and things work then. You can use either nets or filters for this

[tidal lynx]

in general topology

nets* I mean

[ebon galleon]

For some purposes yeah

Convergence spaces (which are a generalization of topological spaces) take convergence of filters to be fundamental, so some people do choose to characterize things lately on these terms

convergence spaces also suck tho

[viral atlas]

Oh true, my bad

[tidal lynx]

Wait, why did we ever need your sequence to be increasing or decreasing? Doesn’t B-W still apply?

Like, Ryxiann’s solution is literally that: pick a (possibly not monotonic) sequence and apply B-W.

[viral atlas]

True, I just had some line of thought in my mind that led me to consider monotonic sequences

It's not necessary

[tidal lynx]

When trying to prove that every infinite subset of a metric space has a monotonic sequence as a subset, I learned that every sequence has a monotone subsequence, which is much stronger!

[cursive tendon]

is it true that for a set $S$, $\operatorname{iso} S^c \subset \partial S$?

[gentle osprey]

Mattuwu

[cursive tendon]

seems true in Euclidean spaces, but can anyone give a counterexample in general?

[steel glen]

what's iso S

[cursive tendon]

the isolated points of the complement of S

[steel glen]

i think it's true

you should be able to show every nbhd of x in isoS^c intersects S and S^c

hm ok i take it back

i have to think about it a little more

ok i think there's a lower limit topology counterexample

let me write it up and see if it works

hm it didn't

ok i think you can just take the discrete topology on any space with more than 1 element

like in R:
S = {x}, then every point of S^c is isolated, but the boundary of S is empty

[ebon galleon]

Let X={0,1} be discrete space on two points, S={0}. Iso S^c={0} since every point in discrete space is isolated, but bd S =Ø

[ebon galleon]

You should check out Counterexamples in Topology. Not sure how much they have on operations like these, but it should give you an idea of how one might prove/disprove stuff like this

[trail charm]

whats a “cross section?” i havent been able to find anything nor in this book unless i missed it

[gritty widget]

just a section of p

i.e. a map B -> X sending every b in B to an element of the fiber of p at b. in other words a right-inverse of p

continuous has the usual meaning

[trail charm]

ohh gotcha

tyty

[gritty widget]

the exercise says that trivializing a principal bundle is the same thing as continuously choosing an element of every fiber over B: once you've done so, you can get to any element of X by just group-acting on the chosen element in the same fiber of p

every fiber is the structure group, but maybe without a distinguished base element (the jargon here is "torsor"). picking a continuous section picks such elements

[trail charm]

[nimble marsh]

Could someone explain this? "Every loop $\gamma:\mathbb{S}^1 \to \mathbb{S}^1$ is homotopic to precisely one $c_n:\mathbb{S}^1 \to \mathbb{S}^1,c_n(z) = z^n, n \in \mathbb{Z}.$"

[gentle osprey]

Scerball

[brittle rapids]

in other words, the z ↦ z^n form representatives for each homotopy class

intuitively you're counting how many times the loop loops around the circle

[gritty widget]

it is nice to have that a map into a product is continuous iff each component is and this is one thing that fails with the box topology

you just have too many open sets

that products of compact spaces are compact is also nice to have

true in the product topology (the tychonoff theorem) and false in the box topology (product of countably many discrete {0, 1}s will do)

the way i see it is that the box topology just has way too many open sets for the properties we desire to hold, while the product topology has just the right amount

[ebon galleon]

The product topology is a type of topology called an "initial" topology: it's the coarsest topology for which a collection of maps is continuous (the projection maps, as you said). It might be easiest to first look at when we have one map, $f \from X \to (Y,\sigma)$. There is a unique topology on $X$ that is the coarsest making $f$ continuous, say $f^{-1}\sigma$. When we want to take the initial topology with a collection of maps ${\pi_a \from \prod_k X_k \to (X_a,\tau_a)}$ then, what we can do it take the topology generated by each $\pi_k$ independently, this will give us a collection of topologies ${ \pi^{-1}\tau_a}$, where $\pi^{-1} \tau_a = { U_a \times \prod_{k \neq a} X_k}$ (i think?), and then take the topology generate by their union. Explicitly, what this means is that we "close" the union $\tau' = \bigcup_k \pi^{-1}\tau_k$ under finite intersections (since this is all we need to form a topology: closing under infinite intersections gives too many open sets). For finite products, then, this is okay, the box and product topology agree. But for infinite products, this is why the product topology has only finitely many that are not the entire space $X_k$

[gritty widget]

maybe thinking about it the other way around will do: the coarsest topology making all of the projections continuous is a natural thing to ask for, and this is the same thing as the definition involving "all but finitely many U_n are X_n"

[gentle osprey]

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

[gritty widget]

you have a bunch of projection functions coming out of the product (at this stage, merely a set). you want them to be continuous, so you need to topologize the product. you don't want to be wasteful - what's the most efficient way to do this?

[ebon galleon]

also, this topology ends op having the nice property that maps into the product space are exactly the maps that are continuous into each coordinate

[gritty widget]

you want the projections p_n to be continuous, so you want p_n^{-1}(U_n) to be open whenever U_n is open in X_n. now you want to take the topology generated by these guys, since that's the smallest topology containing all of these p_n^{-1}(U_n)

maybe noting that p_n^{-1}(U_n) is precisely X_1 x X_2 x ... x X_{n-1} x U_n x X_{n+1} x ... will clear some things up regarding the product notation

[ebon galleon]

yeah i didn't feel like typing that out with the product, sorry 😅

But yeah, each individual pi^{-1}tau_k has only one component that's not the entire space. So when you take finite intersections, of course only finitely many are not the entire space

[urban zinc]

why is this lie algebra called gl? doesn't it include non-invertible matrices?

[dry jolt]

It's the Lie algebra for the Lie group GL(n, R). Usually the Lie algebra of the Lie group G is denoted by g (in fraktur)

[urban zinc]

ohhh

got it, ty!

[obtuse meteor]

Another thing i think about is it’s just computationally the thing that makes sense. Generally you can only check “finitely many things at once” in some way. And that’s what the product topology is telling you to do

[heady skiff]

is anybody here working through/or has worked through armstrong's basic topology

[silver umbra]

is it true that if X is path-connected, then H_0(X; G) = 0 for any abelian group G

[gritty widget]

it contains every n by n matrix you can think of :3

any such matrix A, invertible or not, induces a path t -> exp(tA) through the identity in the general linear group, and the tangent vector of this curve at 0 is just A

exp here is just the normal matrix exponential you probably know from ODEs

[novel acorn]

No

H_0(X;Z) = Z when X is path connected

[silver umbra]

sorry

i mean H_0(X; G) = G

in my first statement i also meant reduced homology

[novel acorn]

I do believe so

[tawdry widget]

Where can I find proof of that any knot is torus or satellite or hyperbolic? Does someone have a reference?

[urban zinc]

this makes sense!

[unreal stratus]

Yes so like

Let C be the singular chain complex of X, note C_0 can be identified with free G-module on points of X and C_1 similarly with module on paths in X.

in the image of the differential d:C_1 -> C_0 you get all the terms of the form x - y, since this is the boundary of the path from x to y

indeed d(C_1) is just generated by those terms

which means we just get C_0/d(C_1) since we are taking free G-module on the points and then identifying all the points with one another

[nimble marsh]

Could someone help me find a proof of the hairy ball theorem following this structure:

Consider $\mathbb{S}^2,$ is there a tangential vector field that is nowhere zero?

On $\mathbb{S}^1:$ Yes.

On $\mathbb{S}^2:$ No. If one could, one could write down a function $\mathbb{S}^1 \to \mathbb{S}^1$ that is both homotopic to $c_0$ and $c_2.$ Contradiction! (Where $c_n:\mathbb{S}^1 \to \mathbb{S}^1,c_n(z) = z^n, n \in \mathbb{Z}.$)

[gentle osprey]

Scerball

[unreal stratus]

is this a homework problem or smth lol seems oddly specific

[nimble marsh]

My lecturer was too lazy to give a full proof of the hairy ball theorem lol

I've graduated now and I'm looking back through my notes

We had just finished working on the fundamental group of the circle IIRC

[gritty widget]

i know a proof using homology

[urban zinc]

Yeah if you know the homology of the sphere and the fact that homotopic maps act the same way on the homology then this isn't too hard, not sure what your lecturer wants though

[coarse night]

If there is a nowhere zero vector field, Euler characteristic must be 0, which is not in case of S²

A bit different argument

(in fact it's an iff. closed manifold have nz vector field iff ec is 0)

(keywords: Poincare Hopf index formula, Euler class, obstruction)

[nimble marsh]

Could you share?

[gritty widget]

proof in dm

[umbral panther]

Analytic Proofs of the "Hairy Ball Theorem" and the Brouwer Fixed Point Theorem
John Milnor 1978
https://www.jstor.org/stable/2320860

[hushed lichen]

I have a really basic question

In set notation, for a set S, what is meant by S^(-1)?

[gritty widget]

context?

if all you know about S is that it is a set, then i'm not sure what S^{-1} could mean

[hushed lichen]

Looking at the topological definition of continuity. For every open set S, S^(-1) is open

[gritty widget]

it's (bad) notation for the pre-image of S under some map

[hushed lichen]

That's the entirety of the definition as it was given in class. I can go back to the book if that isn't sufficient to clear things up. We've seen that notation mean several things depending on context.

Haha thank you!

[gritty widget]

f^{-1}(S) is much clearer (f being your continuous map)

a function between topological spaces is continuous if the pre-image of every open set under it is again open

[high hill]

Let f, g : S -> S. S^-1?

[hushed lichen]

Thank you very much for the clarification! 🙂

[high hill]

maybe S^-f is tolerable but smh

im sure theres a set theory reason why that looks stupid

[languid patrol]

@hushed lichen people often write that in the context of topological groups where it makes more sense

[hushed lichen]

"that" meaning the S^{-1} notation? Or the f^{-1}(S) notation?

[languid patrol]

S^-1 in that context means {s^-1 for all s in S}

[ebon galleon]

f^{-1} is standard for inverse image of sets, which is what is meant for general topology (no groups)

[eager herald]

i'm confused on how these polygons for the torus, proj plane, and klein bottle induce a delta-structure

ok so if i focus on the torus for the moment, what will the 6 maps $\sigma_\alpha$ be?

[gentle osprey]

sean

[eager herald]

is it the $\sigma_\alpha: \Delta^0 \to X$ which sends the $0$-simplex to the point

[gentle osprey]

sean

[eager herald]

and 3 other maps $\Delta^1 \to X$ which send the line to the edges

[gentle osprey]

sean

[eager herald]

and two final maps $\Delta^2 \to X$ which send the triangle with vertices $(0,0),(0,1),(1,0)$ to the two triangles with edges $a,b,c$

[gentle osprey]

sean

[eager herald]

do i have this right?

and similarly for the klein bottle and real projective plane

[tawdry valve]

yup

[eager herald]

i kinda dont see how the real projective plane has two points

i havent worked w the projective spaces much so

the bottom left v is identified with the top right v

when b is identified

and similarly for w

we have a mobius band so far

i dont see how to identify a

my brain can't think four dimensionally right now

[tawdry valve]

I'm not sure if it's helpful but I can give some pictures how I picture RP2. I don't usually try to picture gluing edges successively, like how we usually picture the torus or klein bottle

maybe some people are better at that

[eager herald]

sure

[tawdry valve]

these are some of the ways I'll think about it

rounding out the corners of the square, you get the second picture where it's identifying opposite points on the boundary of the disc

[eager herald]

ah yeah ive seen this construction but this makes it clear

thanks for the great diagrams

[tawdry valve]

yeah for sure

[eager herald]

you're a talented drawer

[tawdry valve]

thank you

[eager herald]

just to make sure, RP^2 has two individual points (vertices)

because each two antipodal points are identified?

[tawdry valve]

you mean in the delta complex structure?

that's true

[eager herald]

yeah

aight i think i get it now

thank you

[broken nacelle]

you could've asked in the reading group 😭

or at least pinged me

[broken nacelle]

this looks so pretty

what did you use to draw that?

[coarse night]

hands?

[broken nacelle]

was "hands" the only tool he used?

did he magic the shit into his screen?

[coarse night]

certainly couldn't have drawn it with their legs

[high hill]

watch out for these hands right here

[broken nacelle]

[broken nacelle]

abelist

[high hill]

why dont u ask how i make my beautiful diagrams huh

the answer is also hands btw

[broken nacelle]

uhhhhh

uhhhhhhhhhhhhh

[winged dagger]

What does it mean for boundaries to be singly connected? Reading this in the context of classification. In the book it says that a certain way of defining boundaries is convex and singly connected. The convexity is clear to me but I have no idea what singly connected means amd neither have any Google searches aided in this understanding

[ebon galleon]

singly or simply

[high hill]

well im going to guess connected by a single point

[high hill]

oh wait we re talking one boundary

[umbral panther]

Sometimes people use “multiply connected” to mean not simply connected (this makes more sense in graph theory than topology). If you tried to invert this, you might end up with singly connected

[winged dagger]

Singly

Maybe this can help

[winged dagger]

Like I'm not familiar with the study of topology, is it possible to explain what singly connected means in simple terms?

[novel acorn]

https://stackoverflow.com/questions/25140075/definition-of-singly-connected-set
There is this

[umbral panther]

I’ve never heard the phrase. I assume it’s an error and should be simply connected

[winged dagger]

@novel acorn I checked that as well but I can't understand the answer

Like all of it

Why is it talking about connecting points in curves and polygons

[languid patrol]

It is an older term for simply connected. It just means that there is, up to homotopy, only one path between two points. Just like simply connected

[winged dagger]

@languid patrol but aren't there multiple path between two points here?

What is meant by path?

BRB 5 mins, will check answer later

[ebon galleon]

A path from $x$ to $y$ in a space $X$ is a continuous function $p \colon [0,1] \to X$ such that $p(0)=x$ and $p(1)=y$. You can think of it like being able to draw a line from point $x$ to point $y$, without ever lifting your pen.

To say that a space $X$ is simply connected means, in some sense, that if I give you a path $p$ and a path $q$ from $x$ to $y$, you can continuously deform $p$ into $q$, without ever breaking the line.

[gentle osprey]

honorable chmonkey fan

[tawdry valve]

I had gotten a wacom graphics tablet to try and get into digital art a while ago, but it turned out to be super useful when everything went online for the pandemic lol. Software-wise, rn I'm using Lorien (https://github.com/mbrlabs/Lorien) since it's infinite canvas/vector graphics. It's nice for doing math with a friend online---I'll get in a call, share my screen, and transcribe. Ideally there would be a "multiplayer" infinite canvas vector graphics software, but I've yet to find a good one. Microsoft whiteboard promises to do this, but in the past it's been janky and doesn't run on Linux

and yes i used my hands too lol

[tiny ridge]

looks super useful

i used to use xournal++ before i got an ipad

but of course thats primarily for math and not digital art

[broken nacelle]

this whole sentence is basedness supreme

open source??
linux???

please do tell if you find the multiplayer canvas btw

I need that too

I only see the online whiteboards

the wacom tablets are affordable af too

[nimble marsh]

Do spheres of n dimensions have k < n dimension great sphere's on their surfaces?

For example, the 2-sphere has a great 1-sphere (an equator)

Maybe the great spheres are of dimensions n-1?

[tough hamlet]

how are you defining great sphere

also this isn't really topology

[nimble marsh]

well, I didn't know where to ask it. I started thinking about this after looking at some borsak-ulam related stuff

[nimble marsh]

Some sort of generalisation of this: https://en.wikipedia.org/wiki/Great_circle

[ebon galleon]

You can probably make sense of it, but it would need a precise definition

[nimble marsh]

I am thinking about applying borsak ulam. I know on a 2-sphere I have infinitely many great circles and hence infinitely many 1-parameter points that are the same. I wonder if this sort of thing generalises

[tough hamlet]

the obvious thing is just to have a greak k-sphere be the intersection with a (k+1)-dimensional subspace

[nimble marsh]

yeah, that makes sense

[tough hamlet]

the case k=n-1 being the most important

[feral copper]

The case k=1 is as important too (at least from a geometric point of vue), because geodesics

In any case, 'great k-spheres' happen to be totally geodesic 🙂

[gaunt linden]

Don't discount k=0 either: The set of great 0-spheres is exactly projective (n-1)-space.

[feral copper]

What are great spheres of tropospheres?

[broken nacelle]

so I'm finally learning about singular homology

and I'm wandering how I should interpret the coefficients in a formal sum of singular n-simplices

[feral copper]

Wdym by 'interpret'?

[broken nacelle]

like, is there any geometric meaning to the coefficients?

[feral copper]

Taking 2*something is taking the union of two distinct copies, taking -something is reversing orientation

Idk if that helps

(I always think of a homology class to be represented by a submanifold)

(although that is not always the case)

But that's for Z-coefficients of course
I never thought of any meaningful way to see what would be homology with, say, Z[t,t^-1]-coefficients

[broken nacelle]

does that really workout tho? like, taking a simple example of mapping 1-simplex to, say, \bR^2

hence the singular 1-simplex is really just a path in \bR^2

and then you reverse the map in the obvious way

does that amount to slapping a negative sign before the singular 1-simplex?

if so, why

or why not

[feral copper]

Think of it on a torus rather, where there is homology

But yeah, taking - a curve is that curve with reversed orientation

[feral copper]

I mean, they will be homologous

[broken nacelle]

but why is that the case?

[broken nacelle]

[feral copper]

Because take the sum of both, it's an obvious boundary

Bounded by a cylinder

[broken nacelle]

one sec, I forgor a definition lmfao

oh ok

[feral copper]

This is bounded by the cylinder in the middle

Because orientations agree

This is not a boundary

[broken nacelle]

interesting

I'm sry I'm just taking a moment to understand why lol

[feral copper]

Sure, that's the point in learning 🙂

Maybe this drawing helps regarding orientations? Idk

[broken nacelle]

I'm trying to understand how the cilinder is an n-chain

[feral copper]

Here it is a 2-chain

It maps the square to the cylinder by gluing two opposide sides together

Like the first step of folding the torus from the square, if you've seen this picture

[broken nacelle]

but the square isn't a 2 simplex, no?

[feral copper]

Yeah well, okay, 2-chains are maps from the triangle to the space, but it's roughly the same

Another drawing (coming)

[broken nacelle]

bruh

you can do that??

[feral copper]

That describes a continuous map, doesn't it?

You don't ask for embeddings

How would you see that this '8'-shaped thingy is a 1-simplex otherwise?

You'd have to glue things in the same way

[broken nacelle]

yea but the boundaries don't cancel out

leaving only the two paths you drew

that's what I'm trying to achieve

[feral copper]

I don't follow!

[broken nacelle]

ok so for them to be a boundary they must be in the image of \del, right?

[languid patrol]

The two are obviously not the same formal sum of simplices, but they are homologous

[feral copper]

It must be del of something

[feral copper]

Yes that was my claim!

[broken nacelle]

yea

lol

[broken nacelle]

so like, if you take the del of a a single singular simplex

the result is gonna be a formal sum of 3 1-simplices

2 of which has to be our boundary circles

[feral copper]

Not necessarily 3

Of 'some' 1-simplexes

[broken nacelle]

huh

wait I don't follow

[feral copper]

Here's a weird 2-simplex whose boundary is 4 thingies:

[broken nacelle]

bruh

I'm sorry I have to go get lunch lol

I'll be back

[feral copper]

Have a nice lunch!

[eager herald]

darq what part of chap 2 r u on

is it hard to read

should i read van kampen first or homology

[feral copper]

Damn, did they unlink the Hopf link? Am I missing something?

[broken nacelle]

singular homology

[broken nacelle]

it has been chill so far

I'm still having trouble understanding singular homology tho so that might be when it gets real

[broken nacelle]

it doesn't really matter

as I said, one isn't a prereq for the other

[broken nacelle]

tbh the actual theorems and definitions aren't super complicated

but making sense of them geometrically is getting me

I've heard that's a common theme over homology

[feral copper]

Personnally I had started with \pi_1 and SVK before homology, because at least you can make drawings of genuine loops

But that's indeed up to preference

[eager herald]

ah i see

fuck i have 0 geometric intuition tho

kinda screwed

bruh i have trouble visualizing like gluing together a mobius strip

ok maybe not to that extent but still

[broken nacelle]

oof

[feral copper]

That'll come with practice 🙂

[broken nacelle]

you might like homology better then!

like, algebraic topologists are very good at abstracting away geometric structure

intro, well, algebraic structure

in my case, I like the geometric part too much to ignore it tho lol

[feral copper]

I wouldn't be so definitive about it, you do need geometric intuition to derive the right algebraic computations

Like if I give you the presentation of a fundamental group, you can do algebra, sure, but you still want to see which generator corresponds to which loop

[eager herald]

idk what'll murder me first the homological algebra with like the big ass diagrams or the geometry

[trim chasm]

it literally just tells you to undo it in the second most obvious way? (i.e. through the wrist loops rather than untie the wrist loops)

[feral copper]

Idk, isn't it unlinking this?

[trim chasm]

but the instructions basically say cheat knot theory

[feral copper]

Ah okay it's cheating by going under the knot between the rope and the wrist?

Ah yes, I dislike that

Thanks!

[broken nacelle]

tbh I'm worried less about those then the geometry lmfao

I dunno, they look like you'd get used to them quickly

[queen prism]

geometry scary

[feral copper]

Eeew geometry > diagrams

[trim chasm]

I find the geometric proofs of section 1 tedious, especially Van Kampen

[feral copper]

Keep in mind geometry =/= fat ass analysis

[trim chasm]

I much prefer Grothendieck's proof using covering theory

[eager herald]

honestly i wouldnt mind analysis if u meant that

[broken nacelle]

what's wrong with analysis?

[eager herald]

true

i want to learn mayer-vietoris

sequences

[broken nacelle]

the what?

oh

I see them in 2.2

why?

[feral copper]

Well, usually it ends up with PDE shenanigans and I don't like those

[eager herald]

idk i read they were like van kampen but better for homology

[broken nacelle]

valid

[tiny ridge]

I have a slightly less stupid version

[eager herald]

also they sound cool

[feral copper]

Ibsen-sempai, sup? Surprise me! 😄

[eager herald]

one step closer to spectral sequences

[broken nacelle]

actually I might get into PDEs coz of geometry

I couldn't care less about physics

but geometry sounds ocol

[feral copper]

As long as it converges on E_2 I can buy that

[tiny ridge]

Say you and your friends left and right hands are L, L' and R, R' and they're standing in front of each other. Consider a rope joining L and L' and R and R' and the ropes braid twice around each other ie it represents sigma^2 in B_2

Then you can do shenanigans to make it sigma^0

Ie trivial braid

Think Dirac belt trick

[feral copper]

Isn't that a different problem? (still related tho, and I can believe that this one has a non-cheaty solution)

[tiny ridge]

Yeah not the same problem

[feral copper]

Ah yeah this

[tiny ridge]

Both the people should be tied to a chair for a nontrivial solution

Else the other person can do a sommersault

[feral copper]

(Call social services)
You mean otherwise one can go through the other's rope entirely?

[tiny ridge]

Well I meant say your friend does a handstand and stands back up again

Sideways

That does the trick

[bleak anvil]

How to prevent yourself from killing?

[feral copper]

Genius

[trim chasm]

but but Van Kampen essentially states that especially nice pushouts diagrams in {*}/Top are preserved by the pi_1 functor

[ebon galleon]

Van Kampen

[feral copper]

Yeah it's true, but it requires a shit-ton of previous knowledge to make sense of something so visual geometrically

[tiny ridge]

What is */Top

[feral copper]

Pointed top spaces

[tiny ridge]

Weird notation

[feral copper]

But like a slanted thingy version

[languid patrol]

undercategory of a point

[feral copper]

I remember seeing this slant thingy in May's cat theory

[ebon galleon]

I mean it's a slice category so it works

[tiny ridge]

Slice is over

[ebon galleon]

Also Top_*

[tiny ridge]

Not under

[tiny ridge]

This is what I use

[ebon galleon]

i can never remember which is slice and which is coslice

[languid patrol]

Cat/x means the overcategory of x

x/Cat seems reasonable for the undercategory

[ebon galleon]

I guess lol, I just forget

[feral copper]

x\Cat

idk

[languid patrol]

It's like something a programmer would come up with

[tiny ridge]

Category theorists are like slightly better than silicon valley codebros

[languid patrol]

Except then Cat/x wouldn't include Id: x \to x and x/Cat would for no explicable reason

[tiny ridge]

Only slightly

[trim chasm]

only basic category theory, and then the proof of the theorem assumes knowledge of covering spaces

[tiny ridge]

Consider three homomorphisms F_3 -> F_2 given by forgetting one of the standard generators. What is the intersection of their kernels?

[feral copper]

But the other folks were struggling with singular homology, I don't think it's necessary to add that vocabulary x') also you don't need covering spaces for a geometrical proof

Just cover opens with disks

And the Lebesgue lemma

[unreal stratus]

Smh good notation

[tiny ridge]

In other words how many ways can you hang a portrait on the wall with three pins such that if you remove one it falls down

[unreal stratus]

Jk the right notation for pointed spaces is Top

[feral copper]

Do you know the solution and ask that as a puzzle?

[languid patrol]

yes it's a famous puzzle

[tiny ridge]

I don't know this time

[languid patrol]

oh i guess not

[tiny ridge]

Two is aba^-1b^-1

And it's powers

[feral copper]

Well, just products of commutators, isn't it?

[tiny ridge]

There's [a, [b, c]]

Why do these generate

[feral copper]

Ah

[hidden crag]

Ooh this is a cool puzzle

[unreal stratus]

Lie algebras

[tiny ridge]

This doesn't work

Forget b you get [a, c]

[[a, [b, c]], [a, c]] works except c. Correct for that by another commutator?

Dunno

No clue

Seems like there should be an exceptional word which is not a product of commutators

[languid patrol]

Are you looking for minimal generators?

Or just any set of elements which generate?

[feral copper]

It does work tho? Forget b, you get [a,1]

[tiny ridge]

Oh cool

Thanks

[feral copper]

I think for n pins you iterate commutators

[tiny ridge]

Both will be nice

[feral copper]

[a,[b,[...]]

[hidden crag]

Yes

That was a homework problem in my alg top course

[tiny ridge]

Oh interesting

Is there some covering space solution

[hidden crag]

Why is that bleaksully

[tiny ridge]

What's the answer for F_m -> F_n?

[hidden crag]

It’s an interesting problem

[languid patrol]

yes it's a very common problem, no there isn't a particularly nice geometric solution that I'm aware of.

[feral copper]

You mean remove k pins among the n to have it fall down?

[tiny ridge]

Yeah

[ebon galleon]

I don't like working with free groups

[feral copper]

Interesting question

You're intersecting k choose n kernels tho!

[ebon galleon]

But it's probably not terrible when you just do it lmao

[tiny ridge]

Free groups is hard

Ok what's the solution

If there is not an obviously geometric one I might as well know the sol

[languid patrol]

I only know the solution for the n-1 out of n or 1 out of n versions. The k out of n version seems like a nightmare.

[trim chasm]

that's too analytic for me

[feral copper]

I think we can try 2 choose n by hand maybe? But I agree I don't wanna dive into this

[languid patrol]

Yeah some low-lying cases will be nice to write down

or at least doable

[tiny ridge]

Seems like a good undergrad project

[feral copper]

(n-1) choose n is 'dual' to 1 choose n in some sense, right? It has to!

[languid patrol]

I think it's been done to death as an undergrad or highschool mathcamp project

[tiny ridge]

One should come up with purely geometric proofs

No algebra allowed

[feral copper]

Yeah but even for the 1C2 version, how do you come up with all possible solutions? You'd need algebra to describe the intersection of kernels, otherwise you could miss on some (possibly)

Let alone the kCn

Like, fiding one should be doable, okay

[languid patrol]

It's actually nontrivial to show that you can find one with a reasonable algorithm

oop I replied to the wrong comment

[feral copper]

😛

[tiny ridge]

Kernel of some map F_3 -> F_2 x F_2 x F_2... Who knows what this means in terms of covering space theory

Mapping wedge of three circles into product of three figure eights

[feral copper]

This should correspond to a covering space at least

[languid patrol]

For 1C2 you can show that it's just the derived subgroup, so depending on what you want that's the answer.

[tiny ridge]

Yeah one has to pin down what the cover is from the geometry of this map

[feral copper]

I can't think of a reasonable way to find the quotient even (for the 'number of sheets' of the covering)

[tiny ridge]

Oh this seems infinite index to me

Any normal subgroup of a free group is

This is a kernel, so

[feral copper]

Yeah I'd have said this too

But still, I found that understanding the quotient helps in understanding the action by deck transformations

Wait

I don't know, whatever x')

[tiny ridge]

Here's a separate puzzle I thought about some time ago and didn't get far

[languid patrol]

hmm okay

I guess maybe it's interesting to consider the corresponding rank 6 local system on S^1 wedge itself 3 times

At least that local system uniquely determines this cover

[tiny ridge]

How so? The number of sheets is infinite

I'm a little confused

Ah ok, you're going to use some standard rep of F_2 in GL_2(C)?

[languid patrol]

Yeah there is the well-known standard rep a, b \to (1 2 | 0 1), (1 0 | 2 1)

[tiny ridge]

Right

Ping pong

[languid patrol]

exakt

[tiny ridge]

Nice idea

[languid patrol]

Not sure it's terribly useful though.

[tiny ridge]

You know what one should classify all group homomorphisms F_n -> F_m, and then pi_1(Sigma_n) -> pi_1(Sigma_m)

That's fine I only care about good ideas

I live in the idea-space

F_n -> F_m is dumb, just choose n more or less random words

What about surface groups

Seems nontrivial

There is this theorem of Edmonds which says all maps between surfaces is a branched cover followed by collapsing some handles to points

Homotopic to one

[trim chasm]

even F_n -> F_n is hard

[tiny ridge]

Maybe you have a point

I'm just worried it's either dumb or unclassifiable depending on what one means

[trim chasm]

is outer space not big enough already?

(I haven't looked at any recent literature tbh)

I'm hopeful taking a module lectured by Karen Vogtmann next year

[tiny ridge]

Out(F_n) is cool

[fading vale]

Out(F_2 hat)

[broken nacelle]

@eager herald

If someone "hands you" two topological spaces, you basically have a useless pile of garbage.

[eager herald]

LMFAO where is this from

[broken nacelle]

https://math.stackexchange.com/questions/40149/intuition-of-the-meaning-of-homology-groups

I haven't read the post completely yet

I'm watching anime

[eager herald]

i see

im kinda struggling with delta complexes rn

so we have this

right

when it says the n simplices e^n_\alpha, is it referring to this fact?

im struggling to see the quotient construction here

so im thinking of $X$, a $\Delta$-complex, as a CW-complex rn

[gentle osprey]

sean

[eager herald]

well kinda but not really

so we have a bunch of $\Delta^n_\alpha$s, which are just $n$-simplices which are characterized into $X$ by corresponding maps $\sigma_\alpha:\Delta^n \to X$.

[gentle osprey]

sean

[eager herald]

now this is great and all, but of course we need to glue everything together

hence, we quotient each face of $\Delta^n_\alpha$, which are obtained by deleting a point from the $n$-simplex, with a corresponding $\Delta^{n-1}_\beta$, and identify them together to "glue" them tgt

[gentle osprey]

sean

[eager herald]

and this $\beta$ is one for the map $\sigma_\beta:\Delta^{n-1} \to X$ which is the appropriate restriction of $\sigma_\alpha$

[gentle osprey]

sean

[eager herald]

ok now we've glued everything together by quotienting

ah wait i think i get it now

[tiny ridge]

Is that the one which contains Gal(Qbar)

[fading vale]

Yes

[tiny ridge]

Ah yeah because CP^1 \ {0, 1, infty} has etale fundamental group F_2hat

[fading vale]

Right

So you apply the etale exact sequence

And then weep

[tiny ridge]

Cool

[fading vale]

I think taking an elliptic curve with some prescribed j function gets you injectivity of the representation

[tiny ridge]

That's interesting

I was going to ask

[fading vale]

I don't remember the details bc it's been like well over a year since I thought about this

[tiny ridge]

Sullivan has this tome article from 1970 where develops completion of spaces

[fading vale]

But it's along those lines

[trim chasm]

that's a weird way to write G_Q

[tiny ridge]

And the homology of these completions of manifolds are acted upon by Gal(Qbar)

[gritty widget]

sullivan

[tiny ridge]

I never read beyond localisation

Lol

[tiny ridge]

Isn't that a fashion magazine

[languid patrol]

No it's a men's lifestyle magazine

[tiny ridge]

Terrible

[trim chasm]

okay I'll accept Gal(Q)

[tiny ridge]

Aktchuyally Q is the one which is redundant because every number field automorphism fixes Q pointwise

I know you want to pretend real bad that Gal(F) is like pi_1 Spec F

[trim chasm]

aktahxkly Qbar is determined from Q so.... and the choice of separable closure is a basepoint

[tiny ridge]

Imagine saying the universal cover is my basepoint

[languid patrol]

Gal(\bar{Q}) is bad notation objectively

[broken nacelle]

lol that was quick

:anyapat:

[tiny ridge]

The notation is as bad as the object

I can't help it

[languid patrol]

In this case the natural object that it would refer to is the trivial group

Anyways this conversation should really be had in #advanced-algebra , or maybe #advanced-advanced-algebra if there was such a thing

[tiny ridge]

I don't subscribe to co-conventions made up by the utterly deranged

I think this should go in #1100503863586455632

[ebon galleon]

or #1136771350518837308

[broken nacelle]

ok, so back to this

now I gotta ask wtf is a simplex?

how is that a triangle?

that can't be homeo a triangle

[tiny ridge]

It is an image of a simplex is probably what they meant

By some random continuous map

[trim chasm]

these are singular simplices

[tiny ridge]

In the context of singular homology a "singular simplex" is a possibly horrendous continuous map from a geometric simplex to your space

I would argue it is important to keep track of the parametrizing space and not just the image but this is overly pedantic

[broken nacelle]

but how could that be a continuous map?

the restriction to the boundaries can't be continuous, right?

coz it's taking a connected space (the 1-simplex) to a disconnected space (2 circles or smth)

[trim chasm]

it's multiple triangles glued together

a singular chain

[broken nacelle]

well, in that case it's not a singular simplex

[broken nacelle]

yea, exactly

[tiny ridge]

There are lots of surjective maps from a disk to that surface

The preimage of the boundary of the surface need not be the boundary of the disk

Identify disk with a 2-simplex

For instance give the surface a cell decomposition

There's one with a unique 2 dimensional cell

Consider the characteristic map of this 2 cell

[trim chasm]

oh yea 🤦‍♀️

[broken nacelle]

I'm confuddled

[broken nacelle]

what do you mean by this?

and the characteristic map is just $\chi_{D^2}$?

[gentle osprey]

chmonkeynumber1enemy

[broken nacelle]

[tiny ridge]

Do you know what a CW complex is

[broken nacelle]

yea

oh

[tiny ridge]

Give that surface a CW decomposition

[broken nacelle]

sure, you can use the fundamental polygon of the 3-torus

and then not glue some of the edges

[tiny ridge]

Yup

Now let the map from the closure of the polygon be your map

Identify polygon with simplex

[broken nacelle]

I see hmm

ok but the del of that isn't just the circles @feral copper

thankyu ibsen btw

[tiny ridge]

Del of that is a 1-chain homologous to the circlee

Circles

[languid patrol]

the other parts of the boundary cancel out because d^{op} ~ -d

where d^{op} is the one-simplex traversed in the opposite direction

[tiny ridge]

Because of this ^^^

[feral copper]

Yeah, sorry for the confusion, I identify the formal sums of continuous maps \Delta^n\to X to the union of their images

[tiny ridge]

Spoken like a topologist

Respect

[broken nacelle]

lol

[tiny ridge]

Chain schmain

I call them wiggly polyhedra

[feral copper]

Chmochain

[tiny ridge]

Cochains I call functions

[languid patrol]

cochains I call them cococochains

[tiny ridge]

Cocochain on a very ashy day fam

(RIP Doom)

[feral copper]

Actually, @broken nacelle, this is a pretty common thing in topology, so whenever you get confused, keep this rule in mind: we like to identify stuff and call them the same way and write them with the same letters

[tiny ridge]

Yes we like equality

We don't draw fifty million commutative diagrams of isomorphisms

A equal to A equal to A

[broken nacelle]

I'm sorry I'm not sure if I was confused about that

or how I was confused by that

[tiny ridge]

Welcome to topology

It is an essential part of mathematics to be confused about what you're confused by

Category theorists have been setting notions on stone to deny this fundamental theological nature of being

[broken nacelle]

are you saying what you meant by "here's a weird 2-simplex" is "here's a the image of a weird n-simplex"?

[tiny ridge]

I call them heretics

[broken nacelle]

yea, ok, I think this is what you meant

[broken nacelle]

now onto why this is true lmfao

they're in the image of del ig?

[tiny ridge]

Wrong perspective, it may or may not be true. Infact it can be simultaneously both

[broken nacelle]

oh yea

[tiny ridge]

Embrace dialectics

Here's the point. Take the 1-chain c which traverses the [0, 1] starting at 0 ending at 1.

Let c' be c with orientation reversed

What is c + c'?

Not zero. But shouldn't it be?

What a conundrum

[broken nacelle]

lmfao

it's certainly homotopic to 0

[tiny ridge]

What does homotopy of chain mean

[broken nacelle]

no clue

[tiny ridge]

Exactly

[broken nacelle]

[feral copper]

Again, my way to visualize simplices is: a k-simplex k-chain is a k-submanifold which you allow to bend over, intersect itself. Basically: an immersion of a k-manifold. Then the boundary of the simplex is the image of the boundary of the k-manifold.
If you allow yourself to think of everything in these terms, you see that formal sum is just union of such thingies (or connected sums even), and taking opposites is reversing orientation of the thingy.
In some cases, this goes beyond a simple frame of thought. For instance, in 4-manifolds, any homology class in H_2 can be represented by an embedded closed connected surface. In the simply-connected case, each class can be represented by an immersed sphere with transverse double points only.

[tiny ridge]

I think it is too much to conflate k-cycle with k-simplex thouhh

Which is what you seem to be doing above

k-cycle is a k dim submanifold, k-chain is a k dim submanifold with boundary

[feral copper]

Sorry, I meant to say k-chain, not k-simplex!

[trim chasm]

I don't think you can call it a connected sum?

[tiny ridge]

Paste manifold with boundaries along boundaries

Not quite connect sum yes

[red yoke]

Is there something missing here?

This seems to imply every space is discrete

[solemn oar]

It does not imply that, but you can choose subsets to produce the discrete topology.

Nvm, indeed it's strange.

[gaunt linden]

No, something looks fishy to me too. For example, cover R by the subsets {a} for each a in R, each with the trivial topology. That seems to satisfy the assumption in the claim: The subspace topology on {a}cap{b} is indeed the same no matter whether we look at it as a subspace of {a} or of {b}, and {a}cap{b} is closed in each of {a} and {b}.
But there are many different topologies on R that induce the trivial topology on each singleton.

[tough hamlet]

I think you'll want like

each X_i to be open as well

[gaunt linden]

Then it will be kind of hard to make Xi cap Xj closed in Xi and Xj, though.

[tough hamlet]

86 the closedness

[gaunt linden]

(Conversely, wanting each Xi to be closed wouldn't even exclude my example).

[tough hamlet]

yeah replacing closed with open at least works

[gaunt linden]

I think the claim could also be repaired by requiring each Xi to be close in X, and assuming the index set J is finite.

[ebon galleon]

Perhaps it's supposed to something like "for every i, there's a j such that X_i \cap X_j intersect"... Not sure that would fix it but I agree it's not right as is

But that doesn't seem quite right either now that I think about it

[opaque scroll]

You could just let the Xi's be the set of all pairs of points in R, and again same problem

[ebon galleon]

just write out colimit properly in terms of diagrams and this is not an issue

[opaque scroll]

I guess you can say that there is a finest such topology

Though, not sure what then would be the point of the intersections being closed

[opaque scroll]

What is the context here, like what is this an "example of"?

[ebon galleon]

You can't even say there's a unique topology for which these sets X_i are closed in X and it induces the topology on them, tropo's example would work for any T_1 space lol

[ebon galleon]

Actually, this makes sense, since that is the definition of colimit topology

[opaque scroll]

Indeed, but is this example suppose to illustrate what the colimit topology is? (If so it's doing a terrible job) or is it trying to say something else and there is just a small mistake somewhere

[ebon galleon]

I am not quite sure, it is written quite poorly/unclear lmao

[high hill]

https://media.discordapp.net/attachments/1116077637299875910/1137841146186448956/Screenshot_2023-08-06-21-14-08-57_40deb401b9ffe8e1df2f1cc5ba480b12.jpg

$$X = {1, 2, 3}$$
$$T = {{1, 2}, {2, 3}, {2}, \varnothing, X}$$

$$B = {{1, 2}, {2, 3}}$$

$2 \in {1, 2}, {2, 3}$ but ${2} \not\in B$

[high hill]

What am I missing with regards to point 2

[ebon galleon]

B is not a basis

[opaque scroll]

The usual definition of basis is that every open can be written as a union of basis sets

[ebon galleon]

Because {2} \notin B, like you said

[opaque scroll]

This thing about finite intersection is not the standard definition

[high hill]

The screenshot gives 2 definitions

B is a basis by the first, but not the second, is that right?

[ebon galleon]

The first should be subbasis

B is a subbasis for this topology, not a basis

[languid patrol]

no the screenshot gives two equivalent definitions

[ebon galleon]

They both generate the same topology, yes

[opaque scroll]

I think it's just a mistake maybe

[high hill]

See my example though?

[languid patrol]

ah yes they are not equivalent I see, one is finer than the other

but the only difference is being closed under finite interesections

[high hill]

What's the intuition behind point 2 of the 2nd defn anyways - that was what I was hunting for

[ebon galleon]

If you close B under finite intersections (i.e. add {2}), these are equivalent

[languid patrol]

I think these terms like basis and subbasis are not very important

[opaque scroll]

It's not necessarily closed under finite intersections

[ebon galleon]

The second definition requires it to be

[high hill]

like why require it.

[opaque scroll]

No, only that the intersection is the union of basis sets

[languid patrol]

I guess technically not since it's not closed under unions

[opaque scroll]

Not that the intersection is a basis set

[queen prism]

say you have a point p in the intersection of two balls, B1 and B2
you’d like to say that there is another ball B in that intersection containing p

[languid patrol]

because you want it to be fine enough that the resulting topology is closed under finite intersections

i.e. is a topology

the actual important object associated to a basis or subbasis or whatever is the set of arbitrary unions of finite intersections of its elements

[ebon galleon]

oh right, i guess it's not quite closure under intersections, my bad

[high hill]

mmm ok ill have another look now that confusion was cleared up

[high hill]

so ill take this

[grim knot]

guys, if X is T2 and Y is not, why does there not exist a homeomorphism?

I thought that a continuous function has no influence on hausdorff property

[paper wedge]

you thought wrong

or is it you think wrong?

haha

suppose f:X-->Y is continuous , X is hausdorff and Y isnt

the preimages of your choice of non-hausdorfness of Y will contradict hausdorfness of X

do u see it

and

if u do the proof you will see

that you do not even need "full" homemorphism

u just need closed and bijective

[grim knot]

it's just that in my notes, I found that "generally T2 is not preserved under continuous functions"

[paper wedge]

what is the difference between a homeommorphism and a continuous function?

[grim knot]

bijectivity(?)

[paper wedge]

yes

inverses exist

u need that to do the proof

[grim knot]

okay, I see it

[paper wedge]

so if ur problem is asking for a homeomorphism

then yes they do preserve

otherwise ( as you sholud try in the proof its just straight-forward ) u will find that u need bijectivity anyways

[grim knot]

I'm gonna try and prove it, thank you

you were really helpful!

[paper wedge]

no need gl

[hidden crag]

Exercise: suppose Y is hausdorff and f:X->Y is a continuous and injective map, is X hausdorff?

[opaque scroll]

The key feature of homeomorphism is that they preserves all topological properties. The same is not true for just arbitrary continuous functions

[hidden crag]

@grim knot

Potato

Shh

Don’t spoil it

[unreal stratus]

I was just responding to an idea

[hidden crag]

Let her cook

[grim knot]

I'd say yes hahaha

[grim knot]

HAHAHAHAH

[hidden crag]

Lass sie kochen

[hidden crag]

That is true, why?

[grim knot]

because we have disjoint open set V1 and V2 in Y for y1!=y2, so, the preimage is also disjoint, and since y1!=y2, so is the preimage of the intersection empty and f^1(V1), f^1(V2) is what we were looking for

it's just a sketch

[hidden crag]

That’s correct

[grim knot]

Thank you for the crown

[grim knot]

I hope the dish was good

[hidden crag]

5 star meal

I was asked this question in my topology oral exam

[grim knot]

Timo I have a question in German, do you have 5 secs?

[hidden crag]

Im in the gym

But I can have a look sure

[grim knot]

I understand the whole exercise here, just don't see why such a x0 has to exist

[hidden crag]

Was ist R,O

Im Domain

[grim knot]

I sent you O in the second picture actually

[hidden crag]

Also was ist da die Topologie

My bad

Das x_0 ist beliebig

Du nimmst irgendeins, bildest das ab und guckst was passiert

[grim knot]

okay gut, ich dachte sie starten by y0 an

darum hats mich verwirrt

[hidden crag]

Y_0 ist einfach der Wert auf den das x_0 abbildet was man random wählt

[grim knot]

I have a written one next week on wednesday

[grim knot]

ja voll macht sinn jetzt, ich hab es eben umgekehrt angeschaut, und da f nicht surjektiv ist, muss y0 nicht unbedingt ein Urbild haben

[hidden crag]

[grim knot]

thanks for the time

have a good pump

[paper wedge]

du bist weisetmaster

[hidden crag]

Thanks

[grim knot]

is it true that finite topological spaces are compact(?)

it depends on the topology we have right?

I think it should not depend

[high hill]

surely all open covers will have a finite open subcover

your topology is finite

all covers are finite

[grim knot]

I'm still confused on how the frick to use Seifert van Kampen

I understand the whole computation, besides from how do you calculate N

[hidden crag]

Have you worked out examples

And what’s N here?

[grim knot]

I literally am doing an exercise right N

the kernel of the grouphomorphism

[hidden crag]

An isomorphism will have trivial kernel

[grim knot]

yeah dumb me

[hidden crag]

Edit made me look stupid

[grim knot]

this is the exercise I'm doing

[hidden crag]

Read examples if you struggle to do it yourself for now

[grim knot]

The problem is that I do not see where they get N from

[lime sable]

i think in a lot of natural cases you have a basis instead of just a subbasis, and in these cases you want to write open sets as unions of basic opens, rather than worry about finite intersections or whatever

[high hill]

makes sense

[lime sable]

by "natural cases" i mean metric topology and zariski topology because that's all i can think about right now

[high hill]

meanwhile in measure u have un, unu, nunun, etc

[hidden crag]

@grim knot why did you delete the solution

[grim knot]

I understood a part of it

like the first part, I can send it again

but still don't see how they come up with the kernel again

[hidden crag]

No harm in resending it if there’s something you don’t understand

I’ll try to have a look later (busy dying to numerics rn) or someone else might

[grim knot]

this is the solution, and they don't even calculate it

Like I don't know how to come up with it

I understand that $\pi_1(A) * \p1_(B)= \mathbb{Z}*\mathbb{Z} $

[hidden crag]

The amalgamation is trivial

[grim knot]

but why is N trivial?

[hidden crag]

Because A cap B is contractible

Hence it’s fundamental group is trivial

[grim knot]

wait a sec

what is the conenction between the fundamental group of the intersection and N

(asking to make it clear in my head)

[hidden crag]

Look at the theorem again

It’s in the definition of amalgamation

This is an isomorphism

And if pi_1(U1 cap U2) is trivial

The amalgamation doesn’t change anything

[grim knot]

does it come from the fact that c is only the neutral element and thus k^N does only contain the element?

[hidden crag]

You basically take modulo 0

[grim knot]

oh okay I see it, this is another definition(equivalent)

[hidden crag]

That’s not a definition that’s SVK

Can you send your version of SVK

[grim knot]

yes

this is the one from my prof hahaha

can you read it?

[hidden crag]

Ye

My handwriting is worse

But yeah the smallest normal subgroup containing that set is just the trivial group

So you take modulo 0

And get the result

[grim knot]

yeah I know that (spammed y'all enough last semester with algebra)

[hidden crag]

Or should I say normal divisor

[grim knot]

If you want to show your german side

then go on

[grim knot]

my problem is mostly gettin the right subgroup

[hidden crag]

Yeah that can be tricky

There’s no way to reliably always do it

Except making smart choices for the spaces that cover your bigger spacer

[hidden crag]

I think I only remember the funktheo questions

[grim knot]

well I'm happy for that, cause I spammed much more for algebra haha

it's gonna come again cause I have an algebra 2 exam

another question about topology, does somebody have an idea on when to use coverings and when to use SVK to compute the fundamentalgroup?

[hidden crag]

Lots of algebra for a german student

Master or bachelor

[hidden crag]

Im not sure if there’s an answer to this

[grim knot]

bachelor

[grim knot]

swiss actually

[hidden crag]

Makes sense then

[grim knot]

it pisses me off somehow, cause we have too many rushed subjects

I love what I study, but we do not really have time to digest the subjects and to fully understand them

I asked before, because I read this

and no idea where to begin with 💔

[hidden crag]

you could start by deforming this into a known space

[languid patrol]

have you ever seen a space before where you remove the origin from R^n and quotient out by x ~ -x?

[grim knot]

yes, I also know which fundamental groups both spaces have

[grim knot]

I thought about real projective spaces, I might be wrnog though

[languid patrol]

So $\mathbb{R}^n - {0} \cong \mathbb{R}_{> 0} \times S^{n-1}$

[gentle osprey]

Topos_Theory_E-Girl

[languid patrol]

how does your equivalence relation work in this coordinate system?

[grim knot]

omg this is ugly

we have the right quarter rght?

[languid patrol]

I don't understand what this means

If (x, y) are coordinates, where $x$ is a positive real and $y$ is a point on the sphere, then $-(x, y) = (x, -y)$

[gentle osprey]

Topos_Theory_E-Girl

[grim knot]

you just mirror your sphere

this is why I meant, for example in R2, we have the quarter of a circle(?)

[trim chasm]

(x,y) is not the same as (x,y)

I think you're getting a quarter of a circle from treating topos's (x,y) as a point in the plane in some way

[trim chasm]

for n=2 you should imagine concentric circles centred at the origin

[broken nacelle]

am I supposed to know why K_\xi is a manifold here except at the (n-3) skeleton?

or is hatcher just stating it without a proof to demonstrate a geometric way of characterizing homology classes?

[cursive tendon]

Is it true that the limit points of a set $S$ is equal to $\operatorname{int} S \cup \partial S$?

[gentle osprey]

Mattuwu

[broken nacelle]

yes

[cursive tendon]

hmmm,, I saw a counterexample...

[broken nacelle]

oh huh

interesting

sorry about that

wait, how're boundary points defined again?

[cursive tendon]

boundary points of a set S are the points whose every neibouhood contains some points both from in S and outside of S.

[broken nacelle]

oh so it includes isolated points

I see

[cursive tendon]

For any set $S \subset \mathbb{R}$, is it true that $\operatorname{iso} S^c \subset L(S)$? Can somebody prove or give conterexamples?

[gentle osprey]

Mattuwu

[cursive tendon]

where $L(S)$ are the limit points of S

[gentle osprey]

Mattuwu

[cursive tendon]

and iso S^c is the isolated points of the complement of S

[ebon galleon]

This is true at least in R

You should try to prove it

[cursive tendon]

thanks! seems like a good exercise

[ebon galleon]

Not true in general btw

*for metric spaces

[cursive tendon]

also, I know for a fact that in R^n, iso S^c \subset \partial S

how does L(S) compare to \partial S?

can we have something nice like: iso S^c \subset L(S) \subset \partial S ?

[ebon galleon]

How is \partial S defined

[cursive tendon]

the boundary points of S

[ebon galleon]

Which are

[cursive tendon]

uhhh... I don't know there are several definition of boundary points in R^n..?

[ebon galleon]

Any works

Oh wakt

[cursive tendon]

points whose open balls around it has non-empty intersection with both S and S^c

[ebon galleon]

No this is not true as written

[cursive tendon]

hmm?

[ebon galleon]

Interior points (in particular, non-isolated interior points) can be limit points

Which are clearly not in the boundary

[cursive tendon]

oHH right!

OK... new question should be..
is L(S) \ int S a subset of \partial S?

[ebon galleon]

I want to say yes at least for Rn but I'm not sure off the top of my head if this one generalizes

You should try this one as an exercise too

[grim knot]

shouldn't these inclusions be reversed?

[tough hamlet]

no

[grim knot]

how does this work?

I don't see it

like the discrete topology has only the empty set and X it self

the other topologies are bigger, or not(?)

[tough hamlet]

no that's the trivial topology

or the indiscrete topology

[grim knot]

OH